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Math Help - Maclaurin Series Problem

  1. #1
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    Maclaurin Series Problem

    Find the maclaurin polynomials of orders n=0,1,2,3, and 4, and then find the nth Maclaurin polynomials for the function in sigma notation.

    cos(pi*x)

    http://img382.imageshack.us/img382/5889/48183237ut4.png

    This is the answer from the solution mannual

    Can anybody tell me how it finds f(k) first? I mean not just for this question because all the solution about maclaurin and taylor, the solution finds f(k) first.

    Another question, I do not understand the NB means? Why is it n/2 ?
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by soleilion View Post
    Find the maclaurin polynomials of orders n=0,1,2,3, and 4, and then find the nth Maclaurin polynomials for the function in sigma notation.

    cos(pi*x)

    http://img382.imageshack.us/img382/5889/48183237ut4.png

    This is the answer from the solution mannual

    Can anybody tell me how it finds f(k) first? I mean not just for this question because all the solution about maclaurin and taylor, the solution finds f(k) first.

    Another question, I do not understand the NB means? Why is it n/2 ?
    I cant open your attachment...but if you mean \sum_{x=0}^{\infty}cos(x\pi)=\sum_{x=0}^{\infty}(-1)^{x} which is divergent
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    I cant open your attachment...but if you mean \sum_{x=0}^{\infty}cos(x\pi)=\sum_{x=0}^{\infty}(-1)^{x} which is divergent



    Can you see this right now????
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    Mathstud28

    are you still online?
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  5. #5
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    Can anybody help me, please?

    Thank you
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  6. #6
    Behold, the power of SARDINES!
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    Quote Originally Posted by soleilion View Post
    Find the maclaurin polynomials of orders n=0,1,2,3, and 4, and then find the nth Maclaurin polynomials for the function in sigma notation.

    cos(pi*x)

    http://img382.imageshack.us/img382/5889/48183237ut4.png

    This is the answer from the solution mannual

    Can anybody tell me how it finds f(k) first? I mean not just for this question because all the solution about maclaurin and taylor, the solution finds f(k) first.

    Another question, I do not understand the NB means? Why is it n/2 ?

    the f^k are the kth derivatives so lets get started

    f(x)=\cos(\pi x)
    f'(x)=-\pi \sin(\pi x)
    f''(x)=-\pi^2 \cos(\pi x)
    f'''(x)=\pi^3 \sin(\pi x)
    f^{(4)}(x)=\pi^4 \cos(\pi x)

    Now we need to evaluate each of these at the center for this series at x=0
    so we get...

    f(0)=\cos(\pi 0)=1
    f'(0)=-\pi \sin(\pi 0)=0
    f''(0)=-\pi^2 \cos(\pi 0)-\pi^2
    f'''(0)=\pi^3 \sin(\pi 0)=0
    f^{(4)}(0)=\pi^4 \cos(\pi 0)=\pi^4

    Now we can use these in taylor theorem for a power series centered at a is

    f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(a)}{n!}(x-a)^n

    so we get

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

    we get this because everyother term is zero.

    I hope this helps
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  7. #7
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    Quote Originally Posted by TheEmptySet View Post
    the f^k are the kth derivatives so lets get started

    f(x)=\cos(\pi x)
    f'(x)=-\pi \sin(\pi x)
    f''(x)=-\pi^2 \cos(\pi x)
    f'''(x)=\pi^3 \sin(\pi x)
    f^{(4)}(x)=\pi^4 \cos(\pi x)

    Now we need to evaluate each of these at the center for this series at x=0
    so we get...

    f(0)=\cos(\pi 0)=1
    f'(0)=-\pi \sin(\pi 0)=0
    f''(0)=-\pi^2 \cos(\pi 0)-\pi^2
    f'''(0)=\pi^3 \sin(\pi 0)=0
    f^{(4)}(0)=\pi^4 \cos(\pi 0)=\pi^4

    Now we can use these in taylor theorem for a power series centered at a is

    f(x)=\sum_{n=0}^{\infty}\frac{f^{n}(a)}{n!}(x-a)^n

    so we get

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

    we get this because everyother term is zero.

    I hope this helps

    Do you see the answer? Do you understand what NB means?
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  8. #8
    Behold, the power of SARDINES!
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    Quote Originally Posted by soleilion View Post
    Do you see the answer? Do you understand what NB means?
    I have no idea what NB stands for. I "think" it is saying Note: that [x] is floor function.
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