Prove that Cexp(x)C\exp(x) is the only set of functions for which f(x)=f′(x)f(x) = f'(x)

I was wondering on the following and I probably know the answer already: NO.

Is there another number with similar properties as e? So that the derivative of exp(x) is the same as the function itself.

I can guess that it’s probably not, because otherwise e wouldn’t be that special, but is there any proof of it?

Answer

Of course Cex has the same property for any C (including C=0). But these are the only ones.

Proposition: Let f:RR be a differentiable function such that f(0)=1 and f(x)=f(x). Then it must be the case that f=ex.

Proof. Let g(x)=f(x)ex. Then

g(x)=f(x)ex+f(x)ex=(f(x)f(x))ex=0

by assumption, so g is constant. But g(0)=1, so g(x)=1 identically.

N.B. Note that it is also true that ex+c has the same property for any c. Thus there exists a function g(c) such that ex+c=g(c)ex=ecg(x), and setting c=0, then x=0, we conclude that g(c)=ec, hence ex+c=exec.

This observation generalizes to any differential equation with translation symmetry. Apply it to the differential equation f and you get the angle addition formulas for sine and cosine.

Attribution
Source : Link , Question Author : Timo Willemsen , Answer Author : Qiaochu Yuan

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