The Student Room Group

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Reply 1
equiv. to

sin(y)sin(2x+y) = sin(x)sin(x+2y)

-1/2 (cos(2x+2y) + cos(2x)) = -1/2 (cos(2x+2y) + cos (2y))

cos(2x) = cos(2y)

solutions 2x = 2y or 2x + 2y = 2pi, i.e.

x = y or x + y = pi

surely?
Reply 2
Addition Formulae perhaps?
Reply 3
Right ho, thanks! I should have spotted that.

As I said, more stupid questions are sure to follow. :smile:
Does the symmetry not imply x = y 2pi*k)?
Reply 5
James_Gurung
0 < x,y < Pi

no
Well, x=y then?
Reply 7
I assumed you were just pointing out the multiples of 2pi, as x=y was noted by ukgea in the second post :tongue:
Speleo
I assumed you were just pointing out the multiples of 2pi, as x=y was noted by ukgea in the second post :tongue:

Well, no, I was pointing out that we should be able to infer this (x=y) from the symmetry, non? :p:
Reply 9
I really don't understand, that doesn't give you all the solutions and doesn't help in obtaining the rest...
Reply 10
Okay, here's another. I'm actually still working on it, but my method is going to involve taking third order terms from the Taylor expansion, so I hope there's an easier trick!

Classify the critical point at (0,0) for f(x,y)=x4+2x3y+x2y2+y4f(x,y)=x^4+2x^3y+x^2y^2+y^4

EDIT: Nope, at this rate it's going to be fourth-order terms! :eek:

EDIT 2: Okay, this is what I have so far:

f(h,k)=24h4+48h3k+24h2k2+24k4f(h,k)=24h^4+48h^3k+24h^2k^2+24k^4

That was by going back to the Taylor Expansion until nonzero derivatives appeared, but it's no nicer to work with than the original expression!
Reply 11
Look at how it behaves along the lines y=mx in R2\mathbb{R}^2.

f(x,mx)=(1+2m+m2+m4)x4=(m4+(m+1)2)x4f(x,mx) = (1+2m+m^2 + m^4)x^4 = (m^4 + (m+1)^2)x^4

So it's a minimum.
Reply 12
SsEe
Look at how it behaves along the lines y=mx in R2\mathbb{R}^2.

f(x,mx)=(1+2m+m2+m4)x4=(m4+(m+1)2)x4f(x,mx) = (1+2m+m^2 + m^4)x^4 = (m^4 + (m+1)^2)x^4

So it's a minimum.

Nice!! I used a similar argument with the line y=x in a previous question, but that was to show a saddle point. I never thought to consider a general line.

Thanks. :smile:
Reply 13
To show it's a saddle you only need to find one line with a max and another with a min but for a max/min you need to look at it in general and show all lines have a max/min.
Reply 14
Okay thanks all for your help so far; here's another I'm struggling with

http://www.maths.ox.ac.uk/current-students/undergraduates/lecture-material/Mods/sets-and-groups/pdf/sets-groups4.pdf
Question 6.

I've tried to play around with the 'hint' a bit, but not quite sure where to go with it.

I'm also having some trouble with stats, but im going to keep working on that for a bit.
Reply 15
James Gurung
Okay thanks all for your help so far; here's another I'm struggling with

http://www.maths.ox.ac.uk/current-students/undergraduates/lecture-material/Mods/sets-and-groups/pdf/sets-groups4.pdf
Question 6.

I've tried to play around with the 'hint' a bit, but not quite sure where to go with it.

I'm also having some trouble with stats, but im going to keep working on that for a bit.


Think about the number of elements in the two sets given in the hint. A little caveat, at least I think I can solve the problem going in that direction, but I'm not completely sure about my solution.
James Gurung
Okay thanks all for your help so far; here's another I'm struggling with

http://www.maths.ox.ac.uk/current-students/undergraduates/lecture-material/Mods/sets-and-groups/pdf/sets-groups4.pdf
Question 6.

I've tried to play around with the 'hint' a bit, but not quite sure where to go with it.

I'm also having some trouble with stats, but im going to keep working on that for a bit.
The key point here is that you have cancellation laws, so the functions fa(b)=(ab),ga(b)=(ba)f_a(b) = (a*b), g_a(b) = (b*a) are 1-1. So what are the sizes of fa(G),ga(G)f_a(G), g_a(G)? What does this tell us about f and g?
Reply 17
Okay, so f_a(G)=f_b(G)=G.
But I guess we need to show that there is an identity element in G, and each element has an inverse. I really don't see how any of this helps! :redface:
James Gurung
Okay, so f_a(G)=f_b(G)=G.
But I guess we need to show that there is an identity element in G, and each element has an inverse. I really don't see how any of this helps! :redface:
Well, what do we know about the identity element, e? (I realise we don't know e exists. Bear with me!).

Well, we know a*e = a. But afa(G)a\in f_a(G), so eG\exists e \in G with ae=aa * e = a. Of course, we need to show the 'e' we get when we look at a is the same as the 'e' we get when we look at b, etc, but from here it's not too hard to start bootstrapping our way up to the full set of group axioms.
Reply 19
Right, I'm not sure if what I'm going is okay. Please don't pick me up on tiny details, but do let me know if it's completely wrong!!!

G      aGG\neq\emptyset\;\therefore\;\exists\;a\in G

By the cancellation laws, fa(b)=abf_a(b) = a*b and ga(b)=bag_a(b) = b*a are injective

So {ab    bG}={ba    bG}=G\{a*b \;|\; b \in G\} = \{b*a\; |\; b \in G\} = G

Thus e1G\exists e_1 \in G such that ae1=aa*e_1=a.

And e2G\exists e_2 \in G such that e2a=ae_2*a=a.

Now a=ae1=a(e2e1)=(ae2)e1=ae2a=a*e_1=a*(e_2*e_1)=(a*e_2)*e_1=a*e_2 and so e1=e2=ee_1=e_2=e

{ba    bG}=G\{b*a\; |\; b \in G\} = G, so given any aGa\in G we can find bGb\in G such that ba=eb*a=e

[Are these last two lines okay??]

Now I assume that if the set is infinite, we can't form a bijection so the argument is invalid. But that may be wrong and I certainly can't corroborate it.