How do you expand √(a+b)?

This is a question that was recently asked on Quora:

it’s easy to expand
(a+b)^2 = a^2+2ab+b^2 or
(a+b)^3=a^3+3ab^2+3a^2b+b^3
or some other (a+b)^n but what about (a+b)^{1/2} aka. \sqrt{a+b}

Here’s my answer:

Just have Wolfram|Alpha do it for you :-).

But if you were on a desert island without access to Wolfram Alpha, here’s how you might think it through:

Are you already comfortable with the Binomial Theorem? Here it is again, but stated in a particular way that I think we’ll like.

\left(x+1\right)^r=1+rx+\frac{r(r-1)}{2!}x^2+\frac{r(r-1)(r-2)}{3!}x^3+\cdots

Look at it and make sure you understand it, and verify that it really is equivalent to the formulation of the Binomial Theorem you know.

Now, for the big trick. It turns out the above statement holds true not for just r=1,2,3,\ldots but for all real r. The only catch is that this often results in an infinite series. (These series results can also be obtained by Taylor expansion.)

In particular, it works for r=1/2:

\left(x+1\right)^{1/2}=1+\frac{1}{2}x+\frac{1/2(1/2-1)}{2!}x^2+\cdots

\left(x+1\right)^{1/2}=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3+\cdots

Now, rewriting your original expression (a + b)^{\frac{1}{2}} as \sqrt{b}\left(a/b+1\right)^{1/2} gives

\sqrt{b}\left(1+\frac{1}{2}\left(\frac{a}{b}\right)-\frac{1}{8}\left(\frac{a}{b}\right)^2+\frac{1}{16}\left(\frac{a}{b}\right)^3+\cdots\right)

=\sqrt{b}+\frac{a}{2\sqrt{b}}-\frac{a^2}{8b^{3/2}}+\frac{a^3}{16b^{5/2}}+\cdots

which is the same result Wolfram Alpha will spit back.

Hope that helps!

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