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    ...I'm stuck on these types of questions:

    How many terms of the series for ln(1+x) are required to find ln.2 to plus or minus 0.001?

    can anyone help out?
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    The error is roughly the absolute value of the first omitted term.

    ln(1 + x) = x - x^2/2 + x^3/3 - x^4/4 + ...

    ln(2) = 1 - 1/2 + 1/3 - 1/4 + ...

    For an error of 0.001 we want the first omitted term to be less than 0.001 in absolute value:

    ln(2) ~= 1 - 1/2 + 1/3 - 1/4 + ... - 1/1000.
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    (Original post by Elle)
    ...I'm stuck on these types of questions:

    How many terms of the series for ln(1+x) are required to find ln.2 to plus or minus 0.001?

    can anyone help out?
    Following on from jonny's post, you can generalise it with,

    ln(1 + x) = x - x^2/2 + x^3/3 - x^4/4 + ...(-1)^(n+1)x^n/n + ...

    let error be δ

    δ = |(-1)^(n+1)x^n/n|
    δ = |x^n|/n

    for δ < 0.001 and x=1,

    |1^n|/n < 0.001
    1/n < 0.001
    n > 1000

    so you need at least 100 terms.
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    Thanks for the replies.. but the answer in the book says 3 :confused:
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    (Original post by Elle)
    Thanks for the replies.. but the answer in the book says 3 :confused:
    Ahh. Just noticed. You have.
    ln.2
    I/we thought it was ln(2) - no wonder so many terms were needed - but then I should have noticed that, shouldn't I
    So, if its ln(0.2) that you want then from ln(1+x), the x should be x=-0.8.

    let error be δ

    δ = |(-1)^(n+1)x^n/n|
    δ = |x^n|/n

    for δ < 0.001 and x=-0.8,

    |-0.8^n|/n < 0.001
    0.8^n < 0.001n
    0.8^n - 0.001n < 0

    Now n=4 (to give only 3 terms needed for series) is not gonna work!
    So looking back on your question - if we take instead ln(1.2) - then,

    δ=0.001, x=0.2, giving

    |0.2^n|/n < 0.001
    0.2^n < 0.001n
    0.2^n - 0.001n < 0

    Now, you can solve this using iterative techniques, or you can just take a guess. Plug in some values and see what happens.

    n = 2: 0.2^2 - 0.001*2 = 0.038 > 0 - no good
    n = 3: 0.2^3 - 0.001*3 = 0.005 > 0 - no good
    n = 4: 0.2^4 - 0.001*4 = -0.0024 < 0 - success!

    Therefore need three terms to give error < 0.001.
 
 
 

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