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.
 A: 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.
A: $$(x^{\frac13})' = \frac13 x^{-\frac23}$$
$$(x^{\frac13})'' = -\frac29 x^{-\frac53}$$
$$(x^{\frac13})''' = \frac{10}{27} x^{-\frac83} $$
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
$$
T_3(1.3) - (1.3)^{\frac13} = 1.1117 - 1.0914 = 0.0203
$$
A: 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.
