converges to 1.5 for any choice of initial approximation.
a) Using Newton-Rapshson method, find an iterative formula to compute the reciprocal of a natural number N .
b) Calculate the nnth divided difference of (1)/(x)\frac{1}{\mathrm{x}}, on the nodal points x_(0),x_(1),dots,x_(n)\mathrm{x}_0, \mathrm{x}_1, \ldots, \mathrm{x}_{\mathrm{n}}.
using LU decomposition method with u_(11)=u_(22)=u_(33)=1\mathrm{u}_{11}=\mathrm{u}_{22}=\mathrm{u}_{33}=1.
b) Use secant method to determine the root of the equation cos x-xe^(x)=0\cos x-x e^x=0. Take the initial approximation as x_(0)=0,x_(1)=1\mathrm{x}_0=0, \mathrm{x}_1=1 and perform two iterations of the method.
a) Using the data sin(0.1)=0.09983\sin (0.1)=0.09983 and sin(0.2)=0.19867\sin (0.2)=0.19867, find an approximate value of sin(0.15)\sin (0.15) by Lagrange’s interpolation. Also obtain a bound on the truncation error.
b) Use the Euler’s method to solve numerically the initial value problem:
y^(‘)=-2xy^(2),y(0)=1y^{\prime}=-2 x y^2, y(0)=1
with h=0.2\mathrm{h}=0.2 on the interval [0,1][0,1].
a) Using Runge-Kutta fourth order method with h=0.1\mathrm{h}=0.1, find an approximation value of y(0.1)\mathrm{y}(0.1) for the initial value problem:
y^(‘)=xy+y^(2),y(0)=1y^{\prime}=x y+y^2, y(0)=1
b) Evaluate the integral:
int_(0.2)^(1.4)(sin x-log _(e)x+e^(x))dx\int_{0.2}^{1.4}\left(\sin x-\log _e x+e^x\right) d x
using composite trapezoidal rule with h=0.2\mathrm{h}=0.2, compare with the exact value.
a) Given the data:
{:[f(3)=168],[f(7)=120],[” and “f(9)=72]:}\begin{aligned}
\mathrm{f}(3) & =168 \\
\mathrm{f}(7) & =120 \\
\text { and } \mathrm{f}(9) & =72
\end{aligned}
If f(k)\mathrm{f}(\mathrm{k}) is estimated as 138 using the Newton’s form of the interpolating polynomial, then find the value of kk.
b) Using the third order Taylor’s series method, find the solution of the initial value problem:
y^(‘)=x-y,y(0)=1y^{\prime}=x-y, y(0)=1
at x=0.1\mathrm{x}=0.1 taking h=0.1\mathrm{h}=0.1.
a) How many terms nn be chosen in Maclaurin’s expansion for e^(x)e^{\mathrm{x}} with an error less than 10^(-5),-1 <= x <= 110^{-5},-1 \leq \mathrm{x} \leq 1 ?
b) Find one root of the equation:
x^(3)-2x-5=0x^3-2 x-5=0
in the interval [2,3][2,3] using Birge-Vieta method. Perform only one iteration.
Answer:
Question:-01
Which of the following statements are true and which are false? Give a short proof or a counterexample in support of your answer:
i) The equation x^(3)-4x-16=0\mathrm{x}^3-4 \mathrm{x}-16=0 has a root in the interval [3, 4].
Answer:
Answer:
To determine if the equation x^(3)-4x-16=0\mathrm{x}^3 – 4\mathrm{x} – 16 = 0 has a root in the interval [3, 4], we can apply the Intermediate Value Theorem. This theorem states that if a continuous function f(x)f(x) changes sign over an interval, there must be at least one root within that interval.
Let f(x)=x^(3)-4x-16f(x) = \mathrm{x}^3 – 4\mathrm{x} – 16. Evaluate f(x)f(x) at the endpoints of the interval:
Since f(3)=-1 < 0f(3) = -1 < 0 and f(4)=32 > 0f(4) = 32 > 0, and f(x)f(x) is continuous (as it is a polynomial), the function must cross zero at least once in [3, 4] by the Intermediate Value Theorem. Thus, the statement is true.
ii) The order of convergence of the secant method is 0.62 .
Answer:
Statement: The order of convergence of the secant method is 0.62.
Answer: False.
Justification: The order of convergence of the secant method is approximately 1.618 (the golden ratio, derived from the characteristic equation of the method’s error recurrence). This is because the secant method uses two previous iterates to approximate the derivative, leading to superlinear convergence with an order given by (1+sqrt5)//2~~1.618(1 + \sqrt{5})/2 \approx 1.618. The value 0.62 is incorrect, as it is less than 1 and does not match the established convergence rate. For comparison, a method with an order of 0.62 would converge sublinearly, which contradicts the secant method’s behavior.
Justification: A matrix is diagonally dominant if for each row, the absolute value of the diagonal element is at least as large as the sum of the absolute values of the off-diagonal elements (|a_(ii)| >= sum_(j!=i)|a_(ij)||a_{ii}| \geq \sum_{j \neq i} |a_{ij}|). The coefficient matrix is:
Justification: Relative stability for a numerical method typically refers to the boundedness of the solution relative to the exact solution or the method’s ability to control error growth. The given method resembles the Taylor series expansion for the solution of the ODE y^(‘)=lambda yy’ = \lambda y, where the exact solution is y(t)=y_(0)e^(lambda t)y(t) = y_0 e^{\lambda t}. The numerical method’s amplification factor is:
1+lambda h+(lambda^(2)h^(2))/(2)1 + \lambda h + \frac{\lambda^2 h^2}{2}
This is the second-order Taylor approximation of e^(lambda h)e^{\lambda h}, since:
For lambda > 0\lambda > 0, the amplification factor is positive and grows with hh, but it closely approximates e^(lambda h)e^{\lambda h}, which is the exact growth factor per step. The method is consistent (order 2, as higher-order terms are truncated) and, for small hh, the error growth is controlled relative to the exact solution. In the context of numerical methods for ODEs, relative stability often implies the method does not introduce excessive error amplification compared to the exact solution. Since the method approximates the exact solution’s behavior well for small hh, it is relatively stable.
So, x=1.5x = 1.5 and x=-1.5x = -1.5 are fixed points. The statement claims convergence to 1.51.5 for any initial approximation. To test this, consider the method’s behavior. The iteration can be rewritten as a function:
Since |f^(‘)(1.5)|=0 < 1|f'(1.5)| = 0 < 1, the fixed point at x=1.5x = 1.5 is attractive, suggesting local convergence for initial guesses near 1.5. However, at x=-1.5x = -1.5:
The iterates approach -1.5-1.5, not 1.51.5. Thus, for initial guesses near -1.5-1.5, the method converges to -1.5-1.5. Additionally, if x_(0)=0x_0 = 0, the term (9)/(8x_(0))\frac{9}{8x_0} is undefined, so the method fails. Since the method does not converge to 1.51.5 for all initial approximations (e.g., negative guesses or zero), the statement is false.