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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? $$(x + dx)^{1/2} = x^{1/2}(1 + \frac{dx}{x})^{1/2} = \sqrt{x} + \frac{1}{2}\frac{dx}{\sqrt{x}} ....$$ Attached is a screenshot for more information.enter image description here

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    $\begingroup$ Where did you see such an expansion? Do you know Taylor's series? $\endgroup$
    – user9464
    Commented Sep 6, 2020 at 0:33
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    $\begingroup$ See en.wikipedia.org/wiki/Binomial_series for a formula to expand $(1+t)^{\alpha}$ for arbitrary $\alpha \in \Bbb{C}$. $\endgroup$
    – Minus One-Twelfth
    Commented Sep 6, 2020 at 0:36
  • $\begingroup$ It's from a book called "Calculus Made Easy" by Silvanus Thompson. I think I heard of Taylor's series but I'm clueless about it. Is that a hint of the answer? $\endgroup$
    – Nemo
    Commented Sep 6, 2020 at 0:37

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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.

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  • $\begingroup$ isn't there a binomial expansion formula for this as well? $\endgroup$
    – C Squared
    Commented Sep 6, 2020 at 0:38
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    $\begingroup$ @CSquared: the definition of "binomial expansion" is from the Taylor series. $\endgroup$
    – user9464
    Commented Sep 6, 2020 at 0:40

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