I know that $(a + b)^2$ can be expanded as $(a + b) * (a + b) = a^2 + 2ab + b^2$.
Is there an equivalent expansion method for the square root of a sum, that is, $(a + b)^{1/2}$?
If there's no method, how could one derive these equalities? $$(x + dx)^{1/2} = x^{1/2}(1 + \frac{dx}{x})^{1/2} = \sqrt{x} + \frac{1}{2}\frac{dx}{\sqrt{x}} ....$$
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The Taylor series of $\sqrt{1+x}$ about $x = 0$ converges for $|x| ≤ 1$, and is given by $$ {\displaystyle {\sqrt {1+x}}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(2n)!}{(1-2n)(n!)^{2}(4^{n})}}x^{n}=1+{\frac {1}{2}}x-{\frac {1}{8}}x^{2}+\cdots ,} $$
This is where the formal identity $(1+\frac{dx}{x})^{1/2}$ from.
The first formal identity is nothing but $$ (a+b)^{1/2}=[a(1+\frac{b}{a})]^{1/2}=a^{1/2}[1+\frac{b}{a}]^{1/2} $$ assuming all the quantities are positive.