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idku
 one year ago
I am not going to approximate sqrt(9.3).
Well, I will attempt.
idku
 one year ago
I am not going to approximate sqrt(9.3). Well, I will attempt.

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idku
 one year ago
Best ResponseYou've already chosen the best response.0Tailor polynomial, of a function f(x) at x=a, of nth degree. \(\displaystyle\large\color{black}{\sum_{n=0}^{\infty}\frac{f^{n}(a)}{n!}(xa)^n}\) 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 \(\displaystyle\large\color{black}{\sum_{n=0}^{2}\frac{f^{n}(a)}{n!}(xa)^n~~{\Huge}_{{\LARGE a=9}}~~~=3+\frac{1}{6}(x9)+\frac{1}{216}(x9)^2}\) \(\displaystyle\large\color{black}{\sqrt{9.3}\approx3+\frac{1}{6}(9.39)+\frac{1}{216}(9.39)^2}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0and 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
 one year ago
Best ResponseYou've already chosen the best response.0the third row when I started to do my derivatives should be f''(x) and f''(9)
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