Section-A
(i). Write down the Riccati’s equation.
|
z
|
<
1
|
z
|
<
1
|z| < 1 |z|<1 | z | < 1
|
z
1
−
z
|
<
1
z
1
−
z
<
1
|(z)/(1-z)| < 1 \left|\frac{z}{1-z}\right|<1 | z 1 − z | < 1
2
F
1
(
a
,
b
;
c
;
1
2
)
2
F
1
a
,
b
;
c
;
1
2
2F_(1)(a,b;c;(1)/(2)) 2 F_1\left(a, b ; c ; \frac{1}{2}\right) 2 F 1 ( a , b ; c ; 1 2 )
Section-B
Find the general solution of the Riccati’s equation.
d
y
d
x
=
2
−
2
y
+
y
2
d
y
d
x
=
2
−
2
y
+
y
2
(dy)/(dx)=2-2y+y^(2) \frac{d y}{d x}=2-2 y+y^2 d y d x = 2 − 2 y + y 2 Whose one particular solution is
(
1
+
tan
x
)
(
1
+
tan
x
)
(1+tan x) (1+\tan x) ( 1 + tan x ) .
Prove that
d
d
x
[
2
F
1
(
a
,
b
;
c
;
x
)
]
=
a
b
c
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
x
)
d
d
x
2
F
1
(
a
,
b
;
c
;
x
)
=
a
b
c
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
x
)
(d)/(dx)[_(2)F_(1)(a,b;c;x)]=(ab)/(c)_(2)F_(1)(a+1,b+1;c+1;x) \frac{d}{d x}\left[{ }_2 F_1(a, b ; c ; x)\right]=\frac{a b}{c}{ }_2 F_1(a+1, b+1 ; c+1 ; x) d d x [ 2 F 1 ( a , b ; c ; x ) ] = a b c 2 F 1 ( a + 1 , b + 1 ; c + 1 ; x )
Solve the following strem Liouville problem.
y
′
′
+
λ
y
=
0
,
y
′
(
−
π
)
=
0
,
y
′
(
π
)
=
0
y
′
′
+
λ
y
=
0
,
y
′
(
−
π
)
=
0
,
y
′
(
π
)
=
0
y^(”)+lambda y=0,quady^(‘)(-pi)=0,quady^(‘)(pi)=0 y^{\prime \prime}+\lambda y=0, \quad y^{\prime}(-\pi)=0, \quad y^{\prime}(\pi)=0 y ′ ′ + λ y = 0 , y ′ ( − π ) = 0 , y ′ ( π ) = 0
Prove that:
(
n
+
1
)
L
n
+
1
(
x
)
=
(
2
n
+
1
−
x
)
L
n
(
x
)
−
n
L
n
−
1
(
x
)
Section – C
(Long Answer Questions)
(
n
+
1
)
L
n
+
1
(
x
)
=
(
2
n
+
1
−
x
)
L
n
(
x
)
−
n
L
n
−
1
(
x
)
Section – C
(Long Answer Questions)
{:[(n+1)L_(n+1)(x)=(2n+1-x)L_(n)(x)-nL_(n-1)(x)],[quad” Section – C “],[” (Long Answer Questions) “]:} \begin{gathered}
(n+1) L_{n+1}(x)=(2 n+1-x) L_n(x)-n L_{n-1}(x) \\
\quad \text { Section – C } \\
\text { (Long Answer Questions) }
\end{gathered} ( n + 1 ) L n + 1 ( x ) = ( 2 n + 1 − x ) L n ( x ) − n L n − 1 ( x ) Section – C (Long Answer Questions) Note :- Answer any one question. Each answer should be given in 800 words. Each question carries 08 marks.
(a) Find the curve with fixed boundary revolves such that its rotation about
x
x
x x x
(b) Solve in series :
x
(
1
−
x
)
d
2
y
d
x
2
+
(
1
+
5
x
)
d
y
d
x
−
4
y
=
0
x
(
1
−
x
)
d
2
y
d
x
2
+
(
1
+
5
x
)
d
y
d
x
−
4
y
=
0
x(1-x)(d^(2)y)/(dx^(2))+(1+5x)(dy)/(dx)-4y=0 x(1-x) \frac{d^2 y}{d x^2}+(1+5 x) \frac{d y}{d x}-4 y=0 x ( 1 − x ) d 2 y d x 2 + ( 1 + 5 x ) d y d x − 4 y = 0
(a) Prove that :
B
(
λ
,
c
−
λ
)
2
F
1
(
a
,
b
;
c
;
z
)
=
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
f
1
(
a
,
b
;
λ
;
z
t
)
d
t
B
(
λ
,
c
−
λ
)
2
F
1
(
a
,
b
;
c
;
z
)
=
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
f
1
(
a
,
b
;
λ
;
z
t
)
d
t
B(lambda,c-lambda)2F_(1)(a,b;c;z)=int_(0)^(1)t^(lambda-1)(1-t)^(c-lambda-1)2f_(1)(a,b;lambda;zt)dt B(\lambda, c-\lambda) 2 F_1(a, b ; c ; z)=\int_0^1 t^{\lambda-1}(1-t)^{c-\lambda-1} 2 f_1(a, b ; \lambda ; z t) d t B ( λ , c − λ ) 2 F 1 ( a , b ; c ; z ) = ∫ 0 1 t λ − 1 ( 1 − t ) c − λ − 1 2 f 1 ( a , b ; λ ; z t ) d t (b) Prove that:
(
2
n
+
1
)
(
x
2
−
1
)
P
n
′
=
n
(
n
+
1
)
(
P
n
+
1
−
P
n
−
1
)
(
2
n
+
1
)
x
2
−
1
P
n
′
=
n
(
n
+
1
)
P
n
+
1
−
P
n
−
1
(2n+1)(x^(2)-1)P_(n)^(‘)=n(n+1)(P_(n+1)-P_(n-1)) (2 n+1)\left(x^2-1\right) P_n^{\prime}=n(n+1)\left(P_{n+1}-P_{n-1}\right) ( 2 n + 1 ) ( x 2 − 1 ) P n ′ = n ( n + 1 ) ( P n + 1 − P n − 1 ) and hence deduce that
∫
−
1
1
(
x
2
−
1
)
P
n
+
1
(
x
)
P
n
′
(
x
)
d
x
=
2
n
(
n
+
1
)
(
2
n
+
1
)
(
2
n
+
3
)
∫
−
1
1
x
2
−
1
P
n
+
1
(
x
)
P
n
′
(
x
)
d
x
=
2
n
(
n
+
1
)
(
2
n
+
1
)
(
2
n
+
3
)
int_(-1)^(1)(x^(2)-1)P_(n+1)(x)P_(n)^(‘)(x)dx=(2n(n+1))/((2n+1)(2n+3)) \int_{-1}^1\left(x^2-1\right) P_{n+1}(x) P_n^{\prime}(x) d x=\frac{2 n(n+1)}{(2 n+1)(2 n+3)} ∫ − 1 1 ( x 2 − 1 ) P n + 1 ( x ) P n ′ ( x ) d x = 2 n ( n + 1 ) ( 2 n + 1 ) ( 2 n + 3 ) Answer:
Question:-01(a)
Write down the Riccati’s equation.
Answer:
Riccati’s Equation: The Riccati equation is a first-order, nonlinear differential equation of the form:
d
y
d
x
=
a
(
x
)
y
2
+
b
(
x
)
y
+
c
(
x
)
,
d
y
d
x
=
a
(
x
)
y
2
+
b
(
x
)
y
+
c
(
x
)
,
(dy)/(dx)=a(x)y^(2)+b(x)y+c(x), \frac{dy}{dx} = a(x) y^2 + b(x) y + c(x), d y d x = a ( x ) y 2 + b ( x ) y + c ( x ) , where:
y
=
y
(
x
)
y
=
y
(
x
)
y=y(x) y = y(x) y = y ( x )
a
(
x
)
,
b
(
x
)
,
c
(
x
)
a
(
x
)
,
b
(
x
)
,
c
(
x
)
a(x),b(x),c(x) a(x), b(x), c(x) a ( x ) , b ( x ) , c ( x )
x
x
x x x
Special Cases:
If
a
(
x
)
=
0
a
(
x
)
=
0
a(x)=0 a(x) = 0 a ( x ) = 0 , the Riccati equation reduces to a linear first-order differential equation:
d
y
d
x
=
b
(
x
)
y
+
c
(
x
)
.
d
y
d
x
=
b
(
x
)
y
+
c
(
x
)
.
(dy)/(dx)=b(x)y+c(x). \frac{dy}{dx} = b(x) y + c(x). d y d x = b ( x ) y + c ( x ) .
If
b
(
x
)
=
0
b
(
x
)
=
0
b(x)=0 b(x) = 0 b ( x ) = 0 , it becomes:
d
y
d
x
=
a
(
x
)
y
2
+
c
(
x
)
,
d
y
d
x
=
a
(
x
)
y
2
+
c
(
x
)
,
(dy)/(dx)=a(x)y^(2)+c(x), \frac{dy}{dx} = a(x) y^2 + c(x), d y d x = a ( x ) y 2 + c ( x ) , which is a Bernoulli equation when
c
(
x
)
=
0
c
(
x
)
=
0
c(x)=0 c(x) = 0 c ( x ) = 0 .
Question:-01(b)
Write down Euler-Lagrange equation.
Answer:
Euler-Lagrange Equation The Euler-Lagrange equation is a fundamental equation in the calculus of variations. It provides the necessary condition for a functional to have an extremum.
Statement: Given a functional of the form:
J
[
y
]
=
∫
x
1
x
2
L
(
x
,
y
(
x
)
,
y
′
(
x
)
)
d
x
,
J
[
y
]
=
∫
x
1
x
2
L
(
x
,
y
(
x
)
,
y
′
(
x
)
)
d
x
,
J[y]=int_(x_(1))^(x_(2))L(x,y(x),y^(‘)(x))dx, J[y] = \int_{x_1}^{x_2} L(x, y(x), y'(x)) \, dx, J [ y ] = ∫ x 1 x 2 L ( x , y ( x ) , y ′ ( x ) ) d x , where:
L
(
x
,
y
,
y
′
)
L
(
x
,
y
,
y
′
)
L(x,y,y^(‘)) L(x, y, y’) L ( x , y , y ′ ) Lagrangian , a function of
x
x
x x x
y
(
x
)
y
(
x
)
y(x) y(x) y ( x )
y
′
(
x
)
y
′
(
x
)
y^(‘)(x) y'(x) y ′ ( x )
y
(
x
)
y
(
x
)
y(x) y(x) y ( x )
y
′
(
x
)
=
d
y
d
x
y
′
(
x
)
=
d
y
d
x
y^(‘)(x)=(dy)/(dx) y'(x) = \frac{dy}{dx} y ′ ( x ) = d y d x
the Euler-Lagrange equation is:
∂
L
∂
y
−
d
d
x
(
∂
L
∂
y
′
)
=
0.
∂
L
∂
y
−
d
d
x
∂
L
∂
y
′
=
0.
(del L)/(del y)-(d)/(dx)((del L)/(dely^(‘)))=0. \frac{\partial L}{\partial y} – \frac{d}{dx} \left( \frac{\partial L}{\partial y’} \right) = 0. ∂ L ∂ y − d d x ( ∂ L ∂ y ′ ) = 0. Derivation Idea: The Euler-Lagrange equation arises from requiring that the first variation of the functional
J
[
y
]
J
[
y
]
J[y] J[y] J [ y ] vanishes, i.e.,
δ
J
[
y
]
=
0
δ
J
[
y
]
=
0
delta J[y]=0 \delta J[y] = 0 δ J [ y ] = 0 .
Example Application: If
L
(
x
,
y
,
y
′
)
=
1
2
(
y
′
)
2
−
V
(
y
)
L
(
x
,
y
,
y
′
)
=
1
2
(
y
′
)
2
−
V
(
y
)
L(x,y,y^(‘))=(1)/(2)(y^(‘))^(2)-V(y) L(x, y, y’) = \frac{1}{2}(y’)^2 – V(y) L ( x , y , y ′ ) = 1 2 ( y ′ ) 2 − V ( y ) , the Euler-Lagrange equation gives:
d
2
y
d
x
2
+
d
V
d
y
=
0.
d
2
y
d
x
2
+
d
V
d
y
=
0.
(d^(2)y)/(dx^(2))+(dV)/(dy)=0. \frac{d^2y}{dx^2} + \frac{dV}{dy} = 0. d 2 y d x 2 + d V d y = 0.
Question:-01(c)
If
|
z
|
<
1
|
z
|
<
1
|z| < 1 |z| < 1 | z | < 1
|
z
1
−
z
|
<
1
z
1
−
z
<
1
|(z)/(1-z)| < 1 \left|\frac{z}{1-z}\right| < 1 | z 1 − z | < 1
2
F
1
(
a
,
b
;
c
;
1
2
)
2
F
1
a
,
b
;
c
;
1
2
2F_(1)(a,b;c;(1)/(2)) 2 F_1\left(a, b ; c ; \frac{1}{2}\right) 2 F 1 ( a , b ; c ; 1 2 )
Answer:
To evaluate
2
F
1
(
a
,
b
;
c
;
1
2
)
2
F
1
(
a
,
b
;
c
;
1
2
)
2F_(1)(a,b;c;(1)/(2)) 2 F_1(a, b; c; \frac{1}{2}) 2 F 1 ( a , b ; c ; 1 2 ) , let’s analyze the given conditions and the structure of the hypergeometric function
2
F
1
2
F
1
_(2)F_(1) {}_2F_1 2 F 1 .
Problem Setup The hypergeometric function
2
F
1
(
a
,
b
;
c
;
z
)
2
F
1
(
a
,
b
;
c
;
z
)
_(2)F_(1)(a,b;c;z) {}_2F_1(a, b; c; z) 2 F 1 ( a , b ; c ; z ) is defined as:
2
F
1
(
a
,
b
;
c
;
z
)
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
z
n
n
!
,
2
F
1
(
a
,
b
;
c
;
z
)
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
z
n
n
!
,
_(2)F_(1)(a,b;c;z)=sum_(n=0)^(oo)((a)_(n)(b)_(n))/((c)_(n))(z^(n))/(n!), {}_2F_1(a, b; c; z) = \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}, 2 F 1 ( a , b ; c ; z ) = ∑ n = 0 ∞ ( a ) n ( b ) n ( c ) n z n n ! , where:
(
a
)
n
(
a
)
n
(a)_(n) (a)_n ( a ) n
(
a
)
n
=
a
(
a
+
1
)
(
a
+
2
)
⋯
(
a
+
n
−
1
)
(
a
)
n
=
a
(
a
+
1
)
(
a
+
2
)
⋯
(
a
+
n
−
1
)
(a)_(n)=a(a+1)(a+2)cdots(a+n-1) (a)_n = a (a+1)(a+2)\cdots(a+n-1) ( a ) n = a ( a + 1 ) ( a + 2 ) ⋯ ( a + n − 1 )
|
z
|
<
1
|
z
|
<
1
|z| < 1 |z| < 1 | z | < 1
Given:
|
z
|
<
1
|
z
|
<
1
|z| < 1 |z| < 1 | z | < 1
|
z
1
−
z
|
<
1
z
1
−
z
<
1
|(z)/(1-z)| < 1 \left| \frac{z}{1-z} \right| < 1 | z 1 − z | < 1
Interpretation and Simplification Step 1: Analyze
z
1
−
z
z
1
−
z
(z)/(1-z) \frac{z}{1-z} z 1 − z
The condition
|
z
1
−
z
|
<
1
z
1
−
z
<
1
|(z)/(1-z)| < 1 \left| \frac{z}{1-z} \right| < 1 | z 1 − z | < 1 implies:
|
z
|
|
1
−
z
|
<
1
⟹
|
z
|
<
|
1
−
z
|
.
|
z
|
|
1
−
z
|
<
1
⟹
|
z
|
<
|
1
−
z
|
.
(|z|)/(|1-z|) < 1quadLongrightarrowquad|z| < |1-z|. \frac{|z|}{|1-z|} < 1 \quad \implies \quad |z| < |1-z|. | z | | 1 − z | < 1 ⟹ | z | < | 1 − z | . This suggests the series for
2
F
1
(
a
,
b
;
c
;
z
)
2
F
1
(
a
,
b
;
c
;
z
)
_(2)F_(1)(a,b;c;z) {}_2F_1(a, b; c; z) 2 F 1 ( a , b ; c ; z ) converges, as
z
z
z z z remains well within the unit disk.
Step 2: Evaluate
2
F
1
2
F
1
_(2)F_(1) {}_2F_1 2 F 1
z
=
1
2
z
=
1
2
z=(1)/(2) z = \frac{1}{2} z = 1 2
The value of
2
F
1
(
a
,
b
;
c
;
1
2
)
2
F
1
(
a
,
b
;
c
;
1
2
)
_(2)F_(1)(a,b;c;(1)/(2)) {}_2F_1(a, b; c; \frac{1}{2}) 2 F 1 ( a , b ; c ; 1 2 ) is derived from the series expansion:
2
F
1
(
a
,
b
;
c
;
1
2
)
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
(
1
/
2
)
n
n
!
.
2
F
1
(
a
,
b
;
c
;
1
2
)
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
(
1
/
2
)
n
n
!
.
_(2)F_(1)(a,b;c;(1)/(2))=sum_(n=0)^(oo)((a)_(n)(b)_(n))/((c)_(n))((1//2)^(n))/(n!). {}_2F_1(a, b; c; \frac{1}{2}) = \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{(1/2)^n}{n!}. 2 F 1 ( a , b ; c ; 1 2 ) = ∑ n = 0 ∞ ( a ) n ( b ) n ( c ) n ( 1 / 2 ) n n ! . For a general evaluation, the exact closed form of this series depends on the parameters
a
,
b
,
c
a
,
b
,
c
a,b,c a, b, c a , b , c . However, in some cases, known results for specific parameter values can be used.
Step 3: Factor of 2
The problem asks for
2
F
1
(
a
,
b
;
c
;
1
2
)
2
F
1
(
a
,
b
;
c
;
1
2
)
2F_(1)(a,b;c;(1)/(2)) 2 F_1(a, b; c; \frac{1}{2}) 2 F 1 ( a , b ; c ; 1 2 ) , so the final value will be twice the sum.
Final Answer Without specific values for
a
,
b
,
c
a
,
b
,
c
a,b,c a, b, c a , b , c , the value of
2
2
F
1
(
a
,
b
;
c
;
1
2
)
2
2
F
1
(
a
,
b
;
c
;
1
2
)
2_(2)F_(1)(a,b;c;(1)/(2)) 2 {}_2F_1(a, b; c; \frac{1}{2}) 2 2 F 1 ( a , b ; c ; 1 2 ) is written as:
2
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
(
1
/
2
)
n
n
!
.
2
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
(
1
/
2
)
n
n
!
.
2sum_(n=0)^(oo)((a)_(n)(b)_(n))/((c)_(n))((1//2)^(n))/(n!). 2 \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{(1/2)^n}{n!}. 2 ∑ n = 0 ∞ ( a ) n ( b ) n ( c ) n ( 1 / 2 ) n n ! .
Question:-01(d)
Write down the Legendre equation.
Answer:
Legendre’s Differential Equation: The Legendre equation is a second-order linear differential equation of the form:
(
1
−
x
2
)
d
2
y
d
x
2
−
2
x
d
y
d
x
+
n
(
n
+
1
)
y
=
0
,
(
1
−
x
2
)
d
2
y
d
x
2
−
2
x
d
y
d
x
+
n
(
n
+
1
)
y
=
0
,
(1-x^(2))(d^(2)y)/(dx^(2))-2x(dy)/(dx)+n(n+1)y=0, (1 – x^2) \frac{d^2y}{dx^2} – 2x \frac{dy}{dx} + n(n+1)y = 0, ( 1 − x 2 ) d 2 y d x 2 − 2 x d y d x + n ( n + 1 ) y = 0 , where:
n
n
n n n
y
=
y
(
x
)
y
=
y
(
x
)
y=y(x) y = y(x) y = y ( x )
x
∈
(
−
1
,
1
)
x
∈
(
−
1
,
1
)
x in(-1,1) x \in (-1, 1) x ∈ ( − 1 , 1 )
General Solution: The solutions to Legendre’s equation are the
Legendre polynomials
P
n
(
x
)
P
n
(
x
)
P_(n)(x) P_n(x) P n ( x ) , which form a complete orthogonal set over
[
−
1
,
1
]
[
−
1
,
1
]
[-1,1] [-1, 1] [ − 1 , 1 ] and are defined recursively or by the Rodrigues formula:
P
n
(
x
)
=
1
2
n
n
!
d
n
d
x
n
(
(
x
2
−
1
)
n
)
.
P
n
(
x
)
=
1
2
n
n
!
d
n
d
x
n
(
x
2
−
1
)
n
.
P_(n)(x)=(1)/(2^(n)n!)(d^(n))/(dx^(n))((x^(2)-1)^(n)). P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} \left( (x^2 – 1)^n \right). P n ( x ) = 1 2 n n ! d n d x n ( ( x 2 − 1 ) n ) . Applications:
Legendre’s equation arises in solving Laplace’s equation in spherical coordinates.
It is used extensively in physics, particularly in potential theory and quantum mechanics.
Question:-02
Find the general solution of the Riccati’s equation:
d
y
d
x
=
2
−
2
y
+
y
2
d
y
d
x
=
2
−
2
y
+
y
2
(dy)/(dx)=2-2y+y^(2) \frac{d y}{d x} = 2 – 2 y + y^2 d y d x = 2 − 2 y + y 2 Whose one particular solution is
(
1
+
tan
x
)
(
1
+
tan
x
)
(1+tan x) (1 + \tan x) ( 1 + tan x ) .
Answer:
To solve the Riccati equation:
d
y
d
x
=
2
−
2
y
+
y
2
,
d
y
d
x
=
2
−
2
y
+
y
2
,
(dy)/(dx)=2-2y+y^(2), \frac{dy}{dx} = 2 – 2y + y^2, d y d x = 2 − 2 y + y 2 , with a known particular solution
y
p
(
x
)
=
1
+
tan
x
y
p
(
x
)
=
1
+
tan
x
y_(p)(x)=1+tan x y_p(x) = 1 + \tan x y p ( x ) = 1 + tan x , we will transform the equation into a linear equation by substituting
y
(
x
)
=
y
p
(
x
)
+
1
z
(
x
)
y
(
x
)
=
y
p
(
x
)
+
1
z
(
x
)
y(x)=y_(p)(x)+(1)/(z(x)) y(x) = y_p(x) + \frac{1}{z(x)} y ( x ) = y p ( x ) + 1 z ( x ) .
Step 1: General Substitution Let:
y
(
x
)
=
y
p
(
x
)
+
1
z
(
x
)
,
y
(
x
)
=
y
p
(
x
)
+
1
z
(
x
)
,
y(x)=y_(p)(x)+(1)/(z(x)), y(x) = y_p(x) + \frac{1}{z(x)}, y ( x ) = y p ( x ) + 1 z ( x ) , where
y
p
(
x
)
=
1
+
tan
x
y
p
(
x
)
=
1
+
tan
x
y_(p)(x)=1+tan x y_p(x) = 1 + \tan x y p ( x ) = 1 + tan x is the given particular solution. Substituting this into the Riccati equation, we aim to derive a linear equation for
z
(
x
)
z
(
x
)
z(x) z(x) z ( x ) .
Step 2: Differentiate
y
(
x
)
y
(
x
)
y(x) y(x) y ( x ) Differentiating
y
(
x
)
=
y
p
(
x
)
+
1
z
(
x
)
y
(
x
)
=
y
p
(
x
)
+
1
z
(
x
)
y(x)=y_(p)(x)+(1)/(z(x)) y(x) = y_p(x) + \frac{1}{z(x)} y ( x ) = y p ( x ) + 1 z ( x ) , we get:
d
y
d
x
=
d
y
p
d
x
−
1
z
2
d
z
d
x
.
d
y
d
x
=
d
y
p
d
x
−
1
z
2
d
z
d
x
.
(dy)/(dx)=(dy_(p))/(dx)-(1)/(z^(2))(dz)/(dx). \frac{dy}{dx} = \frac{dy_p}{dx} – \frac{1}{z^2} \frac{dz}{dx}. d y d x = d y p d x − 1 z 2 d z d x . Substitute
d
y
d
x
d
y
d
x
(dy)/(dx) \frac{dy}{dx} d y d x into the Riccati equation.
Step 3: Simplify Using the Particular Solution: Using the fact that
y
p
(
x
)
=
1
+
tan
x
y
p
(
x
)
=
1
+
tan
x
y_(p)(x)=1+tan x y_p(x) = 1 + \tan x y p ( x ) = 1 + tan x satisfies the Riccati equation, we know:
d
y
p
d
x
=
2
−
2
y
p
+
y
p
2
.
d
y
p
d
x
=
2
−
2
y
p
+
y
p
2
.
(dy_(p))/(dx)=2-2y_(p)+y_(p)^(2). \frac{dy_p}{dx} = 2 – 2y_p + y_p^2. d y p d x = 2 − 2 y p + y p 2 . Now substitute
y
(
x
)
=
y
p
(
x
)
+
1
z
y
(
x
)
=
y
p
(
x
)
+
1
z
y(x)=y_(p)(x)+(1)/(z) y(x) = y_p(x) + \frac{1}{z} y ( x ) = y p ( x ) + 1 z and
d
y
d
x
=
d
y
p
d
x
−
1
z
2
d
z
d
x
d
y
d
x
=
d
y
p
d
x
−
1
z
2
d
z
d
x
(dy)/(dx)=(dy_(p))/(dx)-(1)/(z^(2))(dz)/(dx) \frac{dy}{dx} = \frac{dy_p}{dx} – \frac{1}{z^2} \frac{dz}{dx} d y d x = d y p d x − 1 z 2 d z d x into the equation:
d
y
d
x
=
2
−
2
y
+
y
2
.
d
y
d
x
=
2
−
2
y
+
y
2
.
(dy)/(dx)=2-2y+y^(2). \frac{dy}{dx} = 2 – 2y + y^2. d y d x = 2 − 2 y + y 2 . Simplify step by step:
Substitute
y
=
y
p
+
1
z
y
=
y
p
+
1
z
y=y_(p)+(1)/(z) y = y_p + \frac{1}{z} y = y p + 1 z :
2
−
2
y
+
y
2
=
2
−
2
(
y
p
+
1
z
)
+
(
y
p
+
1
z
)
2
.
2
−
2
y
+
y
2
=
2
−
2
y
p
+
1
z
+
y
p
+
1
z
2
.
2-2y+y^(2)=2-2(y_(p)+(1)/(z))+(y_(p)+(1)/(z))^(2). 2 – 2y + y^2 = 2 – 2\left(y_p + \frac{1}{z}\right) + \left(y_p + \frac{1}{z}\right)^2. 2 − 2 y + y 2 = 2 − 2 ( y p + 1 z ) + ( y p + 1 z ) 2 .
Expand
(
y
p
+
1
z
)
2
y
p
+
1
z
2
(y_(p)+(1)/(z))^(2) \left(y_p + \frac{1}{z}\right)^2 ( y p + 1 z ) 2 :
(
y
p
+
1
z
)
2
=
y
p
2
+
2
y
p
z
+
1
z
2
.
y
p
+
1
z
2
=
y
p
2
+
2
y
p
z
+
1
z
2
.
(y_(p)+(1)/(z))^(2)=y_(p)^(2)+(2y_(p))/(z)+(1)/(z^(2)). \left(y_p + \frac{1}{z}\right)^2 = y_p^2 + \frac{2y_p}{z} + \frac{1}{z^2}. ( y p + 1 z ) 2 = y p 2 + 2 y p z + 1 z 2 .
Collect terms:
2
−
2
(
y
p
+
1
z
)
+
y
p
2
+
2
y
p
z
+
1
z
2
.
2
−
2
y
p
+
1
z
+
y
p
2
+
2
y
p
z
+
1
z
2
.
2-2(y_(p)+(1)/(z))+y_(p)^(2)+(2y_(p))/(z)+(1)/(z^(2)). 2 – 2\left(y_p + \frac{1}{z}\right) + y_p^2 + \frac{2y_p}{z} + \frac{1}{z^2}. 2 − 2 ( y p + 1 z ) + y p 2 + 2 y p z + 1 z 2 .
Substituting
d
y
p
d
x
=
2
−
2
y
p
+
y
p
2
d
y
p
d
x
=
2
−
2
y
p
+
y
p
2
(dy_(p))/(dx)=2-2y_(p)+y_(p)^(2) \frac{dy_p}{dx} = 2 – 2y_p + y_p^2 d y p d x = 2 − 2 y p + y p 2 , the terms involving
y
p
y
p
y_(p) y_p y p cancel out, leaving:
−
1
z
2
d
z
d
x
=
−
2
z
.
−
1
z
2
d
z
d
x
=
−
2
z
.
-(1)/(z^(2))(dz)/(dx)=-(2)/(z). -\frac{1}{z^2} \frac{dz}{dx} = -\frac{2}{z}. − 1 z 2 d z d x = − 2 z . Step 4: Linear Equation for
z
(
x
)
z
(
x
)
z(x) z(x) z ( x ) Simplify the equation:
1
z
2
d
z
d
x
=
2
z
.
1
z
2
d
z
d
x
=
2
z
.
(1)/(z^(2))(dz)/(dx)=(2)/(z). \frac{1}{z^2} \frac{dz}{dx} = \frac{2}{z}. 1 z 2 d z d x = 2 z . Multiply through by
z
2
z
2
z^(2) z^2 z 2 :
d
z
d
x
=
2
z
.
d
z
d
x
=
2
z
.
(dz)/(dx)=2z. \frac{dz}{dx} = 2z. d z d x = 2 z . Step 5: Solve the Linear Equation: This is a first-order linear differential equation for
z
(
x
)
z
(
x
)
z(x) z(x) z ( x ) :
d
z
d
x
=
2
z
.
d
z
d
x
=
2
z
.
(dz)/(dx)=2z. \frac{dz}{dx} = 2z. d z d x = 2 z . The solution is:
z
(
x
)
=
C
e
2
x
,
z
(
x
)
=
C
e
2
x
,
z(x)=Ce^(2x), z(x) = Ce^{2x}, z ( x ) = C e 2 x , where
C
C
C C C is an arbitrary constant.
Step 6: Back-Substitute for
y
(
x
)
y
(
x
)
y(x) y(x) y ( x ) Recall the substitution
y
(
x
)
=
y
p
(
x
)
+
1
z
(
x
)
y
(
x
)
=
y
p
(
x
)
+
1
z
(
x
)
y(x)=y_(p)(x)+(1)/(z(x)) y(x) = y_p(x) + \frac{1}{z(x)} y ( x ) = y p ( x ) + 1 z ( x ) . Substituting
z
(
x
)
=
C
e
2
x
z
(
x
)
=
C
e
2
x
z(x)=Ce^(2x) z(x) = Ce^{2x} z ( x ) = C e 2 x , we get:
y
(
x
)
=
y
p
(
x
)
+
1
C
e
2
x
.
y
(
x
)
=
y
p
(
x
)
+
1
C
e
2
x
.
y(x)=y_(p)(x)+(1)/(Ce^(2x)). y(x) = y_p(x) + \frac{1}{Ce^{2x}}. y ( x ) = y p ( x ) + 1 C e 2 x . Since
y
p
(
x
)
=
1
+
tan
x
y
p
(
x
)
=
1
+
tan
x
y_(p)(x)=1+tan x y_p(x) = 1 + \tan x y p ( x ) = 1 + tan x , the general solution is:
y
(
x
)
=
1
+
tan
x
+
1
C
e
2
x
.
y
(
x
)
=
1
+
tan
x
+
1
C
e
2
x
.
y(x)=1+tan x+(1)/(Ce^(2x)). y(x) = 1 + \tan x + \frac{1}{Ce^{2x}}. y ( x ) = 1 + tan x + 1 C e 2 x . Final Answer: The general solution of the Riccati equation is:
y
(
x
)
=
1
+
tan
x
+
1
C
e
2
x
,
y
(
x
)
=
1
+
tan
x
+
1
C
e
2
x
,
y(x)=1+tan x+(1)/(Ce^(2x)), y(x) = 1 + \tan x + \frac{1}{Ce^{2x}}, y ( x ) = 1 + tan x + 1 C e 2 x , where
C
C
C C C is an arbitrary constant.
Question:-03
Prove that:
d
d
x
[
2
F
1
(
a
,
b
;
c
;
x
)
]
=
a
b
c
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
x
)
d
d
x
2
F
1
(
a
,
b
;
c
;
x
)
=
a
b
c
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
x
)
(d)/(dx)[_(2)F_(1)(a,b;c;x)]=(ab)/(c)_(2)F_(1)(a+1,b+1;c+1;x) \frac{d}{d x}\left[{ }_2 F_1(a, b ; c ; x)\right] = \frac{a b}{c} {_2 F_1(a+1, b+1 ; c+1 ; x)} d d x [ 2 F 1 ( a , b ; c ; x ) ] = a b c 2 F 1 ( a + 1 , b + 1 ; c + 1 ; x ) Answer:
To Prove:
d
d
x
[
2
F
1
(
a
,
b
;
c
;
x
)
]
=
a
b
c
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
x
)
.
d
d
x
2
F
1
(
a
,
b
;
c
;
x
)
=
a
b
c
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
x
)
.
(d)/(dx)[_(2)F_(1)(a,b;c;x)]=(ab)/(c)_(2)F_(1)(a+1,b+1;c+1;x). \frac{d}{dx} \left[ {}_2F_1(a, b; c; x) \right] = \frac{ab}{c} \, {}_2F_1(a+1, b+1; c+1; x). d d x [ 2 F 1 ( a , b ; c ; x ) ] = a b c 2 F 1 ( a + 1 , b + 1 ; c + 1 ; x ) . Proof: Definition of the Hypergeometric Function: The hypergeometric function
2
F
1
(
a
,
b
;
c
;
x
)
2
F
1
(
a
,
b
;
c
;
x
)
_(2)F_(1)(a,b;c;x) {}_2F_1(a, b; c; x) 2 F 1 ( a , b ; c ; x ) is given by the series:
2
F
1
(
a
,
b
;
c
;
x
)
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
x
n
n
!
,
2
F
1
(
a
,
b
;
c
;
x
)
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
x
n
n
!
,
_(2)F_(1)(a,b;c;x)=sum_(n=0)^(oo)((a)_(n)(b)_(n))/((c)_(n))(x^(n))/(n!), {}_2F_1(a, b; c; x) = \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{x^n}{n!}, 2 F 1 ( a , b ; c ; x ) = ∑ n = 0 ∞ ( a ) n ( b ) n ( c ) n x n n ! , where:
(
a
)
n
=
a
(
a
+
1
)
(
a
+
2
)
⋯
(
a
+
n
−
1
)
(
a
)
n
=
a
(
a
+
1
)
(
a
+
2
)
⋯
(
a
+
n
−
1
)
(a)_(n)=a(a+1)(a+2)cdots(a+n-1) (a)_n = a (a+1)(a+2)\cdots(a+n-1) ( a ) n = a ( a + 1 ) ( a + 2 ) ⋯ ( a + n − 1 )
(
c
)
n
=
c
(
c
+
1
)
(
c
+
2
)
⋯
(
c
+
n
−
1
)
(
c
)
n
=
c
(
c
+
1
)
(
c
+
2
)
⋯
(
c
+
n
−
1
)
(c)_(n)=c(c+1)(c+2)cdots(c+n-1) (c)_n = c (c+1)(c+2)\cdots(c+n-1) ( c ) n = c ( c + 1 ) ( c + 2 ) ⋯ ( c + n − 1 )
x
x
x x x
|
x
|
<
1
|
x
|
<
1
|x| < 1 |x| < 1 | x | < 1
Step 1: Differentiate Term by Term Differentiating the series term-by-term:
d
d
x
[
2
F
1
(
a
,
b
;
c
;
x
)
]
=
d
d
x
[
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
x
n
n
!
]
.
d
d
x
2
F
1
(
a
,
b
;
c
;
x
)
=
d
d
x
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
x
n
n
!
.
(d)/(dx)[_(2)F_(1)(a,b;c;x)]=(d)/(dx)[sum_(n=0)^(oo)((a)_(n)(b)_(n))/((c)_(n))(x^(n))/(n!)]. \frac{d}{dx} \left[ {}_2F_1(a, b; c; x) \right] = \frac{d}{dx} \left[ \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{x^n}{n!} \right]. d d x [ 2 F 1 ( a , b ; c ; x ) ] = d d x [ ∑ n = 0 ∞ ( a ) n ( b ) n ( c ) n x n n ! ] . Since differentiation with respect to
x
x
x x x acts only on the
x
n
x
n
x^(n) x^n x n term:
d
d
x
[
2
F
1
(
a
,
b
;
c
;
x
)
]
=
∑
n
=
1
∞
(
a
)
n
(
b
)
n
(
c
)
n
n
x
n
−
1
n
!
.
d
d
x
2
F
1
(
a
,
b
;
c
;
x
)
=
∑
n
=
1
∞
(
a
)
n
(
b
)
n
(
c
)
n
n
x
n
−
1
n
!
.
(d)/(dx)[_(2)F_(1)(a,b;c;x)]=sum_(n=1)^(oo)((a)_(n)(b)_(n))/((c)_(n))(nx^(n-1))/(n!). \frac{d}{dx} \left[ {}_2F_1(a, b; c; x) \right] = \sum_{n=1}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{n x^{n-1}}{n!}. d d x [ 2 F 1 ( a , b ; c ; x ) ] = ∑ n = 1 ∞ ( a ) n ( b ) n ( c ) n n x n − 1 n ! . Simplify the term:
n
n
!
=
1
(
n
−
1
)
!
,
n
n
!
=
1
(
n
−
1
)
!
,
(n)/(n!)=(1)/((n-1)!), \frac{n}{n!} = \frac{1}{(n-1)!}, n n ! = 1 ( n − 1 ) ! , so the series becomes:
d
d
x
[
2
F
1
(
a
,
b
;
c
;
x
)
]
=
∑
n
=
1
∞
(
a
)
n
(
b
)
n
(
c
)
n
x
n
−
1
(
n
−
1
)
!
.
d
d
x
2
F
1
(
a
,
b
;
c
;
x
)
=
∑
n
=
1
∞
(
a
)
n
(
b
)
n
(
c
)
n
x
n
−
1
(
n
−
1
)
!
.
(d)/(dx)[_(2)F_(1)(a,b;c;x)]=sum_(n=1)^(oo)((a)_(n)(b)_(n))/((c)_(n))(x^(n-1))/((n-1)!). \frac{d}{dx} \left[ {}_2F_1(a, b; c; x) \right] = \sum_{n=1}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{x^{n-1}}{(n-1)!}. d d x [ 2 F 1 ( a , b ; c ; x ) ] = ∑ n = 1 ∞ ( a ) n ( b ) n ( c ) n x n − 1 ( n − 1 ) ! . Step 2: Shift the Index Change the index of summation by letting
m
=
n
−
1
m
=
n
−
1
m=n-1 m = n-1 m = n − 1 . When
n
=
1
n
=
1
n=1 n = 1 n = 1 ,
m
=
0
m
=
0
m=0 m = 0 m = 0 , and when
n
→
∞
n
→
∞
n rarr oo n \to \infty n → ∞ ,
m
→
∞
m
→
∞
m rarr oo m \to \infty m → ∞ . Substituting
n
=
m
+
1
n
=
m
+
1
n=m+1 n = m+1 n = m + 1 :
d
d
x
[
2
F
1
(
a
,
b
;
c
;
x
)
]
=
∑
m
=
0
∞
(
a
)
m
+
1
(
b
)
m
+
1
(
c
)
m
+
1
x
m
m
!
.
d
d
x
2
F
1
(
a
,
b
;
c
;
x
)
=
∑
m
=
0
∞
(
a
)
m
+
1
(
b
)
m
+
1
(
c
)
m
+
1
x
m
m
!
.
(d)/(dx)[_(2)F_(1)(a,b;c;x)]=sum_(m=0)^(oo)((a)_(m+1)(b)_(m+1))/((c)_(m+1))(x^(m))/(m!). \frac{d}{dx} \left[ {}_2F_1(a, b; c; x) \right] = \sum_{m=0}^\infty \frac{(a)_{m+1} (b)_{m+1}}{(c)_{m+1}} \frac{x^m}{m!}. d d x [ 2 F 1 ( a , b ; c ; x ) ] = ∑ m = 0 ∞ ( a ) m + 1 ( b ) m + 1 ( c ) m + 1 x m m ! . Step 3: Simplify the Pochhammer Symbols Using the property of the Pochhammer symbol:
(
a
)
m
+
1
=
a
(
a
+
1
)
m
,
(
b
)
m
+
1
=
b
(
b
+
1
)
m
,
(
c
)
m
+
1
=
c
(
c
+
1
)
m
,
(
a
)
m
+
1
=
a
(
a
+
1
)
m
,
(
b
)
m
+
1
=
b
(
b
+
1
)
m
,
(
c
)
m
+
1
=
c
(
c
+
1
)
m
,
(a)_(m+1)=a(a+1)_(m),quad(b)_(m+1)=b(b+1)_(m),quad(c)_(m+1)=c(c+1)_(m), (a)_{m+1} = a (a+1)_m, \quad (b)_{m+1} = b (b+1)_m, \quad (c)_{m+1} = c (c+1)_m, ( a ) m + 1 = a ( a + 1 ) m , ( b ) m + 1 = b ( b + 1 ) m , ( c ) m + 1 = c ( c + 1 ) m , the series becomes:
d
d
x
[
2
F
1
(
a
,
b
;
c
;
x
)
]
=
∑
m
=
0
∞
a
b
(
a
+
1
)
m
(
b
+
1
)
m
c
(
c
+
1
)
m
x
m
m
!
.
d
d
x
2
F
1
(
a
,
b
;
c
;
x
)
=
∑
m
=
0
∞
a
b
(
a
+
1
)
m
(
b
+
1
)
m
c
(
c
+
1
)
m
x
m
m
!
.
(d)/(dx)[_(2)F_(1)(a,b;c;x)]=sum_(m=0)^(oo)(ab(a+1)_(m)(b+1)_(m))/(c(c+1)_(m))(x^(m))/(m!). \frac{d}{dx} \left[ {}_2F_1(a, b; c; x) \right] = \sum_{m=0}^\infty \frac{a b (a+1)_m (b+1)_m}{c (c+1)_m} \frac{x^m}{m!}. d d x [ 2 F 1 ( a , b ; c ; x ) ] = ∑ m = 0 ∞ a b ( a + 1 ) m ( b + 1 ) m c ( c + 1 ) m x m m ! . Factor out
a
b
c
a
b
c
(ab)/(c) \frac{ab}{c} a b c , leaving:
d
d
x
[
2
F
1
(
a
,
b
;
c
;
x
)
]
=
a
b
c
∑
m
=
0
∞
(
a
+
1
)
m
(
b
+
1
)
m
(
c
+
1
)
m
x
m
m
!
.
d
d
x
2
F
1
(
a
,
b
;
c
;
x
)
=
a
b
c
∑
m
=
0
∞
(
a
+
1
)
m
(
b
+
1
)
m
(
c
+
1
)
m
x
m
m
!
.
(d)/(dx)[_(2)F_(1)(a,b;c;x)]=(ab)/(c)sum_(m=0)^(oo)((a+1)_(m)(b+1)_(m))/((c+1)_(m))(x^(m))/(m!). \frac{d}{dx} \left[ {}_2F_1(a, b; c; x) \right] = \frac{ab}{c} \sum_{m=0}^\infty \frac{(a+1)_m (b+1)_m}{(c+1)_m} \frac{x^m}{m!}. d d x [ 2 F 1 ( a , b ; c ; x ) ] = a b c ∑ m = 0 ∞ ( a + 1 ) m ( b + 1 ) m ( c + 1 ) m x m m ! . Step 4: Recognize the Hypergeometric Function The series in the summation is precisely
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
x
)
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
x
)
_(2)F_(1)(a+1,b+1;c+1;x) {}_2F_1(a+1, b+1; c+1; x) 2 F 1 ( a + 1 , b + 1 ; c + 1 ; x ) . Therefore:
d
d
x
[
2
F
1
(
a
,
b
;
c
;
x
)
]
=
a
b
c
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
x
)
.
d
d
x
2
F
1
(
a
,
b
;
c
;
x
)
=
a
b
c
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
x
)
.
(d)/(dx)[_(2)F_(1)(a,b;c;x)]=(ab)/(c)_(2)F_(1)(a+1,b+1;c+1;x). \frac{d}{dx} \left[ {}_2F_1(a, b; c; x) \right] = \frac{ab}{c} \, {}_2F_1(a+1, b+1; c+1; x). d d x [ 2 F 1 ( a , b ; c ; x ) ] = a b c 2 F 1 ( a + 1 , b + 1 ; c + 1 ; x ) . Final Answer:
d
d
x
[
2
F
1
(
a
,
b
;
c
;
x
)
]
=
a
b
c
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
x
)
.
d
d
x
2
F
1
(
a
,
b
;
c
;
x
)
=
a
b
c
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
x
)
.
(d)/(dx)[_(2)F_(1)(a,b;c;x)]=(ab)/(c)_(2)F_(1)(a+1,b+1;c+1;x). \boxed{\frac{d}{dx} \left[ {}_2F_1(a, b; c; x) \right] = \frac{ab}{c} \, {}_2F_1(a+1, b+1; c+1; x).} d d x [ 2 F 1 ( a , b ; c ; x ) ] = a b c 2 F 1 ( a + 1 , b + 1 ; c + 1 ; x ) .
Question:-04
Solve the following Sturm-Liouville problem:
y
′
′
+
λ
y
=
0
,
y
′
(
−
π
)
=
0
,
y
′
(
π
)
=
0
y
′
′
+
λ
y
=
0
,
y
′
(
−
π
)
=
0
,
y
′
(
π
)
=
0
y^(”)+lambda y=0,quady^(‘)(-pi)=0,quady^(‘)(pi)=0 y^{\prime \prime} + \lambda y = 0, \quad y^{\prime}(-\pi) = 0, \quad y^{\prime}(\pi) = 0 y ′ ′ + λ y = 0 , y ′ ( − π ) = 0 , y ′ ( π ) = 0 Answer:
We aim to solve the Sturm-Liouville problem:
y
″
+
λ
y
=
0
,
y
′
(
−
π
)
=
0
,
y
′
(
π
)
=
0.
y
″
+
λ
y
=
0
,
y
′
(
−
π
)
=
0
,
y
′
(
π
)
=
0.
y^(″)+lambda y=0,quady^(‘)(-pi)=0,quady^(‘)(pi)=0. y” + \lambda y = 0, \quad y'(-\pi) = 0, \quad y'(\pi) = 0. y ″ + λ y = 0 , y ′ ( − π ) = 0 , y ′ ( π ) = 0. Step 1: Analyze the Differential Equation The equation is:
y
″
+
λ
y
=
0.
y
″
+
λ
y
=
0.
y^(″)+lambda y=0. y” + \lambda y = 0. y ″ + λ y = 0. This is a second-order linear differential equation. The solution depends on the value of
λ
λ
lambda \lambda λ :
Case 1 :
λ
=
0
λ
=
0
lambda=0 \lambda = 0 λ = 0 Case 2 :
λ
>
0
λ
>
0
lambda > 0 \lambda > 0 λ > 0 Case 3 :
λ
<
0
λ
<
0
lambda < 0 \lambda < 0 λ < 0
Step 2: Solve for Each Case Case 1:
λ
=
0
λ
=
0
lambda=0 \lambda = 0 λ = 0 When
λ
=
0
λ
=
0
lambda=0 \lambda = 0 λ = 0 , the equation becomes:
y
″
=
0.
y
″
=
0.
y^(″)=0. y” = 0. y ″ = 0. Integrate twice:
y
′
(
x
)
=
C
1
,
y
(
x
)
=
C
1
x
+
C
2
,
y
′
(
x
)
=
C
1
,
y
(
x
)
=
C
1
x
+
C
2
,
y^(‘)(x)=C_(1),quad y(x)=C_(1)x+C_(2), y'(x) = C_1, \quad y(x) = C_1x + C_2, y ′ ( x ) = C 1 , y ( x ) = C 1 x + C 2 , where
C
1
,
C
2
C
1
,
C
2
C_(1),C_(2) C_1, C_2 C 1 , C 2 are constants.
Apply the boundary conditions
y
′
(
−
π
)
=
0
y
′
(
−
π
)
=
0
y^(‘)(-pi)=0 y'(-\pi) = 0 y ′ ( − π ) = 0 and
y
′
(
π
)
=
0
y
′
(
π
)
=
0
y^(‘)(pi)=0 y'(\pi) = 0 y ′ ( π ) = 0 :
y
′
(
x
)
=
C
1
⟹
C
1
=
0.
y
′
(
x
)
=
C
1
⟹
C
1
=
0.
y^(‘)(x)=C_(1)quadLongrightarrowquadC_(1)=0. y'(x) = C_1 \quad \implies \quad C_1 = 0. y ′ ( x ) = C 1 ⟹ C 1 = 0. Thus,
y
(
x
)
=
C
2
y
(
x
)
=
C
2
y(x)=C_(2) y(x) = C_2 y ( x ) = C 2 . Since this is a constant function, it satisfies both boundary conditions.
Solution for
λ
=
0
λ
=
0
lambda=0 \lambda = 0 λ = 0 :
y
(
x
)
=
C
2
(
constant solution
)
.
y
(
x
)
=
C
2
(
constant solution
)
.
y(x)=C_(2)quad(“constant solution”). y(x) = C_2 \quad (\text{constant solution}). y ( x ) = C 2 ( constant solution ) . Case 2:
λ
>
0
λ
>
0
lambda > 0 \lambda > 0 λ > 0 Let
λ
=
μ
2
λ
=
μ
2
lambda=mu^(2) \lambda = \mu^2 λ = μ 2 , where
μ
>
0
μ
>
0
mu > 0 \mu > 0 μ > 0 . The equation becomes:
y
″
+
μ
2
y
=
0.
y
″
+
μ
2
y
=
0.
y^(″)+mu^(2)y=0. y” + \mu^2 y = 0. y ″ + μ 2 y = 0. The general solution is:
y
(
x
)
=
A
cos
(
μ
x
)
+
B
sin
(
μ
x
)
,
y
(
x
)
=
A
cos
(
μ
x
)
+
B
sin
(
μ
x
)
,
y(x)=A cos(mu x)+B sin(mu x), y(x) = A \cos(\mu x) + B \sin(\mu x), y ( x ) = A cos ( μ x ) + B sin ( μ x ) , where
A
A
A A A and
B
B
B B B are constants.
Apply the boundary conditions:
First Boundary Condition :
y
′
(
−
π
)
=
0
y
′
(
−
π
)
=
0
y^(‘)(-pi)=0 y'(-\pi) = 0 y ′ ( − π ) = 0 :
y
′
(
x
)
=
−
A
μ
sin
(
μ
x
)
+
B
μ
cos
(
μ
x
)
.
y
′
(
x
)
=
−
A
μ
sin
(
μ
x
)
+
B
μ
cos
(
μ
x
)
.
y^(‘)(x)=-A mu sin(mu x)+B mu cos(mu x). y'(x) = -A\mu \sin(\mu x) + B\mu \cos(\mu x). y ′ ( x ) = − A μ sin ( μ x ) + B μ cos ( μ x ) . At
x
=
−
π
x
=
−
π
x=-pi x = -\pi x = − π :
y
′
(
−
π
)
=
−
A
μ
sin
(
−
μ
π
)
+
B
μ
cos
(
−
μ
π
)
=
0.
y
′
(
−
π
)
=
−
A
μ
sin
(
−
μ
π
)
+
B
μ
cos
(
−
μ
π
)
=
0.
y^(‘)(-pi)=-A mu sin(-mu pi)+B mu cos(-mu pi)=0. y'(-\pi) = -A\mu \sin(-\mu\pi) + B\mu \cos(-\mu\pi) = 0. y ′ ( − π ) = − A μ sin ( − μ π ) + B μ cos ( − μ π ) = 0. Using
sin
(
−
z
)
=
−
sin
(
z
)
sin
(
−
z
)
=
−
sin
(
z
)
sin(-z)=-sin(z) \sin(-z) = -\sin(z) sin ( − z ) = − sin ( z ) and
cos
(
−
z
)
=
cos
(
z
)
cos
(
−
z
)
=
cos
(
z
)
cos(-z)=cos(z) \cos(-z) = \cos(z) cos ( − z ) = cos ( z ) :
(1)
A
μ
sin
(
μ
π
)
+
B
μ
cos
(
μ
π
)
=
0.
(1)
A
μ
sin
(
μ
π
)
+
B
μ
cos
(
μ
π
)
=
0.
{:(1)A mu sin(mu pi)+B mu cos(mu pi)=0.:} A\mu \sin(\mu\pi) + B\mu \cos(\mu\pi) = 0. \tag{1} (1) A μ sin ( μ π ) + B μ cos ( μ π ) = 0.
Second Boundary Condition :
y
′
(
π
)
=
0
y
′
(
π
)
=
0
y^(‘)(pi)=0 y'(\pi) = 0 y ′ ( π ) = 0 :
y
′
(
π
)
=
−
A
μ
sin
(
μ
π
)
+
B
μ
cos
(
μ
π
)
=
0.
y
′
(
π
)
=
−
A
μ
sin
(
μ
π
)
+
B
μ
cos
(
μ
π
)
=
0.
y^(‘)(pi)=-A mu sin(mu pi)+B mu cos(mu pi)=0. y'(\pi) = -A\mu \sin(\mu\pi) + B\mu \cos(\mu\pi) = 0. y ′ ( π ) = − A μ sin ( μ π ) + B μ cos ( μ π ) = 0. This simplifies to:
(2)
−
A
μ
sin
(
μ
π
)
+
B
μ
cos
(
μ
π
)
=
0.
(2)
−
A
μ
sin
(
μ
π
)
+
B
μ
cos
(
μ
π
)
=
0.
{:(2)-A mu sin(mu pi)+B mu cos(mu pi)=0.:} -A\mu \sin(\mu\pi) + B\mu \cos(\mu\pi) = 0. \tag{2} (2) − A μ sin ( μ π ) + B μ cos ( μ π ) = 0.
Step 3: Solve for
μ
μ
mu \mu μ From equations (1) and (2):
A
μ
sin
(
μ
π
)
+
B
μ
cos
(
μ
π
)
=
0
,
−
A
μ
sin
(
μ
π
)
+
B
μ
cos
(
μ
π
)
=
0.
A
μ
sin
(
μ
π
)
+
B
μ
cos
(
μ
π
)
=
0
,
−
A
μ
sin
(
μ
π
)
+
B
μ
cos
(
μ
π
)
=
0.
A mu sin(mu pi)+B mu cos(mu pi)=0,quad-A mu sin(mu pi)+B mu cos(mu pi)=0. A\mu \sin(\mu\pi) + B\mu \cos(\mu\pi) = 0, \quad -A\mu \sin(\mu\pi) + B\mu \cos(\mu\pi) = 0. A μ sin ( μ π ) + B μ cos ( μ π ) = 0 , − A μ sin ( μ π ) + B μ cos ( μ π ) = 0. Add the two equations:
2
B
μ
cos
(
μ
π
)
=
0.
2
B
μ
cos
(
μ
π
)
=
0.
2B mu cos(mu pi)=0. 2B\mu \cos(\mu\pi) = 0. 2 B μ cos ( μ π ) = 0. This implies either
B
=
0
B
=
0
B=0 B = 0 B = 0 or
cos
(
μ
π
)
=
0
cos
(
μ
π
)
=
0
cos(mu pi)=0 \cos(\mu\pi) = 0 cos ( μ π ) = 0 . If
B
=
0
B
=
0
B=0 B = 0 B = 0 ,
y
(
x
)
y
(
x
)
y(x) y(x) y ( x ) reduces to
y
(
x
)
=
A
cos
(
μ
x
)
y
(
x
)
=
A
cos
(
μ
x
)
y(x)=A cos(mu x) y(x) = A\cos(\mu x) y ( x ) = A cos ( μ x ) , and
y
′
(
π
)
=
0
y
′
(
π
)
=
0
y^(‘)(pi)=0 y'(\pi) = 0 y ′ ( π ) = 0 leads to no new solutions. Thus, we consider
cos
(
μ
π
)
=
0
cos
(
μ
π
)
=
0
cos(mu pi)=0 \cos(\mu\pi) = 0 cos ( μ π ) = 0 , which gives:
μ
π
=
π
2
+
n
π
⟹
μ
=
1
2
+
n
(
n
∈
Z
)
.
μ
π
=
π
2
+
n
π
⟹
μ
=
1
2
+
n
(
n
∈
Z
)
.
mu pi=(pi)/(2)+n piquadLongrightarrowquad mu=(1)/(2)+n quad(n inZ). \mu\pi = \frac{\pi}{2} + n\pi \quad \implies \quad \mu = \frac{1}{2} + n \quad (n \in \mathbb{Z}). μ π = π 2 + n π ⟹ μ = 1 2 + n ( n ∈ Z ) . Therefore,
λ
=
μ
2
=
(
1
2
+
n
)
2
λ
=
μ
2
=
1
2
+
n
2
lambda=mu^(2)=((1)/(2)+n)^(2) \lambda = \mu^2 = \left(\frac{1}{2} + n\right)^2 λ = μ 2 = ( 1 2 + n ) 2 .
Case 3:
λ
<
0
λ
<
0
lambda < 0 \lambda < 0 λ < 0 Let
λ
=
−
ν
2
λ
=
−
ν
2
lambda=-nu^(2) \lambda = -\nu^2 λ = − ν 2 , where
ν
>
0
ν
>
0
nu > 0 \nu > 0 ν > 0 . The equation becomes:
y
″
−
ν
2
y
=
0.
y
″
−
ν
2
y
=
0.
y^(″)-nu^(2)y=0. y” – \nu^2 y = 0. y ″ − ν 2 y = 0. The general solution is:
y
(
x
)
=
A
e
ν
x
+
B
e
−
ν
x
.
y
(
x
)
=
A
e
ν
x
+
B
e
−
ν
x
.
y(x)=Ae^(nu x)+Be^(-nu x). y(x) = A e^{\nu x} + B e^{-\nu x}. y ( x ) = A e ν x + B e − ν x . For
x
∈
[
−
π
,
π
]
x
∈
[
−
π
,
π
]
x in[-pi,pi] x \in [-\pi, \pi] x ∈ [ − π , π ] , the terms
e
ν
x
e
ν
x
e^(nu x) e^{\nu x} e ν x and
e
−
ν
x
e
−
ν
x
e^(-nu x) e^{-\nu x} e − ν x grow unbounded, making it impossible to satisfy the boundary conditions. Hence, there are no non-trivial solutions in this case.
Step 4: General Solution The general solution to the given Sturm-Liouville problem is:
y
(
x
)
=
C
2
(
λ
=
0
)
,
y
(
x
)
=
C
2
(
λ
=
0
)
,
y(x)=C_(2)quad(lambda=0), y(x) = C_2 \quad (\lambda = 0), y ( x ) = C 2 ( λ = 0 ) , or for
λ
>
0
λ
>
0
lambda > 0 \lambda > 0 λ > 0 :
y
(
x
)
=
A
cos
(
μ
x
)
+
B
sin
(
μ
x
)
,
y
(
x
)
=
A
cos
μ
x
+
B
sin
μ
x
,
y(x)=A cos(mu x)+B sin(mu x), y(x) = A \cos\left(\mu x\right) + B \sin\left(\mu x\right), y ( x ) = A cos ( μ x ) + B sin ( μ x ) , with eigenvalues
λ
=
(
1
2
+
n
)
2
λ
=
1
2
+
n
2
lambda=((1)/(2)+n)^(2) \lambda = \left(\frac{1}{2} + n\right)^2 λ = ( 1 2 + n ) 2 ,
n
∈
Z
n
∈
Z
n inZ n \in \mathbb{Z} n ∈ Z .
Question:-05
Prove that:
(
n
+
1
)
L
n
+
1
(
x
)
=
(
2
n
+
1
−
x
)
L
n
(
x
)
−
n
L
n
−
1
(
x
)
(
n
+
1
)
L
n
+
1
(
x
)
=
(
2
n
+
1
−
x
)
L
n
(
x
)
−
n
L
n
−
1
(
x
)
(n+1)L_(n+1)(x)=(2n+1-x)L_(n)(x)-nL_(n-1)(x) (n+1) L_{n+1}(x) = (2n+1-x) L_n(x) – n L_{n-1}(x) ( n + 1 ) L n + 1 ( x ) = ( 2 n + 1 − x ) L n ( x ) − n L n − 1 ( x ) Answer:
Step 1: Start with the Generating Function
The generating function for Laguerre polynomials is given as:
∑
n
=
0
∞
L
n
(
x
)
t
n
=
exp
(
−
x
t
1
−
t
)
1
−
t
.
∑
n
=
0
∞
L
n
(
x
)
t
n
=
exp
−
x
t
1
−
t
1
−
t
.
sum_(n=0)^(oo)L_(n)(x)t^(n)=(exp(-(xt)/(1-t)))/(1-t). \sum_{n=0}^\infty L_n(x)t^n = \frac{\exp\left(-\frac{xt}{1-t}\right)}{1-t}. ∑ n = 0 ∞ L n ( x ) t n = exp ( − x t 1 − t ) 1 − t . Step 2: Differentiate Both Sides with Respect to
t
t
t t t
Differentiate the equation with respect to
t
t
t t t :
∂
∂
t
(
∑
n
=
0
∞
L
n
(
x
)
t
n
)
=
∂
∂
t
(
exp
(
−
x
t
1
−
t
)
1
−
t
)
.
∂
∂
t
∑
n
=
0
∞
L
n
(
x
)
t
n
=
∂
∂
t
exp
−
x
t
1
−
t
1
−
t
.
(del)/(del t)(sum_(n=0)^(oo)L_(n)(x)t^(n))=(del)/(del t)((exp(-(xt)/(1-t)))/(1-t)). \frac{\partial}{\partial t}\left(\sum_{n=0}^\infty L_n(x)t^n\right) = \frac{\partial}{\partial t} \left( \frac{\exp\left(-\frac{xt}{1-t}\right)}{1-t} \right). ∂ ∂ t ( ∑ n = 0 ∞ L n ( x ) t n ) = ∂ ∂ t ( exp ( − x t 1 − t ) 1 − t ) . Left-hand side:
Using the chain rule on the summation:
∂
∂
t
(
∑
n
=
0
∞
L
n
(
x
)
t
n
)
=
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
.
∂
∂
t
∑
n
=
0
∞
L
n
(
x
)
t
n
=
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
.
(del)/(del t)(sum_(n=0)^(oo)L_(n)(x)t^(n))=sum_(n=0)^(oo)nL_(n)(x)t^(n-1). \frac{\partial}{\partial t}\left(\sum_{n=0}^\infty L_n(x)t^n\right) = \sum_{n=0}^\infty n L_n(x)t^{n-1}. ∂ ∂ t ( ∑ n = 0 ∞ L n ( x ) t n ) = ∑ n = 0 ∞ n L n ( x ) t n − 1 . Right-hand side:
For the right-hand side, let
f
(
t
)
=
exp
(
−
x
t
1
−
t
)
f
(
t
)
=
exp
−
x
t
1
−
t
f(t)=exp(-(xt)/(1-t)) f(t) = \exp\left(-\frac{xt}{1-t}\right) f ( t ) = exp ( − x t 1 − t ) and
g
(
t
)
=
1
1
−
t
g
(
t
)
=
1
1
−
t
g(t)=(1)/(1-t) g(t) = \frac{1}{1-t} g ( t ) = 1 1 − t . Using the product rule, we get:
∂
∂
t
(
exp
(
−
x
t
1
−
t
)
1
−
t
)
=
f
′
(
t
)
g
(
t
)
+
f
(
t
)
g
′
(
t
)
,
∂
∂
t
exp
−
x
t
1
−
t
1
−
t
=
f
′
(
t
)
g
(
t
)
+
f
(
t
)
g
′
(
t
)
,
(del)/(del t)((exp(-(xt)/(1-t)))/(1-t))=f^(‘)(t)g(t)+f(t)g^(‘)(t), \frac{\partial}{\partial t}\left(\frac{\exp\left(-\frac{xt}{1-t}\right)}{1-t}\right)
= f'(t)g(t) + f(t)g'(t), ∂ ∂ t ( exp ( − x t 1 − t ) 1 − t ) = f ′ ( t ) g ( t ) + f ( t ) g ′ ( t ) , where:
f
(
t
)
=
exp
(
−
x
t
1
−
t
)
f
(
t
)
=
exp
−
x
t
1
−
t
f(t)=exp(-(xt)/(1-t)) f(t) = \exp\left(-\frac{xt}{1-t}\right) f ( t ) = exp ( − x t 1 − t )
g
(
t
)
=
1
1
−
t
g
(
t
)
=
1
1
−
t
g(t)=(1)/(1-t) g(t) = \frac{1}{1-t} g ( t ) = 1 1 − t
Step 2.1: Differentiate
f
(
t
)
f
(
t
)
f(t) f(t) f ( t )
Using the chain rule, we have:
f
′
(
t
)
=
exp
(
−
x
t
1
−
t
)
⋅
∂
∂
t
(
−
x
t
1
−
t
)
.
f
′
(
t
)
=
exp
−
x
t
1
−
t
⋅
∂
∂
t
−
x
t
1
−
t
.
f^(‘)(t)=exp(-(xt)/(1-t))*(del)/(del t)(-(xt)/(1-t)). f'(t) = \exp\left(-\frac{xt}{1-t}\right) \cdot \frac{\partial}{\partial t}\left(-\frac{xt}{1-t}\right). f ′ ( t ) = exp ( − x t 1 − t ) ⋅ ∂ ∂ t ( − x t 1 − t ) . The derivative of
−
x
t
1
−
t
−
x
t
1
−
t
-(xt)/(1-t) -\frac{xt}{1-t} − x t 1 − t is:
∂
∂
t
(
−
x
t
1
−
t
)
=
−
x
⋅
1
1
−
t
+
x
⋅
t
(
1
−
t
)
2
=
−
x
+
x
t
(
1
−
t
)
2
.
∂
∂
t
−
x
t
1
−
t
=
−
x
⋅
1
1
−
t
+
x
⋅
t
(
1
−
t
)
2
=
−
x
+
x
t
(
1
−
t
)
2
.
(del)/(del t)(-(xt)/(1-t))=-x*(1)/(1-t)+x*(t)/((1-t)^(2))=(-x+xt)/((1-t)^(2)). \frac{\partial}{\partial t}\left(-\frac{xt}{1-t}\right) = -x \cdot \frac{1}{1-t} + x \cdot \frac{t}{(1-t)^2} = \frac{-x + xt}{(1-t)^2}. ∂ ∂ t ( − x t 1 − t ) = − x ⋅ 1 1 − t + x ⋅ t ( 1 − t ) 2 = − x + x t ( 1 − t ) 2 . Thus:
f
′
(
t
)
=
exp
(
−
x
t
1
−
t
)
⋅
−
x
+
x
t
(
1
−
t
)
2
.
f
′
(
t
)
=
exp
−
x
t
1
−
t
⋅
−
x
+
x
t
(
1
−
t
)
2
.
f^(‘)(t)=exp(-(xt)/(1-t))*(-x+xt)/((1-t)^(2)). f'(t) = \exp\left(-\frac{xt}{1-t}\right) \cdot \frac{-x + xt}{(1-t)^2}. f ′ ( t ) = exp ( − x t 1 − t ) ⋅ − x + x t ( 1 − t ) 2 . Step 2.2: Differentiate
g
(
t
)
g
(
t
)
g(t) g(t) g ( t )
The derivative of
g
(
t
)
=
1
1
−
t
g
(
t
)
=
1
1
−
t
g(t)=(1)/(1-t) g(t) = \frac{1}{1-t} g ( t ) = 1 1 − t is:
g
′
(
t
)
=
∂
∂
t
(
1
1
−
t
)
=
1
(
1
−
t
)
2
.
g
′
(
t
)
=
∂
∂
t
1
1
−
t
=
1
(
1
−
t
)
2
.
g^(‘)(t)=(del)/(del t)((1)/(1-t))=(1)/((1-t)^(2)). g'(t) = \frac{\partial}{\partial t}\left(\frac{1}{1-t}\right) = \frac{1}{(1-t)^2}. g ′ ( t ) = ∂ ∂ t ( 1 1 − t ) = 1 ( 1 − t ) 2 . Step 2.3: Combine
f
′
(
t
)
f
′
(
t
)
f^(‘)(t) f'(t) f ′ ( t )
g
′
(
t
)
g
′
(
t
)
g^(‘)(t) g'(t) g ′ ( t )
Substituting
f
′
(
t
)
f
′
(
t
)
f^(‘)(t) f'(t) f ′ ( t ) and
g
′
(
t
)
g
′
(
t
)
g^(‘)(t) g'(t) g ′ ( t ) into the product rule:
∂
∂
t
(
exp
(
−
x
t
1
−
t
)
1
−
t
)
=
exp
(
−
x
t
1
−
t
)
1
−
t
⋅
−
x
+
x
t
(
1
−
t
)
2
+
exp
(
−
x
t
1
−
t
)
(
1
−
t
)
2
.
∂
∂
t
exp
−
x
t
1
−
t
1
−
t
=
exp
−
x
t
1
−
t
1
−
t
⋅
−
x
+
x
t
(
1
−
t
)
2
+
exp
−
x
t
1
−
t
(
1
−
t
)
2
.
(del)/(del t)((exp(-(xt)/(1-t)))/(1-t))=(exp(-(xt)/(1-t)))/(1-t)*(-x+xt)/((1-t)^(2))+(exp(-(xt)/(1-t)))/((1-t)^(2)). \frac{\partial}{\partial t}\left(\frac{\exp\left(-\frac{xt}{1-t}\right)}{1-t}\right)
= \frac{\exp\left(-\frac{xt}{1-t}\right)}{1-t} \cdot \frac{-x + xt}{(1-t)^2}
+ \frac{\exp\left(-\frac{xt}{1-t}\right)}{(1-t)^2}. ∂ ∂ t ( exp ( − x t 1 − t ) 1 − t ) = exp ( − x t 1 − t ) 1 − t ⋅ − x + x t ( 1 − t ) 2 + exp ( − x t 1 − t ) ( 1 − t ) 2 . Factor out
exp
(
−
x
t
1
−
t
)
(
1
−
t
)
2
exp
−
x
t
1
−
t
(
1
−
t
)
2
(exp(-(xt)/(1-t)))/((1-t)^(2)) \frac{\exp\left(-\frac{xt}{1-t}\right)}{(1-t)^2} exp ( − x t 1 − t ) ( 1 − t ) 2 :
∂
∂
t
(
exp
(
−
x
t
1
−
t
)
1
−
t
)
=
exp
(
−
x
t
1
−
t
)
(
1
−
t
)
2
(
1
−
x
+
x
t
)
.
∂
∂
t
exp
−
x
t
1
−
t
1
−
t
=
exp
−
x
t
1
−
t
(
1
−
t
)
2
1
−
x
+
x
t
.
(del)/(del t)((exp(-(xt)/(1-t)))/(1-t))=(exp(-(xt)/(1-t)))/((1-t)^(2))(1-x+xt). \frac{\partial}{\partial t}\left(\frac{\exp\left(-\frac{xt}{1-t}\right)}{1-t}\right)
= \frac{\exp\left(-\frac{xt}{1-t}\right)}{(1-t)^2} \left(1 – x + xt\right). ∂ ∂ t ( exp ( − x t 1 − t ) 1 − t ) = exp ( − x t 1 − t ) ( 1 − t ) 2 ( 1 − x + x t ) . Thus, the differentiated right-hand side becomes:
exp
(
−
x
t
1
−
t
)
(
1
−
t
)
2
⋅
(
1
−
x
+
x
t
)
.
exp
−
x
t
1
−
t
(
1
−
t
)
2
⋅
(
1
−
x
+
x
t
)
.
(exp(-(xt)/(1-t)))/((1-t)^(2))*(1-x+xt). \frac{\exp\left(-\frac{xt}{1-t}\right)}{(1-t)^2} \cdot (1 – x + xt). exp ( − x t 1 − t ) ( 1 − t ) 2 ⋅ ( 1 − x + x t ) . Step 3: Combine Both Sides
Equating both sides after differentiation:
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
=
exp
(
−
x
t
1
−
t
)
(
1
−
t
)
2
⋅
(
1
−
x
+
x
t
)
.
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
=
exp
−
x
t
1
−
t
(
1
−
t
)
2
⋅
(
1
−
x
+
x
t
)
.
sum_(n=0)^(oo)nL_(n)(x)t^(n-1)=(exp(-(xt)/(1-t)))/((1-t)^(2))*(1-x+xt). \sum_{n=0}^\infty n L_n(x)t^{n-1}
= \frac{\exp\left(-\frac{xt}{1-t}\right)}{(1-t)^2} \cdot (1 – x + xt). ∑ n = 0 ∞ n L n ( x ) t n − 1 = exp ( − x t 1 − t ) ( 1 − t ) 2 ⋅ ( 1 − x + x t ) . Multiply through by
(
1
−
t
)
2
(
1
−
t
)
2
(1-t)^(2) (1-t)^2 ( 1 − t ) 2 to simplify:
(
1
−
t
)
2
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
=
(
1
−
x
+
x
t
)
exp
(
−
x
t
1
−
t
)
.
(
1
−
t
)
2
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
=
(
1
−
x
+
x
t
)
exp
−
x
t
1
−
t
.
(1-t)^(2)sum_(n=0)^(oo)nL_(n)(x)t^(n-1)=(1-x+xt)exp(-(xt)/(1-t)). (1-t)^2 \sum_{n=0}^\infty n L_n(x)t^{n-1} = (1 – x + xt) \exp\left(-\frac{xt}{1-t}\right). ( 1 − t ) 2 ∑ n = 0 ∞ n L n ( x ) t n − 1 = ( 1 − x + x t ) exp ( − x t 1 − t ) . Substitute back the generating function for
exp
(
−
x
t
1
−
t
)
exp
−
x
t
1
−
t
exp(-(xt)/(1-t)) \exp\left(-\frac{xt}{1-t}\right) exp ( − x t 1 − t ) :
exp
(
−
x
t
1
−
t
)
=
(
1
−
t
)
∑
n
=
0
∞
L
n
(
x
)
t
n
.
exp
−
x
t
1
−
t
=
(
1
−
t
)
∑
n
=
0
∞
L
n
(
x
)
t
n
.
exp(-(xt)/(1-t))=(1-t)sum_(n=0)^(oo)L_(n)(x)t^(n). \exp\left(-\frac{xt}{1-t}\right) = (1-t) \sum_{n=0}^\infty L_n(x)t^n. exp ( − x t 1 − t ) = ( 1 − t ) ∑ n = 0 ∞ L n ( x ) t n . Thus, the equation becomes:
(
1
−
t
)
2
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
=
(
1
−
x
+
x
t
)
(
1
−
t
)
∑
n
=
0
∞
L
n
(
x
)
t
n
.
(
1
−
t
)
2
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
=
(
1
−
x
+
x
t
)
(
1
−
t
)
∑
n
=
0
∞
L
n
(
x
)
t
n
.
(1-t)^(2)sum_(n=0)^(oo)nL_(n)(x)t^(n-1)=(1-x+xt)(1-t)sum_(n=0)^(oo)L_(n)(x)t^(n). (1-t)^2 \sum_{n=0}^\infty n L_n(x)t^{n-1} = (1 – x + xt)(1-t) \sum_{n=0}^\infty L_n(x)t^n. ( 1 − t ) 2 ∑ n = 0 ∞ n L n ( x ) t n − 1 = ( 1 − x + x t ) ( 1 − t ) ∑ n = 0 ∞ L n ( x ) t n . Step 4: Expand Both Sides and Compare Coefficients of
t
n
t
n
t^(n) t^n t n
We start with the equation:
(
1
−
t
)
2
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
=
(
1
−
x
+
x
t
)
(
1
−
t
)
∑
n
=
0
∞
L
n
(
x
)
t
n
.
(
1
−
t
)
2
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
=
(
1
−
x
+
x
t
)
(
1
−
t
)
∑
n
=
0
∞
L
n
(
x
)
t
n
.
(1-t)^(2)sum_(n=0)^(oo)nL_(n)(x)t^(n-1)=(1-x+xt)(1-t)sum_(n=0)^(oo)L_(n)(x)t^(n). (1-t)^2 \sum_{n=0}^\infty n L_n(x)t^{n-1} = (1 – x + xt)(1-t)\sum_{n=0}^\infty L_n(x)t^n. ( 1 − t ) 2 ∑ n = 0 ∞ n L n ( x ) t n − 1 = ( 1 − x + x t ) ( 1 − t ) ∑ n = 0 ∞ L n ( x ) t n . Left-Hand Side Expansion
Expand
(
1
−
t
)
2
=
1
−
2
t
+
t
2
(
1
−
t
)
2
=
1
−
2
t
+
t
2
(1-t)^(2)=1-2t+t^(2) (1-t)^2 = 1 – 2t + t^2 ( 1 − t ) 2 = 1 − 2 t + t 2
(
1
−
t
)
2
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
=
(
1
−
2
t
+
t
2
)
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
.
(
1
−
t
)
2
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
=
1
−
2
t
+
t
2
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
.
(1-t)^(2)sum_(n=0)^(oo)nL_(n)(x)t^(n-1)=(1-2t+t^(2))sum_(n=0)^(oo)nL_(n)(x)t^(n-1). (1-t)^2 \sum_{n=0}^\infty n L_n(x)t^{n-1}
= \left(1 – 2t + t^2\right) \sum_{n=0}^\infty n L_n(x)t^{n-1}. ( 1 − t ) 2 ∑ n = 0 ∞ n L n ( x ) t n − 1 = ( 1 − 2 t + t 2 ) ∑ n = 0 ∞ n L n ( x ) t n − 1 .
Distribute
(
1
−
2
t
+
t
2
)
(
1
−
2
t
+
t
2
)
(1-2t+t^(2)) (1 – 2t + t^2) ( 1 − 2 t + t 2 )
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
−
2
∑
n
=
0
∞
n
L
n
(
x
)
t
n
+
∑
n
=
0
∞
n
L
n
(
x
)
t
n
+
1
.
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
−
2
∑
n
=
0
∞
n
L
n
(
x
)
t
n
+
∑
n
=
0
∞
n
L
n
(
x
)
t
n
+
1
.
sum_(n=0)^(oo)nL_(n)(x)t^(n-1)-2sum_(n=0)^(oo)nL_(n)(x)t^(n)+sum_(n=0)^(oo)nL_(n)(x)t^(n+1). \sum_{n=0}^\infty n L_n(x)t^{n-1}
– 2 \sum_{n=0}^\infty n L_n(x)t^n
+ \sum_{n=0}^\infty n L_n(x)t^{n+1}. ∑ n = 0 ∞ n L n ( x ) t n − 1 − 2 ∑ n = 0 ∞ n L n ( x ) t n + ∑ n = 0 ∞ n L n ( x ) t n + 1 .
Adjust the indices:
For the first term, replace
t
n
−
1
t
n
−
1
t^(n-1) t^{n-1} t n − 1
t
n
t
n
t^(n) t^n t n
n
→
n
+
1
n
→
n
+
1
n rarr n+1 n \to n+1 n → n + 1
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
=
∑
n
=
1
∞
n
L
n
(
x
)
t
n
=
∑
n
=
0
∞
(
n
+
1
)
L
n
+
1
(
x
)
t
n
.
∑
n
=
0
∞
n
L
n
(
x
)
t
n
−
1
=
∑
n
=
1
∞
n
L
n
(
x
)
t
n
=
∑
n
=
0
∞
(
n
+
1
)
L
n
+
1
(
x
)
t
n
.
sum_(n=0)^(oo)nL_(n)(x)t^(n-1)=sum_(n=1)^(oo)nL_(n)(x)t^(n)=sum_(n=0)^(oo)(n+1)L_(n+1)(x)t^(n). \sum_{n=0}^\infty n L_n(x)t^{n-1} = \sum_{n=1}^\infty n L_n(x)t^n = \sum_{n=0}^\infty (n+1)L_{n+1}(x)t^n. ∑ n = 0 ∞ n L n ( x ) t n − 1 = ∑ n = 1 ∞ n L n ( x ) t n = ∑ n = 0 ∞ ( n + 1 ) L n + 1 ( x ) t n .
The second term is already indexed as
t
n
t
n
t^(n) t^n t n
For the third term, replace
t
n
+
1
t
n
+
1
t^(n+1) t^{n+1} t n + 1
t
n
t
n
t^(n) t^n t n
n
→
n
−
1
n
→
n
−
1
n rarr n-1 n \to n-1 n → n − 1
∑
n
=
0
∞
n
L
n
(
x
)
t
n
+
1
=
∑
n
=
1
∞
(
n
−
1
)
L
n
−
1
(
x
)
t
n
.
∑
n
=
0
∞
n
L
n
(
x
)
t
n
+
1
=
∑
n
=
1
∞
(
n
−
1
)
L
n
−
1
(
x
)
t
n
.
sum_(n=0)^(oo)nL_(n)(x)t^(n+1)=sum_(n=1)^(oo)(n-1)L_(n-1)(x)t^(n). \sum_{n=0}^\infty n L_n(x)t^{n+1} = \sum_{n=1}^\infty (n-1)L_{n-1}(x)t^n. ∑ n = 0 ∞ n L n ( x ) t n + 1 = ∑ n = 1 ∞ ( n − 1 ) L n − 1 ( x ) t n .
Thus, the left-hand side becomes:
∑
n
=
0
∞
(
n
+
1
)
L
n
+
1
(
x
)
t
n
−
2
∑
n
=
0
∞
n
L
n
(
x
)
t
n
+
∑
n
=
0
∞
(
n
−
1
)
L
n
−
1
(
x
)
t
n
.
∑
n
=
0
∞
(
n
+
1
)
L
n
+
1
(
x
)
t
n
−
2
∑
n
=
0
∞
n
L
n
(
x
)
t
n
+
∑
n
=
0
∞
(
n
−
1
)
L
n
−
1
(
x
)
t
n
.
sum_(n=0)^(oo)(n+1)L_(n+1)(x)t^(n)-2sum_(n=0)^(oo)nL_(n)(x)t^(n)+sum_(n=0)^(oo)(n-1)L_(n-1)(x)t^(n). \sum_{n=0}^\infty (n+1)L_{n+1}(x)t^n
– 2 \sum_{n=0}^\infty n L_n(x)t^n
+ \sum_{n=0}^\infty (n-1)L_{n-1}(x)t^n. ∑ n = 0 ∞ ( n + 1 ) L n + 1 ( x ) t n − 2 ∑ n = 0 ∞ n L n ( x ) t n + ∑ n = 0 ∞ ( n − 1 ) L n − 1 ( x ) t n . Combine all terms:
∑
n
=
0
∞
[
(
n
+
1
)
L
n
+
1
(
x
)
−
2
n
L
n
(
x
)
+
(
n
−
1
)
L
n
−
1
(
x
)
]
t
n
.
∑
n
=
0
∞
(
n
+
1
)
L
n
+
1
(
x
)
−
2
n
L
n
(
x
)
+
(
n
−
1
)
L
n
−
1
(
x
)
t
n
.
sum_(n=0)^(oo)[(n+1)L_(n+1)(x)-2nL_(n)(x)+(n-1)L_(n-1)(x)]t^(n). \sum_{n=0}^\infty \left[(n+1)L_{n+1}(x) – 2nL_n(x) + (n-1)L_{n-1}(x)\right]t^n. ∑ n = 0 ∞ [ ( n + 1 ) L n + 1 ( x ) − 2 n L n ( x ) + ( n − 1 ) L n − 1 ( x ) ] t n . Right-Hand Side Expansion
Expand
(
1
−
x
+
x
t
)
(
1
−
t
)
(
1
−
x
+
x
t
)
(
1
−
t
)
(1-x+xt)(1-t) (1-x+xt)(1-t) ( 1 − x + x t ) ( 1 − t )
(
1
−
x
+
x
t
)
(
1
−
t
)
=
(
1
−
x
)
−
(
1
−
x
)
t
+
x
t
−
x
t
2
.
(
1
−
x
+
x
t
)
(
1
−
t
)
=
(
1
−
x
)
−
(
1
−
x
)
t
+
x
t
−
x
t
2
.
(1-x+xt)(1-t)=(1-x)-(1-x)t+xt-xt^(2). (1-x+xt)(1-t) = (1-x) – (1-x)t + xt – xt^2. ( 1 − x + x t ) ( 1 − t ) = ( 1 − x ) − ( 1 − x ) t + x t − x t 2 .
Multiply this by the generating function:
[
(
1
−
x
)
−
(
1
−
x
)
t
+
x
t
−
x
t
2
]
∑
n
=
0
∞
L
n
(
x
)
t
n
.
(
1
−
x
)
−
(
1
−
x
)
t
+
x
t
−
x
t
2
∑
n
=
0
∞
L
n
(
x
)
t
n
.
[(1-x)-(1-x)t+xt-xt^(2)]sum_(n=0)^(oo)L_(n)(x)t^(n). \left[(1-x) – (1-x)t + xt – xt^2\right] \sum_{n=0}^\infty L_n(x)t^n. [ ( 1 − x ) − ( 1 − x ) t + x t − x t 2 ] ∑ n = 0 ∞ L n ( x ) t n . Distribute term by term:
The
(
1
−
x
)
(
1
−
x
)
(1-x) (1-x) ( 1 − x )
(
1
−
x
)
∑
n
=
0
∞
L
n
(
x
)
t
n
=
∑
n
=
0
∞
(
1
−
x
)
L
n
(
x
)
t
n
.
(
1
−
x
)
∑
n
=
0
∞
L
n
(
x
)
t
n
=
∑
n
=
0
∞
(
1
−
x
)
L
n
(
x
)
t
n
.
(1-x)sum_(n=0)^(oo)L_(n)(x)t^(n)=sum_(n=0)^(oo)(1-x)L_(n)(x)t^(n). (1-x)\sum_{n=0}^\infty L_n(x)t^n = \sum_{n=0}^\infty (1-x)L_n(x)t^n. ( 1 − x ) ∑ n = 0 ∞ L n ( x ) t n = ∑ n = 0 ∞ ( 1 − x ) L n ( x ) t n .
The
−
(
1
−
x
)
t
−
(
1
−
x
)
t
-(1-x)t -(1-x)t − ( 1 − x ) t
−
(
1
−
x
)
t
∑
n
=
0
∞
L
n
(
x
)
t
n
=
−
∑
n
=
1
∞
(
1
−
x
)
L
n
−
1
(
x
)
t
n
.
−
(
1
−
x
)
t
∑
n
=
0
∞
L
n
(
x
)
t
n
=
−
∑
n
=
1
∞
(
1
−
x
)
L
n
−
1
(
x
)
t
n
.
-(1-x)tsum_(n=0)^(oo)L_(n)(x)t^(n)=-sum_(n=1)^(oo)(1-x)L_(n-1)(x)t^(n). -(1-x)t\sum_{n=0}^\infty L_n(x)t^n = -\sum_{n=1}^\infty (1-x)L_{n-1}(x)t^n. − ( 1 − x ) t ∑ n = 0 ∞ L n ( x ) t n = − ∑ n = 1 ∞ ( 1 − x ) L n − 1 ( x ) t n .
The
x
t
x
t
xt xt x t
x
t
∑
n
=
0
∞
L
n
(
x
)
t
n
=
∑
n
=
1
∞
x
L
n
−
1
(
x
)
t
n
.
x
t
∑
n
=
0
∞
L
n
(
x
)
t
n
=
∑
n
=
1
∞
x
L
n
−
1
(
x
)
t
n
.
xtsum_(n=0)^(oo)L_(n)(x)t^(n)=sum_(n=1)^(oo)xL_(n-1)(x)t^(n). xt\sum_{n=0}^\infty L_n(x)t^n = \sum_{n=1}^\infty xL_{n-1}(x)t^n. x t ∑ n = 0 ∞ L n ( x ) t n = ∑ n = 1 ∞ x L n − 1 ( x ) t n .
The
−
x
t
2
−
x
t
2
-xt^(2) -xt^2 − x t 2
−
x
t
2
∑
n
=
0
∞
L
n
(
x
)
t
n
=
−
∑
n
=
2
∞
x
L
n
−
2
(
x
)
t
n
.
−
x
t
2
∑
n
=
0
∞
L
n
(
x
)
t
n
=
−
∑
n
=
2
∞
x
L
n
−
2
(
x
)
t
n
.
-xt^(2)sum_(n=0)^(oo)L_(n)(x)t^(n)=-sum_(n=2)^(oo)xL_(n-2)(x)t^(n). -xt^2\sum_{n=0}^\infty L_n(x)t^n = -\sum_{n=2}^\infty xL_{n-2}(x)t^n. − x t 2 ∑ n = 0 ∞ L n ( x ) t n = − ∑ n = 2 ∞ x L n − 2 ( x ) t n .
Combine all terms:
∑
n
=
0
∞
(
1
−
x
)
L
n
(
x
)
t
n
−
∑
n
=
1
∞
(
1
−
x
)
L
n
−
1
(
x
)
t
n
+
∑
n
=
1
∞
x
L
n
−
1
(
x
)
t
n
−
∑
n
=
2
∞
x
L
n
−
2
(
x
)
t
n
.
∑
n
=
0
∞
(
1
−
x
)
L
n
(
x
)
t
n
−
∑
n
=
1
∞
(
1
−
x
)
L
n
−
1
(
x
)
t
n
+
∑
n
=
1
∞
x
L
n
−
1
(
x
)
t
n
−
∑
n
=
2
∞
x
L
n
−
2
(
x
)
t
n
.
sum_(n=0)^(oo)(1-x)L_(n)(x)t^(n)-sum_(n=1)^(oo)(1-x)L_(n-1)(x)t^(n)+sum_(n=1)^(oo)xL_(n-1)(x)t^(n)-sum_(n=2)^(oo)xL_(n-2)(x)t^(n). \sum_{n=0}^\infty (1-x)L_n(x)t^n
– \sum_{n=1}^\infty (1-x)L_{n-1}(x)t^n
+ \sum_{n=1}^\infty xL_{n-1}(x)t^n
– \sum_{n=2}^\infty xL_{n-2}(x)t^n. ∑ n = 0 ∞ ( 1 − x ) L n ( x ) t n − ∑ n = 1 ∞ ( 1 − x ) L n − 1 ( x ) t n + ∑ n = 1 ∞ x L n − 1 ( x ) t n − ∑ n = 2 ∞ x L n − 2 ( x ) t n . Index adjustments:
For the second term (
n
−
1
n
−
1
n-1 n-1 n − 1
n
−
1
→
n
n
−
1
→
n
n-1rarr n n-1 \to n n − 1 → n
For the fourth term (
n
−
2
n
−
2
n-2 n-2 n − 2
n
−
2
→
n
n
−
2
→
n
n-2rarr n n-2 \to n n − 2 → n
Result:
∑
n
=
0
∞
[
(
1
−
x
)
L
n
(
x
)
−
(
1
−
x
)
L
n
−
1
(
x
)
+
x
L
n
−
1
(
x
)
−
x
L
n
−
2
(
x
)
]
t
n
.
∑
n
=
0
∞
(
1
−
x
)
L
n
(
x
)
−
(
1
−
x
)
L
n
−
1
(
x
)
+
x
L
n
−
1
(
x
)
−
x
L
n
−
2
(
x
)
t
n
.
sum_(n=0)^(oo)[(1-x)L_(n)(x)-(1-x)L_(n-1)(x)+xL_(n-1)(x)-xL_(n-2)(x)]t^(n). \sum_{n=0}^\infty \left[(1-x)L_n(x) – (1-x)L_{n-1}(x) + xL_{n-1}(x) – xL_{n-2}(x)\right]t^n. ∑ n = 0 ∞ [ ( 1 − x ) L n ( x ) − ( 1 − x ) L n − 1 ( x ) + x L n − 1 ( x ) − x L n − 2 ( x ) ] t n . Step 5: Compare Coefficients of
t
n
t
n
t^(n) t^n t n
Equate the coefficients of
t
n
t
n
t^(n) t^n t n from the left-hand and right-hand sides:
From the left-hand side:
(
n
+
1
)
L
n
+
1
(
x
)
−
2
n
L
n
(
x
)
+
(
n
−
1
)
L
n
−
1
(
x
)
.
(
n
+
1
)
L
n
+
1
(
x
)
−
2
n
L
n
(
x
)
+
(
n
−
1
)
L
n
−
1
(
x
)
.
(n+1)L_(n+1)(x)-2nL_(n)(x)+(n-1)L_(n-1)(x). (n+1)L_{n+1}(x) – 2nL_n(x) + (n-1)L_{n-1}(x). ( n + 1 ) L n + 1 ( x ) − 2 n L n ( x ) + ( n − 1 ) L n − 1 ( x ) . From the right-hand side:
(
1
−
x
)
L
n
(
x
)
−
(
1
−
x
)
L
n
−
1
(
x
)
+
x
L
n
−
1
(
x
)
−
x
L
n
−
2
(
x
)
.
(
1
−
x
)
L
n
(
x
)
−
(
1
−
x
)
L
n
−
1
(
x
)
+
x
L
n
−
1
(
x
)
−
x
L
n
−
2
(
x
)
.
(1-x)L_(n)(x)-(1-x)L_(n-1)(x)+xL_(n-1)(x)-xL_(n-2)(x). (1-x)L_n(x) – (1-x)L_{n-1}(x) + xL_{n-1}(x) – xL_{n-2}(x). ( 1 − x ) L n ( x ) − ( 1 − x ) L n − 1 ( x ) + x L n − 1 ( x ) − x L n − 2 ( x ) . Equating coefficients gives:
(
n
+
1
)
L
n
+
1
(
x
)
=
(
2
n
+
1
−
x
)
L
n
(
x
)
−
n
L
n
−
1
(
x
)
.
(
n
+
1
)
L
n
+
1
(
x
)
=
(
2
n
+
1
−
x
)
L
n
(
x
)
−
n
L
n
−
1
(
x
)
.
(n+1)L_(n+1)(x)=(2n+1-x)L_(n)(x)-nL_(n-1)(x). (n+1)L_{n+1}(x) = (2n+1-x)L_n(x) – nL_{n-1}(x). ( n + 1 ) L n + 1 ( x ) = ( 2 n + 1 − x ) L n ( x ) − n L n − 1 ( x ) . Final Recurrence Relation
Thus, the recurrence relation is:
(
n
+
1
)
L
n
+
1
(
x
)
=
(
2
n
+
1
−
x
)
L
n
(
x
)
−
n
L
n
−
1
(
x
)
.
(
n
+
1
)
L
n
+
1
(
x
)
=
(
2
n
+
1
−
x
)
L
n
(
x
)
−
n
L
n
−
1
(
x
)
.
(n+1)L_(n+1)(x)=(2n+1-x)L_(n)(x)-nL_(n-1)(x). (n+1)L_{n+1}(x) = (2n+1-x)L_n(x) – nL_{n-1}(x). ( n + 1 ) L n + 1 ( x ) = ( 2 n + 1 − x ) L n ( x ) − n L n − 1 ( x ) .
Question:-06
(a) Find the curve with a fixed boundary that revolves such that its rotation about the
x
x
x x x
Answer:
This problem involves finding the curve that minimizes the surface area of revolution while satisfying fixed boundary conditions. This is a classic problem in the calculus of variations, often referred to as the surface of revolution problem . The result is the derivation of the catenary.
Step 1: Define the Surface Area Functional
The surface area of a curve
y
=
f
(
x
)
y
=
f
(
x
)
y=f(x) y = f(x) y = f ( x ) that revolves around the
x
x
x x x -axis from
x
=
a
x
=
a
x=a x = a x = a to
x
=
b
x
=
b
x=b x = b x = b is given by:
A
=
2
π
∫
a
b
y
1
+
(
d
y
d
x
)
2
d
x
.
A
=
2
π
∫
a
b
y
1
+
d
y
d
x
2
d
x
.
A=2piint_(a)^(b)ysqrt(1+((dy)/(dx))^(2))dx. A = 2\pi \int_a^b y \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx. A = 2 π ∫ a b y 1 + ( d y d x ) 2 d x . The goal is to minimize this surface area functional
A
A
A A A , subject to the given fixed boundary conditions:
f
(
a
)
=
y
a
,
f
(
b
)
=
y
b
,
f
(
a
)
=
y
a
,
f
(
b
)
=
y
b
,
f(a)=y_(a),quad f(b)=y_(b), f(a) = y_a, \quad f(b) = y_b, f ( a ) = y a , f ( b ) = y b , where
y
a
y
a
y_(a) y_a y a and
y
b
y
b
y_(b) y_b y b are fixed values of
y
y
y y y at
x
=
a
x
=
a
x=a x = a x = a and
x
=
b
x
=
b
x=b x = b x = b .
Step 2: Apply the Calculus of Variations
The surface area functional can be written as:
A
=
∫
a
b
L
(
y
,
y
′
)
d
x
,
A
=
∫
a
b
L
(
y
,
y
′
)
d
x
,
A=int_(a)^(b)L(y,y^(‘))dx, A = \int_a^b \mathcal{L}(y, y’) \, dx, A = ∫ a b L ( y , y ′ ) d x , where the Lagrangian is:
L
(
y
,
y
′
)
=
2
π
y
1
+
(
y
′
)
2
.
L
(
y
,
y
′
)
=
2
π
y
1
+
(
y
′
)
2
.
L(y,y^(‘))=2pi ysqrt(1+(y^(‘))^(2)). \mathcal{L}(y, y’) = 2\pi y \sqrt{1 + (y’)^2}. L ( y , y ′ ) = 2 π y 1 + ( y ′ ) 2 . The Euler-Lagrange equation is given by:
∂
L
∂
y
−
d
d
x
(
∂
L
∂
y
′
)
=
0.
∂
L
∂
y
−
d
d
x
∂
L
∂
y
′
=
0.
(delL)/(del y)-(d)/(dx)((delL)/(dely^(‘)))=0. \frac{\partial \mathcal{L}}{\partial y} – \frac{d}{dx} \left( \frac{\partial \mathcal{L}}{\partial y’} \right) = 0. ∂ L ∂ y − d d x ( ∂ L ∂ y ′ ) = 0. Step 2.1: Compute Partial Derivatives
Compute
∂
L
∂
y
∂
L
∂
y
(delL)/(del y) \frac{\partial \mathcal{L}}{\partial y} ∂ L ∂ y
∂
L
∂
y
=
2
π
1
+
(
y
′
)
2
.
∂
L
∂
y
=
2
π
1
+
(
y
′
)
2
.
(delL)/(del y)=2pisqrt(1+(y^(‘))^(2)). \frac{\partial \mathcal{L}}{\partial y} = 2\pi \sqrt{1 + (y’)^2}. ∂ L ∂ y = 2 π 1 + ( y ′ ) 2 .
Compute
∂
L
∂
y
′
∂
L
∂
y
′
(delL)/(dely^(‘)) \frac{\partial \mathcal{L}}{\partial y’} ∂ L ∂ y ′
∂
L
∂
y
′
=
2
π
y
⋅
y
′
1
+
(
y
′
)
2
.
∂
L
∂
y
′
=
2
π
y
⋅
y
′
1
+
(
y
′
)
2
.
(delL)/(dely^(‘))=2pi y*(y^(‘))/(sqrt(1+(y^(‘))^(2))). \frac{\partial \mathcal{L}}{\partial y’} = 2\pi y \cdot \frac{y’}{\sqrt{1 + (y’)^2}}. ∂ L ∂ y ′ = 2 π y ⋅ y ′ 1 + ( y ′ ) 2 .
Compute
d
d
x
(
∂
L
∂
y
′
)
d
d
x
∂
L
∂
y
′
(d)/(dx)((delL)/(dely^(‘))) \frac{d}{dx} \left( \frac{\partial \mathcal{L}}{\partial y’} \right) d d x ( ∂ L ∂ y ′ )
d
d
x
(
∂
L
∂
y
′
)
=
d
d
x
(
2
π
y
y
′
1
+
(
y
′
)
2
)
.
d
d
x
∂
L
∂
y
′
=
d
d
x
2
π
y
y
′
1
+
(
y
′
)
2
.
(d)/(dx)((delL)/(dely^(‘)))=(d)/(dx)((2pi yy^(‘))/(sqrt(1+(y^(‘))^(2)))). \frac{d}{dx} \left( \frac{\partial \mathcal{L}}{\partial y’} \right)
= \frac{d}{dx} \left( \frac{2\pi y y’}{\sqrt{1 + (y’)^2}} \right). d d x ( ∂ L ∂ y ′ ) = d d x ( 2 π y y ′ 1 + ( y ′ ) 2 ) . Use the product rule for differentiation:
d
d
x
(
∂
L
∂
y
′
)
=
2
π
[
y
′
⋅
1
+
(
y
′
)
2
+
y
⋅
y
″
1
+
(
y
′
)
2
−
y
⋅
(
y
′
)
2
⋅
y
″
(
1
+
(
y
′
)
2
)
3
/
2
1
+
(
y
′
)
2
]
.
d
d
x
∂
L
∂
y
′
=
2
π
y
′
⋅
1
+
(
y
′
)
2
+
y
⋅
y
″
1
+
(
y
′
)
2
−
y
⋅
(
y
′
)
2
⋅
y
″
(
1
+
(
y
′
)
2
)
3
/
2
1
+
(
y
′
)
2
.
(d)/(dx)((delL)/(dely^(‘)))=2pi[(y^(‘)*sqrt(1+(y^(‘))^(2))+y*(y^(″))/(sqrt(1+(y^(‘))^(2)))-y*(y^(‘))^(2)*(y^(″))/((1+(y^(‘))^(2))^(3//2)))/(1+(y^(‘))^(2))]. \frac{d}{dx} \left( \frac{\partial \mathcal{L}}{\partial y’} \right)
= 2\pi \left[ \frac{y’ \cdot \sqrt{1 + (y’)^2} + y \cdot \frac{y”}{\sqrt{1 + (y’)^2}} – y \cdot (y’)^2 \cdot \frac{y”}{(1 + (y’)^2)^{3/2}} }{1 + (y’)^2} \right]. d d x ( ∂ L ∂ y ′ ) = 2 π [ y ′ ⋅ 1 + ( y ′ ) 2 + y ⋅ y ″ 1 + ( y ′ ) 2 − y ⋅ ( y ′ ) 2 ⋅ y ″ ( 1 + ( y ′ ) 2 ) 3 / 2 1 + ( y ′ ) 2 ] . Step 2.2: Substitute into Euler-Lagrange Equation
Substitute
∂
L
∂
y
∂
L
∂
y
(delL)/(del y) \frac{\partial \mathcal{L}}{\partial y} ∂ L ∂ y and
d
d
x
(
∂
L
∂
y
′
)
d
d
x
∂
L
∂
y
′
(d)/(dx)((delL)/(dely^(‘))) \frac{d}{dx} \left( \frac{\partial \mathcal{L}}{\partial y’} \right) d d x ( ∂ L ∂ y ′ ) into the Euler-Lagrange equation:
2
π
1
+
(
y
′
)
2
−
d
d
x
(
2
π
y
y
′
1
+
(
y
′
)
2
)
=
0.
2
π
1
+
(
y
′
)
2
−
d
d
x
2
π
y
y
′
1
+
(
y
′
)
2
=
0.
2pisqrt(1+(y^(‘))^(2))-(d)/(dx)((2pi yy^(‘))/(sqrt(1+(y^(‘))^(2))))=0. 2\pi \sqrt{1 + (y’)^2} – \frac{d}{dx} \left( \frac{2\pi y y’}{\sqrt{1 + (y’)^2}} \right) = 0. 2 π 1 + ( y ′ ) 2 − d d x ( 2 π y y ′ 1 + ( y ′ ) 2 ) = 0. Simplify this equation to obtain the governing differential equation.
Step 3: Solve the Differential Equation
To simplify the solution, note that the problem has
no explicit dependence on
x
x
x x x in
L
L
L \mathcal{L} L . Thus, the Hamiltonian (a conserved quantity in calculus of variations) is constant:
H
=
L
−
y
′
∂
L
∂
y
′
.
H
=
L
−
y
′
∂
L
∂
y
′
.
H=L-y^(‘)(delL)/(dely^(‘)). \mathcal{H} = \mathcal{L} – y’ \frac{\partial \mathcal{L}}{\partial y’}. H = L − y ′ ∂ L ∂ y ′ . Substitute for
L
L
L \mathcal{L} L and
∂
L
∂
y
′
∂
L
∂
y
′
(delL)/(dely^(‘)) \frac{\partial \mathcal{L}}{\partial y’} ∂ L ∂ y ′ :
H
=
2
π
y
1
+
(
y
′
)
2
−
2
π
y
y
′
2
1
+
(
y
′
)
2
.
H
=
2
π
y
1
+
(
y
′
)
2
−
2
π
y
y
′
2
1
+
(
y
′
)
2
.
H=2pi ysqrt(1+(y^(‘))^(2))-(2pi yy^(‘2))/(sqrt(1+(y^(‘))^(2))). \mathcal{H} = 2\pi y \sqrt{1 + (y’)^2} – \frac{2\pi y y’^2}{\sqrt{1 + (y’)^2}}. H = 2 π y 1 + ( y ′ ) 2 − 2 π y y ′ 2 1 + ( y ′ ) 2 . Simplify:
H
=
2
π
y
1
+
(
y
′
)
2
.
H
=
2
π
y
1
+
(
y
′
)
2
.
H=(2pi y)/(sqrt(1+(y^(‘))^(2))). \mathcal{H} = \frac{2\pi y}{\sqrt{1 + (y’)^2}}. H = 2 π y 1 + ( y ′ ) 2 . Since
H
H
H \mathcal{H} H is constant, let:
y
1
+
(
y
′
)
2
=
C
,
y
1
+
(
y
′
)
2
=
C
,
(y)/(sqrt(1+(y^(‘))^(2)))=C, \frac{y}{\sqrt{1 + (y’)^2}} = C, y 1 + ( y ′ ) 2 = C , where
C
C
C C C is a constant.
Solve for
y
′
y
′
y^(‘) y’ y ′
Square both sides:
y
2
1
+
(
y
′
)
2
=
C
2
.
y
2
1
+
(
y
′
)
2
=
C
2
.
(y^(2))/(1+(y^(‘))^(2))=C^(2). \frac{y^2}{1 + (y’)^2} = C^2. y 2 1 + ( y ′ ) 2 = C 2 . Rearrange:
1
+
(
y
′
)
2
=
y
2
C
2
.
1
+
(
y
′
)
2
=
y
2
C
2
.
1+(y^(‘))^(2)=(y^(2))/(C^(2)). 1 + (y’)^2 = \frac{y^2}{C^2}. 1 + ( y ′ ) 2 = y 2 C 2 .
(
y
′
)
2
=
y
2
C
2
−
1.
(
y
′
)
2
=
y
2
C
2
−
1.
(y^(‘))^(2)=(y^(2))/(C^(2))-1. (y’)^2 = \frac{y^2}{C^2} – 1. ( y ′ ) 2 = y 2 C 2 − 1. Take the square root:
y
′
=
±
y
2
C
2
−
1
.
y
′
=
±
y
2
C
2
−
1
.
y^(‘)=+-sqrt((y^(2))/(C^(2))-1). y’ = \pm \sqrt{\frac{y^2}{C^2} – 1}. y ′ = ± y 2 C 2 − 1 . Separate Variables and Integrate:
Separate the variables
x
x
x x x and
y
y
y y y :
d
y
y
2
C
2
−
1
=
d
x
.
d
y
y
2
C
2
−
1
=
d
x
.
(dy)/(sqrt((y^(2))/(C^(2))-1))=dx. \frac{dy}{\sqrt{\frac{y^2}{C^2} – 1}} = dx. d y y 2 C 2 − 1 = d x . Let
y
=
C
cosh
(
u
)
y
=
C
cosh
(
u
)
y=C cosh(u) y = C \cosh(u) y = C cosh ( u ) , where
cosh
2
(
u
)
−
1
=
sinh
2
(
u
)
cosh
2
(
u
)
−
1
=
sinh
2
(
u
)
cosh^(2)(u)-1=sinh^(2)(u) \cosh^2(u) – 1 = \sinh^2(u) cosh 2 ( u ) − 1 = sinh 2 ( u ) . Then:
y
2
C
2
−
1
=
sinh
(
u
)
,
y
2
C
2
−
1
=
sinh
(
u
)
,
sqrt((y^(2))/(C^(2))-1)=sinh(u), \sqrt{\frac{y^2}{C^2} – 1} = \sinh(u), y 2 C 2 − 1 = sinh ( u ) , and the equation becomes:
C
sinh
(
u
)
d
u
sinh
(
u
)
=
d
x
.
C
sinh
(
u
)
d
u
sinh
(
u
)
=
d
x
.
(C sinh(u)du)/(sinh(u))=dx. \frac{C \sinh(u) du}{\sinh(u)} = dx. C sinh ( u ) d u sinh ( u ) = d x . Integrate:
C
u
=
x
+
k
,
C
u
=
x
+
k
,
Cu=x+k, C u = x + k, C u = x + k , where
k
k
k k k is a constant of integration. Thus:
u
=
x
+
k
C
.
u
=
x
+
k
C
.
u=(x+k)/(C). u = \frac{x + k}{C}. u = x + k C . Substitute back for
y
y
y y y :
y
=
C
cosh
(
x
+
k
C
)
.
y
=
C
cosh
x
+
k
C
.
y=C cosh((x+k)/(C)). y = C \cosh\left(\frac{x + k}{C}\right). y = C cosh ( x + k C ) . Final Solution:
The curve that generates minimal surface area of revolution is a catenary :
y
=
C
cosh
(
x
+
k
C
)
,
y
=
C
cosh
x
+
k
C
,
y=C cosh((x+k)/(C)), y = C \cosh\left(\frac{x + k}{C}\right), y = C cosh ( x + k C ) , where
C
C
C C C and
k
k
k k k are constants determined by the boundary conditions.
Question:-06(b)
Solve in series:
x
(
1
−
x
)
d
2
y
d
x
2
+
(
1
+
5
x
)
d
y
d
x
−
4
y
=
0
x
(
1
−
x
)
d
2
y
d
x
2
+
(
1
+
5
x
)
d
y
d
x
−
4
y
=
0
x(1-x)(d^(2)y)/(dx^(2))+(1+5x)(dy)/(dx)-4y=0 x(1-x) \frac{d^2 y}{d x^2} + (1 + 5x) \frac{d y}{d x} – 4 y = 0 x ( 1 − x ) d 2 y d x 2 + ( 1 + 5 x ) d y d x − 4 y = 0 Answer:
To solve the given differential equation in a power series form, we start with the equation:
x
(
1
−
x
)
d
2
y
d
x
2
+
(
1
+
5
x
)
d
y
d
x
−
4
y
=
0.
x
(
1
−
x
)
d
2
y
d
x
2
+
(
1
+
5
x
)
d
y
d
x
−
4
y
=
0.
x(1-x)(d^(2)y)/(dx^(2))+(1+5x)(dy)/(dx)-4y=0. x(1-x) \frac{d^2 y}{dx^2} + (1 + 5x) \frac{dy}{dx} – 4y = 0. x ( 1 − x ) d 2 y d x 2 + ( 1 + 5 x ) d y d x − 4 y = 0. Step 1: Assume a Power Series Solution
Assume the solution
y
(
x
)
y
(
x
)
y(x) y(x) y ( x ) can be written as a power series around
x
=
0
x
=
0
x=0 x = 0 x = 0 :
y
(
x
)
=
∑
n
=
0
∞
a
n
x
n
.
y
(
x
)
=
∑
n
=
0
∞
a
n
x
n
.
y(x)=sum_(n=0)^(oo)a_(n)x^(n). y(x) = \sum_{n=0}^\infty a_n x^n. y ( x ) = ∑ n = 0 ∞ a n x n . Step 2: Compute the Derivatives
The first derivative of
y
(
x
)
y
(
x
)
y(x) y(x) y ( x ) is:
d
y
d
x
=
∑
n
=
1
∞
n
a
n
x
n
−
1
.
d
y
d
x
=
∑
n
=
1
∞
n
a
n
x
n
−
1
.
(dy)/(dx)=sum_(n=1)^(oo)na_(n)x^(n-1). \frac{dy}{dx} = \sum_{n=1}^\infty n a_n x^{n-1}. d y d x = ∑ n = 1 ∞ n a n x n − 1 . The second derivative of
y
(
x
)
y
(
x
)
y(x) y(x) y ( x ) is:
d
2
y
d
x
2
=
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
−
2
.
d
2
y
d
x
2
=
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
−
2
.
(d^(2)y)/(dx^(2))=sum_(n=2)^(oo)n(n-1)a_(n)x^(n-2). \frac{d^2y}{dx^2} = \sum_{n=2}^\infty n(n-1)a_n x^{n-2}. d 2 y d x 2 = ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 . Step 3: Substitute into the Differential Equation
Substitute
y
(
x
)
y
(
x
)
y(x) y(x) y ( x ) ,
d
y
d
x
d
y
d
x
(dy)/(dx) \frac{dy}{dx} d y d x , and
d
2
y
d
x
2
d
2
y
d
x
2
(d^(2)y)/(dx^(2)) \frac{d^2y}{dx^2} d 2 y d x 2 into the equation:
x
(
1
−
x
)
d
2
y
d
x
2
+
(
1
+
5
x
)
d
y
d
x
−
4
y
=
0.
x
(
1
−
x
)
d
2
y
d
x
2
+
(
1
+
5
x
)
d
y
d
x
−
4
y
=
0.
x(1-x)(d^(2)y)/(dx^(2))+(1+5x)(dy)/(dx)-4y=0. x(1-x) \frac{d^2y}{dx^2} + (1 + 5x) \frac{dy}{dx} – 4y = 0. x ( 1 − x ) d 2 y d x 2 + ( 1 + 5 x ) d y d x − 4 y = 0. Substitution for
x
(
1
−
x
)
d
2
y
d
x
2
x
(
1
−
x
)
d
2
y
d
x
2
x(1-x)(d^(2)y)/(dx^(2)) x(1-x) \frac{d^2y}{dx^2} x ( 1 − x ) d 2 y d x 2
x
(
1
−
x
)
d
2
y
d
x
2
=
x
(
1
−
x
)
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
−
2
.
x
(
1
−
x
)
d
2
y
d
x
2
=
x
(
1
−
x
)
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
−
2
.
x(1-x)(d^(2)y)/(dx^(2))=x(1-x)sum_(n=2)^(oo)n(n-1)a_(n)x^(n-2). x(1-x) \frac{d^2y}{dx^2} = x(1-x) \sum_{n=2}^\infty n(n-1)a_n x^{n-2}. x ( 1 − x ) d 2 y d x 2 = x ( 1 − x ) ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 . Distribute
x
(
1
−
x
)
x
(
1
−
x
)
x(1-x) x(1-x) x ( 1 − x ) :
x
(
1
−
x
)
d
2
y
d
x
2
=
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
−
1
−
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
.
x
(
1
−
x
)
d
2
y
d
x
2
=
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
−
1
−
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
.
x(1-x)(d^(2)y)/(dx^(2))=sum_(n=2)^(oo)n(n-1)a_(n)x^(n-1)-sum_(n=2)^(oo)n(n-1)a_(n)x^(n). x(1-x) \frac{d^2y}{dx^2} = \sum_{n=2}^\infty n(n-1)a_n x^{n-1} – \sum_{n=2}^\infty n(n-1)a_n x^n. x ( 1 − x ) d 2 y d x 2 = ∑ n = 2 ∞ n ( n − 1 ) a n x n − 1 − ∑ n = 2 ∞ n ( n − 1 ) a n x n . Substitution for
(
1
+
5
x
)
d
y
d
x
(
1
+
5
x
)
d
y
d
x
(1+5x)(dy)/(dx) (1+5x)\frac{dy}{dx} ( 1 + 5 x ) d y d x
(
1
+
5
x
)
d
y
d
x
=
∑
n
=
1
∞
n
a
n
x
n
−
1
+
5
∑
n
=
1
∞
n
a
n
x
n
.
(
1
+
5
x
)
d
y
d
x
=
∑
n
=
1
∞
n
a
n
x
n
−
1
+
5
∑
n
=
1
∞
n
a
n
x
n
.
(1+5x)(dy)/(dx)=sum_(n=1)^(oo)na_(n)x^(n-1)+5sum_(n=1)^(oo)na_(n)x^(n). (1+5x)\frac{dy}{dx} = \sum_{n=1}^\infty n a_n x^{n-1} + 5 \sum_{n=1}^\infty n a_n x^n. ( 1 + 5 x ) d y d x = ∑ n = 1 ∞ n a n x n − 1 + 5 ∑ n = 1 ∞ n a n x n . Substitution for
−
4
y
−
4
y
-4y -4y − 4 y
−
4
y
=
−
4
∑
n
=
0
∞
a
n
x
n
.
−
4
y
=
−
4
∑
n
=
0
∞
a
n
x
n
.
-4y=-4sum_(n=0)^(oo)a_(n)x^(n). -4y = -4 \sum_{n=0}^\infty a_n x^n. − 4 y = − 4 ∑ n = 0 ∞ a n x n . Step 4: Combine Terms
Combine all terms into a single summation:
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
−
1
−
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
+
∑
n
=
1
∞
n
a
n
x
n
−
1
+
5
∑
n
=
1
∞
n
a
n
x
n
−
4
∑
n
=
0
∞
a
n
x
n
=
0.
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
−
1
−
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
+
∑
n
=
1
∞
n
a
n
x
n
−
1
+
5
∑
n
=
1
∞
n
a
n
x
n
−
4
∑
n
=
0
∞
a
n
x
n
=
0.
sum_(n=2)^(oo)n(n-1)a_(n)x^(n-1)-sum_(n=2)^(oo)n(n-1)a_(n)x^(n)+sum_(n=1)^(oo)na_(n)x^(n-1)+5sum_(n=1)^(oo)na_(n)x^(n)-4sum_(n=0)^(oo)a_(n)x^(n)=0. \sum_{n=2}^\infty n(n-1)a_n x^{n-1} – \sum_{n=2}^\infty n(n-1)a_n x^n
+ \sum_{n=1}^\infty n a_n x^{n-1} + 5 \sum_{n=1}^\infty n a_n x^n
– 4 \sum_{n=0}^\infty a_n x^n = 0. ∑ n = 2 ∞ n ( n − 1 ) a n x n − 1 − ∑ n = 2 ∞ n ( n − 1 ) a n x n + ∑ n = 1 ∞ n a n x n − 1 + 5 ∑ n = 1 ∞ n a n x n − 4 ∑ n = 0 ∞ a n x n = 0. Group terms with the same powers of
x
x
x x x . For clarity, split the equation into powers of
x
n
−
1
x
n
−
1
x^(n-1) x^{n-1} x n − 1 and
x
n
x
n
x^(n) x^n x n .
Coefficients of
x
n
−
1
x
n
−
1
x^(n-1) x^{n-1} x n − 1
From the first summation, the terms with
x
n
−
1
x
n
−
1
x^(n-1) x^{n-1} x n − 1 come from:
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
−
1
+
∑
n
=
1
∞
n
a
n
x
n
−
1
.
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
−
1
+
∑
n
=
1
∞
n
a
n
x
n
−
1
.
sum_(n=2)^(oo)n(n-1)a_(n)x^(n-1)+sum_(n=1)^(oo)na_(n)x^(n-1). \sum_{n=2}^\infty n(n-1)a_n x^{n-1} + \sum_{n=1}^\infty n a_n x^{n-1}. ∑ n = 2 ∞ n ( n − 1 ) a n x n − 1 + ∑ n = 1 ∞ n a n x n − 1 . Combine these terms:
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
−
1
+
∑
n
=
1
∞
n
a
n
x
n
−
1
=
∑
n
=
1
∞
[
n
(
n
−
1
)
a
n
+
n
a
n
]
x
n
−
1
.
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
−
1
+
∑
n
=
1
∞
n
a
n
x
n
−
1
=
∑
n
=
1
∞
n
(
n
−
1
)
a
n
+
n
a
n
x
n
−
1
.
sum_(n=2)^(oo)n(n-1)a_(n)x^(n-1)+sum_(n=1)^(oo)na_(n)x^(n-1)=sum_(n=1)^(oo)[n(n-1)a_(n)+na_(n)]x^(n-1). \sum_{n=2}^\infty n(n-1)a_n x^{n-1} + \sum_{n=1}^\infty n a_n x^{n-1}
= \sum_{n=1}^\infty \left[n(n-1)a_n + n a_n\right] x^{n-1}. ∑ n = 2 ∞ n ( n − 1 ) a n x n − 1 + ∑ n = 1 ∞ n a n x n − 1 = ∑ n = 1 ∞ [ n ( n − 1 ) a n + n a n ] x n − 1 . This simplifies to:
∑
n
=
1
∞
n
2
a
n
x
n
−
1
.
∑
n
=
1
∞
n
2
a
n
x
n
−
1
.
sum_(n=1)^(oo)n^(2)a_(n)x^(n-1). \sum_{n=1}^\infty n^2 a_n x^{n-1}. ∑ n = 1 ∞ n 2 a n x n − 1 . Coefficients of
x
n
x
n
x^(n) x^n x n
From the other terms, the coefficients of
x
n
x
n
x^(n) x^n x n come from:
−
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
+
5
∑
n
=
1
∞
n
a
n
x
n
−
4
∑
n
=
0
∞
a
n
x
n
.
−
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
+
5
∑
n
=
1
∞
n
a
n
x
n
−
4
∑
n
=
0
∞
a
n
x
n
.
-sum_(n=2)^(oo)n(n-1)a_(n)x^(n)+5sum_(n=1)^(oo)na_(n)x^(n)-4sum_(n=0)^(oo)a_(n)x^(n). -\sum_{n=2}^\infty n(n-1)a_n x^n + 5\sum_{n=1}^\infty n a_n x^n – 4\sum_{n=0}^\infty a_n x^n. − ∑ n = 2 ∞ n ( n − 1 ) a n x n + 5 ∑ n = 1 ∞ n a n x n − 4 ∑ n = 0 ∞ a n x n . Shift indices in the first summation (
n
→
n
+
1
n
→
n
+
1
n rarr n+1 n \to n+1 n → n + 1 ):
−
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
=
−
∑
n
=
1
∞
(
n
+
1
)
n
a
n
+
1
x
n
.
−
∑
n
=
2
∞
n
(
n
−
1
)
a
n
x
n
=
−
∑
n
=
1
∞
(
n
+
1
)
n
a
n
+
1
x
n
.
-sum_(n=2)^(oo)n(n-1)a_(n)x^(n)=-sum_(n=1)^(oo)(n+1)na_(n+1)x^(n). -\sum_{n=2}^\infty n(n-1)a_n x^n = -\sum_{n=1}^\infty (n+1)n a_{n+1} x^n. − ∑ n = 2 ∞ n ( n − 1 ) a n x n = − ∑ n = 1 ∞ ( n + 1 ) n a n + 1 x n . The combined
x
n
x
n
x^(n) x^n x n -terms are:
∑
n
=
1
∞
[
5
n
a
n
−
(
n
+
1
)
n
a
n
+
1
−
4
a
n
]
x
n
.
∑
n
=
1
∞
5
n
a
n
−
(
n
+
1
)
n
a
n
+
1
−
4
a
n
x
n
.
sum_(n=1)^(oo)[5na_(n)-(n+1)na_(n+1)-4a_(n)]x^(n). \sum_{n=1}^\infty \left[5n a_n – (n+1)n a_{n+1} – 4a_n\right] x^n. ∑ n = 1 ∞ [ 5 n a n − ( n + 1 ) n a n + 1 − 4 a n ] x n . Step 5: Recurrence Relation
Equating the coefficients of
x
n
x
n
x^(n) x^n x n to 0, we obtain the recurrence relation:
5
n
a
n
−
(
n
+
1
)
n
a
n
+
1
−
4
a
n
=
0.
5
n
a
n
−
(
n
+
1
)
n
a
n
+
1
−
4
a
n
=
0.
5na_(n)-(n+1)na_(n+1)-4a_(n)=0. 5n a_n – (n+1)n a_{n+1} – 4a_n = 0. 5 n a n − ( n + 1 ) n a n + 1 − 4 a n = 0. Simplify:
a
n
+
1
=
(
5
n
−
4
)
a
n
(
n
+
1
)
n
.
a
n
+
1
=
(
5
n
−
4
)
a
n
(
n
+
1
)
n
.
a_(n+1)=((5n-4)a_(n))/((n+1)n). a_{n+1} = \frac{(5n – 4)a_n}{(n+1)n}. a n + 1 = ( 5 n − 4 ) a n ( n + 1 ) n . Step 6: Solve the Recurrence Relation
a
n
+
1
=
(
5
n
−
4
)
a
n
(
n
+
1
)
n
.
a
n
+
1
=
(
5
n
−
4
)
a
n
(
n
+
1
)
n
.
a_(n+1)=((5n-4)a_(n))/((n+1)n). a_{n+1} = \frac{(5n – 4)a_n}{(n+1)n}. a n + 1 = ( 5 n − 4 ) a n ( n + 1 ) n . Initial Conditions:
For a second-order differential equation, the solution typically involves two free parameters. These parameters come from
a
0
a
0
a_(0) a_0 a 0 and
a
1
a
1
a_(1) a_1 a 1 , which we choose as arbitrary constants (unless boundary or initial conditions specify otherwise).
Compute the Coefficients:
Let
a
0
=
C
0
a
0
=
C
0
a_(0)=C_(0) a_0 = C_0 a 0 = C 0 and
a
1
=
C
1
a
1
=
C
1
a_(1)=C_(1) a_1 = C_1 a 1 = C 1 . Use the recurrence relation to compute higher-order terms:
For
n
=
1
n
=
1
n=1 n=1 n = 1
a
2
=
(
5
(
1
)
−
4
)
a
1
(
2
)
(
1
)
=
(
5
−
4
)
a
1
2
=
a
1
2
.
a
2
=
(
5
(
1
)
−
4
)
a
1
(
2
)
(
1
)
=
(
5
−
4
)
a
1
2
=
a
1
2
.
a_(2)=((5(1)-4)a_(1))/((2)(1))=((5-4)a_(1))/(2)=(a_(1))/(2). a_2 = \frac{(5(1) – 4)a_1}{(2)(1)} = \frac{(5 – 4)a_1}{2} = \frac{a_1}{2}. a 2 = ( 5 ( 1 ) − 4 ) a 1 ( 2 ) ( 1 ) = ( 5 − 4 ) a 1 2 = a 1 2 .
For
n
=
2
n
=
2
n=2 n=2 n = 2
a
3
=
(
5
(
2
)
−
4
)
a
2
(
3
)
(
2
)
=
(
10
−
4
)
a
2
6
=
6
a
2
6
=
a
2
.
a
3
=
(
5
(
2
)
−
4
)
a
2
(
3
)
(
2
)
=
(
10
−
4
)
a
2
6
=
6
a
2
6
=
a
2
.
a_(3)=((5(2)-4)a_(2))/((3)(2))=((10-4)a_(2))/(6)=(6a_(2))/(6)=a_(2). a_3 = \frac{(5(2) – 4)a_2}{(3)(2)} = \frac{(10 – 4)a_2}{6} = \frac{6a_2}{6} = a_2. a 3 = ( 5 ( 2 ) − 4 ) a 2 ( 3 ) ( 2 ) = ( 10 − 4 ) a 2 6 = 6 a 2 6 = a 2 . Substituting
a
2
=
a
1
2
a
2
=
a
1
2
a_(2)=(a_(1))/(2) a_2 = \frac{a_1}{2} a 2 = a 1 2 , we get:
a
3
=
a
1
2
.
a
3
=
a
1
2
.
a_(3)=(a_(1))/(2). a_3 = \frac{a_1}{2}. a 3 = a 1 2 .
For
n
=
3
n
=
3
n=3 n=3 n = 3
a
4
=
(
5
(
3
)
−
4
)
a
3
(
4
)
(
3
)
=
(
15
−
4
)
a
3
12
=
11
a
3
12
.
a
4
=
(
5
(
3
)
−
4
)
a
3
(
4
)
(
3
)
=
(
15
−
4
)
a
3
12
=
11
a
3
12
.
a_(4)=((5(3)-4)a_(3))/((4)(3))=((15-4)a_(3))/(12)=(11a_(3))/(12). a_4 = \frac{(5(3) – 4)a_3}{(4)(3)} = \frac{(15 – 4)a_3}{12} = \frac{11a_3}{12}. a 4 = ( 5 ( 3 ) − 4 ) a 3 ( 4 ) ( 3 ) = ( 15 − 4 ) a 3 12 = 11 a 3 12 . Substituting
a
3
=
a
1
2
a
3
=
a
1
2
a_(3)=(a_(1))/(2) a_3 = \frac{a_1}{2} a 3 = a 1 2 :
a
4
=
11
12
⋅
a
1
2
=
11
a
1
24
.
a
4
=
11
12
⋅
a
1
2
=
11
a
1
24
.
a_(4)=(11)/(12)*(a_(1))/(2)=(11a_(1))/(24). a_4 = \frac{11}{12} \cdot \frac{a_1}{2} = \frac{11a_1}{24}. a 4 = 11 12 ⋅ a 1 2 = 11 a 1 24 .
For
n
=
4
n
=
4
n=4 n=4 n = 4
a
5
=
(
5
(
4
)
−
4
)
a
4
(
5
)
(
4
)
=
(
20
−
4
)
a
4
20
=
16
a
4
20
=
4
a
4
5
.
a
5
=
(
5
(
4
)
−
4
)
a
4
(
5
)
(
4
)
=
(
20
−
4
)
a
4
20
=
16
a
4
20
=
4
a
4
5
.
a_(5)=((5(4)-4)a_(4))/((5)(4))=((20-4)a_(4))/(20)=(16a_(4))/(20)=(4a_(4))/(5). a_5 = \frac{(5(4) – 4)a_4}{(5)(4)} = \frac{(20 – 4)a_4}{20} = \frac{16a_4}{20} = \frac{4a_4}{5}. a 5 = ( 5 ( 4 ) − 4 ) a 4 ( 5 ) ( 4 ) = ( 20 − 4 ) a 4 20 = 16 a 4 20 = 4 a 4 5 . Substituting
a
4
=
11
a
1
24
a
4
=
11
a
1
24
a_(4)=(11a_(1))/(24) a_4 = \frac{11a_1}{24} a 4 = 11 a 1 24 :
a
5
=
4
5
⋅
11
a
1
24
=
44
a
1
120
=
11
a
1
30
.
a
5
=
4
5
⋅
11
a
1
24
=
44
a
1
120
=
11
a
1
30
.
a_(5)=(4)/(5)*(11a_(1))/(24)=(44a_(1))/(120)=(11a_(1))/(30). a_5 = \frac{4}{5} \cdot \frac{11a_1}{24} = \frac{44a_1}{120} = \frac{11a_1}{30}. a 5 = 4 5 ⋅ 11 a 1 24 = 44 a 1 120 = 11 a 1 30 .
Step 7: Write the Series Solution
Using the coefficients derived, the solution is expressed as:
y
(
x
)
=
a
0
+
a
1
x
+
a
1
2
x
2
+
a
1
2
x
3
+
11
a
1
24
x
4
+
11
a
1
30
x
5
+
⋯
.
y
(
x
)
=
a
0
+
a
1
x
+
a
1
2
x
2
+
a
1
2
x
3
+
11
a
1
24
x
4
+
11
a
1
30
x
5
+
⋯
.
y(x)=a_(0)+a_(1)x+(a_(1))/(2)x^(2)+(a_(1))/(2)x^(3)+(11a_(1))/(24)x^(4)+(11a_(1))/(30)x^(5)+cdots. y(x) = a_0 + a_1x + \frac{a_1}{2}x^2 + \frac{a_1}{2}x^3 + \frac{11a_1}{24}x^4 + \frac{11a_1}{30}x^5 + \cdots. y ( x ) = a 0 + a 1 x + a 1 2 x 2 + a 1 2 x 3 + 11 a 1 24 x 4 + 11 a 1 30 x 5 + ⋯ . Substitute
a
0
=
C
0
a
0
=
C
0
a_(0)=C_(0) a_0 = C_0 a 0 = C 0 and
a
1
=
C
1
a
1
=
C
1
a_(1)=C_(1) a_1 = C_1 a 1 = C 1 :
y
(
x
)
=
C
0
+
C
1
x
+
C
1
2
x
2
+
C
1
2
x
3
+
11
C
1
24
x
4
+
11
C
1
30
x
5
+
⋯
.
y
(
x
)
=
C
0
+
C
1
x
+
C
1
2
x
2
+
C
1
2
x
3
+
11
C
1
24
x
4
+
11
C
1
30
x
5
+
⋯
.
y(x)=C_(0)+C_(1)x+(C_(1))/(2)x^(2)+(C_(1))/(2)x^(3)+(11C_(1))/(24)x^(4)+(11C_(1))/(30)x^(5)+cdots. y(x) = C_0 + C_1x + \frac{C_1}{2}x^2 + \frac{C_1}{2}x^3 + \frac{11C_1}{24}x^4 + \frac{11C_1}{30}x^5 + \cdots. y ( x ) = C 0 + C 1 x + C 1 2 x 2 + C 1 2 x 3 + 11 C 1 24 x 4 + 11 C 1 30 x 5 + ⋯ . Step 8: General Solution
The general solution is a linear combination of two independent series solutions,
y
1
(
x
)
y
1
(
x
)
y_(1)(x) y_1(x) y 1 ( x ) and
y
2
(
x
)
y
2
(
x
)
y_(2)(x) y_2(x) y 2 ( x ) , corresponding to the arbitrary constants
C
0
C
0
C_(0) C_0 C 0 and
C
1
C
1
C_(1) C_1 C 1 . These series may converge differently depending on the interval of interest for
x
x
x x x .
Question:-07
(a) Prove that:
B
(
λ
,
c
−
λ
)
2
F
1
(
a
,
b
;
c
;
z
)
=
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
B
(
λ
,
c
−
λ
)
2
F
1
(
a
,
b
;
c
;
z
)
=
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
B(lambda,c-lambda)2F_(1)(a,b;c;z)=int_(0)^(1)t^(lambda-1)(1-t)^(c-lambda-1)2F_(1)(a,b;lambda;zt)dt B(\lambda, c-\lambda) 2 F_1(a, b ; c ; z) = \int_0^1 t^{\lambda-1}(1-t)^{c-\lambda-1} 2 F_1(a, b ; \lambda ; z t) \, dt B ( λ , c − λ ) 2 F 1 ( a , b ; c ; z ) = ∫ 0 1 t λ − 1 ( 1 − t ) c − λ − 1 2 F 1 ( a , b ; λ ; z t ) d t Answer:
To prove the identity:
B
(
λ
,
c
−
λ
)
2
F
1
(
a
,
b
;
c
;
z
)
=
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
,
B
(
λ
,
c
−
λ
)
2
F
1
(
a
,
b
;
c
;
z
)
=
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
,
B(lambda,c-lambda)_(2)F_(1)(a,b;c;z)=int_(0)^(1)t^(lambda-1)(1-t)^(c-lambda-1)_(2)F_(1)(a,b;lambda;zt)dt, B(\lambda, c-\lambda) \, {}_2F_1(a, b; c; z) = \int_0^1 t^{\lambda-1} (1-t)^{c-\lambda-1} \, {}_2F_1(a, b; \lambda; zt) \, dt, B ( λ , c − λ ) 2 F 1 ( a , b ; c ; z ) = ∫ 0 1 t λ − 1 ( 1 − t ) c − λ − 1 2 F 1 ( a , b ; λ ; z t ) d t , where
B
(
λ
,
c
−
λ
)
B
(
λ
,
c
−
λ
)
B(lambda,c-lambda) B(\lambda, c-\lambda) B ( λ , c − λ ) is the Beta function,
2
F
1
(
a
,
b
;
c
;
z
)
2
F
1
(
a
,
b
;
c
;
z
)
_(2)F_(1)(a,b;c;z) {}_2F_1(a, b; c; z) 2 F 1 ( a , b ; c ; z ) is the Gauss hypergeometric function, and the integral is defined over
[
0
,
1
]
[
0
,
1
]
[0,1] [0, 1] [ 0 , 1 ] , we proceed as follows:
Step 1: Write the Beta Function
The Beta function is defined as:
B
(
x
,
y
)
=
∫
0
1
t
x
−
1
(
1
−
t
)
y
−
1
d
t
.
B
(
x
,
y
)
=
∫
0
1
t
x
−
1
(
1
−
t
)
y
−
1
d
t
.
B(x,y)=int_(0)^(1)t^(x-1)(1-t)^(y-1)dt. B(x, y) = \int_0^1 t^{x-1} (1-t)^{y-1} \, dt. B ( x , y ) = ∫ 0 1 t x − 1 ( 1 − t ) y − 1 d t . In this context:
B
(
λ
,
c
−
λ
)
=
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
d
t
.
B
(
λ
,
c
−
λ
)
=
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
d
t
.
B(lambda,c-lambda)=int_(0)^(1)t^(lambda-1)(1-t)^(c-lambda-1)dt. B(\lambda, c-\lambda) = \int_0^1 t^{\lambda-1} (1-t)^{c-\lambda-1} \, dt. B ( λ , c − λ ) = ∫ 0 1 t λ − 1 ( 1 − t ) c − λ − 1 d t . Step 2: Expand the Hypergeometric Function
The hypergeometric function
2
F
1
(
a
,
b
;
λ
;
z
t
)
2
F
1
(
a
,
b
;
λ
;
z
t
)
_(2)F_(1)(a,b;lambda;zt) {}_2F_1(a, b; \lambda; zt) 2 F 1 ( a , b ; λ ; z t ) has the series representation:
2
F
1
(
a
,
b
;
λ
;
z
t
)
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
λ
)
n
n
!
(
z
t
)
n
,
2
F
1
(
a
,
b
;
λ
;
z
t
)
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
λ
)
n
n
!
(
z
t
)
n
,
_(2)F_(1)(a,b;lambda;zt)=sum_(n=0)^(oo)((a)_(n)(b)_(n))/((lambda)_(n)n!)(zt)^(n), {}_2F_1(a, b; \lambda; zt) = \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(\lambda)_n n!} (zt)^n, 2 F 1 ( a , b ; λ ; z t ) = ∑ n = 0 ∞ ( a ) n ( b ) n ( λ ) n n ! ( z t ) n , where
(
a
)
n
=
a
(
a
+
1
)
(
a
+
2
)
⋯
(
a
+
n
−
1
)
(
a
)
n
=
a
(
a
+
1
)
(
a
+
2
)
⋯
(
a
+
n
−
1
)
(a)_(n)=a(a+1)(a+2)cdots(a+n-1) (a)_n = a(a+1)(a+2)\cdots(a+n-1) ( a ) n = a ( a + 1 ) ( a + 2 ) ⋯ ( a + n − 1 ) is the Pochhammer symbol.
Substitute this series expansion into the integral:
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
=
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
λ
)
n
n
!
(
z
t
)
n
d
t
.
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
=
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
λ
)
n
n
!
(
z
t
)
n
d
t
.
int_(0)^(1)t^(lambda-1)(1-t)^(c-lambda-1)_(2)F_(1)(a,b;lambda;zt)dt=int_(0)^(1)t^(lambda-1)(1-t)^(c-lambda-1)sum_(n=0)^(oo)((a)_(n)(b)_(n))/((lambda)_(n)n!)(zt)^(n)dt. \int_0^1 t^{\lambda-1} (1-t)^{c-\lambda-1} \, {}_2F_1(a, b; \lambda; zt) \, dt
= \int_0^1 t^{\lambda-1} (1-t)^{c-\lambda-1} \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(\lambda)_n n!} (zt)^n \, dt. ∫ 0 1 t λ − 1 ( 1 − t ) c − λ − 1 2 F 1 ( a , b ; λ ; z t ) d t = ∫ 0 1 t λ − 1 ( 1 − t ) c − λ − 1 ∑ n = 0 ∞ ( a ) n ( b ) n ( λ ) n n ! ( z t ) n d t . Step 3: Interchange the Sum and the Integral
Interchange the summation and the integral (justified since the series converges uniformly for
|
z
|
<
1
|
z
|
<
1
|z| < 1 |z| < 1 | z | < 1 ):
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
λ
)
n
n
!
z
n
∫
0
1
t
λ
+
n
−
1
(
1
−
t
)
c
−
λ
−
1
d
t
.
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
λ
)
n
n
!
z
n
∫
0
1
t
λ
+
n
−
1
(
1
−
t
)
c
−
λ
−
1
d
t
.
int_(0)^(1)t^(lambda-1)(1-t)^(c-lambda-1)_(2)F_(1)(a,b;lambda;zt)dt=sum_(n=0)^(oo)((a)_(n)(b)_(n))/((lambda)_(n)n!)z^(n)int_(0)^(1)t^(lambda+n-1)(1-t)^(c-lambda-1)dt. \int_0^1 t^{\lambda-1} (1-t)^{c-\lambda-1} \, {}_2F_1(a, b; \lambda; zt) \, dt
= \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(\lambda)_n n!} z^n \int_0^1 t^{\lambda+n-1} (1-t)^{c-\lambda-1} \, dt. ∫ 0 1 t λ − 1 ( 1 − t ) c − λ − 1 2 F 1 ( a , b ; λ ; z t ) d t = ∑ n = 0 ∞ ( a ) n ( b ) n ( λ ) n n ! z n ∫ 0 1 t λ + n − 1 ( 1 − t ) c − λ − 1 d t . Step 4: Evaluate the Inner Integral
The inner integral is:
∫
0
1
t
λ
+
n
−
1
(
1
−
t
)
c
−
λ
−
1
d
t
.
∫
0
1
t
λ
+
n
−
1
(
1
−
t
)
c
−
λ
−
1
d
t
.
int_(0)^(1)t^(lambda+n-1)(1-t)^(c-lambda-1)dt. \int_0^1 t^{\lambda+n-1} (1-t)^{c-\lambda-1} \, dt. ∫ 0 1 t λ + n − 1 ( 1 − t ) c − λ − 1 d t . This is a Beta function:
∫
0
1
t
λ
+
n
−
1
(
1
−
t
)
c
−
λ
−
1
d
t
=
B
(
λ
+
n
,
c
−
λ
)
.
∫
0
1
t
λ
+
n
−
1
(
1
−
t
)
c
−
λ
−
1
d
t
=
B
(
λ
+
n
,
c
−
λ
)
.
int_(0)^(1)t^(lambda+n-1)(1-t)^(c-lambda-1)dt=B(lambda+n,c-lambda). \int_0^1 t^{\lambda+n-1} (1-t)^{c-\lambda-1} \, dt = B(\lambda+n, c-\lambda). ∫ 0 1 t λ + n − 1 ( 1 − t ) c − λ − 1 d t = B ( λ + n , c − λ ) . Using the property of the Beta function:
B
(
x
,
y
)
=
Γ
(
x
)
Γ
(
y
)
Γ
(
x
+
y
)
,
B
(
x
,
y
)
=
Γ
(
x
)
Γ
(
y
)
Γ
(
x
+
y
)
,
B(x,y)=(Gamma(x)Gamma(y))/(Gamma(x+y)), B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}, B ( x , y ) = Γ ( x ) Γ ( y ) Γ ( x + y ) , we have:
B
(
λ
+
n
,
c
−
λ
)
=
Γ
(
λ
+
n
)
Γ
(
c
−
λ
)
Γ
(
c
+
n
)
.
B
(
λ
+
n
,
c
−
λ
)
=
Γ
(
λ
+
n
)
Γ
(
c
−
λ
)
Γ
(
c
+
n
)
.
B(lambda+n,c-lambda)=(Gamma(lambda+n)Gamma(c-lambda))/(Gamma(c+n)). B(\lambda+n, c-\lambda) = \frac{\Gamma(\lambda+n)\Gamma(c-\lambda)}{\Gamma(c+n)}. B ( λ + n , c − λ ) = Γ ( λ + n ) Γ ( c − λ ) Γ ( c + n ) . Step 5: Substitute Back into the Summation
Substitute
B
(
λ
+
n
,
c
−
λ
)
B
(
λ
+
n
,
c
−
λ
)
B(lambda+n,c-lambda) B(\lambda+n, c-\lambda) B ( λ + n , c − λ ) into the summation:
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
λ
)
n
n
!
z
n
Γ
(
λ
+
n
)
Γ
(
c
−
λ
)
Γ
(
c
+
n
)
.
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
λ
)
n
n
!
z
n
Γ
(
λ
+
n
)
Γ
(
c
−
λ
)
Γ
(
c
+
n
)
.
int_(0)^(1)t^(lambda-1)(1-t)^(c-lambda-1)_(2)F_(1)(a,b;lambda;zt)dt=sum_(n=0)^(oo)((a)_(n)(b)_(n))/((lambda)_(n)n!)z^(n)(Gamma(lambda+n)Gamma(c-lambda))/(Gamma(c+n)). \int_0^1 t^{\lambda-1} (1-t)^{c-\lambda-1} \, {}_2F_1(a, b; \lambda; zt) \, dt
= \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(\lambda)_n n!} z^n \frac{\Gamma(\lambda+n)\Gamma(c-\lambda)}{\Gamma(c+n)}. ∫ 0 1 t λ − 1 ( 1 − t ) c − λ − 1 2 F 1 ( a , b ; λ ; z t ) d t = ∑ n = 0 ∞ ( a ) n ( b ) n ( λ ) n n ! z n Γ ( λ + n ) Γ ( c − λ ) Γ ( c + n ) . Step 6: Simplify the Expression
Factor out
Γ
(
c
−
λ
)
Γ
(
c
−
λ
)
Gamma(c-lambda) \Gamma(c-\lambda) Γ ( c − λ ) , which is independent of
n
n
n n n :
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
=
Γ
(
c
−
λ
)
∑
n
=
0
∞
(
a
)
n
(
b
)
n
Γ
(
λ
+
n
)
(
λ
)
n
Γ
(
c
+
n
)
n
!
z
n
.
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
=
Γ
(
c
−
λ
)
∑
n
=
0
∞
(
a
)
n
(
b
)
n
Γ
(
λ
+
n
)
(
λ
)
n
Γ
(
c
+
n
)
n
!
z
n
.
int_(0)^(1)t^(lambda-1)(1-t)^(c-lambda-1)_(2)F_(1)(a,b;lambda;zt)dt=Gamma(c-lambda)sum_(n=0)^(oo)((a)_(n)(b)_(n)Gamma(lambda+n))/((lambda)_(n)Gamma(c+n)n!)z^(n). \int_0^1 t^{\lambda-1} (1-t)^{c-\lambda-1} \, {}_2F_1(a, b; \lambda; zt) \, dt
= \Gamma(c-\lambda) \sum_{n=0}^\infty \frac{(a)_n (b)_n \Gamma(\lambda+n)}{(\lambda)_n \Gamma(c+n) n!} z^n. ∫ 0 1 t λ − 1 ( 1 − t ) c − λ − 1 2 F 1 ( a , b ; λ ; z t ) d t = Γ ( c − λ ) ∑ n = 0 ∞ ( a ) n ( b ) n Γ ( λ + n ) ( λ ) n Γ ( c + n ) n ! z n . Using the relationship between Gamma functions and Pochhammer symbols:
Γ
(
λ
+
n
)
=
(
λ
)
n
Γ
(
λ
)
,
Γ
(
λ
+
n
)
=
(
λ
)
n
Γ
(
λ
)
,
Gamma(lambda+n)=(lambda)_(n)Gamma(lambda), \Gamma(\lambda+n) = (\lambda)_n \Gamma(\lambda), Γ ( λ + n ) = ( λ ) n Γ ( λ ) , this becomes:
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
=
Γ
(
λ
)
Γ
(
c
−
λ
)
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
n
!
z
n
.
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
=
Γ
(
λ
)
Γ
(
c
−
λ
)
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
n
!
z
n
.
int_(0)^(1)t^(lambda-1)(1-t)^(c-lambda-1)_(2)F_(1)(a,b;lambda;zt)dt=Gamma(lambda)Gamma(c-lambda)sum_(n=0)^(oo)((a)_(n)(b)_(n))/((c)_(n)n!)z^(n). \int_0^1 t^{\lambda-1} (1-t)^{c-\lambda-1} \, {}_2F_1(a, b; \lambda; zt) \, dt
= \Gamma(\lambda) \Gamma(c-\lambda) \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n n!} z^n. ∫ 0 1 t λ − 1 ( 1 − t ) c − λ − 1 2 F 1 ( a , b ; λ ; z t ) d t = Γ ( λ ) Γ ( c − λ ) ∑ n = 0 ∞ ( a ) n ( b ) n ( c ) n n ! z n . Step 7: Recognize the Hypergeometric Series
The summation is precisely the series definition of
2
F
1
(
a
,
b
;
c
;
z
)
2
F
1
(
a
,
b
;
c
;
z
)
_(2)F_(1)(a,b;c;z) {}_2F_1(a, b; c; z) 2 F 1 ( a , b ; c ; z ) :
2
F
1
(
a
,
b
;
c
;
z
)
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
n
!
z
n
.
2
F
1
(
a
,
b
;
c
;
z
)
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
n
!
z
n
.
_(2)F_(1)(a,b;c;z)=sum_(n=0)^(oo)((a)_(n)(b)_(n))/((c)_(n)n!)z^(n). {}_2F_1(a, b; c; z) = \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n n!} z^n. 2 F 1 ( a , b ; c ; z ) = ∑ n = 0 ∞ ( a ) n ( b ) n ( c ) n n ! z n . Thus:
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
=
Γ
(
λ
)
Γ
(
c
−
λ
)
2
F
1
(
a
,
b
;
c
;
z
)
.
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
=
Γ
(
λ
)
Γ
(
c
−
λ
)
2
F
1
(
a
,
b
;
c
;
z
)
.
int_(0)^(1)t^(lambda-1)(1-t)^(c-lambda-1)_(2)F_(1)(a,b;lambda;zt)dt=Gamma(lambda)Gamma(c-lambda)_(2)F_(1)(a,b;c;z). \int_0^1 t^{\lambda-1} (1-t)^{c-\lambda-1} \, {}_2F_1(a, b; \lambda; zt) \, dt
= \Gamma(\lambda) \Gamma(c-\lambda) \, {}_2F_1(a, b; c; z). ∫ 0 1 t λ − 1 ( 1 − t ) c − λ − 1 2 F 1 ( a , b ; λ ; z t ) d t = Γ ( λ ) Γ ( c − λ ) 2 F 1 ( a , b ; c ; z ) . Step 8: Express Using the Beta Function
Recall that:
B
(
λ
,
c
−
λ
)
=
Γ
(
λ
)
Γ
(
c
−
λ
)
Γ
(
c
)
.
B
(
λ
,
c
−
λ
)
=
Γ
(
λ
)
Γ
(
c
−
λ
)
Γ
(
c
)
.
B(lambda,c-lambda)=(Gamma(lambda)Gamma(c-lambda))/(Gamma(c)). B(\lambda, c-\lambda) = \frac{\Gamma(\lambda) \Gamma(c-\lambda)}{\Gamma(c)}. B ( λ , c − λ ) = Γ ( λ ) Γ ( c − λ ) Γ ( c ) . Multiply both sides by
Γ
(
c
)
Γ
(
c
)
Gamma(c) \Gamma(c) Γ ( c ) to isolate the Beta function:
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
=
B
(
λ
,
c
−
λ
)
2
F
1
(
a
,
b
;
c
;
z
)
.
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
=
B
(
λ
,
c
−
λ
)
2
F
1
(
a
,
b
;
c
;
z
)
.
int_(0)^(1)t^(lambda-1)(1-t)^(c-lambda-1)_(2)F_(1)(a,b;lambda;zt)dt=B(lambda,c-lambda)_(2)F_(1)(a,b;c;z). \int_0^1 t^{\lambda-1} (1-t)^{c-\lambda-1} \, {}_2F_1(a, b; \lambda; zt) \, dt = B(\lambda, c-\lambda) \, {}_2F_1(a, b; c; z). ∫ 0 1 t λ − 1 ( 1 − t ) c − λ − 1 2 F 1 ( a , b ; λ ; z t ) d t = B ( λ , c − λ ) 2 F 1 ( a , b ; c ; z ) . Final Result:
B
(
λ
,
c
−
λ
)
2
F
1
(
a
,
b
;
c
;
z
)
=
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
.
B
(
λ
,
c
−
λ
)
2
F
1
(
a
,
b
;
c
;
z
)
=
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
F
1
(
a
,
b
;
λ
;
z
t
)
d
t
.
B(lambda,c-lambda)_(2)F_(1)(a,b;c;z)=int_(0)^(1)t^(lambda-1)(1-t)^(c-lambda-1)_(2)F_(1)(a,b;lambda;zt)dt. B(\lambda, c-\lambda) \, {}_2F_1(a, b; c; z) = \int_0^1 t^{\lambda-1} (1-t)^{c-\lambda-1} \, {}_2F_1(a, b; \lambda; zt) \, dt. B ( λ , c − λ ) 2 F 1 ( a , b ; c ; z ) = ∫ 0 1 t λ − 1 ( 1 − t ) c − λ − 1 2 F 1 ( a , b ; λ ; z t ) d t .
Question:-07(b)
Prove that:
(
2
n
+
1
)
(
x
2
−
1
)
P
n
′
=
n
(
n
+
1
)
(
P
n
+
1
−
P
n
−
1
)
(
2
n
+
1
)
x
2
−
1
P
n
′
=
n
(
n
+
1
)
P
n
+
1
−
P
n
−
1
(2n+1)(x^(2)-1)P_(n)^(‘)=n(n+1)(P_(n+1)-P_(n-1)) (2 n+1)\left(x^2 – 1\right) P_n^{\prime} = n(n+1)\left(P_{n+1} – P_{n-1}\right) ( 2 n + 1 ) ( x 2 − 1 ) P n ′ = n ( n + 1 ) ( P n + 1 − P n − 1 ) and hence deduce that:
∫
−
1
1
(
x
2
−
1
)
P
n
+
1
(
x
)
P
n
′
(
x
)
d
x
=
2
n
(
n
+
1
)
(
2
n
+
1
)
(
2
n
+
3
)
∫
−
1
1
x
2
−
1
P
n
+
1
(
x
)
P
n
′
(
x
)
d
x
=
2
n
(
n
+
1
)
(
2
n
+
1
)
(
2
n
+
3
)
int_(-1)^(1)(x^(2)-1)P_(n+1)(x)P_(n)^(‘)(x)dx=(2n(n+1))/((2n+1)(2n+3)) \int_{-1}^1 \left(x^2 – 1\right) P_{n+1}(x) P_n^{\prime}(x) \, dx = \frac{2n(n+1)}{(2n+1)(2n+3)} ∫ − 1 1 ( x 2 − 1 ) P n + 1 ( x ) P n ′ ( x ) d x = 2 n ( n + 1 ) ( 2 n + 1 ) ( 2 n + 3 ) Answer:
Let’s prove the first equation and then use it to deduce the given integral.
Part 1: Proof of the Relation
Statement to Prove:
(
2
n
+
1
)
(
x
2
−
1
)
P
n
′
(
x
)
=
n
(
n
+
1
)
(
P
n
+
1
(
x
)
−
P
n
−
1
(
x
)
)
,
(
2
n
+
1
)
(
x
2
−
1
)
P
n
′
(
x
)
=
n
(
n
+
1
)
P
n
+
1
(
x
)
−
P
n
−
1
(
x
)
,
(2n+1)(x^(2)-1)P_(n)^(‘)(x)=n(n+1)(P_(n+1)(x)-P_(n-1)(x)), (2n+1)(x^2-1)P_n'(x) = n(n+1)\left(P_{n+1}(x) – P_{n-1}(x)\right), ( 2 n + 1 ) ( x 2 − 1 ) P n ′ ( x ) = n ( n + 1 ) ( P n + 1 ( x ) − P n − 1 ( x ) ) , where
P
n
(
x
)
P
n
(
x
)
P_(n)(x) P_n(x) P n ( x ) are the Legendre polynomials.
Step 1: Legendre Polynomial Recurrence Relations
The Legendre polynomials satisfy the following standard recurrence relations:
Rodrigues’ formula:
P
n
(
x
)
=
1
2
n
n
!
d
n
d
x
n
(
x
2
−
1
)
n
.
P
n
(
x
)
=
1
2
n
n
!
d
n
d
x
n
x
2
−
1
n
.
P_(n)(x)=(1)/(2^(n)n!)(d^(n))/(dx^(n))(x^(2)-1)^(n). P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} \left( x^2 – 1 \right)^n. P n ( x ) = 1 2 n n ! d n d x n ( x 2 − 1 ) n .
A three-term recurrence relation:
(
n
+
1
)
P
n
+
1
(
x
)
=
(
2
n
+
1
)
x
P
n
(
x
)
−
n
P
n
−
1
(
x
)
.
(
n
+
1
)
P
n
+
1
(
x
)
=
(
2
n
+
1
)
x
P
n
(
x
)
−
n
P
n
−
1
(
x
)
.
(n+1)P_(n+1)(x)=(2n+1)xP_(n)(x)-nP_(n-1)(x). (n+1)P_{n+1}(x) = (2n+1)xP_n(x) – nP_{n-1}(x). ( n + 1 ) P n + 1 ( x ) = ( 2 n + 1 ) x P n ( x ) − n P n − 1 ( x ) .
Derivative formula:
d
d
x
(
P
n
(
x
)
)
=
n
(
x
P
n
(
x
)
−
P
n
−
1
(
x
)
)
.
d
d
x
P
n
(
x
)
=
n
x
P
n
(
x
)
−
P
n
−
1
(
x
)
.
(d)/(dx)(P_(n)(x))=n(xP_(n)(x)-P_(n-1)(x)). \frac{d}{dx} \left( P_n(x) \right) = n \left( xP_n(x) – P_{n-1}(x) \right). d d x ( P n ( x ) ) = n ( x P n ( x ) − P n − 1 ( x ) ) .
The derivative of
P
n
(
x
)
P
n
(
x
)
P_(n)(x) P_n(x) P n ( x ) is:
P
n
′
(
x
)
=
n
(
x
P
n
(
x
)
−
P
n
−
1
(
x
)
)
.
P
n
′
(
x
)
=
n
x
P
n
(
x
)
−
P
n
−
1
(
x
)
.
P_(n)^(‘)(x)=n(xP_(n)(x)-P_(n-1)(x)). P_n'(x) = n \left(xP_n(x) – P_{n-1}(x)\right). P n ′ ( x ) = n ( x P n ( x ) − P n − 1 ( x ) ) . Multiply both sides by
(
x
2
−
1
)
(
x
2
−
1
)
(x^(2)-1) (x^2 – 1) ( x 2 − 1 ) :
(
x
2
−
1
)
P
n
′
(
x
)
=
n
(
x
2
−
1
)
(
x
P
n
(
x
)
−
P
n
−
1
(
x
)
)
.
(
x
2
−
1
)
P
n
′
(
x
)
=
n
(
x
2
−
1
)
(
x
P
n
(
x
)
−
P
n
−
1
(
x
)
)
.
(x^(2)-1)P_(n)^(‘)(x)=n(x^(2)-1)(xP_(n)(x)-P_(n-1)(x)). (x^2-1)P_n'(x) = n(x^2-1)(xP_n(x) – P_{n-1}(x)). ( x 2 − 1 ) P n ′ ( x ) = n ( x 2 − 1 ) ( x P n ( x ) − P n − 1 ( x ) ) . Distribute the terms:
(
x
2
−
1
)
P
n
′
(
x
)
=
n
[
(
x
2
−
1
)
x
P
n
(
x
)
−
(
x
2
−
1
)
P
n
−
1
(
x
)
]
.
(
x
2
−
1
)
P
n
′
(
x
)
=
n
(
x
2
−
1
)
x
P
n
(
x
)
−
(
x
2
−
1
)
P
n
−
1
(
x
)
.
(x^(2)-1)P_(n)^(‘)(x)=n[(x^(2)-1)xP_(n)(x)-(x^(2)-1)P_(n-1)(x)]. (x^2-1)P_n'(x) = n \left[(x^2-1)xP_n(x) – (x^2-1)P_{n-1}(x)\right]. ( x 2 − 1 ) P n ′ ( x ) = n [ ( x 2 − 1 ) x P n ( x ) − ( x 2 − 1 ) P n − 1 ( x ) ] . Step 3: Simplify Using the Recurrence Relation
Using the recurrence relation:
x
P
n
(
x
)
=
(
n
+
1
)
P
n
+
1
(
x
)
+
n
P
n
−
1
(
x
)
2
n
+
1
,
x
P
n
(
x
)
=
(
n
+
1
)
P
n
+
1
(
x
)
+
n
P
n
−
1
(
x
)
2
n
+
1
,
xP_(n)(x)=((n+1)P_(n+1)(x)+nP_(n-1)(x))/(2n+1), xP_n(x) = \frac{(n+1)P_{n+1}(x) + nP_{n-1}(x)}{2n+1}, x P n ( x ) = ( n + 1 ) P n + 1 ( x ) + n P n − 1 ( x ) 2 n + 1 , substitute
x
P
n
(
x
)
x
P
n
(
x
)
xP_(n)(x) xP_n(x) x P n ( x ) into the first term:
(
x
2
−
1
)
P
n
′
(
x
)
=
n
[
(
x
2
−
1
)
(
n
+
1
)
P
n
+
1
(
x
)
+
n
P
n
−
1
(
x
)
2
n
+
1
−
(
x
2
−
1
)
P
n
−
1
(
x
)
]
.
(
x
2
−
1
)
P
n
′
(
x
)
=
n
(
x
2
−
1
)
(
n
+
1
)
P
n
+
1
(
x
)
+
n
P
n
−
1
(
x
)
2
n
+
1
−
(
x
2
−
1
)
P
n
−
1
(
x
)
.
(x^(2)-1)P_(n)^(‘)(x)=n[(x^(2)-1)((n+1)P_(n+1)(x)+nP_(n-1)(x))/(2n+1)-(x^(2)-1)P_(n-1)(x)]. (x^2-1)P_n'(x) = n \left[(x^2-1) \frac{(n+1)P_{n+1}(x) + nP_{n-1}(x)}{2n+1} – (x^2-1)P_{n-1}(x)\right]. ( x 2 − 1 ) P n ′ ( x ) = n [ ( x 2 − 1 ) ( n + 1 ) P n + 1 ( x ) + n P n − 1 ( x ) 2 n + 1 − ( x 2 − 1 ) P n − 1 ( x ) ] . Factor out
(
x
2
−
1
)
(
x
2
−
1
)
(x^(2)-1) (x^2 – 1) ( x 2 − 1 ) :
(
x
2
−
1
)
P
n
′
(
x
)
=
n
(
x
2
−
1
)
2
n
+
1
[
(
n
+
1
)
P
n
+
1
(
x
)
+
n
P
n
−
1
(
x
)
−
(
2
n
+
1
)
P
n
−
1
(
x
)
]
.
(
x
2
−
1
)
P
n
′
(
x
)
=
n
(
x
2
−
1
)
2
n
+
1
(
n
+
1
)
P
n
+
1
(
x
)
+
n
P
n
−
1
(
x
)
−
(
2
n
+
1
)
P
n
−
1
(
x
)
.
(x^(2)-1)P_(n)^(‘)(x)=(n(x^(2)-1))/(2n+1)[(n+1)P_(n+1)(x)+nP_(n-1)(x)-(2n+1)P_(n-1)(x)]. (x^2-1)P_n'(x) = \frac{n(x^2-1)}{2n+1} \left[ (n+1)P_{n+1}(x) + nP_{n-1}(x) – (2n+1)P_{n-1}(x) \right]. ( x 2 − 1 ) P n ′ ( x ) = n ( x 2 − 1 ) 2 n + 1 [ ( n + 1 ) P n + 1 ( x ) + n P n − 1 ( x ) − ( 2 n + 1 ) P n − 1 ( x ) ] . Simplify the term in brackets:
n
P
n
−
1
(
x
)
−
(
2
n
+
1
)
P
n
−
1
(
x
)
=
−
n
P
n
−
1
(
x
)
.
n
P
n
−
1
(
x
)
−
(
2
n
+
1
)
P
n
−
1
(
x
)
=
−
n
P
n
−
1
(
x
)
.
nP_(n-1)(x)-(2n+1)P_(n-1)(x)=-nP_(n-1)(x). nP_{n-1}(x) – (2n+1)P_{n-1}(x) = -nP_{n-1}(x). n P n − 1 ( x ) − ( 2 n + 1 ) P n − 1 ( x ) = − n P n − 1 ( x ) . Thus:
(
x
2
−
1
)
P
n
′
(
x
)
=
n
(
x
2
−
1
)
2
n
+
1
[
(
n
+
1
)
P
n
+
1
(
x
)
−
n
P
n
−
1
(
x
)
]
.
(
x
2
−
1
)
P
n
′
(
x
)
=
n
(
x
2
−
1
)
2
n
+
1
(
n
+
1
)
P
n
+
1
(
x
)
−
n
P
n
−
1
(
x
)
.
(x^(2)-1)P_(n)^(‘)(x)=(n(x^(2)-1))/(2n+1)[(n+1)P_(n+1)(x)-nP_(n-1)(x)]. (x^2-1)P_n'(x) = \frac{n(x^2-1)}{2n+1} \left[(n+1)P_{n+1}(x) – nP_{n-1}(x)\right]. ( x 2 − 1 ) P n ′ ( x ) = n ( x 2 − 1 ) 2 n + 1 [ ( n + 1 ) P n + 1 ( x ) − n P n − 1 ( x ) ] . Multiply through by
2
n
+
1
2
n
+
1
2n+1 2n+1 2 n + 1 :
(
2
n
+
1
)
(
x
2
−
1
)
P
n
′
(
x
)
=
n
(
n
+
1
)
P
n
+
1
(
x
)
−
n
2
P
n
−
1
(
x
)
.
(
2
n
+
1
)
(
x
2
−
1
)
P
n
′
(
x
)
=
n
(
n
+
1
)
P
n
+
1
(
x
)
−
n
2
P
n
−
1
(
x
)
.
(2n+1)(x^(2)-1)P_(n)^(‘)(x)=n(n+1)P_(n+1)(x)-n^(2)P_(n-1)(x). (2n+1)(x^2-1)P_n'(x) = n(n+1)P_{n+1}(x) – n^2P_{n-1}(x). ( 2 n + 1 ) ( x 2 − 1 ) P n ′ ( x ) = n ( n + 1 ) P n + 1 ( x ) − n 2 P n − 1 ( x ) . Simplify:
(
2
n
+
1
)
(
x
2
−
1
)
P
n
′
(
x
)
=
n
(
n
+
1
)
(
P
n
+
1
(
x
)
−
P
n
−
1
(
x
)
)
.
(
2
n
+
1
)
(
x
2
−
1
)
P
n
′
(
x
)
=
n
(
n
+
1
)
P
n
+
1
(
x
)
−
P
n
−
1
(
x
)
.
(2n+1)(x^(2)-1)P_(n)^(‘)(x)=n(n+1)(P_(n+1)(x)-P_(n-1)(x)). (2n+1)(x^2-1)P_n'(x) = n(n+1)\left(P_{n+1}(x) – P_{n-1}(x)\right). ( 2 n + 1 ) ( x 2 − 1 ) P n ′ ( x ) = n ( n + 1 ) ( P n + 1 ( x ) − P n − 1 ( x ) ) . This proves the first part.
Part 2: Deduce the Integral
We want to compute:
I
=
∫
−
1
1
(
x
2
−
1
)
P
n
+
1
(
x
)
P
n
′
(
x
)
d
x
.
I
=
∫
−
1
1
(
x
2
−
1
)
P
n
+
1
(
x
)
P
n
′
(
x
)
d
x
.
I=int_(-1)^(1)(x^(2)-1)P_(n+1)(x)P_(n)^(‘)(x)dx. I = \int_{-1}^1 (x^2-1)P_{n+1}(x)P_n'(x) \, dx. I = ∫ − 1 1 ( x 2 − 1 ) P n + 1 ( x ) P n ′ ( x ) d x . Step 1: Use the Proven Relation
From the proven relation:
(
2
n
+
1
)
(
x
2
−
1
)
P
n
′
(
x
)
=
n
(
n
+
1
)
(
P
n
+
1
(
x
)
−
P
n
−
1
(
x
)
)
.
(
2
n
+
1
)
(
x
2
−
1
)
P
n
′
(
x
)
=
n
(
n
+
1
)
P
n
+
1
(
x
)
−
P
n
−
1
(
x
)
.
(2n+1)(x^(2)-1)P_(n)^(‘)(x)=n(n+1)(P_(n+1)(x)-P_(n-1)(x)). (2n+1)(x^2-1)P_n'(x) = n(n+1)\left(P_{n+1}(x) – P_{n-1}(x)\right). ( 2 n + 1 ) ( x 2 − 1 ) P n ′ ( x ) = n ( n + 1 ) ( P n + 1 ( x ) − P n − 1 ( x ) ) . Substitute into the integral:
(
2
n
+
1
)
I
=
n
(
n
+
1
)
∫
−
1
1
P
n
+
1
(
x
)
(
P
n
+
1
(
x
)
−
P
n
−
1
(
x
)
)
d
x
.
(
2
n
+
1
)
I
=
n
(
n
+
1
)
∫
−
1
1
P
n
+
1
(
x
)
P
n
+
1
(
x
)
−
P
n
−
1
(
x
)
d
x
.
(2n+1)I=n(n+1)int_(-1)^(1)P_(n+1)(x)(P_(n+1)(x)-P_(n-1)(x))dx. (2n+1)I = n(n+1) \int_{-1}^1 P_{n+1}(x) \left(P_{n+1}(x) – P_{n-1}(x)\right) \, dx. ( 2 n + 1 ) I = n ( n + 1 ) ∫ − 1 1 P n + 1 ( x ) ( P n + 1 ( x ) − P n − 1 ( x ) ) d x . Expand the product:
(
2
n
+
1
)
I
=
n
(
n
+
1
)
[
∫
−
1
1
P
n
+
1
2
(
x
)
d
x
−
∫
−
1
1
P
n
+
1
(
x
)
P
n
−
1
(
x
)
d
x
]
.
(
2
n
+
1
)
I
=
n
(
n
+
1
)
∫
−
1
1
P
n
+
1
2
(
x
)
d
x
−
∫
−
1
1
P
n
+
1
(
x
)
P
n
−
1
(
x
)
d
x
.
(2n+1)I=n(n+1)[int_(-1)^(1)P_(n+1)^(2)(x)dx-int_(-1)^(1)P_(n+1)(x)P_(n-1)(x)dx]. (2n+1)I = n(n+1) \left[\int_{-1}^1 P_{n+1}^2(x) \, dx – \int_{-1}^1 P_{n+1}(x)P_{n-1}(x) \, dx\right]. ( 2 n + 1 ) I = n ( n + 1 ) [ ∫ − 1 1 P n + 1 2 ( x ) d x − ∫ − 1 1 P n + 1 ( x ) P n − 1 ( x ) d x ] . Step 2: Orthogonality of Legendre Polynomials
The Legendre polynomials satisfy the orthogonality relation:
∫
−
1
1
P
m
(
x
)
P
n
(
x
)
d
x
=
2
2
n
+
1
δ
m
n
.
∫
−
1
1
P
m
(
x
)
P
n
(
x
)
d
x
=
2
2
n
+
1
δ
m
n
.
int_(-1)^(1)P_(m)(x)P_(n)(x)dx=(2)/(2n+1)delta_(mn). \int_{-1}^1 P_m(x)P_n(x) \, dx = \frac{2}{2n+1} \delta_{mn}. ∫ − 1 1 P m ( x ) P n ( x ) d x = 2 2 n + 1 δ m n . Using this:
For
∫
−
1
1
P
n
+
1
2
(
x
)
d
x
∫
−
1
1
P
n
+
1
2
(
x
)
d
x
int_(-1)^(1)P_(n+1)^(2)(x)dx \int_{-1}^1 P_{n+1}^2(x) \, dx ∫ − 1 1 P n + 1 2 ( x ) d x :
∫
−
1
1
P
n
+
1
2
(
x
)
d
x
=
2
2
n
+
3
.
∫
−
1
1
P
n
+
1
2
(
x
)
d
x
=
2
2
n
+
3
.
int_(-1)^(1)P_(n+1)^(2)(x)dx=(2)/(2n+3). \int_{-1}^1 P_{n+1}^2(x) \, dx = \frac{2}{2n+3}. ∫ − 1 1 P n + 1 2 ( x ) d x = 2 2 n + 3 .
For
∫
−
1
1
P
n
+
1
(
x
)
P
n
−
1
(
x
)
d
x
∫
−
1
1
P
n
+
1
(
x
)
P
n
−
1
(
x
)
d
x
int_(-1)^(1)P_(n+1)(x)P_(n-1)(x)dx \int_{-1}^1 P_{n+1}(x)P_{n-1}(x) \, dx ∫ − 1 1 P n + 1 ( x ) P n − 1 ( x ) d x :
∫
−
1
1
P
n
+
1
(
x
)
P
n
−
1
(
x
)
d
x
=
0
(since
n
+
1
≠
n
−
1
)
.
∫
−
1
1
P
n
+
1
(
x
)
P
n
−
1
(
x
)
d
x
=
0
(since
n
+
1
≠
n
−
1
)
.
int_(-1)^(1)P_(n+1)(x)P_(n-1)(x)dx=0quad(since (n+1!=n-1)”)”. \int_{-1}^1 P_{n+1}(x)P_{n-1}(x) \, dx = 0 \quad \text{(since \(n+1 \neq n-1\))}. ∫ − 1 1 P n + 1 ( x ) P n − 1 ( x ) d x = 0 (since n + 1 ≠ n − 1 ) .
Substitute these into the equation for
I
I
I I I :
(
2
n
+
1
)
I
=
n
(
n
+
1
)
[
2
2
n
+
3
−
0
]
.
(
2
n
+
1
)
I
=
n
(
n
+
1
)
2
2
n
+
3
−
0
.
(2n+1)I=n(n+1)[(2)/(2n+3)-0]. (2n+1)I = n(n+1) \left[\frac{2}{2n+3} – 0\right]. ( 2 n + 1 ) I = n ( n + 1 ) [ 2 2 n + 3 − 0 ] . Simplify:
(
2
n
+
1
)
I
=
2
n
(
n
+
1
)
2
n
+
3
.
(
2
n
+
1
)
I
=
2
n
(
n
+
1
)
2
n
+
3
.
(2n+1)I=(2n(n+1))/(2n+3). (2n+1)I = \frac{2n(n+1)}{2n+3}. ( 2 n + 1 ) I = 2 n ( n + 1 ) 2 n + 3 . Step 3: Solve for
I
I
I I I
Divide through by
2
n
+
1
2
n
+
1
2n+1 2n+1 2 n + 1 :
I
=
2
n
(
n
+
1
)
(
2
n
+
1
)
(
2
n
+
3
)
.
I
=
2
n
(
n
+
1
)
(
2
n
+
1
)
(
2
n
+
3
)
.
I=(2n(n+1))/((2n+1)(2n+3)). I = \frac{2n(n+1)}{(2n+1)(2n+3)}. I = 2 n ( n + 1 ) ( 2 n + 1 ) ( 2 n + 3 ) . Final Answer:
∫
−
1
1
(
x
2
−
1
)
P
n
+
1
(
x
)
P
n
′
(
x
)
d
x
=
2
n
(
n
+
1
)
(
2
n
+
1
)
(
2
n
+
3
)
.
∫
−
1
1
(
x
2
−
1
)
P
n
+
1
(
x
)
P
n
′
(
x
)
d
x
=
2
n
(
n
+
1
)
(
2
n
+
1
)
(
2
n
+
3
)
.
int_(-1)^(1)(x^(2)-1)P_(n+1)(x)P_(n)^(‘)(x)dx=(2n(n+1))/((2n+1)(2n+3)). \int_{-1}^1 (x^2-1)P_{n+1}(x)P_n'(x) \, dx = \frac{2n(n+1)}{(2n+1)(2n+3)}. ∫ − 1 1 ( x 2 − 1 ) P n + 1 ( x ) P n ′ ( x ) d x = 2 n ( n + 1 ) ( 2 n + 1 ) ( 2 n + 3 ) .
