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Math Help - Taylor Polynomials?

  1. #1
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    Taylor Polynomials?

    Find the Taylor polynomial Tn(x) for the function f at the number a. Graph f and T3 on the same paper.

    f(x)= e^-3x sin(3x), a= 0, n=3

    f(x)= ln x/x, a= 1, x= 3

    Please help me with these two problems. Thanks
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by shah4u19 View Post
    Find the Taylor polynomial Tn(x) for the function f at the number a. Graph f and T3 on the same paper.

    f(x)= e^-3x sin(3x), a= 0, n=3

    f(x)= ln x/x, a= 1, x= 3

    Please help me with these two problems. Thanks
    What have you tried? I'm guessing you don't know the Maclaurin series themselves and can just take the first three terms huh?

    to calculate a third degree (I assume thats what the n means) Taylor polynomial at x=a you must calcuate

    \sum_{n=0}^{3}\frac{f^{(n)}(a)(x-a)^n}{n!}. Does that seem doable?
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    Well, for the first term, I have tried whatever I could. I know maclaurin series for e^-3x is (-3x)^n/n! and maclaurin series for sin (3x) is (-1)^n (3x)^(2n+1)/(2n+1)!. I tried to divide sin(3x)/e^3x upto 3rd degree and submitted the answer I got, but it was wrong. So I don't know what to do now.

    For ln x/x, I dont know where to start.

    Thanks Anyways.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by shah4u19 View Post
    Well, for the first term, I have tried whatever I could. I know maclaurin series for e^-3x is (-3x)^n/n! and maclaurin series for sin (3x) is (-1)^n (3x)^(2n+1)/(2n+1)!. I tried to divide sin(3x)/e^3x upto 3rd degree and submitted the answer I got, but it was wrong. So I don't know what to do now.

    For ln x/x, I dont know where to start.

    Thanks Anyways.
    Ok. Since you know the MaClaurin series you don't have to do the derivatives (whew)

    Merely take \left(\sum_{n=0}^{3}\frac{x^n}{n!}\right)\cdot\lef  t(\sum_{n=0}^{3}\frac{(-1)^nx^{2n+1}}{(2n+1)!}\right).

    For the second one you could to the derivatives or notice that

    \frac{\ln(x)}{x}=\frac{d}{dx}\left[\frac{\ln^2(x)}{2}\right]. The second one is well known...so then all you have to do is...
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