!! HL Maths 2015 - predictions, guesses, Q & A, discussion ...

Broseph

Magnate wrote: »

You're thinking of binomial distribution, that's slightly different.

Paper 2?

Magnate

Broseph wrote: »

Paper 2?

Yep

oktplz

Anyone have any pretty complex maths equations? Wouldn't mind scaring myself before tomorrow

is mise spartacus

can someone simplify the proof of de moivres theorem? even just a few basic steps for attempt marks....

skippy1977

oktplz wrote: »

Anyone have any pretty complex maths equations? Wouldn't mind scaring myself before tomorrow

How about this....the kind of question that would throw people on the day.

Dianabluex

hope maths paper one is easy like the previous years

Peregrine

is mise spartacus wrote: »

can someone simplify the proof of de moivres theorem? even just a few basic steps for attempt marks....

By induction?

It's just like any other proofs by induction. Have a look at this. It's only 5 or 6 lines.

DarraghF197

oktplz wrote: »

Anyone have any pretty complex maths equations? Wouldn't mind scaring myself before tomorrow

Haha do you want complex complex equations or just complex equations, or do you want complex equations?

One that haunted me for Geometric series was one from the book. I tried it there and it finally worked out!

The sum to infinity of a geometric progression is 16 and the sum of the first four terms ie 15. Find the first four terms. Note: there are two possible series.

(Pg 257 ex 9.6)

Another one that proved a challenge was a proof by induction:

Prove that, I don't know how I'll write this. It's page 293 11.2 question 12.

The sum of (1/R*2-1) = 3/4 - [2n + 1]/2n(N+1)]

N is greater than 2, qnd element of natural number.

Enjoy!

Kremin

skippy1977 wrote: »

How about this....the kind of question that would throw people on the day.

That question honestly isn't even that hard, it just looks ridiculously hard.

DarraghF197

skippy1977 wrote: »

How about this....the kind of question that would throw people on the day.

Does it matter if I don't take heed of the w squared and find the roots nonetheless? I just drew a circle and found roots and they added to cancel.

Then for part two, if the first part is zero, then would it to the power of five also be zero? Or am I missing something in this question?

lc2015x

Is it true that the mocks are harder than the real? We had the Examcraft paper, but we didn't do the financial maths question because he hadn't covered it in class yet & I passed by one percent, I wasn't expecting a D1...Was aiming for a C3 at least Hoping a long question on financial maths comes up tomorrow to up my grade now!

Kremin

DarraghF197 wrote: »

Does it matter if I don't take heed of the w squared and find the roots nonetheless? I just drew a circle and found roots and they added to cancel.

Then for part two, if the first part is zero, then would it to the power of five also be zero? Or am I missing something in this question?

This question was in my book and it was like prove the other root is w^2, whats interesting is that w and w squared are actually just the conjugate of eachother which is a rule for complex numbers ofc.

skippy1977

DarraghF197 wrote: »

Does it matter if I don't take heed of the w squared and find the roots nonetheless? I just drew a circle and found roots and they added to cancel.

Then for part two, if the first part is zero, then would it to the power of five also be zero? Or am I missing something in this question?

2nd part can be done without any knowledge of complex numbers.
If 1+w+w^2=0
then (w+w^2)=-1
so (1-w-w^2)^5=(1-(w+w^2))^5
=(1-(-1))^5
=(2)^5
=32

As Kremin said, it looks harder that it is...but remember that will be the case with a lot of the questions tomorrow. People know the stuff (or a lot of it) but at Higher Level it gets asked in a more complicated manner.

DarraghF197

Kremin wrote: »

This question was in my book and it was like prove the other root is w^2, whats interesting is that w and w squared are actually just the conjugate of eachother which is a rule for complex numbers ofc.

That is quite interesting, something that I should be taught about the day before maths paper one!

skippy1977 wrote: »

2nd part can be done without any knowledge of complex numbers.
If 1+w+w^2=0
then (w+w^2)=-1
so (1-w-w^2)^5=(1-(w+w^2))^5
=(1-(-1))^5
=(2)^5
=32

As Kremin said, it looks harder that it is...but remember that will be the case with a lot of the questions tomorrow. People know the stuff (or a lot of it) but at Higher Level it gets asked in a more complicated manner.

Lol my mind did a miny blow up as two different approaches lead to different results! That makes sense, thanks! Just a little confusing overall that something equal to zero squared can be re-arranged not to. I suppose, never be too casual about approaching complex numbers.

The_Mac

For the first part of that do you say they must equal to zero as they are all confirmed roots of the z^3 -1 sum yeah?

pa limerick

Can somebody go through the formula that need to be learnt off again as there not in the log book just want to make sure!?

Dianabluex

pa limerick wrote: »

Can somebody go through the formula that need to be learnt off again as there not in the log book just want to make sure!?

The only one I can t hink of is the f(x)_lim h>x (x+h)^2 +(x+h)/h

YouKnowNothing

Any body have any idea of this one?
" use de moivres theorem to solve for three roots of unity, 1, w, w^2
Hence throw show that the sum of these three roots is zero
Z= -2 + root5 i "
It's edco sample E question 2b

skippy1977

YouKnowNothing wrote: »

Any body have any idea of this one?
" use de moivres theorem to solve for three roots of unity, 1, w, w^2
Hence throw show that the sum of these three roots is zero
Z= -2 + root5 i "
It's edco sample E question 2b

The z you have written is from (a) part and not part of the (b) question.

YouKnowNothing

skippy1977 wrote: »

The z you have written is from (a) part and not part of the (b) question.

ahh I was wondering...
Where does z^3 -1 come from?

skippy1977

YouKnowNothing wrote: »

skippy1977 wrote: »

The z you have written is from (a) part and not part of the (b) question.

ahh I was wondering...
Where does z^3 -1 come from?

So this is not something that can be made up most likely on the day. The syllabus has it that students need to be able to find the roots of unity.
What are the complex numbers that when cubed give 1 (unity).
So we write that equation z^3=1 and try and find the complex number z that when cubed gives one.

So if they ask about roots of unity in the exam...start with z^3=1
1 is the same as saying 1+0i....so put this on an Argand Diagram and express in polar form. Then apply de moivre's to get the cubed roots as above.

Fiona G

pa limerick wrote:

Can somebody go through the formula that need to be learnt off again as there not in the log book just want to make sure!?

Average value of a function. I think that's it?

YouKnowNothing

Thanks! I'd just never heard if unity being one?!

skippy1977

YouKnowNothing wrote: »

Thanks! I'd just never heard if unity being one?!

No problem. Best of luck today anyway.

lc2015x

any statistics or any other definitions that could be asked for p1?
Or should i leave them for the weekend for p2?

pa limerick

Good luck everyone! Here's hoping I pass, the one exam I'm really worried about!

andyman

Howdy guys.

I'll be doing all papers this evening and Monday evening HL and OL. Now, I wouldn't recommend talking about the exam after its finished (especially looking for answers) but if you are desperate to know the answers to any of the questions then pop them up here.

AlfaJack

All the theorem proofs are on paper two arn't they? I'm so worried something from paper two will come up that I haven't revised :P

And are trigonometric proofs on paper two also?

BlueWolf16

AlfaJack wrote: »

And are trigonometric proofs on paper two also?

Yes on both. P2 only, though they may try to confuse you and include some little part. But mostly, no.

Magnate

Ugh I feel sick, really dreading this one