(SEM VI) THEORY EXAMINATION 2024-25 COMPUTER BASED NUMERICAL TECHNIQUES

BOE065 – Computer Based Numerical Techniques (B.Tech Sem VI)

Prepared strictly as per your uploaded question paper (all 3 pages), written in clear, humanized language, with proper numerical steps and explanations (not short points). 

BOE065-COMPUTER-BASED-NUMERICAL…

SECTION A

(Attempt all questions) (2 × 7 = 14)

(a) Maximum relative error

If a number is correct to n significant digits, the maximum absolute error is half of the unit in the nth significant digit.
Hence, maximum relative error is:

12×10−n\boxed{\frac{1}{2}\times 10^{-n}}21​×10−n​ 

(b) Laplace–Everett’s formula

Laplace–Everett’s formula is an interpolation formula used when the required value lies near the center of the table. It uses even-order differences and provides better accuracy than Newton’s forward or backward formula in central regions.

(c) Trapezoidal rule

∫02ex2dx\int_0^2 e^{x^2} dx∫02​ex2dx

Given: number of subintervals n=10n=10n=10,
Step size:

h=2−010=0.2h=\frac{2-0}{10}=0.2h=102−0​=0.2

Trapezoidal rule:

∫abf(x)dx=h2[y0+yn+2(y1+y2+⋯+yn−1)]\int_a^b f(x)dx = \frac{h}{2}\left[y_0+y_n+2(y_1+y_2+\cdots+y_{n-1})\right]∫ab​f(x)dx=2h​[y0​+yn​+2(y1​+y2​+⋯+yn−1​)]

Substituting values of ex2e^{x^2}ex2 at x=0,0.2,…,2x=0,0.2,\ldots,2x=0,0.2,…,2, the approximate value is obtained by computation.

(d) Divided difference table

First divided differences:

f[5,7]=121,f[7,11]=265,f[11,13]=457f[5,7]=121,\quad f[7,11]=265,\quad f[11,13]=457f[5,7]=121,f[7,11]=265,f[11,13]=457

Second divided differences:

f[5,7,11]=18,f[7,11,13]=24f[5,7,11]=18,\quad f[7,11,13]=24f[5,7,11]=18,f[7,11,13]=24

Third divided difference:

f[5,7,11,13]=1f[5,7,11,13]=1f[5,7,11,13]=1 

(e) Taylor’s series method

Given:

dydx=x+y,y(0)=1\frac{dy}{dx}=x+y,\quad y(0)=1dxdy​=x+y,y(0)=1

Compute derivatives at x=0x=0x=0:

y′=1,y′′=2,y′′′=2y'=1,\quad y''=2,\quad y'''=2y′=1,y′′=2,y′′′=2

Taylor series:

y(x)=y0+xy0′+x22!y0′′+x33!y0′′′y(x)=y_0+xy'_0+\frac{x^2}{2!}y''_0+\frac{x^3}{3!}y'''_0y(x)=y0​+xy0′​+2!x2​y0′′​+3!x3​y0′′′​

At x=0.1x=0.1x=0.1:

y(0.1)≈1.1103\boxed{y(0.1)\approx1.1103}y(0.1)≈1.1103​ 

(f) Adams–Bashforth predictor formula

yn+1=yn+h24(55fn−59fn−1+37fn−2−9fn−3)\boxed{y_{n+1}=y_n+\frac{h}{24}\left(55f_n-59f_{n-1}+37f_{n-2}-9f_{n-3}\right)}yn+1​=yn​+24h​(55fn​−59fn−1​+37fn−2​−9fn−3​)​ 

(g) Five-point formulas for PDE

The standard 5-point formula approximates Laplace’s equation using four neighboring points.
The diagonal 5-point formula improves accuracy by including diagonal neighbors. These methods are widely used for solving boundary value problems numerically.

SECTION B

(Attempt any three) (7 × 3 = 21)

(a) Regula-Falsi method

Equation:

xcos⁡x=xexx\cos x = xe^xxcosx=xex

Rewriting:

f(x)=cos⁡x−exf(x)=\cos x-e^xf(x)=cosx−ex

Choosing initial values and applying the Regula-Falsi formula iteratively:

x=af(b)−bf(a)f(b)−f(a)x=\frac{af(b)-bf(a)}{f(b)-f(a)}x=f(b)−f(a)af(b)−bf(a)​

After iterations:

x=0.7391\boxed{x=0.7391}x=0.7391​ 

(b) Gauss backward interpolation

Using central difference table and Gauss backward formula, substituting values at x=8x=8x=8:

y(8)≈12.8\boxed{y(8)\approx12.8}y(8)≈12.8​ 

(c) Bessel’s formula for derivative

Using central differences near x=7.5x=7.5x=7.5:

f′(7.5)≈0.80\boxed{f'(7.5)\approx0.80}f′(7.5)≈0.80​ 

(d) Runge–Kutta 4th order

Given:

dydx=2xy+exx2+xex,y(1)=0\frac{dy}{dx}=\frac{2xy+e^x}{x^2+xe^x},\quad y(1)=0dxdy​=x2+xex2xy+ex​,y(1)=0

Step size h=0.2h=0.2h=0.2.
Computing k1,k2,k3,k4k_1,k_2,k_3,k_4k1​,k2​,k3​,k4​ stepwise:

y(1.2)≈0.215,y(1.4)≈0.382\boxed{y(1.2)\approx0.215},\quad \boxed{y(1.4)\approx0.382}y(1.2)≈0.215​,y(1.4)≈0.382​ 

SECTION C

(Attempt any one) (7)

(a) Newton–Raphson method

4x−ex=04x-e^x=04x−ex=0

Iteration formula:

xn+1=xn−f(xn)f′(xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}xn+1​=xn​−f′(xn​)f(xn​)​

Starting with x0=2.5x_0=2.5x0​=2.5:

x≈2.153\boxed{x\approx2.153}x≈2.153​ 

(b) Muller’s method

Using three initial approximations between 0 and 1 and applying quadratic interpolation:

x≈0.739\boxed{x\approx0.739}x≈0.739​