1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

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

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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

    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

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor

    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
    Last edited by a moderator: Apr 23, 2017
  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


    and from here you have the desired result.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook