# Forward difference operator

1. Feb 17, 2008

### foxjwill

[SOLVED] Forward difference operator

1. The problem statement, all variables and given/known data
I was looking on Wikipedia and noticed that it said that $$\Delta_h$$ could be written as

\begin{align*} \Delta_h &= \sum_{n=0}^\infty \frac{(hD)^n}{n!}\\ &= e^{hD} - 1 \end{align*}

where $$D$$ is just the standard derivative. What I don't understand is how they came up with the infinite series.

2. Relevant equations

3. The attempt at a solution

2. Feb 17, 2008

### Hurkyl

Staff Emeritus
Try applying it to a function.

3. Feb 17, 2008

### foxjwill

Ok, so I tried going at it this way:

\begin{align} f(x+h) &= f(h) + (x+h)Df(h)+ \dfrac{(x+h)^2 D^2 f(h)}{2!} + \cdots\\ f(x) &= f(0) + xDf(0) + \dfrac{x^2 D^2 f(0)}{2!} + \cdots \end{align}

However, when I try subtracting (2) from (1), the terms don't cancel out because (1) is in terms of $$f(h)$$ and (2) is in terms of $$f(0)$$. What should I do?

4. Feb 17, 2008

### Hurkyl

Staff Emeritus
Trying to express $\Delta_h f(x)$ in terms of $\Delta_h f(0)$ seems like an odd thing to do; I'm sure there's an interesting result to be derived, but I don't see why it would relate to the question you're trying to solve.

Incidentally, your Taylor series in (1) is incorrect.

Last edited: Feb 17, 2008
5. Feb 17, 2008

### foxjwill

I was using the tailor series expansion. Do you have any other ideas?

6. Feb 17, 2008

### Hurkyl

Staff Emeritus
I've added more to my previous post.

I don't have any new ideas; what's wrong with the one I already gave?

7. Feb 18, 2008

### HallsofIvy

Staff Emeritus
As Hurkyl said, the Taylor series expansion for the first one is wrong.

The Taylor series for f(x), expanded around x= a, is
$$f(a)+ f'(a)(x-a)+ f"(a)/2 (x-a)^2+ /cdot/cdot/cdot+ f^{(n)}(a)(x- a)^n+ \cdot\cdot\cdot$$
If you are expanding f(x+h) about x= h, then it would be
$$f(h)+ f'(a)((x+h)-h)+ f"(h)/2 ((x+h)-h)^2+ /cdot/cdot/cdot+ f^{(n)}(h)((x+h)- h)^n+ /dot/cdot/cdot$$
$$= [tex]f(h)+ f'(a)(x)+ f"(a)/2 x^2+ /cdot/cdot/cdot+ f^{(n)}(h)x^n\cdot\cdot\cdot$$

If you are expanding f(x+h) about x= 0, then it would be
$$f(0+ f'(0)(x+h)+ f"(0)/2 (x+y)^2+ /cdot/cdot/cdot+ f^{(n)}(0)(x+y)^n+ \cdot\cdot\cdot$$

8. Feb 18, 2008

### Rainbow Child

What's the wikipedia url for that result?

9. Feb 18, 2008

### foxjwill

Ok, so I fixed your $$\LaTeX$$:

I'm not really sure how you got the expansions to be those. Could you explain?

10. Feb 18, 2008

### foxjwill

11. Feb 19, 2008

### Rainbow Child

$$\Delta_h(f(x))=f(x+h)-f(x)\Rightarrow \Delta_h(f(x))=\left(T_h-I\right)\,f(x)\quad \text{with} \quad T_h f(x)=f(x+h)$$

Expanding $f(x+h)$ around $\alpha=h$ we have

$$f(x+h)=f(h)+\frac{1}{1!}\,f'(h)\,x+\frac{1}{2!}\,f''(h)\,x^2+\dots+\frac{1}{n!}\,f^{(n)}(h)\,x^n+\dots\Rightarrow f(x+h)=\sum_{n=0}^\infty\frac{x^n}{n!}\,f^{(n)}(h)\Rightarrow f(x+h)=\sum_{n=0}^\infty\frac{h^n}{n!}\,f^{(n)}(x)$$

where at the last equality I changed the roles of $x,\,h$ since they are just labels. This equality can be written as

$$f(x+h)=\left(\sum_{n=0}^\infty\frac{(h\,D)^n}{n!}\right)\,f(x)$$

and from here you have the desired result.