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Math Help - Maclaurin And Taylor Series

  1. #1
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    Maclaurin And Taylor Series

    For example use sigma notation to write maclaurin seires ofr the function

    e^x

    use sigma notation to write taylor seires ofr the function

    e^x x0=1


    this is very easy function
    my solution step for maclaurin is

    1. find f , f', f'', f''', f'''' when x=0
    2. then find p0, p1, p2, p3, p4
    3. look at the regularity of them, write the sigma notation


    same as taylor seires except first step is when x=1


    I just want to know is this solution ok or not?
    Do anybody have any other solution?


    Thank you!
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by soleilion View Post
    For example use sigma notation to write maclaurin seires ofr the function

    e^x

    use sigma notation to write taylor seires ofr the function

    e^x x0=1


    this is very easy function
    my solution step for maclaurin is

    1. find f , f', f'', f''', f'''' when x=0
    2. then find p0, p1, p2, p3, p4
    3. look at the regularity of them, write the sigma notation


    same as taylor seires except first step is when x=1


    I just want to know is this solution ok or not?
    Do anybody have any other solution?


    Thank you!
    It is well known that e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    It is well known that e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}

    yeah, I know
    this is easy example
    If I have a function is not that easy, not well know
    How can I do?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by soleilion View Post
    yeah, I know
    this is easy example
    If I have a function is not that easy, not well know
    How can I do?
    ok for example how about \frac{-1}{(1+x^2)^2}? since we know that \frac{D\bigg[\frac{1}{1+x^2}\bigg]}{dx}=\frac{-1}{(1+x^2)^2} and since we know using a simple substitution of \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n}x^{n} to get \frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n} so now using that knowledge we see that \frac{-1}{(1+x^2)^2}=\frac{D\bigg[\frac{1}{1+x^2}\bigg]}{dx}=\frac{D[\sum_{n=0}^{\infty}(-1)^{n}x^{2n}\bigg]}{dx}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n-1}\cdot{2n}
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    ok for example how about \frac{-1}{(1+x^2)^2}? since we know that \frac{D\bigg[\frac{1}{1+x^2}\bigg]}{dx}=\frac{-1}{(1+x^2)^2} and since we know using a simple substitution of \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n}x^{n} to get \frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n} so now using that knowledge we see that \frac{-1}{(1+x^2)^2}=\frac{D\bigg[\frac{1}{1+x^2}\bigg]}{dx}=\frac{D[\sum_{n=0}^{\infty}(-1)^{n}x^{2n}\bigg]}{dx}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n-1}\cdot{2n}
    I think d/dx [1/(1+x^2)]= -2x(1+x^2)^-2
    and I dont understand since you are using a simple substitution
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by soleilion View Post
    I think d/dx [1/(1+x^2)]= -2x(1+x^2)^-2
    and I dont understand since you are using a simple substitution
    Wow I am really out of it tonight...well i have no idea how why I did that but I will do another to make more sense... \frac{-1}{(1+x)^2}=\frac{D\bigg[\frac{1}{1+x}\bigg]}{dx}=\frac{D\bigg[\sum_{n=0}^{\infty}(-1)^nx^{n}\bigg]}{dx}=\sum_{n=0}^{\infty}n(-1)^nx^{n-1}
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  7. #7
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    Quote Originally Posted by Mathstud28 View Post
    Wow I am really out of it tonight...well i have no idea how why I did that but I will do another to make more sense... \frac{-1}{(1+x)^2}=\frac{D\bigg[\frac{1}{1+x}\bigg]}{dx}=\frac{D\bigg[\sum_{n=0}^{\infty}(-1)^nx^{n}\bigg]}{dx}=\sum_{n=0}^{\infty}n(-1)^nx^{n-1}
    If I do maclaurin and taylor problem
    Can I follow any regularity?
    I'm still confusing right now.
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  8. #8
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    Quote Originally Posted by Mathstud28 View Post
    ok for example how about \frac{-1}{(1+x^2)^2}? since we know that \frac{D\bigg[\frac{1}{1+x^2}\bigg]}{dx}=\frac{-1}{(1+x^2)^2} and since we know using a simple substitution of \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n}x^{n} to get \frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n} so now using that knowledge we see that \frac{-1}{(1+x^2)^2}=\frac{D\bigg[\frac{1}{1+x^2}\bigg]}{dx}=\frac{D[\sum_{n=0}^{\infty}(-1)^{n}x^{2n}\bigg]}{dx}=\sum_{n=0}^{\infty}(-1)^{n}x^{2n-1}\cdot{2n}
    He was just typing quickly, I belive what he ment is

    let f(x)=\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n

    Then  f'(x)=\frac{1}{(1-x)^2}=\sum_{n=0}^{\infty}nx^{n-1}

    now to get the power series we wanted we would evaluate f'(-x^2)

    so

     f'(-x^2)=\frac{1}{(1+x^2)^2}=\sum_{n=0}^{\infty}n(-x^2)^{n-1}=\sum_{n=0}^{\infty}n(-1)^{n-1}(x)^{2n-2}

    I hope this helps
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  9. #9
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    Quote Originally Posted by TheEmptySet View Post
    He was just typing quickly, I belive what he ment is

    let f(x)=\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n

    Then  f'(x)=\frac{1}{(1-x)^2}=\sum_{n=0}^{\infty}nx^{n-1}

    now to get the power series we wanted we would evaluate f'(-x^2)

    so

     f'(-x^2)=\frac{1}{(1+x^2)^2}=\sum_{n=0}^{\infty}n(-x^2)^{n-1}=\sum_{n=0}^{\infty}n(-1)^{n-1}(x)^{2n-2}

    I hope this helps

    why 1/1-x=sigma x^n???????????????
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by soleilion View Post
    If I do maclaurin and taylor problem
    Can I follow any regularity?
    I'm still confusing right now.
    yeah for example? say you wanted to do arctan\bigg(\frac{3x}{11}\bigg)...this is equal to \frac{1}{11}\int{\frac{3}{1+\bigg(\frac{3}{11}x\bi  gg)^2}}dx so this is equal to \frac{3}{11}\int\sum_{n=0}^{\infty}(-1)^{n}\bigg(\frac{3}{11}x\bigg)^{2n}dx=\sum_{n=0}^  {\infty}\frac{(-1)^{n}\bigg(\frac{3}{11}x\bigg)^{2n+1}}{2n+1}
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  11. #11
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by soleilion View Post
    why 1/1-x=sigma x^n???????????????
    It is just a commonly known power series
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    It is just a commonly known power series
    Maclaurin Series -- from Wolfram MathWorld
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  13. #13
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    Quote Originally Posted by Mathstud28 View Post
    yeah for example? say you wanted to do arctan\bigg(\frac{3x}{11}\bigg)...this is equal to \frac{1}{11}\int{\frac{3}{1+\bigg(\frac{3}{11}x\bi  gg)^2}}dx so this is equal to \frac{3}{11}\int\sum_{n=0}^{\infty}(-1)^{n}\bigg(\frac{3}{11}x\bigg)^{2n}dx=\sum_{n=0}^  {\infty}\frac{(-1)^{n}\bigg(\frac{3}{11}x\bigg)^{2n+1}}{2n+1}
    My HW question is not that hard

    they are only cos(pi*x) or ln(1+x) or xsinx

    If I do my way, I know how to fix them
    but I see the solution mannual
    they are different, I dont understand either

    I just wanna know how to do this kind of problem
    I want to have something like process, so I can follow it.
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  14. #14
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by soleilion View Post
    My HW question is not that hard

    they are only cos(pi*x) or ln(1+x) or xsinx

    If I do my way, I know how to fix them
    but I see the solution mannual
    they are different, I dont understand either

    I just wanna know how to do this kind of problem
    I want to have something like process, so I can follow it.
    The first one you just insert \pi{x} into the powers series of cos(x)...the second one you either know the formula or you use the fact that ln(x+1)=\int\sum_{n=0}^{\infty}(-1)^{n}x^{n}dx and the third you take the regular power series for sin(x) and add one to the exponent of x
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  15. #15
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    Quote Originally Posted by Mathstud28 View Post
    Thank you very much
    I saw this website, it's very good
    I understand #3 to #25
    When I do this kind of problem, I use that method
    For explicit form, I can not use that because my instructor want us to solve step by step
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