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Maclaurin Series and the general term

  1. Mar 16, 2004 #1
    using the sin and cos Maclaurin series, validate each of them using at least 3 values for x and determine how many terms are needed to provide reasonable accuracy.

    Find the General Term (Tn where n = 1, 2, 3, .......) for each expression and show that each correctly generates the terms of the series. Then using the specific and General Terms, determine expressions for the derivative of sin x and the integral of cos x.



    Now I get the part where i must use 3 values for x and how many terms, but what has really got me stumped is the 2nd paragraph. Any help would be greatly appreciated. Thanks
     
  2. jcsd
  3. Mar 16, 2004 #2
    I've never heard of a "general term" before, but it seems that they want you to find an expression for the nth term of the series, i.e.

    [tex]\sin{x} = \sum_{n=1}^\infty T_n[/tex]

    It's not difficult for the sin and cosine series. For instance, for sin(x)

    [tex]T_1 = x[/tex]
    [tex]T_2 = -\frac{x^3}{6}[/tex]

    After that, they want you to use term-by-term differentiation and term-by-term integration of [itex]T_n[/itex] in order to find the derivative and antiderivative of sin and cos, respectively.

    cookiemonster
     
  4. Mar 16, 2004 #3

    matt grime

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    General term usually just means give a formula that describes the n'th power. For sin it is, (-1)^(n)(x^2n+1)/(2n+1)! as n is 0,1,2,3...

    I don't see how you could do the first part of the question without knowing this though.
     
  5. Mar 16, 2004 #4
    i did the first part from a formula i found on a website

    the first part's formula was something like this...

    cos x = 1 - (x^2 / 2!) + (x^4 / 4!) - (x^6 / 6!) - (x^8 / 8!) + ........ (x^n / n!)

    Can someone tell me if this is right?
     
  6. Mar 16, 2004 #5
    Would a website ever lie to you (and if so, would we ever be truthful to you)?

    Not quite. You're not considering that the function skips odd powers and you're not considering the alternating sign.

    cookiemonster
     
  7. Mar 16, 2004 #6
    yeh it had that point on the site aswell, skips odd numbers, but about the alternating sign, please do tell
     
  8. Mar 16, 2004 #7

    matt grime

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    (-1)^n(x^2n)/2n!
     
  9. Mar 16, 2004 #8
    Since we all like LaTeX so much:

    [tex]\cos{x} = \sum_{n=0}^\infty (-1)^n\frac{x^{(2n)}}{(2n)!}[/tex]

    and

    [tex]\sin{x} = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}[/tex]

    Talk about service!

    cookiemonster
     
    Last edited: Mar 20, 2004
  10. Mar 16, 2004 #9
    Then using the specific and General Terms, determine expressions for the derivative of sin x and the integral of cos x.

    Any idea on how to start this part.

    P.S. Thanks guys for helping me out
     
  11. Mar 16, 2004 #10
    Just differentiate the general term with respect to x and add all the differentiated terms together. The same for the antiderivative, except this time integrate.

    cookiemonster
     
  12. Mar 16, 2004 #11

    matt grime

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    There's a sign wrong in your sin term, cookiemonster. It gives -x+x^3/6...
     
  13. Mar 16, 2004 #12
    Oops.

    cookiemonster
     
  14. Mar 16, 2004 #13
    Thanks heaps guys for all your help
     
  15. Mar 20, 2004 #14

    NSX

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    Summation signs...

    :p
     
  16. Mar 20, 2004 #15
    Grr! That was a bad day.

    cookiemonster
     
  17. Mar 10, 2007 #16
    This question is similar to a question i am currently undertaking in my course study...could anyone help me with the assumptions that have been assumed
    Thankyou
    P.S. if anyone has any ideas about the strengths and limitations of this model that would also be of great assistance
     
  18. Mar 10, 2007 #17
    sin x = x – (x3 / 3!) + (x5 / 5!) – (x7 / 7!) + (x9 / 9!) - …
    ‘sin x = 1 – (3x2 / 3!) + (5x4 / 5!) - (7x6 / 7!) + (9x8 / 9!) - …

    cos x = 1 – (x2 / 2!) + (x4 / 4!) – (x6 / 6!) + (x8 / 8!) - …
    COS x = x – (x3 / 3 / 2!) + (x5 / 5 / 4!) – (x7 / 7 / 6!) + (x9 / 9 / 8!) - …

    are these the correct derivatives and integrals?
     
  19. Mar 11, 2007 #18

    Gib Z

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    Do you know that the derivative of sin x is cos x? Note on your 'sin x, you have 3x^2/3!, and 5x^4/5!.

    Do you know what factorial actually means? 5! = 5 times 4 times 3 times 2 times 1. Every number before it multiplied. The First term in the facotial cancels out, it give you the correct series for sin.
     
  20. Mar 13, 2007 #19
    help!

    ive found the maclaurin series for sin x and cos x , and ive tried to validate them, i even went to like the 41st term and the answer is totally wrong. I dont know wat im doing wrong, as im just subbing a number into x so it should be easy????
     
  21. Mar 14, 2007 #20

    Gib Z

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    Are your series the correct ones? If so, check if your not putting brackets on your factions or something. Easiest one to check in sin 1, x is always just 1.
     
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