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-Find the Taylor polynomial of degree $n$ for $f(x) = \sqrt{(1+x)}$, centered at $x = 0$.

-I know how to find taylor polynomial but when it comes to finding the $n^{th}$ term, I'm a little lost on finding what the pattern is.

We have $f^{'}(0) = \frac{1}{2}, f^{''}(0) = -\frac{1}{4}, f^{'''}(0) = \frac{3}{8}, f^{4}(0) = \frac{-3*5}{16}$ and $f^{n}(0) = \frac{-1^{n+1}}{2^n}*$ ?

Thanks!

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  • $\begingroup$ You're right-- it's not an easy pattern-- it has to involve with binomial coefficients. $\endgroup$
    – DUO Labs
    Apr 14, 2020 at 0:24
  • $\begingroup$ Write it as $(1+x)^{1/2}$. Keep differentiating. YOu should see a pattern in terms of binomial coefficients $\endgroup$
    – peek-a-boo
    Apr 14, 2020 at 0:24

1 Answer 1

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It would be easier to apply the bionomial theorem. $$(1+x)^a=\sum_{n=0}^\infty\binom{a}{n} x^n$$ and, if required, truncate where ever you need.

If $a=\frac 12$, remember that $$\binom{a}{n}=\frac{\sqrt{\pi }}{2 \left(\frac{1}{2}-n\right)! n!}$$ which then makes $$\sqrt{1+x}=\frac{\sqrt{\pi }}{2}\sum_{n=0}^\infty\frac {x^n}{\left(\frac{1}{2}-n\right)!\,n!}$$

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