Binomial Expansion with Negative/Rational Powers

[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 statement 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! :)

 

[micromass]

Have you encountered Taylor series before?

 

[Alettix]

Have you encountered Taylor series before?

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

 

[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.

 

[Alettix]

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.

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.

 

[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.

 

[Alettix]

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.

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.

 

[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 $(a+b)^n$?

 

[micromass]

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 $(a+b)^n$?

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 $a/b$.

 

[Alettix]

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 $a/b$.

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".

 

[PeroK]

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".

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.

 

[PeroK]

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".

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.

 

[Alettix]

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.

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.

 

[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?

 

[fresh_42]

Oh, now I understand! So $x \in (-1 , 1)$ doesn't mean that x should be either $1$ or $−1$, but rather that $1 \leq x \leq -1$.

$x \in (-1,1)$ 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 $a=b$).

 

[PeroK]

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?

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

 

[fresh_42]

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

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

However, Micromas said earlier that we do not generally expand (a+b)n(a+b)n(a+b)^n, but rewrite it as an(1+b/a)nan(1+b/a)na^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?

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.

 

[PeroK]

I must confess, I've always reduced it to $(1+x)^r$ without thinking. Good question!

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!

 

[fresh_42]

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.

 

[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?

 

[micromass]

That is correct.

 

[PeroK]

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?

I still think you are misinterpreting what has been said. The sum itself makes no sense unless $\alpha$ is a positive integer.

1) When $\alpha$ is a positive integer, the binomial expansion has a finite number of terms ($\alpha + 1$). And, because of this, there is no constraint on $a$ and $b$.

2) When $\alpha$ is not a positive integer, the binomial expansion has an infinite number of terms. And, because of this, there is a constraint on $a$ and $b$ in order for the resulting infinite series to converge..

Note that, in general:

$(a + b)^{\alpha} = a^\alpha\sum_{k=0}^{\infty}{{\alpha \choose k}(b/a)^k}$

For any $\alpha$ (assuming $|a| > |b|$)

And, when $\alpha$ is a positive integer, the infinite sum reduces to the finite one - if we take all "invalid" binomials to be 0. By invalid, I mean when $k > \alpha$.

 

[Alettix]

I still think you are misinterpreting what has been said. The sum itself makes no sense unless $\alpha$ is a positive integer.

1) When $\alpha$ is a positive integer, the binomial expansion has a finite number of terms ($\alpha + 1$). And, because of this, there is no constraint on $a$ and $b$.

2) When $\alpha$ is not a positive integer, the binomial expansion has an infinite number of terms. And, because of this, there is a constraint on $a$ and $b$ in order for the resulting infinite series to converge..

Note that, in general:

$(a + b)^{\alpha} = a^\alpha\sum_{k=0}^{\infty}{{\alpha \choose k}(b/a)^k}$

For any $\alpha$ (assuming $|a| > |b|$)

And, when $\alpha$ is a positive integer, the infinite sum reduces to the finite one - if we take all "invalid" binomials to be 0. By invalid, I mean when $k > \alpha$.

Thank you for the summary, I believe I understand. However, my question was just more or less if:
$a^\alpha \cdot ($"binominal expansion of $(1+b/a)^\alpha") = ("$binominal expansion of $(a+b)^\alpha")$
where $a>b$ is a valid equality for all $\alpha$ or not.

 

[micromass]

Thank you for the summary, I believe I understand. However, my question was just more or less if:
$a^\alpha \cdot ($"binominal expansion of $(1+b/a)^\alpha") = ("$binominal expansion of $(a+b)^\alpha")$
where $a>b$ is a valid equality for all $\alpha$ or not.

Yes.

 

[Alettix]

Yes.

Okay!
Big thanks to you all who helped me understand this! :)

 

[micromass]

You know, it's a good exercise to try to prove everything we talked about in this thread!