# Binomial expansion for fractional power

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1. Mar 23, 2016

### Paradoxx

1. The problem statement, all variables and given/known data
So, I'm solving a dipole thing and I have these vectors:
|r + d - r'| = (r² + d² - r'²)(1/2)
2. Relevant equations
I want to expand this but I have no idea how! I know I may have an infinite power series, but I may expand at the square terms tops...
Before I needed to do the same with something like (a + b)(1/2) and I got it, I found how to do it. But in this case I can't find a conection...
Can anyone please explain to me how to? Or at least indicate somewhere where I can find it? Because I looked over and couldn't find...

Thanks...

2. Mar 23, 2016

### Ray Vickson

What do you want to expand with respect to? Expand in powers of $r^2$? Expand in powers of $d^2$? Expand in powers of $r'^2$? Expand in something else?

Just expand $(a+x)^{1/2}$ in powers of $x$, then identify $a$ and $x$ for your problem.

3. Mar 23, 2016

### Paradoxx

For example,if I have $(a+b)^n$ , where n can be a fraction...I know I solve like:

$(a+b)^n= b^n +\left( ^n _1 \right)b^{n-1}a +\left( ^n _2 \right)b^{n-2}a^2+...$

If I had $(a-b)^n$ I should alternate the signal + and -...

But in the case $(a+b+c)^n$ or more precisely $(a+b-c)^n$ I don't know how to do it...
I just want to expand like the one above...

4. Mar 23, 2016

### WWGD

(a+b+c)=((a+b)+c) is the original form.

5. Mar 23, 2016

### Ray Vickson

You have still have not answered my original question: expand with respect to what?

You could expand in power of $a$ as
$$(a+b+c)^n = (b+c)^n \sum_{k=0}^{\infty} {n \choose k} \left( \frac{a}{b+c} \right)^k,$$
where
$${n \choose k}= \frac{n(n-1) \cdots (n-k+1)}{k!}$$
for any real $n$ and non-negative integer $k$.

Or, you could try to expand the whole thing in powers like $a^k b^j c^l$, perhaps by re-writing the above as
$$(a+b+c)^n = \sum_{k=0}^{\infty} {n \choose k} a^k (b+c)^{n-k},$$
then expanding each of the $(c+b)^{n-k}$ in powers of $b$ and $c$, and then collecting terms. It would be messy.

Or, you could use the trinomial expansion, which says that
$$(a+b+c)^n = \sum_{k_1,k_2,k_3} c(k_1,k_2,k_3) a^{k_1} b^{k_2} c^{k_3},$$
where
$$c(k_1,k_2,k_3) = {n \choose k_1} {n-k_1 \choose k_2} {n-k_1-k_2 \choose k_3}$$

Last edited: Mar 23, 2016
6. Mar 23, 2016

### vela

Staff Emeritus
What vectors? How are you managing to equate the lefthand side with the righthand side?

7. Mar 23, 2016

### Paradoxx

Sorry, I know I have not been very clear, but I'm trying to understand (and failing) that's why...
Here what I have (in the figure):
I need to expand the first term, conserve only the linear term in "d". I need to get the last equation...but I have no idea about what I'm doing :(

8. Mar 23, 2016

### Ray Vickson

Much better: write $\vec{p} = \frac{1}{2} \vec{d}$ so you have
$$\phi(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \left[ \frac{1}{|\vec{r} - \vec{p}|} - \frac{1}{|\vec{r} + \vec{p}|} \right]$$
Now
$$\frac{1}{|\vec{r} - \vec{p}|}=\frac{1}{r} (1-x_1)^{-1/2}, \;\; x_1 = \frac{2 \vec{r}\cdot \vec{p}}{r^2} - \frac{p^2}{r^2}$$
with as similar expression for $1/|\vec{r} + \vec{p}|$. Now use the binomial expansion for $(1-x_1)^{-1/2}$ in powers of $x_1$, exactly as I had suggested in post #2.

In your case, stopping at the first power ($x_1^1$) will be good enough.

9. Mar 24, 2016

### vela

Staff Emeritus
You have a sign error in your diagram, but I think your expression for the potential is okay for the way I think you're defining $\vec{d}$. Still, you probably should start over and go with Ray's suggestion since $\theta$ is defined in terms of $\vec{r}$ and the dipole axis.

Remember that $\lvert \vec{x} \rvert$ is defined as $\sqrt{\vec{x}\cdot\vec{x}}$. Use that fact and then factor the $r^2$ out of the square root to get the expression Ray wrote above on the righthand side of the equation.

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