Hon. Maths-How to get the general term in the mcLaurin series q's

fatal

In honours maths,how the hell do you get the general term in the further calculus/integration question (paper 2 question 8) when you have done the Mclaurin series??????
I know that there are 3 or 4 rules to it but can never figure out how to get the general term when I get the McLaurin series

whassupp2

I thought you just had to learn them off.... well thats what my grinds teacher told me to do...evedn though i know there is a way but its quite complicated

This didn't help me but check it out any way:


http://www.skoool.ie/asktheexpert/faq.asp?id=1322&subjectid=3

DonaldDuck

I always just look at the values to get it...Never knew there was rules for it

Gileadi

post me up afew sample ones and ill explain it, if i seen it i could explain it but i cba going looking for them myself

brianfiel

U DONT HAVE TO LEARN EM OFF!!!

I have an easy way to do it but i dont want to publish it for the public so if ya want it send me a private message.Trust me its a brilliant method not in any books!!!

NADA

I just do it by trial and error. I can't ecxplain it. Sorry. Learn off the basic ones tan ln(1+x) etc

shane0312

all you have to do is break it up into sections and use the normal arithmetric and geometric rules. For example if it alternates -+-+-+ then multiply top line by (-1)n+1

casanova_kid

You do have to get a general term. Usually, do it bvy trial and error, put a few n-1 and 2n-1's in and you'll get some marks.

brid_m

My friend went to one of those seminar thingys for maths. This was the way they were told to do it... hope it makes sense!

General Term for Mac. Series:

Method:
1. Write out the powers (2,4,6,8)

2. Write out the formula {U1 + (n-1)d} where d is the
difference between the powers, and U is the 1st power.

3. Fill it in.. e.g. 2 + (n-1)2 = 2n-2+2 = 2n {<-result}

4. Un = x ^(result) divided by (result)!
=> Un = (-1)^ n+1 x {x^2n divided by (2n)!}

5. (-1)6n or (-1)^n+1............ basically trial and error

NOTE: ^ means to the power of.. i didnt know how else to show it! :rolleyes:

I really hope this makes sense and helps!

:|

what the hell are yous talking about...oh god....panic...

Funkstard

Here...this isn't really asked on the paper by the way. They either ask one of the 'learn offable ones' ie. cosx, sins, log(1+x) or if it is one where you have to find the general term, it'll be an easy one

fatal

brid_m wrote:

My friend went to one of those seminar thingys for maths. This was the way they were told to do it... hope it makes sense!

General Term for Mac. Series:

Method:
1. Write out the powers (2,4,6,8)

2. Write out the formula {U1 + (n-1)d} where d is the
difference between the powers, and U is the 1st power.

3. Fill it in.. e.g. 2 + (n-1)2 = 2n-2+2 = 2n {<-result}

4. Un = x ^(result) divided by (result)!
=> Un = (-1)^ n+1 x {x^2n divided by (2n)!}

5. (-1)6n or (-1)^n+1............ basically trial and error

NOTE: ^ means to the power of.. i didnt know how else to show it! :rolleyes:

I really hope this makes sense and helps!

thanks for that...i completely understand it now;)

Attractive Nun

A bit late, considering you seem to completely understand it. But, for anyone else (and also for my own revision, I'll try do out an example):

Let's say f(x)=log(1+x), the Maclaurin series will be:

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

(I know that looks fairly confusing, but obviously the "1/2" represents the variable over 2, "1/3" over 3 etc. and the ^ means "to the power of" and the "." means multiplied by.)

The general term is simply a general expression in "n" that can be used to represent every term in the series. For the first term, if you sub in "1" for "n" in the general term, you will get the correct answer, for the second, sub in "2" etc...

To get the general term of the series, simply break up each term into separate parts and try figure out the pattern for each one as the series progresses.

First, notice that every figure has an "x" in it, the only difference being the powers. The powers progress in the sequence 1, 2, 3... so you should see that x^n will get this result for every term. If you can't see that, simply use the formula a+(n-1)d to represent the power of x
i.e. [1]+(n-1)[1] = n
Now forget about this result, and move onto the next part.

Each term is fraction, i.e. each x-value is 'over' a number. This numbers are also in the series 1, 2, 3...., so by the same method, the general expression representing them is simply n.


So far, if we put together what we have, we can say the general term is:
1/n.x^n. Subbing in 1, 2, 3... for n, this gives us the correct value for each term, except for the sign in front of it.

To correct that, simply place (-1)^n+1 in front of the general term. For the first term, this gives (-1)^2, or simply 1, for the second, (-1)^3, or -1 etc. This will give alternating signs for the series, as is required.

If the series starts with a negative, and then alternates, simply put (-1)^n in front of the general term. If it is always positive (or negative), just ignore this step. It is always one of these three options.


Combined together, the general term is (-1)^n+1.1/n.x^n.



I know that was a bit rambly and long, but hope it helps someone.