Why are there factorials in Taylor series?

Question

I looked up this question online and some people mentioned that they are used to cancel out the constants from differentiating $f(x)$. I don't see how this would work when $f(x) = \sin(x)$ or $f(x) = e^x$.

Am I misunderstanding something?

Answer 1

Let's say you want to write $f(x)$ as a power series $f(x) = a_0 + a_1 x + a_2 x^2 + \cdots$. You want to find the coefficients $a_i$.

To do this, set $x=0$, then $a_0=f(0)$.

So $f(x) = f(0)+a_1 x+a_2x^2+\cdots$

To find $a_1$ first differentiate to get $f'(x) = 0+a_1 +2 a_2 x + \cdots$. Now set $x=0$ to get $a_1 = f'(x)$.

To find $a_2$ you need to differentiate twice and set $x=0$. And so on.

In general, in the $n$-th step you'll have $f^{(n)}(0) = n!a_n$. The $n!$ comes from differentiate $n$ times $x^n$. From here, $a_n =f^{(n)}(0)/n!$.

Answer 2

Taylor identifies a smooth function to a polynomial by equating the derivatives at zero.

E.g.

$$\sin(x)=p_0+p_1x+p_2x^2+p_3x^3+\cdots$$

giving

$$\sin(0)=\left.p_0+p_1x+p_2x^2+p_3x^3+\cdots\right|_{x=0}=p_0$$ $$\sin'(0)=\left.p_1+2\,p_2x+3\,p_3x^2+\cdots\right|_{x=0}=p_1$$ $$\sin''(0)=\left.2\,p_2+3\cdot2\,p_3x+\cdots\right|_{x=0}=2\,p_2$$ $$\sin'''(0)=\left.3\cdot2\,p_3+\cdots\right|_{x=0}=3!\,p_3$$ $$\cdots$$