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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 >0\ \exists\ N such that |f(x)|<\epsilon\ \forall x>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.
    The limit does indeed exist.

    Take e>0. For any positive x, |f(x)| <= 2pi/x. Now choose N=2pi/e so that for all x>N, we have | f(x) | <= 2pi/x < e. Done.


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    (Original post by IrrationalRoot)
    This line requires parity considerations I believe.
    Reason being is you can't just arbitrarily pair off terms and use the proven result (ii). You need to consider how the terms pair off in the middle.
    That's why the ms approach was a bit more complicated.
    Thank you for your response, I finally understand the way marksheme uses. but I didn't pair up anything as in the marksheme though. Could you please take a look again since I made it a bit clearer for what i did. Or do I have to do exactly the same as the markscheme to get any mark.
    Spoiler:
    Show
    S=1k+2k+...+nk
    so S=(n-(n-1))k +(n-(n-2))k+...+(n-1)k+(n-0)k

    (n-(n-1))k + (n-1)k is divisible by n by (ii)
    and (n-(n-2))k + (n-2)k is divisible by n as well

    so if we add (n-1)k+(n-2)k+...+1k+0k (S-nk)to S
    it would be like
    (n-(n-1))k + (n-1)k+ (n-(n-2))k + (n-2)k+....+ (n-1)k+1k+ nk
    which should be divisible by n as well since each term(in different colours) is divisible by n

    so 2S-nk is divisible by n, which should be true for both even and odd n

    and if n is odd, S has to be divisible by n because 2 is not a factor of n
    and if n is even, S then would be divisible by n/2

    Could you please explain why it doesn't work again, thank you
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    (Original post by Ecasx)
    The limit does indeed exist.

    Take e>0. For any positive x, |f(x)| <= 2pi/x. Now choose N=2pi/e so that for all x>N, we have | f(x) | <= 2pi/x < e. Done.


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    What's the point of taking |f(x)| <= 2pi/x? I mean, it works... but why not just f(x) <= 1/x?
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    (Original post by Zacken)
    What's the point of taking |f(x)| <= 2pi/x? I mean, it works... but why not just f(x) <= 1/x?
    To exclude the possibility that it tends to a negative limit.
    Just using the definition with L = 0.
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    (Original post by Zacken)
    What's the point of taking |f(x)| <= 2pi/x? I mean, it works... but why not just f(x) <= 1/x?
    Take modulus of f since that's the definition of the limit.

    My mistake, max[sinx] = 1 not 2pi.


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    (Original post by Ecasx)
    Take modulus of f since that's the definition of the limit.

    My mistake, max[sinx] = 1 not 2pi.
    (Original post by EricPiphany)
    To exclude the possibility that it tends to a negative limit.
    Just using the definition with L = 0.
    Yeah, sorry, forgot the modulus. Was wondering why 2pi instead of 1.
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    (Original post by Zacken)
    Yeah, sorry, forgot the modulus. Was wondering why 2pi instead of 1.
    lol didn't notice that. just popped in.
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    (Original post by charlielao)
    Thank you for your response, I finally understand the way marksheme uses. but I didn't pair up anything as in the marksheme though. Could you please take a look again since I made it a bit clearer for what i did. Or do I have to do exactly the same as the markscheme to get any mark.
    Spoiler:
    Show
    S=1k+2k+...+nk
    so S=(n-(n-1))k +(n-(n-2))k+...+(n-1)k+(n-0)k

    (n-(n-1))k + (n-1)k is divisible by n by (ii)
    and (n-(n-2))k + (n-2)k is divisible by n as well

    so if we add (n-1)k+(n-2)k+...+1k+0k (S-nk)to S
    it would be like
    (n-(n-1))k + (n-1)k+ (n-(n-2))k + (n-2)k+....+ (n-1)k+1k+ nk
    which should be divisible by n as well since each term(in different colours) is divisible by n

    so 2S-nk is divisible by n, which should be true for both even and odd n

    and if n is odd, S has to be divisible by n because 2 is not a factor of n
    and if n is even, S then would be divisible by n/2

    Could you please explain why it doesn't work again, thank you
    Oh I see what you're doing; that seems to work. The mark scheme approach just considers S on its own so that's why they considered the pairing up of terms. But here if you're just adding the same series reversed to itself which should avoid any pairing issues.
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    Did STEP II 2015 and failed miserably, makes me realise how bad I am at maths. Probably gonna fail all my STEP exams.
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    (Original post by IrrationalRoot)
    Did STEP II 2015 and failed miserably, makes me realise how bad I am at maths. Probably gonna fail all my STEP exams.
    Dw!
    You gkt this mate, boundaries were low for the 1 , lower then normal.
    Spoiler:
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    Did you manage the integral?
    Try Q2 it is trivial honestly like GCSE level.
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    (Original post by physicsmaths)
    Dw!
    You gkt this mate, boundaries were low for the 1 , lower then normal.
    Spoiler:
    Show

    Did you manage the integral?
    Try Q2 it is trivial honestly like GCSE level.
    I didn't get any fulls and very much doubt I got the 1, worst paper I ever sat.
    Spoiler:
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    Started with integration and was completely fine, subbed y=\pi -x into last integral and it boiled to to solving integral of 1/(1+sinx)^2, which I wasn't able to do since I was too stupid to realise that I could just write this as that sec expression squared.

    I did try Q2 and it was very easy until the last part which I just failed at.

    With Q4 I did everything right but lost half the marks because for some ridiculous reason I had every trig graph correct but flipped in the y-axis.

    Q5 I could do all of it except proving that Sn was arctan(n/(n+1)).

    Q7 was just a failure, probably like 2 marks.

    Q1 simply couldn't do the second part. Probably because I did the first part in a somewhat different way to the ms, but I believe still correctly.

    Not even sure if this performance was even a f***ing 2. Strange how I got S in II 2014 and 2013 though.
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    Q2II 1992 (further discussion about solutions of differential equations)

    Determine the solutions of y = x * dy/dx - cosh( dy/dx ) (*)

    by differentiating we get that either y '' = 0 (1) OR x = sinh ( y' ) (2)

    the solutions of these two SEPARATELY are y =Ax - cosh(A) (3) and y = x * arsinh(x) - sqrt ( x^2+1 ) (4)

    and that's where the TSR solution stops ... well there are other solutions to * as well (branched functions)!

    (1) or (2) means for some x it might be (1) for other (2) only condition : in the branch point y has to be continuous and dy/dx should be defined.

    Thus other functions that satisfy * : y = (3) with A=A1 for x<=sinh(A1) , (4) for sinh(A1)<x<sinh(A2) , (3) with A=A2 for x>=sinh(A2) for all A1<A2
    which is graphically straight line - curve - straight line we can also have straight line-curve or vice versa so with only one branch point but we can't have a line in the middle of the curve (easy to notice in geogebra) so these are all the possible combinations.

    Don't you think that's right ? Why was it omitted ? Please I really want to get an answer !
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    (Original post by Vesniep)
    ...
    Yep when I was doing this Q I also briefly considered the fact that combinations of these solutions could also be solutions. I think the examiner just expected candidates to conclude that there are two separate solutions and ignore that there may be differentiable piecewise-defined functions which combine the two solutions.
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    (Original post by IrrationalRoot)
    Yep when I was doing this Q I also briefly considered the fact that combinations of these solutions could also be solutions. I think the examiner just expected candidates to conclude that there are two separate solutions and ignore that there may be differentiable piecewise-defined functions which combine the two solutions.
    (Original post by Vesniep)
    Q2II 1992 (further discussion about solutions of differential equations)

    Determine the solutions of y = x * dy/dx - cosh( dy/dx ) (*)

    by differentiating we get that either y '' = 0 (1) OR x = sinh ( y' ) (2)

    the solutions of these two SEPARATELY are y =Ax - cosh(A) (3) and y = x * arsinh(x) - sqrt ( x^2+1 ) (4)

    and that's where the TSR solution stops ... well there are other solutions to * as well (branched functions)!

    (1) or (2) means for some x it might be (1) for other (2) only condition : in the branch point y has to be continuous and dy/dx should be defined.

    Thus other functions that satisfy * : y = (3) with A=A1 for x<=sinh(A1) , (4) for sinh(A1)<x<sinh(A2) , (3) with A=A2 for x>=sinh(A2) for all A1<A2
    which is graphically straight line - curve - straight line we can also have straight line-curve or vice versa so with only one branch point but we can't have a line in the middle of the curve (easy to notice in geogebra) so these are all the possible combinations.

    Don't you think that's right ? Why was it omitted ? Please I really want to get an answer !
    I'm pretty sure that sum of solutions to this first order DE are not a solution to the first order DE.

    (Original post by DFranklin)
    Basically because \cosh\left(\frac{dy_1}{dx}+\frac  {dy_2}{dx}\right) \neq \cosh\left(\frac{dy_1}{dx}\right  ) + \cosh\left(\frac{dy_2}{dx}\right  ).
    (Original post by DFranklin)
    Having had a look, subbing in for y=Ax+B you end up finding B = -cosh A, so y=Ax-coshA is one family of solutions.I haven't looked at how the 2nd (nastier) solution behaves.

    As a general comment: If you have a first order DE (no d2y/dx2 terms), then you don't expect more than 1 arbitrary constant. But you may get more than one family of solutions if there are some many-to-one functions in there.E.g. (dy/dx)^2 = 1 has two families of solutions y=A+x and y = A - x.
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    (Original post by Zacken)
    I'm pretty sure that sum of solutions to this first order DE are not a solution to the first order DE.
    I wasn't saying that linear combinations of the solutions would be solutions (obviously they wouldn't) but I was wondering whether piecewise combinations could be solutions.
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    (Original post by Zacken)
    I'm pretty sure that sum of solutions to this first order DE are not a solution to the first order DE.
    We are not talking about the sum of these two solutions , but a change from one to another (branched functions) which is possible I used geogebra to demonstrate it and it works perfectly fine as well (of course I also proved it analytically)
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    (Original post by Vesniep)
    We are not talking about the sum of these two solutions , but a change from one to another (branched functions) which is possible I used geogebra to demonstrate it and it works perfectly fine as well (of course I also proved it analytically)
    Every tangent to y=x\sinh^{-1}x-\sqrt{1+x^2} is of the form y=Ax-\cosh A I believe and so making differentiable piecewise solutions is possible yes.
    This means that there are not just two solutions but infinitely many piecewise ones combining the two.
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    (Original post by IrrationalRoot)
    Yep when I was doing this Q I also briefly considered the fact that combinations of these solutions could also be solutions. I think the examiner just expected candidates to conclude that there are two separate solutions and ignore that there may be differentiable piecewise-defined functions which combine the two solutions.
    Yes I think it's not really required to think that many cases when solving DEs in STEPs.
    Another common mistake is in logarithms when the solutions frequently use ln(f(x)) instead of ln(abs(f(x)) in the end signs and everything are fine but you are not sure if you have found all solutions and it works 'magically' rather than rigorously ... that's why I spend so much time on these questions
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    (Original post by Vesniep)
    Yes I think it's not really required to think that many cases when solving DEs in STEPs.
    Another common mistake is in logarithms when the solutions frequently use ln(f(x)) instead of ln(abs(f(x)) in the end signs and everything are fine but you are not sure if you have found all solutions and it works 'magically' rather than rigorously ... that's why I spend so much time on these questions
    I agree, a lot of the time I get stuck because of technical points like these which most of the time they don't expect you to consider.
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    (Original post by IrrationalRoot)
    I agree, a lot of the time I get stuck because of technical points like these which most of the time they don't expect you to consider.
    Did you study analysis at least a little bit ? I mean I have that problem of rigour because in my school we study analysis and rigour is our god (if we are to solve a dif equation we should consider continuity differentiability etc so that we get all solutions) and our book is full of proofs ( 3 quarters proofs 1 quarter questions about proofs lol ) so I guess that's why I have that problem .
 
 
 
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