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So the task says: Do the Maclaurin expansion of the given term: $$f(x) = \frac{x}{(1-x)^2}$$

what I did was: https://ibb.co/19Ddd8d

The way i tried to solve the task was wrong, I know that i should derive and than multiply with x and that would give me the correct Maclaurin expansion.

But why couldn't I square both sides and then multiply with x instead of deriving and than multiplying with x?

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    $\begingroup$ The second image is incomprehensible. $\endgroup$ – Gae. S. Jul 7 at 15:40
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    $\begingroup$ $(\sum_{m=1}^\infty x^m)^2=\sum_{m=1}^\infty \sum_{n=1}^\infty x^{m+n} \ne \sum_{m=1}^\infty x^{2m}$!!!!! $\endgroup$ – herb steinberg Jul 7 at 15:42
  • $\begingroup$ Especially those $/^2$ and $/\cdot x$ (I guess?) $\endgroup$ – Gae. S. Jul 7 at 15:42
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Perhaps you should sniff around some texts about generating functions to get more comfortable with these calculations. You can certainly square both sides and multiply by $x$. For example:

$$\frac{1}{1-x} = \sum_{m=0}^{\infty} x^m$$

Squaring both sides yields \begin{align} \frac{1}{(1-x)^2} &= \left(\sum_{m=0}^{\infty} x^m\right)^2\\ &= (1+x+x^2+\cdots )(1+x+x^2+\cdots)\\ &= (1+2x+3x^2+4x^3+\cdots)\\ &= \sum_{m=0}^{\infty} (m+1)x^m. \end{align}

Notice that you'd need to calculate the square of the polynomial. Fortunately it is easy enough in this case.

Finally multiplying by $x$ yields \begin{align} \frac{x}{(1-x)^2} &= \sum_{m=0}^{\infty} (m+1)x^{m+1}\\ &= \sum_{m=1}^{\infty} mx^{m} \end{align}

On the other hand, we can take a derivative and be done with it. Taking the derivative of the right hand side is fairly straightforward, as we just take the derivative of every term individually.

\begin{align} \frac{1}{1-x}\frac{d}{dx} &= \sum_{m=0}^{\infty} x^m\frac{d}{dx}\\ \frac{1}{(1-x)^2} &= \sum_{m=0}^{\infty} mx^{m-1}\\ \frac{x}{(1-x)^2} &= \sum_{m=0}^{\infty} mx^{m}\\ \frac{x}{(1-x)^2} &= \sum_{m=1}^{\infty} mx^{m}\\ \end{align}

I forget if one should really take caution while taking the derivative of a sum $\sum_{m=0}^{\infty} x^m$ as the first term without adjusting indices becomes $0(\frac{1}{x})$, but perhaps someone with more experience can speak to this.

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    $\begingroup$ The first term $x^0$ is just the constant $1$, so its derivative is zero. $\endgroup$ – Bungo Jul 7 at 17:39

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