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Math Help - Taylor Series / Power Series

  1. #1
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    Taylor Series / Power Series

    Alright, I'm just a little confused on what my work is asking me. The problem says "Use the definition to find the Taylor series, centered at c, for the function."

    The problem is:

    \cos(x), c = \pi/4

    So I have an answer book that has the answer for this problem. I understand Taylor series / power series, but I'm just a bit confused on what is going on.

    The answer book goes on to do a few derivatives, plugs in the c, etc. Then it forms the sum which is:

    \frac{\sqrt2}{2}\sum{\frac{(-1)^{n(n+1)/2}(x-\frac{\pi}{4})^n}{n!}}


    Now I understand this is in the form of:

    \sum{\frac{f^n(c)(x-c)^n}{n!}}


    Yet... Why does it not follow the "basic list" of power series? Which has:

    \cos(x) = \sum{\frac{(-1)^nx^{2n}}{(2n)!}}



    Why is there only n powers in the answer, and no 2n to be found, etc.? Can someone please explain? Maybe I am mixing two concepts...
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  2. #2
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    Quote Originally Posted by Fire Mage View Post
    Alright, I'm just a little confused on what my work is asking me. The problem says "Use the definition to find the Taylor series, centered at c, for the function."

    The problem is:

    \cos(x), c = \pi/4

    So I have an answer book that has the answer for this problem. I understand Taylor series / power series, but I'm just a bit confused on what is going on.

    The answer book goes on to do a few derivatives, plugs in the c, etc. Then it forms the sum which is:

    \frac{\sqrt2}{2}\sum{\frac{(-1)^{n(n+1)/2}(x-\frac{\pi}{4})^n}{n!}}


    Now I understand this is in the form of:

    \sum{\frac{f^n(c)(x-c)^n}{n!}}


    Yet... Why does it not follow the "basic list" of power series? Which has:

    \cos(x) = \sum{\frac{(-1)^nx^{2n}}{(2n)!}}



    Why is there only n powers in the answer, and no 2n to be found, etc.? Can someone please explain? Maybe I am mixing two concepts...
    the one from the "basic list" is centered at c = 0 ... the odd powered terms cancel out because the odd degree derivatives are of the form \pm \sin(0)
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  3. #3
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    Yes, I also understand that.


    But what I'm trying to get at is: why does it become just n and not 2n? The only thing that changed it was the c. So why does changing the c automatically change the 2n's?
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    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Fire Mage View Post
    Yes, I also understand that.


    But what I'm trying to get at is: why does it become just n and not 2n? The only thing that changed it was the c. So why does changing the c automatically change the 2n's?

    Because when you center at c=0, you have every other term evaluated at sin 0=0. BUT in this case all the derivatives of cosine are either plus and minus cosine or sine. And these four functions evaluated at \pi/4 you'll never get zero. Instead you'll get plus and minus \sqrt{2}/2.
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  5. #5
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    Quote Originally Posted by matheagle View Post
    Because when you center at c=0, you have every other term evaluated at sin 0=0. BUT in this case all the derivatives of cosine are either plus and minus cosine or sine. And these four functions evaluated at \pi/4 you'll never get zero. Instead you'll get plus and minus \sqrt{2}/2.
    Ah thank you for explaining it out. Like I comprehended the problem, understood everything, I can work the problem myself... I just didn't understand WHY.


    But now that I read it, it's like a slap in the face. I totally ignored that fact. Thanks.
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  6. #6
    MHF Contributor matheagle's Avatar
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    Sorry that I had to get physical with you.
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  7. #7
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    Haha, of course.
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