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I'm having some trouble with expanding a binomial when it is a fraction. eg $(a+b)^{-n}$ where $n$ is a positive integer and $a$ and $b$ are real numbers.

I've looked at several other answers on this site and around the rest of the web, but I can't get it to make sense. From what I could figure out from the Wikipedia page on the subject, $(a+b)^n$ where $n>0$

Any help will be much appreciated.

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  • $\begingroup$ Use the generalized binomial theorem. This is equivalent to the Taylor series expansion, which might be a way forward that is more familiar to you. $\endgroup$
    – Mark Viola
    Commented Oct 16, 2016 at 21:26
  • $\begingroup$ @Dr.MV Would it be something like ${\frac {1}{(1-x)^{s}}}=\sum _{k=0}^{\infty }{s+k-1 \choose k}x^{k}\equiv \sum _{k=0}^{\infty }{s+k-1 \choose s-1}x^{k}.$ then? In that case, how would one calculate the other term, e.g. the 1 in this case. Obviously that 1 won't make a difference, but say it was another variable. What would it be called? The series is infinite, so there would be no way to subtract it from the total as per usual. $\endgroup$ Commented Oct 16, 2016 at 21:39

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For $\lvert x\rvert<1$ and a real number $\alpha$, you can write $(1+x)^{\alpha}$ as the convergent series $$(1+x)^{\alpha}=\sum_{k=0}^\infty \binom{\alpha}{k} x^k$$

Were $\binom\alpha k=\frac{\alpha(\alpha-1)(\alpha-2)\cdots (\alpha-k+1)}{k!}$. For instance \begin{align}\binom{1/2}{4}&=\frac{\frac12\cdot\left(-\frac12\right)\cdot\left(-\frac32\right)\cdot\left(-\frac52\right)}{24}=-\frac{15}{16\cdot24}\\\binom{-n}{k}&=\frac{-n\cdot(-n-1)\cdots(-n-k+1)}{k!}=(-1)^k\frac{n(n+1)\cdots(n+k-1)}{k!}=\\&=(-1)^k\frac{(n+k-1)!}{(n-1)!k!}=(-1)^k\binom{n+k-1}{k}\end{align}

Now, to use this effectively when $(a+b)^{-n}$, you need at least one between $a$ and $b$ not to be zero and $\lvert a\rvert\ne\lvert b\rvert$. Then, pick the one with the highest absolute value and factor it out. If it is $a$, this means writing all as

$$a^{-n}\left(1+\frac ba\right)^{-n}=a^{-n}\sum_{k=0}^\infty \binom{-n}{k}\frac{b^k}{a^k}=\sum_{k=0}^\infty \binom{n+k-1}{k}(-b)^ka^{-k-n}$$

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  • $\begingroup$ Thinking about it, $\lvert a\rvert \ne\lvert b\rvert$ already implies that at least one of them is non-zero, but better be safe than sorry... $\endgroup$
    – user228113
    Commented Oct 16, 2016 at 21:43
  • $\begingroup$ Thanks a lot :) It makes a lot more sense now. I didn't consider making the coefficient of "a" 1 $\endgroup$ Commented Oct 16, 2016 at 21:44
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Note that $(a+b)^{-n} = \frac{1}{(a+b)^n}$.

Now, apply $(a+b)^n = \sum_{i=0}^n \binom{n}{i} a^i b^{n-i}$ to calculate the denominator.

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    $\begingroup$ is $\dfrac 1 {\text{something}}$ an "expansion"? $\qquad$ $\endgroup$ Commented Oct 16, 2016 at 21:30
  • $\begingroup$ @MichaelHardy Depends how you look at it. :) $\endgroup$ Commented Oct 17, 2016 at 0:45
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You will always get a series, and for my own part, I find the story easiest to understand when $a=1$. Then, the standard formula applies: $$ (1+b)^t=1+tb+\frac{t(t-1)}2b^2+\sum_{k=3}^\infty\binom tkb^k\,, $$ where $\binom tk=t(t-1)(t-2)\cdots (t-k+1)\big/k!$ . This is the Generalized Binomial Theorem mentioned in the comment of @Dr.MV, and it’s often proved in second-term Calculus. The formula is good and convergent whenever $|b|<1$, no matter whether $t$ is a negative integer or a rational number, or even a real number. When $a\ne1$, you’ll be mentioning $a^t$ everywhere. Note that this series expansion does not treat $a$ and $b$ symmetrically.

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When $n$ is a negative integer and so is $k$, then we can write \begin{align} \binom n k & = \frac{n!}{(n-k)! k!} \tag 1 \\[10pt] & = \frac{n(n-1)(n-2)(n-3)\cdots(n-k+1)}{k!}. \tag 2 \end{align} Note that although line $(1)$ assumes $n$ is a nonnegative integer, line $(2)$ does not. Line $(2)$ makes sense if $n$ is negative or is a non-integer.

Now suppose among $a$ and $b$ the one with the smaller absolute value is $b$, i.e. we have $0 \le |b| < |a|.$ That implies $\left|\dfrac b a\right|<1,$ and we will need that to assure convergence of a series we will see below.

Now suppose we want to expand $(a+b)^m$ as a sum of powers of $a$ and powers of $b$, and $m$ is not necessarily positive and not necessarily an integer. Then $$ (a+b)^m = a^m \left( 1 + \frac b a \right)^m = a^m \sum_{k=0}^\infty \binom m k \left(\frac b a \right)^k = \sum_{k=0}^\infty \binom m k a^{m-k} b^k. $$ This is "Newton's binomial theorem."

Note that

  • All of the powers to which $b$ is raised are nonnegative integers, whereas those of $a$ need not be;
  • If $m$ happens to be a nonnegative integer then all of the terms in which $k>m$ are $0$, since $\dbinom m k=0$ in that case.
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