# Applying a Taylor series "with respect to $a/r$" and "around $0$"

My question is what does it mean

applying a Taylor series with respect to something and around a point.

What is the difference?

Please explain it with the following example:

Apply a Taylor series expansion to $$r'$$ with respect to $$a/r$$ around $$0$$. Only until the squared terms (inclusive).

Where $$r'^2 = a^2 + r^2 +2ar\sin(\theta)\cos(\alpha-\phi)$$

• This is a polynomial in $a$, so its Taylor series is fairly straightforward. Commented Jul 29, 2019 at 21:29
• As far as I know, they mean the same thing. Commented Jul 30, 2019 at 1:22
• @nmasanta: In continuation of our previous conversation, if you're editing, don't forget titles. Commented Jul 30, 2019 at 7:05

The "with respect to" specifies the variable whose powers appear in the series. "around" is the point near which you want an approximation. So your answer will look like $$c_o + c_1(a/r - 0) + c_2(a/r - 0)^2 + \text{ higher order terms}.$$

(Of course the zeroes go away. I just put them in to illustrate "around $$0$$".)

• thank you @ethan-bolker for the example! Commented Jul 29, 2019 at 21:35
• by the way, in your answer is it r/a or you meant a/r ? thanks Commented Jul 29, 2019 at 22:08
• @nachofest Yes,. Fixed thanks. Commented Jul 29, 2019 at 22:49

For an easier example, let $$f(x) = \sin(x)$$. The Taylor series expansion of (or, I guess, in your above language 'to') $$f$$ with respect to $$x$$ around $$0$$ is the familar $$f(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$$

So you want to rewrite $$r'$$ in terms of $$a/r$$ and expand around 0.

• thank you @bounceback for your answer, specially for specifying the idea about rewriting it, that was they key point. Commented Jul 29, 2019 at 21:35