Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm discussing this in another forum but I'd like to run it past you as well. I'm observing a periodic comb function $f(t_i)$ such that all its non-zero values within the period [0,T] are

$f(t_0)=−5$,

$f(t_1)=−2.5$,

$f(t_2)=0$,

$f(t_3)=−2.5$,

$f(t_4)=−5$,

$f(t_5)=−7.5$,

$f(t_6)=−10$,

$f(t_7)=−7.5$,

$f(t_8)=−5$

where $f(t_8)=f(T)$ and I'm looking for the first derivative of that function at each one of the non-zero points -- 9 altogether. Mathematicians, most likely, will not agree that this is a differentiable function (a function should be continuous in order to be differentiable) but it appears to me that I can consider the real change of that function around each point in this particular case as $f(t_{i+1})−f(t_{i−1})$ and, therefore, the first derivative over time at each point can be written as $\frac{f(t_{i+1})−f(t_{i−1})}{t_{i+1}−t_{i−1}}$. If that's not a first derivative then what would you call that ratio?

EDIT: I think the problem boils down to the following:

Let us denote the comb function by f(t) although in our case we know it consists of only 9 values. This is given. Now, the theorem that has to be proved is that no matter what way of obtaining the 9 outcomes from f(t) we may find (note I'm not calling these outcomes by the term derivatives), which we will denote by f'(t), there will always be a condition (offset) whereby the sum of the 9 products f(t)f'(t) will be negative.

EDIT 2:

I'll be trying to refine further the problem:

Let us denote the periodic comb function by $f(t)$, consisting of only $2n+1$ values within a period $[0.T]$, then no matter what way of obtaining the $2n + 1$ difference quotient outcomes from $f(t)$ we may find (for instance, backward, central or forward difference quotients), which we will denote by $f^{dq}(t)$, there will always be a condition (offset) whereby the sum of the $2n + 1$ products $f(t)f^{dq}(t)$ will be negative.

share|improve this question
    
On what domain is $f$ defined? Also, en.wikipedia.org/wiki/Symmetric_derivative –  Jonathan Sep 29 '12 at 16:23
    
As far as I can tell, the function at hand is defined on the domain of all real integer numbers. Also, there is no limiting process characteristic for a derivative of a continuous function. –  ganzewoort Sep 29 '12 at 16:38
    
The ordinary derivative blows up and the symmetric derivative vanishes at the bad points. It sounds like what you're really after is a discrete interpretation. –  user17794 Sep 29 '12 at 17:46
    
What puzzles me is that when I sum up the 9 central difference quotients I mentioned above (to find the value of the integral within the period [0,T]), I get a non-zero sum. The integral over a period of any periodic function, however, is zero. What do you think about this difference in the outcome when integrating comb function compared to continuous function? –  ganzewoort Sep 30 '12 at 6:43
1  
I'm not sure why you call that a "comb function". It's just a train of (equispaced) Dirac deltas with periodic amplitudes? Why not consider it just a discrete function and speak of finite differences instead of derivatives? –  leonbloy Oct 2 '12 at 23:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.