7
$\begingroup$

My theory:

derivative of $\cos x = - \sin x $

derivative of $-\sin x = -\cos x $

derivative of $-\cos x = \sin x.$

cycle occurs three times but then what do you do??

Is there a good way to solve this?

$\endgroup$
  • 1
    $\begingroup$ The period is 4. Divide 99 by four, and see if its remainder is 0, 1, 2, or 3. $\endgroup$ – Pedro Tamaroff Nov 4 '13 at 2:59
12
$\begingroup$

The cycle occurs, not three times but 4 times. Thus every 4 derivatives taken you return to $\cos(x)$. Therefore, since $99=96+3$, you need the $3$rd derivative of $\cos(x)$ which is $\sin(x)$.

$\endgroup$
14
$\begingroup$

We have

\begin{align*} f(x) &= \cos{x} \\ f^{(1)}(x) &= -\sin{x} \\ f^{(2)}(x) &= - \cos{x} \\ f^{(3)}(x) &= \sin{x} \\ f^{(4)}(x) &= \cos{x} \end{align*}

So the cycle has length $4$. In particular, we'll see that

$$f^{(8)}(x) = \left(\cos{x}\right)^{(8)} = \left(\cos{x}\right)^{(4)} = \cos{x}$$

Likewise for $12$, $16$, and so on.

So now use the fact that $$99 = 4 \cdot 24 + 3$$

$\endgroup$
13
$\begingroup$

Notice $\quad\displaystyle \cos(x) = \frac12 (e^{ix} + e^{-ix})\quad$ and $e^{\pm i x}$ are eigenfunctions of the operator of taking derivative against $x$ with eigenvalues $\pm i$. i.e. $\quad\displaystyle\frac{d}{dx} e^{\pm ix} = \pm i e^{\pm ix}.$ We have $$f^{(99)}(x) = \frac{d^{99}}{dx^{99}} \frac12 (e^{ix} + e^{-ix}) = \frac12 ( i^{99} e^{ix} + (-i)^{99} e^{-ix} ) = \frac12 ( -i e^{ix} + i e^{-ix} ) = \sin(x)$$

$\endgroup$
  • 6
    $\begingroup$ yikes!!!!!!!!!!! $\endgroup$ – Jessica Nov 4 '13 at 3:14
  • $\begingroup$ @Jessica It's actually not that "yikes" once you know complex numbers... $\endgroup$ – Tobias Kienzler Nov 4 '13 at 8:58
7
$\begingroup$

$ f = \cos x , f' = -\sin x , f'' = -\cos x , f''' = \sin x , f^{(4)} = \cos x $

Therefore, the derivatives of sine cycle every $4$. In particular since $99 = 3 (mod 4 )$, so $f^{(99)} = f''' = \sin x $

$\endgroup$

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.