# binomial expansion for negative and fractional powers

I understand how a binomial expression can be expanded for positive integer indices by using pascals triangle or combinations to find out the number of ways different terms occur. However, I do not understand why the same logic can be used with negative and fractional powers. Could somebody please provide an intuitive proof or explanation for why the binomial expansion can be applied for negative and fractional indices.

• By indices you mean exponents or numbers inside? – Phicar Apr 12 '17 at 22:33
• I meant exponents as in (a+b)^n where n is an exponent that is negative or or a fraction – Nav Hari Apr 12 '17 at 22:35
• Do you know Taylor expansions? – Phicar Apr 12 '17 at 22:38
• I have not done Taylor expansions yet, does it explain this? – Nav Hari Apr 12 '17 at 23:13
• Yes it does, at least form fractional. For negative is a cauchy multiplication. The answer of David is the way to do it. – Phicar Apr 13 '17 at 5:57

For indices which are not positive integers you look at $(1+x)^a$ for $|x| \lt 1$ and expand as a power series in $x$. When $a$ is a positive integer the coefficient of $x^k$ is $\binom{a}{k}$. This may be written as:
$$P_k(a) = \frac1{k!}a(a-1)...(a-k+1)$$ so that (still with $a$ a positive integer) we have the binomial expansion ($P_0(a)=1$): $$(1+x)^a = \sum_{k=0}^a P_k(a)x^k$$ since $P_k(a) = 0$ if $k \gt a$ we may write this as: $$(1+x)^a = \sum_{k=0}^{\infty} P_k(a)x^k$$ and it turns out that this same form can be used for fractional or negative integer values of $a$ for which $P_k(a) \ne 0$ for an infinite sequence of values of $k$.
To see why this should work let us compute: $$(1+x)^{a+1} = (1+x)(1+x)^a$$ if the expansion is valid we require: $$\sum_{k=0}^{\infty} P_k(a+1)x^k = (1+x)\sum_{k=0}^{\infty} P_k(a)x^k$$ or, for $k \gt 0$ $$P_k(a+1) = P_k(a) + P_{k-1}(a)\tag{1}$$ In other words (leaving questions of convergence aside) we want the polynomials $P_k(a)$ to satisfy the same recurrence relation as the binomial coefficients do for $a$ an integer: $$\binom{a+1}{k}=\binom{a}{k}+\binom{a}{k-1}$$ you may find it instructive to prove (1) directly. Or you could note that for each value of $k$ the relation is a polynomial equation in $a$ of degree $k$, which we already know is satisfied for an infinite set of (positive integer $\gt k$) values of $a$. so it must hold identically.