0

First, there was a question that stated: $g(x) = \frac{1}{1+x}$, with the power series representation being $\sum_{n=0}^\infty (-1)^nx^n$.

The second question was as follows: find the power series representation for $h(x) = \frac{-1}{(1+x)^2}$.

My method was to define $h(x)$ = $g'(x)$, and I got a power series like $-1+2x-3x^2+...$

My classmate's method was to define $h(x) = -g^2(x)$ and he got something along the lines of $-1+x^2-x^4+...$

As you can tell, these are very different results. Whose method is correct (if any) and what mistake was made?

Andrew Wang
  • 181
  • 1
    Both methods are correct. Their answer isn't. – user647486 Apr 11 '19 at 20:14
  • There is a dfference between $-g^2(x)$ and $-g(x^2)$. Or, what is the same thing, between $-1/(1+x)^2$ and $-1/(1+x^2)$. – Somos Apr 11 '19 at 20:20
  • @Somos I notice the difference between $-g^2(x)$ and $-g(x^2)$ seems to be in the signs, but I cannot figure out the reason for the discrepancy with my answer and my classmate's answer, as it is more than a difference in signs. – Andrew Wang Apr 11 '19 at 20:25
  • @Somos I still do not quite get it. It appears my classmate substituted in $x^2$ as opposed to squaring. That I understand. It changes the sign a bit. But this still does not explain the discrepancy between my answer $-1 + 2x+3x^2....$ and his answer. – Andrew Wang Apr 11 '19 at 20:44
  • Do see the difference between $(1+1)^2 = 2^2 = 4$ and $(1+1^2) = (1+1) = 2$? – Somos Apr 11 '19 at 20:52
  • @Somos Yes, I do. – Andrew Wang Apr 11 '19 at 20:55

1 Answers1

1

Further to the comments, let's prove without calculus that $(1-x+x^2-\cdots)^2=1-2x+3x^2-4x^3+\cdots$ as expected. The left-hand side's $x^n$ coefficient is $\sum_{j=0}^n(-1)^j(-1)^{n-j}=\sum_j(-1)^n=(n+1)(-1)^n$, as required.

J.G.
  • 115,835