Generalized Binomial Expansion

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The binomial theorem states that

${(a+b)^n}=\sum\limits_{r=0}^{n} {{n\choose r}a^rb^{n-r}},\quad \quad n\in\mathbb{N}$

where

${n\choose r}=\frac{n!}{r! \left( n - r \right) !}$

This gives the number of ways to select r object from n, disregarding the order of the selected objects. We may call this the choice function. If r>n this would give us 0, since there are 0 ways to select say 10 objects from 5. This enable us to rewrite the theorem as

${(a+b)^n}=\sum\limits_{r=0}^{\infty} {{n\choose r}a^rb^{n-r}},\quad \quad n\in\mathbb{N}$

This is since all r>n will give us terms with the value 0.

Now let us look at the choice function again. We can cancel away the “tail” of n! against (n–r)! to get

${n\choose r}= \frac{{n \cdot (n - 1) \cdot ... \cdot (n - r + 1)}}{{r \cdot (r - 1) \cdot ... \cdot 1}}$

I.e. we keep as many factors in the numerator as in the denominator. So we have, for example, that

${10\choose 4}= \frac{{10 \cdot 9 \cdot 8 \cdot 7}}{{4 \cdot 3 \cdot 2 \cdot 1}}= 210$

This we can easily get with a few cancellations.

OK, so our next step is starting to think a bit outside the box. We will do this in a rather experimental way to later prove it more formally using series expansion. Anyhow, in the binomial theorem, the value of n must be a natural number, but what happens if it is any real number? Let us just try to reshape the binomial theorem to fit this new situation. It will be a bit like trying to fit a square peg in a round hole, but anyhow… Let us see where this will lead us. We get

${(a+b)^c}=\sum\limits_{r=0}^{\infty} {{c\choose r}a^rb^{n-r}},\quad \quad n\in\mathbb{R}$

So what could the choice function mean now? What could choose 3 elements from 3.6 mean? Let us forget that interpretation and just use our formula for calculating the choice function we devised earlier. We will then get

${3.6\choose3}= \frac{{3.6 \cdot 2.6 \cdot 1.6}}{{ 3 \cdot 2 \cdot 1}}= 2.469$

OK, could this be useful in any way whatsoever? Let us use this to try to calculate say

${1.2^{1.3}} = {\left(1 + 0.2\right)^{1.3}}$

We get

${(1 + 0.2)^{1.3}} = {(0.2 + 1)^{1.3}} = \sum\limits_{r = 0}^\infty { {1.3\choose r}{{0.2}^r} \cdot {1^{c - r}}}$

Expanding this we get

${ \sum\limits_{r = 0}^\infty { {1.3\choose r}{{0.2}^r} }} = {1.3\choose 0}{{0.2}^0} + {1.3\choose 1}{{0.2}^1} + {1.3\choose 2}{{0.2}^2} +{1.3\choose 3}{{0.2}^3 + ... }$

$1 + \frac{{1.3}}{1}{0.2^1} + \frac{{1.3 \cdot 0.3}}{{1 \cdot 2}}{0.2^2} + \frac{{1.3 \cdot 0.3 \cdot ( - 0.7)}}{{1 \cdot 2}}{0.2^3} + ...$

After 15 terms we get 1.26746396212711 which is 1.2^1.3 correct to all shown decimals.

So, amazingly enough, this works. You may download this (PowerBin) Excel file where you can try it with various values.

You can change the value of p (the base) and c (the exponent) and then you will, hopefully, get the value of p^c calculated. It will work for 0<p<2 and c>0. In the cell to the right of the cell with the text “p^c=” the value is calculated using Excel’s built-in function, and in the column “Partial sum” the successive partial sums are calculated.

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Last modified: Dec 28, 2023 @ 14:00