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$$f(x) = \frac{1}{4+x^{2}}$$

What I did was notice that this is similar to the geometric series: $$f(x) = \frac{1}{1-x} = 1 + x^{2} + x^{3} + x^{4} + ....$$

So I altered my original function from: $$f(x) = \frac{1}{4 + x^{2}}$$ to: $$f(x) = \frac{1}{4(1 - (-\frac{1}{4}x^{2}))}$$ And so it would now look like: $$f(x) = \frac{1}{4(1 - \left(-\frac{1}{4}x^{2}\right))} = 1 + \left(\frac{-1}{4}x^{2}\right) + \left(\frac{-1}{4}x^{2}\right)^{2} + \left(\frac{-1}{4}x^{2}\right)^{3} + \left(\frac{-1}{4}x^{2}\right)^{4} + .... $$ now to account for the 4 out front: $$ = 4\bigg(1 + \left(\frac{-1}{4}x^{2}\right) + \left(\frac{-1}{4}x^{2}\right)^{2} + \left(\frac{-1}{4}x^{2}\right))^{3} + \left(\frac{-1}{4}x^{2}\right))^{4} + ....\bigg) $$

Why is this wrong? The correct answer is suppose to be: $$= \frac{1}{4}-\frac{1}{16}x^2+\frac{1}{64}x^4-\frac{1}{256}x^6+\frac{1}{1024}x^8+\ldots $$

Thank you

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  • $\begingroup$ $\sum x^n/n!$ is the exponential series, not the geometric series. $\endgroup$ Commented Apr 25, 2017 at 16:31
  • $\begingroup$ There was no factor $4$ out front, but instead a factor $\frac14$. $\endgroup$ Commented Apr 25, 2017 at 16:32
  • $\begingroup$ @LordSharktheUnknown Thanks! that's what I meant to put, the other computation was the geometric series, just wrote that first part wrong. $\endgroup$
    – yre
    Commented Apr 25, 2017 at 16:33
  • $\begingroup$ You are right in principle. It can be a very useful substitution to stuff things in the geometric series. But important to remember radii of convergence. $\endgroup$ Commented Apr 25, 2017 at 16:35
  • $\begingroup$ @MarcvanLeeuwen I know this is probably super basic algebra that is messing me up, but why is it $\frac{1}{4}$ and not 4? I thought with the geo-series we replaced everything with the "new" x's we got then distribute what's in the denominator. But what we are supposed to do is distribute what's in the numerator and denominator? $\endgroup$
    – yre
    Commented Apr 25, 2017 at 16:36

1 Answer 1

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Everything is perfect, just one thing:

You have

$$ \frac{1}{4} \cdot \frac{1}{1-\left(-\frac{1}{4} x^2\right)} $$ so you need to take the series and multiply by $\frac14$, not $4$

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  • $\begingroup$ Why $\frac{1}{4}$ and not 4? $\endgroup$
    – yre
    Commented Apr 25, 2017 at 16:34
  • $\begingroup$ Because you have $\frac{1}{ab}=\frac{1}{a}\frac{1}{b}$. @yre In this case, $a=4$ and $b=1+x^2/4$. $\endgroup$ Commented Apr 25, 2017 at 16:35
  • $\begingroup$ because you have a $4$ in the denominator, which means you divide by it, not multiply: $ \frac{6}{2\cdot 3} = \frac{1}{2}\cdot \frac{6}{3} $ $\endgroup$ Commented Apr 25, 2017 at 16:37
  • $\begingroup$ @ThomasAndrews Thanks! I think what I must remember is we have to distribute the entire fraction as one unit $\frac{1}{4}$ not just the denominator 4, I don't know why I was thinking it was just the denominator that got distributed across. Thanks! $\endgroup$
    – yre
    Commented Apr 25, 2017 at 16:38

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