Taylor Polynomial for $x^{1/3}$

a. Compute the Taylor polynomial $T_3(x)$ for the function $(x)^{1/3}$ around the point $x=1$.
b. Compute an error bound for the above approximation at $x = 1.3$.

I'm having trouble figuring out what to do for this problem.

I have to find the first three derivatives and after that I'm not quite sure what to do.

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Hint to start : expand $(1+t)^a$ with $t:=x-1$ in the specific case $a=\frac 13$. – Raymond Manzoni May 16 '13 at 15:43

Hints:

A Taylor polynomial centered at $a$ has the form: $$T_n(x) = \sum_{k=0}^n\frac{f^{(k)}(a)(x-a)^k}{k!}$$

So, when you compute the first three derivatives, you plug them into that formula: $$T_3(x) = f(a)+ f'(a)(x-a) +\frac{f''(a)(x-a)^2}{2} + \frac{f'''(a)(x-a)^3}{6}$$

In your case, $a=1$.

For the error bound, we use the Taylor series remainder term: $$R_n = \frac{f^{(n+1)}(a)(x-a)^{(n+1)}}{(n+1)!}$$ To find the error at $x=1.3$, plug in that value.

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$$(x^{\frac13})' = \frac13 x^{-\frac23}$$ $$(x^{\frac13})'' = -\frac29 x^{-\frac53}$$ $$(x^{\frac13})''' = \frac{10}{27} x^{-\frac83}$$

Taylor expansion: $$x^{\frac13} = (1)^{\frac13} + \frac13 1^{-\frac23}\frac{x-1}{1!} -\frac29 1^{-\frac53} \frac{(x-1)^2}{2!} + \frac{10}{27} 1^{-\frac83} \frac{(x-1)^3}{3!} + \dots$$ You have $$T_3(x) = 1 + \frac13 (x-1) - \frac29 \frac{(x-1)^2}{2!} + \frac{10}{27} \frac{(x-1)^3}{3!}$$ Plugging in $x=1.3$ you get $$T_3(1.3) = 1.1117$$ Your error is $$T_3(1.3) - (1.3)^{\frac13} = 1.1117 - 1.0914 = 0.0203$$

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Hint: Expand $x^{\frac{1}{3}}$ as a Taylor series till the appropriate term and ignore the remaining terms. Ignoring the remaining terms results in an approximate value of the function for $x=1.3$. Then compare the approximation with the true value.

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what would x be and what would xsub(0) be? – user71317 May 16 '13 at 15:48
I am not sure which definition of the Taylor series you are looking at but the wikipedia has a useful definiton you can follow: en.wikipedia.org/wiki/Taylor_series#Definition – response May 16 '13 at 15:49