## blackstreet23 one year ago Picture Problem using Simpson's Rule Using Simpsons Rule (n = 8) , approximate an integral

1. blackstreet23

Where can I post pictures ?

2. blackstreet23

3. nincompoop
4. SolomonZelman

$$\large\color{black}{ \displaystyle {\rm Simpson's~Rule} \\[0.9em] {\rm Area~~of~f~~over~~[1,3]~~with~n=8}~\\[1.9em]~\displaystyle \normalsize \frac{3-1}{8}\left(f(1)+4f(1.25)+2f(1.50)+4f(1.75)+2f(2) \\ +4f(2.25)+2f(2.50)+4f(2.75)+f(3)\right) }$$

5. SolomonZelman

then the maximum error formula is $$\large\color{black}{ \displaystyle {\rm E}_{S}=\frac{ k(b-a)^5 }{180n^4} }$$ where k is the bound on the 4th derivative.... (if you want this one)

6. mathmate

@blackstreet23 @solomanzelman Do you have everything you need to finish all four parts of the problem?

7. blackstreet23

No I am stuck at part b

8. mathmate

Did you try @SolomonZelman 's formula? (you must have seen that before)

9. mathmate

@blackstreet23

10. blackstreet23

I mean I know that maximum points in a closed interval picture occur at critical points and end points. so do I need to take the fifth derivative ?

11. freckles

isn't the 4th derivative given? (you don't need 5th derivative )

12. blackstreet23

i need the highest point of the fourth derivative and extrema occur at end points and critical points.

13. freckles

I see... The fifth derivative doesn't look too easy to guess highest you could differentiate the 4th derivative to find the 5th derivative

14. blackstreet23

but that is correct right?

15. freckles

I see... The forth derivative doesn't look too easy to guess highest *

16. freckles

http://www.wolframalpha.com/input/?i=f%28x%29%3D-3%285x%5E4-60x%5E2%2B10%29%2F%2816%28x%5E2%2B1%29%5E%2815%2F4%29%29+x%3D-10+to+10+and+y%3D-1+to+1 beautiful finally figured out how to play with zooming options

17. freckles

notice the max is just under 1 for approximately what x value does that occur the 4th derivative is even (doesn't matter if you choose the positive or negative version of this particular number )

18. blackstreet23

but the fifth derivative is still necessary just to show work right?

19. mathmate

@blackstreet23

20. mathmate

I would graph and estimate the max. instead of going through the 5th derivative AND solving the resulting equation for f5(x)=0. Generally the actual error is very much lower than the upper bound estimate, so numerically, it should be ok.

21. mathmate

gtg

22. mathmate

Graphically, it looks like to be at x=0.8. Poking around with a few iterations give x=0.787, and f4(.787) around .7149. That should solve part D. For part C, you don't need the relative max.

23. blackstreet23

but because those values are not within the interval i do not need to worry of them until part d right?

24. blackstreet23

@mathmate

25. mathmate

Exactly. For part B, the maximum is at x=1, and it's a strictly decreasing function, so no need to find the relative maximum.

26. mathmate

For part D, even a few casual guesses will give you the maximum within 3 decimal digits. Have you done part C (find the value of n)?

27. blackstreet23

yes n=6

28. blackstreet23

on d the new value of K4 will be 0.7149345503 right?

29. mathmate

For part C, I have 10.1 or 10.2, which is kind of awkward, because it will bring the n=12 when we know almost sure that even n=10 may be good. Yes, the value of $$k_4$$ is 0.7149 at about x=0.79.

30. mathmate

Correction: Yes, I have n=5.713 => 6 for part C. I must have used the error bound ten times less when I calculated yesterday.