Arghh that's true, but it works out since in both instances, the other part is always equal. I hope I wouldn't lose too many marks, I really need to keep as many as I can :frown:

I will edit the solution in a bit.

(edited 9 years ago)

Reply 43

newblood

19

Original post by Brammer

Here is a solution to STEP III 2014 Q2

did we actually have to use partial fractions in part i or could we not use the result from the formula book for the artanh sub integral?

Reply 44

username937760

16

STEP II - Q13

Spoiler

Reply 45

Brammer

4

Original post by newblood

did we actually have to use partial fractions in part i or could we not use the result from the formula book for the artanh sub integral?

Yes - if the integral was provided in the formula book, then you could have used it without using partial fractions. Sorry if I complicated things unnecessarily.

Reply 46

jtSketchy

6

STEP III - Q1

Spoiler

(edited 9 years ago)

Reply 47

Brammer

4

Here is a solution to STEP II 2014 Q11

STEP II 2014 Q11.pdf52.5KB

Reply 48

edrraa

Original post by metaltron

Arghh that's true, but it works out since in both instances, the other part is always equal. I hope I wouldn't lose too many marks, I really need to keep as many as I can :frown:

I will edit the solution in a bit.

It's ok friend you are not alone in this feeling of regret and sadness.

Reply 49

DFranklin

18

STEP III, Q8:

For any 2 positive integers M < N, we have f(M) > f(M+1) > ... > f(N). The sum has (N+1-M) terms, all terms are >= f(N), and N-M terms are > f(N). So

\sum_{n=M}^N f(n) > \sum_{n=M}^N f(N) = (N+1-M) f(N)

and similarly

\sum_M^N f(n) < (M+1-N) f(M)[

.

Put

M = k^n, N = k^{n+1} - 1

.

(N+1-M) = k^n(k-1)

, so as long as k >2 or n > 0 we have N+1-M > 1. (We can't prove strict inequality in the case n=0, k=2, but if you plug these in, we do actually get equality on the RHS, so it's not actually a strict inequality in this case, which is a minor error in the question).

But assuming we do have (N+1-M) > 1, then M < N, and so by the above we have

k^n(k-1)f(k^{n+1}) < \sum_{r=k^n}^{k^{n+1}-1} f(r) < k^n(k-1)f(k^n)

as desired.

(i) Let f(n) = 1/n and k =2. Then we have

\displaystyle 2^n \dfrac{1}{2^{n+1}} < \sum_{r=2^n}^{2^{n+1}-1} \frac{1}{r} < 2^n \dfrac{1}{2^n}

, so

\displaystyle \dfrac{1}{2} < \sum_{r=2^n}^{2^{n+1}-1} \frac{1}{r} < 1

.

This gives us:

\displaystyle \sum_{n=0}^N \dfrac{1}{2} <\sum_{n=0}^N \sum_{r=2^n}^{2^{n+1}-1} \frac{1}{r} < \sum_{n=0}^N 1

and so

\displaystyle \frac{N+1}{2} < \sum_1^{2^{N+1}-1} \frac{1}{r} < N+1

as desired.(note that again we actually get equality when N = 0).

(ii) Now take f(n) = 1/n^3, k = 2. We get

\sum_{r=2^n}^{2^{n+1}-1} \frac{1}{r^3} < 2^n \dfrac{1}{2^{3n}} = 1/4^n

and so

\sum_{n=0}^N \sum_{r=2^n}^{2^{n+1}-1} \frac{1}{r^3} < \sum_{n=0}^N 1/4^n

so

\sum_{r=1}^{2^{N+1}-1} \frac{1}{r^3} < \sum_0^N (1/4)^n < \sum_0^\infty (1/4)^n = 4/3

and so finally

\sum_{r=1}^\infty \frac{1}{r^3} < 4/3

.

(iii) Assume we write all integers 0, 1, ..., 999 using leading zeros (e.g. "009" for 9). Then

S(1000) \cup \{0\}

is the set of all numbers where all 3 digits are NOT equal to 2. So there are 9 choices for each digit, so a total of 9^3 possibilities. Removing the "000" case (since 0 is not a positive integer) leaves 9^3 -1 as required.

Now for k>0

\rho(10^{k})-\rho(10^{k-1})

is the sum of the reciprocals of all k digit numbers without a 2. There are 8 choices for the leading digit, 9 choices for each of the other k-1 digits, so there are

8 \cdot 9^{k-1}

numbers of this form, and each number is at least

10^{k-1}

(and at least one is larger). So

\rho(10^{k})-\rho(10^{k-1}) < 8\cdot 9^{k-1}/(10^{k-1}) = 8\left(\frac{9}{10}\right)^{k-1}

.

So

\rho(10^k) < 8 \sum_1^k \left(\frac{9}{10}\right)^{k-1} < 8 \frac{1}{1-9/10} = 80

. Since

\rho(10^k) \geq \rho(n)

whenever 10^k > n, the result follows.

Note: I thought it more intuitive to do the end this way - it's probably not what they wanted, but it's essentially the same underneath. Also this was LaTeX hell, so there may well be typos.

(edited 9 years ago)

Reply 50

DFranklin

18

Original post by Khallil

II/4Maybe I'm missing something, but I don't see how you got that

\displaystyle \int_0^\infty \dfrac{1}{(1+u^2)^2}\,du = \dfrac{\pi}{4}

(and this does seem to be only a linear substitution away from the original integral, so is dangerously close to circular, IMHO).

Reply 51

user2020user

16

Original post by DFranklin

Maybe I'm missing something, but I don't see how you got that

\displaystyle \int_0^\infty \dfrac{1}{(1+u^2)^2}\,du = \dfrac{\pi}{4}

(and this does seem to be only a linear substitution away from the original integral, so is dangerously close to circular, IMHO).

DJ pointed it out to me a short while ago. I'll edit it in a few minutes with the method that he recommended.

(edited 9 years ago)

Reply 52

username219321

14

Poincare's inequality in II Q2, Method of characteristics in II Q5 and (vaguely) calculus of variations in III Q4 :O - there must have been a PDE theorist on the STEP-setting committee this year, gutted I didn't get to answer them before other people!

(edited 9 years ago)

Reply 53

mikelbird

8

Step 2 question 1 solution

Step2014Paper2Question1.pdf79.3KB

Reply 54

mikelbird

8

Original post by Stray

STEP II Q3

i)

Spoiler

ii)
a)

Spoiler

b)

Spoiler

iii)

Spoiler

I think (ii) (b) is a bit unsatisfactory....see file

Step2014Paper2Question3.pdf77.6KB

Reply 55

Stray

6

Original post by mikelbird

I think (ii) (b) is a bit unsatisfactory....see file

Thanks - I've improved that. For some reason my brain had decided that the variables couldn't be separated... duh.

Step iii q 3

STEP II, Q6:

sin(r+1/2)x - sin(r-1/2)x = sin rx cos x/2 + cos rx sin x/2 - sin rx cos x/2 + cos rx sin x/2 = 2 cos rx sin x/2.

So 2sin(x/2) (cos x + cos 2x + ... + cos nx) = sin(n+1/2)x - sin(n-1/2)x + sin(n-1/2)x - sin(n-3/2)x + ... + sin x/2 - sin x/2. All terms on RHS except first and last cancel, and then dividing by 2sin x/2 (since we are given this is non zero) we get

\displaystyle \sum_1^n \cos rx = \dfrac{\sin(n+\frac{1}{2})x - sin(\frac{1}{2}x)}{2\sin \frac{x}{2}}

as desired.

S_2(x) = sin(x) + sin(2x) / 2, so

\dfrac{d}{dx}S_2(x) = \cos x + \cos 2x = 2\cos\frac{3x}{2}\cos\frac{x}{2}

. cos(x/2) is never zero for 0<=x<=pi, so the only root is where cos(3x/2) =0, i.e. 3x/2 = pi/2 + 2mpi, so x = pi/3 + 4/3m pi. If 0<=x<=pi the only possibilities is x = pi/3.

http://www5a.wolframalpha.com/Calculate/MSP/MSP2431d8ahdi9gcfi21ci000036di7dcag60dg904?MSPStoreType=image/gif&s=53&w=325.&h=150.&cdf=RangeControl

\displaystyle \dfrac{d}{dx} s_n(x) = \sum_1^n \frac{1}{r} r \cos rx = \sum_1^n \cos rx = \dfrac{\sin(n+\frac{1}{2})x - sin(\frac{1}{2}x)}{2\sin \frac{x}{2}}

This has stationary points when sin(n+1/2)x - sin(1/2)x = 0. We can rewrite this as sin nx cos x/2 + cos nx sin x/2 - sin x/2 = 0, or sin(x/2)[(cos nx -1) + sin nx cot x / 2 ]= 0. We are given 0 < x < pi and so sin(x/2) is non-zero, so we can divde by it. Rearranging we get (1-cosnx) = sin nx cot x/2 and then multiplying by tan x/2 gives

(1 -\cos nx)\tan \frac{x}{2} = \sin nx

. So if x = x_0 is a stationary point we have

(1-\cos n x_0) \tan \frac{x_0}{2} = \sin nx_0

as desired.

Note that since

(1-\cos n x_0)\geq 0

and

\tan(x_0/2)\geq 0

we must have

\sin(nx_0) \geq 0

and so

S_n(x_0)\geq S_{n-1}(x_0)

(since

S_n(x_0) = S_{n-1}(x_0) + \frac{1}{n}\sin n x_0

).

Suppose

S_{n-1}(x) .> 0

for all x in (0, pi). Now consider S_n. Since S_n(0)=S_n(pi) =0, S attains a minimum value at some point x_0 in (0, pi). Then by the argument above,

S_n(x_0) \geq S_n-1(x_0) > 0

as desired.

Finally, suppose (for contradiction) that we can find

n \geq 1

and

x\in[0,\pi]

s.t. S_n(x) < 0. Wlog, we can choose the smallest such n. Note that n cannot be 1, since sin(x) >=0 for all x in [0,pi]. So n > 1, and since n is chosen minimal, we must have