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Proof using taylor series

  1. Oct 25, 2006 #1
    I am supposed to prove using taylor series the following:

    [tex]\frac{d^2\Psi}{dx^2} \approx \frac{1}{h^2}[\Psi (x+h) - 2\Psi(x) + \Psi (x-h)][/tex] where x is the point where the derivative is evaluated and h is a small quantity.

    what i have done is used:
    [itex] f(x+h)= f(x) + f'(x) h + f''(x)\frac{h^2}{2!}+...[/itex]

    and solved so that

    [tex]f''(x)=\frac{2}{h^2}[f(x+h) - f(x) - f'(x) h][/tex]

    i am not sure how to get this into the required form..
    I noticed that solving the given equation for [tex]\Psi(x)[/tex] gives a term that looks like [tex]\frac{\Psi(x+h) + \Psi(x-h)}{2}[/tex] i.e. average value on the interval, can this be somehow used to write as a derivative or something?

    Last edited: Oct 25, 2006
  2. jcsd
  3. Oct 25, 2006 #2


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    First, you should be clearer about what you mean by "[itex]\approx[/itex]". Once you sort that out, write out the first few terms of the Taylor series for f(x), f(x+h), f(x-h). Now can you find a linear combination of these expressions such that the coefficient of f(x) and f'(x) is zero, while that of f''(x) is 1?

    If you don't care about deriving the identity, only verifiying it, then just plug in the first few terms of the taylor series into the RHS of the first expression.
  4. Oct 28, 2006 #3
    i have proven this but the second part says, find the highest valued term representing the error in this.

    i do not understand what this refers to or what this means, any clarification?
  5. Oct 29, 2006 #4


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    When you plug in the taylor series to that approximation of f''(x), there will be extra terms that don't cancel. What is the smallest power of h left over? (this will be the biggest error term, since h is very small)
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