I'm learning Taylor series and I am currently trying to figure how to estimate the value of $e$ using the Lagrange remainder.
In school we rewrote $e^x$ as $1+\frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^n}{n!} + Rn(x)$ and we used the Lagrange remainder as Rn(x). Thus $Rn(x)=\frac{e^{\theta x}*x^{n+1}}{(n+1)!}$.
What I don't understand is how we got the $Rn(x)$. I know that the Lagrange remainder is $Rn(x)=\frac{(x-x_0)^{n+1}}{(n+1)!}*f^{n+1}(x_0+\theta(x-x_0))$. Can someone explain to me how did we get from the formula to the $\frac{e^{\theta x}*x^{n+1}}{(n+1)!}$?
Thanks
You have developed the Taylor series around $x_0=0$, i.e. to get the Lagrange remainder, you input $x_0=0$ and get \begin{align} Rn(x)&=\frac{(x-0)^{n+1}}{(n+1)!}*f^{(n+1)}(0+\theta(x-0)) \\ &=\frac{x^{n+1}e^{\theta x}}{(n+1)!} \end{align} Where we have used that $f^{(n)}$ is the $n$'th derivative of $f$, and that $e^x$ is its own derivative.