Forward difference operator [SOLVED] Forward difference operator

[foxjwill]

[SOLVED] Forward difference operator

Homework Statement

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.

Homework Equations

The Attempt at a Solution

 

[Hurkyl]

Try applying it to a function.

 

[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?

 

[Hurkyl]

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.

 

[foxjwill]

Trying to express $\Delta_h f(x)$ in terms of f(0) and f(h) seems like an odd thing to do...

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

 

[Hurkyl]

I've added more to my previous post.

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

 

[HallsofIvy]

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?

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
$$f(0+ f'(0)(x+h)+ f"(0)/2 (x+y)^2+ /cdot/cdot/cdot+ f^{(n)}(0)(x+y)^n+ \cdot\cdot\cdot$$

 

[Rainbow Child]

What's the wikipedia url for that result?

 

[foxjwill]

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

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

The Taylor series for f(x), expanded around x= a, is

$$f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \cdots + f^{(n)}(a)(x-a)^n + \cdots$$

If you are expanding f(x+h) about x= h, then it would be

$$f(h) + f'(a)((x+h)-h) + \frac{f''(h)}{2} ((x+h)-h)^2 + \cdots + f^{(n)}(h) ((x+h)-h)^n + \cdots$$

$$= f(h) + f'(a)(x) + \frac{f''(a)}{2} x^2 + \cdots + f^{(n)}(h) x^n + \cdots$$

If you are expanding f(x+h) about x= 0, then it would be

$$f(0) + f'(0)(x+h) + \frac{f''(0)}{2}(x+y)^2 + \cdots + f^{(n)}(0)(x+y)^n + \cdots$$

$$f(a) + f'(a)(x-a) + \frac{f''(a)}{2}(x-a)^2 + \cdots + f^{(n)}(a)(x-a)^n + \cdots$$

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

 

[foxjwill]

What's the wikipedia url for that result?

http://en.wikipedia.org/wiki/Finite_difference#Calculus_of_finite_differences"

 

[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.