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Math Help - Series

  1. #1
    Junior Member
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    Series

    ok i really hate series, they are confusing and i almost have no idea how to do them. These questions are really driving me up the wall.
    Any help would be greatly appreciated.




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  2. #2
    Grand Panjandrum
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    First problem:

    Lets look at the first few terms of the Taylor series for f(x)=ln(1+x) centred
    at 0.

    f(0)=0

    df/dx=1/(1+x),

    so df/dx(0) = 1

    d^2f/dx^2=-1/(1+x)^2

    so d^2f/dx^2(0)=-1.

    So the series starts:

    f(x)=0+x-(1/2)x^2+...

    Now look at your candidate answers. The first two can be ruled out as
    the absolute value of the coefficient of x^2 is 2.

    That leaves us with the second two. The fourth gives a -ve coefficient for
    x, so it is ruled out, which leaves us with (c), which agrees with the terms
    of the series that we know.

    RonL
    Last edited by CaptainBlack; September 25th 2006 at 11:27 PM.
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  3. #3
    Grand Panjandrum
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    Second Problem:

    Look at the Lagrange form of the remainder for a truncated Taylor series.

    In this case it is R(x)=f''''(zeta)(x-1)^4/4!, for some zeta in [0.9,1.1].

    f''''(x)=120x,

    so:

    |R(x)|< 120 (1.1) (0.1)^4 /4!=5 (1.1) (0.1)^4

    RonL
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  4. #4
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    He have,
    f(0)=0

    f'(0)x/1!=x

    f''(0)x^2/2!=0

    f'''(0)x^3/3!=x^3/3!

    And so one,

    Thus,
    b is the answer.

    Also, the hyperbolic sine gets its name because it is in ways similar to the sine function. For example,
    sin x=x-x^3/3!+x^5/5!-...
    But the hyperbolic sine has all positive terms,
    sinh x=x+x^3/3!+x^5/5!+...
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