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I am having trouble finding the correct binomial expansion for $\dfrac{1}{\sqrt{1-4x}}$:

Simplifying the radical I get: $(1-4x)^{-\frac{1}{2}}$

Now I want to find ${n\choose k} = {\frac{-1}{2}\choose k}$

\begin{align} {\frac{-1}{2}\choose k} &= \dfrac{\frac{-1}{2}(\frac{-1}{2}-1)(\frac{-1}{2}-2)\ldots(\frac{-1}{2}-k+1)}{k!} \\ &= (-1)^k\dfrac{\frac{1}{2}(\frac{1}{2}+1)(\frac{1}{2}+2)\ldots(\frac{1}{2}+k-1)}{k!} \\ &=(-1)^k{\frac{1}{2}+k-1\choose k} \end{align}

Now, applying Newton's Generalized Binomial Theorem I get:

\begin{equation}\sum\limits_{k=0}^\infty(-1)^k{\frac{1}{2}+k-1\choose k}(4x)^k \end{equation}

Is this correct? it seems odd to me since the top number of the combination is always less than the bottom number.

Furthermore, I would like to write $\left(\dfrac{1}{\sqrt{1-4x}}\right)^2$ as an infinite sum like:

\begin{equation}\sum\limits_{k=0}^\infty\sum\limits_{i=0}^k(expression)x^k \end{equation} I should be able to square the entire sum for what I derived above, and get what I know from calculus: \begin{equation} \left(\dfrac{1}{\sqrt{1-4x}}\right)^2 = \dfrac{1}{1-4x} = \sum\limits_{k=0}^\infty(4x)^k \end{equation}

The trouble is finding the expression for the double sum.

All help is greatly appreciated!

EDIT: With the help from below we should get the double sum to be: $\sum\limits_{k=0}^\infty\sum\limits_{i=0}^k{k\choose i}^2x^k$

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2 Answers 2

up vote 3 down vote accepted

It is useful to work out the terms in say $x$, $x^2$, $x^3$, and $x^4$ carefully. We will simplify the coefficient of $x^3$. If you do the same thing for $x^4$, everything will be clear.

The coefficient of $x^3$ is $$\frac{(-4)^3}{3!} \left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right).$$ Simplify. First note that the total number of minus signs is even. Indeed it is always even, for when we advance $n$ by $1$, we are multiplying by two additional negative terms, one from the $-4$, and the other from the binomial coefficient.

Borrow three $2$'s from $4^3$ to get rid of fractions. We now have $$\frac{2^3}{3!}(1)(3)(5).$$ Fill in the "missing" numbers $2,4,6$, and divide by $2^33!$ to compensate. We end up with $\dbinom{6}{3}$.

Edit: For the added question about squaring the expression, you will need a binomial coefficient identity.

Now note that the same sort of simplification works with the general term.

Remark: There is no need to worry about the fact that in $\dbinom{\frac{1}{2}-k+1}{k}$, the term on top is less than the one at the bottom. In this generalized binomial coefficient, we are no longer "choosing." But the generalized binomial coefficient has formal properties that are close to those of the ordinary binomial coefficient, so the same notation is used for both.

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Altogether we then get: $\dfrac{1}{\sqrt{1-4x}} = \sum\limits_{k=0}^{\infty}{2k\choose k}$ Is this correct? I am seeing something wrong, give me a few more minutes. –  CodeKingPlusPlus Feb 10 '13 at 16:46
    
There is no $k$ outside the binomial coefficient. You should get $\sum \binom{2k}{k}x^k$. –  André Nicolas Feb 10 '13 at 16:52
    
Thanks, I put the final answer I found in the edit above. I hope the lack of parens in the expression doesn't lead to confusion. –  CodeKingPlusPlus Feb 10 '13 at 17:25

Also, generalizing what Andre Nicolas wrote,

\begin{align} {\frac{-1}{2}\choose k} &= \dfrac{\frac{-1}{2}(\frac{-1}{2}-1)(\frac{-1}{2}-2)\ldots(\frac{-1}{2}-k+1)}{k!} \\ &= (-1)^k\dfrac{\frac{1}{2}(\frac{1}{2}+1)(\frac{1}{2}+2)\ldots(\frac{1}{2}+k-1)}{k!} \\ &=(-1)^k \frac{1\ 3\ 5 ... (2k-1)}{2^k k!} \\ &=(-1)^k \frac{1\ 2\ 3\ 4\ 5 ... (2k-1)(2k)}{2^k k!(2\ 4\ ...(2k))} \\ &=(-1)^k \frac{(2k)!}{2^k k!2^k(1\ 2\ ...k)} \\ &=(-1)^k \frac{(2k)!}{4^k k!^2} \\ &= (-1)^k {2k \choose k} \big/4^{k} \\ \end{align}

This is a standard bit of manipulation that, combined with Stirling's factorial approximation, allows good approximations for this expression to be derived.

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