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Series for Cosh^1.822

  1. Dec 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Use an appropriate series to compute the value of cosh(1.822)


    2. Relevant equations
    coshx=e^x/2+e^(-x)/2


    3. The attempt at a solution

    I follow the series 1+x+x^2/(2! )…. etc and 1-x+x^2/(2! )…. then add the answers together and divide by 2. However the answers im coming out at is 3.11 rather than 3.17 which i get on the calculator.
    Thanks so much in advance.
     
  2. jcsd
  3. Dec 15, 2013 #2

    jedishrfu

    Staff: Mentor

    how many terms have you extended it to?
     
  4. Dec 15, 2013 #3
    I've gone to 14 so far. Do I just need to keep going further? Thanks a lot for replying
     
  5. Dec 15, 2013 #4

    jedishrfu

    Staff: Mentor

    well you could compare your iterations and see if they are approaching the calculator answer and that would give you confidence that you didn't make a mistake.

    you could write a program in python or whatever language you're comfortable in and show each iteration and compare to what you've done.

    Calculators will go many iterations to get a result. They do it so fast its had to imagine how many they did in the short time it took. They may also use some sort of shortcut to get the result like interpolation from stored tables of values...
     
  6. Dec 15, 2013 #5

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    You can use the series
    [tex]\cosh x=\sum_{k=0}^{\infty} \frac{x^{2k}}{(2k)!},[/tex]
    which follows from your ansatz.

    Of course, you can only calculate a finite partial series. You can estimate the error made by using Taylor's theorem which says
    [tex]\cosh x=\sum_{k=0}^{n} \frac{x^{2k}}{(2k)!}+\frac{1}{(2n+1)!}\sinh \xi x^{2n+1},[/tex]
    where [itex]\xi[/itex] is between [itex]0[/itex] and [itex]x[/itex]. We can thus estimate the error made, when just stopping at [itex]=k=2n[/itex] by
    [tex]\left | \frac{1}{(2n+1)!} \sinh \xi x^{2n+1} \right | \leq \left |\frac{1}{(2n+1)!} \sinh x x^{2n+1} \right|.[/tex]
     
  7. Dec 15, 2013 #6
    Thanks so much to both of you, I think with what both of you have said I should be fine to get a decent report out of it! Thanks!
     
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