Expanding $(a+b)^\frac{1}{2}$

Question

I was wondering if it's possible to expand $(a+b)^\frac{1}{2}$.For example, $(a+b)^2 = a^2 + 2ab + b^2$. But what is the expansion of $(a+b)^\frac{1}{2}$? I've learned about binomial theorem but I can't figure it out.

Answer 1

Only asymptotically. There are ways to approximate it in terms of $a^{1/2}$ and $b^{1/2}$. There are also some expansion formulas, for example, the expression can be written as a McLaurin series, such as, $$(a + b)^{1/2} = a^{1/2} + \frac {1} {2} b^{1/2} + \cdots.$$ Also, if we make use of the Binomial expansion, we will have $$(a + b)^n = \sum_{k = 0}^{n} {n \choose k} a^k b^{n - k},$$ hence, by putting $n = 1/2$, $$(a + b)^{1/2} = \sqrt {a} + \frac {1} {2} \sqrt {b} + \cdots.$$ But there is not an exact rational expression as you might wish.

Answer 2

$(a+b)^\frac{1}{2}=a^\frac{1}{2}(1+\frac{b}{a})^\frac{1}{2}=a^\frac{1}{2}[1+\frac{b}{2a}+\frac{(\frac{1}{2})(\frac{1}{2}-1)b^{2}}{2a^{2}}+\frac{(\frac{1}{2})(\frac{1}{2}-1)(\frac{1}{2}-2)b^{3}}{6a^{3}}+...$

This is from the Maclaurin series, known as the binomial series, of the function

$f(x)=(1+x)^{n}$ where $n$ is an arbitrary real number.

Answer 3

Let us suppose $a>b$. Then $$(a+b)^{\frac 12}=\sqrt a\,(1+\frac b a)^{\frac 12}$$ Now, call $x=\frac ba$ and use $$(1+x)^{\frac 12}= 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \frac{7}{256}x^5 - \cdots$$