Another question about $x_0$ in the Taylor series 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.
 A: 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).
A: 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.
