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Numerical analysis (composite numerical integration)

  1. Dec 28, 2006 #1
    using composite trapezoidal rule with n=4 how can i get a bound for the error of I=integration tan(x) from x=0 to x=pi/2

    i know that the term of error in comp trapezoidal rule is (b-a)/12 h^2 f''(eita)
    i got the second derevative of tanx to be 2sec^2 x tanx then i don't know with what value exactly i need to substitute in this function to get the bound.
     
  2. jcsd
  3. Jan 2, 2007 #2

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    Since tan(pi/2) is infinite, you cannot perform this integration.
     
  4. Jan 2, 2007 #3
    sorry it is to x=pi/4
    should i substitute by sec^2x by (1+tan^2x)
     
  5. Jan 2, 2007 #4

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    You must have [tex]0 < \eta < \pi/4[/tex]

    If you want a worst case error, differentiate again and find the value of [tex]\eta[/tex] that maximizes f''.
     
  6. Jan 2, 2007 #5
    ok i did and the second derivative is 2sec^2 x tanx now is the maximum value of this function is 4 as sec^2(x)=tan^2(x) + 1 ??????????????
     
  7. Jan 2, 2007 #6

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    Taking the derivative of f'' and setting it to zero gives an immaginary root. So, there is no point of maximum.
    The value of f'' at [tex]\pi/4[/tex] is 1, so we take it as the worst case.
    [tex]h = \pi/4/4 = \pi/16[/tex]
    Then your error will be less then [tex]\frac{\pi/4}{12}\cdot \frac{\pi^2}{16^2}[/tex].
     
  8. Jan 3, 2007 #7
    why we set the value of f'' to zero???
    we set it to a value that gives us the maximum of the function so we substitute in 2sec^2(x)tan(x) which is equal to 2(1+tan^2(x))tan(x) by PI/4 so we get 2(1+1) which is 4.
    right??
     
  9. Jan 3, 2007 #8

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    I have set the value of f''' to zero. The maximum of f'' happens when its derivative is zero.
    Since f''' has no real roots the maximum of f'' must be at one of its boundary points 0 or [tex]\pi/4[/tex]. The value of the funtion at 0 is 0 and at [tex]\pi/4[/tex] is 4 (and not 1 as I wrote previously). So in the expression for the error, you replace [tex]f''(\eta)[/tex] by 4, getting [tex]\frac{\pi/4}{12}\cdot \frac{\pi^2}{16^2}\cdot 4[/tex]
     
  10. Jan 4, 2007 #9
    ok what about f(x)=exp x sinx using composite trapezoidal rule what is the bound of error for integration from 1 to 3

    the second derevative will be 2exp x cos x then to get the bound of error which value should i substitute with??
     
  11. Jan 4, 2007 #10

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    Since exp x is monotone and grows very fast, the maximum value of f'' is at x = 3. This is the value to substitute in the expression.
     
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