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.

Does there exist a twice differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

$$f'(x) = f(x+1)-f(x)$$ for all $x$ and $f''(0) \ne 0$.

Some things that I have tried is that obviously any linear function satisfies the first condition but not the second. Any hints/motivation would be helpful. Thanks.

share|improve this question
    
I'm afraid that's unnacceptable behaviour ssilwa. You'll end up getting yourself arrested one of these days. –  Eulerian Adventurer Sep 15 at 18:25
    
@EulerianAdventurer Ok, cool story bro. –  Sandeep Silwal Sep 15 at 18:48

2 Answers 2

Since we are just looking for existence, this will do.

Assume $f(z)$ is a function on complex plane instead, and assume it have the form $f(z)=e^{cz}$ for some constant $c$. The condition $f''(z)\not=0$ simply means $c\not=0$, while the other condition give: $ce^{cz}=(e^{c}-1)e^{cz}$ which means we need to solve for $e^{c}-c-1=0$. Applying standard technique to reduce to: $-(c+1)e^{-(c+1)}=-\frac{1}{e}$ which allow us to use Lambert W-function to get $c=-1-W(-\frac{1}{e})$. Hence found the solution in complex number.

To reduce this back to real number just take the real part, so we get: $f(x)=e^{Re(c)x}\cos(Im(c)x)$ which satisfy all condition.

share|improve this answer
    
Thanks but I am looking for a more "elementary" approach. –  Sandeep Silwal Jun 9 at 4:57

Since the question is about existence: you can think of the right side as a difference quotient. $$f^{'}(x) =\frac{f(x+1)-f(x)}{(x+1)-x}.$$ The right hand side is a slope of the secant line between the points with $x-$coordinates $x$ and $x+1$. So any affine function (with non-zero slope and non-zero constant term) will satisfy this relation. For sure they are twice differentiable.

By affine I mean: $f(x)=ax+b$

share|improve this answer
    
@user88595 Thanks for being so prompt with pointing out mistakes. I had read the question but due to bad usage of wording I wrote linear instead of saying straight line functions. –  Anurag A Jun 2 at 22:18
    
Sorry, but I had one of the conditions wrong. Please look again. –  Sandeep Silwal Jun 2 at 22:34

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.