I was reading this question and there was something in the first answer (adfriedman's answer) that I didn't understand.

It states that: $$ \int_0^1(1+(-x^2))^{1/2}dx = \int_0^1\sum_{k=0}^\infty\binom{1/2}{k}(-x^2)^k dx $$ Which I assume gives: $$ (1+(-x^2))^{1/2} = \sum_{k=0}^\infty\binom{1/2}{k}(-x^2)^k $$ I have never seen this, and I couldn't find anything about it online, so I was wondering where this comes from. Could it possibly be generalised to: $$ (1+a)^{b} = \sum_{k=0}^\infty\binom{b}{k}(a)^k $$

If so, where does this come from? What's the intuition behind it? Why does it make sense?

  • 4
    $\begingroup$ The binomial series is a generalization of the binomial theorem from algebra. $\endgroup$ Jun 22 '21 at 1:21
  • 1
    $\begingroup$ This is called Newton's binomial theorem. It is also possible to obtain it as a Maclauren series. $\endgroup$
    – GEdgar
    Jun 22 '21 at 1:21
  • $\begingroup$ So this only works when $|x|<1$? $\endgroup$ Jun 22 '21 at 1:27
  • $\begingroup$ Yeah, only $|x|<1.$ But here, that is also the only real domain of $(1-x^2)^{1/2}$ $\endgroup$ Jun 22 '21 at 1:43
  • $\begingroup$ What about $(1-0^2)^{1/2} = 1$. Would it still work? $\endgroup$ Jun 22 '21 at 1:49

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