Find the Taylor polynomial $T_5(x)$ = of order 5 about $x = 0$ for $f(x) = \sqrt{1+x}$. Write down the remainder term $R_5(x)$ and estimate the size of the error if $T_5(1)$ is used as an approximation to $f(1)$.

My attempt at the question: I think this is the correct Taylor polynomial: $$T_5(x) = 1 + \frac{x}2 - \frac{x^2}8 + \frac{x^3}{12} - \frac{5x^4}{128}+\frac{7x^5}{256}$$ I'm not sure if this fully answers the remainder term part: $$R_5(x) = \frac{f^{(6)}(c)}{6!}x^6$$

Could someone please give some pointers on how to estimate the size of the error and just check if the other parts are correct. Thank you for any help.

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    $\begingroup$ FAQ section + Reading directions to use LaTeX to write mathematicas + using those directions in posts here = nice looking and appealing questions. $\endgroup$ – DonAntonio May 5 '13 at 10:19
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    $\begingroup$ You should specify if you need the Peano (little-o reminder) or the Lagrange reminder, which is exactly the one you wrote, only with $c \in [o,x]$. $\endgroup$ – Kore-N May 5 '13 at 10:39
  • $\begingroup$ @Cornelis sorry I forgot to specify. I meant to specify Lagrange remainder $\endgroup$ – Joe S May 5 '13 at 10:46

Recall that $$(1+x)^\alpha=1+\sum_{k=1}^n\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}x^k+o(x^n)$$ so take $\alpha=\frac{1}{2}$ and try to simplify $\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}$ with this value.

Now for the remainder: by Taylor-Lagrange formula there's $\xi \in (0,x)$ $$R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1}$$ so if there's $M>0 $ s.t. $$|f^{(n+1)}(y)|\leq M \quad \forall y\in[0,x]$$ then we have an estimate for the error: $$|R_n(x)|\leq\frac{M}{(n+1)!}x^{n+1}$$

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  • $\begingroup$ You're welcome. $\endgroup$ – user63181 May 5 '13 at 10:50

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