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Math Help - sumation of cos x

  1. #1
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    sumation of cos x

    Write the summation notation of the maclaurin polynomial of order n for cos x
    I worked out what the maclaurin of cos x by doing the derivations, plug in 0 and divide by n!. I get this

    1 - \frac {x^2} {2!} + \frac {x^4} {4!} - \frac {x^6} {6!} ...

    then for the sumation i got this

    \sum_{i = 0}^{??} \frac {(-1)^i x^{2i}} {(2i)!}

    My question is what goes on top of the sigma? The top part is the stopping point when computing the sum, correct? I need to find the Nth so the top is not infinity.

    Let's say I want the thrid maclaurin, n =3. would that be

    1 - \frac {x^2} {2!} + \frac {x^4} {4!}

    or

    1 + 0 * \frac {x^1} {1!} - \frac {x^2} {2!}  + 0 * x^3
    counting on zero but the 0 coeffients remove those odd powered x's so the above is just the same as the previous.


    With x^2i that will take into acount the 0 coeffients, for when sine at 0. It out puts 1, x^2, x^4 ..., so that forms the first example of the thrid maclaurin. Does that mean the top of sigma is just n?

    The thing is my teacher wrote something like int[n/2 + 1.5] as the top of sigma but said he wasnt sure of that. and after runing different forms n/2 +.5, n/2 +1 it doesnt make sense to me why you would need something like that. the top part is just the stoping point so why the heck do you even need a little equation on the top ?!.

    grrr
    Last edited by diroga; May 2nd 2009 at 10:35 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    You left out the factorial (2i)! in the denomiator.
    And what's wrong with the upper limit of your sum being n?
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  3. #3
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    I fixed the factorial

    I think im stuck on the fact that in class my instructor wrote int[n/2 + 1.5]. I just don't under stand why that would be needed. I do think that just n would work.
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  4. #4
    MHF Contributor matheagle's Avatar
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    that integer value is ok if you use different terms, but unnecessary.
    write exactly what your instructor placed on the board
    and it's (2i)!, 2i! isn't clear at all
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  5. #5
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    What the instructor wrote in class

    <br />
\sum_{i = 0}^{int[\frac {n} {2} + 1.5]} \frac {(-1)^i x^{2i}} {2i!}<br />
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  6. #6
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by diroga View Post
    What the instructor wrote in class

    <br />
\sum_{i = 0}^{int[\frac {n} {2} + 1.5]} \frac {(-1)^i x^{2i}} {(2i)!}<br />
    I assume he/she meant the integer part....
    Now, are you sure of the 1.5 and not .5?
    Because if I let n=0, I get the first two term and not just the first term.
    Most people say S_1=1, S_2=1-{x^2\over 2},...

    This claims that S_0=1-{x^2\over 2} which seems odd.
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  7. #7
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    Quote Originally Posted by matheagle View Post
    I assume he/she meant the integer part....
    do you mean the int[n/2 + 1.5] part?

    Quote Originally Posted by matheagle View Post
    Now, are you sure of the 1.5 and not .5?
    Because if I let n=0, I get the first two term and not just the first term.
    I forggot there was talk about this in the math center. I think .5 is correct. If i want the third term then it would be

    <br />
\sum_{i = 0}^{int[\frac {n} {2} + 1.5]} \frac {(-1)^i x^{2i}} {(2i)!}<br />

    int[\frac {3} {2} + .5] = 2

    i = 0,1,2

    <br /> <br />
1 - \frac {x^2} {2!} + \frac {x^4} {4!}<br />
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