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Year 13 Maths Help Thread

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Original post by Ayman!
Better not post it here - the TSR police will arrest me. :mob:

Check your PM soon, I'm looking for the link :lol:


Aye aye. Local bookshop only stocks stuff that are usable for just maths, not so much for FM ;-;
Original post by Imperion
Aye aye. Local bookshop only stocks stuff that are usable for just maths, not so much for FM ;-;


There used to be a school website who had them all, the website is no longer up, I've got the livetext for each book but none that are just the PDF themselves


edit: if anyone has the pdf msg me it :P
(edited 7 years ago)
Original post by metrize
There used to be a school website who had them all, the website is no longer up, I've got the livetext for each book but none that are just the PDF themselves


edit: if anyone has the pdf msg me it :P


Tbf I wouldn't mind livetext either

Spoiler

Reply 63
I'll have to insist that the above discussion about infringing copyright will have to stop. :smile:
:security:
Reply 65
Original post by SeanFM
:security:


:elefant:
All in the best interest of education.
Reply 67
Prove that 12(1+i3)4+12(1i3)4122(1+i)3+122(1i)3 \displaystyle \sqrt[4]{\frac{1}{2}(1+ i\sqrt 3 )} + \sqrt[4]{\frac{1}{2}(1-i \sqrt 3)}\equiv \sqrt[3]{\frac{1}{2}\sqrt 2 (1+i)} + \sqrt[3]{\frac{1}{2}\sqrt 2 (1-i)}.

Is this really easy for a FM question?
(edited 7 years ago)
Original post by Ano123
Prove that 12(1+i3)4122(1+i)3 \displaystyle \sqrt[4]{\frac{1}{2}(1+ i\sqrt 3 )} \equiv \sqrt[3]{\frac{1}{2}\sqrt 2 (1+i)}.

Is this really easy for a FM question?


Yes.
Reply 69
Original post by Ano123
Is this really easy for a FM question?


Doesn't make sense, there isn't really a thing as a principal root for nn-th roots where n>2n > 2 in the complexes, so for example, the LHS has 4 fourth roots whilst the RHS has 3 third roots and you're claiming the two are equivalent.
Reply 70
Original post by Zacken
Doesn't make sense, there isn't really a thing as a principal root for nn-th roots where n>2n > 2 in the complexes, so for example, the LHS has 4 fourth roots whilst the RHS has 3 third roots and you're claiming the two are equivalent.


I've tried to edit it because it's not fully what I mean to put. I will try and edit it in a minute again.
Reply 71
Original post by RDKGames
Yes.


I've edited it now.
Original post by Ano123
I've edited it now.


That's still straight forward. Maybe it's just me, never had a problem with roots of unity in FM. The only thing to do here is to convert it into reiθre^{i\theta} form, factor out indices and maybe draw yourself a complex plane so there isn't much challenge.
(edited 7 years ago)
Reply 73
Original post by RDKGames
That's still straight forward. Maybe it's just me, never had a problem with roots of unity in FM. The only thing to do here is to convert it into reiθre^{i\theta} form so there isn't much challenge.


I just play around with a few concepts and see if I can make any interesting questions from it. Obviously not here.
Original post by Ano123
I just play around with a few concepts and see if I can make any interesting questions from it. Obviously not here.


Try linking these types to De Moivre's perhaps? Those can sometimes be challenging and can lead onto roots of polynomials.
Reply 75
Original post by RDKGames
Try linking these types to De Moivre's perhaps? Those can sometimes be challenging and can lead onto roots of polynomials.


Yeah. I like those questions. Let me see what I can come up with.
Reply 76
Bleurgh, no. All these are boring and tedious questions. For a given complex number aa, call aa a primitive nnth root of unity if and only if an=1a^n = 1 and there is no integer mm such that am=1a^m = 1 where 0<m<n.0< m < n.

Let Cn(x)C_n(x) be the (cyclotomic) polynomial such that the roots of the equations of CnC_n are the primitive nth roots of unity, the coefficient of the highest power of xx is 1 and all roots have multiplicity 1.

Find Cp(x)C_p(x) where pp is a given prime and prove that there are no positive integers q,rq,r and ss such that Cq(x)Cr(x)Cs(x)C_q(x) \equiv C_r(x)C_s(x).
Original post by Zacken
Bleurgh, no. All these are boring and tedious questions. For a given complex number aa, call aa a primitive nnth root of unity if and only if an=1a^n = 1 and there is no integer mm such that am=1a^m = 1 where 0<m<n.0< m < n.

Let Cn(x)C_n(x) be the (cyclotomic) polynomial such that the roots of the equations of CnC_n are the primitive nth roots of unity, the coefficient of the highest power of xx is 1 and all roots have multiplicity 1.

Find Cp(x)C_p(x) where pp is a given prime and prove that there are no positive integers q,rq,r and ss such that Cq(x)Cr(x)Cs(x)C_q(x) \equiv C_r(x)C_s(x).


No thanks. I don't think many Y13's will know what to do here lol
Reply 78
What would a graph of xxx^x look like when x0x \leq 0?

Secondly, what is limx>xxlim_{x->-\infty} x^x (apologies for the TeX)?
(edited 7 years ago)
Original post by Palette
What would a graph of xxx^x look like when x0x \leq 0?

Secondly, what is limx>xxlim_{x->-\infty} x^x (apologies for the TeX)?


1. You cant draw it in the real plane.

2. Limit does not exist by continuity

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