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The Binomial Series

April 4th, 2009 |

Author: Merrin

The binomial theorem for positive integers, \textit{n}, states that

$(a+b)^{n} =\displaystyle \sum _{k=0}^{n}\cfrac{n!}{k!(n-k)!} a^{k} b^{n-k}$

For example,

$(a+b)^{3} =a^{3} +3a^{2} b+3ab^{2} +b^{3}$

The set of numbers (1, 3, 3, and 1) are the fourth line of Pascal’s Triangle. We will use our new tool of Maclaurin series to find the expansion of the function

$(1+x)^{\alpha }$

for any exponent. Many commonly occurring expressions in applications take the form of a binomial series with a general exponent, so this expansion will be very useful.

Example 1. Find the infinite series for the function by the Maclaurin series method.

$f(x)=\cfrac{1}{\sqrt{1-x} } =(1-x)^{-1/2}$

Solution 1.The series does not terminate for a power of -1/2.

The coefficients are found by $\cfrac{f^{(n)}(0)}{n!}$.

$= {1-\cfrac{1}{2} \cfrac{(-x)}{1!} +\left(-\cfrac{1}{2} \right)\left(-\cfrac{3}{2} \right)\cfrac{(-x)^{2} }{2!} +\left(-\cfrac{1}{2} \right)\left(-\cfrac{3}{2} \right)\left(-\cfrac{5}{2} \right)\cfrac{(-x)^{3} }{3!} +...}$

$= {1+\cfrac{1}{2} x+\cfrac{1\cdot 3}{2\cdot 4} x^{2} +\cfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6} x^{3} ...}$

The coefficients follow a pattern where

$(1-x)^{1/2} =1+\displaystyle \sum _{n=1}^{\infty }\cfrac{(2n-1)!!}{(2n)!!} x^{n}$

The double factorial is defined as

$(2n-1)!!=(2n-1)(2n-3)...(1)$

$(2n)!!=(2n)(2n-2)...(4)(2)$

Here is a good point to comment on what is possible with Taylor series. We can take the previous series and replace x with x squared. Then we have an infinite series fo

$\cfrac{1}{\sqrt{1-x^{2} } } =1+\cfrac{1}{2} x^{2} +\cfrac{1\cdot 3}{2\cdot 4} x^{4} +\cfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6} x^{6} ...$

Now what we can do is integrate both sides to get a new infinite series. On the left, we recognize the integral as $\arcsin x$ and on the right we have its Taylor series without computing anything much new.

$\arcsin x=C+x+\cfrac{1}{3} \cfrac{1}{2} x^{3} +\cfrac{1}{5} \cfrac{1\cdot 3}{2\cdot 4} x^{5} +\cfrac{1}{7} \cfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6} x^{7} ...$

We can do this for another example. For example take the geometric series

$\cfrac{1}{1+x} =1-x+x^{2} -x^{3} +...$

If we integrate both sides of this equation

$\ln \left|1+x\right|=C+x-\cfrac{x^{2} }{2} +\cfrac{x^{3} }{3} -\cfrac{x^{4} }{4} +...$

It is a simple matter to find the integration constant and then calculate a series for ln2 for example.

Another application of this method is to integrate arctangent.

$\int \cfrac{dx}{1+x^2} = \int (1-x^2+x^4-x^6+\cdots)dx$

Integrating the right hand side from zero to 1 we can find an expression for pi.

$\cfrac{\pi}{4}=1-\cfrac{1}{3}+\cfrac{1}{5}-...$

Instead of integrating series we can also differentiate them instead to get new power series

$\cfrac{d}{dx} \sin x=\cfrac{d}{dx} (x-\cfrac{1}{3!} x^{3} +\cfrac{1}{5!} x^{5} -...)=1-\cfrac{x^{2} }{2!} +\cfrac{x^{4} }{4!} -...=\cos x$

Example 3. Generalize the binomial series to an arbitrary exponent alpha.

$f(x)=(1+x)^{\alpha }$

Solution 3. We can find the coefficients by studying the derivatives of this function.

$f(x)= (1+x)^{\alpha}$

$f^{(1)} (x) = \alpha (1+x)^{\alpha -1}$

${f^{(2)} (x)}={\alpha (\alpha -1)(1+x)^{\alpha -2} \quad }$

${f^{(3)} (x)} = {\alpha (\alpha -1)(\alpha -2)(1+x)^{\alpha -3} }$

Plugging these calculations into the formula for the Maclaurin expansion, we can generate the binomial series.

$f(x)=\displaystyle \sum _{k=0}^{\infty }\cfrac{f^{(k)} (0)x^{k} }{k!}$

$(1+x)^{\alpha}=\cfrac{1}{0!} +\cfrac{\alpha x}{1!} +\cfrac{\alpha (\alpha -1)}{2!} x^{2} +...$

Notice that if alpha is a positive integer then it terminates when it gets to the term with a coefficient containing alpha -alpha.

Example 1. When does the binomial series converge?

Solution 1. We apply the ratio test to the binomial series.

${r} = {\mathop{\lim }\limits_{n\to \infty } \left|\cfrac{x^{n+1} }{x^{n} } \cfrac{n!}{n+1!} \cfrac{\alpha !}{(\alpha -n-1)!} \cdot \cfrac{(\alpha -n)!}{\alpha !} \right|}$

$= {\mathop{\lim }\limits_{n\to \infty } \left|x\cfrac{\alpha -n}{n+1} \right|=x}$

The binomial series converges for absolute values of x<1 since r=x. The ratio test is inconclusive for x=1. The binomial series diverges for x>1 by the ratio test.

When alpha is not an integer we need an extension of factorials to fractional powers. There is a therefore relation between the binomial coefficients containing $\alpha $ and the gamma function.

${\left(\begin{array}{l} {n} \\ {k} \end{array}\right)=\cfrac{n!}{k!\left(n-k\right)!} =\cfrac{n(n-1)\cdots (n-k+1)}{k!} } \\ {\left(\begin{array}{l} {n} \\ {k} \end{array}\right)\to \cfrac{\Gamma (\alpha +1)}{\Gamma (k+1)\Gamma (\alpha -k+1)} }$

The gamma function obeys a recursion relation just like factorial, but can be non integer

$\Gamma (s+1)=s\Gamma (s)$

We do not need to know how to find the gamma function itself at this point, only the ratio of two gamma functions whose arguments differ by an integer. Just use the recursion relation until the two remaining gamma functions cancel out.

Definition 1. To find a gamma function directly then one definition is the following integral.

$\displaystyle \Gamma (s+1)=s!=\int _{0}^{\infty }x^{s} e^{-x} dx$