# Fractional Binomial Theorem

The binomial theorem for integer exponents can be generalized to fractional exponents. The associated Maclaurin series give rise to some interesting identities (including generating functions) and other applications in calculus.

\( f(x) = \sqrt{1+x}=(1+x)^{1/2} \) is not a polynomial. While positive integer powers of \( 1+x \) can be expanded into polynomials (e.g. \( (1+x)^3 = 1+3x+3x^2+x^3\)), this does not make \( f(x) \) a polynomial, so there cannot be a finite sum of monomial terms that equals \( f(x) \). But there is a way to recover the same type of expansion if infinite sums are allowed.

As a first approximation, since \( f'(0) = 1/2 \), the tangent line to \( \sqrt{1+x} \) at \( x=0 \) is \( y = 1+\frac12x\). So for \( x \) small, \[ \sqrt{1+x} \approx 1+\frac12 x. \] This approximation is already quite useful, but it is possible to approximate this function more carefully using series.

Expand \( \sqrt{1+x} \) as a Maclaurin series.

Solution:The power rule in calculus can be generalized to fractional exponents using the chain rule: the derivative of \( x^{p/q} \) is \( \frac{p}{q}x^{p/q-1} \).Now let \( f(x) = \sqrt{1+x}. \) Then \[ \begin{align} f(x) = (1+x)^{1/2} &\Rightarrow f(0) = 1 \\ f'(x) = \frac12(1+x)^{-1/2} &\Rightarrow f'(0) = \frac12 \\ f''(x) = \frac12 \cdot -\frac12(1+x)^{-3/2} &\Rightarrow f''(0) = \frac12 \cdot -\frac12 \\ &\vdots\\ f^{(k)}(x) = \frac12 \cdot -\frac12 \cdots -\frac{2k-3}2 (1+x)^{-(2k-1)/2} &\Rightarrow f^{(k)}(0) = \frac12 \cdot -\frac12 \cdots -\frac{2k-3}2 \end{align} \]

So the Maclaurin series equals \[ (1+x)^{1/2} = 1 + \frac12 x + \frac{\frac12 \cdot -\frac12}{2!} x^2 + \cdots + \frac{\frac12 \cdot -\frac12 \cdots -\frac{2k-3}2}{k!} x^k + \cdots \] It can be shown by the ratio test that the series converges for \( |x|<1 \).

## Binomial theorem for fractional exponent

The above example generalizes immediately for all fractional exponents \( \alpha \). Let \( \alpha \) be a real number and \(k \) a positive integer. Define \[ \binom{\alpha}{k} = \frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}; \] then the same analysis as in the example gives:

Let \( \alpha = p/q \) be a rational number (\(p,q\) integers). Then \[ (1+x)^\alpha = 1 +\alpha x + \frac{(\alpha)(\alpha-1)}{2!} x^2 + \cdots = \sum_{k=0}^{\infty} \binom{\alpha}{k} x^k \] which converges for \( |x| < 1 \).

## Examples

The binomial coefficients \( \binom{p/q}{k}\) can sometimes be rewritten in interesting ways.

Let \( k \) be a positive integer. Then \[ \begin{align} \binom{-1/2}{k} &= \frac{-\frac12 \cdot -\frac32 \cdots -\frac{2k-1}2}{k!} \\ &= (-1)^k \frac{1 \cdot 3 \cdots (2k-1)}{2^k k!} \\ &= (-1)^k \frac{1 \cdot 3 \cdots (2k-1)}{2 \cdot 4 \cdots (2k)} \\ &= (-1)^k \frac{(2k)!}{(2 \cdot 4 \cdots (2k))^2} \\ &= (-1/4)^k \binom{2k}{k} \\ \end{align} \] Alternatively, this can be written with a double factorial simply as \[ (-1)^k \frac{(2k-1)!!}{(2k)!!} \] So \[ \frac1{\sqrt{1+x}} = \sum_{k=0}^\infty \binom{2k}{k} (-x/4)^k \] for \( |x|<1 \). Substituting \( -4x \) for \( x \) gives the result that the generating function for the central binomial coefficients is \[ \frac1{\sqrt{1-4x}} = \sum_{k=0}^\infty \binom{2k}{k} x^k. \]

\[ 1 + \frac {2}{3} \left ( \frac 1 2 \right )^1 + \frac {2 \times 5}{3 \times 6} \left ( \frac 1 2 \right )^2 + \frac {2 \times 5 \times 8}{3 \times 6 \times 9} \left ( \frac 1 2 \right )^3 + \ldots \]

If the series above can be stated in terms of form of \( a^{1/b} \) for positive integers \(a,b\), what is the value of \(a+b\)?

**Cite as:**Fractional Binomial Theorem.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/fractional-binomial-theorem/