DFranklin
Badges: 18
Rep:
?
#41
Report 11 years ago
#41
I believe STEP III is the only paper specifically for the FM syllabus.

As to which is easier, everyone agrees STEP III is hardest by some way. Most people reckon the AEA is also easier than STEP II and STEP I. My feeling is that the difference isn't great, and the biggest factor is probably which exam format you get on with best. There's a fair degree of personal perference: some will find AEA much easier than STEP, others will find STEP easier.
0
reply
johnheyes91
Badges: 0
Rep:
?
#42
Report 11 years ago
#42
Thanks Franklin, do you think doing STEP before going to uni would improve/prepare me better for maths at uni? I only did a levels maths, but doing maths in uni this fall..

also, whats the difference between STEP, A level and University maths like?

cheers!
0
reply
Zhen Lin
Badges: 10
Rep:
?
#43
Report 11 years ago
#43
(Original post by johnheyes91)
Zhen just curious, how do you type maths and transfer it to a pdf? Cheers!
LaTeX. The other thread probably explains better than I could.
0
reply
johnheyes91
Badges: 0
Rep:
?
#44
Report 11 years ago
#44
ah, thanks!
0
reply
speedy_s
Badges: 0
Rep:
?
#45
Report 11 years ago
#45
(Original post by DFranklin)
Re: Q8. When it comes to solving the recurrence relation, I think you've approached it the wrong way around. Essentially, you've derived a formula for the solution, but you haven't shown it's the only possible solution.

In fact, to answer the question you don't need to derive it. Consider

\displaystyle c_n = \frac{\alpha^{2n-1} + \alpha^{-(2n-1)}}{\sqrt 5}

All you need is to show c_0 = c_1 = 1, and c_n = 3c_{n-1}-c_{n-2}. Then write b_n = c_n - a_n; you have b_0 = b_1 = 0 and b_n = 3b_{n-1}-b_{n-2}, so trivially b_n = 0 for all n by induction.
Sorry, but why is b_n = 3b_{n-1}-b_{n-2}?
0
reply
Square
Badges: 12
Rep:
?
#46
Report 11 years ago
#46
If someone could take a look at my solution to 2000/III Q6 I would be thankful!

Spoiler:
Show


r=bc \\ q=ab-ac \\ p=c-a^2+b

f(u)=u^3+2pu^2+(p^2-4r)u-q^2=0

If a^2 root then f(a^2)=0

f(a^2)=a^6+2pa^4+(p^2-4r)a^2-q^2=0 Substituting in the equations above and a lot of algebra later (which I don't really fancy latexing) you do get f(a^2)=0 which implies a^2 is a root.

For the non-negative root proof, I just wrote a small note saying that since a^2 will always be greater than or equal to zero for any value of a, f(u) must always have a non-negative root.




f(9)=9^3+2p(81)+(p^2-4r)9-q^2 \\ p=-1, q=-6, r=15 \\ f(9)=9^3-2(81)-59(9)-36 \\ f(9)=(81)(9)-(18)(9)-59(9)-4(9) \\ f(9)=9(81-18-59-4)=9(0)=0

Therefore 9 a root of f(u) when p=-1 q=-6 r=15



y^4-8y^3+23y^2-34y+39=0

Substitute y=x+2 - this will get rid of the y^3 term and make the quartic similar to that in the first part of the question

Several lines of algebra later:

x^4-x^2-6x+15=(x-ax+b)(x+ax+c)

When you compare to the first part of the question you get p=-1 q=-6 and r=15 which is the same as the second part of the question.

This implies that the cubic f(u) from the previous part holds. Since we showed that f(u) has a root at u=9 for p=-1 q=-6 and r=15 AND that f(u) has a root a^2:

a^2=9
a=3

a=3 implies b=3 and c=5

Therefore x^4-x^2-6x+15=(x^2-3x+3)(x^2+3x+5).

Subsitute back for x=y-2

y^4-8y^3+23y^2-34y+39=((y-2)^2-3(y-2)+3)((y-2)^2+3(y-2)+5)

=(y^2-7y+13)(y^2-y+3)

I hope that makes sense, I've condensed most of the algebra and my actual solution has about 3 and a half sides of A4.
0
reply
DFranklin
Badges: 18
Rep:
?
#47
Report 11 years ago
#47
(Original post by speedy_s)
Sorry, but why is b_n = 3b_{n-1}-b_{n-2}?
Because a_n = 3a_{n-1}-a_{n-2} and c_n = 3c_{n-1}-c_{n-2} and b_n = c_n - a_n (for all n).
0
reply
DFranklin
Badges: 18
Rep:
?
#48
Report 11 years ago
#48
Square: I'm pretty sure you've misunderstood a (small) part of the question.

When it says "explain why this equation always has a non-negative root", I think you're supposed to be looking at the equation by itself, without assuming that you know about a^2 being a root. In which case:

Spoiler:
Show
if g(u) = u^3 + 2pu^2+(p^2-4r)u - q^2, then g(0) <=0. if g(0) = 0, 0 is a non-negative root. If g(u)<0, then as g(u) -> +inf as u->inf, g changes sign for some positive u and so has a non-negative root.


I don't think this could cost you many marks, and I'm not even sure I'm right in my interpretation here.

For the rest of it, it's very hard to comment when you've included so little working; assuming your working is right, you should be OK.

For what it's worth, this really doesn't feel like it should be a "3 sides of A4" question, but that's easy for me to say.
0
reply
Square
Badges: 12
Rep:
?
#49
Report 11 years ago
#49
(Original post by DFranklin)
Square: I'm pretty sure you've misunderstood a (small) part of the question.

When it says "explain why this equation always has a non-negative root", I think you're supposed to be looking at the equation by itself, without assuming that you know about a^2 being a root. In which case:

Spoiler:
Show
if g(u) = u^3 + 2pu^2+(p^2-4r)u - q^2, then g(0) <=0. if g(0) = 0, 0 is a non-negative root. If g(u)<0, then as g(u) -> +inf as u->inf, g changes sign for some positive u and so has a non-negative root.


I don't think this could cost you many marks, and I'm not even sure I'm right in my interpretation here.

For the rest of it, it's very hard to comment when you've included so little working; assuming your working is right, you should be OK.

For what it's worth, this really doesn't feel like it should be a "3 sides of A4" question, but that's easy for me to say.
Aha, I see where you're getting at there.

I do have quite large writing (especially when it comes to doing maths) so maybe slightly less with averaged sized writing.

I can put in some more working if you like but some of it was quite long winded and tedious and I really could not be bothered to LaTeX it.

Maybe you have some better algebra skills than I do (which wouldn't be very hard!)
0
reply
Oh I Really Don't Care
Badges: 15
Rep:
?
#50
Report 10 years ago
#50
I / 7

If  \displaystyle f(x) = ax - \frac{x^3}{1 + x^2} , show that  \displaystyle f'(x) \ge 0  \displaystyle\forall x iff  \displaystyle a \ge \frac{9}{8}}

Note that f(x) is a quotient and an inequality is involved. While tempting to use something such as the product rule, by using the quotient rule the denominator will always be > 0.

 \displaystyle f'(x) = a - \frac{3x^2(1+x^2) - 2x(x^3)}{(1 + x^2)^2} = a + \frac{x^4 - 3x^2}{(1+x^2)^2} = \frac{(a+1)x^4 + (2a-3)x^2 + a}{(1+x^2)^2}

 \displaystyle (a+1)x^4 + (2a-3)x^2 + a \ge 0 \Rightarrow B^2 \le 4AC

 \displaystyle (2a - 3)^2 \le 4a(a-1) \Rightarrow a \ge \frac{9}{8}
0
reply
Oh I Really Don't Care
Badges: 15
Rep:
?
#51
Report 10 years ago
#51
II / 1

Guess a solution to  \displaystyle \frac{1}{N} = \frac{1}{a} + \frac{1}{b} you are provided with  \displaystyle \frac{1}{2} = \frac{1}{3} + \frac{1}{6} and also  \displaystyle \frac{1}{3} = \frac{1}{4} + \frac{1}{12}

The solution is clear -  \displaystyle \frac{1}{N} = \frac{1}{N + 1} + \frac{1}{N^2 + N}

To prove this is so;  \displaystyle \frac{1}{N} \equiv \frac{1}{N + 1} + \frac{1}{N^2 + N} = \frac{N+1}{N(N+1)} = \frac{1}{N}

Looking at  \displaystyle (a-N)(b-N) = N^2 for N is a prime it is apparent that the only factors of N^2 are  \displaystyle N^2 = N \times N = N^2 \times 1

This would imply that if distinct natural numbers (a,b) are sought for the equation  \frac{1}{N} = \frac{1}{a} + \frac{1}{b} for N prime then there is only one pair of solutions.

wip for  \displaystyle \frac{2}{N}
0
reply
Oh I Really Don't Care
Badges: 15
Rep:
?
#52
Report 10 years ago
#52
II / 2

 \displaystyle p[x] = (x-a)^2q[x] \Rightarrow p'[x] = (x-a)(2q[x] + (x-a)q'[x])

 \displaystyle g[x] = (x-a)^4q[x] \Rightarrow p'[x] = (x-a)^3(4q[x] + (x-a)q'[x]) *

Given p(x) we differentiate to find;

The roots of the equation  (x-\alpha)^3(4q[x] + (x-\alpha)q'[x])  = 6x^5 + 20x^4 - 20x^3 - 120x^2 - 80x + 32

Inspection reults in  \displaystyle \alpha = -2 So if p[x] denotes the polynomial given then p[-2] = 0. Using the factor and remainder theorem; k = 48
1
reply
Oh I Really Don't Care
Badges: 15
Rep:
?
#53
Report 10 years ago
#53
II / 4

 \displaystyle e^{i\theta} = cos(\theta) + isin(\theta) \Rightarrow e^{i\theta}e^{i\alpha} = e^{i(\theta + \alpha)} = cos(\theta + \alpha) + isin(\theta + \alpha)

 \displaystyle (e^{i\theta})^n = e^{ni\theta} = cos(n\theta) + isin(n\theta)

 \displaystyle Z_1 = (5-i)^2(1+i) = 34 + 14i

 \displaystyle Z_2 = (5-i)

 \displaystyle Z_3 = (1+i)

 \displaystyle arg(Z_1) = 2arg(Z_2) + arg(Z_3) \Rightarrow arctan(\frac{7}{17}) = -2arctan(\frac{1}{5}) + arctan(1) \Rightarrow arctan(\frac{7}{17}) + 2arctan(\frac{1}{5}) = \frac{\pi}{4}

The second part is alot easier if you just have a hunch.

By sheer observation the complex number  \displaystyle \beta = (1+i)(4-i)^3(-20+i) shows the fact;  \displaystyle 3arctan(\frac{1}{4}) + arctan(\frac{1}{20}) + arctan(\frac{1}{1985}) = \frac{\pi}{4}
0
reply
Unbounded
Badges: 12
Rep:
?
#54
Report 10 years ago
#54
Attempt at I/11
part 1:
Letting the normal reaction force from the ground be R and the normal reaction force from the wall be T, and the friction from the ground be  F_1 , and from the wall  F_2 (see if diagram 1).

I attempted to approach the question with proof by contradiction. Assume the friction is limiting at both the ground and the wall

 \therefore F_1 = 0.2R \ and \ F_2 = R

Resolving verticaly,  R+T = 5g

and resolving horizontally,  0.2R = T \therefore R = 5T

 \implies 6T = 5g \implies T = \frac{5g}{6}

 \therefore R = \frac{25g}{6}

From taking moments about A, the point of contact with the ground:

 5g \cdot (0.21)\cos 45 - 2T\cdot (0.81)\cos 25 = 0

 \implies 1.05g = 1.62T \implies T = \frac{1.05g}{1.62} \not=\frac{5g}{6} \ \square

so we have a contradiction, and so our assumption that the friction is limiting at both the contact points with the ground and the wall is false.

part 2:
Letting the distance of the particle from point A be X metres (see the diagram 2).

Resolving vertically  T+R = 10g
and resolving horizontally  R = 5T

 \implies 6T = 10g \implies T = \frac{5g}{3}

Taking moments about A, the point of contact between the rod and the ground:

 5g\cdot (0.21)\cos 45 + 5g\cdot X\cos 45 = 2T\cdot (0.81)\cos 45

 \implies 1.05g + 5gX = 2.7g \implies 5X = 1.65 \implies X = \boxed{0.33\mathrm{m}}
Attached files
0
reply
Oh I Really Don't Care
Badges: 15
Rep:
?
#55
Report 10 years ago
#55
II / 8

 \displaystyle - \int \frac{dy}{(y+3)^\frac{1}{2}} = -2 \int -2xe^{-x^2} dx \implies -2(y+3)^{\frac{1}{2}} = -2e^{-x^2} + C \implies y = e^{-2x^2} + 2ke^{-x^2} + (k^2 - 2)

After subbing (0,6) ;  \displaystyle (k+4)(k-2) = 0

 \displaystyle \therefore y = e^{-2x^2} - 8e^{-x^2} + 13 and case 2  \displaystyle  y = e^{-2x^2} + 4e^{-x^2} + 1

There is a vlaue of k such that as x tends to  \displaystyle \infty y tends to  \displaystyle 1

(ii) wip. Got
 \displaystyle  y^2e^{-3x^2} but the answer requires  \displaystyle ye^{-3x^2}
0
reply
Glutamic Acid
Badges: 16
#56
Report 10 years ago
#56
III/12:

P(3 or fewer | Holding a winning ticket) = P(3 or fewer AND holding a winning ticket)/P(holding a winning ticket).

3 or fewer and you win: you win and two others win; you win and one other wins; you win and no others win. You winning and other people winning is independent, so P(3 or fewer AND holding a winning ticket)/P(holding a winning ticket) = P(2 others or fewer win).

The others can be modelled as a binomial distribution, X ~ B(2n-1, 1/n)
P(X <= 2) = P(X = 2) + P(X = 1) + P(X = 0)
= \dbinom{2n-1}{2}\dfrac{1}{n^2} \left(1 - \dfrac{1}{n}\right)^{2n-3} + \dbinom{2n-1}{1}\dfrac{1}{n} \left(1 - \dfrac{1}{n}\right)^{2n-2} + \left(1 - \dfrac{1}{n}\right)^{2n-1}

= \left(1 - \dfrac{1}{n}\right)^{2n-3}\left[\dfrac{(2n-1)(2n-2)}{2!n^2} + \dfrac{(2n-1)}{n}\left(1 - \dfrac{1}{n}\right) + \left(1 - \dfrac{1}{n}\right)^2 \right]

= \left(1 - \dfrac{1}{n}\right)^{2n-3}\left[5 - \dfrac{8}{n} + \dfrac{3}{n^2}\right] = \left(1 - \dfrac{1}{n}\right)^{2n}(5 - \dfrac{8}{n} + \dfrac{3}{n^2}) \div \left(1 - \dfrac{1}{n}\right)^3 = \left(1 - \dfrac{1}{n}\right)^{2n} \dfrac{5n^2 - 3n}{n^2 - 2n + 1} = \left(1 - \dfrac{1}{n}\right)^{2n} \dfrac{5 - 3/n}{1 - 2/n + 1/n^2}

As n gets large, \left(1 - \dfrac{1}{n}\right)^{2n} \approx e^{-2} \text{ and }  \dfrac{5 - 3/n}{1 - 2/n + 1/n^2} \approx 5/1 = 5 So the probability is approximately 5e^{-2}.

5e^{-2} \approx 5(1 - 2 + \frac{4}{2} - \frac{8}{6} + \frac{16}{24} - \frac{32}{120} + \frac{64}{720} - \frac{128}{5040}) = \frac{41}{63} \approx \frac{2}{3}.

The last part is slightly dodgy, plus the arithmetic was rather tedious. I've been light on the algebra at several parts; none of it's particularly difficult.

- Human selections of numbers won't be uniform, and some will be more popular than others. So given that I have a winning ticket, it's likelier that more others will have a winning ticket so this probability will decrease. (Not sure if this is what they want or even correct.)

Consider the other people playing: X ~ B(2N - 1, 1/N) E(X) = "np" = 1/N*(2N-1) = 2 - N-1. So in total, given that you have a winning ticket, the expectation is 3 - N-1.
1
reply
Glutamic Acid
Badges: 16
#57
Report 10 years ago
#57
III/13:

If the first die to show a six does so on the rth roll then there'll (r-1) rolls where all dice fail to show a six = (q^{r-1})^n = q^{rn - n}

On the rth roll: P(1 or more shows a six) = 1 - P(none shows a six) = 1 - q^n. Overall probability = q^{rn-n}(1 - q^n) = q^{rn - n} - q^{rn} = q^{rn}(q^{-n}-1), as required.

Let G(t) denote the probability generating function. r can take values 1, 2, 3, ... so \text{G}(t) = q^n(q^{-n}-1)t + q^{2n}(q^{-n} - 1)t^2 + ... Geometric series, a = (1-q^n)t, r = q^nt. Sum to infinity = \dfrac{(1-q^n)t}{1-q^nt}.

\text{G'}(t) = (1-q^n)\left[\dfrac{tq^n}{(1-q^nt}^2} + \dfrac{1}{1-q^nt} \right] \text{ and } E(R) = G(1) = \dfrac{q^n}{1-q^n} + 1
When n = 2, p = 1/6, q = 5/6 so E(R) = \dfrac{25/36}{1 - 25/36} + 1 = \frac{25}{11} + 1 = \frac{36}{11}.

P(S = r) = P(S \le r) - P(S \le r - 1).
Less than or equal to r for one die is 1 - q^r; not n failures. So for n-dice: (1-q^r)^n, and similarly for r - 1: (1-q^{r-1})^n. Overall probability: (1-q^r)^n - (1-q^{r-1})^n

(1-q^r)^2 - (1-q^{r-1})^2 = q^{2r} - q^{2r-2} + 2q^{r-1} - 2q^r r can take values 1, 2, 3, ...

=> \displaystyle \text{G}(t) = \sum_{r=1}^{\infty}(q^{2r} - q^{2r-2} + 2q^{r-1} - 2q^r)t^r = \sum_{r=1}^{\infty} \left[ t^rq^{2r} - t^rq^{2r-2} + 2t^rq^{r-1} - 2q^rt^r \right]

= \dfrac{tq^2}{1-tq^2} - \dfrac{t}{1-tq^2} + \dfrac{2t}{1-tq} - \dfrac{2qt}{1-qt} = \dfrac{tq^2 - t}{1 - tq^2} + \dfrac{2t-2qt}{1-qt}.

\text{G'}(1) = \text{E}(S) = \dfrac{2}{1-q} = \dfrac{2}{p} = 12 when p = 1/6. (I haven't LaTeXed the derivative of G(t), it's too late and it's just tedious.)
0
reply
Oh I Really Don't Care
Badges: 15
Rep:
?
#58
Report 10 years ago
#58
... many of the questions for 2000 are from 1987 and 1990 papers ...
0
reply
Unbounded
Badges: 12
Rep:
?
#59
Report 10 years ago
#59
STEP I, Question 9, 2000
The value of tan(2theta)
Let the point at which the child fires the toy cannon from be P. Let us consider the situation relative to point P. Let d be the distance the shell travels horizontally. Let T be the time for the shell to land after being fired.

SUVAT vertically for the toy shell:
 s = ut + \frac{1}{2} at^2
 0 = VT\sin \theta - \frac{g}{2}T^2
 \implies T = \dfrac{2V\sin \theta}{g}

SUVAT horizontally for the toy shell:
 d = VT\cos \theta + \frac{a}{2}T^2

 \implies gd = V^2\sin 2\theta + \dfrac{2aV^2\sin^2 \theta}{g}

differentiating with respect to  \theta :

 g\dfrac{\mathrm{d}d}{\mathrm{d}\  theta} = 2V^2\cos 2\theta + \dfrac{2aV^2}{g}\sin 2\theta

For maximum range, d is max, so  \frac{\mathrm{d}d}{\mathrm{d}\th  eta} = 0

 \therefore 2V^2\cos 2\theta + \dfrac{2aV^2}{g}\sin 2\theta = 0

 \iff \cos 2\theta + \frac{a}{g} \sin 2\theta = 0

 \iff \boxed{\tan 2\theta = \frac{-g}{a}}
Proof for theta
 \theta &gt; 0

\tan 2\theta = \dfrac{-g}{a} \implies \dfrac{2\tan \theta}{1-\tan^2 \theta} = \dfrac{-g}{a}

 \implies \tan^2 \theta - \dfrac{2a}{g}\tan \theta - 1 = 0

 \implies \tan \theta = \dfrac{\frac{2a}{g} \pm \sqrt{\frac{4a^2}{g^2} + 4}}{2}

\implies \tan \theta = \frac{a}{g} + \sqrt{\frac{a^2}{g^2} +1}

As a is small compared to g,  \frac{a^2}{g^2} will be even smaller, so  \implies \tan \theta = \frac{a}{g} + 1
Lemma 1
Lemma 1:  \boxed{\tan \theta = 1 + y \implies \theta \approx \frac{\pi}{4} + \frac{y}{2} \ \mathrm{for \ small \ y.}}

Proof:
 \theta = \frac{\pi}{4} + \frac{y}{2} \iff\tan \theta = \tan (\frac{\pi}{4} + \frac{y}{2})

 \iff \tan \theta = \dfrac{1 + \tan \frac{y}{2}}{1-\tan \frac{y}{2}}

For small  \alpha ,  \tan \alpha \approx \alpha

 \iff \tan \theta \approx  \dfrac{1 + \frac{y}{2}}{1-\frac{y}{2}}

 \iff \tan \theta \approx \dfrac{2+y}{2-y}

 \iff \tan \theta \approx 1 + \dfrac{2y}{2-y}

So we need to show that  \dfrac{2y}{2-y} \approx y for small y:

 \dfrac{2y}{2-y} \approx y \iff \frac{2}{2-y} \approx 1 which is certainly true for small y.
Therefore we have  \tan \theta = 1 + \dfrac{a}{g} \implies \boxed{\theta \approx \dfrac{\pi}{4} + \dfrac{a}{2g}} \ \ \ \square
Proof for the range
 \theta \approx \frac{\pi}{4} + \frac{a}{2g}

 \implies \sin \theta \approx \dfrac{\sqrt{2}}{2}

 \implies \cos \theta \approx \dfrac{\sqrt{2}}{2}

 \implies \sin 2\theta \approx 1

from the original equations in the first part,  gd = V^2\sin 2\theta + \dfrac{2aV^2\sin^2 \theta}{g}

 \iff gd \approx V^2 + \dfrac{V^2a}{g}

\iff \boxed{d \approx \dfrac{V^2}{g} + \dfrac{V^2a}{g^2}} \ \ \ \square
2
reply
overclocked
Badges: 12
Rep:
?
#60
Report 10 years ago
#60
:lolwut:
1
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of Surrey
    Postgraduate Open Afternoon Postgraduate
    Wed, 23 Oct '19
  • University of Bristol
    Undergraduate Open Afternoon Undergraduate
    Wed, 23 Oct '19
  • University of Exeter
    Undergraduate Open Day - Penryn Campus Undergraduate
    Wed, 23 Oct '19

Have you made up your mind on your five uni choices?

Yes I know where I'm applying (162)
58.91%
No I haven't decided yet (66)
24%
Yes but I might change my mind (47)
17.09%

Watched Threads

View All
Latest
My Feed