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I've been studying electronics, where they make great use of the relationship between the sine and exponential functions ($e^{i \omega t} = \cos{\omega t} + i \sin \omega t)$. This relationship is confusing to me, so I started digging into it, and thinking about how they have similar definitions, in terms of differential equations.

$f(x) = e^x$ is the solution to this differential equation:

$$ f'(x) = f(x) $$

and $f(x) = \sin x$ is a solution to this similar equation:

$$ f'(x) = f(x + \pi/2) $$

I wanted to see solutions to the following, for other values of constant $k$.

$$ f'(x) = f(x + k) $$

but my differential equation solving skills are non-existent. So my main question is: What are these functions, and what do their graphs look like? A secondary question is: Do you know how to write the bit of code necessary to solve that third DE (for some value of $k$) using sage or wolfram alpha? I have sage but don't know what to write.

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See my answer here. –  Mhenni Benghorbal Nov 23 '12 at 20:44
    
There is a typo in the last equation, $k\mapsto x$. –  NikolajK Nov 23 '12 at 21:09
    
Thanks, fixed it. –  Rob N Nov 23 '12 at 21:50
    
Another interesting feature, following from you differential equation, that I realized when thinking about your problem is the implication for $f^{(n)}(x)$, namely shift to $f(x+nk)$. –  NikolajK Nov 23 '12 at 23:12

3 Answers 3

up vote 5 down vote accepted

They are all solutions of some $$ f'' + A f' + B f = 0 $$ with constants $A,B.$ The constants can be real numbers for $e^x, \sin x,$ but the full story allows them complex as needed.

Don't get sidetracked by the first order delay differential equations $f'(x) = f(x+k).$ The reason that appears is that there are identities for $\sin (x+k)$ and $\cos(x+k).$

As a simple example, $$ f'' + 2 f' + 2 f = 0 $$ has solutions $$ e^{-x} \sin x , \; \; e^{-x} \cos x $$ as (real-valued) solutions. So your $f(x)$ could be $$ f(x) = C e^{-x} \cos x + D e^{-x} \sin x $$ with real constants $C,D.$ Note that damping effect of the $e^{-x},$ which says that any such $f$ oscillates but goes fairly rapidly to $0.$ This is the type of phenomenon worked with in an automobile spring/shock absorber system.

The similar but unusual $$ f'' + 2 f' + f = 0 $$ has solutions $$ e^{-x} , \; \; x e^{-x} $$ as (real-valued) solutions. So your $f(x)$ could be $$ f(x) = C e^{-x} + D x e^{-x} $$ with real constants $C,D.$

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If I could I would accept more than one answer; all were helpful. The others address the delay differential equation; but as you said, that would be a sidetrack given my immediate goal of understanding the electric circuits. Thanks everyone! –  Rob N Nov 30 '12 at 0:08

There's a good reason why you're finding yourself unable to find solutions in the latter cases. Those are examples of delay differential equations and they have a theory that's substantially more complicated than that of ODEs.

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A year ago I asked a similar question here. So let me generalize GEdgars nice answer. Choose a $k$ and consider

$$f(x):=\sum_n a_n\ \text{e}^{M_n x/k},$$

with some magic numbers $M_n$ fulfilling the relation $\text{e}^{M_n}=M_n/k$. These are related to the Lambert W function as explained in the question I linked you to above. Then

$$f(x+k)=\sum_n a_n\ \text{e}^{M_n x/k}\ \text{e}^{M_n}=\sum_n a_n\ \text{e}^{M_n x/k}\ M_n/k=f'(x).$$

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