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Let

$f:$ ($-1 \over 2$, $1 \over 2$) $\rightarrow \Bbb R$,

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

be a function.

I am searching for a power series representation of this function.

I solved this the following way:

We know that

$1 \over {1 - x}$ $= \sum_{n=0}^\infty x^n.$

Therefore,

$1 \over {1 + 4x^2}$ $=$ $1 \over {1 - (-4x^2)}$ $=$ $\sum_{n=0}^\infty (-4x^2)^n$ $=$ $\sum_{n=0}^\infty (-1)^n 4x^n 2^n$

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user351609
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1 Answers1

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Almost:

$$(-4x^2)^n=(-(2x)^2)^n=(-1)^n(2x)^{2n}=(-1)^{n}x^{2n}2^{2n}=(-1)^nx^{2n}4^n$$

Ahmed S. Attaalla
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  • Thanks! What mathematical law tells me that I have to pull out $(-4)^n$ first and not, like I did, $(-1)^n$? – user351609 Jul 02 '16 at 13:45
  • Oh, and you changed the $4$ into a $2$? – user351609 Jul 02 '16 at 13:47
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    @user351609 He was just saying that since $4$ equals $2^{2}$, you can write $4x^{2}$ as $2^{2}x^{2}$, and now since they have the same power you can "pull the power out" as $(2x)^{2}$. – layman Jul 02 '16 at 13:50
  • @user351609 Also, to your first question, you don't have to pull out the $(-4)^{n}$ first, but to skip writing this down is to skip writing a step down. And when you don't write every step, you can easily make mistakes. We know $(-4)^{n}$ is $(-1 \cdot 4)^{n}$, which gives us $(-1)^{n}4^{n}$. Finally, in your original work, you forgot that $(x^{2})^{n} = x^{2n}$. When you raise something with a power to another power, you can multiply the powers together. – layman Jul 02 '16 at 13:52
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    @user351609: This is a good answer, but it isn't quite correct about one thing. The power series converges wherever $$\bigl\lvert-4x^2\bigr\rvert<1,$$ or equivalently, $4x^2<1.$ One can show the interval of convergence to be precisely the domain of $f$. – Cameron Buie Jul 02 '16 at 14:03
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    Thanks that's what I was struggling in and so I kept editing my answer @CameronBuie – Ahmed S. Attaalla Jul 02 '16 at 14:05