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Taylor Series Extrapolation

  1. May 3, 2014 #1
    Hi could anyone give me pointer as to where to go with part iii please?
     

    Attached Files:

  2. jcsd
  3. May 3, 2014 #2

    lurflurf

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    second order accurate means
    $$\begin{cases}
    A|_{h=0}=B|_{h=0}
    \\
    \left. \dfrac{dA}{dh}\right|_{h=0}=\left. \dfrac{dB}{dh}\right|_{h=0}
    \end{cases}$$
    so require
    $$\begin{cases}
    \mathrm{f}(x+2h)|_{h=0}=a \, \mathrm{f}(x+h)+b \, \mathrm{f}(x)|_{h=0}
    \\
    \left. \dfrac{d}{dh}\mathrm{f}(x+2h)\right|_{h=0}=\left. \dfrac{d}{dh}(a \, \mathrm{f}(x+h)+b \, \mathrm{f}(x))\right|_{h=0}
    \end{cases}$$
    then solve for a and b
     
  4. May 3, 2014 #3
    Hi, I dont really understand the notation you have used there with the lines, but does that basically mean that the equation holds true when differentiated once and in its original form?

    so just to check my understanding, if it was supposed to be 3rd order accurate, there would be a third equation which would be the second differential of the original and this would have to be satisfied as well by the values of a and b?
     
  5. May 3, 2014 #4

    lurflurf

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    The line means let the variable take a certain value
    in this case let h=0 after taking the derivative

    yes third order accurate would be
    \begin{cases}
    A|_{h=0}=B|_{h=0}
    \\
    \left. \dfrac{dA}{dh}\right|_{h=0}=\left. \dfrac{dB}{dh}\right|_{h=0}
    \\
    \left. \dfrac{d^2A}{dh^2}\right|_{h=0}=\left. \dfrac{d^2B}{dh^2}\right|_{h=0}
    \end{cases}
    and so on
     
  6. May 3, 2014 #5

    lurflurf

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    did you find a and b?
     
  7. May 3, 2014 #6
    Ok thats great thank you! I am going to attempt it now that I understand it! I'll place my answer on here afterwards, I gather you have found them then?
     
  8. May 3, 2014 #7

    lurflurf

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    ^I could not resist
     
  9. May 3, 2014 #8
    I have found a as 1 is that correct? and using that value I get b as

    b=((((hdf(x)/dx + 3/2*h^2*d2f(x)/dx))))/f(x)
     
    Last edited: May 3, 2014
  10. May 3, 2014 #9

    lurflurf

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    Remember h=0 so it should not appear. Since we are second order we should not have h^2.

    Here is another equivalent approach

    $$\text{if}
    \\
    \mathrm{f}(x+2h)=\mathrm{f}(x)+2h\, \dfrac{df}{dx}(x)+\mathrm{O}(x^2)
    \\
    \mathrm{f}(x+h)=\mathrm{f}(x)+h\, \dfrac{df}{dx}(x)+\mathrm{O}(x^2)
    \\
    \mathrm{f}(x+2h)=a\, \mathrm{f}(x+h)+b\, \mathrm{f}(x)
    \\
    \text{then}
    \\
    \mathrm{f}(x)+2h\, \dfrac{df}{dx}(x)+\mathrm{O}(x^2)=a\left[\mathrm{f}(x)+h\, \dfrac{df}{dx}(x)\right]+b\, \mathrm{f}(x)+\mathrm{O}(x^2)
    \\
    \mathrm{f}(x)+2h\, \dfrac{df}{dx}(x)=a\left[\mathrm{f}(x)+h\, \dfrac{df}{dx}(x)\right]+b\, \mathrm{f}(x)$$
    What do a and b need to be?
     
  11. May 3, 2014 #10
    that makes so much more sense! as you say I included the h^2 accidentally so that must be where i went wrong, I'll try and solve it the original way again and see if I get the same answer!

    From what I can see a needs to be 2 and b has to be -1
     
  12. May 3, 2014 #11

    lurflurf

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    ^Good job
     
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