# Math Help - when is nth differential non zero

1. ## when is nth differential non zero

>>>>>>> Question <<<<<<<<<

it is easy to find it generally but i was thinking that what if in some question ... say 100th differential is 0 ... there has to be a shortcut ... some help please

i even found out the general term of what differential would be after the third differential (when $2/3 x^{3}$ goes away)...

its $f_{n}(x) = e^{x} + (-1)^{n-1}e^{-x} -2 sin[n\pi/2 + (-1)^{n}x]$

But i dont know how to find when $f_{n}(x)$ will be non-zero

2. Use Taylor series. In my calculations, the series for f(x) starts with $4x^7/7!$, so the answer is 7.

3. Originally Posted by cupid
>>>>>>> Question <<<<<<<<<

it is easy to find it generally but i was thinking that what if in some question ... say 100th differential is 0 ... there has to be a shortcut ... some help please

i even found out the general term of what differential would be after the third differential (when $2/3 x^{3}$ goes away)...

its $f_{n}(x) = e^{x} + (-1)^{n-1}e^{-x} -2 sin[n\pi/2 + (-1)^{n}x]$

But i dont know how to find when $f_{n}(x)$ will be non-zero
Focus on that last term.
$\displaystyle \left ( \frac{2}{3}x^3 \right ) ^{\prime} = 2x^2$

$( 2x ^2 ) ^{\prime} = 4x$

$( 4x ) ^{\prime} = 4$
definitely non-zero.

-Dan

4. Originally Posted by topsquark
Focus on that last term.
$\displaystyle \left ( \frac{2}{3}x^3 \right ) ^{\prime} = 2x^2$

$( 2x ^2 ) ^{\prime} = 4x$

$( 4x ) ^{\prime} = 4$
definitely non-zero.
You can't focus on the last term only because it may be canceled by other terms.

5. And that is what happens .. 1,2,3 differentials are not 0

6. I have no idea what you are talking about. You can check in WolframAlpha that $\displaystyle\frac{d^7f(x)}{dx^7}(0)=4$ and $\displaystyle\frac{d^nf(x)}{dx^n}(0)=0$ for $n=1,\dots,6$.

7. sorry i meant that they are 0

typing mistake

but isnt there is any general method to do these kind of problems

8. Originally Posted by emakarov
You can't focus on the last term only because it may be canceled by other terms.
Good point.

-Dan

9. but isnt there is any general method to do these kind of problems
I don't know the answer because I don't know the exact class of problems you are referring to. For an arbitrary function f(x) given by its formula, I think the easiest way is to compute the nth derivative and evaluate it at 0. This particular problem seems to me to be designed to be solved using Taylor series. However, concerning problems that you may encounter during tests, etc., I don't know how similar to this one they will be.