[insparato]

I wouldnt even know about matrices if i didnt do FP3.

[Swayum]

(Original post by DFranklin)
Huh? Why do you think there's an r^(3n+1) term in S_n(r)?

Well their series was:

S_n(r) = r + r^2 + r^4 + r^5 + r^7 + r^8 + r^10 + ... + r^(3n-1)

Let n = 3. Then the last term would be r^(3*3 - 1) = r^8

Consider the series U_n = 3n - 1, it gives:

u_1 = 2
u_2 = 5
u_3 = 8

If you then take r^(U_n) (so basically r^2, r^5, r^8), the terms left out from S_n(r) are r, r^4 and r^7.

Consider the series U2_n = 1, 4, 7, 10.... Its nth term would be 3n + 1 (if you include n = 0). If you then took r^(U_n) (which would give r, r^4, r^7) you'd get all the terms that the r^(U_n) series missed out.

Which is why I think S_n(r) can be expressed as r^(U_n) + r^(U2_n).

And the last term in the series is given by r^(3n-1). So if n was 3, S_n(r) would be everything up to and including r^8 which is essentially:

S_3(r) = r + r^2 + r^4 + r^5 + r^7 + r^8.

And that sum is equal to the formula I gave.

[Swayum]

(Original post by nota bene)
edit: DeathAwaitsU, S_n(r) is just a silly way of writing (like f(x) for a function of x this is a sum of r) as far as I know...

I don't think it is though. If you do the iterations from n = 1 to n = 3, using the formula r^(3n - 1), you get:

r^(3*1 - 1) = r^2
r^(3*2 - 1) = r^5
r^(3*3 - 1) = r^8

Where as their series suggests that it should give r + r^2 + r^4 + r^5 + r^7 + r^8 or maybe r + r^2 + r^4.

Although actually I think I'm totally misunderstanding the question.

[DFranklin]

(Original post by nota bene)
Okay, do you think it is better if I prove it is impossible (which I accidentally did in my working as I slipped on an argument and went two steps wrong which I quickly realised). Because I tried to see an obvious way of dismissing the possibility of 2+2 but didn't find any information under the points that were dealt with . (Well if you chose 2+2 strategy you'll end up with 112 on one die and 2 on the other and then trying to move on with this proves impossible when we add the 3 to 112 (can't add a 4 etc.) which makes it impossible to get what we want).

Having pondered this a bit, I couldn't find a nice way without basically solving the whole thing at the same time. (The big problem I had was showing you can't have two 1s - maybe I missed something).

First make the easy progress at the "edges" of the freq table:

Since we can't get a 2, we know that there can't be a 1 on both dice. But since we can get a 3, we know exactly one die has a 1. Call that dice A and the other one B. Then B must have a 2, so we know 1 \in A and 2 \in B.

Now, there are 6 ways of getting 11 and no ways of getting 12. So there isn't a 6 on both dice. So we either have 2 5's on 1 die and 3 6's on the other, or 2 6's on one die and 3 5's on the other. (Where the die with 6's can also have 5's).

Now it gets harder. Since we have 2 ways of getting 3, either 1,1 \in A, 2 \in B or 1 \in A, 2,2 \in B.

Suppose 1,1 \in A, 2 \in B. We know there's exactly one way of getting a 4, so we must have 2 \in A (as if 3\in B we get 2 ways). So in fact we have 1,1,2 \in A, 2 \in B. Then there's exactly one way of getting a 5, so either 3 \in A or 3\in B (but not both). Now consider how we can find 4 ways of making 6. We can't combine 2 3's. We can have at most one 4 \in A (because A must contain at least 2 sixes or 2 fives). So we need some 5's \in B. But then each 5 combines with 2 1's, so each 5 contributes 2 ways. So as the number of ways is even, 4 \notin A. So then B contains two 5's, and so A must contain 3 sixes. So A = 1,1,2,6,6,6 and 2,5,5 \in B (and 1,6 \notin B). But then there are only 2 ways of making 7.

So we must have 1 \in A, 2,2 \in B. So then there's one way of getting a 4, so 3 \in B, 2 \notin A. And then 1 way of getting a 5, so 4 \in B, 3 \notin A. So 1\in A, 2,3 \notin A, 2,2,3,4 \in B. Suppose B is the die with the sixes, then B = 2,2,3,4,6,6. Then 1,5,5,5 \in A, 2,3,6 \notin A. So A = 1,4,4,5,5,5. But then we choosing 2 from B and 5 from A gives 6 ways of making 7.

So B must contain the 5's, so B = 2,2,3,4,5,5. So 1,6,6,6\in A, 2,3 \not in A. So we can now reduce to the possibilities A = {1,4,4,6,6,6}, A = {1,4,5,6,6,6} or A = {1,5,5,6,6,6}. If A = {1,4,4,6,6,6} there are only 2 ways of making 7, if A = {1,5,5,6,6,6] there are 4 ways. So the only possibility is A = {1,4,5,6,6,6}.

(I've still been a bit sloppy about explicitly saying what can't be in the sets).

[DFranklin]

(Original post by DeathAwaitsU)
Well their series was:

S_n(r) = r + r^2 + r^4 + r^5 + r^7 + r^8 + r^10 + ... + r^(3n-1)

Let n = 3. Then the last term would be r^(3*3 - 1) = r^8

Consider the series U_n = 3n - 1, it gives:

u_1 = 2
u_2 = 5
u_3 = 8

If you then take r^(U_n) (so basically r^2, r^5, r^8), the terms left out from S_n(r) are r, r^4 and r^7.

Consider the series U2_n = 1, 4, 7, 10.... Its nth term would be 3n + 1 (if you include n = 0).

If you then took r^(U_n) (which would give r, r^4, r^7) you'd get all the terms that the r^(U_n) series missed out.

No, you'd get r^4, r^7, r^10. You have to start from n=1 if the r^(3n-1) part of what you're doing is going to make sense.

[generalebriety]

STEP I question 11.

Introduce cartesian axes into the problem, such that the x-axis lies along BC and the y-axis is perpendicular and passes through the midpoint. Let BC = 2a, such that A is the point (-a, 2a), B is the point (-a, 0), C is the point (a, 0) and D is the point (a, 2a). Model the wire as three particles, each of mass m, located at (-a, a), (0, 0) and (a, a). The centre of mass has coordinates (0, y) (by symmetry). Then the total mass M = 3m, and My = ma + ma = 2ma, so y = 2/3 a.

(See diagram for the next part.)

The angle required (between directions of BC) is θ. Since γ = 180 - α - β, and θ = 180 - γ (or simply by vertically opposite angles), θ = α + β. From the diagram,
tan α = 4/3
tan β = 2/3
and so tan θ = (4/3 + 2/3) / (1 - (4/3)(2/3)) = 2 / (1 - 8/9) = 2 / (1/9) = 18.

[generalebriety]

Also Q12...

Let the angle between the vertical and the rod be θ. Take GPE=0 at the horizontal line through B.

Let the extension of the spring CD be x. Then the length of CD is (a+x).
a+x = 2a sin(θ/2) --> x = a(2 sin (θ/2) - 1).

Unparseable or potentially dangerous latex formula. Error 5: Image dimensions are out of bounds: 423x651

\displaystyle \text{EPE} = \frac{\lambda a^2 (2 \sin \frac{\theta}{2} - 1)^2}{2a} = \frac{1}{2} \lambda a(2 \sin \frac{\theta}{2} - 1)^2. \\

\text{GPE} = 2mga \cos \theta. \\ \\



\text{Potential energy of the system:} \\

V = 2mga \cos \theta + \frac{1}{2} \lambda a(4 \sin^2 \frac{\theta}{2} - 4 \sin \frac{\theta}{2} + 1) \\

= mga(2 - 4 \sin^2 \frac{\theta}{2}) + 2\lambda a \sin^2 \frac{\theta}{2} - 2\lambda a\sin \frac{\theta}{2} + \frac{1}{2}\lambda a \\

= 2(\lambda a - 2mga)\sin^2 \frac{\theta}{2} - 2 \lambda a \sin \frac{\theta}{2} + \text{constant} \\

\frac{\text{d}V}{\text{d}\theta} = 2(\lambda a - 2mga) \sin \frac{\theta}{2} \cos \frac{\theta}{2} - \lambda a\cos \frac{\theta}{2} \\

= a\cos \frac{\theta}{2} (2(\lambda - 2mg)\sin \frac{\theta}{2} - \lambda). \\

\frac{\text{d}V}{\text{d}\theta} = 0 \Rightarrow \cos \frac{\theta}{2} = 0 \text{ (n/a)}, \; \sin \frac{\theta}{2} = \frac{\lambda}{2(\lambda - 2mg)}. \\

\lambda > 4mg \Rightarrow \lambda - 2mg > 2mg \\

\Rightarrow 0 < \frac{mg}{\lambda - 2mg} < \frac{1}{2} \\

\Rightarrow \frac{1}{2} < \sin \frac{\theta}{2} < 1 \\

\Rightarrow 30 < \frac{\theta}{2} < 90 \\

\Rightarrow 60 < \theta < 180.

[generalebriety]

And STEP II Q7:

a (mod 5) | 2a^2 (mod 5)
0 | 0
1 | 2
2 | 3
3 | 3
4 | 2

b (mod 5) | b^2 (mod 5)
0 | 0
1 | 1
2 | 4
3 | 4
4 | 1

So 2a^2 + b^2 is divisible by 5 if and only if 2a^2 + b^2 = 0 (mod 5), which happens if and only if a = 0 (mod 5), b = 0 (mod 5).

2a^2 + b^2 = 5c^2. (*)

Assume a = 0 (mod 5), b = 0 (mod 5). Then a = 5p, b = 5q.
Then 2a^2 + b^2 = 5c^2
25(2p^2 + q^2) = 5c^2
5(2p^2 + q^2) = c^2
therefore c^2 = 0 (mod 5), and so c = 0 (mod 5). Then c = 5r.
Then 5(2p^2 + q^2) = c^2
5(2p^2 + q^2) = 25r^2
2p^2 + q^2 = 5r^2.
But this is equivalent to (*).

Hence, by the method of infinite descent, a, b and c can be divided by 5 infinitely many times and still remain integers.

Therefore the only solution is a = b = c = 0.


Ok, that's me done for the day. Are these solutions going on the wiki? If not, would it be a nice idea to do so?

[Speleo]

Did that last one already g.e.

DFranklin, is that question finished now (the dice one)?

[DFranklin]

(Original post by Speleo)
Did that last one already g.e.

DFranklin, is that question finished now (the dice one)?

Unless someone finds a mistake!

[*bobo*]

10 STEP I)modulous of elascticity=x and extension=e
<BCD=<BDC= Y
tension in wire= xe/a

for equib: moments about B:
xesinY= 2amgsin(180-2Y)=2amgsin2Y

e=a(2-2cos(180-2Y))^0.5 -a
=a(2+2cos2Y)^0.5 -a
=a(2 + 4cos^2 Y -2)^0.5 -a
=a(2cosY -1)

a(2cosY-1)xsinY=4amgsinYcosY
(2cosY-1)x=4mgcosY
cosY= x/(2x-4mg)

if rod is in equib away from wall (ie not against it):
Y>0 so cosY<1
x/(2x-4mg)<1

=> x>4mg

[*bobo*]

13 STEP I)
i)time taken to reach speed u:
dv/dt= F/M
v=Ft/M => t= uM/F

after it reaches speed u:
Mv dv/dt= Fu
=> (M/2)v^2= Fut +C
when t=0, v=u
(M/2)(v^2 -u^2)=Fut
=> t= (M/2Fu)(v^2-u^2)

so time taken to reach speed v=
(M/2Fu)(v^2-u^2) + uM/F (P=Fu)
=(M/2P)(v^2-u^2) + Mu^2/P
=(M/2P)(v^2+u^2)

[*bobo*]

13 ii)

distance travelled before speed reaches u:
initial speed=0, v=u, a=F/M
using v^2= u^2 +2as
s= u^2M/2F= Mu^3/2P

distance travelled after speed reaches u:
Mv^2 dv/dx=Fu

(M/3)( v^3- u^3)= Fux (if x=0 when v=u)

=> x= (M/3P)(v^3-u^3)

total distance travelled= (M/3P)(v^3- u^3) + Mu^3/2P
=(M/6P)(2v^3 + u^3)

[Rabite]

I think I did Q2 in the Further Maths A.
But it seems quite easy so I've probably made a mistake. It's still red on the front page, so if no one else has typed it out already, I'll do so~

[edit] Here it is anyway.

By the product/sum formulae that no one remembers.


But if m=±n, one of the fractions explodes. So in that case the question is:









If m=n=0, the integral turns to .

As for the second bit.
Let x = sinh²t

dx = 2sinhtcosht dt









Ignoring the +c for now


Which you can rewrite using the log form of arsinh.

[*bobo*]

12 STEP III)
i)T= (kmg/2a)e
e= 2(9a^2 + a^2/16)^0.5= a(145)^0.5/2

2Tsin(PDA)=mg
sin(PDA)=(145)^-0.5

=>2k(145)^0.5x0.25(145)^-0.5=1
=> k= 2

conservation of energy:
0.5(2mg/2a)(4a^2(p^2 +9))= (p+3)mga + 0.5(2mg/2a)64a^2

=>2p^2 + 18= p+3 +32
=>2p^2 -p -17=0

[*bobo*]

12 ii)
at Q: e=2a(p^2 + 9)^0.5
at R: e= a(4^2 +3^2)^0.5 - 2= 8a

2p^2 -p -17=0
p= (1 + 137^0.5)/4
applying p it can be seen that e at Q is greater than at R
hence tension in string is greater at Q than at R

[nota bene]

(Original post by DFranklin)
Having pondered this a bit, I couldn't find a nice way without basically solving the whole thing at the same time. (The big problem I had was showing you can't have two 1s - maybe I missed something).

The same problem I had, so I don't think we are missing anything. In your solution you had to disprove the (1,1,2) as well, although you did it in a much better way by starting to consider the ways to combine the sum of 11, which cuts down a bit on the testing of possibilities.

(I think I'll stay away from any further tedious questions on testing possibilities until I learn to explain, or I'll scan up scribbled unreadable notes...)

Now to the next STEP question...
STEP I Q 15
Probability an Ascii gives the same answer twice is TT+FF (T=True, F=False) Hence
So, given that an Ascii tells the same answer twice in succession the probability of lying is and therefore the probability of truth is 1-P(lying) = 9/10.

Given that a tribe member tells the same answer twice (as in the scenario) the probability of him being Ascii is possible to calculate via Bayes theorem, and Biscii is 1-P(A). By Bayes, P(A|two consequtive answers)=
Now we know P(A)=11/16, P(two|A)=5/8, P(A')=5/16 and P(two|A')=1
Calculating this gives and therefore Probability of Biscii given two consecutive answers is . Now the probability of lie is Therefore probability of truth is 1-P(lie) and thus as desired.

Following the scenario we are now given three consecutive answers, working with the same principle as above we calculate P(three|A) to be FFF+TTT =
(The other components of the earlier calculation stays the same)
So, by Bayes, P(A|three consequtive answers)=

Therefore probability of Biscii giving three consecutive answers is .
Now we need the probability for lie given it is an Ascii giving three consecutive answers. This is given by

So, now P(truth)=1-P(lie) and P(lie)=
So P(truth) is . Therefore the man shall not follow the given answer (i.e. left); he shall go to the right, with the probability of coming to the town.

[DFranklin]

(Original post by nota bene)
The same problem I had, so I don't think we are missing anything. In your solution you had to disprove the (1,1,2) as well, although you did it in a much better way by starting to consider the ways to combine the sum of 11, which cuts down a bit on the testing of possibilities.

It looked easy, but in hindsight, I think this was actually a pretty nasty question. It's very tricky to express all the reasoning without ending up with pages upon pages of cases.

Now to the next STEP question...

It's so much nicer when you get a "show..." question isn't it?

[*bobo*]

12 Stepii:

[*bobo*]

part 2 to question12 STEP II:
take moments about Q(2), (where A is the angle bwteen the two walls):
rMgcosA= mg sin0.5A. r(2-2cosA)^0.5

(McosA)^2=m^2sin^2 0.5A(2-2cosA)=m^2(2-2cosA)(0.5-0.5cosA)m^2(1-cosA)^2
=> McosA=m(1-cosA)
m= McosA/(1-cosA)= M/(secA-1)