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I am trying to find a Taylor series for the following function: ${1\over 1-9x}$ centered at c = 7

I browsed through my Calc II book and found that I can use the general formula for a Taylor series expansion.

$$T(x) = \sum_{n=0}^\infty {f^n (c)\over n!} (x-c)^n$$

The book said that $f^n$ is a representation of the various derivatives of the given function. Do I try to compute a few derivatives and than represent them in a $f^n$ format?

Can someone help explain the initial steps to solving this problem, and just nudge me in the right direction please? Thank You

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I would try taking a few derivatives of $f(x) = \frac{1}{1-9x}$, i.e., compute $f'(x)$, $f''(x)$, and so on, until you can recognize a pattern. You should be able to find a fairly simple formula for the $n$th derivative $f^{(n)}(x)$ in terms of $n$. To get the coefficients in the Taylor series, just plug in $x=7$ into $f^{(n)}(x)$.

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We have a smooth function $\frac{1}{1-9x}$, by statement of Taylor's theorem, each analytical function could be written as series inovling countinue derivatives of initial function, let find first derivative $$f^{'}(x)=\frac{d}{dx}(1-9x)^{-1}=-(1-9x)^{-2}\cdot(1-9x)^{'}=-9(1-9x)^{-2}$$ generalize above expression we receive: $$f^{(n)}(x)=\frac{d^n}{dx^n}(1-9x)^{-1}=-9n(1-9x)^{-(n+1)}$$ It follows that Taylor series of such function at $c=7$ is $$f(x)=\sum_{n=0}^{\infty}\frac{-9n(1-9c)^{-(n+1)}}{n!}(x-7)^{n}$$

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  • $\begingroup$ Almost correct. Notice, though, that $(1-9x)' = -9$, so $f'(x) = 9 (1-9x)^{-2}$. You might want to check the other signs as well (I believe that you should have some $(-1)^{n-1}$ in $f^{(n)}$). $\endgroup$ – Alex M. Nov 22 '17 at 17:48

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