Examples lv 1 suppose that f x is a polynomial of

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Examples LV. 1. Suppose that f ( x ) is a polynomial of degree r . Then f ( n ) ( x ) is identically zero when n > r , and the theorem leads to the algebraical identity f ( a + h ) = f ( a ) + hf 0 ( a ) + h 2 2! f 00 ( a ) + · · · + h r r ! f ( r ) ( a ) . 2. By applying the theorem to f ( x ) = 1 /x , and supposing x and x + h positive, obtain the result 1 x + h = 1 x - h x 2 + h 2 x 3 - · · · + ( - 1) n - 1 h n - 1 x n + ( - 1) n h n ( x + θ n h ) n +1 . [Since 1 x + h = 1 x - h x 2 + h 2 x 3 - · · · + ( - 1) n - 1 h n - 1 x n + ( - 1) n h n x n ( x + h ) , we can verify the result by showing that x n ( x + h ) can be put in the form ( x + θ n h ) n +1 , or that x n +1 < x n ( x + h ) < ( x + h ) n +1 , as is evidently the case.] 3. Obtain the formula sin( x + h ) = sin x + h cos x - h 2 2! sin x - h 3 3! cos x + . . . + ( - 1) n - 1 h 2 n - 1 (2 n - 1)! cos x + ( - 1) n h 2 n 2 n ! sin ( x + θ 2 n h ) ,
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[VII : 147] ADDITIONAL THEOREMS IN THE CALCULUS 322 the corresponding formula for cos( x + h ), and similar formulae involving powers of h extending up to h 2 n +1 . 4. Show that if m is a positive integer, and n a positive integer not greater than m , then ( x + h ) m = x m + m 1 x m - 1 h + · · · + m n - 1 x m - n +1 h n - 1 + m n ( x + θ n h ) m - n h n . Show also that, if the interval [ x, x + h ] does not include x = 0, the formula holds for all real values of m and all positive integral values of n ; and that, even if x < 0 < x + h or x + h < 0 < x , the formula still holds if m - n is positive. 5. The formula f ( x + h ) = f ( x )+ hf 0 ( x + θ 1 h ) is not true if f ( x ) = 1 /x and x < 0 < x + h . [For f ( x + h ) - f ( x ) > 0 and hf 0 ( x + θ 1 h ) = - h/ ( x + θ 1 h ) 2 < 0; it is evident that the conditions for the truth of the Mean Value Theorem are not satisfied.] 6. If x = - a , h = 2 a , f ( x ) = x 1 / 3 , then the equation f ( x + h ) = f ( x ) + hf 0 ( x + θ 1 h ) is satisfied by θ 1 = 1 2 ± 1 18 3. [This example shows that the result of the theorem may hold even if the conditions under which it was proved are not satisfied.] 7. Newton’s method of approximation to the roots of equations. Let ξ be an approximation to a root of an algebraical equation f ( x ) = 0, the actual root being ξ + h . Then 0 = f ( ξ + h ) = f ( ξ ) + hf 0 ( ξ ) + 1 2 h 2 f 00 ( ξ + θ 2 h ) , so that h = - f ( ξ ) f 0 ( ξ ) - 1 2 h 2 f 00 ( ξ + θ 2 h ) f 0 ( ξ ) . It follows that in general a better approximation than x = ξ is x = ξ - f ( ξ ) f 0 ( ξ ) . If the root is a simple root, so that f 0 ( ξ + h ) 6 = 0, we can, when h is small enough, find a positive constant K such that | f 0 ( x ) | > K for all the values of x which we are considering, and then, if h is regarded as of the first order of smallness,
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[VII : 147] ADDITIONAL THEOREMS IN THE CALCULUS 323 f ( ξ ) is of the first order of smallness, and the error in taking ξ - { f ( ξ ) /f 0 ( ξ ) } as the root is of the second order. 8. Apply this process to the equation x 2 = 2, taking ξ = 3 / 2 as the first approximation. [We find h = - 1 / 12, ξ + h = 17 / 12 = 1 . 417 . . . , which is quite a good approximation, in spite of the roughness of the first. If now we repeat the process, taking ξ = 17 / 12, we obtain ξ + h = 577 / 408 = 1 . 414 215 . . . , which is correct to 5 places of decimals.] 9. By considering in this way the equation x 2 - 1 - y = 0, where y is small, show that 1 + y = 1 + 1 2 y - { 1 4 y 2 / (2 + y ) } approximately, the error being of the fourth order.
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