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    (Original post by IrrationalRoot)
    I believe that one of the 4k+1 and 4k+3 cases was somewhat elementary to prove and the other one was a lot more difficult.
    3 is easy, 1 is harder
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    (Original post by gasfxekl)
    3 is easy, 1 is harder
    do you know how to prove it for all n?
    It's not a step level question I'm so sure about this .
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    (Original post by Vesniep)
    do you know how to prove it for all n?
    It's not a step level question I'm so sure about this .
    yah
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    (Original post by Vesniep)
    I just don't remember about the 4l+1 proof . Elementary yes but to understand the proof not to think of it even the euclidean proof which is trivial to understand it's not that trivial to think of it .
    Also did you understand the limits thing you said 0/0 it is 0/infinity ok?
    Yeah no I only thought it was 0/0 because for some stupid reason I took n to be 0 when sinn was 0. It should be obvious now what I'm on about when I say I ruin my mocks with really stupid mistakes.
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    (Original post by IrrationalRoot)
    Yeah no I only thought it was 0/0 because for some stupid reason I took n to be 0 when sinn was 0. It should be obvious now what I'm on about when I say I ruin my mocks with really stupid mistakes.
    You should still consider the squeeze theorem because sinx doesn't have a limit to infinity of course.
    use for positive x that -1<sinx<1 thus -1/x <sinx/x <1/x , + -1/x tends to 0 so does your function
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    (Original post by gasfxekl)
    yah
    ok tell us , it's beyond me
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    (Original post by Vesniep)
    ok tell us , it's beyond me
    think intensively
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    (Original post by gasfxekl)
    think intensively
    Oh not now it's exams period not the best time for creative maths.
    I'm not even gonna study any number theory at all in university (physics and maths offerholder here) .
    I might think of it more in summer
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    (Original post by IrrationalRoot)
    STEP II 1992 Q1

    (iii) Don't get this one. Solutions say it's 0, which yeah ok at first glance you'd think that, but I thought \dfrac{\sin x}{x} becomes a \dfrac{0}{0} indeterminate form every 2\pi and so it appears that we do not have:

    'for every \epsilon &gt;0\ \exists\ N such that |f(x)|&lt;\epsilon\ \forall x&gt;N.

    Now I don't know whether there is a subtle extension to this limit to infinity definition that takes into account the domain of the function but I would think that this shows that the limit doesn't exist, which is what I put.

    Also I have to say the last limit (vi) was really tough especially after than many easy limits.
    Part vi is not that hard for someone who has done practice in limits .
    When you have to find limits with square roots you try to use ( srt(x) - sqrt (a) ) * ( sqrt(x) + sqrt (a)) = x-a so you have to multiply with conjugate so that you get rid of the sqrts I know that will leave other sqrts but they would have positive signs so no problem in finding them .
    the expression inside the limit is equal to 1/2 * ( (sqrt(1+2/n) +1) / (sqrt (1+1/n)+1 ) but sqrts tend to 1 so the fraction tends to 2/2 thus the answer is 1/2
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    IrrationalRoot

    Does Q4 of that paper reminds you of something ............
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    (Original post by Vesniep)
    Part vi is not that hard for someone who has done practice in limits .
    When you have to find limits with square roots you try to use ( srt(x) - sqrt (a) ) * ( sqrt(x) + sqrt (a)) = x-a so you have to multiply with conjugate so that you get rid of the sqrts I know that will leave other sqrts but they would have positive signs so no problem in finding them .
    the expression inside the limit is equal to 1/2 * ( (sqrt(1+2/n) +1) / (sqrt (1+1/n)+1 ) but sqrts tend to 1 so the fraction tends to 2/2 thus the answer is 1/2
    Of course the first thing I tried was multiplying top and bottom by the conjugate of the bottom but that didn't work.
    It took me a while but I did get the whole binomial expansion thing, it's just that it took more time than any of the other limits to see.

    No idea how you got the bit in bold though.
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    (Original post by Vesniep)
    IrrationalRoot

    Does Q4 of that paper reminds you of something ............
    Yep I noticed and it was extremely easy as well so definitely not doing that again.
    I swear this is the third time I've seen blatant duplicates of Qs. Considering how far back 2006 is I would think a large number of candidates would have done 1992 as practice and will have done that Q, which is pretty bad.
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    (Original post by IrrationalRoot)
    Of course the first thing I tried was multiplying top and bottom by the conjugate of the bottom but that didn't work.
    It took me a while but I did get the whole binomial expansion thing, it's just that it took more time than any of the other limits to see.

    No idea how you got the bit in bold though.
    Multiplying (sqrt(n+1)+sqrt(n))/(sqrt(n+2)+2) in the first expression so that we get that 1/2 which is (n+1-n)/(n+2-n)
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    (Original post by IrrationalRoot)
    Yep I noticed and it was extremely easy as well so definitely not doing that again.
    I swear this is the third time I've seen blatant duplicates of Qs. Considering how far back 2006 is I would think a large number of candidates would have done 1992 as practice and will have done that Q, which is pretty bad.
    You see it was quiet difficult back then to get all past papers so perhaps that's why they didn't care that much about repetition.
    Why do you think they repeat questions it's hard to think of 8 new pure questions restricted to the a level syllabus.
    Repetition is more common in applied maths I think the reason is obvious
    Edit : it's not exactly the same they used a triangle in 2006
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    Is it possible to predict Siklos the mad man's question choice for this year?

    Also, do you think some question may be repeated and how far back?


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    On STEP Easter School this year, Siklos presented a Hitler video of STEP III 2013 which was infested with complex numbers. He went into detail about how to calm down during exam IF we get complex numbers on this year's paper. Coincidence? I think not. The devil is in the details. Complex numbers is 8 questions this year. Watch.


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    (Original post by Insight314)
    On STEP Easter School this year, Siklos presented a Hitler video of STEP III 2013 which was infested with complex numbers. He went into detail about how to calm down during exam IF we get complex numbers on this year's paper. Coincidence? I think not. The devil is in the details. Complex numbers is 8 questions this year. Watch.


    Posted from TSR Mobile
    What advice did he give?
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    (Original post by KFazza)
    What advice did he give?
    "Relax"


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    (Original post by Insight314)
    "Relax"


    Posted from TSR Mobile
    That would make a difference even if you didn't practice complex numbers for STEPs ...
    In his booklets he tries to convince you to relax and just observe the question for a while so much that I can consider it a mild form of propaganda .
    I think he plans to introduce complex numbers in mechanics and statistics in all five questions as well
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    Insight314
    Link for the video ?
 
 
 
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