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I am aware what the taylor expansion is when doing it at points such as $f(x_0 +kh)$ and others in this format where k is just some constant, but what is it when you just have $f(x_0)$. Is it just

$$ f(x_0) = \sum_{i=0}^n \frac{f^{(i)}(x_0)}{i!} $$

Oppose to when you would have some h value if you had to find the expansion at a certain point such as $x_0 + h$ would produce

$$ f(x_0 + h) = \sum_{i=0}^n \frac{ h^i f^{(i)}(x_0) }{i!} $$

I looked up the formula, but I just want to make sure I'm applying it correctly. Thanks!

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The correct expansion is $$ f(x)=\overbrace{\sum_{n=0}^{N}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n}^{\text{N-th taylor polynomial}}+\overbrace{\frac{f^{(N+1)}(c)}{(N+1)!}(x-x_0)^{N+1}}^{\text{$R_{N}(x)$ remainder}} $$ where c is between $x$ and $x_0$. If $\lim_{N\to\infty}R_N(x)=0$, then $$ f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n $$

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    $\begingroup$ so would it look like $f(x_0) = f(x_0)$, as if I have $x=x_0$ then all of the iterations past 0 would result in the exponent $(x_0- x_0)^n$ to become 0. $\endgroup$ – Hawaiian Rolls May 6 '17 at 17:42
  • $\begingroup$ Both the equations you wrote are wrong. $\endgroup$ – boaz May 6 '17 at 17:54
  • $\begingroup$ Explain please. (Disregard error term as that isn't important right now as it will change on depending which derivative we are approximating for). $\endgroup$ – Hawaiian Rolls May 6 '17 at 18:08
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    $\begingroup$ You first equation is $f(x_0)=f(x_0)+f^{(1)}(x_0)+\frac{1}{2!}f^{(2)}(x_0)+\ldots+\frac{1}{n!}f^{(n)}(x_0)$. This is obviously wrong. e.g., take $f(x)=x^2$, $x_0=1$. $\endgroup$ – boaz May 6 '17 at 18:38
  • $\begingroup$ but from the equation you yourself gave, if we have $f(x) = \sum_{n=0}^N \frac{f^{(n)}(x_0)}{n!} (x - x_0)^n$, if we allow $x = x_0$, in all instances past $n=0$, we will have $(x_0-x_0)^n = 0$ which would leave $f(x_0) = f(x_0) $ $\endgroup$ – Hawaiian Rolls May 6 '17 at 19:11

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