2
$\begingroup$

I have been trying to understand why the binomial theorem can work for negative and fractional indices.

I understand that when raising binomials to positive integral indices, each coefficient is simply the number of ways that you can pick each term (e.g. for $(x+y)^5$, if you want to make up $xy^4$ there are 5 brackets from which to pick the $x$, so this term will come up 5 times in the full expansion).

I am not sure if a way to understand the infinite expansion for negative and fractional indices exists, but if it does I would very much like to know! Otherwise, I haven't been able to find a proof that shows that the result of the expansion for positive integer powers is valid for negative or fractional indices. I would be very grateful if someone could point me in the direction of such a proof.

$\endgroup$
  • 1
    $\begingroup$ Hm, you might want to be careful with the negative values, since binomial expansion often doesn't make sense for negative values. See that $$(1+1)^{-2}=1-2+3-4+\dots$$ which doesn't have much meaning here. $\endgroup$ – Simply Beautiful Art Sep 13 '16 at 21:00
  • 1
    $\begingroup$ But when it does converge, binomial expansion makes sense even for complex exponents. $\endgroup$ – Simply Beautiful Art Sep 13 '16 at 21:05
  • 2
    $\begingroup$ Algebraically, one can think of it like this: We can express a binomial coefficient as $\binom{n}{k}=\frac{n!}{k!(n-k)!}=\frac{n(n-1)\cdots (n-k+1)}{k!}$. This last expression still makes sense if $n$ is no longer a positive integer, and that's what allows the binomial expansion to still be computed. (Why that expansion still is valid for $(1+x)^{n}$ when $n$ isn't a positive integer is a much harder question.) $\endgroup$ – Semiclassical Sep 13 '16 at 21:17
  • 1
    $\begingroup$ @Semiclassical that is the question for me! It seems too me that we find a formula for computing combinations- this formula came from an idea very much grounded in the real world (how many ways you can make a term) and yet then we try out the formula for numbers which no longer have a physical meaning, and the formula still works in calculating things in the real world... It frustrates me that i can't see why. $\endgroup$ – 21joanna12 Sep 14 '16 at 12:10
  • 2
    $\begingroup$ If one wants some intuition, though, a formal proof for $(1+x)^{1/2}$ is possible. Assume $(1+x)^{1/2}=a_0+a_1x+\cdots$ for some infinite set of coefficients $a_k$ and square both sides to get $$1+x=(a_0+a_1x+\cdots)^2=a_0^2+(2a_0 a_1)x+\cdots.$$ Matching coefficients on both sides gives an infinite set of equations to solve for $(a_0,a_1,\cdots)$. One can show that $a_n=\binom{1/2}{n}$ solves these equations. So everything works out formally, though this isn't a rigorous proof of convergence. $\endgroup$ – Semiclassical Sep 14 '16 at 13:55
2
$\begingroup$

This is probably the wrong proof for you, but I will post it anyways. (requires calculus)

Note that $f(x)=(a+x)^n$ is an analytic function in $x$ for arbitrary $a,n$ since on its own, it is a power series with one term.

If it is an analytic function, then it should follow Taylor's theorem.

Now, if we take the expansion around $x=0$, we get

$$(a+x)^n=a^n+na^{n-1}x+\frac{n(n+1)}2a^{n-2}x^2+\dots$$

Since $f(0)=a^n$, $f'(0)=na^{n-1}$, $\dots f^{(k)}(0)=n(n+1)(n+2)\dots(n+k-1)a^{n-k}$

or

$$(a+x)^n=\sum_{k=0}^\infty\frac{n(n+1)(n+2)\dots(n+k-1)}{k!}a^{n-k}x^n$$

$$(a+x)^n=\sum_{k=0}^\infty\binom nka^{n-k}x^n$$

where $f'(x)$ is the first derivative of $f(x)$, $f''(x)$ the second derivative, etc. $f^{(k)}(x)$ is the $k$th derivative of $f(x)$.

$\endgroup$
1
$\begingroup$

I don't know if this is what you are looking for but I feel like it warrants a mention, if you do not understand binomial expansion for indices say $n \lt 0$ then it helps to think of it in this way. Note, this is only for $n \in \mathbb{I}$. Anyway I am assuming you know that we define an infinite geometric progression $A_n$ like this, $$a + ar + ar^2 + ar^3 + \ldots + ar^{n} + \ldots \infty$$ Now if we define the sum of this progression $$\sum ^{\infty} _ 1 A_n = S$$ then we can say that $$\sum ^{\infty} _ 1 A_n = S = \dfrac{a}{1-r}$$ Now, let us say the first term $a$ is $1$ and the common ratio $r=x$ then we get, the series as $$A_x = 1 + x + x^2 + x^3 + \ldots + x^{n} + \ldots \infty$$ and we can define the sum as, $$\sum ^{\infty} _ 1 A_x = S = \dfrac{1}{1-x}$$ which can be written as, $$\left(1-x\right)^{-1}$$ so, you can now see what really goes on using an $\infty$ geometric progression $A_x$.
For higher powers obviously, you can define it as $$\left(1 + x + x^2 + x^3 + \ldots + x^{n} + \ldots \infty \right) \cdot \left(1 + x + x^2 + x^3 + \ldots + x^{n} + \ldots \infty \right)= \left(1-x\right)^{-2}$$ and for a concise formula you can refer to the Taylor series, which has been discussed in the previous question. Obviously if we wanted to do this for any expansion we could define it using a geometric progression by changing the common ratio. For the fractional indices, $n \in \mathbb{R}$ you can refer to the proof given above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.