STEP Prep Thread 2016 (Mark. II)

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?

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.

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.

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.

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.

...

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.

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 !

Basically because $\cosh\left(\frac{dy_1}{dx}+\frac{dy_2}{dx}\right) \neq \cosh\left(\frac{dy_1}{dx}\right) + \cosh\left(\frac{dy_2}{dx}\right)$.

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.

I'm pretty sure that sum of solutions to this first order DE are not a solution to the first order DE.

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)

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

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 .