A TEXTBOOK OF COMPUTER BASED NUMERICAL AND STATISCAL TECHNIQUES PART 37 DOC

dyydx=− with y (0) = 0 in the range 00.2x≤≤(take 0.1h =). [Ans. y(0.1) = 0.09524, y(0.2) = 0.1814076]7.6 RUNGE-KUTTA METHODThe method is very simple. It is named after two german mathematicians Carl Runge (1856-1927)and Wilhelm Kutta (1867-1944). These methods are well-known as Runge-Kutta Me[r]

A TEXTBOOK OFCOMPUTER BASED NUMERICALANDSTATISTICAL TECHNIQUESThis pageintentionally leftblankA TEXTBOOK OFCOMPUTER BASED NUMERICALANDSTATISTICAL TECHNIQUESAnju Khandelwal M.Sc., Ph.D.Department of MathematicsSRMS College of Engineering &Techn[r]

xfx fx fx∆∆−INTERPOLATION WITH UNEQUAL INTERVAL255∴f(x) = 40.7 + (x – 0) (–2.6) + x(x – 1) (1.9) = 1.9x2 – 4.5x + 40.7For f(x) to be maximum or minimum, we have f′(x) = 03.8x – 4.5 = 0 ⇒x = 1.184Again, f′′(x) = 3.8 > 0∴ f(x) is minimum at x = 1.184Again, unit 1 ≡ 30 days∴ 1.184 = 30×1.184 = 3[r]

35° = 0.7002, tan 38° = 0.7813. [Ans. 0.64942084]14. Apply Lagrange’s formula to find f(5) and f(6) given that f(2) = 4, f(1) = 2, f(3) = 8, f(7) =128. Explain why the result differs from those obtained by completing the series of powersof 2? [Ans. 38.8, 74; 2x is not a polynomi[r]

f(x) = 281690. 56⇒f(x) = 2.8169056⇒ log10 656 = 2.8169056PROBLEM SET 5.31. By means of Newton’s divided difference formula,Find the value of f(8) and f(15) from the following table:(): 4 5 7 10 11 13: 48 100 294 900 1210 2028xfx[Ans. 448, 3150]2. Using Newton’s divided differenc[r]

()()()22 22 22b c fa c a fb a b fcfa b c fb c a fc a b−+−+−−+ −+ −. Proved.Example 5. Applying Lagrange’s formula, find a cubic polynomial which approximate the followingdatax : –2 –1 2 3y(x) : –12 –8 3 5Sol. Now, using Lagrange’s formula, we havef(x)=()()()()()()([r]

Gauss ForwardStirlingBessel’sLaplace Evertt’sFIG. 5.15.7.2 Approximation of FunctionTo evaluate most mathematical functions, we must first produce computable approximations tothem. Functions are defined in a variety of ways in applications, with integrals and infinite ser[r]

[Ans. 0.39199981]214COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUES7. Apply Bessel’s interpolation formula, show that tan160° = 0.2867, Given that:0 5 10 15 20 25 30tan 0 0.0875 0.1763 0.2679 0.3640 0.4663 0.5774xx°° °°°°°8. Apply Bessel’s formula to find a<[r]

− × 0.06928+()()()0.2 0.8 1.2 1.824− × 0.02346INTERPOLATION WITH EQUAL INTERVAL195f(u) = 3.20714 – 0.162116 – 0.021932 + 0.002216 + 0.00033782= 3.0256458 (Approx.)Example 4. Using Gauss backward formula, Estimate the no. of persons earning wages between Rs.60 and Rs. 70 from the follow[r]

°°°°°[Ans. 0.61981013]3. The population of a town in the years are as follows:()1931 1941 1951 1961 197115 20 27 39 52YearPopulation in thousandsFind the population of the town in 1946 by applying Gauss’s backward formula.[Ans. 22898]4. Interpolate by means of Gauss’s bac[r]

of x. Therefore, when the values of the argument are not at equally spaced then we use two suchformulae for interpolation.1. Lagrange’s Interpolation formula2. Newton’s Divided difference formula.The main advantage of these formulas is, they can also be used in case of eq[r]

–.5–1Chebyshev polynomialsFIG. 5.2. Chebbyshev polynomiats T0(x) through T6(x). Note that Tj has j roots in the interval (–1,1)and that all the polynomials are bounded between +1.(iii) Economization of power series: To describe the process of economization, whichis essential due[r]

1.0000 0.198013=− = 0.9801986 ≅ 0.9802The exact value of y(0.2) is 0.9802.Example 3. Solve the equation ()yxy′=+with y0 = 1 by Runge-Kutta rule from x = 0 to x = 0.4with h = 0.1.Sol. Here ()=+ =, , 0.1fxy x yh, given 01y = when 00.x =We have, ()100,khfxy

to 4.4x = we obtain (4.3) 1.014256y =and (4.4) 1.018701y =. Ans.Example 5. Solve the equation 21,yxy′=+ given that 0,y = at 0,x = by the use of Taylor series,taking 0.2h = and going as far as 4.x=Sol. The first few derivatives and their values at 00,x = 00,y =

3 upto and including term in x6.We can use the relation y = 1222+x with the knowledge that the error is approximately –x65.Thus, we can find y1 and y2 correct to 4 decimal places with h = 0.1.7.4 EULER’S METHODThe oldest and simplest method was derived by Euler. In this method,[r]

n approaches the exact solution y(xn) at h tends to zero provided therounding error arising from the initial conditions approach zero. This means that as a method iscontinually refined by taking smaller and smaller step-sizes, the sequence of approximate solutionsmust converge t[r]

376COMPUTER BASED NUMERICAL AND STATISTICAL TECHNIQUESlargest coefficient of y. We continue this process till last equation. This procedure is known aspartial pivoting. In general, the rearrangement of equation is done even if pivot element isnon-zero to imp[r]

=∆ + ∆∴ () ( )[]51.1 25 0.288 0.5 1 0.0481 25 0.288 0.024f′′=+−×=−Hence,()1.1 6.6.f′′= Ans.Example 7. Find f′(1.5) from the following table:x 0.0 0.5 1.0 1.5 2.0f(x) 0.3989 0.3521 0.2420 0.1245 0.0540Sol. Here we want to find the derivatives of f (x) at x = 1.5, which is near the end[r]

==GGGCHAPTER 6Numerical Differentiationand Integration6.1 INTRODUCTIONThe differentiation and integration are losely linked processes which are actually inversely related.For example, if the given function y(t) represents an objects position as a function of time, itsdi[r]

2]10. The table given below reveals the velocity ‘v’ of a body during the time ‘t’ specified. Find itsacceleration at t = 1.1.1.0 1.1 1.2 1.3 1.443.1 47.7 52.1 56.4 60.8tv[Ans. 44.92]NUMERICAL DIFFERENTIATION AND INTEGRATION315 6.3 NUMERICAL INTEGRATIONLike nume[r]