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Forward difference operator

  1. Feb 17, 2008 #1
    [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 [tex]\Delta_h[/tex] could be written as

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

    where [tex]D[/tex] 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. jcsd
  3. Feb 17, 2008 #2

    Hurkyl

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    Try applying it to a function.
     
  4. Feb 17, 2008 #3
    Ok, so I tried going at it this way:

    [tex]
    \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}
    [/tex]

    However, when I try subtracting (2) from (1), the terms don't cancel out because (1) is in terms of [tex]f(h)[/tex] and (2) is in terms of [tex]f(0)[/tex]. What should I do?
     
  5. Feb 17, 2008 #4

    Hurkyl

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    Trying to express [itex]\Delta_h f(x)[/itex] in terms of [itex]\Delta_h f(0)[/itex] 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
  6. Feb 17, 2008 #5
    I was using the tailor series expansion. Do you have any other ideas?
     
  7. Feb 17, 2008 #6

    Hurkyl

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    I've added more to my previous post.

    I don't have any new ideas; what's wrong with the one I already gave?
     
  8. Feb 18, 2008 #7

    HallsofIvy

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

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

    If you are expanding f(x+h) about x= 0, then it would be
    [tex]f(0+ f'(0)(x+h)+ f"(0)/2 (x+y)^2+ /cdot/cdot/cdot+ f^{(n)}(0)(x+y)^n+ \cdot\cdot\cdot[/tex]
     
  9. Feb 18, 2008 #8
    What's the wikipedia url for that result? :confused:
     
  10. Feb 18, 2008 #9
    Ok, so I fixed your [tex]\LaTeX[/tex]:

    I'm not really sure how you got the expansions to be those. Could you explain?
     
  11. Feb 18, 2008 #10
  12. Feb 19, 2008 #11
    [tex]\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)[/tex]

    Expanding [itex]f(x+h)[/itex] around [itex]\alpha=h[/itex] we have

    [tex]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)[/tex]

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

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

    and from here you have the desired result.
     
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