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Math Help - Just need clarification on a power series question

  1. #1
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    Just need clarification on a power series question

    Hello, a question on my homework is in this form:

    f(x) = 1/X about Xo = 3. (Xo is x naught, btw)

    I'm confused what "about x naught" means, because the form I've been shown in class is about a, aka, the series:

    Sum from 0 to infinity of Cn(X-a)^n

    is called a power series about. Does x naught take the place of a, or do I use it to find the first term (since the power n would be 0) and therefore the c value, or some other way?

    If you know what my prof means by this it would be wonderful.
    Thank you
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  2. #2
    Member Mentia's Avatar
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    Bellingham, WA
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    Yep, you're right. Basically he means X0 = a. You are taking the series about X0 = 3. I.e:

    f(x) = f(3)/0! + f'(3)*(x-3)/1! + f''(3)*(x-3)^2/2! + f'''(3)*(x-3)^3/3! + ...

    so in your particular case you have:

    f(x) = 1/x, f'(x) = -1/x^2, f''(x) = 2/x^3, f'''(x) = -6/x^4, ... fn(x) = (n!*(-1)^n)/x^(n+1)

    which gives

    f(x) (about x = a) = 1/a - (1/a^2)*(x-a) + (1/a^3)*(x-a)^2 - (1/a^4)*(x-a)^3 + - + - ...

    about x=3 this is:

    f(x) = 1/3 - (1/9)*(x-3) + (1/27)*(x-3)^2 - (1/81)*(x-3)^3 + ...


    plug y = 1/x and y = 1/3 - (1/9)*(x-3) + (1/27)*(x-3)^2 - (1/81)*(x-3)^3 into your calculator and open the table. Put in values near 3, like 3.001 or 2.999 and you'll find that both y's return nearly the same value.
    Last edited by Mentia; March 5th 2009 at 02:29 PM.
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  3. #3
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    Thank you very much
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