ratio test for convergence? - help!!

ayapatrick

Anyone know of a good website or somewhere that would basically show you this step from step!

The ratio test i can understand, but when your asked to test a Maclaurin series for convergence thats the problem! Some of the series alternate from negative and positive - and sometimes its un and then for others its u(n+1)

ps. I understand the maclaurin and can work out the series etc, but to use the ratio test on them - its gibberrish to me basically!

any help appreciated!;)

Jam-Fly

Do the probability option...


I know you shouldn't really 'learn off' anything for maths, but it is kinda reccommended to learn this thing off for Further Integration



I don't think that answers your question, but hopefully it'll be of some help

Fringe

You get the absolute value so you always make it positive. Put the thing in those two vertical lines like l -1^n etc l

JSK 252

Learn the generators off as it will save you time in the exam. There is only 5 of them or so so they are easy enough to remember.

Jam-Fly there should be a ( -1)^k for the sin x and cos x generator because of the +-+-+-+-+-+-+-+-+-+-+

ayapatrick

Jam-Fly wrote: »

Do the probability option...


I know you shouldn't really 'learn off' anything for maths, but it is kinda reccommended to learn this thing off for Further Integration



I don't think that answers your question, but hopefully it'll be of some help

This is exactly what i was on about! Ah the teacher kinda said to learn it! , but all i had in the book was the one for sinx and cosx! (they actually write it in the book as a point to learn?)

My worries about doing it without fully understanding though was that you could get caught out some other where!

Or is it only on these five functions that you can ever be asked in the exam?

Thing about probability....... Never done it in my life! We have a new teacher at the minute(and will till summer) so shes going doing probability with us from the beginning, whereas the other fella taught us the calculus option!

My question - Which is easier?

xOxSinéadxOx

is there anyway you can find them without learning them off?

ayapatrick

xOxSinéadxOx wrote: »

is there anyway you can find them without learning them off?

im sure there is, but maybe it was just me, but i think the teacher just skipped by it!

Jam-Fly

ayapatrick wrote: »

This is exactly what i was on about! Ah the teacher kinda said to learn it! , but all i had in the book was the one for sinx and cosx! (they actually write it in the book as a point to learn?)

My worries about doing it without fully understanding though was that you could get caught out some other where!

Or is it only on these five functions that you can ever be asked in the exam?

Thing about probability....... Never done it in my life! We have a new teacher at the minute(and will till summer) so shes going doing probability with us from the beginning, whereas the other fella taught us the calculus option!

My question - Which is easier?

The thing about probability is, it can be very hard to understand unless you have an aptitude for it. Like, alot of people are just 'good' at probability, so they do that option.
If you're good at it, definitely do it. If you'd rather rather do a bit of work and learning, do integreation.

And yes, I think you'll basically be asked on these functions only (like, these or a variation off these, so if you know these, you should be fairly sorted [I think! Don't quote me though])

Jam-Fly

xOxSinéadxOx wrote: »

is there anyway you can find them without learning them off?

Yeah there is...


if we want to find the MacLaurin Series for loge(x+1) for example:


f(0) = f(0) + (f'(0)x)/1! + (f''(0)x^2)/2! + (f'''(0)x^3)/3! + (f''''(0)x^4)/4!...etc.

is the general term

so

f(x) = loge(1+x) then f(0)= loge(1+0)= loge1 = 0

f'(x) = 1/(1+x) = (1+x)^-1 then f'(0) = 1/1 = 1

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

f'''(x) = 2(1+x)^-3(1) = 2/(1+x)^3 then f'''(0)= 2

f''''(x) = -6(1+x)^-4(1) = -6/(1+x)^4 then f''''(0)= -6/1

etc.


then, subbing these back into our original general term...

0 + 1x/1! - 1x^2/2! + 2x^3/3! - 6x^4/4! .....etc.

PurpleFistMixer

Ratio test for convergence? Not that tricky.

Now if only boards.ie had support for LaTeX...

You just take the limit as n goes to infinity of the absolute value of An+1 over An. If this resolves to be less than 1, then your series converges. If it's greater than 1, it diverges, and if it's equal to one, no conclusion can be drawn.

Also... where did that image of the various functions come from? sinx and cosx are missing a somewhat important (-1)^k... Otherwise they wouldn't be alternating with positive and negative signs! (I see someone already pointed this out, but no harm in doing it again. >.>)

Also, the probability option also includes further statistics.

And hurray, I have an exam on this stuff today.

ayapatrick

PurpleFistMixer wrote: »

Ratio test for convergence? Not that tricky.

Now if only boards.ie had support for LaTeX...

You just take the limit as n goes to infinity of the absolute value of An+1 over An. If this resolves to be less than 1, then your series converges. If it's greater than 1, it diverges, and if it's equal to one, no conclusion can be drawn.

Also... where did that image of the various functions come from? sinx and cosx are missing a somewhat important (-1)^k... Otherwise they wouldn't be alternating with positive and negative signs! (I see someone already pointed this out, but no harm in doing it again. >.>)

Also, the probability option also includes further statistics.

And hurray, I have an exam on this stuff today. :quote

cheers. ah the ratio test i can understand, but it was just when you were asked to test a maclaurin series for convergence that i got caught!

I sse, that would be kind of important al right!

ayapatrick

ayapatrick wrote: »

cheers. ah the ratio test i can understand, but it was just when you were asked to test a maclaurin series for convergence that i got caught!

I sse, that would be kind of important al right!

You just have to remember that a Maclaurin Series is no different from any other arithmetic series, so there shouldn't really be a problem using the ratio test on one.

I think your problem might lie in finding the general term (Un), does it? Because, that can be the trickiest part of it for some people. If you have to, learn the general terms for the five functions that can be asked, but, I'd recommend actually learning how to do it. If your problem doesn't lie in finding the Nth term, it more than likely lies in the division of the (n+1)th term by the general term. That, again, can be tricky. Know how to divide indices and factorials, as that's all that really comes up. After that you just have to take the limit as n tends to infinity of the absolute value of the result of the division.

If you can't find any guides online for the Maclaurin Series online, just look for some guides on the Taylor series, as they're basically the same thing.

Oh, and unless you're very good at probability and statistics, stay away from the option. It's the type of question that can fool you into thinking that you're good at it, when actually you are terrible.

ayapatrick

-JammyDodger- wrote: »

You just have to remember that a Maclaurin Series is no different from any other arithmetic series, so there shouldn't really be a problem using the ratio test on one.

I think your problem might lie in finding the general term (Un), does it? Because, that can be the trickiest part of it for some people. If you have to, learn the general terms for the five functions that can be asked, but, I'd recommend actually learning how to do it. If your problem doesn't lie in finding the Nth term, it more than likely lies in the division of the (n+1)th term by the general term. That, again, can be tricky. Know how to divide indices and factorials, as that's all that really comes up. After that you just have to take the limit as n tends to infinity of the absolute value of the result of the division.

If you can't find any guides online for the Maclaurin Series online, just look for some guides on the Taylor series, as they're basically the same thing.

Oh, and unless you're very good at probability and statistics, stay away from the option. It's the type of question that can fool you into thinking that you're good at it, when actually you are terrible.

Ya thats where i get stuck! Finding the general term!

I dont have a problem with the maclaurin, its just to ratio test it!

Ya my last maths teacher gave me the same advice on that so i think il stick at the calculus!

thanks for the help, its helped make it clearer for me!

Challenged

http://www.studentxpress.ie/maclaurin.pdf

JSK 252

PurpleFistMixer wrote: »

Also... where did that image of the various functions come from? sinx and cosx are missing a somewhat important (-1)^k... Otherwise they wouldn't be alternating with positive and negative signs! (I see someone already pointed this out, but no harm in doing it again. >.>)

That would be moi!

MathsManiac

From my perspective as a former teacher, who also corrected this paper for a number of years, I'd make a few points:

I would NOT recommend that anyone try to deal with this topic by learning off the various series by heart. Here's why:

1. On principle, it's a very poor way to try and cope with ANY topic in maths. Understanding is always the key, without exception.
2. It's not much use to you in the exam anyway. You are rarely, (if ever) asked to WRITE DOWN the Maclaurin series; you are usually asked to DERIVE it. This means you have to produce it in the way that Jam-Fly has described. If you just write down the correct series, you'd get the attempt mark only. And believe me, there are easier ways to pick up an attempt mark than learning a bunch of MacLaurin series off by heart. (The only advantage to knowing the answer is that you'll know if you've made a mistake and so, if you have time, you could go back and try to find the mistake. I don't think that this benefit justifies the effort.)

Identifying the general term is certainly one of the things that students have trouble with. I would point out the following strategies that I believe are useful:
For a start, once you've derived and simplified the first few terms, this is basically a pattern-spotting exercise. The basic shape of all these terms is the same - it's always going to be x to the power of something divided by something (leaving aside the +/- issue for a minute). So just look at the pattern:

• if the first term has a 1, and the second term has a 2 and the third term has a 3, etc., then the nth term is obviously going to have an n.
• if the first term has a 2, and the second term has a 4 and the third term has a 6, etc., then the nth term is obviously going to have a 2n.
• (slightly less obvious):if the first term has a 1, and the second term has a 3 and the third term has a 5, etc., then you need to be able to see that you're doubling it and subtracting 1, so the nth term is going to have a 2n-1.

NB: the above are not "rules" to be learned; they're just illustrations of how you turn a pattern into a statement of a general term.

If the series has an alternating sign, then the "trick", as has already been pointed out, is to put in a (-1) to the power of n or n+1 in front, depending on wheter you want the even or the odd terms to be minus.

Having written down what you think is the general term, sub in one or two of the values for n to check it, and fix it if it's wrong.

By the way, life gets a bit complicated if you're looking up answers in different places, or talking to different people. Some people call the first term u0 and other people call it u1. If you don't realise this, you might think that people are giving you conflicting answers to the same question. For example, if I'm in the habit of starting my series at u0, then I would say that the general term in e^x is un = (x^n)/n!. However, if I were in the habit of starting at u1, I would say that un = (x^(n-1))/(n-1)! By and large, the version that starts at 0 has a simpler expression for un, but there's the disadvantage that un is not then the nth term. Unless you're quite clear about what you're doing, I would recommend you stick to the way your teacher does it.

xOxSinéadxOx

I still don't really get how you find the general term, am I just stupid?

MathsManiac

xOxSinéadxOx wrote: »

I still don't really get how you find the general term, am I just stupid?

You're doing higher level maths. Therefore, almost by definition, you are not stupid.

Don't be too hard on yourself. The Chief Examiners' reports on the Exams Commission website indicate that this is a problem area for an awful lot of students.

Challenged

• if the first term has a 1, and the second term has a 2 and the third term has a 3, etc., then the nth term is obviously going to have an n.
• if the first term has a 2, and the second term has a 4 and the third term has a 6, etc., then the nth term is obviously going to have a 2n.
• (slightly less obvious):if the first term has a 1, and the second term has a 3 and the third term has a 5, etc., then you need to be able to see that you're doubling it and subtracting 1, so the nth term is going to have a 2n-1.

NB: the above are not "rules" to be learned; they're just illustrations of how you turn a pattern into a statement of a general term.

Please stop. I'm getting more offended by the minute. The formula for the general term of an arithmetic sequence is: Tn = a + (n - 1)d where a is the first term and d is the common difference.
Pattern 1: 1, 2, 3, ......
a = 1, d = 1
Tn = 1 + (n - 1)1 = n

Pattern 2: 2, 4, 6, .....
a = 2, d = 2
Tn = 2 + (n - 1)2 = 2n

Pattern 3: 1, 3, 5,.....
a = 1, d = 2
Tn = 1 + (n - 1)2 = 2n - 1

MathsManiac

Indeed. But the patterns that appear in MacLaurin series are not always arithmetic, so the skill that is required is the ability to see a pattern and write an expression in n that represents it.

(Not sure what you find so offensive about it!)

Challenged

All the functions on the LC course give rise to series that are arithmetic.

Des23

I know this is slightly off topic, but does anyone know where I could find material for the further probability option? It's not in the book I use and I'd like to have a go off doing it on my own.