STEP III 2015 Solutions Thread

Paper (sorry it has my scribbles over it from the exam):
https://www.dropbox.com/s/ljag5a221el8wsh/step%20iii%202015.pdf?dl=0

Q1) Brammer
Q2) DFranklin
Q3)
Q4) morgan8002
Q5) charles98
Q6) DFranklin
Q7) DFranklin
Q8) DomStaff
Q9) Brammer
Q10) Brammer ; Alt - mikelbird
Q11) mikelbird
Q12) DFranklin
Q13) Brammer

Paper (Google Drive link courtesy of jneill)

(edited 8 years ago)

Reply 1

Gawain

OP

6

Q3)

Here is a very rough copy until I LaTeX the solution:

https://www.dropbox.com/s/hlaid84re8n6jag/step%20iii%202015%20q3.pdf?dl=0

This may be wrong, I did it quickly after the exam.

Reply 2

shamika

16

What on earth was Q5? :s-smilie:

interested if people had difficulties. I haven't tried any of the others but it looks like a very reasonable paper. Hope everyone did well :smile:

Reply 3

Gawain

OP

6

Original post by shamika

What on earth was Q5? :s-smilie:

interested if people had difficulties. I haven't tried any of the others but it looks like a very reasonable paper. Hope everyone did well :smile:

No idea, I thought Q2 was equally simple.

What did you think of the stats? Q13 looks quite doable but I didn't have the confidence (having done no stats modules) to actually attempt it. That and there were plenty of other good questions to pick.

Reply 4

shamika

16

Original post by Gawain

No idea, I thought Q2 was equally simple.

What did you think of the stats? Q13 looks quite doable but I didn't have the confidence (having done no stats modules) to actually attempt it. That and there were plenty of other good questions to pick.

Don't like Q13, on the assumption that few people will know about bivariate distributions and will be out off. However, it's a standard type of question for first year stats. i also expect Q12 to be unpopular because generating functions are on so few syllabuses now.

We should direct people onto this thread for solutions. It's unlikely that I'll be able to sit in front of a computer over the weekend but if I do, I'll put up some solutions. Some of these questions look like they're fun.

Reply 5

DFranklin

18

"Sort of" solution to Q12. I feel there should be a way of doing the "reduction mod 6" by dividing by (1-x^6) or something, but it wasn't coming to me. This is an approach where "anyone can see this works, but to prove it works would be nasty" . (And the LaTeX was yuck, too)

Q12: (i)

G_1 = \frac{1}{6}(1+x+x^2+x^3+x^4+x^5)

.
Now, if we just wanted the sum of the terms, we would have

G_2 = (G_1)^2

. To deal with the fact that we only care about the result mod 6 we need to coalesce all the coefficients whose orders are equal modulo 6. We will denote this by "mod x^6", so that, for example, x^7 = x mod x^6. [Note, this is not the standard meaning of "mod x^6"].

But then

x G_1 \mod x^6 = \frac{1}{6}( x+x^2+x^3+x^4+x^5+x^6) =\frac{1}{6}( x+x^2+...+x^5 + 1) = G_1

and similarly

x^k G_1 \mod x^6 = G_1

for k = 1,2,3...

But then

G_1^2 \mod x^6 = \frac{1}{6}( 1 + x+x^2+x^3+x^4+x^5) G_1 = \frac{1}{6}(G_1+G_1+G_1+G_1+G_1+G_1) = G_1

And then of course G_3 =

G_1 (G_1)^2 = G_1 G_1 = G_1

and so on.
So G_k = G_1 and so p(S_k = 0) = G_1(0) = 1/6.

(ii) P(T_1 = 1) = 2/6, P(T_1 = k (k=/=1)) = 1/6. So

T_1(x) = \frac{1}{6}(1+2x+x^2+x^3+x^4) = \frac{1}{6}(x + y)

.
Again,

T_2 = (T_1)^2 mod x^5

and again

xy = y \mod x^5

, while

y^2 = (1+x+x^2+x^3+x^4)y = y+y+y+y+y = 5y \\mod x^5

.

So T_2 =

\frac{1}{36}(x+y)^2 = \frac{1}{36}(x^2+2xy+y^2) = \frac{1}{36}(x^2 + 2y + 5y) = \frac{1}{36}(x^2+7y)

and T_3 =

\frac{1}{216}(x+y)(x^2+7y) = \frac{1}{216}(x^3+7xy+x^2y + 7y) = \frac{1}{216}(43y + x^3)

Claim:

T_k = \frac{1}{6^k}(\frac{6^k-1}{5}y + x^{k \mod 5})

True for k = 2, 3.
But if true for k =n, then

T_{n+1} = \frac{1}{6^{k+1}} (x+y) (\frac{6^k-1}{5}y + x^{k \mod 5} \mod x^5)

= \frac{1}{6^{k+1}} (\frac{6^k-1}{5}xy + x^{(k+1) \mod 5} + \frac{6^k-1}{5}y^2 + x^{k \mod 5}) y)

= \frac{1}{6^{k+1}} (\frac{6^k-1}{5}y + x^{(k+1) \mod 5} + 5\frac{6^k-1}{5} y +y)

= \frac{1}{6^{k+1}} (\frac{6^{k+1}-1}{5}y + x^{(k+1) \mod 5})

and so the result follows by induction.

So when k=/=0 mod 5,

P(T_k = 0) = \frac{1}{6^k}(6^k - 1) / 5 = \frac{1}{5}(1-1/6^k)

as desired.
When k = 0 mod 5,

P(T_k = 0) = \frac{1}{6^k}((6^k-1)/5 + 1= \frac{1}{5}(1+4/6^k)

Reply 6

DFranklin

18

Q2: (i) True. Take m = 1000. Then

n \geq m \implies n \geq 1000 \implies n^2 \geq 1000n

.
(ii) False. Take p_n = (-1)^n, q_n = 0. For any m, setting n = 2m we have p_n = 1 > q_n. And so it is not true that {p_n <= q_n}. But taking n = 2m+1 we have p_n = -1 < q_n and so it is not true that {q_n <= p_n}.
(iii) True. By defn of {}, we can find m_1, m_2 s.t.

n > m_1 \implies s_n \leq t_n

and

n > m_2 \implies t_n \leq u_n

. Then taking m = max(m_1, m_2) we have

n > m \implies s_n \leq t_n

and t_n \leq u_n and so implies

s_n \leq u_n

.
(iv) True. Take m = 5. Claim n^2 < 2^n for n >=5. True for n = 5. And if true when n = k, n>4, then (n+1)^2 = n^2 + 2n+1 < 2n^2 (since n > 3) < 2^(n+1) by induction hypothesis. So true for n>=5 by induction.

Part (iii) feels out-of-place relative to the other parts. I think expecting someone to find the "max(m_1, m_2)" argument is a little unfair here (but anyone who's done a bit of analysis reading would find it easy).

Reply 7

raff97

8

Original post by DFranklin

Q2: (i) True. Take m = 1000. Then

n \geq m \implies n \geq 1000 \implies n^2 \geq 1000n

.
(ii) False. Take p_n = (-1)^n, q_n = 0. For any m, setting n = 2m we have p_n = 1 > q_n. And so it is not true that {p_n <= q_n}. But taking n = 2m+1 we have p_n = -1 < q_n and so it is not true that {q_n <= p_n}.
(iii) True. By defn of {}, we can find m_1, m_2 s.t.

n > m_1 \implies s_n \leq t_n

and

n > m_2 \implies t_n \leq u_n

. Then taking m = max(m_1, m_2) we have

n > m \implies s_n \leq t_n

and

t_n \leq u_n and so implies

s_n \leq u_n

.
(iv) True. Take m = 5. Claim n^2 < 2^n for n >=5. True for n = 5. And if true when n = k, n>4, then (n+1)^2 = n^2 + 2n+1 < 2n^2 (since n > 3) < 2^(n+1) by induction hypothesis. So true for n>=5 by induction.

Part (iii) feels out-of-place relative to the other parts. I think expecting someone to find the "max(m_1, m_2)" argument is a little unfair here (but anyone who's done a bit of analysis reading would find it easy).

I think 4 is also included in (iv)

Reply 8

DomStaff

2

Q8

Using your definitions

x=rcos(\theta), y =rsin(\theta)

and differentiating implicitly we arrive at

\dfrac{dy}{dx} = \dfrac{ \frac{dr}{d\theta} sin(\theta) + rcos(\theta)}{\frac{dr}{d\theta} cos(\theta) - rsin(\theta)}

Sub these into the DE.

Cancelling r from each side, multiplying out the denominator, then cancelling:

\frac{dr}{d\theta}sin^2 (\theta) + rcos^2 (\theta) = - \frac{dr}{d\theta} cos^2 (\theta) -rsin^2 (\theta)

Which rearranges to:

\frac{dr}{d\theta} + r = 0

Separating varaibles:

\frac{dr}{d\theta} = -r

Integrating:

r=Ae^{\theta}

For the second part, which would be a nightmare to LaTeX, you apply the same change of variable, cancel r again and multiply out the denominator, ending up with:

\frac{dr}{d\theta} = r^3 -r

\displaystyle\int \dfrac{1}{r^3 - r}\ dr = \displaystyle\int \ 1 d\theta

Partial fractions:

\displaystyle\int \dfrac{1/2}{r+1} + \dfrac{1/2}{r-1} - \dfrac{1}{r}\ dr = \dfrac{1}{2} ln(\dfrac{r^2 -1}{r^2}) = \theta + c

\dfrac{r^2 -1}{r^2} = ke^{2\theta}

Which rearranges to give the result.

Has anyone got any good drawing software for the graphs?

(edited 8 years ago)

Reply 9

DomStaff

2

Original post by shamika

What on earth was Q5? :s-smilie:

interested if people had difficulties. I haven't tried any of the others but it looks like a very reasonable paper. Hope everyone did well :smile:

It looks like I chose my questions terribly :smile:

Reply 10

DFranklin

18

Q4: just (iii), the rest should be routine:

Define

z_k = r_k e^{i\theta_k}

and S_1, S_2, S_3 as in the earlier part.
Then the 3 equations given correspond to the statements that S_1, S_2, S_3 are all real.
But we also know we can write z_1, z_2, z_3 are the roots of a cubic whose coefficents can be written in terms of S_1, S_2 and S_3. And so they are the roots of a cubic with real roots.

So at least one of z_1, z_2, z_3 must be real, call in z_k. Then

\sin \theta_k = 0

and given the restrictions on the ranges for theta, this implies theta_k = 0.
If theta_1 = 0, then z_1 is real. and then z_2, z_3 must be conjugate pairs, which iimplies

\theta_2 = - \theta_3

.

Reply 11

DFranklin

18

Original post by raff97

I think 4 is also included in (iv)

Yes, but it doesn't matter. You need to show that you can find an m that works, not that it is the smallest possible m.

(I was going to set m = 10 to make the point, but decide the mental arithmetic got sufficiently harder that it was a bad idea).

Reply 12

username1162972

18

Original post by DFranklin

Yes, but it doesn't matter. You need to show that you can find an m that works, not that it is the smallest possible m.

(I was going to set m = 10 to make the point, but decide the mental arithmetic got sufficiently harder that it was a bad idea).

My god I think I got Q2 fully correct. Ooh. Literally made my day.

Posted from TSR Mobile

Reply 13

DomStaff

2

I'll write up Q7 at some point today.

Reply 14

shamika

16

Original post by DFranklin

Q2: (i) True. Take m = 1000. Then

n \geq m \implies n \geq 1000 \implies n^2 \geq 1000n

.
(ii) False. Take p_n = (-1)^n, q_n = 0. For any m, setting n = 2m we have p_n = 1 > q_n. And so it is not true that {p_n <= q_n}. But taking n = 2m+1 we have p_n = -1 < q_n and so it is not true that {q_n <= p_n}.
(iii) True. By defn of {}, we can find m_1, m_2 s.t.

n > m_1 \implies s_n \leq t_n

and

n > m_2 \implies t_n \leq u_n

. Then taking m = max(m_1, m_2) we have

n > m \implies s_n \leq t_n

and

t_n \leq u_n and so implies

s_n \leq u_n

.
(iv) True. Take m = 5. Claim n^2 < 2^n for n >=5. True for n = 5. And if true when n = k, n>4, then (n+1)^2 = n^2 + 2n+1 < 2n^2 (since n > 3) < 2^(n+1) by induction hypothesis. So true for n>=5 by induction.

Part (iii) feels out-of-place relative to the other parts. I think expecting someone to find the "max(m_1, m_2)" argument is a little unfair here (but anyone who's done a bit of analysis reading would find it easy).

Part (ii) needs examples where the sequences are positive, but the argument is identical. I'll let someone give alternative example sequences which do work. Big big hint:

Spoiler

Reply 15

Doones

21

Original post by physicsmaths

My god I think I got Q2 fully correct. Ooh. Literally made my day.

Posted from TSR Mobile

Posted from TSR Mobile

Reply 16

DFranklin

18

Original post by shamika

Part (ii) needs examples where the sequences are positive, but the argument is identical. I'll let someone give alternative example sequences which do work. Big big hint:

Spoiler

Meh. That's what comes from not being able to write solution and see question simultaneously. Any thoughts on Q12 - I wasn't exactly "happy" with my solution.

Reply 17

newblood

19

Original post by shamika

Part (ii) needs examples where the sequences are positive, but the argument is identical. I'll let someone give alternative example sequences which do work. Big big hint:

Spoiler

sn=|sin(npi/10)|, tn=1/2?

Reply 18

newblood

19

Original post by DFranklin

Q4: just (iii), the rest should be routine:

Define

z_k = r_k e^{i\theta_k}

and S_1, S_2, S_3 as in the earlier part.
Then the 3 equations given correspond to the statements that S_1, S_2, S_3 are all real.
But we also know we can write z_1, z_2, z_3 are the roots of a cubic whose coefficents can be written in terms of S_1, S_2 and S_3. And so they are the roots of a cubic with real roots.

So at least one of z_1, z_2, z_3 must be real, call in z_k. Then

\sin \theta_k = 0

and given the restrictions on the ranges for theta, this implies theta_k = 0.
If theta_1 = 0, then z_1 is real. and then z_2, z_3 must be conjugate pairs, which iimplies