idku
  • idku
I am not going to approximate sqrt(9.3). Well, I will attempt.
Mathematics
jamiebookeater
  • jamiebookeater
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idku
  • idku
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 \(\displaystyle\large\color{black}{\sum_{n=0}^{2}\frac{f^{n}(a)}{n!}(x-a)^n~~{\Huge|}_{{\LARGE a=9}}~~~=3+\frac{1}{6}(x-9)+\frac{1}{216}(x-9)^2}\) \(\displaystyle\large\color{black}{\sqrt{9.3}\approx3+\frac{1}{6}(9.3-9)+\frac{1}{216}(9.3-9)^2}\)
idku
  • idku
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.
idku
  • idku
right ?

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idku
  • idku
the third row when I started to do my derivatives should be f''(x) and f''(9)

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