STEP Prep Thread 2017

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Reply 100

Zacken

Original post by RDKGames

Sorry in advance if this is the wrong thread to ask this question.

Of course it's the right thread.

My solution:

Spoiler

(I) I'm not sure what you're doing here. Sure, you end up with 1/xn > 0 so 1 + 1/xn > 0 for all n. But through a very long route. Why not just

xi > 0 \Rightarrow \frac{1}{x_i} > 0 \Rightarrow 1 + \frac{1}{x_i} > 1

.

(ii) is fine.

(iiii) is fine.

(iii) going for dinner rn so can;t check this, but why does it hold for some nk and nk+1 that xnk - xnk+1 = 0? Surely that's assuming the result? Are you trying a proof by induction? Doesn't look like it. I dunno, can you clarify?

there's a slightly better way to do it, kee writing x1-x2 = f(x2, x3) = f(x2, x3, x4) = ... till you get back to x1.

Reply 101

Zacken

Original post by solC

Thanks a bunch for making this thread :biggrin:

, gonna start working through the Siklos booklet next week i think. Hopefully i'll be comfortable with STEP 1 by around December.
How difficult are the questions asked in interview compared to say STEP 1?

They vary. One of my interview questions was straight from a STEP III paper (prove that e is irrational) except with all the guiding parts removed.

You'll be fine! There's no need to be fully comfortable with STEP I by december, but being able to attempt 2/3 questions from a given paper would definitely be a good idea.

Reply 102

RDKGames

Study Forum Helper

Original post by Zacken

(I) I'm not sure what you're doing here. Sure, you end up with 1/xn > 0 so 1 + 1/xn > 0 for all n. But through a very long route. Why not just

xi > 0 \Rightarrow \frac{1}{x_i} > 0 \Rightarrow 1 + \frac{1}{x_i} > 1

But how do I know that

x_i>0

? Or does that come from their information that these numbers are all positive? Because if that's the case then I did not think of it that way when I did this question.

(iii) going for dinner rn so can;t check this, but why does it hold for some nk and nk+1 that xnk - xnk+1 = 0? Surely that's assuming the result? Are you trying a proof by induction? Doesn't look like it. I dunno, can you clarify?

there's a slightly better way to do it, kee writing x1-x2 = f(x2, x3) = f(x2, x3, x4) = ... till you get back to x1.

I wasn't quite sure about this entirely but it made sense to me the way I did it. It is not by induction, though, and I thought about induction but couldn't construct a proper proof out of it so I disregarded it. Now that you mention it I'm not quite sure how I can go about it showing that it would hold for

n_k

and

n_{k+1}

. Can you explain the slightly better way?

Reply 103

solC

Original post by Zacken

They vary. One of my interview questions was straight from a STEP III paper (prove that e is irrational) except with all the guiding parts removed.

You'll be fine! There's no need to be fully comfortable with STEP I by December, but being able to attempt 2/3 questions from a given paper would definitely be a good idea.

Oh wow

What years should i start with when going through papers? I was thinking about first doing 2000-2008/9 and then the older ones whilst leaving the latest ones for timed practise, or should i start way back and just go from there?
Thanks again :smile:

Reply 104

Zacken

Original post by RDKGames

But how do I know that

x_i>0

? Or does that come from their information that these numbers are all positive? Because if that's the case then I did not think of it that way when I did this question.

Yes, a number being positive means that it is > 0. So if all

x_i

are positive then all

x_i > 0

n_k

and

n_{k+1}

. Can you explain the slightly better way?

You write

x_1 - x_2 = \frac{x_3 - x_2}{x_2x_3}

but

x_2 - x_3 =-\frac{x_4 - x_3}{x_3 x_4}

x_1 - x_2 = \frac{x_3 - x_4}{x_2 x_3^2 x_4}

and keep going till you get

x_1 - x_2 = \pm \frac{x_1 -x_2}{x_2 x_3^2 x_4^2 \cdots x_n^2 x_1^2 x_2}

frmo which you can get

x_1 = x_2

and you can then continue that to get

x_2 = x_3

, etc...

Reply 105

Zacken

Original post by solC

1998 is a good paper to start with (easy). Then space out STEP I 2000-2013 from now till mid-Jan. You'll start II and III in mid-Jan assuming you get an offer. Then you can do STEP I 2014-16 as proper timed mocks at the end of May/beginning of June assuming you're sitting it.

Reply 106

ValerieKR

Original post by Zacken

I would say give a few STEP 2 and 3 questions a go before the interview too - some different techniques are required and may come in handy in that interview.

Reply 107

11234

step III q1 2004 for the second bit with coshx/1+2sinh^2x, I got an expression involving complex numbers, can anyone tell me if I'm on the right lines,(I don't want to look at any solutions)

Reply 108

solC

Original post by ValerieKR

I would say give a few STEP 2 and 3 questions a go before the interview too - some different techniques are required and may come in handy in that interview.

Alright, will do.
Thanks :smile:

Reply 109

Zacken

Original post by 11234

step III q1 2004 for the second bit with coshx/1+2sinh^2x, I got an expression involving complex numbers, can anyone tell me if I'm on the right lines,(I don't want to look at any solutions)

Nope, you're woefully off-kilter. You're overthinking it - it's meant to be done in one line.

Reply 110

ValerieKR

Original post by 11234

step III q1 2004 for the second bit with coshx/1+2sinh^2x, I got an expression involving complex numbers, can anyone tell me if I'm on the right lines,(I don't want to look at any solutions)

How did you get to complex numbers? Through trying a partial fractions approach?

I don't know if there's an easier way but I seem to remember doing it with:
(not an answer but a correct approach)

Spoiler

Reply 111

11234

Is there any relationship between coshx and sinhx like cos(90-x)=sinx I know theres one with cosh^2x-sinh^2x=1 but are there any others like a hyperbolic addition formula/factor formula out of interest.

Reply 112

ValerieKR

Original post by 11234

cosh(x) = 1/2(e^x+e^-x)
sinh(x) = 1/2(e^x-e^-x)
There is a cosh(x)=sinh(f(x)) (I think f(x) = cosh(arcsinh(x)) is the only one (you can find out what it is by plugging the logarithmic form of arcsinh into cosh)) :s but generally the same macro expressions will hold as do between sin and cos (except sin^2 or tan^2 becomes -sin^2 or - tan^2)

cosh(x)+sinh(x) = e^x
cosh(x)-sinh(x) = e^-x is another one I guess?

also cosh, sinh, cos and sin are all related through 'i' in various ways

(edited 7 years ago)

Reply 113

Zacken

Original post by 11234

For your question, remember that the derivative of sinh is sitting in the numerator, so a suitable substitution like u = s... would make things v. tidy.

Reply 114

ValerieKR

Original post by Zacken

******

sssshhh ^.^ (or at least spoiler tag it!)

(edited 7 years ago)

Reply 115

11234

Original post by Zacken

For your question, remember that the derivative of sinh is sitting in the numerator, so a suitable substitution like u = s... would make things v. tidy.

Thanks, I feel so silly now -_-

Reply 116

username1162972

Hello.
Good luck with step.
Just do loads of maths be it step MAT and olympiads as you will get a mix at interview. I had something from a Numbers and sets example sheet both times I was interviewed!
Practice is alot of it.

Posted from TSR Mobile

Reply 117

ValerieKR

Anybody else feverently doing STEP at weird times in nervous hysteria?
(I have just spent the hours of 10pm-1am doing questions without realising that it's not actually daytime?)

Reply 118

DylanJ42

Original post by ValerieKR

Anybody else feverently doing STEP at weird times in nervous hysteria?
(I have just spent the hours of 10pm-1am doing questions without realising that it's not actually daytime?)