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Math Help - finding coefficient of nth term

  1. #1
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    finding coefficient of nth term

    Dear sir,
    i would be glad if someone can help me in the below problem
    thanks


    Find the coefficient of x to the power of n in the expansion of
    cos 2x.
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  2. #2
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    The expansion about what value of x?

    The coefficient of x^n in the Taylor series expansion of f(x) about x= a is \frac{f^{(n)}(a)}{n!} where f^{(n)}(a) is the nth derivative of f at x= a. Can you find the nth derivative of cos(2x)?
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  3. #3
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    Thanks for the reply but the problem is that i got something times 2x to the power of 2n instead of what i am requiring to find is x to the power of n. I need some help starting from here.

    thanks you
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  4. #4
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    Quote Originally Posted by kingman View Post
    Thanks for the reply but the problem is that i got something times 2x to the power of 2n instead of what i am requiring to find is x to the power of n. I need some help starting from here.

    thanks you
    Please show all your work - every step - so that we can help you, "starting from here" ....
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  5. #5
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    here is my answer

    here i wonder how get coefficient of x to the power of n but not 2n
    Attached Thumbnails Attached Thumbnails finding coefficient of nth term-co.bmp  
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  6. #6
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    Dear Sir,
    I would be most grateful if someone can help me in the posted question
    thanks
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  7. #7
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    Quote Originally Posted by kingman View Post
    Dear Sir,
    I would be most grateful if someone can help me in the posted question
    thanks
    I cannot see your attachment, which might explain why you have had no replies. Please type your working.
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  8. #8
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    Quote Originally Posted by kingman View Post
    Dear Sir,
    I would be most grateful if someone can help me in the posted question
    thanks
    People have. You have refused to do what you have been asked. You say you can get this "to the power of n but not 2n" but your original problem says nothing about a power of 2n. Do you mean you could find the coefficient of x^n if the problem were to expand cos(x) rather than cos(2x)?
    (Around, I assume, x= 0. You still haven't answered the very first question asked of you.)

    If so, please show us what you would get if this were cos(x) rather than cos(2x).
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  9. #9
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    reply

    I have got my working inserted but need to find coefficient of x to the power of n
    Attached Thumbnails Attached Thumbnails finding coefficient of nth term-bin.bmp  
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  10. #10
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    Hello, kingman!

    \text{Find the coefficient of }x^n\text{ in the expansion of }\cos 2x

    \displaystyle \cos2x \;=\;1 - \frac{(2x)}{2!} + \frac{(2x)^2}{4!} - \frac{(2x)^6}{6!} + \frac{(2x)^8}{8!} - \hdots

    . . . . . \displaystyle =\;1 - \frac{2^2}{2!}x^2 + \frac{2^4}{4!}x^4 - \frac{2^6}{6!}x^6 + \frac{2^8}{8!}x^8 - \hdots


    \,n is an even nonnegative integer.
    . . . The coefficient of \,x^n is: . (\text{-}1)^{\frac{n}{2}}\dfrac{2^n}{n!}

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  11. #11
    MHF Contributor Unknown008's Avatar
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    Don't you mean instead:

    \displaystyle \cos2x \;=\;1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + \frac{(2x)^8}{8!} - \hdots



    My physics teacher once showed us how to do it (we are not supposed to know this however) but I tend to forget. It was nice seeing it again.
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