Question

I am not going to approximate sqrt(9.3).
Well, I will attempt.

Answer 1

Tailor polynomial, of a function f(x) at x=a, of nth degree.

\(\displaystyle\large\color{black}{\sum_{n=0}^{\infty}\frac{f^{n}(a)}{n!}(x-a)^n}\)


starting to work my f of a and nth order derivatives evaluated at x=a. (my a is 9)

\(\displaystyle\large\color{black}{f(x)=\sqrt{x}~~~~~~~~\color{blue}{f(9)=3}}\)

\(\displaystyle\large\color{black}{f'(x)=\frac{1}{2\sqrt{x}}~~~~~~~~\color{blue}{f'(9)=1/6}}\)

\(\displaystyle\large\color{black}{f'(x)=-\frac{1}{4x\sqrt{x}}~~~~~~~~\color{blue}{f'(9)=1/216}}\)

so my first 3 terms I get

\(\displaystyle\large\color{black}{\sum_{n=0}^{2}\frac{f^{n}(a)}{n!}(x-a)^n~~{\Huge|}_{{\LARGE a=9}}~~~=3+\frac{1}{6}(x-9)+\frac{1}{216}(x-9)^2}\)

\(\displaystyle\large\color{black}{\sqrt{9.3}\approx3+\frac{1}{6}(9.3-9)+\frac{1}{216}(9.3-9)^2}\)

Answer 2

and then whatever that last line is going to give me that is an approximation of sqrt(9.3) using a Taylor polynomial of degree 3, at x=9.

Answer 3

right ?

Answer 4

the third row when I started to do my derivatives should be

f''(x) and f''(9)