We know that the binomial theorem and expansion extends to powers which are non-integers.

For integer powers the expansion can be proven easily as the expansion is finite. However what is the proof that the expansion also holds for fractional powers?

A simple an intuitive approach would be appreciated.

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    $\begingroup$ You know that this extension makes you cross the boundary between algebra (without topology) to analysis (with topology creeping into the scene) just because binomial theorem with, for example, exponent $1/3$ means expanding $(1+x)^{1/3}=1+(1/3)x+...$ into a series, and there are convergence issues for the proof (radius of convergence= ?). With a pseudo like yours, this should sound clear, isn't it ? $\endgroup$ – Jean Marie Nov 3 '16 at 9:13
  • $\begingroup$ Hope my answer was helpful:) $\endgroup$ – SirXYZ Nov 3 '16 at 19:08
  • $\begingroup$ Is the Taylor Series proof too sophisticated? $\endgroup$ – Cheerful Parsnip Jun 6 '18 at 0:29
  • $\begingroup$ @CheerfulParsnip - is there a proof using a similar approach as that for integer powers? $\endgroup$ – hypergeometric Jun 6 '18 at 15:13

You could calculate, for example, $(1+x)^{1/2}=a_0+a_1x+a_2x^2+\cdots$ by squaring both sides and comparing coefficients. For example we can get the first three coefficients by ignoring all degree $3$ terms and higher: $$1+x=a_0^2+2a_0a_1x+2a_0a_2x^2+a_1^2x^2+\cdots$$

From here we can conclude that $a_0=\pm1$ (we'll take $+1$ to match what happens when $x=0$). Then comparing coefficients of $x$ we have $2a_1=1$, so $a_1=1/2$. Finally, comparing coefficients of $x^2$, we have $2a_0a_2+a_1^2=0$, so $2a_2+1/4=0$ and $a_2=-1/8$.

You can definitely get as many coefficients as you want this way, and I trust that you can even derive the binomial coefficient formula. However, this is not any easier than the Taylor series, where you take $(1+x)^{1/2}=a_0+a_1x+a_2x^{2}+\cdots$ and find the coefficients by saying the $n$th derivatives on both sides have to be equal at $0$.

For example, plugging in $0$ on both sides we conclude $a_0=1$. Calculating the first derivative of both sides, we have $$\frac{1}{2}(x+1)^{-1/2}=a_1+2a_2x+\cdots$$ Plugging in $0$, we get $a_1=1/2$. Taking the derivative one more time, we see $$(-1/2)(1/2)(1+x)^{-3/2}=2a_2+\cdots$$ Plugging in $x=0$, we have $(-1/2)(1/2)=2a_2$, or $a_2=-1/8$. The advantage to this way, is that it is much easier to see the pattern of coefficients!

Unfortunately, there is a big hole in both arguments. They will give you what the coefficients have to be, but they won't prove that the series expansion converges in the first place. We started off by assuming you could write $1+x$ as an infinite power series, but there is no guarentee that this exists, and actually it doesn't converge unless $|x|<1$, which we never used. So you need to estimate the error in Taylor's formula to complete the proof rigorously.


Heres a proof:

Let $f(a)$ denote the binomial expansion of $(1+x)^a$ and $f(b)$ denote the same for $(1+x)^b$. Where

$(1+x)^a = 1 + ax.....$

$(1+x)^b = 1 + bx....$

On multiplying the two binomial expansions together....the product will be another series in ascending powers of $x$ and will remain unaltered irrespective of $a$ and $b$.

To determine this invariable form of the product we may give to $a$ and $b$ positive integral values for convenience.



But when a and b are positive integers.... the expansion is

$1 + (a+b)x +.....$

This then is the form of the product of $f(a)×f(b)$ in all cases, whatever the values of a and b be; and in agreement with our previous notation it may be denoted by $f(a+b)$; therefore for all values of $a$ and $b$.



$f(a)×f(b)×f(p) = f(a+b+p)...$


$f(a)×f(b)×f(p)... \text{to k factors} = f(a+b+p.... \text{to k terms}).$

Let each of these quantities a,b,p,.... be equal to $(c/k)$, where c and k are positive integers.


$f(c/k)^k = f(c)$

But since c is a positive integer,


$(1+x)^c = f(c/k)^k$


$(1+x)^{c/k} = f(c/k)$


$f(c/k) = 1 + (c/k)x ....$

Hence we get

$(1+x)^{c/k} = 1 + (c/k)x .....$

This proves the binomial theorem for any positive fractional index.

  • $\begingroup$ Can you please format your equations in TeX? This is difficult to read. $\endgroup$ – Pantelis Sopasakis May 14 at 16:04

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