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Maclaurins Series- accuracy (AQA- P3) watch

1. ...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?
2. 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.
3. (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.
4. Thanks for the replies.. but the answer in the book says 3
5. (Original post by Elle)
Thanks for the replies.. but the answer in the book says 3
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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Updated: June 18, 2004
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