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StevensElectronicAccount 
Posted: September 30, 2011 12:02 am

Jr. Member Group: Trusted Members Posts: 73 Member No.: 31,855 Joined: September 28, 2010 
How do I solve the equation f'(t+h)+f(t)=0 where h is a constant?
I know that one solution for f'(t+1)+f(t)=0 is f(t) = k*m^t where m is some weird number such that m^m = e that natural growth number. This question has been confusing me because it combines the difficulty of differential equations with the difficulty of equations like f(t+h)f(t)=0. I know the equation f(t+h)f(t)=0 solves for periodic functions. The equation f'(t+h)+f(t)=0 probably solves for something even weirder. Edit: Just remembered the word. This has the difficulty of a recurrence relation multiplied with the difficulty of a differential equation. 
sherlock ohms 
Posted: September 30, 2011 01:02 pm

Forum Addict ++ Group: Spamminator Taskforce Posts: 2,789 Member No.: 26,125 Joined: September 10, 2009 
h=0 works algebraically. K.I.S.S.
 "Quotation marks make sentences appear more meaningful."

StevensElectronicAccount 
Posted: September 30, 2011 11:12 pm

Jr. Member Group: Trusted Members Posts: 73 Member No.: 31,855 Joined: September 28, 2010 
You seem to be referencing the second equation I listed f(t+h)f(t)=0.
I was giving that equation as an similar example I already know all about. I was asking how do I solve the equation f'(t+h)+f(t)=0 where h is a constant for possible functions f(t)? 
Sch3mat1c 
Posted: October 02, 2011 04:13 pm

Forum Addict ++ Group: Moderators Posts: 20,583 Member No.: 73 Joined: July 24, 2002 
To get the two points to sum to zero, you'll be looking at, for example, the half period of an *odd* periodic function, for instance sin(t) at h = (2N + 1) * pi (for all integer N). There are numerous functions which satisfy this, for example square waves (which are simply built of components of odd harmonics, i.e. all values of N, with decreasing amplitudes).
I'm not sure if it's necessary for the function to be periodic, though if it is, I'm pretty sure it must be a period of 2h. If you do the good old chain rule, you get: df(t) / dt + df(t + h) / d(t + h) * d(t + h) / dt = 0 and of course we know that d(t+h) = dt, since h is a constant, so the differential equation of the original equation tells us nothing we did not already know. In other words, this will help us constrain our choice of functions (because we know the derivative has to follow the same rule), but it won't help us choose or create such functions, because it's just as general. Tim  Answering questions is a tricky subject to practice. Not due to the difficulty of formulating or locating answers, but due to the human inability of asking the right questions; a skill that, were one to possess, would put them in the "answering" category.

migueloreles 
Posted: October 11, 2011 02:37 am

Newbie Group: Members+ Posts: 5 Member No.: 3,248 Joined: September 20, 2005 
hi,
i think you can solve this by fourier analysis. don't know if you're familiar with it, but simply put, you solve it for a generic function as f(t)=A exp(i w t) and then your final function will be a sum (actually an integral) over your solutions. the w's are the frequencies, so you're basically decomposing a function in its spectrum. since the differentiating operator is linear you can find several solutions, add them together and you still have a solution. so in this case f'(t) = i w A exp(i w t) f'(t+h) = A i w exp(i w t) exp( i w h) = iw exp(iwh) f(t) and you want f'(t+h)=f(t) so iw exp(iwh) f(t) = f(t) which means i w exp(i w h)=1 now you have a way of finding which "h" is satisfied for each w: h(w)=log (i/w)/(iw) now you can build any function from a sum of these functions or an integral (a Fourier transform). hope this helped miguel 
StevensElectronicAccount 
Posted: October 19, 2011 04:04 pm

Jr. Member Group: Trusted Members Posts: 73 Member No.: 31,855 Joined: September 28, 2010 
Thank you very much.
I was trying to solve this via taylor series. Using a fourier series seems like it would make a lot more sense as this function has some kind of periodic stuff but I don't know how fourier series work yet so I couldn't solve it that way. 
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