how can I find the values for which a mclaurin series is valid?

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thebrahmabull
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For example ln(1+x) is valid for -1<x<1 (less than or equal one, can't type symbol.) I know they can be graphically found . but that's not possible in an exam or generally is time consuming. Any other way?
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atsruser
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(Original post by thebrahmabull)
For example ln(1+x) is valid for -1<x<1 (less than or equal one, can't type symbol.) I know they can be graphically found . but that's not possible in an exam or generally is time consuming. Any other way?
At what level are you studying? There are various tests that you can apply to find the range of convergence of series (i.e. the range in which they are valid as you have described it) but these aren't taught at A level.
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thebrahmabull
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(Original post by atsruser)
At what level are you studying? There are various tests that you can apply to find the range of convergence of series (i.e. the range in which they are valid as you have described it) but these aren't taught at A level.
I am studying at alevel. So should I just memorize ranges of convergence of some common functions ? Can you recommend some functions whose range I should memorize?
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Zacken
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(Original post by thebrahmabull)
I am studying at alevel. So should I just memorize ranges of convergence of some common functions ? Can you recommend some functions whose range I should memorize?
They're all given in the formula booklet.
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Student403
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I thought this was only taught in IB
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Zacken
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(Original post by Student403)
I thought this was only taught in IB
It's FP2 McLaurin series, isn't it? Unless you mean finding the radius of convergence? In which case, yeah, it's in one of the options in IB, but not many people take that option. :dontknow:
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Student403
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(Original post by Zacken)
It's FP2 McLaurin series, isn't it? Unless you mean finding the radius of convergence? In which case, yeah, it's in one of the options in IB, but not many people take that option. :dontknow:
Yeah Maths HL - Nah I didn't mean FP2 Chapter 6 :lol: We're doing that right now, as a matter of fact! Lovely topic
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Zacken
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(Original post by Student403)
Yeah Maths HL - Nah I didn't mean FP2 Chapter 6 :lol: We're doing that right now, as a matter of fact! Lovely topic
Yeah, you get four "options" to choose one from at the end of your Maths HL course and one of them is calculus, which contains things like radius of convergence, etc.. :yep:

It's lovely!! Wait till you solve DE's using them.
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Student403
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(Original post by Zacken)
Yeah, you get four "options" to choose one from at the end of your Maths HL course and one of them is calculus, which contains things like radius of convergence, etc.. :yep:

It's lovely!! Wait till you solve DE's using them.
Oh wasn't aware - that's pretty cool! What are the other options?

Don't make me nerdgasm!
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Zacken
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(Original post by Student403)
Oh wasn't aware - that's pretty cool! What are the other options?

Don't make me nerdgasm!
Stats, discrete maths and abstract algebra. Not that cool to be honest, there's so little in the Maths HL course

Such an engineer!
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Student403
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(Original post by Zacken)
Stats, discrete maths and abstract algebra. Not that cool to be honest, there's so little in the Maths HL course

Such an engineer!
Ohh no mechanics? :eek:

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the bear
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(Original post by Student403)
I thought this was only taught in IB
FP2... some boards do both Taylor's and MacLaurin's,,, some only do MacLaurin's
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Student403
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(Original post by the bear)
FP2... some boards do both Taylor's and MacLaurin's,,, some only do MacLaurin's
Yeah I'm learning it right now in FP2 I was referring specifically to range between which a series converges. Sorry should have made it clearer
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the bear
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(Original post by Student403)
Yeah I'm learning it right now in FP2 I was referring specifically to range between which a series converges. Sorry should have made it clearer
:spank:
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Student403
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(Original post by the bear)
:spank:
:run:
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Indeterminate
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If you write

\ln(1+x) = \displaystyle \sum_{n \geqslant 1} \dfrac{(-1)^{n+1}x^n}{n}

then you could use the ratio test to find the range of x values for which it converges.

Not something you need to worry about at A-level, though
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the bear
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(Original post by Student403)
:run:
:bl:
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Student403
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(Original post by Indeterminate)
If you write

\ln(1+x) = \displaystyle \sum_{n \geqslant 1} \dfrac{(-1)^{n+1}x^n}{n}

then you could use the ratio test to find the range of x values for which it converges.

Not something you need to worry about at A-level, though
That looks really cool - thanks!
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Mihael_Keehl
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Amazing I was going to post the exact same question...

Obviousuly -1 has to be the lower limit since ln(0) n/a but why is the upper limit 1 here?

isnt the function for values of x greater than 1 still yield values?
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16characterlimit
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For most of them they are valid for all real numbers given enough terms.

The reason ln(x+1) applies for |x|<1 is because it doesn't have the factorial term below it (well it does but gets cancelled out if you derive it), since the increasing powers on top are always going to get bigger and bigger at a much faster rate than the numbers on the bottom any x that you substitute will diverge towards infinity, rather than tend to 0 if there was a factorial term on the bottom.

So basically, check if your bottom numbers are growing at a factorial/exponential rate and they aren't getting cancelled by your f(0)'s.
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