Question

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

Answer 1

If you think of it as approximating `f(x)=sqrtx` at `x=128`, you don't.

The Taylor polynomial, of degree `n+1`, to approximate `f(x)` 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(x)=sqrt(x)` whose derivatives are:
`f'(x)=1/(2sqrt(x))`, `f''(x)=-1/(4sqrt(x)^3)`, `f'''(x)=3/(8sqrt(x)^5)`,

Consequently, if `f(x)=sqrtx`, then `f'(a)` does not exist for `a=0` (nor do any higher-order derivatives).

However, approximating `f(b)` 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 `sqrt128` 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(x)=x^(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.