Ap Calculus BC 2009 Question 6?

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!