Timeline for Finding solutions to $f'(x) = f(x + k)$
Current License: CC BY-SA 2.5
5 events
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| Sep 24, 2010 at 18:23 | comment | added | sleepless in beantown | Choli, you are right. I will change my answer accordingly. | |
| Sep 24, 2010 at 14:49 | comment | added | Choli | Yes, for me the sine is a non-trivial solution (trivial solution would be $f(x)=0$). I understand your point (transforming the sine so its period matches the delay ($k$) we want). However, I still don't get it. For example: $f(x) = a \sin (bx+c)$ Then $f'(x) = ab \cos (bx+c) = ab \sin (bx+c+\frac{\pi}{2})$ $f(x+k) = a \sin(bx+bk+c$ Therefore, if $f'(x)=f(x+k)$: $a=ab$ (1) and $bk-\frac{\pi}{2} = 2\pi n$ (2) From (1), either $a=0$ or $b=1$. We choose $b=1$ ($a=0 \to f(x)=0$). Then, in (2): $k-\frac{\pi}{2} = 2\pi n$, therefore the only solution appears when $k=2\pi n+\frac{\pi}{2}$. Am I wrong? | |
| Sep 24, 2010 at 13:22 | history | edited | sleepless in beantown | CC BY-SA 2.5 |
added 21 characters in body
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| Sep 24, 2010 at 13:17 | history | edited | sleepless in beantown | CC BY-SA 2.5 |
fixed it so that the derivative is valid, and pointed out that cosine with phase shift is a valid solution; deleted 66 characters in body
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| Sep 24, 2010 at 13:05 | history | answered | sleepless in beantown | CC BY-SA 2.5 |