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Math Help - power series centered at zero

  1. #1
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    power series centered at zero

    find a power series for the given function, centered at a=0.

    a)F(x)=ln(3+x)

    b) G(x)=1/(1+2x)^(1/4)

    c) Integral, 0 to x, (1-cos(t)/t^2) dt
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  2. #2
    MHF Contributor matheagle's Avatar
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    You can use Taylor's formula, but I'd use the geometric series for the first one.

    {1\over 3+x}={1/3\over 1-(-x/3)}={1\over 3}\sum_{n=0}^{\infty}(-x/3)^n

    =\sum_{n=0}^{\infty}{(-1)^nx^n\over 3^{n+1}}

    now integrate wrt x.
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  3. #3
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    Quote Originally Posted by matheagle View Post
    You can use Taylor's formula, but I'd use the geometric series for the first one.

    {1\over 3+x}={1/3\over 1-(-x/3)}={1\over 3}\sum_{n=0}^{\infty}(-x/3)^n

    =\sum_{n=0}^{\infty}{(-1)^nx^n\over 3^{n+1}}

    now integrate wrt x.
    although i didnot have mathematics as honours in college i didnot understand one thing: How can ln(3+x) be expressed as 1/(3+x).
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  4. #4
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    Quote Originally Posted by Pulock2009 View Post
    although i didnot have mathematics as honours in college i didnot understand one thing: How can ln(3+x) be expressed as 1/(3+x).
    \frac{d}{dx}[\ln{(3 + x)}] = \frac{1}{3 + x}.


    So \int{\frac{1}{3 + x}\,dx} = \ln{(3 + x)}.


    So if you have a power series for \frac{1}{3 + x}, integrate it, then you will have a power series for \ln{(3 + x)}.
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    Quote Originally Posted by twofortwo View Post
    find a power series for the given function, centered at a=0.

    a)F(x)=ln(3+x)

    b) G(x)=1/(1+2x)^(1/4)

    c) Integral, 0 to x, (1-cos(t)/t^2) dt
    a)F(x)=ln(3+x)
    =ln(3(1+x/3))
    =ln3+ln(1+x/3)
    =ln3+(the standard logarithmic expansion)
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    Quote Originally Posted by Prove It View Post
    \frac{d}{dx}[\ln{(3 + x)}] = \frac{1}{3 + x}.


    So \int{\frac{1}{3 + x}\,dx} = \ln{(3 + x)}.


    So if you have a power series for \frac{1}{3 + x}, integrate it, then you will have a power series for \ln{(3 + x)}.
    that makes a lot of sense!!!thanks!!
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    Quote Originally Posted by twofortwo View Post
    find a power series for the given function, centered at a=0.

    a)F(x)=ln(3+x)

    b) G(x)=1/(1+2x)^(1/4)

    c) Integral, 0 to x, (1-cos(t)/t^2) dt
    b)G(x) can be expanded binomially to get a power series if i am not wrong.
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  8. #8
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    Quote Originally Posted by Pulock2009 View Post
    b)G(x) can be expanded binomially to get a power series if i am not wrong.
    Yes, as long as you write it as

    (1 + 2x)^{-\frac{1}{4}}.
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  9. #9
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by Pulock2009 View Post
    although i didnot have mathematics as honours in college i didnot understand one thing: How can ln(3+x) be expressed as 1/(3+x).
    I clearly said you should now integrate.
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  10. #10
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    thank you!! i understand all of them except the integral one.
    can i get more help on that one please!
    Thanx so much!
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  11. #11
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    Quote Originally Posted by matheagle View Post
    I clearly said you should now integrate.
    Yes you did, the OP just didn't understand WHY you should integrate, hence my fill-in post.
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