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50 years of hurt this summer 1966-2016

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Reply 20
Original post by saule1116
They don't look too scary. :biggrin:


naturally they do not
(but of course you provide no solution ...)
Reply 21
Original post by Compost
That's a bit dramatic, it's only an S level paper for A level - so the equivalent of an AEA.


My comment is dramatic indeed but merely to attract attention to the thread.

However your reply is at best "silly" and at worst "utter nonsense".
If this paper was equivalent to an AEA the distinction grade boundary would have been well under 50%.

I do not know what your expertise is but I can tell you in recent years I have taught a handful of candidates which took STEP or AEA with excellent results.
I can assure you this paper (which you only see part of) would have made mince meat of any of them.
Reply 22
Original post by physicsmaths
That is a step question from step I. It has come up twice!


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This is not a STEP question
this is from 1966 which predates STEP
Original post by TeeEm
This is not a STEP question
this is from 1966 which predates STEP


Yes I do know that, just pointing out it came up on step after. But if it came up in step it is a step question.


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Reply 24
Original post by physicsmaths
Yes I do know that, just pointing out it came up on step after. But if it came up in step it is a step question.


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still no actual attempt on the red ones apart from atsruser's answer (still have not done it myself)
Original post by TeeEm
still no actual attempt on the red ones apart from atsruser's answer (still have not done it myself)


Yes, I have not had time to try them yet, I probably won't because I am ona break for abit. Maybe I can't do it Maybe I can, I won't be bothered if I can't.


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Reply 26
Original post by physicsmaths
Yes, I have not had time to try them yet, I probably won't because I am ona break for abit. Maybe I can't do it Maybe I can, I won't be bothered if I can't.


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Do you expect to get into Cambridge with this attitude?
What if the admission tutor is here (undercover) and reads your last post?
Original post by TeeEm
Do you expect to get into Cambridge with this attitude?
What if the admission tutor is here (undercover) and reads your last post?


I don't expect to get in again. If these are horrible beast questions, why would be bothered if I can't do one??????



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Reply 28
Original post by physicsmaths
I don't expect to get in again. If these are horrible beast questions, why would be bothered if I can't do one??????



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you don't expect to get in?

did you have a bad interview?
Original post by TeeEm
you don't expect to get in?

did you have a bad interview?

Kind of, it was harder then I thought and chose the wrong questions in the test so my fault. The other interview went much better so hopefully that saves me. But for someone on a gap year it might not be enough. But im hoping for the best, preparing for the worst. It might not hurt as much if I prepare myself for not getting in. I just don't want it to consume my mind for the next month. Sorry if I come across as rude, I am just stressed from yesterday due to my stupid mistakes.


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Reply 30
Original post by physicsmaths
Kind of, it was harder then I thought and chose the wrong questions in the test so my fault. The other interview went much better so hopefully that saves me. But for someone on a gap year it might not be enough. But im hoping for the best, preparing for the worst. It might not hurt as much if I prepare myself for not getting in. I just don't want it to consume my mind for the next month. Sorry if I come across as rude, I am just stressed from yesterday due to my stupid mistakes.


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Well you did not across as rude and I do not know what happened yesterday.
(Most of the time I show logged in but rarely I am behind the screen to follow what is going on)
Original post by TeeEm
However your reply is at best "silly" and at worst "utter nonsense"


AEAs were the replacement for S levels, hence, in some sense at least, it is equivalent to an AEA. I accept that doesn't take into account the fact that A levels have been considerably simplified over the past 50 years.
Original post by TeeEm
Well you did not across as rude and I do not know what happened yesterday.
(Most of the time I show logged in but rarely I am behind the screen to follow what is going on)


That is fine, I will try the questions though. Especially the tan one, that seems very interesting.


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Reply 33
Original post by physicsmaths
That is fine, I will try the questions though. Especially the tan one, that seems very interesting.


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sure it is

the first (in red) I wonder if there is a typo.
then again I have no idea what breadth of knowledge you are meant to have 50 years ago when entered for such paper
Original post by TeeEm
In the summer of 1966 England won the World Cup.
A few days before the final the best mathematicians in the country had to face this test
Any thoughts particularly on the "red" questions?


They are interesting questions. My solution for Q2 (i) is as follows.

Observe that tanh1x2=n=0(x2)2n+12n+1=n=1(x2)2n12n1 \displaystyle \tanh^{-1}x^2 = \sum^{\infty}_{n=0} \dfrac{(x^2)^{2n+1}}{2n+1} = \sum^{\infty}_{n=1} \dfrac{(x^2)^{2n-1}}{2n-1} .

Hence we have that

2n=112n1(pqp+q)4n2=2tanh1((pqp+q)2) \displaystyle 2 \sum^{\infty}_{n=1} \dfrac{1}{2n-1}\left(\dfrac{\sqrt{p}-\sqrt{q}}{\sqrt{p}+\sqrt{q}} \right)^{4n-2} = 2 \tanh^{-1} \left( \left(\dfrac{\sqrt{p}-\sqrt{q}}{\sqrt{p}+\sqrt{q}} \right)^2\right) .

Recall that tanh1x=12ln1+x1x \tanh^{-1} x = \dfrac{1}{2} \ln\dfrac{1+x}{1-x} . Hence with some algebraic manipulation we arrive at

2n=112n1(pqp+q)4n2=ln(p+q2pq)=ln(p+q2)(lnp+lnq2) \displaystyle 2 \sum^{\infty}_{n=1} \dfrac{1}{2n-1}\left(\dfrac{\sqrt{p}-\sqrt{q}}{\sqrt{p}+\sqrt{q}} \right)^{4n-2} = \ln \left( \dfrac{p+q}{2 \sqrt{pq}} \right) = \ln \left(\dfrac{p+q}{2}\right) - \left( \dfrac{\ln p + \ln q}{2} \right) as required.
Reply 35
Original post by ThatPerson
They are interesting questions. My solution for Q2 (i) is as follows.

Observe that tanh1x2=n=0(x2)2n+12n+1=n=1(x2)2n12n1 \displaystyle \tanh^{-1}x^2 = \sum^{\infty}_{n=0} \dfrac{(x^2)^{2n+1}}{2n+1} = \sum^{\infty}_{n=1} \dfrac{(x^2)^{2n-1}}{2n-1} .

Hence we have that

2n=112n1(pqp+q)4n2=2tanh1((pqp+q)2) \displaystyle 2 \sum^{\infty}_{n=1} \dfrac{1}{2n-1}\left(\dfrac{\sqrt{p}-\sqrt{q}}{\sqrt{p}+\sqrt{q}} \right)^{4n-2} = 2 \tanh^{-1} \left( \left(\dfrac{\sqrt{p}-\sqrt{q}}{\sqrt{p}+\sqrt{q}} \right)^2\right) .

Recall that tanh1x=12ln1+x1x \tanh^{-1} x = \dfrac{1}{2} \ln\dfrac{1+x}{1-x} . Hence with some algebraic manipulation we arrive at

2n=112n1(pqp+q)4n2=ln(p+q2pq)=ln(p+q2)(lnp+lnq2) \displaystyle 2 \sum^{\infty}_{n=1} \dfrac{1}{2n-1}\left(\dfrac{\sqrt{p}-\sqrt{q}}{\sqrt{p}+\sqrt{q}} \right)^{4n-2} = \ln \left( \dfrac{p+q}{2 \sqrt{pq}} \right) = \ln \left(\dfrac{p+q}{2}\right) - \left( \dfrac{\ln p + \ln q}{2} \right) as required.


that is absolutely ludicrous...
I mean what possessed you to start from artanhx2?
I tried 5 minutes ago but from the Arithmetic means and tried in vain expansions such ln(1+x)

you must be truly amazing to see this
(edited 8 years ago)
Original post by TeeEm
that is absolutely ludicrous...
I mean what possessed you to start from artanhx2?
I tried 5 minutes ago but from the Arithmetic means and tried in vain expansions such ln(1+x)

you must be truly amazing to see this


At first I started the same as you, but then I couldn't see where the 2n-1 on the denominator came from. Then I thought of the A-Level functions to see where I could get such a series expansion and artanh(x) seemed to fit :biggrin:.
Reply 37
Original post by ThatPerson
At first I started the same as you, but then I couldn't see where the 2n-1 on the denominator came from. Then I thought of the A-Level functions to see where I could get such a series expansion and artanh(x) seemed to fit :biggrin:.


excellent!

It now retrospective since I have now seen how it is done but I still think maybe a student in those days had seen similar questions.

I cannot fathom how you could do in 3 hours 8 questions (Paper has choice out of 10) of which this question is only a part of a question.

I think I am getting too old for this ...
Original post by ThatPerson
At first I started the same as you, but then I couldn't see where the 2n-1 on the denominator came from. Then I thought of the A-Level functions to see where I could get such a series expansion and artanh(x) seemed to fit :biggrin:.


I was just about to post some hints for this question, but you beat me to it! Well done.

As one who did "S" level maths (ten years after this particular paper), I would claim that this sort of question was quite typical of the genre.

It isn't actually all that hard, provided you know some stuff. In this case, if you know the series expansions for the trig, inverse trig, hyperbolic and inverse hyperbolic functions, then you simply recognize the series and the rest is straightforward algebraic manipulation.

But the other point to recognize is that the series is there simply to catch those who haven't memorized their series! In those days of yore, one simply learned these sorts of series expansion off by heart.
Original post by Gregorius

It isn't actually all that hard, provided you know some stuff. In this case, if you know the series expansions for the trig, inverse trig, hyperbolic and inverse hyperbolic functions, then you simply recognize the series and the rest is straightforward algebraic manipulation.

But the other point to recognize is that the series is there simply to catch those who haven't memorized their series! In those days of yore, one simply learned these sorts of series expansion off by heart.


You've made the point that I intended to.

Many of these older questions look forbidding due to the lack of practice that we get with them these days. Back in the day, for example, there was far more emphasis on the coordinate geometry of conic sections, which aren't too tricky when you've spent 2 years working similar problems, but would be considered utterly incomprehensible by most students now - that's simply a lack of familiarity - they are no more difficult.

In addition, anyone who was sitting an S paper would have seen similar questions before, and would have picked up a few tricks, rather like these days all the young bloods taking STEP can do the xπ2xx \to \frac{\pi}{2}-x trick in their heads when integrating.

A lot of it comes down to changing tastes in what is taught. In the 1940s, most students would not have coped with a tricky problem in vectors, for example, due to the fact that it hadn't really entered the syllabuses.

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