IGNOU MMT-007 Solved Assignment | 1st January, 2025 to 31st December, 2025 | Differential Equations and Numerical Solutions | MSc MACS

Next Lesson

Sample Solution

IGNOU MMT-007 Solved Assignment | 1st January, 2025 to 31st December, 2025 | Differential Equations and Numerical Solutions | MSc MACS Sample Solution

MMT-007 Solved Assignment 2025

  1. a) Solve the differential equation:
x 2 y ′ ′ + 6 x y ′ + ( 6 + x 2 ) y = 0 x 2 y ′ ′ + 6 x y ′ + 6 + x 2 y = 0 x^(2)y^(”)+6xy^(‘)+(6+x^(2))y=0x^2 y^{\prime \prime}+6 x y^{\prime}+\left(6+x^2\right) y=0x2y′′+6xy′+(6+x2)y=0
in series about x = 0 x = 0 x=0\mathrm{x}=0x=0.
b) Express f ( x ) = x 4 + 3 x 3 + 4 x 2 − x + 2 f ( x ) = x 4 + 3 x 3 + 4 x 2 − x + 2 f(x)=x^(4)+3x^(3)+4x^(2)-x+2f(x)=x^4+3 x^3+4 x^2-x+2f(x)=x4+3x3+4x2−x+2 in terms of Legendre polynomials.
  1. a) Using method of Ferobenius, find the solution of the differential equation:
x 2 d 2 y d x 2 + ( x + x 2 ) d y d x + ( x − 9 ) y = 0 x 2 d 2 y d x 2 + x + x 2 d y d x + ( x − 9 ) y = 0 x^(2)(d^(2)y)/(dx^(2))+(x+x^(2))(dy)/(dx)+(x-9)y=0x^2 \frac{d^2 y}{d x^2}+\left(x+x^2\right) \frac{d y}{d x}+(x-9) y=0x2d2ydx2+(x+x2)dydx+(x−9)y=0
near x − 0 x − 0 x-0\mathrm{x}-0x−0.
b) Find:
L − 1 { s ( s 2 + 4 ) 2 } L − 1 s s 2 + 4 2 L^(-1){((s))/((s^(2)+4)^(2))}\mathrm{L}^{-1}\left\{\frac{\mathrm{~s}}{\left(\mathrm{~s}^2+4\right)^2}\right\}L−1{ s( s2+4)2}
c) Find L { F ( t ) } L { F ( t ) } L{F(t)}L\{F(t)\}L{F(t)}, if:
F ( t ) = { sin ⁡ ( t − π 4 ) , t > π 4 0 , t < π 4 F ( t ) = sin ⁡ t − π 4 , t > π 4 0 , t < π 4 F(t)={[sin(t-(pi)/(4))”,”,t > (pi)/(4)],[0,”,”t < (pi)/(4)]:}\mathrm{F}(\mathrm{t})=\left\{\begin{array}{cc} \sin \left(\mathrm{t}-\frac{\pi}{4}\right), & \mathrm{t}>\frac{\pi}{4} \\ 0 & , \mathrm{t}<\frac{\pi}{4} \end{array}\right.F(t)={sin⁡(t−π4),t>π40,t<π4
  1. a) Find the Fourier transform of e − 9 x 2 e − 9 x 2 e^(-9x^(2))e^{-9 x^2}e−9x2.
b) Find the solution of the heat conduction equation subject to the given initial and boundary conditions:
∂ u ∂ t = ∂ 2 u ∂ x 2 , 0 ≤ x ≤ 1 u ( x , 0 ) = sin ⁡ ( π x ) for 0 ≤ x ≤ 1 u ( 0 , t ) = 0 = u ( 1 , t ) ∂ u ∂ t = ∂ 2 u ∂ x 2 , 0 ≤ x ≤ 1 u ( x , 0 ) = sin ⁡ ( π x ) for 0 ≤ x ≤ 1 u ( 0 , t ) = 0 = u ( 1 , t ) {:[(delu)/(delt)=(del^(2)u)/(delx^(2))”,”0 <= x <= 1],[u(x”,”0)=sin(pix)” for “0 <= x <= 1],[u(0″,”t)=0=u(1”,”t)]:}\begin{aligned} & \frac{\partial \mathrm{u}}{\partial \mathrm{t}}=\frac{\partial^2 \mathrm{u}}{\partial \mathrm{x}^2}, 0 \leq \mathrm{x} \leq 1 \\ & \mathrm{u}(\mathrm{x}, 0)=\sin (\pi \mathrm{x}) \text { for } 0 \leq \mathrm{x} \leq 1 \\ & \mathrm{u}(0, \mathrm{t})=0=\mathrm{u}(1, \mathrm{t}) \end{aligned}∂u∂t=∂2u∂x2,0≤x≤1u(x,0)=sin⁡(πx) for 0≤x≤1u(0,t)=0=u(1,t)
Using Laasonen method with λ = 1 6 λ = 1 6 lambda=(1)/(6)\lambda=\frac{1}{6}λ=16 and h = 1 3 h = 1 3 h=(1)/(3)\mathrm{h}=\frac{1}{3}h=13. Integrate for two levels.
  1. a) Find the solution of the initial boundary value problem:
∂ u ∂ t = ∂ 2 u ∂ x 2 ∂ u ∂ t = ∂ 2 u ∂ x 2 (delu)/(delt)=(del^(2)u)/(delx^(2))\frac{\partial \mathrm{u}}{\partial \mathrm{t}}=\frac{\partial^2 \mathrm{u}}{\partial \mathrm{x}^2}∂u∂t=∂2u∂x2
subject to given initial and boundary conditions:
u ( x , 0 ) = 2 x for x ∈ [ 0 , 1 2 ] u ( x , 0 ) = 2 x for x ∈ 0 , 1 2 u(x,0)=2x” for “x in[0,(1)/(2)]u(x, 0)=2 x \text { for } x \in\left[0, \frac{1}{2}\right]u(x,0)=2x for x∈[0,12]
and 2 ( 1 − x ) 2 ( 1 − x ) 2(1-x)2(1-x)2(1−x) for [ 1 2 , 1 ] 1 2 , 1 [(1)/(2),1]\left[\frac{1}{2}, 1\right][12,1]
u ( 0 , t ) = 0 = u ( 1 , t ) u ( 0 , t ) = 0 = u ( 1 , t ) u(0,t)=0=u(1,t)\mathrm{u}(0, \mathrm{t})=0=\mathrm{u}(1, \mathrm{t})u(0,t)=0=u(1,t)
You may use step length along x -axis, h = 0.2 h = 0.2 h=0.2\mathrm{h}=0.2h=0.2 and solve by Schmidt method with mesh ratio λ = 1 6 λ = 1 6 lambda=(1)/(6)\lambda=\frac{1}{6}λ=16.
b) Show that the method.
y i + 1 = 4 3 y i − 1 3 y i − 1 + 2 h 3 y i + 1 y i + 1 = 4 3 y i − 1 3 y i − 1 + 2 h 3 y i + 1 y_(i+1)=(4)/(3)y_(i)-(1)/(3)y_(i-1)+(2h)/(3)y_(i+1)y_{i+1}=\frac{4}{3} y_i-\frac{1}{3} y_{i-1}+\frac{2 h}{3} y_{i+1}yi+1=43yi−13yi−1+2h3yi+1
is A-stable when applied to test equation y ′ = λ y , λ < 0 y ′ = λ y , λ < 0 y^(‘)=lambday,lambda < 0\mathrm{y}^{\prime}=\lambda \mathrm{y}, \lambda<0y′=λy,λ<0.
  1. a) Use Fourier transforms to solve the boundary value problem:
∂ u ∂ t = 4 ∂ 2 u ∂ x 2 , − ∞ < x < ∞ , t > 0 ∂ u ∂ t = 4 ∂ 2 u ∂ x 2 , − ∞ < x < ∞ , t > 0 (delu)/(delt)=4(del^(2)u)/(delx^(2)),-oo < x < oo,t > 0\frac{\partial \mathrm{u}}{\partial \mathrm{t}}=4 \frac{\partial^2 \mathrm{u}}{\partial \mathrm{x}^2},-\infty<\mathrm{x}<\infty, \mathrm{t}>0∂u∂t=4∂2u∂x2,−∞<x<∞,t>0
subject to the conditions:
i) u , ∂ u ∂ x → 0 u , ∂ u ∂ x → 0 u,(del u)/(delx)rarr0u, \frac{\partial u}{\partial \mathrm{x}} \rightarrow 0u,∂u∂x→0 as x → ± ∞ x → ± ∞ xrarr+-oo\mathrm{x} \rightarrow \pm \inftyx→±∞
ii) u ( x , 0 ) = f ( x ) u ( x , 0 ) = f ( x ) u(x,0)=f(x)u(x, 0)=f(x)u(x,0)=f(x)
b) Solve the initial value problem y ′ = x 2 + y 2 , y ( 0 ) = 1 y ′ = x 2 + y 2 , y ( 0 ) = 1 y^(‘)=x^(2)+y^(2),y(0)=1\mathrm{y}^{\prime}=\mathrm{x}^2+\mathrm{y}^2, \mathrm{y}(0)=1y′=x2+y2,y(0)=1, upto x = 0.2 x = 0.2 x=0.2\mathrm{x}=0.2x=0.2 using third order Taylor series method with h = 0.1 h = 0.1 h=0.1\mathrm{h}=0.1h=0.1.
  1. a) Using Laplace transform, solve the equation:
∂ 2 u ∂ t 2 = 9 ∂ 2 u ∂ x 2 ∂ 2 u ∂ t 2 = 9 ∂ 2 u ∂ x 2 (del^(2)u)/(delt^(2))=9(del^(2)u)/(delx^(2))\frac{\partial^2 u}{\partial t^2}=9 \frac{\partial^2 u}{\partial x^2}∂2u∂t2=9∂2u∂x2
given that:
u ( 0 , t ) = u ( 2 , t ) = 0 , u t ( x , 0 ) = 0 u ( 0 , t ) = u ( 2 , t ) = 0 , u t ( x , 0 ) = 0 u(0,t)=u(2,t)=0,u_(t)(x,0)=0\mathrm{u}(0, \mathrm{t})=\mathrm{u}(2, \mathrm{t})=0, \mathrm{u}_{\mathrm{t}}(\mathrm{x}, 0)=0u(0,t)=u(2,t)=0,ut(x,0)=0
and u ( x , 0 ) = 10 sin ⁡ 2 π x − 20 sin ⁡ 5 π x u ( x , 0 ) = 10 sin ⁡ 2 π x − 20 sin ⁡ 5 π x u(x,0)=10 sin 2pi x-20 sin 5pi xu(x, 0)=10 \sin 2 \pi x-20 \sin 5 \pi xu(x,0)=10sin⁡2πx−20sin⁡5πx.
b) Using second order finite difference method, solve the boundary value problem:
d 2 y dx 2 = 3 2 y 2 with y ( 0 ) = 4 , y ( 1 ) = 1 d 2 y dx 2 = 3 2 y 2 with y ( 0 ) = 4 , y ( 1 ) = 1 (d^(2)y)/(dx^(2))=(3)/(2)y^(2)” with “y(0)=4,y(1)=1\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\frac{3}{2} \mathrm{y}^2 \text { with } \mathrm{y}(0)=4, \mathrm{y}(1)=1d2ydx2=32y2 with y(0)=4,y(1)=1
Using step length h = 1 3 h = 1 3 h=(1)/(3)\mathrm{h}=\frac{1}{3}h=13.
  1. Find the solution of ∇ 2 u = 0 ∇ 2 u = 0 grad^(2)u=0\nabla^2 \mathrm{u}=0∇2u=0 in R subject to the boundary conditions:
u ( x , y ) = x 2 − y 2 on x = 0 , y = 0 , y = 1 u + ∂ u ∂ x = x 2 + 2 x − y 2 on x = 1 u ( x , y ) = x 2 − y 2 on x = 0 , y = 0 , y = 1 u + ∂ u ∂ x = x 2 + 2 x − y 2 on x = 1 {:[u(x”,”y)=x^(2)-y^(2)” on “x=0″,”y=0″,”y=1],[u+(del u)/(del x)=x^(2)+2x-y^(2)” on “x=1]:}\begin{aligned} & u(x, y)=x^2-y^2 \text { on } x=0, y=0, y=1 \\ & u+\frac{\partial u}{\partial x}=x^2+2 x-y^2 \text { on } x=1 \end{aligned}u(x,y)=x2−y2 on x=0,y=0,y=1u+∂u∂x=x2+2x−y2 on x=1
where R is the square 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 0 ≤ x ≤ 1 , 0 ≤ y ≤ 1 0 <= x <= 1,0 <= y <= 10 \leq \mathrm{x} \leq 1,0 \leq \mathrm{y} \leq 10≤x≤1,0≤y≤1, using the five point formula. Use central difference approximation in the boundary condition. Assume uniform step length h = 1 / 2 h = 1 / 2 h=1//2h=1 / 2h=1/2 along the axes.
  1. a) Use finite Fourier transform to solve:
∂ u ∂ t = k ∂ 2 u ∂ x 2 , 0 < x < 4 , t > 0 ∂ u ∂ t = k ∂ 2 u ∂ x 2 , 0 < x < 4 , t > 0 (delu)/(delt)=k(del^(2)u)/(delx^(2)),0 < x < 4,t > 0\frac{\partial \mathrm{u}}{\partial \mathrm{t}}=\mathrm{k} \frac{\partial^2 \mathrm{u}}{\partial \mathrm{x}^2}, 0<\mathrm{x}<4, \mathrm{t}>0∂u∂t=k∂2u∂x2,0<x<4,t>0
Subject to the conditions:
u ( x , 0 ) = 2 x , 0 < x < 4 and u ( 0 , t ) = u ( 4 , t ) = 0 u ( x , 0 ) = 2 x , 0 < x < 4 and u ( 0 , t ) = u ( 4 , t ) = 0 {:[u(x”,”0)=2x”,”quad0 < x < 4],[” and “u(0″,”t)=u(4”,”t)=0]:}\begin{aligned} u(x, 0) & =2 x, \quad 0<x<4 \\ \text { and } u(0, t) & =u(4, t)=0 \end{aligned}u(x,0)=2x,0<x<4 and u(0,t)=u(4,t)=0
b) Solve the boundary value problem:
y ′ ′ + y + f ( x ) = 0 y ′ ( 0 ) = 0 , y ( 1 ) = 0 y ′ ′ + y + f ( x ) = 0 y ′ ( 0 ) = 0 , y ( 1 ) = 0 {:[y^(”)+y+f(x)=0],[y^(‘)(0)=0″,”y(1)=0]:}\begin{aligned} & y^{\prime \prime}+\mathrm{y}+\mathrm{f}(\mathrm{x})=0 \\ & \mathrm{y}^{\prime}(0)=0, \mathrm{y}(1)=0 \end{aligned}y′′+y+f(x)=0y′(0)=0,y(1)=0
by determining the appropriate Green’s function by using the method of variation of parameters and expressing the solution as a definite intergral.
  1. Solve the boundary value problem:
y ′ ′ − 3 y ′ + 2 y = 2 y ′ ′ − 3 y ′ + 2 y = 2 y^(”)-3y^(‘)+2y=2y^{\prime \prime}-3 y^{\prime}+2 y=2y′′−3y′+2y=2
with
y ( 0 ) − y ′ ( 0 ) = − 1 y ( 1 ) + y ′ ( 1 ) = 1 y ( 0 ) − y ′ ( 0 ) = − 1 y ( 1 ) + y ′ ( 1 ) = 1 {:[y(0)-y^(‘)(0)=-1],[y(1)+y^(‘)(1)=1]:}\begin{aligned} & y(0)-y^{\prime}(0)=-1 \\ & y(1)+y^{\prime}(1)=1 \end{aligned}y(0)−y′(0)=−1y(1)+y′(1)=1
using the second order finite difference method with step length h = 1 2 h = 1 2 h=(1)/(2)\mathrm{h}=\frac{1}{2}h=12.
  1. a) Using the generating function J n ( x ) J n ( x ) J_(n)(x)J_n(x)Jn(x), prove that J n − 1 ( x ) + J n + 1 ( x ) = 2 n x J n ( x ) J n − 1 ( x ) + J n + 1 ( x ) = 2 n x J n ( x ) J_(n-1)(x)+J_(n+1)(x)=(2n)/(x)J_(n)(x)J_{n-1}(x)+J_{n+1}(x)=\frac{2 n}{x} J_n(x)Jn−1(x)+Jn+1(x)=2nxJn(x), for integer values of n n nnn.
b) Evaluate:
∫ − 1 1 P n ( x ) 1 − 2 x t + t 2 d x ∫ − 1 1   P n ( x ) 1 − 2 x t + t 2 d x int_(-1)^(1)(P_(n)(x))/(sqrt(1-2xt+t^(2)))dx\int_{-1}^1 \frac{P_n(x)}{\sqrt{1-2 x t+t^2}} d x∫−11Pn(x)1−2xt+t2dx
where P n ( x ) P n ( x ) P_(n)(x)P_n(x)Pn(x) is Legendre polynomial.

Question:-1(a)

Solve the differential equation:

x 2 y ′ ′ + 6 x y ′ + ( 6 + x 2 ) y = 0 x 2 y ′ ′ + 6 x y ′ + 6 + x 2 y = 0 x^(2)y^(”)+6xy^(‘)+(6+x^(2))y=0x^2 y^{\prime \prime} + 6 x y^{\prime} + \left(6 + x^2\right) y = 0x2y′′+6xy′+(6+x2)y=0
in series about x = 0 x = 0 x=0\mathrm{x}=0x=0.

Answer:

To solve the differential equation x 2 y ″ + 6 x y ′ + ( 6 + x 2 ) y = 0 x 2 y ″ + 6 x y ′ + ( 6 + x 2 ) y = 0 x^(2)y^(″)+6xy^(‘)+(6+x^(2))y=0x^2 y” + 6x y’ + (6 + x^2) y = 0x2y″+6xy′+(6+x2)y=0 using a series solution about x = 0 x = 0 x=0x = 0x=0, we recognize that x = 0 x = 0 x=0x = 0x=0 is a regular point (though the equation resembles a form that might suggest a singular point due to the x 2 x 2 x^(2)x^2x2 coefficient, we’ll proceed with a power series and adjust if necessary). We assume a Frobenius-type series solution of the form:
y ( x ) = ∑ n = 0 ∞ a n x n + r , y ( x ) = ∑ n = 0 ∞   a n x n + r , y(x)=sum_(n=0)^(oo)a_(n)x^(n+r),y(x) = \sum_{n=0}^\infty a_n x^{n+r},y(x)=∑n=0∞anxn+r,
where r r rrr is the indicial exponent to be determined, and a n a n a_(n)a_nan are the coefficients, with a 0 ≠ 0 a 0 ≠ 0 a_(0)!=0a_0 \neq 0a0≠0.

Step 1: Compute the derivatives

First, compute the first and second derivatives of y y yyy:
y ′ = ∑ n = 0 ∞ ( n + r ) a n x n + r − 1 , y ′ = ∑ n = 0 ∞   ( n + r ) a n x n + r − 1 , y^(‘)=sum_(n=0)^(oo)(n+r)a_(n)x^(n+r-1),y’ = \sum_{n=0}^\infty (n + r) a_n x^{n + r – 1},y′=∑n=0∞(n+r)anxn+r−1,
y ″ = ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − 2 . y ″ = ∑ n = 0 ∞   ( n + r ) ( n + r − 1 ) a n x n + r − 2 . y^(″)=sum_(n=0)^(oo)(n+r)(n+r-1)a_(n)x^(n+r-2).y” = \sum_{n=0}^\infty (n + r)(n + r – 1) a_n x^{n + r – 2}.y″=∑n=0∞(n+r)(n+r−1)anxn+r−2.

Step 2: Substitute into the differential equation

Substitute y y yyy, y ′ y ′ y^(‘)y’y′, and y ″ y ″ y^(″)y”y″ into the given equation x 2 y ″ + 6 x y ′ + ( 6 + x 2 ) y = 0 x 2 y ″ + 6 x y ′ + ( 6 + x 2 ) y = 0 x^(2)y^(″)+6xy^(‘)+(6+x^(2))y=0x^2 y” + 6x y’ + (6 + x^2) y = 0x2y″+6xy′+(6+x2)y=0.
  • Term 1: x 2 y ″ = x 2 ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r − 2 = ∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r x 2 y ″ = x 2 ∑ n = 0 ∞   ( n + r ) ( n + r − 1 ) a n x n + r − 2 = ∑ n = 0 ∞   ( n + r ) ( n + r − 1 ) a n x n + r x^(2)y^(″)=x^(2)sum_(n=0)^(oo)(n+r)(n+r-1)a_(n)x^(n+r-2)=sum_(n=0)^(oo)(n+r)(n+r-1)a_(n)x^(n+r)x^2 y” = x^2 \sum_{n=0}^\infty (n + r)(n + r – 1) a_n x^{n + r – 2} = \sum_{n=0}^\infty (n + r)(n + r – 1) a_n x^{n + r}x2y″=x2∑n=0∞(n+r)(n+r−1)anxn+r−2=∑n=0∞(n+r)(n+r−1)anxn+r,
  • Term 2: 6 x y ′ = 6 x ∑ n = 0 ∞ ( n + r ) a n x n + r − 1 = ∑ n = 0 ∞ 6 ( n + r ) a n x n + r 6 x y ′ = 6 x ∑ n = 0 ∞   ( n + r ) a n x n + r − 1 = ∑ n = 0 ∞   6 ( n + r ) a n x n + r 6xy^(‘)=6xsum_(n=0)^(oo)(n+r)a_(n)x^(n+r-1)=sum_(n=0)^(oo)6(n+r)a_(n)x^(n+r)6x y’ = 6x \sum_{n=0}^\infty (n + r) a_n x^{n + r – 1} = \sum_{n=0}^\infty 6 (n + r) a_n x^{n + r}6xy′=6x∑n=0∞(n+r)anxn+r−1=∑n=0∞6(n+r)anxn+r,
  • Term 3: ( 6 + x 2 ) y = ( 6 + x 2 ) ∑ n = 0 ∞ a n x n + r = 6 ∑ n = 0 ∞ a n x n + r + ∑ n = 0 ∞ a n x n + r + 2 ( 6 + x 2 ) y = ( 6 + x 2 ) ∑ n = 0 ∞   a n x n + r = 6 ∑ n = 0 ∞   a n x n + r + ∑ n = 0 ∞   a n x n + r + 2 (6+x^(2))y=(6+x^(2))sum_(n=0)^(oo)a_(n)x^(n+r)=6sum_(n=0)^(oo)a_(n)x^(n+r)+sum_(n=0)^(oo)a_(n)x^(n+r+2)(6 + x^2) y = (6 + x^2) \sum_{n=0}^\infty a_n x^{n + r} = 6 \sum_{n=0}^\infty a_n x^{n + r} + \sum_{n=0}^\infty a_n x^{n + r + 2}(6+x2)y=(6+x2)∑n=0∞anxn+r=6∑n=0∞anxn+r+∑n=0∞anxn+r+2.
The full equation becomes:
∑ n = 0 ∞ ( n + r ) ( n + r − 1 ) a n x n + r + ∑ n = 0 ∞ 6 ( n + r ) a n x n + r + 6 ∑ n = 0 ∞ a n x n + r + ∑ n = 0 ∞ a n x n + r + 2 = 0. ∑ n = 0 ∞   ( n + r ) ( n + r − 1 ) a n x n + r + ∑ n = 0 ∞   6 ( n + r ) a n x n + r + 6 ∑ n = 0 ∞   a n x n + r + ∑ n = 0 ∞   a n x n + r + 2 = 0. sum_(n=0)^(oo)(n+r)(n+r-1)a_(n)x^(n+r)+sum_(n=0)^(oo)6(n+r)a_(n)x^(n+r)+6sum_(n=0)^(oo)a_(n)x^(n+r)+sum_(n=0)^(oo)a_(n)x^(n+r+2)=0.\sum_{n=0}^\infty (n + r)(n + r – 1) a_n x^{n + r} + \sum_{n=0}^\infty 6 (n + r) a_n x^{n + r} + 6 \sum_{n=0}^\infty a_n x^{n + r} + \sum_{n=0}^\infty a_n x^{n + r + 2} = 0.∑n=0∞(n+r)(n+r−1)anxn+r+∑n=0∞6(n+r)anxn+r+6∑n=0∞anxn+r+∑n=0∞anxn+r+2=0.

Step 3: Align the indices

The powers of x x xxx are x n + r x n + r x^(n+r)x^{n + r}xn+r and x n + r + 2 x n + r + 2 x^(n+r+2)x^{n + r + 2}xn+r+2. To combine terms, shift the index in the last sum. Let k = n + 2 k = n + 2 k=n+2k = n + 2k=n+2 in the last term, so n = k − 2 n = k − 2 n=k-2n = k – 2n=k−2, and when n = 0 n = 0 n=0n = 0n=0, k = 2 k = 2 k=2k = 2k=2. Adjust the summation:
∑ n = 0 ∞ a n x n + r + 2 = ∑ k = 2 ∞ a k − 2 x k + r . ∑ n = 0 ∞   a n x n + r + 2 = ∑ k = 2 ∞   a k − 2 x k + r . sum_(n=0)^(oo)a_(n)x^(n+r+2)=sum_(k=2)^(oo)a_(k-2)x^(k+r).\sum_{n=0}^\infty a_n x^{n + r + 2} = \sum_{k=2}^\infty a_{k-2} x^{k + r}.∑n=0∞anxn+r+2=∑k=2∞ak−2xk+r.
Now, write the equation, separating the first two terms ( n = 0 n = 0 n=0n = 0n=0 and n = 1 n = 1 n=1n = 1n=1) where the x n + r + 2 x n + r + 2 x^(n+r+2)x^{n + r + 2}xn+r+2 term doesn’t contribute yet:
  • For n = 0 n = 0 n=0n = 0n=0:
( 0 + r ) ( 0 + r − 1 ) a 0 + 6 ( 0 + r ) a 0 + 6 a 0 = [ r ( r − 1 ) + 6 r + 6 ] a 0 x r . ( 0 + r ) ( 0 + r − 1 ) a 0 + 6 ( 0 + r ) a 0 + 6 a 0 = [ r ( r − 1 ) + 6 r + 6 ] a 0 x r . (0+r)(0+r-1)a_(0)+6(0+r)a_(0)+6a_(0)=[r(r-1)+6r+6]a_(0)x^(r).(0 + r)(0 + r – 1) a_0 + 6 (0 + r) a_0 + 6 a_0 = [r(r – 1) + 6r + 6] a_0 x^r.(0+r)(0+r−1)a0+6(0+r)a0+6a0=[r(r−1)+6r+6]a0xr.
  • For n = 1 n = 1 n=1n = 1n=1:
( 1 + r ) ( 1 + r − 1 ) a 1 + 6 ( 1 + r ) a 1 + 6 a 1 = [ ( 1 + r ) r + 6 ( 1 + r ) + 6 ] a 1 x r + 1 . ( 1 + r ) ( 1 + r − 1 ) a 1 + 6 ( 1 + r ) a 1 + 6 a 1 = [ ( 1 + r ) r + 6 ( 1 + r ) + 6 ] a 1 x r + 1 . (1+r)(1+r-1)a_(1)+6(1+r)a_(1)+6a_(1)=[(1+r)r+6(1+r)+6]a_(1)x^(r+1).(1 + r)(1 + r – 1) a_1 + 6 (1 + r) a_1 + 6 a_1 = [(1 + r)r + 6(1 + r) + 6] a_1 x^{r + 1}.(1+r)(1+r−1)a1+6(1+r)a1+6a1=[(1+r)r+6(1+r)+6]a1xr+1.
  • For n ≥ 2 n ≥ 2 n >= 2n \geq 2n≥2:
∑ n = 2 ∞ [ ( n + r ) ( n + r − 1 ) a n + 6 ( n + r ) a n + 6 a n ] x n + r + ∑ n = 2 ∞ a n − 2 x n + r . ∑ n = 2 ∞   [ ( n + r ) ( n + r − 1 ) a n + 6 ( n + r ) a n + 6 a n ] x n + r + ∑ n = 2 ∞   a n − 2 x n + r . sum_(n=2)^(oo)[(n+r)(n+r-1)a_(n)+6(n+r)a_(n)+6a_(n)]x^(n+r)+sum_(n=2)^(oo)a_(n-2)x^(n+r).\sum_{n=2}^\infty [(n + r)(n + r – 1) a_n + 6 (n + r) a_n + 6 a_n] x^{n + r} + \sum_{n=2}^\infty a_{n-2} x^{n + r}.∑n=2∞[(n+r)(n+r−1)an+6(n+r)an+6an]xn+r+∑n=2∞an−2xn+r.
Combine all terms:
[ r ( r − 1 ) + 6 r + 6 ] a 0 x r + [ ( 1 + r ) r + 6 ( 1 + r ) + 6 ] a 1 x r + 1 + ∑ n = 2 ∞ [ ( n + r ) ( n + r − 1 ) + 6 ( n + r ) + 6 ] a n + a n − 2 x n + r = 0. [ r ( r − 1 ) + 6 r + 6 ] a 0 x r + [ ( 1 + r ) r + 6 ( 1 + r ) + 6 ] a 1 x r + 1 + ∑ n = 2 ∞   [ ( n + r ) ( n + r − 1 ) + 6 ( n + r ) + 6 ] a n + a n − 2 x n + r = 0. [r(r-1)+6r+6]a_(0)x^(r)+[(1+r)r+6(1+r)+6]a_(1)x^(r+1)+sum_(n=2)^(oo)[(n+r)(n+r-1)+6(n+r)+6]a_(n)+a_(n-2)x^(n+r)=0.[r(r – 1) + 6r + 6] a_0 x^r + [(1 + r)r + 6(1 + r) + 6] a_1 x^{r + 1} + \sum_{n=2}^\infty [(n + r)(n + r – 1) + 6 (n + r) + 6] a_n + a_{n-2} x^{n + r} = 0.[r(r−1)+6r+6]a0xr+[(1+r)r+6(1+r)+6]a1xr+1+∑n=2∞[(n+r)(n+r−1)+6(n+r)+6]an+an−2xn+r=0.

Step 4: Indicial equation

For the coefficient of x r x r x^(r)x^rxr to be zero:
r ( r − 1 ) + 6 r + 6 = r 2 − r + 6 r + 6 = r 2 + 5 r + 6 = 0. r ( r − 1 ) + 6 r + 6 = r 2 − r + 6 r + 6 = r 2 + 5 r + 6 = 0. r(r-1)+6r+6=r^(2)-r+6r+6=r^(2)+5r+6=0.r(r – 1) + 6r + 6 = r^2 – r + 6r + 6 = r^2 + 5r + 6 = 0.r(r−1)+6r+6=r2−r+6r+6=r2+5r+6=0.
Solve the quadratic:
r 2 + 5 r + 6 = ( r + 2 ) ( r + 3 ) = 0 , r 2 + 5 r + 6 = ( r + 2 ) ( r + 3 ) = 0 , r^(2)+5r+6=(r+2)(r+3)=0,r^2 + 5r + 6 = (r + 2)(r + 3) = 0,r2+5r+6=(r+2)(r+3)=0,
r = − 2 or r = − 3. r = − 2 or r = − 3. r=-2″ or “r=-3.r = -2 \text{ or } r = -3.r=−2 or r=−3.
The roots differ by an integer (1), suggesting a second solution may involve a logarithmic term, but let’s first try the power series for both roots.

Step 5: Recurrence relation

From the general term ( n ≥ 2 n ≥ 2 n >= 2n \geq 2n≥2):
[ ( n + r ) ( n + r − 1 ) + 6 ( n + r ) + 6 ] a n + a n − 2 = 0. [ ( n + r ) ( n + r − 1 ) + 6 ( n + r ) + 6 ] a n + a n − 2 = 0. [(n+r)(n+r-1)+6(n+r)+6]a_(n)+a_(n-2)=0.[(n + r)(n + r – 1) + 6 (n + r) + 6] a_n + a_{n-2} = 0.[(n+r)(n+r−1)+6(n+r)+6]an+an−2=0.
Simplify the coefficient:
( n + r ) ( n + r − 1 ) + 6 ( n + r ) + 6 = n 2 + 2 n r + r 2 − n − r + 6 n + 6 r + 6 = n 2 + ( 2 r + 5 ) n + ( r 2 + 5 r + 6 ) . ( n + r ) ( n + r − 1 ) + 6 ( n + r ) + 6 = n 2 + 2 n r + r 2 − n − r + 6 n + 6 r + 6 = n 2 + ( 2 r + 5 ) n + ( r 2 + 5 r + 6 ) . (n+r)(n+r-1)+6(n+r)+6=n^(2)+2nr+r^(2)-n-r+6n+6r+6=n^(2)+(2r+5)n+(r^(2)+5r+6).(n + r)(n + r – 1) + 6 (n + r) + 6 = n^2 + 2nr + r^2 – n – r + 6n + 6r + 6 = n^2 + (2r + 5)n + (r^2 + 5r + 6).(n+r)(n+r−1)+6(n+r)+6=n2+2nr+r2−n−r+6n+6r+6=n2+(2r+5)n+(r2+5r+6).
Since r 2 + 5 r + 6 = 0 r 2 + 5 r + 6 = 0 r^(2)+5r+6=0r^2 + 5r + 6 = 0r2+5r+6=0 for r = − 2 r = − 2 r=-2r = -2r=−2 or r = − 3 r = − 3 r=-3r = -3r=−3, this becomes:
n 2 + ( 2 r + 5 ) n = n [ n + ( 2 r + 5 ) ] . n 2 + ( 2 r + 5 ) n = n [ n + ( 2 r + 5 ) ] . n^(2)+(2r+5)n=n[n+(2r+5)].n^2 + (2r + 5)n = n [n + (2r + 5)].n2+(2r+5)n=n[n+(2r+5)].
So the recurrence is:
a n = − a n − 2 n ( n + 2 r + 5 ) , n ≥ 2. a n = − a n − 2 n ( n + 2 r + 5 ) , n ≥ 2. a_(n)=-(a_(n-2))/(n(n+2r+5)),quad n >= 2.a_n = -\frac{a_{n-2}}{n (n + 2r + 5)}, \quad n \geq 2.an=−an−2n(n+2r+5),n≥2.
For n = 1 n = 1 n=1n = 1n=1:
( 1 + r ) r + 6 ( 1 + r ) + 6 = r + r 2 + 6 + 6 r + 6 = r 2 + 7 r + 12 = ( r + 3 ) ( r + 4 ) , ( 1 + r ) r + 6 ( 1 + r ) + 6 = r + r 2 + 6 + 6 r + 6 = r 2 + 7 r + 12 = ( r + 3 ) ( r + 4 ) , (1+r)r+6(1+r)+6=r+r^(2)+6+6r+6=r^(2)+7r+12=(r+3)(r+4),(1 + r)r + 6(1 + r) + 6 = r + r^2 + 6 + 6r + 6 = r^2 + 7r + 12 = (r + 3)(r + 4),(1+r)r+6(1+r)+6=r+r2+6+6r+6=r2+7r+12=(r+3)(r+4),
  • If r = − 3 r = − 3 r=-3r = -3r=−3: ( − 3 + 3 ) ( − 3 + 4 ) = 0 ⋅ 1 = 0 ( − 3 + 3 ) ( − 3 + 4 ) = 0 ⋅ 1 = 0 (-3+3)(-3+4)=0*1=0(-3 + 3)(-3 + 4) = 0 \cdot 1 = 0(−3+3)(−3+4)=0⋅1=0,
  • If r = − 2 r = − 2 r=-2r = -2r=−2: ( − 2 + 3 ) ( − 2 + 4 ) = 1 ⋅ 2 = 2 ( − 2 + 3 ) ( − 2 + 4 ) = 1 ⋅ 2 = 2 (-2+3)(-2+4)=1*2=2(-2 + 3)(-2 + 4) = 1 \cdot 2 = 2(−2+3)(−2+4)=1⋅2=2.
This suggests a 1 = 0 a 1 = 0 a_(1)=0a_1 = 0a1=0 when r = − 3 r = − 3 r=-3r = -3r=−3, but we need to check consistency.

Step 6: Solve for r = − 2 r = − 2 r=-2r = -2r=−2

Recurrence: a n = − a n − 2 n ( n + 2 ( − 2 ) + 5 ) = − a n − 2 n ( n + 1 ) a n = − a n − 2 n ( n + 2 ( − 2 ) + 5 ) = − a n − 2 n ( n + 1 ) a_(n)=-(a_(n-2))/(n(n+2(-2)+5))=-(a_(n-2))/(n(n+1))a_n = -\frac{a_{n-2}}{n (n + 2(-2) + 5)} = -\frac{a_{n-2}}{n (n + 1)}an=−an−2n(n+2(−2)+5)=−an−2n(n+1).
  • n = 1 n = 1 n=1n = 1n=1: ( r + 3 ) ( r + 4 ) a 1 = ( − 2 + 3 ) ( − 2 + 4 ) a 1 = 2 a 1 = 0 ( r + 3 ) ( r + 4 ) a 1 = ( − 2 + 3 ) ( − 2 + 4 ) a 1 = 2 a 1 = 0 (r+3)(r+4)a_(1)=(-2+3)(-2+4)a_(1)=2a_(1)=0(r + 3)(r + 4) a_1 = (-2 + 3)(-2 + 4) a_1 = 2 a_1 = 0(r+3)(r+4)a1=(−2+3)(−2+4)a1=2a1=0 only if a 1 = 0 a 1 = 0 a_(1)=0a_1 = 0a1=0,
  • n = 2 n = 2 n=2n = 2n=2: a 2 = − a 0 2 ( 2 + 1 ) = − a 0 6 a 2 = − a 0 2 ( 2 + 1 ) = − a 0 6 a_(2)=-(a_(0))/(2(2+1))=-(a_(0))/(6)a_2 = -\frac{a_0}{2 (2 + 1)} = -\frac{a_0}{6}a2=−a02(2+1)=−a06,
  • n = 3 n = 3 n=3n = 3n=3: a 3 = − a 1 3 ( 3 + 1 ) = 0 a 3 = − a 1 3 ( 3 + 1 ) = 0 a_(3)=-(a_(1))/(3(3+1))=0a_3 = -\frac{a_1}{3 (3 + 1)} = 0a3=−a13(3+1)=0 (since a 1 = 0 a 1 = 0 a_(1)=0a_1 = 0a1=0),
  • n = 4 n = 4 n=4n = 4n=4: a 4 = − a 2 4 ( 4 + 1 ) = − − a 0 6 20 = a 0 120 a 4 = − a 2 4 ( 4 + 1 ) = − − a 0 6 20 = a 0 120 a_(4)=-(a_(2))/(4(4+1))=-(-(a_(0))/(6))/(20)=(a_(0))/(120)a_4 = -\frac{a_2}{4 (4 + 1)} = -\frac{-\frac{a_0}{6}}{20} = \frac{a_0}{120}a4=−a24(4+1)=−−a0620=a0120.
Series: y 1 = x − 2 ( a 0 − a 0 6 x 2 + a 0 120 x 4 − ⋯ ) = a 0 x − 2 ( 1 − x 2 6 + x 4 120 − ⋯ ) y 1 = x − 2 a 0 − a 0 6 x 2 + a 0 120 x 4 − ⋯ = a 0 x − 2 1 − x 2 6 + x 4 120 − ⋯ y_(1)=x^(-2)(a_(0)-(a_(0))/(6)x^(2)+(a_(0))/(120)x^(4)-cdots)=a_(0)x^(-2)(1-(x^(2))/(6)+(x^(4))/(120)-cdots)y_1 = x^{-2} \left( a_0 – \frac{a_0}{6} x^2 + \frac{a_0}{120} x^4 – \cdots \right) = a_0 x^{-2} \left( 1 – \frac{x^2}{6} + \frac{x^4}{120} – \cdots \right)y1=x−2(a0−a06x2+a0120x4−⋯)=a0x−2(1−x26+x4120−⋯).

Step 7: Solve for r = − 3 r = − 3 r=-3r = -3r=−3

Recurrence: a n = − a n − 2 n ( n + 2 ( − 3 ) + 5 ) = − a n − 2 n ( n − 1 ) a n = − a n − 2 n ( n + 2 ( − 3 ) + 5 ) = − a n − 2 n ( n − 1 ) a_(n)=-(a_(n-2))/(n(n+2(-3)+5))=-(a_(n-2))/(n(n-1))a_n = -\frac{a_{n-2}}{n (n + 2(-3) + 5)} = -\frac{a_{n-2}}{n (n – 1)}an=−an−2n(n+2(−3)+5)=−an−2n(n−1).
  • n = 1 n = 1 n=1n = 1n=1: ( r + 3 ) ( r + 4 ) a 1 = 0 ⋅ 1 a 1 = 0 ( r + 3 ) ( r + 4 ) a 1 = 0 ⋅ 1 a 1 = 0 (r+3)(r+4)a_(1)=0*1a_(1)=0(r + 3)(r + 4) a_1 = 0 \cdot 1 a_1 = 0(r+3)(r+4)a1=0⋅1a1=0, a 1 a 1 a_(1)a_1a1 is free,
  • n = 2 n = 2 n=2n = 2n=2: a 2 = − a 0 2 ( 2 − 1 ) = − a 0 2 a 2 = − a 0 2 ( 2 − 1 ) = − a 0 2 a_(2)=-(a_(0))/(2(2-1))=-(a_(0))/(2)a_2 = -\frac{a_0}{2 (2 – 1)} = -\frac{a_0}{2}a2=−a02(2−1)=−a02,
  • n = 3 n = 3 n=3n = 3n=3: a 3 = − a 1 3 ( 3 − 1 ) = − a 1 6 a 3 = − a 1 3 ( 3 − 1 ) = − a 1 6 a_(3)=-(a_(1))/(3(3-1))=-(a_(1))/(6)a_3 = -\frac{a_1}{3 (3 – 1)} = -\frac{a_1}{6}a3=−a13(3−1)=−a16,
  • n = 4 n = 4 n=4n = 4n=4: a 4 = − a 2 4 ( 4 − 1 ) = − − a 0 2 12 = a 0 24 a 4 = − a 2 4 ( 4 − 1 ) = − − a 0 2 12 = a 0 24 a_(4)=-(a_(2))/(4(4-1))=-(-(a_(0))/(2))/(12)=(a_(0))/(24)a_4 = -\frac{a_2}{4 (4 – 1)} = -\frac{-\frac{a_0}{2}}{12} = \frac{a_0}{24}a4=−a24(4−1)=−−a0212=a024.
Series: y 2 = x − 3 ( a 0 + a 1 x − a 0 2 x 2 − a 1 6 x 3 + a 0 24 x 4 − ⋯ ) y 2 = x − 3 a 0 + a 1 x − a 0 2 x 2 − a 1 6 x 3 + a 0 24 x 4 − ⋯ y_(2)=x^(-3)(a_(0)+a_(1)x-(a_(0))/(2)x^(2)-(a_(1))/(6)x^(3)+(a_(0))/(24)x^(4)-cdots)y_2 = x^{-3} \left( a_0 + a_1 x – \frac{a_0}{2} x^2 – \frac{a_1}{6} x^3 + \frac{a_0}{24} x^4 – \cdots \right)y2=x−3(a0+a1x−a02x2−a16x3+a024x4−⋯).

Final Answer

The general solution is:
y ( x ) = c 1 x − 2 ( 1 − x 2 6 + x 4 120 − ⋯ ) + c 2 x − 3 ( 1 + a x − 1 2 x 2 − a 6 x 3 + 1 24 x 4 − ⋯ ) , y ( x ) = c 1 x − 2 1 − x 2 6 + x 4 120 − ⋯ + c 2 x − 3 1 + a x − 1 2 x 2 − a 6 x 3 + 1 24 x 4 − ⋯ , y(x)=c_(1)x^(-2)(1-(x^(2))/(6)+(x^(4))/(120)-cdots)+c_(2)x^(-3)(1+ax-(1)/(2)x^(2)-(a)/(6)x^(3)+(1)/(24)x^(4)-cdots),y(x) = c_1 x^{-2} \left( 1 – \frac{x^2}{6} + \frac{x^4}{120} – \cdots \right) + c_2 x^{-3} \left( 1 + a x – \frac{1}{2} x^2 – \frac{a}{6} x^3 + \frac{1}{24} x^4 – \cdots \right),y(x)=c1x−2(1−x26+x4120−⋯)+c2x−3(1+ax−12x2−a6x3+124x4−⋯),
where c 1 c 1 c_(1)c_1c1 and c 2 c 2 c_(2)c_2c2 are constants, and a = a 1 / a 0 a = a 1 / a 0 a=a_(1)//a_(0)a = a_1/a_0a=a1/a0 is an arbitrary constant for the second solution. Note that x = 0 x = 0 x=0x = 0x=0 is a singular point, and these solutions are valid for x ≠ 0 x ≠ 0 x!=0x \neq 0x≠0.

Question:-1(b)

Express f ( x ) = x 4 + 3 x 3 + 4 x 2 − x + 2 f ( x ) = x 4 + 3 x 3 + 4 x 2 − x + 2 f(x)=x^(4)+3x^(3)+4x^(2)-x+2f(x) = x^4 + 3 x^3 + 4 x^2 – x + 2f(x)=x4+3x3+4x2−x+2 in terms of Legendre polynomials.

Answer:

To express the polynomial f ( x ) = x 4 + 3 x 3 + 4 x 2 − x + 2 f ( x ) = x 4 + 3 x 3 + 4 x 2 − x + 2 f(x)=x^(4)+3x^(3)+4x^(2)-x+2f(x) = x^4 + 3x^3 + 4x^2 – x + 2f(x)=x4+3x3+4x2−x+2 in terms of Legendre polynomials, we need to write it as a linear combination of the form:
f ( x ) = c 0 P 0 ( x ) + c 1 P 1 ( x ) + c 2 P 2 ( x ) + c 3 P 3 ( x ) + c 4 P 4 ( x ) , f ( x ) = c 0 P 0 ( x ) + c 1 P 1 ( x ) + c 2 P 2 ( x ) + c 3 P 3 ( x ) + c 4 P 4 ( x ) , f(x)=c_(0)P_(0)(x)+c_(1)P_(1)(x)+c_(2)P_(2)(x)+c_(3)P_(3)(x)+c_(4)P_(4)(x),f(x) = c_0 P_0(x) + c_1 P_1(x) + c_2 P_2(x) + c_3 P_3(x) + c_4 P_4(x),f(x)=c0P0(x)+c1P1(x)+c2P2(x)+c3P3(x)+c4P4(x),
where P n ( x ) P n ( x ) P_(n)(x)P_n(x)Pn(x) are the Legendre polynomials, and since f ( x ) f ( x ) f(x)f(x)f(x) is a polynomial of degree 4, we only need terms up to P 4 ( x ) P 4 ( x ) P_(4)(x)P_4(x)P4(x). The Legendre polynomials are orthogonal on [ − 1 , 1 ] [ − 1 , 1 ] [-1,1][-1, 1][−1,1] with respect to the weight function 1, and we can find the coefficients c n c n c_(n)c_ncn using the orthogonality property:
c n = ⟨ f , P n ⟩ ⟨ P n , P n ⟩ = ∫ − 1 1 f ( x ) P n ( x ) d x ∫ − 1 1 [ P n ( x ) ] 2 d x . c n = ⟨ f , P n ⟩ ⟨ P n , P n ⟩ = ∫ − 1 1   f ( x ) P n ( x ) d x ∫ − 1 1   [ P n ( x ) ] 2 d x . c_(n)=((:f,P_(n):))/((:P_(n),P_(n):))=(int_(-1)^(1)f(x)P_(n)(x)dx)/(int_(-1)^(1)[P_(n)(x)]^(2)dx).c_n = \frac{\langle f, P_n \rangle}{\langle P_n, P_n \rangle} = \frac{\int_{-1}^{1} f(x) P_n(x) \, dx}{\int_{-1}^{1} [P_n(x)]^2 \, dx}.cn=⟨f,Pn⟩⟨Pn,Pn⟩=∫−11f(x)Pn(x)dx∫−11[Pn(x)]2dx.
The standard Legendre polynomials (normalized such that P n ( 1 ) = 1 P n ( 1 ) = 1 P_(n)(1)=1P_n(1) = 1Pn(1)=1) are:
  • P 0 ( x ) = 1 P 0 ( x ) = 1 P_(0)(x)=1P_0(x) = 1P0(x)=1,
  • P 1 ( x ) = x P 1 ( x ) = x P_(1)(x)=xP_1(x) = xP1(x)=x,
  • P 2 ( x ) = 1 2 ( 3 x 2 − 1 ) P 2 ( x ) = 1 2 ( 3 x 2 − 1 ) P_(2)(x)=(1)/(2)(3x^(2)-1)P_2(x) = \frac{1}{2}(3x^2 – 1)P2(x)=12(3x2−1),
  • P 3 ( x ) = 1 2 ( 5 x 3 − 3 x ) P 3 ( x ) = 1 2 ( 5 x 3 − 3 x ) P_(3)(x)=(1)/(2)(5x^(3)-3x)P_3(x) = \frac{1}{2}(5x^3 – 3x)P3(x)=12(5x3−3x),
  • P 4 ( x ) = 1 8 ( 35 x 4 − 30 x 2 + 3 ) P 4 ( x ) = 1 8 ( 35 x 4 − 30 x 2 + 3 ) P_(4)(x)=(1)/(8)(35x^(4)-30x^(2)+3)P_4(x) = \frac{1}{8}(35x^4 – 30x^2 + 3)P4(x)=18(35x4−30x2+3).
The norm of each is:
⟨ P n , P n ⟩ = ∫ − 1 1 [ P n ( x ) ] 2 d x = 2 2 n + 1 . ⟨ P n , P n ⟩ = ∫ − 1 1   [ P n ( x ) ] 2 d x = 2 2 n + 1 . (:P_(n),P_(n):)=int_(-1)^(1)[P_(n)(x)]^(2)dx=(2)/(2n+1).\langle P_n, P_n \rangle = \int_{-1}^{1} [P_n(x)]^2 \, dx = \frac{2}{2n + 1}.⟨Pn,Pn⟩=∫−11[Pn(x)]2dx=22n+1.
However, a more practical approach for a finite polynomial is to match coefficients by expressing the Legendre polynomials in terms of powers of x x xxx and solving a system of equations, since f ( x ) f ( x ) f(x)f(x)f(x) is a degree-4 polynomial.

Step 1: Write the Legendre polynomials as power series

  • P 0 ( x ) = 1 P 0 ( x ) = 1 P_(0)(x)=1P_0(x) = 1P0(x)=1,
  • P 1 ( x ) = x P 1 ( x ) = x P_(1)(x)=xP_1(x) = xP1(x)=x,
  • P 2 ( x ) = 1 2 ( 3 x 2 − 1 ) = 3 2 x 2 − 1 2 P 2 ( x ) = 1 2 ( 3 x 2 − 1 ) = 3 2 x 2 − 1 2 P_(2)(x)=(1)/(2)(3x^(2)-1)=(3)/(2)x^(2)-(1)/(2)P_2(x) = \frac{1}{2}(3x^2 – 1) = \frac{3}{2}x^2 – \frac{1}{2}P2(x)=12(3x2−1)=32x2−12,
  • P 3 ( x ) = 1 2 ( 5 x 3 − 3 x ) = 5 2 x 3 − 3 2 x P 3 ( x ) = 1 2 ( 5 x 3 − 3 x ) = 5 2 x 3 − 3 2 x P_(3)(x)=(1)/(2)(5x^(3)-3x)=(5)/(2)x^(3)-(3)/(2)xP_3(x) = \frac{1}{2}(5x^3 – 3x) = \frac{5}{2}x^3 – \frac{3}{2}xP3(x)=12(5x3−3x)=52x3−32x,
  • P 4 ( x ) = 1 8 ( 35 x 4 − 30 x 2 + 3 ) = 35 8 x 4 − 30 8 x 2 + 3 8 = 35 8 x 4 − 15 4 x 2 + 3 8 P 4 ( x ) = 1 8 ( 35 x 4 − 30 x 2 + 3 ) = 35 8 x 4 − 30 8 x 2 + 3 8 = 35 8 x 4 − 15 4 x 2 + 3 8 P_(4)(x)=(1)/(8)(35x^(4)-30x^(2)+3)=(35)/(8)x^(4)-(30)/(8)x^(2)+(3)/(8)=(35)/(8)x^(4)-(15)/(4)x^(2)+(3)/(8)P_4(x) = \frac{1}{8}(35x^4 – 30x^2 + 3) = \frac{35}{8}x^4 – \frac{30}{8}x^2 + \frac{3}{8} = \frac{35}{8}x^4 – \frac{15}{4}x^2 + \frac{3}{8}P4(x)=18(35x4−30x2+3)=358x4−308x2+38=358x4−154x2+38.

Step 2: Set up the linear combination

Let:
f ( x ) = c 0 P 0 ( x ) + c 1 P 1 ( x ) + c 2 P 2 ( x ) + c 3 P 3 ( x ) + c 4 P 4 ( x ) . f ( x ) = c 0 P 0 ( x ) + c 1 P 1 ( x ) + c 2 P 2 ( x ) + c 3 P 3 ( x ) + c 4 P 4 ( x ) . f(x)=c_(0)P_(0)(x)+c_(1)P_(1)(x)+c_(2)P_(2)(x)+c_(3)P_(3)(x)+c_(4)P_(4)(x).f(x) = c_0 P_0(x) + c_1 P_1(x) + c_2 P_2(x) + c_3 P_3(x) + c_4 P_4(x).f(x)=c0P0(x)+c1P1(x)+c2P2(x)+c3P3(x)+c4P4(x).
Substitute and expand:
f ( x ) = c 0 ( 1 ) + c 1 ( x ) + c 2 ( 3 2 x 2 − 1 2 ) + c 3 ( 5 2 x 3 − 3 2 x ) + c 4 ( 35 8 x 4 − 15 4 x 2 + 3 8 ) . f ( x ) = c 0 ( 1 ) + c 1 ( x ) + c 2 3 2 x 2 − 1 2 + c 3 5 2 x 3 − 3 2 x + c 4 35 8 x 4 − 15 4 x 2 + 3 8 . f(x)=c_(0)(1)+c_(1)(x)+c_(2)((3)/(2)x^(2)-(1)/(2))+c_(3)((5)/(2)x^(3)-(3)/(2)x)+c_(4)((35)/(8)x^(4)-(15)/(4)x^(2)+(3)/(8)).f(x) = c_0 (1) + c_1 (x) + c_2 \left( \frac{3}{2}x^2 – \frac{1}{2} \right) + c_3 \left( \frac{5}{2}x^3 – \frac{3}{2}x \right) + c_4 \left( \frac{35}{8}x^4 – \frac{15}{4}x^2 + \frac{3}{8} \right).f(x)=c0(1)+c1(x)+c2(32x2−12)+c3(52x3−32x)+c4(358x4−154x2+38).
Collect terms by powers of x x xxx:
  • x 4 x 4 x^(4)x^4x4: c 4 ⋅ 35 8 c 4 ⋅ 35 8 c_(4)*(35)/(8)c_4 \cdot \frac{35}{8}c4⋅358,
  • x 3 x 3 x^(3)x^3x3: c 3 ⋅ 5 2 c 3 ⋅ 5 2 c_(3)*(5)/(2)c_3 \cdot \frac{5}{2}c3⋅52,
  • x 2 x 2 x^(2)x^2x2: c 2 ⋅ 3 2 + c 4 ⋅ ( − 15 4 ) c 2 ⋅ 3 2 + c 4 ⋅ − 15 4 c_(2)*(3)/(2)+c_(4)*(-(15)/(4))c_2 \cdot \frac{3}{2} + c_4 \cdot \left( -\frac{15}{4} \right)c2⋅32+c4⋅(−154),
  • x 1 x 1 x^(1)x^1x1: c 1 + c 3 ⋅ ( − 3 2 ) c 1 + c 3 ⋅ − 3 2 c_(1)+c_(3)*(-(3)/(2))c_1 + c_3 \cdot \left( -\frac{3}{2} \right)c1+c3⋅(−32),
  • x 0 x 0 x^(0)x^0x0: c 0 + c 2 ⋅ ( − 1 2 ) + c 4 ⋅ 3 8 c 0 + c 2 ⋅ − 1 2 + c 4 ⋅ 3 8 c_(0)+c_(2)*(-(1)/(2))+c_(4)*(3)/(8)c_0 + c_2 \cdot \left( -\frac{1}{2} \right) + c_4 \cdot \frac{3}{8}c0+c2⋅(−12)+c4⋅38.
Equate to f ( x ) = x 4 + 3 x 3 + 4 x 2 − x + 2 f ( x ) = x 4 + 3 x 3 + 4 x 2 − x + 2 f(x)=x^(4)+3x^(3)+4x^(2)-x+2f(x) = x^4 + 3x^3 + 4x^2 – x + 2f(x)=x4+3x3+4x2−x+2:
  1. x 4 x 4 x^(4)x^4x4: 35 8 c 4 = 1 35 8 c 4 = 1 (35)/(8)c_(4)=1\frac{35}{8} c_4 = 1358c4=1,
  2. x 3 x 3 x^(3)x^3x3: 5 2 c 3 = 3 5 2 c 3 = 3 (5)/(2)c_(3)=3\frac{5}{2} c_3 = 352c3=3,
  3. x 2 x 2 x^(2)x^2x2: 3 2 c 2 − 15 4 c 4 = 4 3 2 c 2 − 15 4 c 4 = 4 (3)/(2)c_(2)-(15)/(4)c_(4)=4\frac{3}{2} c_2 – \frac{15}{4} c_4 = 432c2−154c4=4,
  4. x x xxx: c 1 − 3 2 c 3 = − 1 c 1 − 3 2 c 3 = − 1 c_(1)-(3)/(2)c_(3)=-1c_1 – \frac{3}{2} c_3 = -1c1−32c3=−1,
  5. Constant: c 0 − 1 2 c 2 + 3 8 c 4 = 2 c 0 − 1 2 c 2 + 3 8 c 4 = 2 c_(0)-(1)/(2)c_(2)+(3)/(8)c_(4)=2c_0 – \frac{1}{2} c_2 + \frac{3}{8} c_4 = 2c0−12c2+38c4=2.

Step 3: Solve the system

  • From (1): 35 8 c 4 = 1 35 8 c 4 = 1 (35)/(8)c_(4)=1\frac{35}{8} c_4 = 1358c4=1, so c 4 = 8 35 c 4 = 8 35 c_(4)=(8)/(35)c_4 = \frac{8}{35}c4=835.
  • From (2): 5 2 c 3 = 3 5 2 c 3 = 3 (5)/(2)c_(3)=3\frac{5}{2} c_3 = 352c3=3, so c 3 = 3 ⋅ 2 5 = 6 5 c 3 = 3 ⋅ 2 5 = 6 5 c_(3)=3*(2)/(5)=(6)/(5)c_3 = 3 \cdot \frac{2}{5} = \frac{6}{5}c3=3⋅25=65.
  • From (3): 3 2 c 2 − 15 4 ⋅ 8 35 = 4 3 2 c 2 − 15 4 ⋅ 8 35 = 4 (3)/(2)c_(2)-(15)/(4)*(8)/(35)=4\frac{3}{2} c_2 – \frac{15}{4} \cdot \frac{8}{35} = 432c2−154⋅835=4, 15 4 ⋅ 8 35 = 120 140 = 6 7 , 15 4 ⋅ 8 35 = 120 140 = 6 7 , (15)/(4)*(8)/(35)=(120)/(140)=(6)/(7),\frac{15}{4} \cdot \frac{8}{35} = \frac{120}{140} = \frac{6}{7},154⋅835=120140=67, 3 2 c 2 − 6 7 = 4 , 3 2 c 2 − 6 7 = 4 , (3)/(2)c_(2)-(6)/(7)=4,\frac{3}{2} c_2 – \frac{6}{7} = 4,32c2−67=4, 3 2 c 2 = 4 + 6 7 = 28 7 + 6 7 = 34 7 , 3 2 c 2 = 4 + 6 7 = 28 7 + 6 7 = 34 7 , (3)/(2)c_(2)=4+(6)/(7)=(28)/(7)+(6)/(7)=(34)/(7),\frac{3}{2} c_2 = 4 + \frac{6}{7} = \frac{28}{7} + \frac{6}{7} = \frac{34}{7},32c2=4+67=287+67=347, c 2 = 34 7 ⋅ 2 3 = 68 21 c 2 = 34 7 ⋅ 2 3 = 68 21 c_(2)=(34)/(7)*(2)/(3)=(68)/(21)c_2 = \frac{34}{7} \cdot \frac{2}{3} = \frac{68}{21}c2=347⋅23=6821.
  • From (4): c 1 − 3 2 ⋅ 6 5 = − 1 c 1 − 3 2 ⋅ 6 5 = − 1 c_(1)-(3)/(2)*(6)/(5)=-1c_1 – \frac{3}{2} \cdot \frac{6}{5} = -1c1−32⋅65=−1, 3 2 ⋅ 6 5 = 9 5 , 3 2 ⋅ 6 5 = 9 5 , (3)/(2)*(6)/(5)=(9)/(5),\frac{3}{2} \cdot \frac{6}{5} = \frac{9}{5},32⋅65=95, c 1 − 9 5 = − 1 , c 1 − 9 5 = − 1 , c_(1)-(9)/(5)=-1,c_1 – \frac{9}{5} = -1,c1−95=−1, c 1 = − 1 + 9 5 = − 5 5 + 9 5 = 4 5 c 1 = − 1 + 9 5 = − 5 5 + 9 5 = 4 5 c_(1)=-1+(9)/(5)=-(5)/(5)+(9)/(5)=(4)/(5)c_1 = -1 + \frac{9}{5} = -\frac{5}{5} + \frac{9}{5} = \frac{4}{5}c1=−1+95=−55+95=45.
  • From (5): c 0 − 1 2 ⋅ 68 21 + 3 8 ⋅ 8 35 = 2 c 0 − 1 2 ⋅ 68 21 + 3 8 ⋅ 8 35 = 2 c_(0)-(1)/(2)*(68)/(21)+(3)/(8)*(8)/(35)=2c_0 – \frac{1}{2} \cdot \frac{68}{21} + \frac{3}{8} \cdot \frac{8}{35} = 2c0−12⋅6821+38⋅835=2, 1 2 ⋅ 68 21 = 34 21 , 1 2 ⋅ 68 21 = 34 21 , (1)/(2)*(68)/(21)=(34)/(21),\frac{1}{2} \cdot \frac{68}{21} = \frac{34}{21},12⋅6821=3421, 3 8 ⋅ 8 35 = 3 35 , 3 8 ⋅ 8 35 = 3 35 , (3)/(8)*(8)/(35)=(3)/(35),\frac{3}{8} \cdot \frac{8}{35} = \frac{3}{35},38⋅835=335, c 0 − 34 21 + 3 35 = 2 , c 0 − 34 21 + 3 35 = 2 , c_(0)-(34)/(21)+(3)/(35)=2,c_0 – \frac{34}{21} + \frac{3}{35} = 2,c0−3421+335=2, 34 21 = 170 105 , 3 35 = 9 105 , 34 21 = 170 105 , 3 35 = 9 105 , (34)/(21)=(170)/(105),quad(3)/(35)=(9)/(105),\frac{34}{21} = \frac{170}{105}, \quad \frac{3}{35} = \frac{9}{105},3421=170105,335=9105, c 0 − 170 105 + 9 105 = 2 , c 0 − 170 105 + 9 105 = 2 , c_(0)-(170)/(105)+(9)/(105)=2,c_0 – \frac{170}{105} + \frac{9}{105} = 2,c0−170105+9105=2, c 0 − 161 105 = 2 , c 0 − 161 105 = 2 , c_(0)-(161)/(105)=2,c_0 – \frac{161}{105} = 2,c0−161105=2, c 0 = 2 + 161 105 = 210 105 + 161 105 = 371 105 c 0 = 2 + 161 105 = 210 105 + 161 105 = 371 105 c_(0)=2+(161)/(105)=(210)/(105)+(161)/(105)=(371)/(105)c_0 = 2 + \frac{161}{105} = \frac{210}{105} + \frac{161}{105} = \frac{371}{105}c0=2+161105=210105+161105=371105.

Step 4: Verify

f ( x ) = 371 105 P 0 + 4 5 P 1 + 68 21 P 2 + 6 5 P 3 + 8 35 P 4 . f ( x ) = 371 105 P 0 + 4 5 P 1 + 68 21 P 2 + 6 5 P 3 + 8 35 P 4 . f(x)=(371)/(105)P_(0)+(4)/(5)P_(1)+(68)/(21)P_(2)+(6)/(5)P_(3)+(8)/(35)P_(4).f(x) = \frac{371}{105} P_0 + \frac{4}{5} P_1 + \frac{68}{21} P_2 + \frac{6}{5} P_3 + \frac{8}{35} P_4.f(x)=371105P0+45P1+6821P2+65P3+835P4.
Compute:
  • x 4 x 4 x^(4)x^4x4: 8 35 ⋅ 35 8 = 1 8 35 ⋅ 35 8 = 1 (8)/(35)*(35)/(8)=1\frac{8}{35} \cdot \frac{35}{8} = 1835⋅358=1,
  • x 3 x 3 x^(3)x^3x3: 6 5 ⋅ 5 2 = 3 6 5 ⋅ 5 2 = 3 (6)/(5)*(5)/(2)=3\frac{6}{5} \cdot \frac{5}{2} = 365⋅52=3,
  • x 2 x 2 x^(2)x^2x2: 68 21 ⋅ 3 2 − 8 35 ⋅ 15 4 = 102 21 − 120 140 = 340 70 − 60 70 = 280 70 = 4 68 21 ⋅ 3 2 − 8 35 ⋅ 15 4 = 102 21 − 120 140 = 340 70 − 60 70 = 280 70 = 4 (68)/(21)*(3)/(2)-(8)/(35)*(15)/(4)=(102)/(21)-(120)/(140)=(340)/(70)-(60)/(70)=(280)/(70)=4\frac{68}{21} \cdot \frac{3}{2} – \frac{8}{35} \cdot \frac{15}{4} = \frac{102}{21} – \frac{120}{140} = \frac{340}{70} – \frac{60}{70} = \frac{280}{70} = 46821⋅32−835⋅154=10221−120140=34070−6070=28070=4,
  • x x xxx: 4 5 − 6 5 ⋅ 3 2 = 4 5 − 9 5 = − 1 4 5 − 6 5 ⋅ 3 2 = 4 5 − 9 5 = − 1 (4)/(5)-(6)/(5)*(3)/(2)=(4)/(5)-(9)/(5)=-1\frac{4}{5} – \frac{6}{5} \cdot \frac{3}{2} = \frac{4}{5} – \frac{9}{5} = -145−65⋅32=45−95=−1,
  • Constant: 371 105 − 68 21 ⋅ 1 2 + 8 35 ⋅ 3 8 = 371 105 − 34 21 + 3 35 = 371 105 − 170 105 + 9 105 = 210 105 = 2 371 105 − 68 21 ⋅ 1 2 + 8 35 ⋅ 3 8 = 371 105 − 34 21 + 3 35 = 371 105 − 170 105 + 9 105 = 210 105 = 2 (371)/(105)-(68)/(21)*(1)/(2)+(8)/(35)*(3)/(8)=(371)/(105)-(34)/(21)+(3)/(35)=(371)/(105)-(170)/(105)+(9)/(105)=(210)/(105)=2\frac{371}{105} – \frac{68}{21} \cdot \frac{1}{2} + \frac{8}{35} \cdot \frac{3}{8} = \frac{371}{105} – \frac{34}{21} + \frac{3}{35} = \frac{371}{105} – \frac{170}{105} + \frac{9}{105} = \frac{210}{105} = 2371105−6821⋅12+835⋅38=371105−3421+335=371105−170105+9105=210105=2.
All coefficients match.

Final Answer

f ( x ) = 371 105 P 0 ( x ) + 4 5 P 1 ( x ) + 68 21 P 2 ( x ) + 6 5 P 3 ( x ) + 8 35 P 4 ( x ) . f ( x ) = 371 105 P 0 ( x ) + 4 5 P 1 ( x ) + 68 21 P 2 ( x ) + 6 5 P 3 ( x ) + 8 35 P 4 ( x ) . f(x)=(371)/(105)P_(0)(x)+(4)/(5)P_(1)(x)+(68)/(21)P_(2)(x)+(6)/(5)P_(3)(x)+(8)/(35)P_(4)(x).f(x) = \frac{371}{105} P_0(x) + \frac{4}{5} P_1(x) + \frac{68}{21} P_2(x) + \frac{6}{5} P_3(x) + \frac{8}{35} P_4(x).f(x)=371105P0(x)+45P1(x)+6821P2(x)+65P3(x)+835P4(x).
Back to Course
Next Lesson
Scroll to Top
Scroll to Top