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# STEP Prep Thread 2016 (Mark. II) watch

1. (Original post by jjsnyder)
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Did you use Binet's formula, I got a very weird answer :P What did you get?

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No just multiplied the series by 3 and added. Got 3/5. If you haven't already you might like to generalise it .
2. (Original post by IrrationalRoot)
Yeah and I also noticed that this argument was very easy to extend to m^n.
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The expression you get is really nifty because the denominator has roots the golden ratio and it's conjugate .
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root(5)/5? Nvm will check my algebra!

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3. (Original post by IrrationalRoot)
Yeah and I also noticed that this argument was very easy to extend to m^n.
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The expression you get is really nifty because the denominator has roots the golden ratio and it's conjugate .
(Original post by jjsnyder)
That's what I did

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nice. and lol yep I also looked at the general summation too. The whole reason fibonnaci numbers are 'neat' on the whole is that it all relies on the conjugate terms cancelling giving integers yea
4. (Original post by IrrationalRoot)
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No just multiplied the series by 3 and added. Got 3/5. If you haven't already you might like to generalise it .
and yep got the same luckily!
5. (Original post by EnglishMuon)
and yep got the same luckily!
In case you want to know this comes from a Mandelbrot competition.
There are some IMO problems in these books though. Like wtf isn't that a bit overkill?!
6. (Original post by IrrationalRoot)
In case you want to know this comes from a Mandelbrot competition.
There are some IMO problems in these books though. Like wtf isn't that a bit overkill?!
well it is a problem solving book
7. (Original post by EnglishMuon)
and yep got the same luckily!
I read this as 3/2 and was so confused why my answer was wrong! Got it, didn't use any geometric series though, had a slightly different method

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8. (Original post by jjsnyder)
I read this as 3/2 and was so confused why my answer was wrong! Got it, didn't use any geometric series though, had a slightly different method

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ah nice, what did u look at?
9. Last year's grade boundaries were pretty lower than usual for all I/II/III papers. Do you think they would try to increase the boundaries back to the usual norm by making the papers easier?

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10. (Original post by drandy76)
Remember someone(dfranklin?) made a really good post about necessary and sufficient conditions a while back, can any one link it to me?

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(Original post by DFranklin)
I've gone through this (briefly) before, and I'm not sure going over the ground directly again is going to help, but I'll try. Note first that I'm assuming you're working broadly follows the examiners hints/solutons - it might not make much sense otherwise.

By comparing coefficients to find relations between a,b,c,d,p,q,r,s and then eliminating p,q,r,s to get a condition on a,b,c,d, you've shown that if the quartic can be written as f(g(x)), then a,b,c,d must satisfy a given condition. In other words the condition on a,b,c,d is necessary.But you have not shown that just because a,b,c,d satisfy this condition, then it's possible to find p,q,r,s that make f(g(x)) = the quartic. In particular, the derivation of that condition doesn't even involve the constant coefficient, so for all you know, it's impossible to choose p,q,r,s such that the constant coefficient "works". So you need to prove it the other way (and the easiest way is to demonstrate values of p,q,r,s that work - particularly as for the end part you're probably going to need to find p,q,r,s for that specific polynomial anyhow).

I don't know if the following "modification" to the problem helps to make the distinction clearer.Suppose you had the extra restriction that r had to equal 0. (i.e. g(x) had to have the form x^2+s instead of x^2 +rx + s). If you do the same process of equating coefficients (starting from the highest power), we quickly find that we must have a = 0. So this is a necessary condition on the coefficients, but it's not sufficient because we haven't checked all the coefficients to find out that we need c = 0 as well.
This is III 2005, Q3.
11. (Original post by Insight314)
Last year's grade boundaries were pretty lower than usual for all I/II/III papers. Do you think they would try to increase the boundaries back to the usual norm by making the papers easier?

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Don't bother second guessing them.
12. One thing about the limits of integration, if you use the substitution sin^2x=(t^2)/(t^2+1). and the integral is in the limits pi/2 and 0 with respect to x. Can I say that sinx=t/(sqroot(t^2+1)) without the modulus sign as the limits considered of sinx are > or equal to 0?
13. (Original post by Zacken)
Don't bother second guessing them.
Yeah, I guess I shouldn't.

It's just that they normally increase boundaries after a large decrease.

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14. (Original post by Insight314)
Yeah, I guess I shouldn't.

It's just that they normally increase boundaries after a large decrease.

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STEP I 2014 and 2015 have approximately the same boundaries?
15. (Original post by Zacken)
STEP I 2014 and 2015 have approximately the same boundaries?
Supports my point of higher boundaries this year (for STEP I) even more. Check the chart with the grade boundaries in the OP. It seems like they usually increase the boundaries after 1,2,3 (only once happened) max lows.

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16. 2015 Paper 1 Q12 part ii:

According to the marking scheme, (sum m from 0 to inf) (6^m / m!) = e^6

Is the binomial expansion of e^x even part of the syllabus? Or can this be done without knowing it?
17. (Original post by lol456)
2015 Paper 1 Q12 part ii:

According to the marking scheme, (sum m from 0 to inf) (6^m / m!) = e^6

Is the binomial expansion of e^x even part of the syllabus? Or can this be done without knowing it?
You mean series expansion of e^x? Yes, it is in the spec for STEP I.

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18. (Original post by lol456)
2015 Paper 1 Q12 part ii:

According to the marking scheme, (sum m from 0 to inf) (6^m / m!) = e^6

Is the binomial expansion of e^x even part of the syllabus? Or can this be done without knowing it?
It's not the binomial expansion, it's the McLaurin/Taylor expansion, and it is part of the syllabus.
19. (Original post by jjsnyder)
I read this as 3/2 and was so confused why my answer was wrong! Got it, didn't use any geometric series though, had a slightly different method

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Yeah you didn't need geometric series, the way I did it was multiplying the series by 3 and then adding it to the original series and the result follows after some algebra.
20. (Original post by Insight314)
Supports my point of higher boundaries this year (for STEP I) even more. Check the chart with the grade boundaries in the OP. It seems like they usually increase the boundaries after 1,2,3 (only once happened) max lows.

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But according to you, STEP I 2015 should have had higher boundaries because STEP I 2014 bottomed out?

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