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Riemann sums

  1. Mar 22, 2006 #1
    hi, is it possible to find the riemann sum of (cos1)^x?

    it looks divergent to me

    can someome please help me... even if it is convergent, i dont know how to find the sum of a trigonometric function
     
  2. jcsd
  3. Mar 22, 2006 #2

    arildno

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    Eeh, do you mean (cos(1))^x?
     
  4. Mar 22, 2006 #3
    Did you really mean the riemann sum? Or did you mean the sum of the infinite series? Well since cos(0) = 1 and cos(pi) = -1 etc.. Then for any x that is not a multiple of pi |cos(x)| will be less than 1. Since this absolute value is less than 1 the geometric series will converge, specifically it will converge to a/(1-r) where a is the first term, and r is the common ratio.
     
  5. Mar 22, 2006 #4
    thanks alot for the input guys

    can someone please tell me if 1.175 is right? thanks
     
    Last edited: Mar 22, 2006
  6. Mar 22, 2006 #5
    Where is the sum from? From 0 to infinity or what?
     
  7. Mar 22, 2006 #6
    this is what it looks like:

    [​IMG]
     
  8. Mar 22, 2006 #7

    arildno

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    1.175 is definitely not "right". What you've got is a geometric series, and you should be able to get a simple, exact expression for the sum, rather than an approximation.
     
  9. Mar 22, 2006 #8

    nrqed

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    yes, that is correct.
     
  10. Mar 22, 2006 #9
    ah i think i got it

    i think the sum would be .540
     
  11. Mar 22, 2006 #10

    nrqed

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    well, you are right. But his answer is correct within 4 sig figs. I assumed he wanted a numerical, approximate value.
     
  12. Mar 22, 2006 #11
    thanks so much for the help guys :!!)


    but now im kinda confused:rofl:
     
  13. Mar 22, 2006 #12

    arildno

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    Since there isn't said anywhere whether cos(1) is the cosine to 1 degree or to 1 radian, for example, the numerical answers are just meaningless.
     
  14. Mar 22, 2006 #13
    o, its to the radians
     
  15. Mar 22, 2006 #14

    arildno

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    And why is it "right" just to keep 4 significant figures? Why not 15, or just 1?
    Nothing in what you posted gives any suggestion of what appoximation scheme you ought to use.
     
  16. Mar 22, 2006 #15
    ok heres what i did to get the 1.7... figure:

    you said that it converges to a/(1-r)

    A. a1= cos1
    r= cos1 (right?)

    so then: cos1/(1-cos1)

    this comes out to 1.7



    B. this is how i got the .5... answer:
    a/(1-r)
    a1= cos1
    r= (cos1)^k

    cos1/(1-(cos1)^k)

    then i took the limits (cos1)^k becomes 0

    so its: cos1/(1)= cos1 or .5...
     
  17. Mar 22, 2006 #16

    arildno

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    To be to the point:
    Why are you so reluctant to give your answer as the exact expression
    cos(1)/(1-cos(1)) ??
     
  18. Mar 22, 2006 #17
    o ok

    hmm i just felt like rounding. thanks guys/girls
     
  19. Mar 22, 2006 #18
    Well for case B, if you even just write out the first few terms of the eries you should see that that definitely cannot be teh answer because you have

    cos(1) + cos2(1) + cos3(1) + ...

    As you can see the first term in that series is about .54... and since none of those terms are ever negative the sum of the series must be larger than the first term about .54...

    Case A gives you an approximation of the correct sum of this series.
     
  20. Mar 22, 2006 #19

    arildno

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    I'm glad you finally posted this, since finally, we can get rid of some misconceptions you show here.
    A geometric series is of the form:
    [tex]S=a*r^{0}+ar^{2}+++ar^{k}+[/tex]
    where a and r are real numbers, and |r|<1
    In your case, your series is:
    [tex]\cos(1)*(\cos(1))^{0}+\cos(1)*(\cos(1))^{1}+++\cos(1)*(\cos(1))^{k}+++[/tex]
    Thus, a=r=cos(1)
     
  21. Mar 22, 2006 #20
    thanks everyone :)

    and also,

    i dont want to flood the board with topics, so can you guys please help with something else?

    how is arctan(2n) divergent when, if you make n a big number, it clearly has a limit of 1.57?
     
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