1. 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 = x + + x2 2 x3 6 xn
n !
= (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
= (x-1)2 +
e(x-1)2

2. Part (b) + . . . + + . . . + + . . . + + . . . + (x-1)2n n ! (x-1)4 2
(x-1)2n
n ! +
(x-1)4 2 6
(x-1)6
= (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) !

3. 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 (-∞, ∞).

4. 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.