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.
On
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 $A= \tfrac1{e- 1}$.
On
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 $1$.
$$(1) \quad \Delta f(x)=e^x$$ 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.