# Numerical Methods and Statistical Techniques Using C

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## TIT-ALLC |
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Contents Chapter Pages 1. 1�24 INTRODUCTION 1.1. 1.2. 1.3. 1.4. 1.5. 1.6. 1.7. 1.8. 1.9. 1.10. 1.11. 1.12. 1.13. 1.14. 1.15. 1.16. 1.17. 1.18. 2. Introduction to Computers ........................................................................................ 4 Definitions ................................................................................................................... 4 Introduction to ‘‘C’’ Language ................................................................................... 6 Advantages/Features of ‘C’ Language ...................................................................... 6 ‘C’ Character Set ......................................................................................................... 6 ‘C’ Constants ............................................................................................................... 7 ‘C’ Variables ................................................................................................................ 8 ‘C’ Key Words .............................................................................................................. 9 |
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## ALLC1-1 |
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1 Introduction The limitations of analytical methods in practical applications have led mathematicians to evolve numerical methods. We know that exact methods often fail in drawing plausible inferences from a given set of tabulated data or in finding roots of transcendental equations or in solving non-linear differential equations. Even if analytical solutions are available, they are not amenable to direct numerical interpretation. The aim of numerical analysis is therefore to provide constructive methods for obtaining answers to such problems in a numerical form. With the advent of high speed computers and increasing demand for numerical solution to various problems, numerical techniques have become indispensible tools in the hands of engineers and scientists. e.g., we can solve equations x2 – 5x + 6 = 0, ax2 + bx + c = 0, y″ + 3y′ + 2y = 0 by analytical methods, but transcendental equations such as a cos2 x + bex = 0 cannot be solved by analytical methods or is impossible. Such equations are solved by numerical analysis. |
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## ALLC2-1 |
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2 Error Analysis and Estimation 2.1. ERRORS AND THEIR ANALYSIS 2.1.1. Sources of Errors Following are the broad sources of errors in numerical analysis : (1) Input errors. The input information is rarely exact since it comes from the experiments and any experiment can give results of only limited accuracy. Moreover, the quantity used can be represented in a computer for only a limited number of digits. (2) Algorithmic errors. If direct algorithms based on a finite sequence of operations are used, errors due to limited steps don’t amplify the existing errors but if infinite algos are used, ideally exact results are expected only after an infinite number of steps. As this cannot be done in practice, the algorithm has to be stopped after a finite number of steps and as a consequence thereof the results are not exact. (3) Computational errors. Even when elementary operations, such as multiplication and division are used, the number of digits increases greatly so that the results cannot be held fully in a register available in a given computer. In such cases, a certain number of digits must be discarded. Furthermore, the errors here accumulate one after another from operation to operation, changing during the process and producing new errors. |
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## ALLC2-2 |
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43 ERROR ANALYSIS AND ESTIMATION Rounding off all other numbers to two decimal digits, we have 0.15, 15.45, 0.00, 8.12, 0.02, 0.64 and 0.17. The sum S is given by S = 305.1 + 143.3 + 0.15 + 15.45 + 0.00 + 8.12 + 0.02 + 0.64 + 0.17 = 472.59 = 472.6. To determine absolute error, we note that first two numbers have each an absolute error of 0.05 and remaining seven numbers have an absolute error of 0.005 each. ∴ Absolute error in all 9 numbers = 2(0.05) + 7(0.005) = 0.1 + 0.035 = 0.135 ≈ 0.14. In addition to the above absolute error, we have to take into account rounding error which is 0.01. Hence the total absolute error in S = 0.14 + 0.01 = 0.15 Thus, S = 472.6 ± 0.15. Example 17. 5.5 = 2.345 and 6.1 = 2.470 correct to four significant figures. Find the relative error in taking the difference of these numbers. 1 × 0.001 = 0.0005 2 ∆x1 ∆x2 ∆x1 0.0005 + ∴ Relative error < =2 =2 = 0.008 . X X 0.125 X Example 18. 10 = 3.162 and e ~ – 2.718 correct to three decimal places. Find the percentage error in their difference. 0.0005 |
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## ALLC3-1 |
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3 Algebraic and Transcendental Equations Consider the equation of the form f(x) = 0 If f(x) is a quadratic, cubic or biquadratic expression then algebraic formulae are available for expressing the roots. But when f(x) is a polynomial of higher degree or an expression involving transcendental functions e.g., 1 + cos x – 5x, x tan x – cos hx, e–x – sin x etc., algebraic methods are not available. In this chapter, we shall describe some numerical methods for the solution of f(x) = 0, where f(x) is algebraic or transcendental or both. 3.1. BISECTION METHOD f(x ) Let the function f (x) be continuous between a and b. For definiteness, let f (a) be (–) ve and f(b) be (+)ve. Then 1 (a + b). the first approximation to the root is x1 = 2 If f (x1) = 0, then x1 is a root of f(x) = 0 otherwise, the root lies between a and x1 or x1 and b according as f(x1) is (+) ve or (–) ve. Then we bisect the interval as before and continue the process until the root is found to desired accuracy. Y y= This method is based on the repeated application of intermediate value property. |
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## ALLC3-2 |
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85 ALGEBRAIC AND TRANSCENDENTAL EQUATIONS FG ax + b IJ which is of the form x H x K LM d RS − (ax + b) UVOP < 1 N dx T x WQ n The formula xn + 1 = – n+1= n f(xn) will converge to x = α if x = xn b <1 xn 2 ⇒ ⇒ or | xn2 | > | b | or xn2 > | b | | α |2 > | b | as | α |2 > | α | | β | |α|>|β| or or Similarly, xn + 1 = (3 αβ = b) <1 x = xn b <1 ( xn + a) 2 or or or xn → α −b will converge to x = α if xn + a LM d FG − b IJ OP N dx H x + a K Q or Using condition of iteration method (xn + a)2 > | b | or β2 > | b | | β |2 > | α | | β | (α + a)2 > | b | as xn → α (3 α + a = – β) | β | > | α | or | α | < | β |. Example 5. The equation x4 + x = e where e is a small number has a root close to e. Computation of this root is done by the expression α = e – e4 + 4e7. or (i) Find an iterative formula xn+1 = F(xn), x0 = 0 for the computation. Show that, we get the above expression after three iterations when neglecting terms of higher order. (ii) Give a good estimate (of the form Nek, where N and k are integers) of the maximum error when the root is estimated by the above expression. |
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## ALLC3-3 |
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103 ALGEBRAIC AND TRANSCENDENTAL EQUATIONS = .56 − FG .56716 − .56 IJ (− .01121) H .00002619 + .01121K = .567143 Since, x2 and x3 agree upto four decimal places, the required root correct to three decimal places is 0.567. (ii) Let f(x) = xex – 2 Since, f(.852) = – .00263 and f(.853) = .001715 Hence, root lies between .852 and .853. Let x0 = .852 and x1 = .853 Using method of false position, RS x − x UV f (x ) T f (x ) − f (x ) W R .853 − .852 UV (− .00263) = .852 − S T.001715 − (− .00263) W 1 x2 = x 0 − 0 1 0 0 = .852605293 Now f(x2) = – .00000090833 Hence, root lies between .852605293 and .853 Using method of false position, RS x − x UV f (x ) | Replacing x by x T f (x ) − f (x ) W R . 853 − . 852605293 UV (− .00000090833) = (.852605293) – S T.001715 − (−.00000090833) W x3 = x 2 − 1 2 1 2 2 0 2 = 0.852605501 Since, x2 and x3 agree upto 6 decimal places, hence the required root correct to 4 decimal places is 0.8526. Example 8. (i) Solve x3 – 5x + 3 = 0 by using Regula-Falsi method. (ii) Use the method of false position to solve x3 – x – 4 = 0. |
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## ALLC3-4 |
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129 ALGEBRAIC AND TRANSCENDENTAL EQUATIONS 3.32. HORNER�S METHOD This is the best method of finding of real root of a numerical polynomial equation. The method of working is as follows. Let a positive root of f(x) = 0 lie in between α and α + 1, where α is an integer. Then the value of the root be α . d1d2d3 ...... where α is the integral part and d1, d2, d3, ...... are the digits in its decimal part. Finding d1. First diminish the roots of f(x) = 0 by α so that the roots of the transformed equation lies between 0 and 1. i.e., the root of the transformed equation is 0 . d1d2d3 ...... Now multiply the roots of the transformed equation by 10 so that the root of the new equation is d1 . d2d3 ...... . Thus the first figure after the decimal place is d1. Again diminish the root by d1 and multiply the roots of the resulting equation by 10 so that the root is d2 . d3 ...... i.e., the second figure after the decimal place is d2. Continue the process to obtain the root to any desired degree of accuracy digit by digit. |
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## ALLC3-5 |
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149 ALGEBRAIC AND TRANSCENDENTAL EQUATIONS 3.40. CHEBYSHEV METHOD Consider f(x) as a polynomial of degree two. f(x) = a0x2 + a1x + a2 = 0 ; a0 ≠ 0 ...(1) where a0, a1, a2 are arbitrary parameters to be determined by prescribing three appropriate conditions on f(x) or its derivatives. We determine a0, a1, a2 in (1) using the conditions fk = a0xk2 + a1xk + a2 fk′ = 2a0xk + a1 U| V| W ...(2) fk″ = 2a0 Eliminating a0, a1 and a2 from (1) and (2), we get 1 (x – xk)2 fk″ = 0 ...(3) 2 which is the Taylor series expansion of f(x) about x = xk such that terms of order (x – xk)3 and higher powers are neglected. In order to get the next approximation to the correct root, we write (3) as fk + (x – xk) fk′ + xk+1 – xk = – fk f ″ 1 − (xk+1 – xk)2 k fk ′ 2 fk ′ ...(4) By Newton-Raphson formula xk+1 = xk – fk fk ′ ...(5) Using (5) in (4), we get xk+1 = xk – fk 1 fk 2 f″ − fk ′ 2 fk ′ 3 k which is called the Chebyshev method. Rate of convergence of this method is 3. EXAMPLE Example. Perform two iterations of Chebyshev method to find the smallest positive root of x3 – 5x + 1 = 0. Take the initial approximation as x0 = 0.5. |
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## ALLC4-1 |
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4 Interpolation 4.1. INTRODUCTION According to Theile, ‘Interpolation is the art of reading between the lines of the table’. It also means insertion or filling up intermediate terms of the series. Suppose we are given the following values of y = f(x) for a set of values of x : x : x0 x1 x2 ...... xn y: y0 y1 y2 ...... yn Thus the process of finding the value of y corresponding to any value of x = xi between x0 and xn is called interpolation. Hence interpolation is the technique of estimating the value of a function for any intermediate value of the independent variable while the process of computing the value of the function outside the given range is called extrapolation. 4.2. ASSUMPTIONS FOR INTERPOLATION 1. There are no sudden jumps or falls in the values during the period under consideration. 2. The rise and fall in the values should be uniform. e.g. if we are given data regarding deaths in various years in a particular town and some of the observations are for the years in which epidemic or war overtook the town then interpolation methods are not applicable in such cases. |
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## ALLC4-2 |
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181 INTERPOLATION Sol. (i) The difference table is : x f(x) 0 –3 1 6 ∆2f(x) ∆f(x) ∆3f(x) 9 –7 2 2 8 3 12 9 2 4 f(0 + 6) = E6f(0) = (1 + ∆)6f(0) = (1 + 6∆ + 15∆2 + 20∆3) f(0) = – 3 + 6 (9) + 15 (– 7) + 20 (9) = – 3 + 54 – 105 + 180 = 126. (ii) Max. power of x in polynomial will be 10 and coeff. of x10 will be abcd. Here k = abcd, h = 1, n = 10 ∴ Expression = k hn n ! = abcd 10 !. Example 20. (i) Prove that if m is a (+)ve integer then x (m) x (m−1) (x + 1)(m) + = m! (m − 1) ! m! (ii) Given u0 + u8 = 1.9243, u1 + u7 = 1.9590 u2 + u6 = 1.9823, u3 + u5 = 1.9956. Find u4. Sol. (i) R.H.S. = = x( x − 1) ...... ( x − m + 1) x( x − 1) ...... ( x − m + 2) + m! (m − 1) ! x( x − 1) ( x − 2) ...... ( x − m + 2) [(x – m + 1) + m] m! ( x + 1) x( x − 1)( x − 2) ...... ( x − m + 2) ( x + 1) ( m) = = L.H.S. m! m! (ii) Taking ∆8 u0 = 0 ⇒ (E – 1)8 u0 = 0 ⇒ u8 – 8c1u7 + 8c2u6 – 8c3u5 + 8c4u4 – 8c5u3 + 8c6u2 – 8c7u1 + 8c8u0 = 0 ⇒ (u0 + u8) – 8(u1 + u7) + 28(u2 + u6) – 56(u3 + u5) + 70 u4 = 0 ⇒ u4 = 0.99996. (After putting the values) Example 21. Evaluate : = |
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## ALLC4-3 |
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192 NUMERICAL METHODS AND STATISTICAL TECHNIQUES USING ‘C’ Sol. R.H.S. = = LM MN FG IJ − FG 1 ∆IJ H K H2 K 1 1 1 ∆ 1− ∆ + 2 2 2 1 . 2 FG H 2 + ...... u0 FG H 1 1 1 1+ ∆ u0 = 1 2 2 1+ ∆ 2 IJ K OP PQ 3 IJ K −1 u0 = (2 + ∆)–1 u0 = (1 + E)–1 u0 = (1 – E + E2 – E3 + ...) u0 = u0 – u1 + u2 – u3 + ...... = L.H.S. Example 4. Using method of separation of symbols, show that : n(n − 1) ux–2 + ...... + (– 1)n ux–n. 2 n(n − 1) –2 R.H.S. = ux – nE–1 ux + E ux + ...... + (– 1)n E–n ux 2 ∆n ux–n = ux – nux–1 + Sol. LM N F 1I = G1 − J H EK −1 = 1 − nE + n ux = Example 5. Show that : F GH FG E − 1IJ H E K ex u0 + x ∆ u0 + Sol. F GH L.H.S. = ex 1 + x∆ + F GH = 1 + xE + OP Q n(n − 1) −2 E + ...... + (− 1) n E − n ux = (1 – E–1)n ux 2 n ux = ∆n En ux = ∆n E–n ux = ∆n ux–n = L.H.S. I JK x2 2 x2 ∆ u0 + ....... = u0 + u1x + u2 + ...... . 2! 2! I JK x 2 ∆2 + ...... u0 = ex . ex∆ u0 = ex(1+∆) u0 = exE u0 2! I JK I JK F GH x2 x2E2 u2 + ....... = R.H.S. + ...... u0 = u0 + xu1 + 2! 2! Example 6. Prove the following identity : x2 x u1 + ∆u1 + ..... 1− x (1 − x) 2 L.H.S. = xu1 + x2 E u1 + x3 E2u1 + ...... = x (1 + xE + x2E2 + ......) u1 u1x + u2x2 + u3x3 + ...... = |
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## ALLC4-4 |
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208 NUMERICAL METHODS AND STATISTICAL TECHNIQUES USING ‘C’ The difference table is as under : Marks obtained less than No. of students = y 40 25 50 60 60 82 70 93 80 100 ∆y 35 22 11 7 ∆2 y – 13 – 11 –4 ∆3 y 2 7 ∆4 y 5 Applying Newton’s forward difference formula, u(u − 1)(u − 2)(u − 3) 4 u(u − 1) 2 u(u − 1)(u − 2) 3 y55 = y40 + u ∆ y40 + ∆ y40 + ∆ y40 ∆ y40 + 4! 2! 3! (1.5)(.5) (1.5)(.5)(− .5) (1.5)(.5)(− .5)(− 1.5) (− 13) + = 25 + (1.5)(35) + (2) + (5) 2! 3! 4! = 71.6171875 ≈ 72 There are 72 students who got less than 55 marks. ∴ No. of students who got more than 55 marks = 100 – 72 = 28 (ii) To calculate the number of students securing marks between 36 and 45, take the difference of y45 and y36. x − a 36 − 40 = u= = – .4 h 10 45 − 40 Also, u= = .5 10 By Newton’s forward difference formula. u(u − 1) 2 u(u − 1)(u − 2) 3 ∆ y40 y36 = y40 + u ∆ y40 + ∆ y40 + 2! 3! u(u − 1)(u − 2)(u − 3) 4 + ∆ y40 4! (− .4)(− 1.4) (− .4)(− 1.4)(− 2.4) (− 13) + = 25 + (– .4)(35) + (2) 2! 3! (− .4)(− 1.4)(− 2.4)(− 3.4) + (5) = 7.864 ≈ 8 4! u(u − 1) 2 u(u − 1)(u − 2) 3 ∆ y40 + Also, y45 = y40 + u ∆ y40 + |
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## ALLC4-5 |
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224 NUMERICAL METHODS AND STATISTICAL TECHNIQUES USING ‘C’ Answers 1. 0.3057 4. (a) (i) .5543 2. 34.22007, 34.30866 (ii) .2662 (b) 2x2 + x + 1.28, 2x2 + x + 1.28; 1.655 6. (i) 1.6751 (ii) 1.7081 9. 219 10. 8666 3. 0.9063, 0.2923 (iii) .4241 5. 421.875 7. 0.783172 8. 194 11. 6.36, 11.02 12. 0.1496. 4.17. CENTRAL DIFFERENCE INTERPOLATION FORMULAE We shall study now the central difference formulae most suited for interpolation near the middle of a tabulated set. 4.18. GAUSS� FORWARD DIFFERENCE FORMULA Newton’s Gregory forward difference formula is u (u − 1) 2 u (u − 1)(u − 2) 3 ∆ f (a) + ∆ f (a) f(a + hu) = f(a) + u∆f(a) + 2! 3! + Put a = 0, u (u − 1)(u − 2)(u − 3) 4 ∆ f (a) + ...... 4! h = 1, we get f(u) = f(0) + u∆f(0) + u (u − 1) 2 u (u − 1)(u − 2) 3 ∆ f (0) + ∆ f (0) 2! 3! u (u − 1)(u − 2)(u − 3) 4 ∆ f (0) + ...... 4! ⇒ ∆2f(0) = ∆3f(– 1) + ∆2f(– 1) + Now, ∆3f(– 1) = ∆2f(0) – ∆2f(– 1) Also, ∆4f(– 1) = ∆3f(0) – ∆3f(– 1) ⇒ ∆3f(0) = ∆4f(– 1) + ∆3f(– 1) ∆5f(– 1) = ∆4f(0) – ∆4f(– 1) ∆4f(0) = ∆5f(– 1) + ∆4f(– 1) and so on. and ∴ (1) ⇒ ...(2) From (2), f(u) = f(0) + u∆f(0) + |
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## ALLC4-6 |
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243 INTERPOLATION = 1330 + (– .2)(682) + (.8)(− .2) (.8)(− .2)(− 1.2) (436) + (132) 2 6 = 1162.944 f(5.8) = 1162.944. ∴ ASSIGNMENT 1. 2. 3. 4. The population of a town in the years 1931, ......, 1971 are as follows : Year : 1931 1941 1951 1961 1971 Population : 15 20 27 39 52 (in thousands) Find the population of the town in 1946 by applying Gauss’ backward formula. Apply Gauss’ backward formula to find the value of (1.06)19 if (1.06)10 = 1.79085, (1.06)15 = 2.39656, (1.06)20 = 3.20714, (1.06)25 = 4.29187 and (1.06)30 = 5.74349. Given that x : 50 51 52 53 54 tan x : 1.1918 1.2349 1.2799 1.3270 1.3764 Using Gauss’ backward formula, find the value of tan 51° 42′. Interpolate by means of Gauss’ backward formula, the population of a town KOSIKALAN for the year 1974 given that : Year : 1939 1949 1959 1969 1979 1989 Population : 12 15 20 27 39 52 (in thousands) 5. 6. Apply Gauss’ backward formula to find sin 45° from the following table : θ° : 20 30 40 50 60 70 80 sin θ : 0.34202 0.502 0.64279 0.76604 0.86603 0.93969 0.98481 Using Gauss’ backward formula, estimate the no. of persons earning wages between Rs. 60 and |
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## ALLC4-7 |
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260 NUMERICAL METHODS AND STATISTICAL TECHNIQUES USING ‘C’ Using Bessel’s formula, RS f (0) + f (1) UV + FG u − 1IJ ∆f (0) + u (u − 1) RS ∆ 2 T 2 W H 2K T 2 f(u) = FG H (u − 1) u − + ⇒ f(.25) = UV W f (− 1) + ∆2 f (0) 2 3! IJ K 1 u 2 ∆3 f (− 1) FG 32 + 35IJ + (.25 – .5) (3) + (.25) (.25 − 1) RS − 5 + 2 UV H 2 K 2 T 2 W + (.25 − 1) (.25 − .5) (.25) (7) 3! = 32.9453125 Hence y25 = 32.9453125. Example 2. Apply Bessel’s formula to find the value of f(27.4) from the table : x : 25 26 27 28 29 30 f(x) : 4.000 3.846 3.704 3.571 3.448 3.333. Sol. Taking origin at 27 and h = 1 x = a + uh ⇒ 27.4 = 27 + u × 1 ∴ u = 0.4 The difference table is as follows : u 103f(u) –2 4000 –1 3847 103 ∆f(u) 103 ∆2f(u) 103 ∆3f(u) 103 ∆4f(u) 103 ∆5f(u) – 154 12 – 142 0 –3 3704 9 – 133 1 3571 2 3448 3 3333 4 1 –7 10 – 123 –3 –2 8 – 115 Bessel’s formula is RS f (0) + f (1) UV + FG u − 1IJ ∆f (0) + u (u − 1) RS ∆ f (0) + ∆ f (− 1) UV 2! 2 T 2 W H 2K W T F 1I (u − 1) G u − J u H 2 K ∆ f (− 1) + (u + 1) u (u − 1) (u − 2) RS ∆ f (− 1) + ∆ f (− 2) UV + 3! 4! 2 W T F 1I (u − 2) (u − 1) G u − J u (u + 1) H 2K ∆ f (− 2) + 2 2 f(u) = |
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## ALLC4-8 |
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272 NUMERICAL METHODS AND STATISTICAL TECHNIQUES USING ‘C’ By Laplace Everett’s formula, RS T f(.4) = (.4) (3571) + RS(.6) (3704) + (1.6) (.6) (− .4) (9) + (2.6) (1.6) (.6) (− .4) (− 1.4) (4)UV 3! 5! T W + UV W (1.4) (.4) (− .6) (2.4) (1.4) (.4) (− .6) (− 1.6) (10) + (− 3) + ..... 3! 5! = 3649.678336. Hence f(27.4) = 3649.678336. ASSIGNMENT 1. 2. 3. 4. 5. 6. 7. 8. 9. Given the table x : 21 22 23 24 25 26 log x : 1.3222 1.3424 1.3617 1.3802 1.3979 1.4150 Apply Laplace-Everett’s formula to find the value of log 2375. From the following table of values of x and y = ex, interpolate the value of y when x = 1.91 x : 1.7 1.8 1.9 2.0 2.1 2.2 y = ex : 5.4739 6.0496 6.6859 7.3891 8.1662 9.0250. Given the table : x : 310 320 330 340 350 360 log x : 2.49136 2.50515 2.51851 2.53148 2.54407 2.55630. Find the value of log 337.5 by Laplace Everett’s formula. Find y12 if y0 = 0, y10 = 43214, y20 = 86002 and y30 = 128372. Obtain the values of y25, given that y20 = 2854, y24 = 3162, y28 = 3544 and y32 = 3992 –x Find the value of e when x = 1.748 from the following : x : 1.72 1.73 1.74 |
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## ALLC4-9 |
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290 NUMERICAL METHODS AND STATISTICAL TECHNIQUES USING ‘C’ Proceeding in this manner, we get y = f(x) = y0 + (x – x0) [x0, x1] + (x – x0) (x – x1) [x0, x1, x2] + (x – x0) (x – x1) (x – x2) [x0, x1, x2, x3] + ..... + (x – x0) (x – x1) (x – x2) ..... (x – xn–1) [x0, x1, x2, x3, ......, xn] + (x – x0) (x – x1) (x – x2) ..... (x – xn) [x, x0, x1, x2, ......, xn] which is called Newton’s general interpolation formula with divided differences, the last term being the remainder term after (n + 1) terms. Newton’s divided difference formula can also be written as y = y0 + (x – x0) ∆| y0 + (x – x0) (x – x1) ∆| 2y0 + (x – x0) (x – x1) (x – x2) ∆| 3y0 + (x – x0) (x – x1) (x – x2) (x – x3) ∆| 4y0 + ..... + (x – x0) (x – x1) ..... (x – xn–1) ∆| ny0 4.31. RELATION BETWEEN DIFFERENCES DIVIDED DIFFERENCES AND ORDINARY Let the arguments x0, x1, x2, ....., xn be equally spaced such that x1 – x0 = x2 – x1 = ... = xn – xn–1 = h ∴ x1 = x0 + h x2 = x0 + 2h ......... xn = x0 + nh ∆| f(x0) = Now x1 ∆ f ( x0 ) f ( x1 ) − f ( x0 ) f ( x0 + h) − f ( x0 ) |

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- Numerical Methods and Statistical Techniques Using C
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