GPODT
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For the function ln (1+x), the validity is -1<x≤1.

I was watching examsolutions earlier and the guy narrating said that if x = 1 in the above example then we would need thousands of terms before we could get an accurate approximation? Why is this so? I mean x=1 is WITHIN the range ?

Thanks
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brianeverit
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(Original post by GPODT)
For the function ln (1+x), the validity is -1<x≤1.

I was watching examsolutions earlier and the guy narrating said that if x = 1 in the above example then we would need thousands of terms before we could get an accurate approximation? Why is this so? I mean x=1 is WITHIN the range ?

Thanks
Yes but the series would converge VERY slowly. Try working out the first few summations and you will see.
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davros
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(Original post by GPODT)
For the function ln (1+x), the validity is -1<x≤1.

I was watching examsolutions earlier and the guy narrating said that if x = 1 in the above example then we would need thousands of terms before we could get an accurate approximation? Why is this so? I mean x=1 is WITHIN the range ?

Thanks
Just because something's "within the range" doesn't necessarily say anything about the speed of convergence - you can have series that converge slowly and series that converge quickly!

But you should note that if you have a power series (i.e. one that involves powers of x) then 1 doesn't decrease when you raise it to a power, whereas a fraction (e.g. 1/100) will decrease very rapidly when you square it etc.
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GPODT
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(Original post by brianeverit)
Yes but the series would converge VERY slowly. Try working out the first few summations and you will see.
(Original post by davros)
Just because something's "within the range" doesn't necessarily say anything about the speed of convergence - you can have series that converge slowly and series that converge quickly!

But you should note that if you have a power series (i.e. one that involves powers of x) then 1 doesn't decrease when you raise it to a power, whereas a fraction (e.g. 1/100) will decrease very rapidly when you square it etc.
If we let x=2 (a value that is outside the range) would this ever yield us an accurate approximation?
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davros
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(Original post by GPODT)
If we let x=2 (a value that is outside the range) would this ever yield us an accurate approximation?
No, it will just diverge horribly!

But there are other techniques available that can help you out - for example combining the series for ln(1-x) and ln(1+x)
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GPODT
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(Original post by davros)
No, it will just diverge horribly!
Ah thought so. Thanks!


(Original post by davros)

But there are other techniques available that can help you out - for example combining the series for ln(1-x) and ln(1+x)
In what way would that help me out?
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davros
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(Original post by GPODT)
Ah thought so. Thanks!




In what way would that help me out?
one series converges for -1 < x <= 1 and the other for -1 <= x < 1 so they both converge for -1 < x < 1.

If you choose a particular fraction e.g. x = 2/3 you get 2 series that converge and you can subtract them to get a series for ln((1 + x) / (1 - x)) which helps you approximate other values for ln that you couldn't do with just one series
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GPODT
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(Original post by davros)
one series converges for -1 < x <= 1 and the other for -1 <= x < 1 so they both converge for -1 < x < 1.

If you choose a particular fraction e.g. x = 2/3 you get 2 series that converge and you can subtract them to get a series for ln((1 + x) / (1 - x)) which helps you approximate other values for ln that you couldn't do with just one series
Cheers. To summarise, for a Maclaurin series expansion:

1) If x is not within the validity range, we will never gain an accurate approximation no matter how many terms we use
2) The closer x is to 0 the faster the speed of convergence

Correct?
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davros
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(Original post by GPODT)
Cheers. To summarise, for a Maclaurin series expansion:

1) If x is not within the validity range, we will never gain an accurate approximation no matter how many terms we use
2) The closer x is to 0 the faster the speed of convergence

Correct?
That's about it
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