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Math Help - taylor series

  1. #1
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    taylor series

    find exactly the taylor polynomial of degree 4 for


    f(x) = e^(2x)
    about x = -3

    f(x) = e^2x e^-6
    f'(x) = 2e^2x 2e^-6
    f"(x) = 4e^2x 4e^-6
    f"'(x) = 8e^2x 8e^-6
    f""(x) = 16e^2x 16e^-6

    right?

    so to write it out its like

    p4(x, -3) = e^-6 + (x-3)/1!*2e^-6 + (x-3)/2!*4e^-6 and so on is that right or since its -3 should it be
    p4(x, -3) = e^-6 + (x+3)/1!*2e^-6 + (x+3)/2!*4e^-6

    ???
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by helmszee View Post
    find exactly the taylor polynomial of degree 4 for


    f(x) = e^(2x)
    about x = -3

    f(x) = e^2x e^-6
    f'(x) = 2e^2x 2e^-6
    f"(x) = 4e^2x 4e^-6
    f"'(x) = 8e^2x 8e^-6
    f""(x) = 16e^2x 16e^-6

    right?

    so to write it out its like

    p4(x, -3) = e^-6 + (x-3)/1!*2e^-6 + (x-3)/2!*4e^-6 and so on is that right or since its -3 should it be
    p4(x, -3) = e^-6 + (x+3)/1!*2e^-6 + (x+3)/2!*4e^-6

    ???
    no. you want:

    T_4(x) = f(-3) ~+ ~ \frac {f'(-3)(x + 3)}{1!} ~+~ \frac {f''(-3)(x + 3)^2}{2!} +~ \frac {f'''(-3)(x + 3)^3}{3!} ~+~ \frac {f''''(-3)(x + 3)^4}{4!}
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