# B Binomial Expansion with Negative/Rational Powers

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1. Sep 5, 2016

### Alettix

Hello!

When studying binominal expansion:
$(a+b)^n = \sum_{k=0}^{n}{{n \choose k}a^{n-k}b^k}$
in high school, we proved this formula with combinatorics considering that "you can choose either a or b each time you multiply with a binom". Probably, this is not a real mathematical proof at all, but at least we developed an understanding about the concept and found it logical.

Now I have often come across the statment that this formula is valid "for any value of n, whether positive, negative, integer or non-integer" and I have also used it when approximating things in physics. However, I lack an understanding of why the formula works for negative and/or non-integer exponents. How is even ${n \choose k}$ defined for non-integer or negative n?

I would be really happy if you could help me understand this, either with a real proof (hopefully not involving maths I haven't learned yet) or with a logic-argument such as that one I encountered in high school (or both!). Thank you in advance! :)

2. Sep 5, 2016

### micromass

Have you encountered Taylor series before?

3. Sep 5, 2016

### Alettix

Not in full depth I belive, but to a basic extent yes.

4. Sep 5, 2016

### micromass

Cool! Can you find the Taylor series of $(1+x)^{1/2}$ for me? Just find the first few terms and see if you notice a pattern.

5. Sep 5, 2016

### Alettix

I believe the Taylor expansion at x = a should be:
$T_a = (1+a)^{1/2} + \frac{x-a}{2(1+x)^{1/2}} - \frac{(x-a)^2}{4(1+x)^{3/2}} = (1+a)^{1/2} + \frac{(1+a)^{1/2}}{2} - \frac{(1+a)^{1/2}}{4}$

I see that the factor $(1+a)^{1/2}$ is repeated, and that the denominators seem to follow $2^k$ (and also change sign?), but I am not yet sure that I have discovered the pattern we are looking for.

6. Sep 5, 2016

### micromass

Oh, but you should do a Taylor expansion around $a=0$. What do you get then? You should get as coefficients a generalization of the binomial coefficient, whatever that generalization is.

7. Sep 5, 2016

### Alettix

Oh wait, in my previous post, what comes after the second equivalence sign is totally wrong!

So if I continue from the first equality with a = 0, we get:
$(1+x)^{1/2} = 1 + \frac{x}{2} - frac{x^2}{4} = 1 + \frac{1}{2} x +\frac{1}{2} (\frac{1}{2}-1) x^2$
which is what one might expect from the binomial theorem if we write it like:
$(a+b)^n = a^n + na^{n-1}*b + \frac{n(n-1)}{2!} a^{n-2}*b^2+ ... +b^n$
instead of using the sum and ${n \choose k}$ notation.

8. Sep 5, 2016

### Alettix

However, this implies futher questions! Because if I am not totally wrong, we will never reach $b^n$ if n is not a positive integer, which means that the binomial expansion is an infinite series and more of an approximation and not an exact formula if n is negative and/or rational. Am right? And if it is just an approximation, for which values of x (or a and b) is it valid? I remember from physics that $(1+x)^n = 1+nx$ require x<<1, but how does this change when more terms are included?

Secondly, how can one genearilize what we have just done? Should I try to find the MacLaurin series for $(a+b)^n$?

9. Sep 5, 2016

### micromass

OK, very good questions. It's true that it's an infinite series, and thus you can never write down an exact formula with finitely many terms.

For the right formula, we put $\binom{\alpha}{k} = \frac{\alpha(\alpha-1)(\alpha-2)...(\alpha-k+1)}{k!}$. Then we have
$$(1+x)^\alpha = 1 + \binom{\alpha}{1} x + \binom{\alpha}{2}x^2 + ... + \binom{\alpha}{n} x^n + ...$$

There are some constraints on this formula. First for all, we want $\alpha\in \mathbb{C}$, so $\alpha$ can be any complex number. For $x$ however, the formula only holds in general for $x\in (-1,1)$ unless $\alpha$ is a positive integer (in which case it holds for any $x$). More comprehensive conditions on $x$ can be found on wiki: https://en.wikipedia.org/wiki/Binomial_series

Thus in general, we do not expand $(a+b)^\alpha$, but only $(1+x)^\alpha$. If you want to expand $(a+b)^\alpha$, then you'll have to do it as follows:
$$(a+b)^\alpha = a^\alpha ( 1 + (b/a))^\alpha =a^\alpha \sum \binom{\alpha}{k}\left(\frac{b}{a}\right)^k = \sum \binom{\alpha}{k} b^k a^{\alpha-k}$$
PROVIDED THAT $b/a\in (-1,1)$. If it is not, then we'll have to look at $a/b$.

10. Sep 6, 2016

### Alettix

Hmm, I am really sorry but I am not entierly sure I can follow you. Where does $x\in (-1,1)$ and $b/a\in (-1,1)$ come from? If we have $x=1$ that's just $2^\alpha$ and if $x=-1$ it's $0^\alpha$, which doesn't make sense really so I must have missunderstood you.

Also, Is ${a \choose k}$ defined for negative and/or non-integer $a$? I cannot imaging choosing 5 elements out of -4, but maybe I am just stuck in the "combinatorics box".

11. Sep 6, 2016

### PeroK

It's best, perhaps, to take an example:

$(1 + x)^{-1} = 1 -x + x^2 -x^3 \dots$

But, this is actually a geometric series, so you can see quite easily that it converges only when $|x| < 1$

This means the Taylor/Binomial expansion is only valid for a limited range of $x \in (-1 , 1)$. This is called the Radius of Convergence (R), and applies generally to Taylor series. In some cases $R = +\infty$, which means the expansion is valid for all $x$. In other cases, like this one, R is finite.

If you start with $(a + b)^{-1}$ and if, say, $a > b$, then $b/a < 1$ so:

$(a + b)^{-1} = a^{-1}(1 + b/a)^{-1} = a^{-1}(1 - (b/a) + (b/a)^2 - (b/a)^3 \dots$

And, if $a < b$, then you would do:

$(a + b)^{-1} = b^{-1}(1 + a/b)^{-1} = b^{-1}(1 - (a/b) + (a/b)^2 - (a/b)^3 \dots$

In order to get a convergent expansion in each case.

12. Sep 6, 2016

### PeroK

Yes, although I don't know if anyone would still view this as "combinatorial". It's just a generalisation, which has the same recursive pattern as the normal binomial coefficients.

13. Sep 6, 2016

### Alettix

Oh, now I understand! So $x \in (-1 , 1)$ doesn't mean that x should be either $1$ or $-1$, but rather that $1 \leq x \leq -1$. Otherwise the series would diverge and could not be approximated as a sum of low degree terms.

14. Sep 6, 2016

### Alettix

However, Micromas said earlier that we do not generally expand $(a+b)^n$, but rewrite it as $a^n(1+b/a)^n$. Does this mean that the formula is not valid for the first case, or simply that we like to keep the structure of the second case?

15. Sep 6, 2016

### Staff: Mentor

$x \in (-1,1)$ actually means $-1 < x < 1$ and to obtain this you could either write $a^n (1+b/a)^n$ or $b^n (1+ a/b)^n$ because one of the two quotients will be in this range (unless $a=b$).

16. Sep 6, 2016

### PeroK

I must confess, I've always reduced it to $(1+x)^r$ without thinking. Good question! Maybe @fresh_42 knows the answer?

17. Sep 6, 2016

### Staff: Mentor

Sorry, I didn't get this. I only wanted to explain why we have to distinguish between $a/b$ and $b/a$.

Of course there is no need to rewrite $(a+b)^n$ for natural $n$. In post #9 in which it has been used, $n=\alpha$ has been complex. But in any case $(1+x)^{\alpha}$ will do.

18. Sep 6, 2016

### PeroK

If we take the same example:

$(a + b)^{-1} = a^{-1} - a^{-2}b + a^{-3}b^2 - a^{-4}b^3 \dots = a^{-1}(1 - (b/a) + (b/a)^2 - (b/a)^3 \dots)$

So, that's just another way of doing it when $|a| > |b|$. I didn't know you could do that!

19. Sep 6, 2016

### Staff: Mentor

I think I understood you now. Indeed a good question. $a,b$ real could help.

Edit: IMO the basic question is, when does the closed formula $(a+b)^{\alpha}$ equals the expansion into a series? Therefore the quotient $|a / b|$ has to be less than $1$ (or vice versa). The closed formulas might be equal but this doesn't guarantee their formal expansions into a series are.

Last edited: Sep 6, 2016
20. Sep 7, 2016

### Alettix

So, if I understand you right the equality
$\sum_{k=0}^{\alpha}{{\alpha \choose k}a^{\alpha-k}b^k} =a^\alpha\sum_{k=0}^{\alpha}{{\alpha \choose k}1^{\alpha-k}(b/a)^k}$
does not hold for any value of $\alpha$, but rather positive integers only?