If $f(x)=e^{ax}$, show that $\Delta^n f(x)=(e^{ax}-1)^ne^{ax}$

Question

Given as an assignment for Interpolation. I, first of all, doubt whether the question is correct or not, because $$\Delta f(x)=e^{ax+ah}-e^{ax}=e^{ax} (e^{ah}-1)$$ $$\Delta^2 f(x)=[e^{ax+ah} (e^{ah}-1)]-[e^{ax} (e^{ah}-1)]=e^{ax} (e^{ah}-1)^2$$ So, $\Delta^n f(x)=e^{ax} (e^{ah}-1)^n$.

And if the question is correct how should I begin solving it. I do understand that Stack Exchange don't do homeworks, but I need help for beginning the solution

What is the question? This is unclear from your post. – parsiad Apr 03 '16 at 16:06 — parsiad, Apr 03 '16 at 16:06

score 1 · Answer 1 · answered Apr 03 '16 at 17:20

Your computation is correct, the task description is dubious at best.

Hint: Without further qualification, and if it is not the Laplace operator, the implied step size is $h=1$, so that $$ (Δf)(x)=f(x+1)-f(x). $$ $Δ_hf$ is used for the more general difference operator.

If $f(x)=e^{ax}$, show that $\Delta^n f(x)=(e^{ax}-1)^ne^{ax}$

1 Answers1