STEP Maths I, II, III 1993 Solutions

STEP III Q4:

Unparseable latex formula:

\displaystyle[br]\sum_{r=0}^\infty \frac{1}{2r+1} \bigg( \frac{1}{2} \bigg)^{2r} = \sum_{r=0}^\infty \frac{2}{2r+1} \bigg( \frac{1}{2} \bigg)^{2r+1}. \\[br]\text{Consider } \sum_{r=0}^\infty 2x^{2r}. \\[br]\int \sum_{r=0}^\infty 2x^{2r} \text{d}x = \sum_{r=0}^\infty \frac{2}{2r+1} x^{2r+1} = \sum_{r=0}^\infty \frac{1}{2r+1} x^{2r}. \\[br]\text{But } \sum_{r=0}^\infty 2x^{2r} = 2 + 2x^2 + 2x^4 + \dots = \frac{2}{1-x^2} = \frac{1}{1-x} + \frac{1}{1+x}. \\[br]\therefore \int \sum_{r=0}^\infty 2x^{2r} \text{d}x = \int \bigg( \frac{1}{1-x} + \frac{1}{1+x} \bigg) \text{d}x = \ln \bigg( \frac{1+x}{1-x} \bigg). \\[br]\therefore \sum_{r=0}^\infty \frac{2}{2r+1} x^{2r+1} = \ln \bigg( \frac{1+x}{1-x} \bigg). \\[br]\text{Put }x = \frac{1}{2}:\; \sum_{r=0}^\infty \frac{1}{2r+1} \bigg( \frac{1}{2} \bigg)^{2r} = \ln 3.

(ii)

\displaystyle \sum_{r=0}^\infty (2x^{4r} - x^{4r+1} - x^{4r+3}) = (2 - x - x^3)(1 + x^4 + x^8 + \dots) = \frac{2 - x - x^3}{1 - x^4}.

(iii)

Unparseable latex formula:

\displaystyle S = \sum_{r=2}^\infty \frac{r2^{r-2}}{3^{r-1}} = \frac{1}{3}\sum_{r=2}^\infty r\bigg( \frac{2}{3} \bigg)^{r-2}.\\[br]\text{Let } u_n = \sum_{r=n}^\infty \bigg( \frac{2}{3} \bigg)^{r-2} = 3\bigg( \frac{2}{3} \bigg)^{n-2}.\\[br]\text{Then } S = \frac{1}{3}(2u_2+u_3+u_4+u_5+\dots). \\[br]u_2 = 3\\[br]u_2+u_3+u_4+\dots = \frac{3}{1 - \frac{2}{3}} = 9.\\[br]\therefore S = \frac{1}{3}(3+9) = 4.

(iv)

Unparseable latex formula:

\displaystyle S = \sum_{r=2}^\infty \frac{2}{r(r^2-1)} = \sum_{r=2}^\infty \frac{2}{(r-1)r(r+1)} \\[br]= \sum_{r=2}^\infty \bigg( \frac{1}{r-1} - \frac{2}{r} + \frac{1}{r+1} \bigg) \text{ (partial fractions)} \\[br]= (1/1 - 2/2 + 1/3) + (1/2 - 2/3 + 1/4) + (1/3 - 2/4 + 2/5) + \dots \\[br]= 1 - \frac{1}{2} \text{ (difference method)} \\[br]= \frac{1}{2}.

STEP I Question 6

13 STEP II)
i)equilibrium:
kQ(1)Q(2)/a^2= kQ(2)Q(3)/(d-a)^2
=> Q(1)^0.5(d-a)=aQ(2)^0.5
=>a= dQ(1)^0.5/(Q(1)^0.5 + Q(2)^0.5)

ii)where x is the distance of P(3) from P(1) and x=a+y
d^2y/dt^2= ( kQ(3)/m)((Q(1)/x^2) - Q(2)/(d-x)^2 )

=(kQ(3)/m)((Q(1)/(a+y)^2 - Q(2)/((d-a)-y)^2)

use approximation that: (b+z)^-2= b^-2 - 2z(b^-3) for small z.

=(kQ(3)/m)( Q(1)/a^2 - 2Q(1)y/a^3 - Q(2)/(d-a)^2 - 2yQ(2)/(d-a)^3 )
use part (i) to cancel out some parts:

= -2(kQ(3)/m)((Q(1)/a^3) +(Q(2)/(d-a)^3))y (SHM)
so accel.y= -w^2y
and T=2pi/w

apply w taken from the equation achieved and you will derive the required answer. end of question.

Reply 65

Finished STEP III Q4, post #63.

Reply 66

Trying STEP III Q7, no promises though...

Edit: abandoned for a while, might come back to it. Until I say otherwise I'm not gonna be doing it though, so anyone else is free to steal it off me. :p:

Reply 67

nota bene

15

Okay, Speleo I think STEP II question 1 is okay (post 3) I edited the working I had on paper, so DFranklin's comments should now be okay, David can you confirm this? (I can't see what more I'd possibly need to do explicitly...)

edit: Okay looked at II Q 16 and here is what I did but it looks wrong (I'm not using the info n=5 in part b and c which doesn't seem correct but can't get my head around what to do about it (I tried something weird with Bayes theorem, didn't work...)).

a) If the probability of arriving at minute 1 is 1/4 and the probability to arrive at any minute is independent means that the probability to arrive at minute n is 1/4 as well. Anyway, the probability that the first bus arrives after n minutes is distributed with binomial distribution (I think). So then we have

\begin{pmatrix} n \\ 1 \end{pmatrix}(\frac{1}{4})^1(\frac{3}{4})^{n-1}=\frac{n}{4}\frac{3}{4}^{n-1}

(This part a doesn't convince me, so I'm likely wrong!)
edit: read generalebriety's post :p:

b) Since the mean (aka Expected value) is 4 we can write P(X=m) as

\frac{20^me^{-20}}{m!}

c) We can find the probability that a bus arriving at 8.05 not can take all passangers by considering 1-P(X=0)-P(X=1)-P(X=2)-...-P(X=25). (as the bus can take up to 25 passengers) The question doesn't require calculating this so I think I'll avoid it :p:

Reply 68

nota bene

Okay, Speleo I think STEP II question 1 is okay (post 3) I edited the working I had on paper, so DFranklin's comments should now be okay, David can you confirm this? (I can't see what more I'd possibly need to do explicitly...)

edit: Okay looked at II Q 16 and here is what I did but it looks wrong (I'm not using the info n=5 in part b and c which doesn't seem correct but can't get my head around what to do about it (I tried something weird with Bayes theorem, didn't work...)).

a) If the probability of arriving at minute 1 is 1/4 and the probability to arrive at any minute is independent means that the probability to arrive at minute n is 1/4 as well. Anyway, the probability that the first bus arrives after n minutes is distributed with binomial distribution (I think). So then we have

\begin{pmatrix} n \\ 1 \end{pmatrix}(\frac{1}{4})^1(\frac{3}{4})^{n-1}=\frac{n}{4}\frac{3}{4}^{n-1}

P(N=1) = 1/4
P(N=2) = 3/4 * 1/4
P(N=3) = (3/4)^2 * 1/4
P(N=4) = (3/4)^3 * 1/4
...
P(N=n) = (3/4)^(n-1) * 1/4

I'm not sure where you got the extra n from. I think you've worked out the probability that exactly one bus arrives within n minutes. :smile:

Reply 69

nota bene

Okay, Speleo I think STEP II question 1 is okay (post 3) I edited the working I had on paper, so DFranklin's comments should now be okay, David can you confirm this? (I can't see what more I'd possibly need to do explicitly...)No that's fine. Probably a bit more than necessary now; it's a slightly odd question where you probably need to do more under "exam conditions" than you'd actually need in a formal proof. (In a formal proof you could just say "by explicit calculation of all 45 possibilities ..." without writing down what happens in each case. But in an exam, I think you need to show a bit more proof that you've actually done the calculations. Basically because otherwise the examiner can't tell if you're bluffing).

edit: Okay looked at II Q 16 and here is what I did but it looks wrong (I'm not using the info n=5 in part b and c which doesn't seem correct but can't get my head around what to do about it (I tried something weird with Bayes theorem, didn't work...)).

Shouldn't the mean for (b) be 20? (4 passengers per minute x 5 minutes)?

And part (c) is somewhat ambiguous, but I think you are no longer supposed to be assuming the 1st bus is at 8:05 (since otherwise the answer is trivial, as you observe).

Edit: And I think the General is right as well...

Reply 70

nota bene

15

DFranklin

No that's fine. Probably a bit more than necessary now; it's a slightly odd question where you probably need to do more under "exam conditions" than you'd actually need in a formal proof. (In a formal proof you could just say "by explicit calculation of all 45 possibilities ..." without writing down what happens in each case. But in an exam, I think you need to show a bit more proof that you've actually done the calculations. Basically because otherwise the examiner can't tell if you're bluffing).

Yes, the question is a bit odd, yet somehow I found it a lot more straightforward than the dice one...

DFranklin

Shouldn't the mean for (b) be 20? (4 passengers per minute x 5 minutes)?

And part (c) is somewhat ambiguous, but I think you are no longer supposed to be assuming the 1st bus is at 8:05 (since otherwise the answer is trivial, as you observe).

For b) Yes, obviously, had it that way on paper, can't even copy what I have written properly (maybe because my notes are unreadable :tongue:

).
c) I'll look at it with that starting point and see where I get

generalebriety, of course that's the way - as I said I spent too little time on this one...

Reply 71

nota bene

Yes, the question is a bit odd, yet somehow I found it a lot more straightforward than the dice one...I found the dice question a real pig, to be honest. I certainly didn't get it out inside 30 minutes.

There's a fairly obvious style change between the 1993 papers and the later ones, and I think you've been unlucky to pick some of the questions where the differences are most acute. (The stats questions seem generally nasty as well; I do prefer it when you have a "show that" with stats - so easy to make mistakes otherwise).

Reply 72

*bobo*

8

STEP I 2)
i)1+r^2 + r^4+..+r^2n
t(x)=r^(2(x-1)) so sum up to term x=n+1
S(n+1) = (1-r^(2(n+1))/(1-r^2)

ii) split into two terms
S1(r)=r+r^4+r^7+ r^10+.... a=r, d=r^3 t(x)=r^(3(x-1) +1)
S2(r)= r^2 + r^5 + r^8 +...r^(3n-1) a=r^2 d=r^3 t(x)=r^(3(x-1)+2)up to term x=n

S1(r)= r(1- r^3n)/(1-r^3)
S2(r)=r^2(1-r^3n)/(1-r^3)
Sn(r)= r(1+r)(1-r^3n)/(1-r^3)
Sn(infinity)= S1(infinity)+S2(infinity)= r/(1-r^3) + r^2/(1-r^3)
=r(1+r)/(1-r^3)= 1/(1-r) - 1/(1-r^3)
so as n approaches infinity, Sn(r) approaches 1/(1-r) - 1/(1-r^3)

iii)again split into two terms
T1(r)=1+r^3+r^6+r^9+..... a=1 d=r^3,t(x)=r^3(x-1)
T2(r)=r^2 +r^4+r^8+r^10+...+r^6n a=r^2 d=r^2 t(x)=r^(2(x-1)+2) up to x=3n
Tn(r)= (1-r^9n)/(1-r^3) + r^2(1-r^6n)/(1-r^2)

Tn(r)infinity= 1/(1-r^3) +r^2/(1-r^2)= (1-r^5)/((1-r^2)(1-r^3))
as n approaches infinity ,if r=1 Tn(r) approaches 0
if r=-1 Tn(r) approaches infinity
(^wrong)

Reply 73

*bobo*

Tn(r)infinity= 1/(1-r^3) +r^2/(1-r^2)= (1-r^5)/((1-r^2)(1-r^3))
as n approaches infinity ,if r=1 Tn(r) approaches 0
if r=-1 Tn(r) approaches infinity

I haven't followed your argument closely, but this doesn't look right. After all, when r=1, it's fairly obvious that

T_{6n}(r) = 5n

,

Reply 74

Swayum

*bobo*

STEP I 2)
i)1+r^2 + r^4+..+r^2n
t(x)=r^(2(x-1)) so sum up to term x=n+1
S(n+1) = (1-r^(2(n+1))/(1-r^2)

ii) split into two terms
S1(r)=r+r^4+r^7+ r^10+.... a=r, d=r^3 t(x)=r^(3(x-1) +1)
S2(r)= r^2 + r^5 + r^8 +...r^(3n-1) a=r^2 d=r^3 t(x)=r^(3(x-1)+2)up to term x=n

S1(r)= r(1- r^3n)/(1-r^3)
S2(r)=r^2(1-r^3n)/(1-r^3)
Sn(r)= r(1+r)(1-r^3n)/(1-r^3)

Oh yay, so my formula was right 2 pages ago :biggrin:

. That's the first time I've gotten anywhere with STEP <_<.

Reply 75

*bobo*

8

Tn(r) represents the sum of the entire series as n is approaching infinity, using the sum to infinity and substituting the value of r. i probably havn't made my terms very clear,sorry. don't know whereyou got T6n(r)=5n from?

Reply 76

*bobo*

Tn(r) represents the sum of the entire series as n is approaching infinity, using the sum to infinity and substituting the value of r. i probably havn't made my terms very clear,sorry. don't know whereyou got T6n(r)=5n from?

Firstly, I'm partly wrong - I hadn't paid enough attention to their notation. T_n(1) = 4n+1. Because T_n is just the sum of 4n+1 '1's when r=1 (edit: oops! I can't count. Doesn't change the main point though).

The sum is not convergent at r=1, so you can't simply take the formula for r<1 and put r=1. (Well you can, but you there's no reason to expect to get the right answer).

What you are effectively doing is interchanging two limiting processes. You have calculated

Unparseable latex formula:

\displaymath \lim_{r\to 1} \lim_{n\to \infty} T_n(r)

, but the question is asking for

Unparseable latex formula:

\displaymath \lim_{n \to \infty} \lim_{r\to 1} T_n(r)

(where the r->1 limit in the second case is trivial as every term is just a power of r).

(for a much simpler example, consider 1+r+r^2+... = 1/(1-r). What is the RHS when r = 2? Does this make sense when you look at the LHS?)

Reply 77

*bobo*

8

oh right, completely forgot about r not being able to equal 1 for that. i get it now thanks.

Reply 78