0
$\begingroup$

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 $x = 0$ ?

$\endgroup$
  • $\begingroup$ 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 $$\sum_{k \ge 0} \frac{1}{k!} x^{k}$$ $\endgroup$ – mattos May 15 '17 at 7:18
2
$\begingroup$

$\frac {df}{dx}$ 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 $\frac1{3!}\cdot e^0\cdot x^3 = \frac{x^3}{6}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.