Differential equation $\Delta^k f(x)$.

Question

Let $f(x)$ be a function of real variable and let $\Delta f$ be the function $\Delta f=f(x+1)-f(x)$. For $k>1$, put $\Delta^k f=\Delta(\Delta^{k-1}f)$. Then $\Delta^k f(x)$ equals:

$$\text{A) }\sum_{j=0}^{k} (-1)^{j} \binom{k}{j}f(x+j)$$ $$\text{B) }\sum_{j=0}^{k} (-1)^{j+1} \binom{k}{j}f(x+j)$$ $$\text{C) }\sum_{j=0}^{k} (-1)^{j} \binom{k}{j}f(x+k-j)$$ $$\text{D) }\sum_{j=0}^{k} (-1)^{j+1} \binom{k}{j}f(x+k-j)$$

Please tell me how to go about this question, and which topic should i study to be able to solve similar questions, thanks in advance!

Answer 1

Leaving $f(x)$ unknown, just start computing.

You already have $\Delta f(x)$.

Use the definition of $\Delta$ to get ${\Delta}^2f(x)$.

Next get ${\Delta}^3f(x)$.

And so on ...

Stop when all but one of the choices are eliminated.

Had there been no multiple choices offered, simply continue until a pattern becomes clear. You can then prove the validity of the discovered pattern by induction.

Answer 2

Let $S$ be the shift of the function by $1$, $(Sf)(x)=f(x+1)$, $(S^mf)(x)=f(x+m)$. Then $Δ=S-1$ and by the binomial theorem $$ Δ^k=(S-1)^k=\sum_{j=0}^k\binom{k}{j}(-1)^j\,S^{k-j}. $$ Now apply to $f$ at the point $x$.