Find $f(x)$ if $\Delta f(x)=e^x$, where $\Delta f(x)$ is the first order forward difference of $f(x)$, step size $=h=1$.

Attempt: We have the definition $\Delta f(x)=f(x+h)-f(x)=f(x+1)-f(x)$

Given $\Delta f(x)=e^x$ i.e $f(x)=\Delta^{-1}e^x=(E-1)^{-1}f(x)$ where $E$ is the shift operator (i.e $Ef(x)=f(x+h)=f(x+1)$).

But it is very troublesome to get the answer.

Answer is given as $f(x)=\frac{e^x}{e-1}$

Please help.

  • $\begingroup$ The function $f$ is not unique. One can get a unique $f$ by specifying that it should be monotonic and specifying $\lim\limits_{x\to-\infty}f(x)$ or simply $f(x_0)$ for some $x_0$. Otherwise, you can add any function with period $1$ to $f$. $\endgroup$ – robjohn Oct 22 '15 at 15:02
  • $\begingroup$ Pretty sure that $f(E)e^x=f(e)e^x$ for all functions $f$. For example, above, the answer was $(E-1)^{-1}e^x=(e-1)^{-1}e^x$. $\endgroup$ – Akiva Weinberger Oct 22 '15 at 15:14
  • $\begingroup$ @AkivaWeinberger please explore your.answer. Why $f(E)e^x=f(e)e^x$? Why $(E-1)^{-1}e^x=(e-1)^{-1}e^x$ is obtained simply replacing $E$ by $e$? $\endgroup$ – user1942348 Oct 22 '15 at 15:28
  • 1
    $\begingroup$ @user1942348 Well, $Ee^x=e^{x+1}=ee^x$, right? And $E^2e^x=e^{x+2}=e^2e^x$. I'm just recognizing patterns, basically. $\endgroup$ – Akiva Weinberger Oct 22 '15 at 15:30

$$(1) \quad \Delta f(x)=e^x$$ Which is equivalent to, $$(2) \quad f(x+1)=f(x)+e^x$$

Assume that an initial condition for $f(0)$ holds. We then have,

$$(3) \quad f(x)=g(x)+\sum_{n=0}^{x-1} e^n$$

Where $x \ge 1$. The nature of $g(x)$ will be shown momentarily. To prove $(3)$, we'll substitute back into $(2)$,

$$(2.1) \quad \color{red}{f(x+1)}=\color{blue}{f(x)}+\color{green}{e^x}$$

$$(4) \quad \color{red}{g(x+1)+\sum_{n=1}^{x} e^n}=\color{blue}{g(x)+\sum_{n=1}^{x-1} e^n}+\color{green}{e^x}$$

Which is obviously true, as long as $g(x)=g(x+1)$. This implies that $g(x)$ must be periodic. The sum in $(3)$ is geometric, and may be evaluated to be,

$$(5) \quad \sum_{n=0}^{x-1} e^n=\cfrac{e^x-1}{e-1}$$

So we have as the final solution,

$$(6) \quad f(x)=g(x)+\cfrac{e^x-1}{e-1}$$

Where $g(x)$ is any periodic function with period $1$ with $g(0)=f(0)$. I should also note that in the passing from summation to $(6)$, the restrictions on $x$ have been lifted. Assuming $g(x)=f(0)$ for all $x$, $x$ may now be any real number and still satisfy $(1)$. If $g(x)$ is non-constant, then $x$ must still be an integer.

  • $\begingroup$ Unless I am mistaken, the intermediate equations make sense only for integral $x$. $\endgroup$ – Martin R Oct 22 '15 at 14:32
  • $\begingroup$ Actually $(6)$ is general, and satisfies $(1)$ for any real $x$. The intermediate equations are just tools to get to $(6)$. $\endgroup$ – Zach466920 Oct 22 '15 at 14:34
  • $\begingroup$ Filling a detail omitted in the above: let $y \in [0,1)$, then consider $x=y+n$ for nonnegative integers $n$. Following the argument above you get $f(x)=f(y)+\frac{e^n-1}{e-1}$. Note that the values of $f(y)$ can be chosen completely arbitrarily. $\endgroup$ – Ian Oct 22 '15 at 14:40
  • $\begingroup$ And instead of $f(0)$ you could put an arbitrary function of the floor of $x$. $\endgroup$ – GEdgar Oct 22 '15 at 14:41
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    $\begingroup$ @user1942348 I edited the answer. The solution is more general, and will work with any periodic function $g(x)$, with period 1, and $g(0)=f(0)$. Just pick an appropriate $g(x)$, or constant, to satisfy your initial condition. $\endgroup$ – Zach466920 Oct 22 '15 at 15:05

Since $f$ is known up to a function with period $1$, let's try to find a monotonically increasing $f$.

Since $f(x-k+1)-f(x-k)=e^{x-k}$, we have that $\lim\limits_{x\to-\infty}f(x)$ exists. Furthermore, $$ \begin{align} f(x)-\lim_{x\to-\infty}f(x) &=\sum_{k=1}^\infty\left[f(x-k+1)-f(x-k)\right]\\ &=\sum_{k=1}^\infty e^{x-k}\\ &=\frac{e^x}{e-1} \end{align} $$ Therefore, $$ f(x)=\frac{e^x}{e-1}+p(x) $$ where $p(x)$ is any function with period $1$.


So you want to solve $$f(x+ 1)- f(x)= e^x.$$ It should be obvious that $f$ must be of the form $$f(x)= Ae^{bx}.$$ Then $$f(x+ 1)= Ae^{bx+ b}= (Ae^b)e^{bx}$$ so that $$f(x+ 1)- f(x)= (Ae^b)e^{bx}- Ae^{bx}= (Ae^b- A)e^{bx}= e^x.$$ We can take $b= 1$ and that reduces to $$Ae^b- A= A(e- 1)e^x= e^x$$ or $A(e- 1)= 1$, thus $A= \tfrac1{e- 1}$.

  • $\begingroup$ Is it not more general to assume $$f(x)= Ae^{bx}+C$$ $\endgroup$ – user1942348 Oct 22 '15 at 15:00
  • $\begingroup$ Why you have not taken $f(x)= Ae^{bx}+C$ $\endgroup$ – user1942348 Oct 22 '15 at 15:08

Taking the first order forward difference of the exponential, you get

$$\Delta e^x=(e-1)e^x.$$

  • $\begingroup$ How does this answer the question? $\endgroup$ – Calle Oct 22 '15 at 17:21
  • $\begingroup$ Actually, this does provide one answer to the question: since $\Delta$ is linear, the above implies that $\Delta \left ( \frac{e^x}{e-1} \right ) = e^x$. Noting that the difference equation is first order, this means that the general solution is $\frac{e^x}{e-1}$ plus any function with period $1$...which is Zach466920's answer. :) $\endgroup$ – Ian Oct 22 '15 at 18:30
  • $\begingroup$ @Ian: I dropped the homogenous solution on purpose, to match the question (Answer is given as ...). $\endgroup$ – Yves Daoust Oct 22 '15 at 19:41
  • $\begingroup$ @calle: by linearity of the first order difference operator. You obviously have $\Delta f=e^x$ when $f=e^x/(e-1)$. $\endgroup$ – Yves Daoust Oct 22 '15 at 19:42
  • $\begingroup$ I see. You could have mentioned that in the answer. $\endgroup$ – Calle Oct 24 '15 at 23:42

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