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 :smile:

Reply 1003

EnglishMuon

Original post by IrrationalRoot

Spoiler

and yep got the same luckily!

Reply 1004

IrrationalRoot

Original post by EnglishMuon

and yep got the same luckily!

In case you want to know this comes from a Mandelbrot competition. :smile:

There are some IMO problems in these books though. Like wtf isn't that a bit overkill?!

Reply 1005

gasfxekl

Original post by IrrationalRoot

In case you want to know this comes from a Mandelbrot competition. :smile:

There are some IMO problems in these books though. Like wtf isn't that a bit overkill?!

well it is a problem solving book

Reply 1006

jjsnyder

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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Reply 1007

EnglishMuon

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?

Reply 1008

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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Reply 1009

Zacken

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.

Reply 1010

Zacken

Original post by Insight314

Don't bother second guessing them. :tongue:

Reply 1011

Geraer100

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? :smile:

Reply 1012

Insight314

Original post by Zacken

Don't bother second guessing them. :tongue:

Yeah, I guess I shouldn't. :tongue:

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

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Reply 1013

Zacken

Original post by Insight314

Yeah, I guess I shouldn't. :tongue:

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?

Reply 1014

Insight314

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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Reply 1015

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?

Reply 1016

Insight314

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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Reply 1017

Zacken

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.

Reply 1018

IrrationalRoot

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.