PureMath.pdf

# In i divide the range of integration into the two

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[In (i) divide the range of integration into the two parts [ - a, 0], [0 , a ], and put x = - y in the first. In (ii) use the substitution x = 1 2 π - y to obtain the first equation: to obtain the second divide the range [0 , π ] into two equal parts and use the substitution x = 1 2 π + y . In (iii) divide the range into m equal parts and use the substitutions x = π + y , x = 2 π + y , . . . . ] 15. Prove that Z b a F ( x ) dx = Z b a F ( a + b - x ) dx. 16. Prove that Z 1 2 π 0 cos m x sin m x dx = 2 - m Z 1 2 π 0 cos m x dx. 17. Prove that Z π 0 (sin x ) dx = 1 2 π Z π 0 φ (sin x ) dx.

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[VII : 162] ADDITIONAL THEOREMS IN THE CALCULUS 367 [Put x = π - y .] 18. Prove that Z π 0 x sin x 1 + cos 2 x dx = 1 4 π 2 . 19. Show by means of the transformation x = a cos 2 θ + b sin 2 θ that Z b a p ( x - a )( b - x ) dx = 1 8 π ( b - a ) 2 . 20. Show by means of the substitution ( a + b cos x )( a - b cos y ) = a 2 - b 2 that Z π 0 ( a + b cos x ) - n dx = ( a 2 - b 2 ) - ( n - 1 2 ) Z π 0 ( a - b cos y ) n - 1 dy, when n is a positive integer and a > | b | , and evaluate the integral when n = 1, 2, 3. 21. If m and n are positive integers then Z b a ( x - a ) m ( b - x ) n dx = ( b - a ) m + n +1 m ! n ! ( m + n + 1)! . [Put x = a + ( b - a ) y , and use Ex. 6.] 162. Proof of Taylor’s Theorem by Integration by Parts. We shall now give the alternative form of the proof of Taylor’s Theorem to which we alluded in § 147 . Let f ( x ) be a function whose first n derivatives are continuous, and let F n ( x ) = f ( b ) - f ( x ) - ( b - x ) f 0 ( x ) - · · · - ( b - x ) n - 1 ( n - 1)! f ( n - 1) ( x ) . Then F 0 n ( x ) = - ( b - x ) n - 1 ( n - 1)! f ( n ) ( x ) , and so F n ( a ) = F n ( b ) - Z b a F 0 n ( x ) dx = 1 ( n - 1)! Z b a ( b - x ) n - 1 f ( n ) ( x ) dx.
[VII : 163] ADDITIONAL THEOREMS IN THE CALCULUS 368 If now we write a + h for b , and transform the integral by putting x = a + th , we obtain f ( a + h ) = f ( a ) + hf 0 ( a ) + · · · + h n - 1 ( n - 1)! f ( n - 1) ( a ) + R n , (1) where R n = h n ( n - 1)! Z 1 0 (1 - t ) n - 1 f ( n ) ( a + th ) dt. (2) Now, if p is any positive integer not greater than n , we have, by Theo- rem (9) of § 160 , Z 1 0 (1 - t ) n - 1 f ( n ) ( a + th ) dt = Z 1 0 (1 - t ) n - p (1 - t ) p - 1 f ( n ) ( a + th ) dt = (1 - θ ) n - p f ( n ) ( a + θh ) Z 1 0 (1 - t ) p - 1 dt, where 0 < θ < 1. Hence R n = (1 - θ ) n - p f ( n ) ( a + θh ) h n p ( n - 1)! . (3) If we take p = n we obtain Lagrange’s form of R n ( § 148 ). If on the other hand we take p = 1 we obtain Cauchy’s form , viz. R n = (1 - θ ) n - 1 f ( n ) ( a + θh ) h n ( n - 1)! . * (4) 163. Application of Cauchy’s form to the Binomial Series. If f ( x ) = (1 + x ) m , where m is not a positive integer, then Cauchy’s form of the remainder is R n = m ( m - 1) . . . ( m - n + 1) 1 · 2 . . . ( n - 1) (1 - θ ) n - 1 x n (1 + θx ) n - m . * The method used in § 147 can also be modified so as to obtain these alternative forms of the remainder.

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[VII : 164] ADDITIONAL THEOREMS IN THE CALCULUS 369 Now (1 - θ ) / (1 + θx ) is less than unity, so long as - 1 < x < 1, whether x is positive or negative; and (1 + θx ) m - 1 is less than a constant K for all values of n , being in fact less than (1+ | x | ) m - 1 if m > 1 and than (1 -| x | ) m - 1 if m < 1. Hence | R n | < K | m | m - 1 n - 1 | x n | = ρ n , say. But ρ n 0 as n → ∞ , by Ex. xxvii . 13, and so R n 0. The truth of the Binomial Theorem is thus established for all rational values of m and all values of x between - 1 and 1. It will be remembered that the difficulty in using Lagrange’s form, in Ex. lvi . 2, arose in connection with negative values of x .
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