The max value of the fourth derivative

[dpb613]

Homework Statement

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[itex]^{\frac{x^{2}}{2}}[/itex] and the interval is from 0 to 2

Homework Equations

|E|[itex]\leq[/itex][itex]\frac{(b-a)^{5}}{180n^{4}}[/itex][max|f[itex]^{(4)}[/itex](x)|]

The Attempt at a Solution

The fourth derivative of 2.718281828[itex]^{\frac{x^{2}}{2}}[/itex] is approximately
(x[itex]^{4}[/itex]+6x[itex]^{2}[/itex]+3)*2.718281828[itex]^{\frac{x^{2}}{2}}[/itex]
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?

 

[Mark44]

Homework Statement

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[itex]^{\frac{x^{2}}{2}}[/itex] and the interval is from 0 to 2

Homework Equations

|E|[itex]\leq[/itex][itex]\frac{(b-a)^{5}}{180n^{4}}[/itex][max|f[itex]^{(4)}[/itex](x)|]

The Attempt at a Solution

The fourth derivative of 2.718281828[itex]^{\frac{x^{2}}{2}}[/itex] is approximately
(x[itex]^{4}[/itex]+6x[itex]^{2}[/itex]+3)*2.718281828[itex]^{\frac{x^{2}}{2}}[/itex]
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?

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?

 

[dpb613]

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?

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)?

 

[Mark44]

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

If you'll recall, I said "ignored for the time being."

, but for y = x⁴ + 6x² + 3 I find a max value of 43 @ x=2.

So, which is it 43 or 43e²(317.729)?

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.

 

[dpb613]

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.

OK Now I understand. Thank You.