2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

$$(1) \quad \Delta f(x)=e^x$$ 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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
4
$\begingroup$

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 $1$.

$\endgroup$
1
$\begingroup$

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 $A= \tfrac1{e- 1}$.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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