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blackstreet23

  • one year ago

Picture Problem using Simpson's Rule Using Simpsons Rule (n = 8) , approximate an integral

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  1. blackstreet23
    • one year ago
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    Where can I post pictures ?

  2. blackstreet23
    • one year ago
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  3. nincompoop
    • one year ago
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    http://www.intmath.com/integration/6-simpsons-rule.php

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

  5. SolomonZelman
    • one year ago
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    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
    • one year ago
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    @blackstreet23 @solomanzelman Do you have everything you need to finish all four parts of the problem?

  7. blackstreet23
    • one year ago
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    No I am stuck at part b

  8. mathmate
    • one year ago
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    Did you try @SolomonZelman 's formula? (you must have seen that before)

  9. mathmate
    • one year ago
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    @blackstreet23

  10. blackstreet23
    • one year ago
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    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
    • one year ago
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    isn't the 4th derivative given? (you don't need 5th derivative )

  12. blackstreet23
    • one year ago
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    i need the highest point of the fourth derivative and extrema occur at end points and critical points.

  13. freckles
    • one year ago
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    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
    • one year ago
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    but that is correct right?

  15. freckles
    • one year ago
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    I see... The forth derivative doesn't look too easy to guess highest *

  16. freckles
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    but the fifth derivative is still necessary just to show work right?

  19. mathmate
    • one year ago
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    @blackstreet23

  20. mathmate
    • one year ago
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    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
    • one year ago
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    gtg

  22. mathmate
    • one year ago
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    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
    • one year ago
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    but because those values are not within the interval i do not need to worry of them until part d right?

  24. blackstreet23
    • one year ago
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    @mathmate

  25. mathmate
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    yes n=6

  28. blackstreet23
    • one year ago
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    on d the new value of K4 will be 0.7149345503 right?

  29. mathmate
    • one year ago
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    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
    • one year ago
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    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.

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