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The 2012 STEP Results Discussion Thread

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Original post by GreenLantern1
Eigentlich, wisst du, die ich jetzt sage und was ich hatte schon gesagt? Du bist glueck, da ich nicht was du kannst nicht verstanden gesagt waere.


Yeah Google translate didn't do a great job for me there, so I still don't really know what you said (even though I think what you said suggested that I will know). But it also suggests I know who you are so who knows.
I'm with Hassi and Ben. This isn't a grime thread so take the grime talk elsewhere please.
Original post by f1mad
Come at me bros :colone:.

:rofl:


RIP Fmad :colone:
Haha, did worse in this STEP II paper than I did in the same year's STEP III... 2004 so the grade boundaries were still OK :smile:

...but this marks my return to STEP after my glorious, glorious week off :frown:
Original post by fruktas
Has anyone got a hold of a decent STEP II 2002 Question 5 solution. The one submitted is only partial, and I cannot seem to find it anywhere else :/


I can give you mine...give me a minute to find it

Found it...which parts do you need help with?
Reply 1765
Avatar for gff
gff
5
Original post by Farhan.Hanif93
I'm with Hassi and Ben. This isn't a grime thread so take the grime talk elsewhere please.


Agreed. I don't think this is funny any more -- if you have some arguments going, address them on private.

Original post by TheMagicMan
I too have a rather nice STEP-like question :tongue:

Determine all functions f:Rf:R to RR which have the following properties:

i)f(x)f(x) is continuous and differentiable everywhere except at x=0x=0

ii)f(−1)=1f(-1)=1

iii)(f(x))2+(f(y))2=f(x2+y2)(f(x))^2+(f(y))^2=f(x^2+y^2)


Let's have a go with it.

Spoiler



Did it with quick thoughts, so if you have any comments on improvements, don't keep them for yourself. :tongue:
(edited 12 years ago)
Original post by gff
Agreed. I don't think this is funny any more -- if you have some arguments going, address them on private.



Let's have a go with it.

Spoiler



Did it with quick thoughts, so if you have any comments on improvements, don't keep them for yourself. :tongue:


I'm quite tired right now so I'll have a look at it tomorrow morning but it looks pretty good to me...very similar to my own approach...perhaps you might like to consider some of the rather exciting things that come out of it if you remove condition i)?
Original post by hassi94
Yeah Google translate didn't do a great job for me there, so I still don't really know what you said (even though I think what you said suggested that I will know). But it also suggests I know who you are so who knows.


It is german and:

Eigentlich, wisst du, die ich jetzt sage und was ich hatte schon gesagt? Du bist glueck, da ich nicht was du kannst nicht verstanden gesagt waere.

translates as:

Actually do you know what I am saying now and what I have already said. You are lucky, as I wouldn't have said anything which you can't understand.

Oh and as the great Hilbert said:

Wir muessen wissen. Wir werden wissen :smile: That sexy beast
Reply 1768
Avatar for gff
gff
5
Original post by TheMagicMan
..perhaps you might like to consider some of the rather exciting things that come out of it if you remove condition i)?


Might do, but have to finish quite a few assignments before Monday -- so, maybe tomorrow afternoon. :tongue:
Original post by TheMagicMan
I too have a rather nice STEP-like question :tongue:

Determine all functions f:Rf:R to RR which have the following properties:

i)f(x)f(x) is continuous and differentiable everywhere except at x=0x=0

ii)f(−1)=1f(-1)=1

iii)(f(x))2+(f(y))2=f(x2+y2)(f(x))^2+(f(y))^2=f(x^2+y^2)


Spoiler



Realised i could cut the soln to half its size, looking at Gff's soln- however part 4 and even part 1 may be worth looking at as an alternative to the mean value theorem.
(edited 12 years ago)
Reply 1770
Avatar for fruktas
fruktas
2
Original post by TheMagicMan
I can give you mine...give me a minute to find it

Found it...which parts do you need help with?


Thanks, but I thought I deleted that post haha! I only wanted to check my answer, and I found the the very last bit posted on the 2002 solution thread.

Thansk anyway :smile:
Hey guys just doing this question I want to see if what I'm doing is okay.

Say it asks you to show that xy≤0.

I have reduced the equation to x=-3y

So then I said therefore either x or y (but not both) must be negative to give positive = positive or negative = negative. Or they must both be equal to zero.

Is this sufficient?
Original post by hassi94
Hey guys just doing this question I want to see if what I'm doing is okay.

Say it asks you to show that xy≤0.

I have reduced the equation to x=-3y

So then I said therefore either x or y (but not both) must be negative to give positive = positive or negative = negative. Or they must both be equal to zero.

Is this sufficient?

What you're saying is true (that x and y must therefore have different signs unless their both zero). Another way you could have easily gotten there is to note that x=−3y  ⟹  xy=−3y2≤0x=-3y \implies xy=-3y^2\leq 0.
(edited 12 years ago)
Original post by Farhan.Hanif93
What you're saying is true (that x and y must therefore have different signs unless their both zero). Another way you could have easily gotten there is to note that x=−3y  ⟹  xy=−3y2≤0x=-3y \implies xy=-3y^2\leq 0.


Ah I don't know how I missed that haha. I even did what you said but failed to notice the rhs was negative :tongue: Thanks a lot.
For STEP II 1997 Q4 (ii), I did it with basically no lines of working; I just wrote the polynomial P(x) down after a few minutes of thinking about it, with a short explanation of why it works. For such a question, will this be okay in the exam?
Original post by safmaster
For STEP II 1997 Q4 (ii), I did it with basically no lines of working; I just wrote the polynomial P(x) down after a few minutes of thinking about it, with a short explanation of why it works. For such a question, will this be okay in the exam?


I'm pretty certain that's fine. As long as you do a short explanation, saying that a is a factor of the first term (the brackets) if it's between 0 and n and then so you get a remainder of a :smile:
Reply 1776
Avatar for gff
gff
5
Original post by Blutooth

Realised i could cut the soln to half its size, looking at Gff's soln- however part 4 and even part 1 may be worth looking at as an alternative to the mean value theorem.


I like alternatives :tongue: -- the MVT is a bit of a sledgehammer there :colondollar:, but didn't think of your idea at the time.
Also, I have made changes on mine explaining where things come from, and more notably fixed the induction step.

Edit: PRSOM, btw.

Original post by TheMagicMan
...perhaps you might like to consider some of the rather exciting things that come out of it if you remove condition i)?


I can cook up few functions using the modulus, or very similar constructions to it, but I guess you have something quite exotic in mind. Hints?
(edited 12 years ago)
Original post by gff
I can cook up few functions using the modulus, or very similar constructions to it, but I guess you have something quite exotic in mind. Hints?


I'm not sure what hints I can give you, considering i can barely remember myself what you're looking for...I seem to remember that you could somehow use complex numbers (unusual, considering that the function has a real domain and range) to find some exotic functions that satisfy it, although i would need to find my own solutions to confirm this
Original post by Blutooth

Spoiler



Realised i could cut the soln to half its size, looking at Gff's soln- however part 4 and even part 1 may be worth looking at as an alternative to the mean value theorem.


This is a very nice elementary (in method, not in insight) solution. PRSOM
Original post by gff
....


I have to put away the solutions to the problem below, so before I forget I'm going to put up the solution spoilered because it is really nice

Spoiler



SOLUTION:

[spoiler]First setting x=0,1x=0,1 in ii) gives f(1)=f(−1)=f(0)f(1)=f(-1)=f(0) (*)

Let h(θ)=f(cos⁡θ)sin⁡θ.So[tex]h(θ+π)=h(θ)[/tex]and[tex]h(−θ)=−h(θ)[/tex](∗∗)[br][br][tex]h(2θ)=f(2cos⁡2θ−1)sin⁡2θ[br][br]=2cos⁡θf(cos⁡θ)2sin⁡θcos⁡θ[br][br]=h(θ)[/tex][br][br]Soforany[tex]a,b∈Z[/tex],wehave[tex]h(1+aπ2b=h(1)[/tex][br][br]But[tex]{1+aπ2b∣a,b∈Z}[/tex]isadensesubsetoftherealssohisconstantonit′sdomainasitiscontinuous.By(∗∗)[tex]h( theta)=0[/tex]so[tex]f(x)=0[/tex][br][br]By(∗)[tex]f(−1)=f(1)=0[/tex][br][br]Hencetheonlyfunctionis[tex]f(x)=0[/tex].[br][br]h( \theta)= \frac{f( \cos \theta)}{\sin \theta}. So [tex]h( \theta + \pi)=h( \theta)[/tex] and [tex]h(- \theta)=-h( \theta)[/tex] (**)[br][br][tex]h(2 \theta)= \frac{f(2 \cos^2 \theta -1)}{ \sin 2 \theta}[br][br]= \frac {2 \cos \theta f( \cos \theta)}{2 \sin \theta \cos \theta}[br][br]= h ( \theta)[/tex][br][br]So for any [tex]a,b \in Z[/tex], we have [tex] h(1+ \frac{a \pi}{2^b}=h(1)[/tex][br][br]But [tex] \{ 1+ \frac{a \pi}{2^b}|a,b \in Z \}[/tex] is a dense subset of the reals so h is constant on it's domain as it is continuous. By (**) [tex]h( \ theta)=0[/tex] so [tex]f(x)=0[/tex][br][br]By (*) [tex]f(-1)=f(1)=0[/tex][br][br]Hence the only function is [tex]f(x)=0[/tex].[br][br]
(edited 12 years ago)

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