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When we talk about Taylor series, we say it's around point $x_0$. It's in the Taylor series formula:

$$f(x) = f(x_0) + (x-x_0)f'(x_0) + \frac{(x-x_0)^2f''(x_0)}{2} + \frac{(x-x_0)^3f'''(x_0)}{6} + + ...$$

What I don't understand what is this point $x_0$?

We're given Taylor formulas for $\sin(x)$, $\cos(x)$, etc, and they don't have this point $x_0$ in the formula? It's somehow chosen to be $0$ so it disappears, but what does that mean?

Doesn't that mean the Taylor series works only around $0$? But then we somehow can use the formula for any $x$, even $100$ or $100,000$, and it still gives the right result?

What happens if we choose $x_0$ to be $1$? or $25$? That would change the Taylor formula for the primitive functions but would it still work for any $x$?

I'm very confused.

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In Taylor's formula, we have to choose an initial point $x_0$, in which the sum actually converges to the function. In any other point the sum may converge or diverge, depending on the function.

For example, by evaluating Taylor's formula for $\sin(x)$, $\cos(x)$ and also $e^x$, it doesn't matter which point $x_0$ you choose, and the sum will converge everywhere to the function (if I am not mistaken, there is a theorem that if the sum converges at $x$, it converges to the value of the function at $x$).

However, in other functions (like $\ln(1+x)$), the sum may not converge always; Taylor's formula for $\ln(1+x)$ converges $\Leftrightarrow$ $|x|<1$ (and also $x=1$, if I remember correctly).

To sum up, you may choose any point $x_0$ you want; but it is usually convinient to choose it as $0$ (because it is easier to calculate the derivatives).

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    $\begingroup$ Even if a Taylor series converges everywhere, the limit can be different from the given function. $\endgroup$ – Martin R Jan 22 '15 at 18:24
  • $\begingroup$ Very interesting. $\endgroup$ – bodacydo Jan 22 '15 at 18:30
  • $\begingroup$ However if the Taylor series around some $x_0$ converges to the function everywhere on $\mathbb R$, then the Taylor series around any other point will also converge to the function everywhere. $\endgroup$ – hmakholm left over Monica Jan 22 '15 at 18:33
  • $\begingroup$ Weird and interesting as f*ck. Thanks. $\endgroup$ – bodacydo Jan 23 '15 at 7:34
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You can develop a Taylor series of a function at any point of the domain/interval where it is nice enough (continuously n-differentiable...). The formulae well known for sine cosine exp are at $x_0=0$ but you could using the general formula get a Taylor expansion of sine at $x_0=\pi$ for example or any other real by the way.

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  • $\begingroup$ I'm sorry, I don't understand your answer. You didn't answer the questions that I've. $\endgroup$ – bodacydo Jan 22 '15 at 18:28
  • $\begingroup$ I meant that the Taylor formula is the general one you wrote down involving derivatives of $f$ and factorials. You could develop sine for example anywhere and this would give a different expansion from the well known expansion at zero $x-x^3/3+\ldots$ $\endgroup$ – marwalix Jan 22 '15 at 18:32
  • $\begingroup$ Thanks. But what does it mean to have a different expansion? Will it still produce the same values for sin(x)? Or they will be shifted by $x_0$? Or does it mean the same function can have many different expansions that all produce the same values? $\endgroup$ – bodacydo Jan 22 '15 at 18:36
  • $\begingroup$ The function is unchanged it is the expansion meaning the formula that is different. Value at a given point is unique but two different formulaes give that value $\endgroup$ – marwalix Jan 22 '15 at 18:44

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