Engineering Mathematics through Applications

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2 80 Engineering Mathematics through Applications

3 Finite Differences, Numerical Differentiations and Integrations 8

4 8 Engineering Mathematics through Applications

5 Finite Differences, Numerical Differentiations and Integrations 8

6 84 Engineering Mathematics through Applications

7 Finite Differences, Numerical Differentiations and Integrations 85

8 86 Engineering Mathematics through Applications

9 Finite Differences, Numerical Differentiations and Integrations 87

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27 Finite Differences, Numerical Differentiations and Integrations 855 Example : Given: log 00 =, log 0 =.004, log 0 =.08, log 04 =.070, find log 0. Solution: Take the missing entry (or the entry under question) as f(0) = A and write the data as under x i : f(x i ) : A = f(0) Since the four values of the variate are given and hence the function is a polynomial of degree. Therefore, the third differnces are constant and consequently fourth differences are zero. i.e. 4 f(x) = 0 for all x. 4 f(00) = 0 (E ) 4 f(00) = 0 (E 4 4E + 6E 4E + )f(00) = 0 f(04) 4f(0) + 6f(0) 4f(0) + f(00) = f(0) = f (0) = = or f(log 0) = Alternately: Let the missing entry be y, then the difference table is TABLE. x i y i = log x i y y y 4 y y.0086 y y 0 y 4.07 y 6y y y y As the fourth differences are zero by above stated assumptions, we have 6y.054 = log 0 = y = =.0086 approx. 6 Example 4: Without forming the difference table find the missing entry in the following table x : 0 4 y : 9 8 Explain why the resulting value differs from.

28 856 Engineering Mathematics through Applications Solution: Since the four values of the variate are given and hence the function is a polynomial of degree, and consequently fourth differences are zero. 4 y 0 = 0 (E ) 4 y 0 = 0 i.e. (E 4 4E + 6E 4E + )y 0 = 0 y 4 4y + 6y 4y + y 0 = 0 8 4y = 0 y = Since f(x) = x is not a polynomial and hence y. Example 5: Given y 0, y, y,, y 5 (fifth differences constant), prove that y ( c b) ( a c) = +, 56 where a = (y 0 + y 5 ), b = (y + y 4 ), c = (y + y ). 5 + c 5/ 0 0 y E y y Solution: = = ( + ) ( ) 5 (5/)(/) (5/)(/)(/) (5/)(/)(/) / = = y + y + y + y y + y ( )( ) 5 (5/)(/)(/) / / + y = y + ( E ) y + ( E ) y + ( E ) y ( E ) y + E y ( ) = y + y y + y y + y + y y + y y 8 6 ( ) ( ) ( ) ( y 4y 6y 4y y ) ( y 5y 0y 0y 5y y ) = + y y y y y + y

29 Finite Differences, Numerical Differentiations and Integrations 857 = y 5 y + 75 y + 75 y 5 y + y = ( y y ) ( y y ) ( y y ) y a b c = + Also, R.H.S. ( ) + ( ) a c 5 c b a c 5 5 = c + = c + + c b = a b+ + c = a b + c Hence the result. Alternately: Taking n = in the Bessel s Formula for Interpolation, where ( ) ( ) nn n nn y = + δ + µδ + n y n y y δ y!! From above, Ist two terms with 0 / / / ( )( ) ( )( ) nn n nn n n µδ y/ + δ y/ + 4! 5! n = become, y + nδ y = y + δ y = y + y y = y + y ( ) ( ) 0 / 0 / Similarly, we have ( ) nn y + y0 µδ y / =! 8 and δ y 0 = 0 and so on,! ( ) n nn 4 4 y + y0 y + y y/ = ( y0 + y ) 8 + 8

30 858 Engineering Mathematics through Applications y = y + y y + y + y + y 6 56 Shifting origin by, 4 4 ( ) ( ) ( ) / ( ) ( ) ( ) y = y + y y + y + y + y Example 6: Given u =, u + u = 5.4, u 4 + u 5 + u 6 = 8.47, Find the value of u x for all integral values of x from to. x x =7u = 90.6 Solution: Let u x = F() = u + (u + u ) + (u 4 + u 5 + u 6 ) + (u 7 + u u ) F() = = 5.4 () Similary: u x = F(4) = (u 4 + u 5 + u 6 ) + (u 7 + u u ) and = = 08.8 () u x = F(7) = u 7 + u u = 90.6 () 7 The difference table with h = is given below TABLE.5 x F(x) F(x) F(x) Now + δ ( + ) / δ + (4). 9 δ F() = F() F() 9 = ( 6.4) (.06) ; ; 0.80, (5) 9 δ F( 4) = F( 4) F( 4) 9

31 Finite Differences, Numerical Differentiations and Integrations 859 = ( 8.47) (.06) ; ; 4.8 (6) 9 Further, δ 9 δ = =.06 = F() F() ( ) Second differences being constant i.e., δ F() = δ F() = =. 4 The differences table for unit interval is as below: TABLE.6 x F(x) δf(x) δ F(x) u x = F(x) F(x + ) 5.4 u = F() F() = u = F() F() = u = F() F(4) = u 4 = u 5 = u 6 = u 7 = u 8 = u 9 = u 0 = u = u =.76 It is being given that u = F() F() = F() = F() = = 4.4. Values within boxes are known. With the help of these values, the table has been completed. e.g. δ F() =.4 δf() = δf() + δ F() = =.4 Similarly: δf() = δf() + δ F() =.4.4 =.48 and so on. F() = F() + δf() = =.0 and so on.

32 860 Engineering Mathematics through Applications e.g. In the last column values u, u,, u have been obtained. u = F( x) F( x+ ) = u u x x x x x+ u = F() F() = u u etc. x x= x= Thus, the values required are,.4,.7,,.76. x ASSIGNMENT. Prove, y 4 = y + y + y + y. [Hint. y 4 y = y ]. y 4 = y y y + 0 y. [Hint. y 4 = E 5 y ]. u 0 + n C u x + n C u x + = ( + x) n u 0 + n C ( + x) n x u 0 + n C ( + x) n x u 0. [Hint. R.H.S. = ( + x + x ) n u 0 ] 4. u0 u + u = u0 u0 + u0 u u u + u u +. = u u + u 4 6 x x x x x+ x+ 4 x+ Hint....5 x x x x m m m 4 m + + / / R.H.S. + E ux = E ( + ) ux. = L.H.S. 4 m = x + x m. ux ux x x 7. u = e u0 + x u0 + u [Hint. e!!! xe = e x( + ) = e x e x ] 8. Evaluate (i) ( + )(E + )(x + ) (ii) ( + ) (x + x + ) (iii) ( )( + )(x + x + )..9. INTERPOLATION WITH EQUAL INTERVAL OF ARGUMENT Here we find suitable polynomials for replacing any given function over a given interval by two ways, using forward and backward differences.. Newton-Gregory Forward Interpolation Formula Let y = f(x) denote a function which takes values y 0, y, y,, y n for equidistant values x 0, x, x,, x n of the independent variable x, and let f n (x) denote a polynomial of nth degree. Thus, this polynomial may be written as:

33 Finite Differences, Numerical Differentiations and Integrations 86 f n (x) = a 0 + a (x x 0 ) + a (x x 0 )(x x ) + a (x x 0 )(x x )(x x ) + a 4 (x x 0 )(x x )(x x )(x x ) + + a n (x x 0 )(x x )(x x ) (x x n ) () Now, our object is to determine the unknown coefficients a 0, a, a,, a n Substituting in (), f(x 0 ) = y 0, f(x ) = y, f(x ) = y,, f(x n ) = y n to successive values x 0, x, x,, x n with x x 0 = h, x x 0 = h, etc., we get y 0 = a 0 or a 0 = y 0 () y = a 0 + a (x x 0 ) = y 0 + a h or a y y y h h 0 0 = = ; () Likewise, y = a 0 + a (x x 0 ) + a (x x 0 )(x x ) or 0 0 a y y0 = y0 + h + a h h h ( ) ( )( ) y y + y y = = (4) h h Again, y = a 0 + a (x x 0 ) + a (x x 0 )(x x ) + a (x x 0 )(x x )(x x ) or 0 0 a y y0 y y + y0 = y0 + h + h h + a h h h h h ( )( ) ( )( )( ) y y + y y y = = 6h h ; (5) By continuing this process for calculating the coefficients, we see that 4 5 n y0 y0 y0 a4 =, a 4 5 =,, a ; 5 n = n 4h 5h nh (6) y0 y0 y0 fn( x) = y0 + ( x x0) + ( x x 0)( x x) + ( x x 0)( x x)( x x) + h h h Now, if we take x x 0 h n + ( )( )( )..( ) (7) 0 nh x x x x x x x x n n = p, this formula takes the simple form as: pp ( ) pp ( )( p ) fn() x = y0 + p y0 + y0 + y0 + + pp ( )( p ) ( p n ) n yn + (8) n

34 86 Engineering Mathematics through Applications This is known as Newtons-Gregory formula for forward interpolation. Working Rule: For any real number p, y p = f(x 0 + ph) = E p f(x 0 ) = ( + ) p y 0 pp ( ) pp ( )( p ) = + p y0!! (Binomial Expansion) pp ( ) pp ( )( p ) = y0 + p y0 + y0 + y 0 + (9)!! Note: This formula is particularly useful for interpolating the values of f(x) near the beginning of the set of given values.. Newton-Gregory Backward Interpolation Formula. For deriving the formula in this case, we take the polynomial f n (x) = a 0 + a (x x n ) + a (x x n )(x x n ) + a (x x n )(x x n )(x x n ) + + a n (x x n )(x x n )(x x n ) (x x ) () On substituting values f(x n ) = y n, f(x n ) = y n -, f(x n ) = y n corresponding to the successive values of x = x n, x = x n, x = x n, respectively, we get y n = a 0 or a 0 = y n () Similarly, y n = a 0 + a (x n x n ) = y n + a ( h) or yn yn yn a = = ; h h () Again, y n = a 0 + a (x n x n ) + a (x n x n )(x n x n ) yn yn = yn + ( h) + a( h)( h) h or yn yn + yn yn a = = ; h h (4) By continuing this process for calculating the coefficient, we see that 4 n yn yn yn a =, a4 = 4,, an = n ;! h 4! h n! h (5) Substituting these values of a 0, a, a,etc. in (), we have yn y y n n f( x) = yn + ( x xn) + ( x x )( n x xn ) + ( x x n)( x x n )( x x n ) h! h! h or n y + + n ( n)( n )( n ) ( ) n x x x x x x x x nh! ( ) x xn yn x xn x xn fx ( ) yn x xn x xn x xn = yn + yn + + h h h! h h h (6) n y + + n x x n x x n x x n x x n n! h h h h (7)

35 Finite Differences, Numerical Differentiations and Integrations 86 x x Now if we take, n = p, then the formula takes the simple form h ( + ) pp f( x) = f ( xn + ph) = yn + p yn + yn +! x x Since if we put, n = p or x = x h n + ph, then x n = x n h, x n = x n h, etc. and Similarly ( ) x xn x xn h x xn + h x xn = = = + = p+ h h h h ( ) x xn x xn h x xn + h x xn = = = + = p + h h h h Using (8), ( ) ( ) x x x n h x x + n h h h h x n n = = = + (8) ( p n ) pp ( + ) pp ( + )( p+ ) fx () = fx ( n + ph) = yn + p yn + yn + yn +!! pp ( + )( p+ ) ( p+ n ) n + y n (9) n! This is called Newton-Gregory backward interpolation formula because it contains values of the tabulated function from y n backward to the left and none to the right. Note: This formula is useful for interpolating values near the end of a set of tabular values. Alternately: y p = f(x n + ph) = E p y n = ( ) p y n pp ( + ) pp ( + )( p+ ) = yn + p yn + yn + yn +.!! Example 7: Given sin45 = 0.707, sin50 = , sin55 = 0.89, sin60 =.8660, find sin5, using Newton s Forward formula. Solution: Write 5 = x 0 + nh, where x 0 = 45 be the base values so that 5 = 45 + n.5 or 7 n = =.4 5

36 864 Engineering Mathematics through Applications The required difference table on taking 0 4 common to all y is as follows: TABLE.7 x i 0 4 y i y i y i y i x 0 = 45 y 0 = 707 y 0 = 589 x = 50 y = 7660 y 0 = 57 y = 5 y 0 = 7 x = 55 y = 89 y = 64 y = 468 x = 60 y = 8660 The value 45 is chosen as the base value, since it will have inclusion of forward differences of highest order, leading to the maximum accuracy in results. Now 0 4 y 5 = y 45 + n C y 45 + n C y 45 + n C y 45 + n C 4 4 y 45 + (.4) (.4)(.4) (.4)(.4)(.6) = ( 57) + ( 7)!!!.9.5 = = y 5 = 7880 y 5 = Example 8: Find the number of men getting wages between Rs and Rs from the follwing data. Wages in Rs.: Frequency : Solution: x i (no. of persons getting Rs. less y i (cumulative frequency ) than the mention amount) Less than (upto) 0 9,, 0 9,, 0 74,, 40 6 We have to find first, the persons getting wages less then Rs. 5 i.e., form the difference table as follows:

37 Finite Differences, Numerical Differentiations and Integrations 865 TABLE.8 x i y i y i y i y i x 0 = 0 9 y 0 = 0 x = 0 9 y 0 = 5 y = 5 y 0 = x = 0 74 y = 7 y = 4 x = 40 6 For x n = x 0 + nh, 5 = 0 + n 0 or 5 n = = 0 n nn ( ) nn ( )( n ) y5 = y( x0 + nh) = y0 + y0 + y0 + y0 +!!! = !! = = = =.50 = 4 approximately. Thus, the persons getting wages less than or upto Rs. 5 are 4 and, the persons getting wages in between Rs. 0 to 5 are (4 9) = 5. Example 9: Given u = 40, u = 45, u 5 = 54, find u and u 4. Solution: Take TABLE.9 x i y i y i y i y i 4 y i x i y = 40 A 40 x y =? = A 85 A 45 A (A + B 75) x y = 45 B + A 90 0 B 45 (89 B A) x 4 y 4 =? = B 99 B 54 B x 5 y 5 = 54

38 866 Engineering Mathematics through Applications With three known values, polynomial will be of degree and the differences of rd and higher order will be zero. i.e. A + B 75 = 0 A + B = 75 B A + 89 = 0 A + B = 89 On solving the above equations, we find, A = 4 and B = 49. Example 0: Given y 0 =, y =, y = 8, y = 00, y 4 = 00, y 5 = 8, without forming the difference table, find 5 y 0. Solution: We know that 5 = (E ) 5 5 y 0 = (E ) 5 y 0 = ( 5 C 0 E 5 5 C E C E 5 C E + 5 C 4 E 5 C 5 E 0 )y 0 = 5 C 0 E 5 y 0 5 C E 4 y C E y 0 5 C E y C 4 Ey 0 5 C 5 y 0 = y 5 5y 4 + 0y 0y + 5y y 0 = = = Example : Given fx ( ) = 50046, fx ( ) = 940, fx ( ) = 75 Find f(). 4 7 and f(0) = Solution: Taken F f ( x) 4 = = F = f( x) = F = f( x) = 75 F 0 = f(0) = 4065 Form the difference table: 7 TABLE.0 x i y i y i y i y i F = F = F 4 = 940 F = 758 F 4 = 5408 F = 0 7 F 7 = 75 F 4 = 98 F 7 = F 0 = 4065

39 Finite Differences, Numerical Differentiations and Integrations 867 Now, we are to find Write where 0 0 f() = f( x) f( x) = F F F() = F(a + nh), a+ nh = + n = n = nn ( ) nn ( )( n ) F() = Fa ( + nh) = Fa ( ) + n Fa ( ) + Fa ( ) + Fa ( )!! = !!! = Hence f() = F() F() = = Alternately: TABLE. x i y i F i F i F i 0 x F0 0 r = 0 0 x F7 7 r = 7 0 x x = = = = F = = F = = Now, we are to find f() = F F with h = Here, x n = x 0 + nh = 0 + n ( ) 8 n = So that nn ( ) nn ( )( n ) F() = Fa ( ) + n Fa ( ) + Fa ( ) + Fa ( )!! = !!! ( 0)

40 868 Engineering Mathematics through Applications = = f() = F() F() = = ASSIGNMENT. If u = 0, u = 8, u = 0, u 4 = 50 find u 0 and u.. Estimate the values of f() and f(4) from the following data: x : y : Apply Newton s backward difference formula to the data below, to obtain a polynomial of degree 4 in x: x : 4 5 y : 4. In the following table, the values of y are consecutive terms of a series of which.5 is the 5th term. Find the first and tenth term of the series ; x : y : From the following data, estimate the number of persons having income between 000 and 500: Income : Below No. of Persons : INTERPOLATION WITH UNEQUAL INTERVALS In the proceeding articles, we have discussed formula for interpolation when functions are given at equally spaced independent variable or arguments. Very oftenly it is inconvenient and some time even impossible to interpolate functions at unequal values of arguments. For such cases, two interpolation formula viz. Lagrange s and Newton s formula for unequal intervals have been developed, the difference used in Newton s formula are called divided differences which are obtained in the usual manner and then divided by certain differences of the agruments... DIVIDED DIFFERENCES Let y 0, y,, y n denote the functional values corresponding to the arguments x 0, x,, x n where the intervals x x 0, x x, x n x n are not necessarily equally spaced. Then the divided differences of y in the ascending order are defined as follows: First order divided differences:

41 Finite Differences, Numerical Differentiations and Integrations 869 y y y y y y [ x0 x] x x [ x x] 0, =,, =,, =,etc. x x 0 x x x x Second order divided differences: x, x x, x x, x x, x x, x, x, x, x, x,etc. 0 0 = = x x0 x x Third order divided differences: x, x, x x0, x, x x0, x, x, x =,etc. x x nth order divided differences: 0 x, x,, xn x0, x,, xn x0, x, x, x n =,etc. x x n 0 Note: Order of any divided difference is less by unity than the number of values of the arguments in it. Thus, if (x 0,y 0 ), (x,y ) etc. are points on a curve, the divided difference is the slope of the secant line through any of these two points. Table of Divided Differences: TABLE. Argument Entry First Second Third Fourth Div. diff. Div. diff Div. diff. Div. diff. Div. diff. x 0 y 0 [x 0,x ] x y [x 0, x, x ] [x, x ] [x 0, x, x, x ] x y [x, x, x ] [x 0, x, x, x, x 4 ] [x, x ] [x, x, x, x 4 ] x y [x, x, x 4 ] [x, x 4 ] x 4 y 4 Alternately: Divided Difference Table is Comparable to an ordinary Differences table as follows: TABLE. Argument Entry First Second Third Fourth Div. diff Div. diff Div. diff Div. diff x 0 y 0 y y 0 h y y + y. hh x y 0 y y h y y + y y.. h hh 0

42 870 Engineering Mathematics through Applications y x y y + y. hh y y h y x y 4 y + y. hh y 4 y h x 4 y 4 y y + y y.. h hh 4 y 4y + 6y 4y + y 4... h h hh 4 0 Hence from this table we see that nth divided difference for any function y k y = nh! i.e. [ x0, x,, xn] n k n.. PROPERTIES OF DIVIDED DIFFERENCES Theorem : The value of a divided difference is independent of the arguments. Proof: The first divided difference [x 0, x ] is given by [ x, x ] 0 The second divided difference y y0 y y0 y y0 = = = + x x x x x x x x x x ( ) ( ) () y y y y = x x0 x x x x0 x x0 0 [ x0, x, x] y y y0 y = + + ( x x0) x x x x x x0 x0 x x x0 y 0 = ( x x )( x x ) ( x x )( x x ) ( x x )( x x ) ( )( ) + y y ( x x )( x x ) 0 ( x x0) y y ( )( )( ) ( )( ) y0 = + + x x x x x x x x x x x x x x y y y = ( x x )( x x ) ( x x )( x x ) ( x x )( x x ) () Results () and () show that divided differences [x 0, x ], [x 0, x, x ] are symmetrical i.e., if the arguments are changed in any order, the values of divided differences remain to be the same. Thus, [, 5, 9] = [5, 9, ] = [5,, 9], etc. We may thus write, [x n, x n,, x, x, x 0 ] = [x 0, x, x,, x n, x n ], etc. It can be proved by mathematical induction that

43 Finite Differences, Numerical Differentiations and Integrations 87 [ xx, 0, x,, xn] y = + ( x x )( x x ) ( x x ) ( x x)( x x ) ( x x ) 0 n y y x x x x x x x x x x x x n n n n 0 n n ( )( ) ( ) ( )( ) ( ) Theorem : The nth divided differences of a polynomial of nth degree are constant. Theorem : The nth divided differences can be expressed as the quotient of two determinants, each of order n +. Theorem 4: The nth divided differences can be expressed as the product of multiple integrals i.e., u u u n n n = n n f( x, x,, x ) du du. du f ( v ) dv where v n = ( u )x + (u u )x + + (u n u n )x n + u n x n, u, u,, u n are independent variables and f n means the (n ) th derivative of f = f(x,x,, x n ). (Proof of all these theorems are not under consideration).. NEWTON'S DIVIDED DIFFERENCE FORMULA Let y 0, y,, y n be the values of y = f(x) corresponding to the variables x 0, x,, x n, then by definition of divided differences, Again y y x x [ xx, 0] = 0 ie.. y= y0 + ( x x0)[ xx, 0] x, x0 x0, x x, x0, x = x x which implies [x, x 0 ] = [x 0, x ] + (x x )[x, x 0, x ] 0 On substituting the above value of [x, x 0 ] in (), we get y = y 0 + (x x 0 ){[x 0 x ] + (x x )[x, x 0, x ]} y 0 n. () () or y = y 0 + (x x 0 )[x 0, x ] + (x x 0 )(x x )[x, x 0, x ] () Further, x, x0, x x0, x, x x, x0, x, x = x x which implies [x, x 0, x ] = [x 0, x, x ] + (x x )[x, x 0, x, x ] (4) On substituting the above value of [x, x 0, x ] in equation (), we get y = y 0 + (x x 0 )[x, x 0 ] + (x x 0 )(x x )[x 0, x, x ] + (x x 0 )(x x )(x x )[x, x 0, x, x ] (5) On continuing in this way, we get f(x) = y = y 0 + (x x 0 )[x 0, x ] + (x x 0 )(x x )[x 0, x, x ]

44 87 Engineering Mathematics through Applications + (x x 0 )(x x )(x x )[x 0, x, x, x ] + + (x x 0 )(x x )(x x ) (x x n )[x 0, x, x, x n ] (6) Observations:. All the terms on the right hand side of (6), have dimensions of y regardless of the nature of x and its units of measurement. Since dimensions of x cancels out in each terms on the right-hand side of (6). For example see the nd term, ( ) ( ) ( y ) y0 x x0 x0, x = x x0 ( x x ) 0 Here dimensions of 'x' in numerator and denominator cancels out.. The last term in the formula (6), is the remainder term after n + terms and is written as: R n + = (x x 0 )(x x )(x x ) (x x n )[x 0, x, x,, x n ] Example : Using divided difference formula, find cosec 5.67 from the following data: x: 5º.05 5º. 5º.5 5º. 5º.4 f(x): Solution: The divided difference table is as follows: TABLE.4 x cosec x First Div. diff. Second Div.diff. Third Div. diff Now employ the formula y = y 0 + (x x 0 )[x 0, x ] + (x x 0 )(x x )[x 0, x, x ] + (x x 0 )(x x )(x x )[x 0, x, x, x ] With x = 5.67, y 0 = y = ( ) ( 0.08)( ) ( 0.08)( 0.5)( ) = =.55. Example 6.: By means of Newton s Divided difference formula (for unequal intervals), find value of f(8) and f(5) from the following table: [Madurai 996, Calcutta 995] x : f(x) = y :

45 Finite Differences, Numerical Differentiations and Integrations 87 Solution: The divided difference table is as follows: TABLE.5 x i y i Ist divided IInd divided IIIrd divided IVth divided difference difference difference difference x 0 = 4 48 x = 5 00 x = 7 94 fx ( ) fx ( 0) ( x x0) (00 48) = (5 4) = 5 (97 5) = 5 (7 4) fx ( ) fx ( ) ( x x) (94 00) = = 97 ( 5) = (7 5) (0 4) 0 97 = f ( x) f ( x) x x = ( ) ( ) ( 0 7) = 0 7 = 5 x = = 7 7 fx ( 4) fx ( ) ( x4 x) (0 900) = ( 0) = 0 7 = 7 0 x 4 = 0 fx ( 5) fx ( 4) ( x x ) 5 4 (08 0) = = 409 ( ) = 0 x 5 = 08 Now, y(x) = y 0 + (x x 0 )[x 0, x ] + (x x 0 )(x x )[x 0, x, x ] + (x x 0 )(x x )(x x )[x 0, x, x, x ] + + (x x 0 )(x x ) (x x n )[x 0, x,, x n ]

46 874 Engineering Mathematics through Applications y(x) = y 0 + (x 4) 5 f(4) + (x 4)(x 5) 5, 7 f(4) + (x 4)(x 5) (x 7) 5, 7, 0 f(4) + (x 4)(x 5)(x 7)(x 0) 4 5, 7, 0, f(4) + (x 4)(x 5)(x 7)(x 0)(x ) 5 5, 7, 0,, y (8) = 48 + (8 4)5 + (8 4)(8 5)5 + (8 4)(8 5)(8 7) + (8 4)(8 5)(8 7)(8 0) = = 448. Example 4: The observed value of a function are respectively 68, 0, 7 and 6 at the four positions, 7, 9 and 0 out of the independent variable. What is the best estimate you can give for value of the function at the position 6 of the independent variable. Solution: Here we employ divided difference to find f(6). The desired table is as follows` : TABLE.6 x i y i Ist Divided diff. nd Divided diff. rd Divided diff. f(4) = = = = = = 0 Here x = 6 f(x) = y = y 0 + (x x 0 )[x 0, x ] + (x x 0 )(x x )[x 0, x, x ] + (x x 0 )(x x )(x x )[x 0, x, x, x ] f(6) = 68 + (x )( ) + (x )(x 7)( ) + (x )(x 7)(x 9) = 68 + ()( )( ) + ()( )( )() = = 47. Example 5: Find the polynomial of the lowest degree which assumes the values 45,, 5, 9 and 5 at x = 4,, 0, and 5. Also find the nature of the polynomial at abscissa x =.

47 Finite Differences, Numerical Differentiations and Integrations 875 Solution: The desired divided differences for the arguments 4,, 0, and 5 are as follows: TABLE.6 x i y i Ist Divided diff. nd Divided diff. rd Divided diff. 4th Divided diff = = = = = = = = = = 44 5 f(x) = y 0 + (x x 0 )[x 0, x ] + (x x 0 )(x x )[x 0, x, x ] + (x x 0 )(x x )(x x )[x 0, x, x, x ] + (x x 0 )(x x )(x x )(x x )[x 0, x, x, x, x 4 ] = 45 + (x + 4)( 404) + (x + 4)(x + )(94) + (x + 4)(x + )(x)( 4) + (x + 4)(x + )(x)(x )() = (x + 4) + 94(x + 5x + 4) 4(x + 5x + 4x) + (x 4 + x 6x 8x) f(x) = x 4 5x + 6x 4x + 5 which is the required polynomial. Now f() = = 5.4. LAGRANGE'S INTERPOLATION FORMULA Let f(x) denotes a polynomial of nth degree which takes the values y 0, y, y,, y n when x takes the values x 0, x, x,, x n respectively. Then the (n + )th differences of the polynomial are zero and hence [x, x 0, x,, x n ] = 0. Thus,

48 876 Engineering Mathematics through Applications y y0 + ( x x )( x x )( x x ) ( x x ) ( x x)( x x )( x x ) ( x x ) 0 n y + + ( x x)( x x )( x x ) ( x x ) 0 yn + = 0 ( xn x)( xn x0)( xn x) ( xn xn ) () On solving for y and then cancelling common factors (x x 0 ), (x x ), (x x ),, (x x n ) in all terms, we get ( x x)( x x) ( x xn) ( x x0)( x x) ( x xn) y = y0 + y ( x x )( x x ) ( x x ) ( x x )( x x ) ( x x ) n 0 This is Lagrange's formula and is seen to give y = y 0, y, y n when x = x 0, x,, respectively. The Lagrange's formula can also be written in the form n Pn( x) y = yi, ( x x) P ( x) i = 0,,, n i= 0 i n n ( x x0)( x x) ( x xn ) + + y ( x x )( x x ) ( x x ) n 0 n n n n n n () d where P n (x) = (x x 0 )(x x ) (x x n ), and P n (x) = P n() x dx Observations. Lagrange's formula is mearly a relation between two variables, either of which may be taken as the independent variable and hence on interchanging x and y in (), we get ( y y)( y y) ( y yn) ( y y0)( y y) ( y y ) x = x n 0 + x ( y0 y)( y0 y) ( y0 yn) ( y y0)( y y) ( y yn) ( y y )( y y ) ( y y ) + + x ( y y )( y y ) ( y y ) 0 n n 0 n n n. It can be proved that the sum of the Lagrangian coefficients is unity. Example 6: Use Lagrange s interpolation formula to find the value of y when x = 7, if the following values of x and y are given: [V.T.U. 000, Madras 000] x: y: 4 6 Solution: If f(x) be a function which takes the values (x 0, y 0 ), (x, y ), (x, y ),, (x n, y n ), then ( x x)( x x) ( x xn) ( x x0)( x x) ( x xn) y = f() x = y0 + y ( x x )( x x ) ( x x ) ( x x )( x x ) ( x x ) n 0 n n

49 Finite Differences, Numerical Differentiations and Integrations 877 ( x x0)( x x) ( x xn ) + + y ( x x )( x x ) ( x x ) n 0 n n n (7 6)(7 9)(7 ) (7 5)(7 9)(7 ) fx () = f(7) = + (5 6)(5 9)(5 ) (6 5)(6 9)(6 ) n (7 5)(7 6)(7 ) (7 5)(7 6)(7 9) (9 5)(9 6)(9 ) ( 5)( 6)( 9) = = = Example 7: Given log00 =.000, log0 =.004, log0 =.08, log04 =.07; find log0. Solution: Consider this problem as a case of unequal interval, then by Lagrange s formula for unequal intervals, we have y = f(x) for different given values of y = y 0, y,, y n corresponding to x = x 0, x,, x n. ( x x)( x x) ( x xn) ( x x0)( x x) ( x xn) y = f() x = y0 + y ( x x )( x x ) ( x x ) ( x x )( x x ) ( x x ) n 0 ( x x0) ( x xn ) + + y ( x x )( x x ) ( x x ) n 0 n n n (0 0)(0 0)(0 04) y(0) = (00 0)(00 0)(00 04) (0 00)(0 0)(0 04) (0 00)(0 0)(0 04) (0 00)(0 0)(0 04) +.08 (0 00)(0 0)(0 04) (0 00)(0 0)(0 0) +.07 (04 00)(04 0)(04 0) ()( )( ) ()( )( ) = ( )( )( 4) ()( )( ) ()()( ) ()()( ) ()()( ) (4)()() = n n

50 878 Engineering Mathematics through Applications () ( ) () () ( ) () () ( ) (4) () () = = Example. 8: The mode of a certain frequency curve is very near to x = 9 and the values of the frequence density f(x) for x = 8.6, 9, 9. are respectively 0.0, 0.5 and 0.5, calculate the approximate value of the mode. [NIT Kurukshetra, 005] Solution: Using Langranges formula, define the polynomial as: ( x x )( x x ) ( x x )( x x ) ( x x )( x x ) f() x = y + y + y ( x x )( x x ) ( x x )( x x ) ( x x )( x x ) with (x 0, y 0 ) = (8.6, 0.0), (x, y ) = (9, 0.5), (x, y ) = (9., 0.5) f(x) x x( x + x) + xx x x( x0 + x) + x0x y 0 y ( x0 x)( x0 x) ( x x0)( x x) = + x x( x0 + x) + x0x + y ( x x )( x x ) 0 Here x ( x + x ) x ( x + x ) x ( x + x ) f ( x) = y + y + y ( x x )( x x ) ( x x )( x x ) ( x x )( x x ) For mode, f (x) = 0 i.e. x 8. x 7.9 x 7.6 = x x x = 0 6.x = or x = Here the mode of the frequency curve will have mode value at x = 9.. ASSIGNMENT 4. If y() =, y() = 9, y(4) = 0, y(6) =, find the Lagrange's interpolation polynomial that takes the same values as y at the given points.. Given f(0) = 8, f() = 0, f() = 0, f(5), = 48, f(6) = 0, f(9) = 04, find f(x).. Certain corresponding values of x and log 0 x are given below : x: f(x): Find log 0 0 by (i) Lagrange's formula (ii) Newton's Divided Diff..5 CENTRAL DIFFERENCE INTERPOLATION FORMULA In the preceeding sections, we derived Newton-Gregory Forward and Backward formulae for equal interval of arguments, Newton s Divided difference and Langrange s formulae for unequal interval of arguments.

51 Finite Differences, Numerical Differentiations and Integrations 879 Now, we shall develop central difference formulae which are most suitable for interpolation near the middle of the difference table. Central Difference Table: If x takes x 0 h, x 0 h, x 0, x 0 + h, x + h with corresponding y as y, y, y 0, y, y, then the corresponding central difference table becomes TABLE.8 x i y i δy i Ist diff. δ y i nd diff. δ y i rd diff. δ 4 y i 4th diff. x 0 h y y = δy / x 0 h y y = δ y y = δy / y = δ y / x 0 y 0 y = δ y 0 4 y = δ 4 y 0 y 0 = δy / y = δ y / x 0 + h y y 0 = δ y y = δy / x 0 + h y Clearly, the expression y y = y is called the first order central difference of y / and is written as: δy / = y y and in general δy n = y n+/ y n /.. Gauss Forward Interpolation Formula Newton's forward formula: pp ( ) pp ( )( p ) yp = y0 + p y0 + y0 + y 0 + ()!! We know, y 0 y = y i.e. y 0 = ( y + y ) () Similarly y 0 = ( y + 4 y ) and so on () Also, y y = 4 y i.e. y = ( y + 4 y ) (4) Similarly, 4 y = 4 y + 5 y and so on. (5) On substituting values of y 0, y 0, etc., we get pp ( ) pp ( )( p ) 4 yp = y0 + p y0 + ( y + y ) + ( y + y )!! on using y, 4 y etc we get pp ( )( p )( p ) ( y + y ) 4! pp ( )( p )( p )( p 4) ( y + y ) + (6) 5!

52 880 Engineering Mathematics through Applications pp ( ) pp ( )!! or yp = y0 + p y0 + y + y pp ( )( p ) 4 pp ( )( p ) 5 + y + y + (7) 4! 5! This is the Gauss Forward formula. Note: In the central differences natation this result will be pp ( ) = + δ + δ pp ( ) + δ pp ( )( p ) y + δ4 p y p y y y y!! 4! 0 / 0 / 0 pp ( )( p ) + δ 5y / + (8) 5! and this formula is employed to interpolate the value of y for p (0 < p < ) measured forward from the origin.. Gauss Backward Interpolation Formula By Newton s forward formula pp ( ) pp ( )( p ) yp = y0 + p y0 + y0 + y0 + () We know, y 0 y = y i.e. y 0 = y + y () Similarly y0 = y + y 4 y0 = y + y Also y y = 4 y and so on () 4 Similarly y = y + y and so on y = y + y On substituting values y 0, y 0, y 0 etc. in (), we get pp ( ) yp = y0 + p( y + y ) + ( y + y )! (4) pp ( )( p ) 4 pp ( )( p )( p ) ( y + y ) + ( y + y ) + (5)! 4! pp ( + ) ( p+ ) pp ( ) 4 = y0 + p y + y + ( y + y )!! ( p + ) p( p )( p ) ( y + y ) + (6) 4! Further, using values y, 4 y etc.

53 Finite Differences, Numerical Differentiations and Integrations 88 ( p+ ) p p( p ) yp = y + p y + y + y!! 0 pp ( )( p+ ) 4 pp ( )( p ) 5 + y + y + (7) 4! 5! which is Gauss Backward formula. Note: In the central differences notation, this result will become ( p+ ) p p( p ) 0 / 0 / yp = y + pδ y + δ y + δ y!! / pp ( )( p ) pp ( )( p ) + δ y + δ y + (8) 4! 5! and it is employed for negative value of 'p' lying between and 0. Though Gauss's forward and backward formulae are not of much practical use, but, these serve as an entermediate steps for obtaining stirling s, Bessel's and Everett's formulas.. Stirling s Formula By Newton-Gregory forward formula pp ( ) pp ( )( p ) yp = y0 + p y0 + y0 + y 0 +!! () Here, y 0 = y y 0 = E / y / E / y / = (E / E / )y / = δy / () Similarly, y =δy, y =δy and so on / Again, y 0 = y y 0 = δy / δy / = δ(y / y / ) = δ(δy ) = δ y () Similarly y 0 = δ y /, 4 y 0 = δ 4 y and so on. (4) Thus, pp ( ) pp ( )( p ) yp = y0 + pδ y/ + δ y + δ y / +!! (5) δ y = [( δ y +δ y )] + [( δy δ y )], by rearrangement of terms Now, / / / / / = δ + + δ ( y / y/ ) ( y/ y / ) / / / / ( ) = δ E + E y + δ( E E ) y 0 0 =µδ y0 + δ y0 (6) Similarly, δ y = δ y 0 + (δ y δ y 0 ) = δ y 0 + δ (y y 0 ) = δ y 0 + δ (δy / )

54 88 Engineering Mathematics through Applications =δ y0 +δ y0 y µδ + δ 0, using (6) 4 =δ y0 +µδ y0 + δ y0, and so on. (7) Using (6), (7), we get pp = + µδ + δ + ( ) 4 y δ +µδ + δ p y0 p y + 0 y0 y 0 y0 y0! p p( p ) p ( p ) 4 = y + pµδ y + δ y + µδ y + δ y!! 4! This is Known as Stirling's Interpolation Formula Further, in forward difference form it becomes, pp ( )( p ) 5 + µδ y 0 + (8) 5! y + y p p( p ) y + y yp = y + p + y +!! p ( p ) 4 p( p )( p ) y + y + y + (9) 4! 5! Which is the mean of Gauss's Forward The path of Stirling s formula across a diagonal difference table is shown in the difference table with entries that occur in the formula are printed in dark. TABLE.9 y y y y 4 y 5 y 6 y y y y y y y y y 4 y y y 5 y y 0 y 4 y 6 y y 0 y 5 y y y 0 4 y y y 0 y y y y Observations: Clearly from (9), the formula involves means of the odd differences just above and below the central line as shown in the difference table.8.

55 Finite Differences, Numerical Differentiations and Integrations Bessel s Formula As we have already derived δ y/ =µδ y0 + δ y0, (See equation (6) in Stirling s Formula) Implies δ y = µδ y/ + δ y/ (Shift y to y / ), Further, δ y = µδ y/ + δ y/ (0) Again, 4 δ y = δ y0 +µδ y0 +δ y/, (Equation (7) in Stirling's Formula) Implies δ y / = δ y / + µδ y / + δ 4 y / Further, δ y / = δ y / + µδ 4 y / + δ 5 y / () By using (0) and (), equation (5) of derivation becomes which is the desired expression. pp ( ) p pp ( ) y = + δ + µδ + p y p y y δ y!! 0 / / / pp ( )( p ) p pp ( )( p ) µδ y/ + δ y/ + 4! 5! Example 9: Interpolate by means of Gauss's backward formula, the population of a town for the year 974, given that: Year : Population : (in thousands) (Kottayam, 005; Madras, 00) Solution: Taking x 0 = 969, h = 0, x 0 + ph = x p i.e. the central difference table as: xp x p = = = 0.5, h 0 Form

56 884 Engineering Mathematics through Applications TABLE.0 x i y i y i y i y i 4 y i 5 y i x = 99 x = x = x 0 = x = x = Now, using Gauss Backward formula, ( p+ ) p ( p+ ) p( p ) yp = y0 + p y + y + y +!! y 5 ( p + )( p + ) p( p ) 4 ( p+ )( p+ ) p( p )( p ) 5 y + y + 4! 5! (.5)(0.5)(5) (.5)( 0.5) = 7 + (0.5)(7) + + () 6 (.5)(.5)( 0.5) (.5)(.5)(0.5)( 0.5)(.5) + ( 7) + ( 0) 4 0 = =.45 thousands approximate Example.40: From following table x: f(x): find f() using stirling s formula. We know that the Stirling s formula is n nn ( ) yx ( o + nh) = y + nµδ y + δ y + µδ y!! Given table is n ( n ) 4 nn ( )( n ) 5 + δ y0 + µδ y 0 + () 4! 5! x i : y i :

57 Finite Differences, Numerical Differentiations and Integrations 885 Now, we need to find the value of y at x =. Here,, 0, 5, x = x0 = h = yn yn = x x0 0 = = = = 0.4 n h 5 5 The corresponding difference table is as follows: () TABLE. x y x 0 = Now from the table, we calculate the following values: y 0 =.57, µδ y0 = ( 0.5) = 0.47, () (4) As y0 y /, δ = and and δ y0 = y = 0.06 / / E + E y + y µδ y0 = y / = 0 (5) As ( y + y ) y + y µδ y0 = δ ( µδ y0) = δ = 0 µδ y = = ( 0.066) = 0.0 (6) and 4 4 δ y0 = y = 0.04 (7) Subtituting the values of (), (4), (5), (6), (7) and h in Stirling's formula stated in equation (), we will get y 0

58 886 Engineering Mathematics through Applications (0.4) y =.57 + (0.4)( 0.47) + ( 0.06) ((0.4) ) ((0.4) ) 0.4 (0.4) + ( 0.0) + (0.04)! 4! =.57 + ( 0.88) + ( ) + ( ) = = ¾.06 Example 4: The pressure p of wind corresponding to the velocity v is given by the following data. Estimate p, when v = 5. v: p: Solution: The desired difference table is as follows: TABLE. v i y i δy i δ y i δ y i v = 0 y =. δy / = 0.9 v 0 = 0 y 0 =.0 δ y 0 =.5 δy / =.4 δ y / = 0.4 v = 0 y = 4.4 δ y =. δy / =.5 v = 40 y = 7.9 v v 5 0 Taking 0 0, 0, o v = h = p= = = h 0 As, p = lies between 4 and, therefore, it is appropriate to use Bessel's Formula, 4 given by Here, y 0 =.0, pp ( ) p pp ( ) y = 0 + δ / + µδ / + p y p y y δ y/ +...!! δ y / =.4, µδ y/ = ( δ y + δ y0 ) = (.5+.) = (.6) =.,

59 Finite Differences, Numerical Differentiations and Integrations 887 δ y / = 0.4 y 5 =.0 + (.4) + (.) + ( 0.4)!!. = = = Example 4: Use Bessel s formula to obtain y 5, given y 0, = 4, y 4 =, y 8 = 5, y = 40. Solution: By definition, Bessel's formula is given as: nn ( ) n nn ( ) y = ( + ) = + δ + µδ + n y x nh y n y y δ y!! 0 0 / / / nn ( )( n ) n nn ( )( n ) µδ y/ + δ y/ ! 5! Now interpolate using the table, taking y 4 as the base value, with an interval length h = 4. TABLE. x i y i δy i δ y i δ y i x y = y 0 = 4 y δy / = 8 x 0 y 0 = y 4 = y = δ y 0 = 5 y 0 = δy / = y = δ y / = 7 x y = y 8 = 5 y o = δ y = y = δy / = 5 x y = y = 40 Here and δ y / = 7 y =, δ y =, µδ y = δ µ y 0 / / = ( δ y+δ y0) = ( 5+ ) = δ / / δ = ( E y / + E y / ) = ( y + y 0 ) Subtituting these values in y n = y 0 + nh implies 5 = 4 + 4n n = 4 y 5 = !! (7)

60 888 Engineering Mathematics through Applications 9 7 = = = Example 4: Given the table x i : logx i : Find the value of log 7.5 by Stirling s and Bessel s formula. Solution: Let us take log0 as the base value so that x n = x 0 + nh when h = or 7.5= 0+ n 0 n = =.75 0 Thus, write the difference table as: TABLE.4 x i y i y i y i y i 4 y i 5 y i x y =.494 y = 0.08 x y =.505 y = y = 0.0 y = x 0 y 0 =.558 y = y - = y 0 = 0.00 y = y = x y =.55 y 0 = y = y = 0.06 y 0 = 0.00 x y =.544 y = y = 0.0 x y =.556 Stirling's formula is Now, (i) 4 n nn ( ) n( n ) δ y yn= y0+ nµδ y0+ δ y0+ µδ y0+!! 4! / / δ ( E y ( 0 + E y0) δ y/ + y /) µδ y0 = δµ y0 = = = ( δ y/ +δ y /) = ( y0 + y ) 0 nn ( )( n ) 5 + µδ y 0 + 5! = ( ) = (0.06) = 0.0 (ii) δ y0= y = (iii) y + y / / µδ y0=δ µ y0=δ

61 Finite Differences, Numerical Differentiations and Integrations 889 (iv) 4 4 δ y0= y = Therefore, = δ +δ y/ y = / y/ / + y / / = y y + = = ( y + y ) µδ =δ = δ + δ 5 5 / / 5 5 y 0 y/ y / y n 5 5 = y y + = + = (.75) ( 0.000) (.75)(.75 ) = !! (.75) (.75 ).75(.75 )(.75 ) + ( 0.000) ! 5! = =.58 Example 44: Employ Bessel s formula to find to the value of y at x =.95, given that [NIT Kurukshetra, 005; 009] x: y: Which other intepolation formula can be used here? Which is more appropriate? Give reasons. Solution: Construct the table as: TABLE.5 x y y y y 4 y 5 y 6 y (x 0 ).8(y 0 ) (δy / ) (δ y / ) 0.50(δ 5 y / )

62 890 Engineering Mathematics through Applications Using Bessel s formula, pp ( ) pp ( )( p /) yn = y + pδ y + µδ y + δ y!! 0 / / / pp ( )( p ) 4 pp ( )( p )( p /) 5 + µδ y/ + δ y/ ! 5! p= = = δy / = µδ y / = = 0.05 δ y / = µδ y / = = 0.55 δ 5 y / = 0.50 (0.5)(0.5 ) y(.95) =.8 + (0.5)(0.08) ( 0.05) (0.5)(0.5 )( ) (0.5)(0.5 )(0.5 ) + ( 0.005) (0.5)(0.5 )(0.5 )( ) (0.5)(0.5) (0.5)(0.75)(.5)(0.55) = (0.05) =.4708 =.47 ASSIGNMENT 5. Using Stirling s formula find y 5, given y 0 = 5, y 0 = 49, y 40, = 46, y 50 = 4.. Apply Bessel s formula to obtain y 5, given y 0 = 854, y 4 = 6, y 8 = 544, y = 99.. Given θ: tan θ : Using Stirling's formula, show that tan 6 = From the data: x: cosx: obtain cos, using Bessel s formula. 5. Use Gauss s forward formula to evaluate y 0 given that y = , y 5 = 7.844, y 9 =7.070, y = 6.4 and y 7 =5.554.

63 Finite Differences, Numerical Differentiations and Integrations INVERSE INTERPOLATION Inverse interpolation is the process of finding the value of the argument corresponding to a given value of the function when the function is intermediate between two tabulated values. Out of several methods of inverse interpolation, only two are explained.. Lagrange s Formula This technique has already been explained while explaining unequal interpolation by Lagrange s method, where x (the argument) can be expressed as a function of y.. Reversion of Series The most obvious method of solving the problem of inverse interpolation is by reversion of series, for all the interpolation formulas developed in the form of power series. Thus the power series y = c 0 + c x + c x + c x c n x n +... () when reverted becomes 4 n y c0 y c0 y c0 y c0 x= + a + a + a y c0 + + an + c c c c c () c Where a =, c c c = + c c a a, c cc c = + 5 5, 4 c c c 4 c5 cc 4 c c c c 4= + + c c + c c c a 6 4, c 6 cc5 + cc 4 c c4 + cc c c c a5 = c c + c () c c In this way, compute all c 's and then substitute their values in (), to find x. We write inverse interpolation formula for Newton s, Stirling s and Bessel s formulae in the power series form. (a) Newton s Formula: pp ( ) pp ( )( p ) pp ( )( p )( p ) 4 y = y0+ p y0 + y0 + y0 + y 0 + (4)!! 4! (Limiting the expression upto 4th difference, however, it could be extended desirably) 5 y 4 y y y y y = y0 + y0 + p+ + y0 4 4 p y y y + p + p (5)

64 89 Engineering Mathematics through Applications On comparison of ()and (4), we get c = y c = +, c c = y0 y0 y0 y0 4 y0 y0 4 = + y0, 4 4 y0 y0, y 0 4 =. c 4 (b) Stirling s Formula (6) y + y0 p p( p ) y + y p ( p ) 4 y = y0+ p + y + + y (7) 6 4 y + y y + y y y = y0 + p + p On comparing (7) with (), we get y + y y + p p (8) c0 = y0 c = ( y + y0) ( y + y ) 4 c = y y 4 c = ( y + y ) 4 c4 = y 4 (c) Bessel s Formula:...(9) p p p y y = + p y + + y y 4 y + y p p y + y + 4 (0)

65 Finite Differences, Numerical Differentiations and Integrations = ( y0 + y) ( y + y0) + ( y + y ) y y p+ ( y + y ) ( y + y ) p On comparing () with (), we get y p + ( y + y ) p () c0 = ( y0 + y) ( y + y0) + ( y + y ) 6 56 c = y0 y c = ( y + y0) ( y + y ) 4 96 c = y c4 = ( y + y ) 48 () Example 45: Using given table of values of the Probability integral the value of x for which integral value equals to. x x e dx, find x 0 x i y i y i y i y i 4 y i (0 7 ) (0 7 ) (0 7 ) (0 7 ) x = x = x 0 = x = x = x = Solution: Here, ( ) , 0 y + y = ( y + y0) = ; ( ) y + y = ()

66 894 Engineering Mathematics through Applications Hence, Using result (), of Reversion of series (Inverse Interpolation) we get Now c = + = , c = = , c = = , c = = , 6 c = 4 0, practically c = = , c c = ( ) = , c c = ( ) = , c c = = c For finding the desired argument, say, () () We find 0 y c0 0 y c y c x = + a + a +... c c c (4) c = = a , c c c a = + = ( ) c c = , c 4 cc c c = c c c = 0 + 5( )( ) 5( ) = y c = = = c x = (0.946) (0.946) = = (5)

67 Finite Differences, Numerical Differentiations and Integrations 895 Example 46: Using stirling's inverse interpolation formula, find x for given sinh x = 6, employing data below: x : y = sinh x: Solution: Using equations (9) Artical.6 (Inverse Interpolution using Reversion of series) c0 = y0 = , c = 0.699, c = = 0.00, c = 0 = c 4 For finding x, employ equation () Here y = 6, y c = = , c c c = = Hence a c = = c a , c c = + ( ) , c = = c (as c = ) With above values of a, a, we get p = (0.0468) (0.0468) = x = (x 0 + ph) = (0.047) = 4.80 Note: The problem of inverse interpolation by reversion of series is employed when the number of digits involved is large. ASSIGNMENT 6. If cosh x =.85, find x by inverse interpolation, using data in the follwing table x : f(x)

68 896 Engineering Mathematics through Applications.7. NUMERICAL DIFFERENTIATION Numerical differentiation is the process of evaluating derivative of the interpolating polynomial at some particular value of the variate. The choice of the interpolating formula for derivative employing differences, are the same as in the case of interpolation. For equidistant value of the independent variable, generally, we employ Newton's Forward Difference formula and Newton BackwardDifference formula for finding the value of the derivative at the beginning and at the end of the interval respectively, whereas, in such cases where the value of the derivative is desired in the middle of the table, Stirling s and Bessel s formula of central differences are employed. However, in case of unequidistant value of independent variables, Lagrange s or Hermite interpolation techniques are preferably employed for finding the derivative of a function. Consider the function y = f(x) tabulated for various x = a + ph, p = 0,,,... (a) Derivative Using Newton's Forward Difference Formula pp ( ) pp ( )( p ) y = y0 + p y0 + y0 + y ()!! where x = x 0 + ph and y 0 = f(a) dy dy dp So that = dx dp dx p p 6 p = y0 y0 y0... h () This formula is used to calculate the value of dy for non-tabulated value of x. dx However, for tabular values, the formula becomes simpler by taking x = x 0 at., p = 0 i.e., dy = y y + y y + dx x= x h () For obtaining IInd order derivative, differentiate () again with respect to p. dy 6p 6 p 6p + 4 = y 0 y0 y0... dx h (4) and thus, dy = y0 y0 y0 y0 y0..., ( p = 0) dx h 6 80 x= x0...(5) Higher order derivatives may be obtained by successive differentiation. (b) Derivative Using Newton s Backward Formula: pp ( + ) pp ( + )( p+ ) y = yn + p yn + yn + y n (6)!!

69 Finite Differences, Numerical Differentiations and Integrations 897 giving dy 4 yn yn yn yn... = dx h 4,...(7) x= xn and dy yn yn yn yn... dx = h 6 x= xn...(8) (c) Derivative Using Stirling s Interpolation Formula: p p( p ) p ( p ) 4 fa ( + ph) = fa () + pµδ fa () + δ fa () + µδ fa () + δ fa () +...!! 4! On sucessively differentiating both sides with respect to p, we get (9) and 4 hf '( a + ph) = µδ f( a) + pδ f( a) + (p ) µδ f( a) + (4p p) δ f( a) +... (0) 6 4 ( + ) =δ ( ) + µδ ( ) + ( ) δ ( ) () 4 4 hf a ph fa p fa p fa However, at x = x 0 (i.e., p = 0) these results become and f a = a f a f a f a h µδ µδ + µδ ( ) ( ) ( ) ( ) ( ) f () a= fa () fa () fa ()... h δ δ + δ 90...()...() Note: Derivation of rest of the interpolation techniques for finding differentiation of a function has been left to the readers. Example 47: Given that x : y : Find dy dx and dy dx at x =. and x =.6. Solution: As the value of y(x) is required at the ends, so we follow forward difference formula, for derivative at x =. and backward difference formula, for derivative at x =.6. The difference table is:

70 898 Engineering Mathematics through Applications TABLE.7 x y x 0 = y 0 = y 0 = y 0 = Thus dy 4 y y0 y0 y0... = + + dx h 4 x0...() and y0 y0 y0 y0... = dy dx x h 0 Here x 0 =., h = 0., y 0 = 0.78, y 0 = 0.0 and so on....() dy 0.78 ( 0.0) (0.004) (0) ( 0.00) = + + =.94, dx () Now, dy 5 dx. (0.) 6 = 0.0 (0.004) + (0) ( 0.00) =.67...(4) and and dy = yn yn yn 4 yn... dx h 4...(5) dy xn yn yn yn yn... dx = h 6 x n Here x n =.6, h = 0., y n = 0.8, y n = 0.08 and so on. dy 0.8 ( 0.08) (0.005) ( 0.00) ( 0.00) = =.7 dx x n dy = ( 0.00) ( 0.00) =.475 dx (0.) 6 x n...(6)...(7)...(8)

71 Finite Differences, Numerical Differentiations and Integrations 899 Example 48: A slider in a machine moves along a fixed straight rod. Its distance x cm along the rod is given below for various values of the time t seconds. Find the velocity and acceleration of the slider when t = 0. seconds. t : x : Solution: As the derivatives are required near the middle of variate range, therefore, we use Stirling s formula. The corresponding difference table is as follows: TABLE.8 t f(t) = x t 0 = ( ) d f t = ft ( ) = µδft ( ) µδ ft ( ) + µδ ft ( ) dt t t h = d 4 6 and f ( t0) = ft ( 0) t= t = ft ( 0 0) ft ( 0) ft ( 0)... dt h δ δ + δ 90 Here t 0 = 0., h = 0., f(t 0 ) =.64 x + x ( ) 0 µδ ft ( 0) = = = 0.54, x + x ( ) ft ( 0) 0.05, µδ = = = x x ( ) µδ ft ( 0) = = = 0.5 δ ft ( 0) = 0.46, 4 δ ft ( 0) = 0.0, 6 ft ( 0) 0.9 δ = f (0.) = 0.54 (0.05) ( 0.5) 5.4 cm/sec = 6 0 () () () (4) (5)