STEP Prep Thread 2017

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    (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:
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    (i) Here I can see that I can substitute the expression for x_n into x_{n-1}. This would give me:

    \displaystyle x_{n-1}=1+\frac{1}{1+\frac{1}{x_1}}=1  +\frac{x_1}{1+x_1}=\frac{2x_1+1}  {x_1+1}

    Now I am told that x_1 is a positive number, therefore I can deduce that x_{n-1}>1

    Therefore: 1+\frac{1}{x_n}>1 \longrightarrow \frac{1}{x_n}>0 for any n (though I get confused here because originally it says n is greater than or equal to 3. Maybe I should've used a different variable such as m where 1\leq m\leq n?)

    From here it is evident that x_1 , x_2 , ... , x_n > 1 as for each one you are adding something greater than 0 onto 1. So as far as I am concerned this is now shown?


    (ii) This just proved to be some simple algebraic manipulation. I knew I can express x_1 in terms of x_2, and x_2 in terms of x_3

    So;

    \displaystyle x_1-x_2=(1+\frac{1}{x_2})-(1+\frac{1}{x_3})

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

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

    (iii) Here I use my answer to (ii) and rearrange it to get:

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

    and for what they're asking to hold true I must show that the numerator on the fraction is 0.

    Using the fact that x_n=1+\frac{1}{x_1} I know it should hold for some n_k and n_{k+1} where 1\leq k \leq n

    Since x_k and x_{k+1} equate to the same expression in terms of x_1 then they must be equal to one another.

    Therefore x_k=x_{k+1} \longrightarrow x_k-x_{k+1}=0

    If we set k=2 then our fraction will indeed be equal to 0 and we get x_1=x_2

    We can play the same game and say that k is any number from 1 to n and that would mean:

    x_1=x_2=x_3=...=x_n

    So I think this is now proven?


    (iiii) Solving for x_1 is quite straight forward as x_k=1+\frac{1}{x_1}

    Setting k=1 we find that (x_1)^2-x_1-1=0 which leads us to:

    x_1=\frac{1+\sqrt5}{2} after disregarding the negative root as that would make x_1 negative. Interestingly enough this whole question is about the golden ratio. Nice stuff.
    (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.
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    (Original post by solC)
    Thanks a bunch for making this thread, 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.
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    (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?
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    (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 :eek:

    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
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    (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.

    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?
    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} so 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...
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    (Original post by solC)
    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

    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.
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    (Original post by Zacken)
    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.
    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.
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    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)
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    (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
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    (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.
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    (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)
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    a substitution to turn it into a standard integral
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    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.
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    (Original post by 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.
    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
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    (Original post by 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.
    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.
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    (Original post by Zacken)
    ******
    sssshhh ^.^ (or at least spoiler tag it!)
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    (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 -_-
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    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.



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    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?)
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    (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?)
    When I was doing FM I would do a good bit of work from the hours of 10pm onwards. Something about working at night is so relaxing :laugh:
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    I'm sure it doesn't matter that much but you can include the grade boundaries for the 2016 papers as well in the main bit of the post.
 
 
 
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