Finding the second derivative using central difference formula

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SUMMARY

The discussion focuses on calculating the second derivative of the function f(x) = π(16x² - y⁴) at y=2 using the second-order Central Difference Formula. The program developed by the user encounters issues with crashing and incorrect output values, particularly when using small step sizes (h). The correct implementation of the formula is confirmed, but the user needs to ensure that the program operates in double precision to avoid numerical instability and errors. The analytical second derivative is established as -16π.

PREREQUISITES
  • Understanding of the Central Difference Formula for numerical differentiation
  • Familiarity with programming concepts, particularly in C or similar languages
  • Knowledge of numerical error analysis and convergence criteria
  • Experience with double precision floating-point arithmetic
NEXT STEPS
  • Implement double precision in the existing program to improve accuracy
  • Research numerical differentiation techniques, focusing on the Central Difference Formula
  • Explore error analysis methods to evaluate convergence in numerical computations
  • Study the implications of function definitions in calculus, particularly in relation to derivatives
USEFUL FOR

Mathematicians, computer scientists, and students involved in numerical analysis or programming who are interested in accurate computation of derivatives and error management in numerical methods.

Kanashii
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Thread moved from the technical forums, so no Homework Help Template is shown.

Homework Statement


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

Homework Equations


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

The Attempt at a Solution


Code: 
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:
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Hi Kanashii, :welcome:

Does your program work in double precision ? Excel does, and it has no problem
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...
 
Perhaps I'm being a bit slow today, but can you explain how "pi(16 x^2 - y^4)" defines a function?
 
Pick 16x^2-x^4
 
BvU said:
Pick 16x^2-x^4
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?
 
Bit slow all right ...: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##
 
BvU said:
Bit slow all right ...: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##
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...
 
How is Kanashii doing ?