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Math Help - Taylor series question

  1. #1
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    Taylor series question

    srry guys i need some help, been stuck on this problem for like half an hour now lol

    anyway,

    find the taylor series about x of zero= pi, corrresponding to the function of F(x)=x+sinx and determine its interval of convergence

    we never learned em in class, so its liek self taught and im not sure how to work this problem thanks
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by willowofwisp View Post
    srry guys i need some help, been stuck on this problem for like half an hour now lol

    anyway,

    find the taylor series about x of zero= pi, corrresponding to the function of F(x)=x+sinx and determine its interval of convergence

    we never learned em in class, so its liek self taught and im not sure how to work this problem thanks
    What's this "Taylor series about x of zero= pi" business? The last I heard was that \pi is not equal to 0. I will assume you mean an expansion about 0.

    According to Taylor's theorem, a function f(x) expanded about the point x = a is:
    f(x) \approx f(a) + \frac{1}{1!}f^{\prime}(a)(x - a) + \frac{1}{2!}f^{\prime \prime}(a)(x - a)^2 + \frac{1}{3!}f^{\prime}(a)(x - a)^3 +~...

    So we need the derivatives of f(x) = x + sin(x) evaluated at x = 0.

    f(x) = x + sin(x) \implies f(0) = 0 + sin(0) = 0

    f^{\prime}(x) = 1 + cos(x) \implies f^{\prime}(0) = 1 + cos(0) = 2

    f^{\prime \prime}(x) = -sin(x) \implies f^{\prime \prime}(0) = -sin(0) = 0

    etc.

    In fact what we can do is this:
    f(x) = x + sin(x)
    so we can write the Taylor series of x about x = 0 and the Taylor series of sin(x) about x = 0 and add them:
    f(x) = (x) + \left ( \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n + 1)!}x^{2n + 1} \right )

    (I'll leave it to you to do enough terms to verify that the sin(x) series is correct.)

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    What's this "Taylor series about x of zero= pi" business? The last I heard was that \pi is not equal to 0. I will assume you mean an expansion about 0.

    According to Taylor's theorem, a function f(x) expanded about the point x = a is:
    f(x) \approx f(a) + \frac{1}{1!}f^{\prime}(a)(x - a) + \frac{1}{2!}f^{\prime \prime}(a)(x - a)^2 + \frac{1}{3!}f^{\prime}(a)(x - a)^3 +~...

    So we need the derivatives of f(x) = x + sin(x) evaluated at x = 0.

    f(x) = x + sin(x) \implies f(0) = 0 + sin(0) = 0

    f^{\prime}(x) = 1 + cos(x) \implies f^{\prime}(0) = 1 + cos(0) = 2

    f^{\prime \prime}(x) = -sin(x) \implies f^{\prime \prime}(0) = -sin(0) = 0

    etc.

    In fact what we can do is this:
    f(x) = x + sin(x)
    so we can write the Taylor series of x about x = 0 and the Taylor series of sin(x) about x = 0 and add them:
    f(x) = (x) + \left ( \sum_{n = 0}^{\infty} \frac{(-1)^n}{(2n + 1)!}x^{2n + 1} \right )

    (I'll leave it to you to do enough terms to verify that the sin(x) series is correct.)

    -Dan
    well its writtin Xsubscript zero =Pi
    not sure if that changes anything
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  4. #4
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    Quote Originally Posted by willowofwisp View Post
    well its writtin Xsubscript zero =Pi
    not sure if that changes anything
    It changes it somewhat. All of Dan's derivatives are still correct since it is still centered about zero. However, the values of the derivatives should be at \pi rather than zero.

    Now, only the third, seventh, eleventh, etc... derivatives will hold values since sine values at pi is still zero, and the first derivative which used to be 2 is now 0 since the value of cosine changes from 1 to -1.
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