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Q. 3.1. What is forward difference? Forward difference table. Q. 3.2. What is Backward Difference? Backward difference table. Q. 3.3. What is central difference ? Sol. In this system 6 is called central difference operator and Higher central differences are defined as Q. 3.4. Define shift and Average operator. Ans. Shift operator F. It is the operation of increasing the argument by ii so that Q. 3.5. Show that Q. 3.6, Find first term of series whose second and subsequent terms are 8,3,0, —1, 0 by using forward difference method. We construct a table Q. 3.7. Assuming that the following values of y belong to a polynomial of degree 4 compute next three terms. Sol. Since y is polynomial of degree 3 so z3y will be constant and A4y will be zero . Q. 3.10. A second degree polynomial passes through (0,1), (1,3), (2,7) (3,13). Find the polynomial. Q. 3.11. Evaluate Maximum power of x in polynomial will be 10 The coefficient of highest power of x remains unchanged while transforming a polynomial to factor notation. Q 3.15. Prove Newton’s forward interpolation formula. It is called Newton’s forward interpolation formula which contains y0 and forward difference of y0. This formula is used to interpolate value near starting of table. Q. 3.16. Prove Newton’s Backward interpolation formula. It is called Newton’s Backward interpolation formula which contains y, and backward difference of y. This formula is used for interpolating the values of y near end of table. Q. 3.17. Using Newton’s forward interpolation formula, find value of f(1.6) If: By Newton’s forward interpolation formula Q. 3.18. From the following table find the number of students who obtained less than 45 marks. Q. 3.19. Write down the advantage of using central difference method in interpolation. Sol . The values in central difference formulae are smaller and converge faster than those in Newton’s formulae. The central differences formulae are used to interpolate value in the middle of table where as Newton’s formulae are used to find values in starting and end of table. Wherever possible central differenece formulae are used instead of Newton’s formulae. Moreover in central differences we have right choice of interpolation depending upon the value of p. Q. 3.20. The pressure p of wind corresponding to ve!3dty v is given by following data. Find value of p when v 25 Applying Bessel’s formula Q. 3.21. Calculate the no. of addition and multiplication needed to implement Lagrange’s interpolation of n degree polynomial. Sol. Let x0 .... x be n-i distinct points y = f(x). We want to construct a polynomial J(x) of degree n which interpolates f(x) at the points x0 x,, that it satisfies The above formula needs at least 2(n+1) multiplications (divisions) and (2n+l) addition and subtraction after the denominators of Lagrange’s polynomials have been calculated once. This is to be compared with n multiplication and n additions for evaluation of polynomial of degree n. Q. 3.22. Using iterative interpolations method, give 2” app. for Newton’s forward interpolation formula. Sol. Newton’s forward interpolation formula Neglecting second and higher differenece we obtain first app. to p. To find second app. retain second difference in (1) and replace p by p1 and To find value of x we use stirling formula. If tabulated function is of flth degree polynomial then A’y0 will be constant. Hence the flt divided difference will be constant. Q. 3.26. Using Newton’s divided difference formula, calculatef (6) from data: Using Newton’s Divided Difference formula Q. 3.27. Evaluate f(9) for the following function by the Newton’s divided difference formula method. Ans. The divided difference table is By Newton’s Divided difference formula Q. 3.28. Derive Newton’s Forward difference interpolation formula for a polynomal y = f (x) and find the value of fi2.15) for the following function by the same method. Which is Newton’s forward Interpolation formula, This formula is used to finding the value of y near the beginning of set of tabutated values. Using thi formula we find f(2.15).]

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