# How do you approximate the sqrt(128) using a taylor polynomial centered at 0?

Mar 7, 2015

If you think of it as approximating $f \left(x\right) = \sqrt{x}$ at $x = 128$, you don't.

The Taylor polynomial, of degree $n + 1$, to approximate $f \left(x\right)$ centered at $a$ is:

f(a)+f'(a)(x-a)+ (f''(a))/(2!)(x-a)^2+(f'''(a))/(3!)(x-a)^3+ . . . +(f^(n+1)(a))/((n+1)!)(x-a)^(n+1)

The function in the question is $f \left(x\right) = \sqrt{x}$ whose derivatives are:
$f ' \left(x\right) = \frac{1}{2 \sqrt{x}}$, $f ' ' \left(x\right) = - \frac{1}{4 {\sqrt{x}}^{3}}$, $f ' ' ' \left(x\right) = \frac{3}{8 {\sqrt{x}}^{5}}$,

Consequently, if $f \left(x\right) = \sqrt{x}$, then $f ' \left(a\right)$ does not exist for $a = 0$ (nor do any higher-order derivatives).

However, approximating $f \left(b\right)$ by a Taylor Polynomial centered at some $a$ that is far from $b$ isn't going to yield a good approximation anyway.

We need a way to think of $\sqrt{128}$ as a function of a number much closer to $0$.

Some thought and some experimentation and some arithmetical exploration are called for.

Eventually one may note that $128 = {2}^{7}$. What we are looking for then, is an approximation of the function $f \left(x\right) = {x}^{\frac{7}{2}}$. The first $3$ derivatives of this function do exist at $0$.

Note that if $a = 0$ is not specified by the exercise, choose $a = 1$.

You can probably finish from here.