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I've always been taught how to perform a taylor series or maclaurin series expansion, i've also had to remember anyone.. but what about if i just have a series. What tips or tricks do you suggest could be used to identify it an expansion from its series.. if that makes any sense

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  • $\begingroup$ Is your question: given a series how to know to which function does it belong? $\endgroup$ – caverac May 5 '17 at 0:02
  • $\begingroup$ yes thankyou, that's what i meant $\endgroup$ – Numbers1234 May 5 '17 at 0:33
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Look at the general form of a Taylor Series - $$ \sum \frac{(x-\alpha)^n}{n!} f^{(n)}(\alpha) $$

First identify the powers of $x$. If there is no $(x-\alpha)^n$ (note: $\alpha$ could be $0$), then it won't be a Taylor series and can be ruled out.

The rest is tricky. The $n!$ may not always be present, in the case that the n-th derivative of $f$ has a factor of $n!$. It is very difficult to determine what the n-th derivative at a point is so we need a better approach...

Better Approach

A better approach would be to look for transformations of well-known series. For example, we know that $ e^x = \sum \frac{x^n}{n!} $, so the series

$$ \sum \frac{(x-2)^{2n}}{xn!} = \frac{e^{(x-2)^2}}{x} $$

This equivalence results from a substitution of $x$ to $(x-2)^2$ and multiplying the series by $\frac{1}{x}$.

Other well-known series: $\sin(x)=\sum \frac{(-1)^nx^{2n+1}}{(2n+1)!}$, $\cos(x) = \sum \frac{(-1)^n x^{2n}}{(2n)!}$, $\frac{1}{1-x} = \sum x^n \, \forall x \in (-1,1)$.

For another example look at my post here, where it is shown that $x^3 \sum \frac{(-x^4)^k}{(4k+3)(2k+1)!}$ is the series for $\int \sin(x^2) \mathrm{d}x$.

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