Doubt about Taylor's polynomial to approximate f(x)

Question

Use the degree 2 Taylor polynomial of $f(x) =$ $\sqrt[3]{1728 + x}$ to approximate $\sqrt[3]{1731}$ and give a bound for the error.

To obtain the degree 2 Taylor's polynomial, I computed the second derivative and so on, and I end up with a formula Pn(f, x, a) = .... But I dont know how to proceed; I dont know what is the meaning of 'a' in the formula when it appears as (x-a) to some power. I suppose that by 'error' it means to use Taylor's remainder, but not sure either...

Really need your help, Thanks a lot!

Answer 1

The value $a$ I believe would be where you centre the expansion, due to the nature of the derivatives (the first few involve $(1728+x)^n$ for some $n$), you could take $a=-1727$ would make things simpler. Then to find $(1731)^\frac{1}{3}$ (approx) sub in $x=3$

Answer 2

Let $f(x)=\sqrt[3]{x}$. We have by Taylor expansion $$f(1731)=f(1728)+3f'(1728)+\frac{3^2}{2!}f''(1728+h),\quad h\in(0,3)$$ so $$\sqrt[3]{1731}=\underbrace{12+3\times\frac{12}{3\times 1728}}_{=12+ \frac1{144}}-\frac{3^2}{2!}\times\frac{2\times\sqrt[3]{1728+h}}{9\times(1728+h)^2}$$

$$\text{bound of the error}:\quad 0\le\frac{1}{(1728+h)^{5/3}}\le \frac{1}{(1728)^{5/3}}=\frac1{(12)^5}\approx 4.10^{-6}$$