# Finding the second derivative using central difference formula

Tags:
1. Oct 24, 2016

### Kanashii

• 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. Oct 24, 2016

### BvU

Hi Kanashii,

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...

3. Oct 24, 2016

### Staff: Mentor

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

4. Oct 24, 2016

### BvU

Pick 16x^2-x^4

5. Oct 24, 2016

### Staff: Mentor

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?

6. Oct 24, 2016

### BvU

Bit slow allright ....
• 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 $-16\pi$

7. Oct 24, 2016

### Staff: Mentor

Ah 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...

8. Oct 25, 2016

### BvU

How is Kanashii doing ?