It was not until recently (why don't they teach it in secondary school?) that I've come across the Generalised Binomial Theorem, which from what I can tell is basically the same as the regular Binomial Theorem, except that the finite sum is replace by an infinite series:

$$ (x+y)^n=\sum^{n}_{r=0}\binom{n}{r}x^{n-r}y^r=\sum^{\infty}_{r=0}\binom{n}{r}x^{n-r}y^r $$

Unfortunately, I wasn't able to find any clear explanation of how to get from the regular theorem to the generalised one, the only proof I found being based on some obscure mathematics, while my math book entirely skips the explanation.

Hence, my question is how do you prove the generalised theorem by deriving it from the regular theorem or otherwise?

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    $\begingroup$ Set $x = 1$ and expand in a Taylor series in $y$ by differentiating repeatedly. (There is also a proof which proceeds by deriving it from the ordinary binomial theorem but it works formally and is a bit hard to explain unless you are very comfortable with formal power series.) $\endgroup$ – Qiaochu Yuan Apr 23 '12 at 17:15
  • $\begingroup$ I find it so unappropriate to call any mathematics obscure. $\endgroup$ – Pedro Tamaroff Apr 23 '12 at 17:49
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    $\begingroup$ The series version of the Binomial Theorem that you quote here is valid only when $|y| < |x|$ or $n$ is a nonnegative integer. Best to do as @Qiaochu does and restrict it to the case $x=1$, which has the additional advantage that the (presumably nonintegral) powers of $x$ all evaluate to $1$. $\endgroup$ – Lubin Apr 23 '12 at 17:59
  • $\begingroup$ possible duplicate of Generalized binomial theorem $\endgroup$ – Pedro Tamaroff Apr 23 '12 at 18:06
  • $\begingroup$ "why don't they teach it in secondary school?" - That's what math.SE is for. $\endgroup$ – Mitch Jun 6 '14 at 14:39

Here is another approach: The formula $$(1+x)^\alpha=\sum_{k=0}^\infty {\alpha\choose k}\, x^k\qquad(1)$$ is true for $\alpha\in{\mathbb N}$; so maybe its true for arbitrary $\alpha$. To find out fix an $\alpha\in{\mathbb R}$ and consider the function $$f(x):=\sum_{k=0}^\infty {\alpha\choose k}\, x^k\qquad(|x|<1)\ .$$ Using termwise differentiation one obtains $$f'(x)=\sum_{k=1}^\infty {\alpha\choose k}\,k\, x^{k-1}=\sum_{k'=0}^\infty {\alpha\choose k'+1}\,(k'+1)\, x^{k'}=\sum_{k=0}^\infty {\alpha\choose k}\,(\alpha -k)\, x^k\ .$$ Therefore, using the third and the first expression for $f'(x)$, we get $$(1+x)f'(x)=\sum_{k=0}^\infty {\alpha\choose k}\,(\alpha -k)\, x^k+\sum_{k=0}^\infty {\alpha\choose k}\,k\, x^k=\alpha f(x)\ ,$$ or $$(1+x)f'(x)-\alpha f(x)\equiv0\ .$$ It follows that ${d\over dx}\bigl((1+x)^{-\alpha} f(x)\bigr)\equiv0$, or $(1+x)^{-\alpha} f(x)={\rm const.}$, and as $f(0)=1$ one concludes that $f(x)\equiv (1+x)^\alpha$; whence $(1)$ is indeed true for $|x|<1$.

  • $\begingroup$ How do you go from $(1+x)f'(x)-\alpha f(x)\equiv0\,$ to $\,(1+x)^{-\alpha}f(x)\equiv1$? $\endgroup$ – Milosz Wielondek Apr 24 '12 at 9:01
  • $\begingroup$ Again, could you expand on how one easily checks that $(1+x)f'(x)-\alpha f(x)\equiv0$? I just can't arrive at that without assuming that $f(x)=(1+x)^a$ from the start. Thanks. $\endgroup$ – Milosz Wielondek Apr 27 '12 at 0:06
  • $\begingroup$ @Milosz: See my edit. $\endgroup$ – Christian Blatter Apr 27 '12 at 17:45

In addition (and for future reference) to everybody else's answers I have found the following to be true:

$$ \begin{align} (x+y)^n&=\sum^{n}_{r=0}\binom{n}{r}x^{n-r}y^r &\qquad(1)\\ &=\sum^{\infty}_{r=0}\binom{n}{r}x^{n-r}y^r &\qquad(2) \end{align} $$

  • Let $x,y\in\mathbb{C}\setminus\{0\}$. Eq. $(1)$ and $(2)$ are both true if and only if $n\in\mathbb N$.
  • For $n\in\mathbb C$ and for $(2)$ to hold, the values of $x$ and $y$ must satisfy $|x|>|y|$.

A quick look at the case $y=1$ tells us that: $$ \begin{align} (1+x)^n&=\sum^{\infty}_{r=0}\binom{n}{r}x^r &\qquad(3)\\ &=\sum^{\infty}_{r=0}\binom{n}{r}x^{n-r} &\qquad(4) \end{align} $$

  • When $n\in\mathbb{C}$ and $|x|<1$ we'd use $(3)$
  • When $n\in\mathbb{C}$ and $|x|>1$ we'd use $(4)$
  • When $n\in\mathbb{N}$ or $|x|=1$ either equation works

In case you don't know the Taylor-expansion yet:
1) For $n \in \mathbb{N}$ and $n < m$ we have $\binom{n}{m}$ = 0. Therefore further summation does not change the result.
2) For $n$ a fraction like -1/2 (and $y = 1, |x| < 1$) you can use the sum of the geometric series to get an infinite series $$ \sqrt{\dfrac{1}{1+x}}\sqrt{\dfrac{1}{1+x}} = 1 - 1x^1 + 1x^2 +- ... = $$ To solve this, put $$ \sqrt{\dfrac{1}{1+x}}= a_0 + a_1x^1 + a_2x^2 + ... $$ and multiply this with itself, ordering the result by exponents of $x$ and comparing with the coefficients 1 and -1, respectively, of the upper series.

  • $\begingroup$ I suppose you mean 2) For $n$ to be a fraction like ${\bf -}1/2$? $\endgroup$ – Milosz Wielondek Apr 23 '12 at 21:16
  • $\begingroup$ Of course you are right. Thanks $\endgroup$ – user29503 Apr 24 '12 at 6:57

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