Taylor series of $\sqrt{1 + x^2}$

Question

I want to know what is the Taylor series of $\sqrt{1+x^2}$ while $x \to 0$.

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I really don't like this one: this site is not a calculator! Why can't you compute it yourself? As this is stated, this is not even a question... I want coffee!

Ross Millikan · Accepted Answer · 2010-10-23 17:40:08Z

up vote 4 down vote accepted

If you just want the answer, you can find it at WolframAlpha

edited Oct 23 '10 at 17:40

answered Oct 23 '10 at 16:58

Ross Millikan
83.3k555117

	link is broken – anonymous Oct 23 '10 at 17:24
	I tried putting the link in using the link command and it didn't take. So I put the full link in, but it breaks at the ^. If I copy the whole thing and paste it into the browser, it works. Best I seem to be able to do. I have posted links in successfully before. – Ross Millikan Oct 23 '10 at 17:28
	You need to use `%5e` for `^`. See meta.stackoverflow.com/questions/60211/…. – KennyTM Oct 23 '10 at 17:34
	@KennyTM Fixed now. Thanks. – Ross Millikan Oct 23 '10 at 17:40

Agustí Roig · Answer 2 · 2011-12-22 14:37:56Z

If you want to work a little bit, you could learn about the binomial series:

$$ (1+x)^\alpha = \sum_{n=0}^\infty \begin{pmatrix}\alpha \\ n\end{pmatrix} x^n $$

and then:

You'll get:

$$ \sqrt{1 + x^2} = \sum_{n=0}^\infty \begin{pmatrix} 1/2 \\ n\end{pmatrix} x^{2n} $$

2 Answers