# Lets apply Taylor series to the function $e^x$

I'm taking a single variable calculus course and following is given :

The Taylor Series of a function f at an input 0 is :

$\sum \limits_{k=0}^\infty \frac {f^k(0)} {k!} x^k = f(0) + \left.\frac {df}{dx}\right|_0 x+ \frac 1{2!} \left. \frac {{d^2}f} {d{x^2}}\right|_0x^2 + \dots$

$\left.\frac {df}{dx}\right|_0 x$ : f at 0 + the derivative at 0 times x

$\frac 1{2!} \left. \frac {{d^2}f} {d{x^2}}\right|_0x^2$ : 1 over 2 factorial times the second derivative at 0 times $x^2$

That is $C_k = \frac {f^k(0)} k! = \frac 1 {k!} \left. \frac {d^k} {dx^k}\right|_0$

That is the k'th coefficient is equal of the k'th derivative of f evaluated at the input 0 and then divided by k factorial.

The instructor then applies Taylor series to the function $e^x$ :

Lets apply Taylor series to the function $e^x. f(x)=e^x$ In order to compute we need to know the derivatives $e^x$ and evaluate them at x = 0. The derivative of $e^x$ is $e^x$. So all derivatives in Taylor series evaluates to $f(x) = e^x = 1 + x + \frac 1 {2!}x^2 + \frac 1 {3!}x^3 + \frac 1 {4!}x^4 + \dots$ which is the series for $e^x$

How is $f(x) = e^x = 1 + x + \frac 1 {2!}x^2 + \frac 1 {3!}x^3 + \frac 1 {4!}x^4 + \dots$ arrived at ?

$f(0) = e^0 = 1$

$\left.\frac {df}{dx}\right|_0 x = e^?$ ?

Does $\left.\frac {df}{dx}\right|_0 x$ mean change in $f$ divided by change in $x$ where $x = 0$ ?

• Due to the fact that the exponential is it's own derivative, we have $\forall k, f^{k}(0) = 1$. So the coefficients are given by $a_{k} = 1/k!$. So the Taylor series for $e^{x}$ is given by $$\sum_{k \ge 0} \frac{1}{k!} x^{k}$$ – mattos May 15 '17 at 7:18

$\frac {df}{dx}$ means the derivative of $f$. In our case the derivative of $e^x$, which is $e^x$.
$\left.\frac {df}{dx}\right|_0$ means the derivative of $f$ at the point $x = 0$ (also known as $f'(0)$, if that's more familiar to you). In our case, $e^0 = 1$. Note that this is always a number, not a function.
$\left.\frac {df}{dx}\right|_0 x$ means the derivative of $f$ at the point $x = 0$, multiplied by $x$. In our case $e^0\cdot x = 1x = x$.
The higher order terms are the same, except you differentiate multiple times instead of just one, and you divide by a factorial. For instance, the third order term $\frac1{3!}\left.\frac {d^3f}{dx^3}\right|_0 x^3$, which means $\frac1{3!}$ multiplied by (the third derivative of $f$ at the point $x = 0$) multiplied by $x^3$. In our case, the third derivative of $f$ is still $e^x$, so we get $\frac1{3!}\cdot e^0\cdot x^3 = \frac{x^3}{6}$.