further calculus option

sully88

Can anyone tell me the last time a binomaial mclauren series was asked, because i cant find one in the papers? Also i cant find a example of when you have to use integration by parts twice in the past papers.

cossyx

Im not sure about Maclaurin Series but integration by parts hasn't been up for a long time! It's a definate for tomorrow Id say!:o

carlowboy

Ah integration by parts is a lovely question. There was a tricky one in my mock paper though. It was integrate lnx dx

you have to have u= lnx and dv=1dx

ninebeanrows

tomorrows paper is much more time consuming than paper 1 i think, so i might have to do a Prob question....... mmmmmm

Its going to be interesting!!

carlowboy

Does anyone know how to work out the general term from any MacLaurin series?
Its wrecking my head.

Weathercheck, 6 is a nice question. Unless you're really good with standard deviation, based on previous years, 7 has been horrible imo.

Nehpets

General term hurts my brain too

carlowboy

I have a sheet from the teacher for learning of by heart of anything that can come up in general terms but its impossible to learn off by heart!

ninebeanrows

yep it wrecks my head tooo

You decide whether its arithmetic or geometric

then use Un = a +(n-1)d

or Un = ar^n-1 (?)

Work out the top if its like 1,3,5,7
then sub in......

then same for bottom and throw on the ! if you have too..

i'm not 100% with them but alright..

then if its + - + - multiply by (-1)^n+1

carlowboy

Ahhh.....maybe I had a block on it having not done sequences and series at the time. Thanks.

Your Man

1996 and 1997 binomial maclauren was asked

Nehpets

carlowboy wrote:

Ahhh.....maybe I had a block on it having not done sequences and series at the time. Thanks.

I haven't done s&s lol

JC 2K3

How do people find general term hard?

You just look everywhere a term is changing in the series, be it a power, divisor or multiplier.

If it's 1,2,3,4.... then write n
If it's 1,3,5,7.... then write 2n-1
If it's 2,4,6,8.... then write 2n

If it alternates between 1 and -1 then multply it by either (-1)^n or (-1)^n+1 depending on if the first term is positive or negative.

easy, no?

analyse this

For the most part, it is relatively easy, but, there are some questions that I find myself just staring at for hours and just can't see what to do!:o

melsman

we were actually never taught any of those general terms, im just gonna have to pray that they dont come up!
what do ye think of inverse tan to get a valuation for pi coming up?

colm-ccfc84

Hopefully the pie valuation comes up, along with two step integration by parts and max/min. I am not even looking at the general term, don't have a clue about it. Not feeling confident about tomorrow at all.

JC 2K3

I've just realised that by using tan^-1(x) + tan^-1(y) = tan^-1((x+y)/1-xy)) you get a very handy way of solving 6(c)(iii) from paper 1.....

3 days too late, damn....

MathsManiac

JC 2K3 wrote:

I've just realised that by using tan^-1(x) + tan^-1(y) = tan^-1((x+y)/1-xy)) you get a very handy way of solving 6(c)(iii) from paper 1.....

3 days too late, damn....

...except for the inconvenient fact that that formula doesn't hold when xy=1.

JC 2K3

The answer was pi/2

MathsManiac

JC 2K3 wrote:

The answer was pi/2

Indeed it was, but it takes some fancy footwork to conclude that logically from the formula you quoted,
(for example, you would need to exclude the possibility that it was -Pi/2.)

Much easier to just sub in any positive number you like (preferably an easy one).

JC 2K3

How do you exclude the possibility of other values by subbing any number in?

For example, sub in 2 in place of x and you get:

arctan(1) + arctan(1) = pi/4 + pi/4 = pi/2
OR
3pi/4 + 3pi/4 = 3pi/2 (=-pi/2?)

...

Thomas_S_Hunterson

JC 2K3 wrote:

I've just realised that by using tan^-1(x) + tan^-1(y) = tan^-1((x+y)/1-xy)) you get a very handy way of solving 6(c)(iii) from paper 1.....

3 days too late, damn....

That's what I did, I only hope i get the marks.

The problem is you're left with tan^-1(undefined), which i took to be pi/2 because tan(pi/2) is undefined. However I don't think it's mathematically sound, but maybe I'll get the marks.

JC 2K3

I don't see how it isn't mathematically sound.

Surely arctan(x/0) = pi/2 when x is not equal to zero....

I mean there are questions you can get on the LC where you get the slope of a line = (x/0) and it's perfectly acceptable to assume that the angle with the horizontal is pi/2 in that case...

MathsManiac

JC 2K3 wrote:

How do you exclude the possibility of other values by subbing any number in?

For example, sub in 2 in place of x and you get:

arctan(1) + arctan(1) = pi/4 + pi/4 = pi/2
OR
3pi/4 + 3pi/4 = 3pi/2 (=-pi/2?)

...

Because arctan(1) is not 3pi/4. It is uniquely defined as pi/4.

MathsManiac

JC 2K3 wrote:

I don't see how it isn't mathematically sound.

Surely arctan(x/0) = pi/2 when x is not equal to zero....

I mean there are questions you can get on the LC where you get the slope of a line = (x/0) and it's perfectly acceptable to assume that the angle with the horizontal is pi/2 in that case...

arctan(x/0) is not pi/2. It's undefined. x/0 is undefined, so it can't have an arctan. arctan is a function on the real numbers.

(By the way, I'm not saying necessarily that you won't get the marks. Complete precision is often not demanded at LC.)

MathsManiac

JC 2K3 wrote:

How do you exclude the possibility of other values by subbing any number in?

For example, sub in 2 in place of x and you get:

arctan(1) + arctan(1) = pi/4 + pi/4 = pi/2
OR
3pi/4 + 3pi/4 = 3pi/2 (=-pi/2?)

...

...and by the way, I also forgot to mention that 3pi/2 isn't equal to -pi/2. (If it were, one could easily prove that all numbers are equal to each other.)

Fobia

cossyx wrote:

Im not sure about Maclaurin Series but integration by parts hasn't been up for a long time! It's a definate for tomorrow Id say!:o

It was asked in 2005..

Nehpets

JC 2K3 wrote:

How do people find general term hard?

You just look everywhere a term is changing in the series, be it a power, divisor or multiplier.

If it's 1,2,3,4.... then write n
If it's 1,3,5,7.... then write 2n-1
If it's 2,4,6,8.... then write 2n

If it alternates between 1 and -1 then multply it by either (-1)^n or (-1)^n+1 depending on if the first term is positive or negative.

easy, no?

Not really!

1 - (1/2)(1/3) + (1/3)(1/3)^2 - (1/4)(1/3)^3 + (1/5)(1/3)^4 - (1/6)(1/3)^5 +....

Un= ((-1)^n+1 / n)(1/3)^n

Why is the 1/3 to the n? because it goes up once each time? Is that the same for why the first part is over n? (in bold) because it goes 1/2 1/3 1/4 1/5 1/6? and the -1^n+1 is because it varies from + to -? But why is it not just to the n since the powers go up one?

E92

when dealing with questions like that only 'principal values' are used i.e.

-pi/2< x < pi/2 in rads, which is the only acceptable unit to be used here or -90<x<90 in degrees, but degrees aren't a proper unit for Calculus or Complex numbers.

Values outside this are not allowed.
Otherwise we'd never know what to do eg sin(5pi/6) is the same as sin(pi/6) (1/2).

rjt

Nehpets wrote:

Not really!

1 - (1/2)(1/3) + (1/3)(1/3)^2 - (1/4)(1/3)^3 + (1/5)(1/3)^4 - (1/6)(1/3)^5 +....

Un= ((-1)^n+1 / n)(1/3)^n

Why is the 1/3 to the n? because it goes up once each time? Is that the same for why the first part is over n? (in bold) because it goes 1/2 1/3 1/4 1/5 1/6? and the -1^n+1 is because it varies from + to -? But why is it not just to the n since the powers go up one?

Should it be:
Un= ((-1)^n+1 / n)(1/3)^(n-1)? (because otherwise we'd have (1)(1/3) - (1/2)(1/3)^2 +...

The thing is, you not only have to work out what the pattern is, in this case +,-,+,-,... is (-1)^n, but also where it starts. If you were to have:
Un= ((-1)^n / n)(1/3)^(n-1)
Then the sequence would be: -1 + (1/2)(1/3) -..., with n=1,2,3,...
We have to make sure that it starts with an even power so that we end up with a +1, so we use n+1 (as n+1 is even when n=1). We could use n-1 instead, and the result would be the exact same.

JC 2K3

Nehpets wrote:

Not really!

1 - (1/2)(1/3) + (1/3)(1/3)^2 - (1/4)(1/3)^3 + (1/5)(1/3)^4 - (1/6)(1/3)^5 +....

Un= ((-1)^n+1 / n)(1/3)^n

Why is the 1/3 to the n? because it goes up once each time? Is that the same for why the first part is over n? (in bold) because it goes 1/2 1/3 1/4 1/5 1/6? and the -1^n+1 is because it varies from + to -? But why is it not just to the n since the powers go up one?

Because (-1)^n = -1 and the first term is +1.

And rjt is right, it should be (1/3)^n-1.

What you should do if you find it difficult is examine every number/variable in every term like:

(1/1), (1/2), (1/3), (1/4).....
(1/3)^0, (1/3)^1, (1/3)^2, (1/3)^3.....
1, -1, 1, -1.....

Those are the three things that are changing in every term in that series.

So, the first one is just 1 over the corresponding number of the term, ie. n, so it's (1/n)

The second one is (1/3) to the power of one less than n, so it'd be (1/3)^n-1

The third one is just alternating 1 and -1. The first term is 1, so that's (-1)^2, and since n is 1 there, it's (-1)^n+1.

Multiply em all together and you get (1/n)((1/3)^n-1)(-1)^n+1.