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I'm trying to do the binomial expansion of $(x-2)^{1/2}$.

How do you do it? As far as I'm aware the expansion only works for $(1+x)^n$. How could I get it in that form?

Thanks.

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migrated from mathoverflow.net May 11 '14 at 23:41

This question came from our site for professional mathematicians.

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    $\begingroup$ You need to say where you want the binomial expansion. I suppose you want a power series in $x$, i.e., an expansion around $x=0$. But there the square root is not real. $\endgroup$ – Marc van Leeuwen May 14 '14 at 6:32
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One way to do this is to rewrite the expression as $(1+(x-3))^{1/2}$, and then substitute $u=x-3$ and expand $(1+u)^{1/2}$ using the Binomial Series $\displaystyle(1+u)^r=\sum_{n=0}^{\infty}\binom{r}{n}u^n$.

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  • $\begingroup$ This does not converge for (or near) $x=0$, so $u=-3$. $\endgroup$ – Marc van Leeuwen May 14 '14 at 6:33
  • $\begingroup$ You're right - I wasn't sure where the OP wanted the series centered, so I was giving one way to get a convergent Taylor series using the Binomial formula. $\endgroup$ – user84413 May 14 '14 at 18:14
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The basic binomial formula $(1+x)^\alpha=\sum_k\binom\alpha kx^k$ is valid as a formal power series in $x$, or for concrete numbers$~x$ with $|x|<1$ (supposing that like is the case here, $\alpha$ is not a natural number). This cannot be used directly in the example, but you could bring the expression into a similar form by extracting a factor $-2$ from the expression $x-2$, writing $(x-2)^\alpha=(-2(1-\frac x2))^\alpha=(-2)^\alpha(1-\frac x2)^\alpha$ for $\alpha=\frac12$.

But this brings to the front a question that one should have considered at the beginning: what is the value $(-2)^\frac12$ that your formula gives at $x=0$, and will have to be the constant term of any reasonable expansion? The value $\sqrt{-2}$ is not very well defined; it does not exist as real number, and as complex number it it could be either $\def\i{\mathbf i}\sqrt2\i$ or $-\sqrt2\i$. So a formula with real numbers valid near $x=0$ is just not possible; if you want a formula that describes one of the complex square roots of $x-2$ near $x=0$ you can choose $r$ to be one of the mentioned square roots of $-2$, and write $$(x-2)^\frac12=r\sqrt{1-\frac x2} =r\sum_k\binom{1/2}k\Bigl(-\frac1{2^k}\Bigr)x^k. $$ This series (which has purely imaginary coefficients due to the factor$~r$) converges for $|x|<2$.

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  • $\begingroup$ So if we replaced $\alpha=\frac12$ with, say, $\alpha=\frac13$ then everything is hunky-dory? Also, I should say that I was initially very confused with your pre-occupation with the series at $x=0$, so I think your post might be improved if you pointed out that the binomial expansion is the MacLaurin series of $(a+bx)^{\alpha}$? +1 though :-) $\endgroup$ – user1729 May 15 '14 at 8:30
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If what you meant to ask about was a binomial expansion, then the approach is as follows:

First, note that the binomial theorem states that \begin{align*} (a+b)^n = \sum_{k=0}^{\infty}\binom{n}{k}a^kb^{n-k} \end{align*} Thus, by substituting for the above values, we have \begin{align*} (x-2)^\frac{1}{2} = \sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}x^k(-2)^{\frac{1}{2}-k} \end{align*}

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  • $\begingroup$ What is $(-2)^{\frac12-k}$, say for $k=1$? $\endgroup$ – Did May 14 '14 at 6:18
  • $\begingroup$ I take it you believe my answer to be wrong. Perhaps you could explain to me why that is the case? To answer your question, Wolframalpha approximated it to be 0.353553*i. $\endgroup$ – Pavelshu May 14 '14 at 6:22
  • $\begingroup$ Expansion of a real valued function which involves imaginary coefficients? At the least, this begs for an explanation... (To be fair, note that this remark applies to the question itself.) $\endgroup$ – Did May 14 '14 at 6:29
  • $\begingroup$ I wish I could explain it further, but I am unsure of why this occurs either. $\endgroup$ – Pavelshu May 14 '14 at 6:38

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