Question

Ap Calculus BC 2009 Question 6?

Answer 1

a) Simply replace `x` with `(x- 1)^2`.

`e^((x- 1)^2) = 1 + (x- 1)^2 + ((x - 1)^2)^2/2 + (((x -1)^2)^3)/6 + ...`

`e^((x -1)^2) = 1 + (x- 1)^2 + (x - 1)^4/2 + (x - 1)^6/6 + ...`

`e^((x - 1)^2) = sum_(n = 0)^oo (x - 1)^(2n)/(n!)`

b) Once again more algebraic manipulations.

`(e^((x - 1)^2) - 1)/(x- 1)^2 = (1 + (x - 1)^2 + (x - 1)^4/2 + (x - 1)^6/6 - 1)/(x- 1)^2`

`(e^((x -1)^2) - 1)/(x- 1)^2 = ((x-1)^2 + (x- 1)^4/2 + (x- 1)^6/6)/(x- 1)^2`

`(e^((x -1)^2) - 1)/(x- 1)^2 = 1 + (x- 1)^2/2 + (x- 1)^4/6 + (x- 1)^6/24`

Rewriting in the general term we get

`(e^((x- 1)^2) - 1)/(x- 1)^2 = sum_(n =0)^oo (x- 1)^(2n)/((n + 1)!)`

c)

`L = lim_(n-> oo) ((x - 1)^(2(n + 1) - 1)/((n + 1 + 1)!))/(((x- 1)^(2n))/((n + 1)!)`

`L = lim_(n->oo) x/(n +2)`

`L = |x| lim_(n-> oo) 1/(n+ 2)`

`L = 0`

Since this is less than `1`, this converges for all values of `x`. Therefore, the interval of convergence is `(-oo, oo)`.

d) Take the second derivative of `f`.

`f''(x) = 1 + 2(x - 1)^2 + 5/4(x -1)^4 + ...`

Since each term in this expansion will be positive, there will be no point of inflection (inflection points occur when `f''(x)` changes sign).

Hopefully this helps!