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Finding the second derivative using central difference formula

  1. Oct 24, 2016 #1
    • Thread moved from the technical forums, so no Homework Help Template is shown.
    1. The problem statement, all variables and given/known data
    Develop aprogram that will determine the second derivative of pi(16 x^2 - y^4) at y=2 with step sizes of 0.1, 0.01, 0.001…. until the absolute error (numerical-analytical) converges to 0.00001. Use the 2nd order Central Difference Formula.
    User Input: y, tolerance
    Output: h, second derivative, error

    2. Relevant equations
    [f(x+h) - 2f(x) + f(x-h)]/h^2

    3. The attempt at a solution
    Code (Text):

    do
        {
            n[0]= h;
            n[i+1]=n/10;
            f= function (y,n);
            error[0]= error_function(true_value,f);
            error[i+1]= error_function(f,f[i-1]);
            printf("%lf        %lf       %lf\n",n,f,error);
            i++;
        }
        while (error > tolerance || error != tolerance);
    When I input 0.00001 (tolerance) and 2 (y) into the program I created, the program crashes but it got the values of h, f`` right and also some values for the error. I do know what to change.
    Thank you.

    Output:

    h ----------------------------f``----------------------------error
    0.100000--------------- -50.328314 --------------- 0.062832
    0.010000-------------- -50.266111 --------------- 50.328314
    0.001000--------------- -50.265489 --------------- 0.062204
    0.000100--------------- -50.265481--------------- 0.000622
    0.000010--------------- -50.265498--------------- 0.000007
    0.000001--------------- -50.249582--------------- 0.000017
    0.000000--------------- -54.001248--------------- 0.015916
    0.000000--------------- -284.217094--------------- 3.751666
    0.000000--------------- 0.000000--------------- 230.215846

    This table would also go on and the values do not make any sense.
     
    Last edited: Oct 24, 2016
  2. jcsd
  3. Oct 24, 2016 #2

    BvU

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    Hi Kanashii, :welcome:

    Does your program work in double precision ? Excel does, and it has no problem
    Code (Fortran):

           0.1 -50.3283143105095 -0.06283185
          0.01 -50.2661107763487 -0.00062832
         0.001 -50.2654887593678 -0.00000630
        0.0001 -50.2654776874780  0.00000477
       0.00001 -50.2654866164213 -0.00000416
      0.000001 -50.2699510880988 -0.00446863
         1E-07 -53.5736601294126 -3.30817767
     
    But, as you see, it runs out of steam only a little bit later after hitting 4e-5 error. Phew...
     
  4. Oct 24, 2016 #3

    gneill

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    Perhaps I'm being a bit slow today, but can you explain how "pi(16 x^2 - y^4)" defines a function?
     
  5. Oct 24, 2016 #4

    BvU

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    Pick 16x^2-x^4
     
  6. Oct 24, 2016 #5

    gneill

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    Thanks. Still not ringing a bell. I've not come across this before (at least not in this form). Do you happen to have a reference I can have a look at?
     
  7. Oct 24, 2016 #6

    BvU

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    Bit slow allright ....:rolleyes:
    • I'm not the OP
    • his (or her) function as object of study is ##\ \pi(16x^2-x^4)\ ##. The second derivative at ##x=2## is ##-16\pi##
     
  8. Oct 24, 2016 #7

    gneill

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    Ah :doh: So there was a typo in the original (no y in the expression) and the function was meant to be ##f(x) = \pi(16x^2-x^4)##. Thanks for the clarification. My mind was going places like implicit differentiation...
     
  9. Oct 25, 2016 #8

    BvU

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    How is Kanashii doing ?
     
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