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I was wondering if there was any particular reason that power series are centred around 0.

$$\displaystyle f(x)=\sum_{k=0}^{\infty}c_{k}x^{k}\qquad \text{ vs }\qquad\displaystyle f(x)=\sum_{k=0}^{\infty}c_{k}(x-a)^{k}$$

when using them to solve differential equations. I've tried looking online for a reason but so far have had no luck.

This seems like a rather arbitrary choice that only conveniently simplifies our computation - eg. what if the function is not well-defined at 0 or deviates significantly away from 0?

Apologies if this seems like a trivial question - I have not done a differential equations course yet that involves power series. Solutions involving only first-year university maths would be much appreciated.

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    $\begingroup$ Do you have some evidence to support your claim that power series used to solve differential equations are always centred around $0$? $\endgroup$ – bof Jul 12 '18 at 3:46
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    $\begingroup$ If $f(0)$ causes problems and you want to develop around $a$, you could as well consider the function $g(x)=f(x+a)$ and develop that around $0$ $\endgroup$ – Hagen von Eitzen Jul 12 '18 at 3:47
  • $\begingroup$ It's not necessary that the power series must converge in an open ball centered at zero. There are various power series that converges outside of that. For example consider this series $s(z)=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^z}$ (en.wikipedia.org/wiki/Riemann_zeta_function) this series converges for all $z$ with real part $> 1$. $\endgroup$ – Sujit Bhattacharyya Jul 12 '18 at 4:11
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    $\begingroup$ If the initial values are given at $0$, then expanding at $0$ makes a lot of sense; for any other $a$, even finding $f(a)$ will be potentially difficult and messy. Similarly, if the initial values are given at $a \neq 0$, then it would be silly to expand at $0$. $\endgroup$ – Theo Bendit Jul 12 '18 at 6:20
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    $\begingroup$ It's just a convention. You can center it around any point. It just so happens that initial conditions are usually given at $0$ $\endgroup$ – Dylan Jul 12 '18 at 16:44
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Imagine you have the following differential equation: $$\frac{d^2y}{dx^2}+\frac{dy}{dx}+x+c=0 \tag{*}$$ Subjected to some boundary conditions.

You can say that your solution $y$ may be expanded as a polynomial series: $$y=\sum_k^N c_k(x-x_0)^k$$ With $c_k$ and $x_0$ constants.

since the solution of the problem $(*)$ is unique, the polynomial $(**)$ is also unique, and you can solve for $c_k$ or for other coefficients: $$y=\sum_k^N c_k(x-x_0)^k=\sum_k^N \hat{c}_kx^k$$ where there is a correspondence between coefficients: $$c_N=\hat{c}_N, \quad c_{N-1}=c_{N-1}-c_Nx_0,...$$

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Some choices in mathematics are selected because they reflect an intrinsic characteristic, whilst others are chosen because they are very convenient. As you state, although real life applications may require a power series expansion around a point not equal to zero, you should not discount how much it can simplify/speed computation to focus on expansions around zero. Also do not discount the general rule that the more compact an expression is, the more likely it can be written, read, understood and digested more readily, and patterns discerned more easily.

Furthermore, what happens very often in maths, is that if one finds that the point of interest is at $x=a$, then for purposes of clarity (to the writer and readers!) it is often useful to define a new variable, say $u$, where $u=x-a$, then do all the complex calculations, and then at the end transform back via $x=u+a$.

From a science perspective, there is often freedom to set the measurement's zero point (eg., displacement of object from equilibrium). In most cases, it is set to coincide with region of most interest. Thus, it follows that a power series expansion around zero is a natural consequence of choosing where how you recorded your measurements.

Hope that helps!

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  • $\begingroup$ Your assertion that "power series are centered around 0" is simply not true in general. If we had, for example, the differential equation $(x- 1)^2\frac{d^2y}{dx^2}+ 3x^2\frac{dy}{dx}+ 4x^3y= 0$ with initial condition y(1)= 3, y'(1)= 4 then, because the initial condition is at x= 1, we would write a series solution as $\sum a_n(x- 1)^n$. More generally, given an initial condition at x= b, we would use $\sum a_n (x- b)^n$. $\endgroup$ – user247327 Jul 12 '18 at 12:13
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Power series do not have to be centred at $0$. $\ln(x)$ is not defined at $x=0$, but it has a nice power series expansion around $x=1$:

$\ln(x) = \sum_{n=1}^{\infty}(-1)^{n-1}\frac{(x-1)^n}{n}$

If we differentiate this formally term by term we get:

$\frac{d(\ln(x))}{dx} = \sum_{n=1}^{\infty}(-1)^{n-1}(x-1)^{n-1} = \sum_{n=0}^{\infty}(1-x)^{n} = \frac{1}{1-(1-x)}=\frac{1}{x}$

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