# Really basic Taylor expansion

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!

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 $$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$
• 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. – Hawaiian Rolls May 6 '17 at 17:42
• 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 $f(x)=x^2$, $x_0=1$. – boaz May 6 '17 at 18:38
• 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 $f(x_0) = f(x_0)$ – Hawaiian Rolls May 6 '17 at 19:11
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