8
$\begingroup$

There is a general formula for the derivative of a function:

$$\frac{d^n}{dx^n}f(x)=\lim_{\epsilon\to0}\frac{1}{\epsilon^n}\sum_{j=0}^n{((-1)^j\frac{\Gamma(n+1)}{j!\Gamma{(n+1-j)}}f(x-j\epsilon))}$$

Where $\Gamma(x) $ is the Gamma function

I tried using the formula to evaluate the 3rd derivative of $\cos(x)$, but I get confused quickly. It would be very appreciated if someone could show a step by step solution to this problem.

I'm totally aware the answer is $\sin(x)$, but what's the process to get to that solution?

$\endgroup$
  • 8
    $\begingroup$ Recommended tip: Don't use that definition. Just take the first derivative 3 times. $\endgroup$ – Simply Beautiful Art Oct 16 '17 at 18:36
  • $\begingroup$ I know of that method, but I’m looking for an answer utilizing the formula I stated. $\endgroup$ – Tom Himler Oct 16 '17 at 18:39
  • 3
    $\begingroup$ "I tried using the formula... but I get confused quickly." In order to receive the best possible answers, it would be helpful to know what you can and can't do, so please include your own efforts. $\endgroup$ – Simply Beautiful Art Oct 16 '17 at 18:40
  • 1
    $\begingroup$ Good exericse to warm up the brain] $\endgroup$ – Guy Fsone Oct 16 '17 at 18:47
  • $\begingroup$ That formula cries desperately for refactoring. $\endgroup$ – Thom Smith Oct 16 '17 at 22:02
11
$\begingroup$

$\Gamma(n+1)$ is an affected way of writing $n!$. So the RHS is $$\newcommand{\ep}{\epsilon}\lim_{\ep\to0}\frac1{\ep^n} \sum_{j=0}^n(-1)^j\frac{n!}{j!(n-j)!}f(x+j\ep)= \lim_{\ep\to0}\frac1{\ep^n} \sum_{j=0}^n(-1)^j\binom{n}{j}f(x+j\ep).$$ This isn't quite correct: there's a sign error, it should be $$(-1)^n\lim_{\ep\to0}\frac1{\ep^n} \sum_{j=0}^n(-1)^j\binom{n}{j}f(x+j\ep) =\lim_{\ep\to0}\frac1{\ep^n} \sum_{j=0}^n(-1)^{n-j}\binom{n}{j}f(x+j\ep).$$ For $n=3$ it should be $$\lim_{\ep\to0}\frac{f(x+3\ep)-3f(x+2\ep)+3f(x+\ep)-f(x)}{\ep^3}.$$

When in addition $f(x)=\cos(x)$ then we get $$f(x+\ep)=\cos \ep\cos x-\sin \ep\sin x,$$ $$f(x+2\ep)=(\cos^2 \ep-\sin^2\ep)\cos x-2\sin\ep\cos\ep\sin x,$$ etc. You eventually get $$\lim_{\ep\to 0}\frac{G(\ep)\cos x+H(\ep)\sin x}{\ep^3}$$ where $G(\ep)/\ep^3$ and $H(\ep)/\ep^3$ are functions that should tend to $0$ and $1$ as $\ep\to0$. I don't want to go into the details though $\ddot\frown$.

$\endgroup$
  • 2
    $\begingroup$ Does this formula for the n-th derivative have a name or does anyone know it's originator ? Just curious. $\endgroup$ – StephenG Oct 17 '17 at 1:34
  • 1
    $\begingroup$ To answer my own question I believe it's the Grünwald–Letnikov derivative although please correct me if someone knows that's wrong. $\endgroup$ – StephenG Oct 17 '17 at 4:13
3
$\begingroup$

We want $$ \lim_{h \to 0} \frac{\cos{(x+3h)}-3\cos{(x+2h)+3\cos{(x+h)}-\cos{x}}}{h^3}. $$ Then $$ \cos{(x+3h)}-\cos{(x+2h)} = -2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{5h}{2}\right)} \\ -\cos{(x+2h)}+\cos{(x+h)} = 4\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{3h}{2}\right)} \\ \cos{(x+h)}-\cos{x} = -2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{h}{2}\right)}, $$ then $$ -2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{5h}{2}\right)} + 2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{3h}{2}\right)} = -\left( 2\sin{\left(\frac{h}{2}\right)}\right)^2 \cos{\left( x + 2h \right)} \\ 2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{3h}{2}\right)} -2\sin{\left(\frac{h}{2}\right)}\sin{\left(x+\frac{h}{2}\right)} = \left( 2\sin{\left(\frac{h}{2}\right)}\right)^2 \cos{\left( x + h \right)}, $$ and finally $$ -\left( 2\sin{\left(\frac{h}{2}\right)}\right)^2 \cos{\left( x + 2h \right)} + \left( 2\sin{\left(\frac{h}{2}\right)}\right)^2 \cos{\left( x + h \right)} = \left( 2\sin{\left(\frac{h}{2}\right)}\right)^3\sin{\left( x + \frac{3h}{2} \right)}. $$ Then $$ \lim_{h \to 0} \frac{1}{h^3}\left( 2\sin{\left(\frac{h}{2}\right)}\right)^3 = 1, $$ and the other term tends to $\sin{x}$, so the whole thing converges to $\sin{x}$. Exactly the same argument works for any number of derivatives, as can be shown by induction.

$\endgroup$

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.