Error of Taylor series too large

Question

I am trying to approximate $$\sqrt[3]{1+10^{12}}$$ using the Taylor series of the function $$h(x)=\sqrt[3]{1+x}$$ I calculated the Taylor series of order $2$ at $x=0$ of $f(x)$ as: $$T_2=1 + \frac{1}{3}x-\frac{1}{9}x^2$$ Substituting $x=10^{12}$ in the Taylor series results in a very large error and my answer is nowhere near to the correct value. Using Taylor series of a higher order is not helpful since the error increases when I do that. For example, $R_2\approx 6.2\times 10^{34}$ and $R_3\approx 4.1\times 10^{46}$

Why does that happen, and how do I approximate $\sqrt[3]{1+10^{12}}$ using a Maclaurin series of order $2$?

Answer 1

Consider $h(x)$ when $x\to +\infty$

$h(x)=\sqrt[3]{1+x}=\sqrt[3]{x\left(1+\dfrac{1}{x}\right)}=x^{\frac{1}{3}}\sqrt[3]{1+\dfrac{1}{x}}$

Since $\dfrac{1}{x} \to 0$ as $x\to +\infty$, you can use $T_2$

$h(x)=x^{\frac{1}{3}}\sqrt[3]{1+\dfrac{1}{x}}\sim x^{\frac{1}{3}}\left(1+\dfrac{1}{3x}-\dfrac{1}{9x^2}\right)\quad (*)$

Now try $(*)$ using $x=10^{12}$.

Answer 2

That works for numbers close to $0$, not for numbers close to $10^{12}$.

Take $f(x)=\sqrt[3]x$. It's Taylor polynomial of order $2$ centered at $10^{12}$ is$$10^4+\frac{x-10^{12}}{3\times10^8}-\frac{\left(x-10^{12}\right)^2}{9\times10^{20}}.$$It's value at $10^{12}+1$ is$$10^4+\frac1{3\times10^8}-\frac1{9\times10^{20}},$$which is indeed very close to $\sqrt[3]{10^{12}+1}$.