Getting to Cambridge: STEP by STEP!

Reply 160

Insight314

7

Original post by Zacken

'innit. I'm putting a break on STEP till I'm done with FP3, so... :tongue:

I've got a few more pages and I'll be done with the whole of integration. :yep:

Oh yeah, 'cause you are doing one STEP paper a day right? I've been doing one question a day for months now (I think) so I don't think I need to stop now. :biggrin:

You could do one question a day as well, if you want. It will probably only take you 20-30 minutes, haha! :biggrin:

Reply 161

Zacken

OP

22

Original post by Insight314

Oh yeah, 'cause you are doing one STEP paper a day right? I've been doing one question a day for months now (I think) so I don't think I need to stop now. :biggrin:

You could do one question a day as well, if you want. It will probably only take you 20-30 minutes, haha! :biggrin:

Nah, I do one STEP paper a week. Like I said, my work ethic is shite. :lol:

Reply 162

Zacken

OP

22

Original post by 雷尼克

because i want to go there and this thread is making me cringe

You probably won't meet your offer for Warwick anyway tbh, not sure why you're worried.

(edited 8 years ago)

Reply 163

Gregorius

14

Original post by Zacken

Took me far too long to spot the

-x^2 = 1 - x^2 - 1

trick. Anyways, this requires two bouts of integration, the first one yields, with

u = (1-x^2)^n

and

\displaystyle v = \frac{1}{\alpha}\sin \alpha x

:

Unparseable latex formula:

\displaystyle [br]\begin{equation*} \frac{\alpha I_n}{2n} = \int_{-1}^{1} x(1-x^2)^{n-1} \sin \alpha x \, \mathrm{d}x[br]\end{equation}

Now, we bash it with IBP again, this time with

u = x(1-x^2)^{n-1}

and

Unparseable latex formula:

\displaystlye v = -\frac{1}{\alpha} \cos \alpha x

which gives us:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\frac{\alpha^2}{2n}I_n = I_{n-1} + 2(n-1)\int_{-1}^{1} \cos \alpha x (1-x^2)^{n-2} (-x^2) \, \mathrm{d}x \end{equation*}

Clever bit:

Unparseable latex formula:

\displaystyle [br]\begin{equation*} \frac{\alpha^2}{2n}I_n = I_{n-1} + 2(n-1)\int_{-1}^{1} \cos \alpha x (1-x^2)^{n-2} (1-x^2 - 1) \, \mathrm{d}x \end{equation*}

So:

Unparseable latex formula:

\displaystyle [br]\begin{equation*} \frac{\alpha^2}{2n}I_n = (2n-1)I_n - 2(n-1)I_{n-2} \iff \alpha^2 I_n = 2n(2n-1)I_{n-1} - 4n(n-1)I_{n-2} \end{equation*}

as required. Lovely.

Bravo! Now, show that this implies that

\alpha^{2n+1} I_n = n! (P_n \sin \alpha + Q_n \cos \alpha)

where

P_n, Q_n

are polynomials in

\alpha

of degree < 2n+1 with integer coefficients.

Hence, show that

\pi

is irrational. Hint: put

\alpha = \pi/2

and by assuming that

\pi = b/a

show that

\displaystyle J_n = \frac{b^{2n+1}I_n}{n!}

is an integer that tends to zero as

n \rightarrow \infty

Reply 164

Zacken

OP

22

Original post by Gregorius

Bravo! Now, show that this implies that

\alpha^{2n+1} I_n = n! (P_n \sin \alpha + Q_n \cos \alpha)

where

P_n, Q_n

are polynomials in

\alpha

of degree < 2n+1 with integer coefficients.

Hence, show that

\pi

is irrational. Hint: put

\alpha = \pi/2

and by assuming that

\pi = b/a

show that

\displaystyle J_n = \frac{b^{2n+1}I_n}{n!}

is an integer that tends to zero as

n \rightarrow \infty

Just kidding, looks lovely, prime procrastination material. :yep:

Reply 165

Zacken

OP

22

Update:

I've put away Greg's problem/extension for a bit, I'm far too groggy to think through it properly. In any case, I think I've finished the integration chapter in FP3, I just need to work through a few more exercises now.

Reply 166

Kevin De Bruyne

21

love the title, and inspirational back story. Must take some time to read through the 9 pages.

Good luck! (not that you'll need it...)

:rave:

Reply 167

Zacken

OP

22

Original post by SeanFM

love the title, and inspirational back story. Must take some time to read through the 9 pages.

Good luck! (not that you'll need it...)

:rave:

Woot! Thanks. Good to have you onboard. :biggrin:

Reply 168

Ayman!

15

Original post by SeanFM

zacken zacken, he's our man if he can't do it, no one can!

https://www.youtube.com/watch?v=emKHcHZ_A2E

Reply 169

Zacken

OP

22

Original post by aymanzayedmannan

zacken zacken, he's our man if he can't do it, no one can!

https://www.youtube.com/watch?v=emKHcHZ_A2E

Brilliant. :rofl:

Reply 170

FriendlyPenguin

20

Original post by Zacken

Nah, I do one STEP paper a week. Like I said, my work ethic is shite. :lol:

That's still a question a day... and you know what they say... a question a day keeps Warwick away

Original post by Gregorius

Bravo! Now, show that this implies that

\alpha^{2n+1} I_n = n! (P_n \sin \alpha + Q_n \cos \alpha)

where

P_n, Q_n

are polynomials in

\alpha

of degree < 2n+1 with integer coefficients.

Hence, show that

\pi

is irrational. Hint: put

\alpha = \pi/2

and by assuming that

\pi = b/a

show that

\displaystyle J_n = \frac{b^{2n+1}I_n}{n!}

is an integer that tends to zero as

n \rightarrow \infty

Nice problem

Reply 171

Zacken

OP

22

Original post by Mathemagicien

That's still a question a day...

Well... not really. I do the STEP paper in one go, so nothing for 6 days then 7 questions in one day.

and you know what they say... a question a day keeps Warwick away

Nice problem

Have you done it?

Reply 172

Zacken

OP

22

Update:

I'll do a proper update one of these days, with a nice long blog post. Anywho, think I'm going to get my FP2 June 2011 mock going now. Wish me luck!

(edited 8 years ago)

Reply 173

Ayman!

15

Original post by Zacken

Update:

I'll do a proper update one of these days, with a nice long blog post. Anywho, think I'm going to get my FP2 June 2011 mock going now. Wish me luck!

Good luck

Reply 174

Zacken

OP

22

FP2 Mock (June 2011):

So let's do some analysis (the boring kind, not the maths kind):

Total time: 59:20 Total raw: 75 Total UMS: 100

Q1: inequalities. Took longer than I should because I did it in two different ways to check my answer. I'll need to improve my speed here. 9:32
Q2: Taylor series solutions to DE's took far longer than I should have fixing a silly mistake. 6:05
Q3: Easy first order DE, happy with my time. 3:45
Q4: Telescoping sums, easy, took me longer than I'd like because I work slowly to avoid silly mistakes, it was an easy one as well, need to cut down on my time. 7:22
Q5: Complex loci, my solution was far longer than need be, not pleased at all. 10:07
Q6: Polar co-ordinates, mildly enjoyable. Did well on this one. 07:03
Q7: Easy C3 stuff, took me long because I graphed to see if I got all my solutions. 8:07
Q8: Easy second order DE, surprised I didn't make a silly slip, pleased with my time. 07:04

All-around, I think I did okay. Certainly lots of space for improvement with regards to my timing issue.