Part (a) ex = 1 + x + + + . . . + + . . . x2 2 x3 6 xn n ! + + . . . + We can re-write ex as e(x-1)2 by substituting directly into the formula. ex =

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Presentation transcript:

Part (a) ex = 1 + x + + + . . . + + . . . x2 2 x3 6 xn n ! + + . . . + We can re-write ex as e(x-1)2 by substituting directly into the formula. ex = 1 + x + + + . . . + + . . . x2 2 x3 6 xn n ! = 1 + (x-1)2 + + + . . . + + . . . ((x-1)2)2 2 6 n ! ((x-1)2)3 ((x-1)2)n e(x-1)2 + . . . + + . . . (x-1)2n n ! + (x-1)4 2 6 (x-1)6 = 1 + (x-1)2 + e(x-1)2

Part (b) + . . . + + . . . + + . . . + + . . . + (x-1)2n n ! (x-1)4 2 + . . . + + . . . (x-1)2n n ! + (x-1)4 2 6 (x-1)6 = 1 + (x-1)2 + e(x-1)2 Answer from Part (a): f(x) = e(x-1)2 - 1 (x – 1)2 = (x – 1)2 (x-1)4 2 1 + (x-1)2 + + (x-1)6 6 + … + (x-1)2n n ! + … - 1 + . . . + + . . . + (x-1)2 2 6 (x-1)4 24 (x-1)6 = 1 + f(x) (x-1)2n (n+1) !

Part (c) + . . . + + . . . + Answer from Part (b): (x-1)2 2 6 (x-1)4 + . . . + + . . . + (x-1)2 2 6 (x-1)4 24 (x-1)6 = 1 + f(x) (x-1)2n (n+1) ! lim n∞ (x-1)2n (n+1) ! (x-1)2n+2 (n+2) ! (x-1)2 lim n∞ = (x-1)2n+2 (n+2) ! (x-1)2n (n+1) ! = < 1 (n+2) Since 0 is always less than 1, the interval of convergence is (-∞, ∞).

We know that these values of “n” are all positive. Part (d) Taylor series answer from Part (b): + . . . + + . . . + (x-1)2 2 6 (x-1)4 24 (x-1)6 = 1 + f(x) (x-1)2n (n+1) ! + . . . + + . . . + 2(x-1)1 2 6 4(x-1)3 24 6(x-1)5 = f’(x) 2n (x-1)2n-1 (n+1) ! Since f”(x) always remains positive, it can’t ever change signs. Therefore, f(x) doesn’t have any points of inflection. + ... + + ... + 1 6 12 (x-1)2 24 30 (x-1)4 = f”(x) (2n) (2n-1) (x-1)2n-2 (n+1) ! Since all these quantities have EVEN exponents, they will be positive as well. We know that these values of “n” are all positive.