Can someone check this over or me? It's a calculus II power series problem.

Question

anonymous · Answer

attachment

amistre64 · Answer

great, now im blind ... :/

amistre64 · Answer

youve got some extra jargon in there, doesnt hurt the outcome, just extra work

amistre64 · Answer

4 = A(x+2) + B(x-2) ; when x=2 we get

4 = A(4) + B(0) ;  A = 1 
..........................................

4 = A(x+2) + B(x-2) ; when x=-2 we get

4 = A(0) + B(-4) ;  B = -1

anonymous · Answer

ok so it works (: how would I simplify that answer?

anonymous · Answer

the final answer?

amistre64 · Answer

"simplify" is a subjective term; but you can pull back the SUM notation and factor out the (-1/2) if need be

amistre64 · Answer

and (-x/2)^n = (-1)^n (x/2)^n .... but that might just be trivial

it looks fine to me

anonymous · Answer

I was going to add my own feedback but I'm trying to black out my Calc 2 experience. Just adding that the approach from Amu was how I approached those sort of problems.

anonymous · Answer

*Ami

amistre64 · Answer

its good to know both ways ;)

phi · Answer

For what it's worth, we can simplify by
(1) use only one summation sign, we are adding term by term
(2) factor out (-1/2) 
(3) now a bit of a trick, but sometimes you see people use it with no explanation:
    (-x)^n = x^n if n is even, and it is -x^n if n is odd
    so for even n we can add to get
$$2x ^{n}$$
and zero for odd n
(4)  we get for only the even values of n
$$(-1/2)(2x/2) ^{n}= -(x/2)^{n}$$
which we would write using the summation as
$$ -\sum_{0}^{\infty}(x/2) ^{2n}$$

anonymous · Answer

thanks to all of you! would the interval of convergence be [-2,2]? I am confused because the result stems from the geometric power series which always has an open interval.

phi · Answer

This is just a guess, but notice that the original expression is infinity at x=2 and x=-2, and so is your geometric power series.  So they match (sort of...) I'm a bit leery saying infinity = infinity.