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# 10 show that the error in taking the root to be ξ ff

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10. Show that the error in taking the root to be ξ - ( f/f 0 ) - 1 2 ( f 2 f 00 /f 0 3 ), where ξ is the argument of every function, is in general of the third order. 11. The equation sin x = αx , where α is small, has a root nearly equal to π . Show that (1 - α ) π is a better approximation, and (1 - α + α 2 ) π a better still. [The method of Exs. 7–10 does not depend on f ( x ) = 0 being an algebraical equation, so long as f 0 and f 00 are continuous.] 12. Show that the limit when h 0 of the number θ n which occurs in the general Mean Value Theorem is 1 / ( n +1), provided that f ( n +1) ( x ) is continuous. [For f ( x + h ) is equal to each of f ( x )+ · · · + h n n ! f ( n ) ( x + θ n h ) , f ( x )+ · · · + h n n ! f ( n ) ( x )+ h n +1 ( n + 1)! f ( n +1) ( x + θ n +1 h ) , where θ n +1 as well as θ n lies between 0 and 1. Hence f ( n ) ( x + θ n h ) = f ( n ) ( x ) + hf ( n +1) ( x + θ n +1 h ) n + 1 . But if we apply the original Mean Value Theorem to the function f ( n ) ( x ), taking θ n h in place of h , we find f ( n ) ( x + θ n h ) = f ( n ) ( x ) + θ n hf ( n +1) ( x + θθ n h ) , where θ also lies between 0 and 1. Hence θ n f ( n +1) ( x + θθ n h ) = f ( n +1) ( x + θ n +1 h ) n + 1 , from which the result follows, since f ( n +1) ( x + θθ n h ) and f ( n +1) ( x + θ n +1 h ) tend to the same limit f ( n +1) ( x ) as h 0.]

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[VII : 148] ADDITIONAL THEOREMS IN THE CALCULUS 324 13. Prove that { f ( x +2 h ) - 2 f ( x + h )+ f ( x ) } /h 2 f 00 ( x ) as h 0, provided that f 00 ( x ) is continuous. [Use equation (2) of § 147 .] 14. Show that, if the f ( n ) ( x ) is continuous for x = 0, then f ( x ) = a 0 + a 1 x + a 2 x 2 + · · · + ( a n + x ) x n , where a r = f ( r ) (0) /r ! and x 0 as x 0. * 15. Show that if a 0 + a 1 x + a 2 x 2 + · · · + ( a n + x ) x n = b 0 + b 1 x + b 2 x 2 + · · · + ( b n + η x ) x n , where x and η x tend to zero as x 0, then a 0 = b 0 , a 1 = b 1 , . . . , a n = b n . [Making x 0 we see that a 0 = b 0 . Now divide by x and afterwards make x 0. We thus obtain a 1 = b 1 ; and this process may be repeated as often as is necessary. It follows that if f ( x ) = a 0 + a 1 x + a 2 x 2 + · · · + ( a n + x ) x n , and the first n derivatives of f ( x ) are continuous, then a r = f ( r ) (0) /r !.] 148. Taylor’s Series. Suppose that f ( x ) is a function all of whose differential coefficients are continuous in an interval [ a - η, a + η ] surrounding the point x = a . Then, if h is numerically less than η , we have f ( a + h ) = f ( a ) + hf 0 ( a ) + · · · + h n - 1 ( n - 1)! f ( n - 1) ( a ) + h n n ! f ( n ) ( a + θ n h ) , where 0 < θ n < 1, for all values of n . Or, if S n = n - 1 X 0 h ν ν ! f ( ν ) ( a ) , R n = h n n ! f ( n ) ( a + θ n h ) , we have f ( a + h ) - S n = R n . * It is in fact sufficient to suppose that f ( n ) (0) exists . See R. H. Fowler, “The elementary differential geometry of plane curves” ( Cambridge Tracts in Mathematics , No. 20, p. 104).
[VII : 148] ADDITIONAL THEOREMS IN THE CALCULUS 325 Now let us suppose, in addition, that we can prove that R n 0 as n → ∞ . Then f ( a + h ) = lim n →∞ S n = f ( a ) + hf 0 ( a ) + h 2 2! f 00 ( a ) + . . . . This expansion of f ( a + h ) is known as Taylor’s Series . When a = 0 the formula reduces to f ( h ) = f (0) + hf 0 (0) + h 2 2! f 00 (0) + . . . , which is known as Maclaurin’s Series . The function R n is known as Lagrange’s form of the remainder .

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