# The max value of the fourth derivative

1. May 8, 2013

### dpb613

1. The problem statement, all variables and given/known data
I am trying to write a program to calculate an integral using Simpson's Rule. My program also needs to calculate the error. In the formula for error I need The max value of the fourth derivative. The function is 2.718281828$^{\frac{x^{2}}{2}}$ and the interval is from 0 to 2

2. Relevant equations
|E|$\leq$$\frac{(b-a)^{5}}{180n^{4}}$[max|f$^{(4)}$(x)|]

3. The attempt at a solution
The fourth derivative of 2.718281828$^{\frac{x^{2}}{2}}$ is approximately
(x$^{4}$+6x$^{2}$+3)*2.718281828$^{\frac{x^{2}}{2}}$
When evaluated at 2 this gives 317.729... Since there is a lot of math involved I would appreciate if someone can review this especially since my program is returning strange results. Also, is there anything I can learn by evaluating the fifth derivative?

2. May 8, 2013

### Staff: Mentor

It would be better to write your function exactly, as e(1/2)x2. The fourth derivative is e(1/2)x2(x4 + 6x2 + 3).

The first factor is always positive and is increasing, so can be ignored for the time being. What about y = x4 + 6x2 + 3? What does the graph of this function look like on the interval [0, 2]? Can you determine where its max. value is?

3. May 9, 2013

### dpb613

It is certainly helpful to keep it in the form e(1/2)x2 and now I see that the max value of the fourth derivative is 43e2.
I am not sure why the first factor can be ignored, but for y = x4 + 6x2 + 3 I find a max value of 43 @ x=2.

So, which is it 43 or 43e2(317.729)?

Last edited: May 9, 2013
4. May 9, 2013

### Staff: Mentor

If you'll recall, I said "ignored for the time being."
For the max value of f(4)(x), use 43 * e2 ≈ 318.

The reasoning here is that e(1/2)x2 is an increasing function, so the maximum value on an interval is attained at the right endpoint. Presumably your graph of y = x4 + 6x2 + 3 showed that this function also is increasing. As that seems the case, the max value of f(4)(x) occurs when x = 2.

5. May 9, 2013

### dpb613

OK Now I understand. Thank You.