# Ap Calculus BC 2009 Question 6?

## Apr 30, 2018

a) Simply replace $x$ with ${\left(x - 1\right)}^{2}$.

${e}^{{\left(x - 1\right)}^{2}} = 1 + {\left(x - 1\right)}^{2} + {\left({\left(x - 1\right)}^{2}\right)}^{2} / 2 + \frac{{\left({\left(x - 1\right)}^{2}\right)}^{3}}{6} + \ldots$

${e}^{{\left(x - 1\right)}^{2}} = 1 + {\left(x - 1\right)}^{2} + {\left(x - 1\right)}^{4} / 2 + {\left(x - 1\right)}^{6} / 6 + \ldots$

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

b) Once again more algebraic manipulations.

$\frac{{e}^{{\left(x - 1\right)}^{2}} - 1}{x - 1} ^ 2 = \frac{1 + {\left(x - 1\right)}^{2} + {\left(x - 1\right)}^{4} / 2 + {\left(x - 1\right)}^{6} / 6 - 1}{x - 1} ^ 2$

$\frac{{e}^{{\left(x - 1\right)}^{2}} - 1}{x - 1} ^ 2 = \frac{{\left(x - 1\right)}^{2} + {\left(x - 1\right)}^{4} / 2 + {\left(x - 1\right)}^{6} / 6}{x - 1} ^ 2$

$\frac{{e}^{{\left(x - 1\right)}^{2}} - 1}{x - 1} ^ 2 = 1 + {\left(x - 1\right)}^{2} / 2 + {\left(x - 1\right)}^{4} / 6 + {\left(x - 1\right)}^{6} / 24$

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)!)

$L = {\lim}_{n \to \infty} \frac{x}{n + 2}$

$L = | x | {\lim}_{n \to \infty} \frac{1}{n + 2}$

$L = 0$

Since this is less than $1$, this converges for all values of $x$. Therefore, the interval of convergence is $\left(- \infty , \infty\right)$.

d) Take the second derivative of $f$.

$f ' ' \left(x\right) = 1 + 2 {\left(x - 1\right)}^{2} + \frac{5}{4} {\left(x - 1\right)}^{4} + \ldots$

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

Hopefully this helps!