# How toexpress $V=\frac{kq}{x-a}-\frac{kq}{x+a}$ in terms of $k,q,x,u$ in Taylor Series for the following condition?

The question calls $u=\frac{a}{x}$ and $u$ is the variable.

So for Taylor Series, we express it in $f(x)=\sum^{\infty}_{k=0}\frac{f^k(0)}{k!}x^k$

However, one hint says all we need is geometric series so we don't need to take derivatives. What does that mean.

HINT:

$$\frac1{x-a}=-\frac1a\cdot\frac1{1-\left(\frac{x}a\right)}=-\frac1a\sum_{n\ge 0}\left(\frac{x}a\right)^n\;,$$

and

$$x+a=a-(-x)=a\left(1-\left(-\frac{x}a\right)\right)\;.$$

• Thanks a lot, that refreshed my memory Apr 3 '16 at 19:46
• @CoolKid: You’re welcome. Apr 3 '16 at 19:46
• But since $u=\frac{a}{x}$, so $a=ux$. Why do you put $\frac{1}{a}$ outside the summation? Apr 3 '16 at 20:06
• @CoolKid: Is $u=\frac{a}x$, or is $u=\frac{x}a$? Apr 3 '16 at 20:08
• @CoolKid: So far as I can tell from what you’ve written $a$ is a constant. Whether your goal is a power series in $x$ or one in $u$, $a$ can be manipulated just as if it were explicitly $2$, or $\pi$, or any other constant. Whether you take the sum and then multiply it by $\frac1a$, or multiply each term by $\frac1a$ and then take the sum, you get the same result. Apr 3 '16 at 20:16