Hey there! Sign in to join this conversationNew here? Join for free
Turn on thread page Beta
    Offline

    22
    ReputationRep:
    Specimen Test 1, Question 9

    Warning: this is going to be fairly rigorous, but it's the best I can do, if anybody can suggest how I can formalise/improve my solution, do let me know!
    Spoiler:
    Show
    We have e^x = Mx, a quick sketch comes in useful: Name:  MSP5921efh59964572di8700000hid1bc6d91ca652.gif
Views: 226
Size:  4.5 KB

    We can note that we will have a root between 0 < x < 1 for M > e (which we can safely assume since M is large) by writing the equation as f(x) = e^x - Mx and noting that f(0) = 1 > 0 whilst f(1) = e - M < 0 since M > e hence there must be a root in the interval (0,1).

    So for 1 < x < w the line Mx is above e^x but because we know that the exponential function must quickly 'catch up' and over take the polynomial Mx that the curves must intersect at w.

    We can formalise this somewhat by noting that e^x has gradient e^x and Mx has gradient M and there must exist some a \in \mathbb{R} such that \forall x > a we have e^x > M and hence the exponential function catches up to the polynomial term forcing an intersection at w.

    Another way of showing this would be to note that we have for 1 < x < w that e^x - Mx < 0 but since \lim_{x \to \infty} e^x - Mx = \infty that eventually e^x - Mx \geq 0 with equality at x = w.

    We can say that e^x intersects Mx for x > 1 roughly close to when the gradient of e^x surpasses the gradient of Mx (given the exponential growth of the gradient versus constant growth) hence a reasonable approximation is when e^x = M \implies x = \log M

    If we write w = \log M + y then we have e^w - Mw = 0 \implies \exp{(\log M + y)} - M(\log M + y) = 0 hence M(e^{y} - \log M - y) = 0 so e^y - y = \log M - we can use the Maclauren series for e^y whilst ignoring terms of order 3 or higher to get (approximately)

    1 + y + \frac{y^2}{2} - y \approx \log M \implies y^2 \approx \log M^2 - 2 \implies y \approx \sqrt{\log M^2 - 2}.

    That's as far as I get. I can't help but feel that my approximation for y is a little hack-y and inelegant (i.e not nice looking)
    DFranklin - any thoughts?

    Edit: This reminds me of something... the Lambert W function(?)

    Edit2: The last part isn't quite right, see Joostan's solution on page 3.
    Offline

    8
    ReputationRep:
    (Original post by -Gifted-)
    Thanks a ton, it makes sense to me. However, I used conservation of mechanical energy, but that yielded a different result where the velocity doesn't cancel out. Let me scan it.
    Unfortunately energy isn't conserved.
    Offline

    2
    ReputationRep:
    Here it is,
    Spoiler:
    Show


    I fail at LATEX so i apologise. But yeah, initial ke= final ke+ GPE

    0.5Mu^2= ((u^2)/8)*(M+PAx)+Mgx

    solving for x, actually yields: (3Mu^2)/(PAu^2+Mg)
    Offline

    2
    ReputationRep:
    (Original post by Renzhi10122)
    Unfortunately energy isn't conserved.
    Why is energy not conserved? I feel like I am missing a main concept here :/
    Offline

    22
    ReputationRep:
    (Original post by StrangeBanana)
    ah, some competition for me, then :P

    What made you choose Pembroke? I hear the food's fantastic.



    We may see each other at the interviews. ^^ Why King's, if you don't mind me asking?
    Ah - I swear I've seen you around quite a bit.

    King's just looked so impressive and spacious, it just gave off this certain feeling of grandeur that I appreciated.
    Offline

    8
    ReputationRep:
    (Original post by -Gifted-)
    Why is energy not conserved? I feel like I am missing a main concept here :/
    Because the dust sticks, it's like a coliision between two particles where they stick together.
    • Thread Starter
    Offline

    4
    ReputationRep:
    (Original post by StrangeBanana)
    What made you choose Pembroke? I hear the food's fantastic.
    That is pretty much the reason I chose it, I'm not gonna lie :lol:
    Offline

    22
    ReputationRep:
    (Original post by Renzhi10122)
    Because the dust sticks, it's like a coliision between two particles where they stick together.
    Non-elastic collision so coefficient of restitution e \neq 1 hence energy isn't conserved as kinetic energy is lost.
    Offline

    19
    ReputationRep:
    (Original post by -Gifted-)
    Thanks a ton, it makes sense to me. However, I used conservation of mechanical energy, but that yielded a different result where the velocity doesn't cancel out. Let me scan it.
    You're welcome! Like Renzhi and Zacken have said, energy isn't conserved; that's only when the collision is perfectly elastic, i.e. the coefficient of restitution is 1.

    (Original post by Zacken)
    Ah - I swear I've seen you around quite a bit.

    King's just looked so impressive and spacious, it just gave off this certain feeling of grandeur that I appreciated.
    :ninja:

    Yeah, same here. That chapel is amazing.

    (Original post by Jordan\)
    That is pretty much the reason I chose it, I'm not gonna lie :lol:
    It's a good enough a reason as any, to be honest.
    Offline

    2
    ReputationRep:
    (Original post by Renzhi10122)
    Because the dust sticks, it's like a coliision between two particles where they stick together.
    I see, so regardless of whether I take into account of the combined mass when the particles stick together energy is never conserved whenever they combine?
    Offline

    8
    ReputationRep:
    (Original post by -Gifted-)
    I see, so regardless of whether I take into account of the combined mass when the particles stick together energy is never conserved whenever they combine?
    Correct, but momentum always is, thus the approach by strangebanana..
    Offline

    2
    ReputationRep:
    (Original post by Renzhi10122)
    Correct, but momentum always is, thus the approach by strangebanana..
    It makes sense now, thank you guys
    Online

    18
    ReputationRep:
    (Original post by atsruser)
    You're right. Damn. sin^2 breaks this. Haven't got time to think about it in detail now. I'll delete that post forthwith.
    It's not sin^2x that breaks it - it would work fine for \int x \sin^2x \,dx. It's the x^2 term that fails.

    It works for multiplying by x because when you pair x with pi - x and add you get a constant (pi).

    But when you pair x^2 with (pi-x)^2 it doesn't work so nicely.
    Offline

    11
    ReputationRep:
    (Original post by DFranklin)
    It's not sin^2x that breaks it - it would work fine for \int x \sin^2x \,dx. It's the x^2 term that fails.
    I meant to write xsin^2x i.e you can't write the integral as \int x \cdot x\sin^2 x dx = \int xf(\sin x) dx which is where my application of the result fails, I think. Not sure I followed your earlier comment re: arcsin BTW.

    BTW I stole the (incorrect) approach from this post: http://www.thestudentroom.co.uk/show...1&postcount=19

    I just did the second part of this, and IMHO, it's beautiful - I think you'll like it.

    I can't see any easy way to get it though, apart from deriving the identity for x^3 from scratch though, which seems pretty challenging for a STEP candidate.
    Offline

    11
    ReputationRep:
    (Original post by Zacken)
    Specimen Test 1, Question 9

    Warning: this is going to be fairly rigorous, but it's the best I can do, if anybody can suggest how I can formalise/improve my solution, do let me know!
    Spoiler:
    Show
    I get the impression that they are looking for  y = \log \log M but I'd be interested in seeing what other people have to say too
    Offline

    22
    ReputationRep:
    (Original post by atsruser)
    I meant to write xsin^2x i.e you can't write the integral as \int x \cdot x\sin^2 x dx = \int xf(\sin x) dx which is where my application of the result fails, I think. Not sure I followed your earlier comment re: arcsin BTW.

    BTW I stole the (incorrect) approach from this post: http://www.thestudentroom.co.uk/show...1&postcount=19

    I just did the second part of this, and IMHO, it's beautiful - I think you'll like it.

    I can't see any easy way to get it though, apart from deriving the identity for x^3 from scratch though, which seems pretty challenging for a STEP candidate.
    The tried exploiting the fact that it was an even function and the symmetry between sine and cosine, but to no avail - brute forcing it using IBP turned out to be the easiest solution. Unless, there's some kind of reduction formulae by considering I_n = \int_0^{\pi} (x \sin x)^n \, \mathrm{d}x although I can't see that being any simpler.
    Offline

    11
    ReputationRep:
    (Original post by DFranklin)
    From memory, the x->pi-x trick works (the numerator of the integrand has been chosen carefully by the examiners).
    It does - that's how I did it. But is that really expected in STEP without a hint?

    I must admit, the form of the numerator didn't tell me much until I was almost finished with \int x^3 f(\sin x) dx so if they're expecting people to recognise the numerator, *and* see the substitution, I feel that they're angling for the budding von Neumanns only.
    Offline

    11
    ReputationRep:
    (Original post by Zacken)
    The tried exploiting the fact that it was an even function and the symmetry between sine and cosine, but to no avail - brute forcing it using IBP turned out to be the easiest solution. Unless, there's some kind of reduction formulae by considering I_n = \int_0^{\pi} (x \sin x)^n \, \mathrm{d}x although I can't see that being any simpler.
    Yeah, I want to think about the reduction formulae a bit more later. No time now. Haven't done this kind of stuff in detail for years so I'm rusty.
    Online

    18
    ReputationRep:
    (Original post by atsruser)
    Spoiler:
    Show
    I get the impression that they are looking for  y = \log \log M but I'd be interested in seeing what other people have to say too
    (Original post by Zacken)
    ..
    My recollection of looking at this was that this question was in a completely different league from the other items on the list and even a half-way rigourous answer was extremely tricky to obtain.

    I'm really not sure what they are after on this one.
    Offline

    22
    ReputationRep:
    (Original post by atsruser)
    Yeah, I want to think about the reduction formulae a bit more later. No time now. Haven't done this kind of stuff in detail for years so I'm rusty.
    I'll investigate a bit when I get some free time tonight, I haven't covered reduction formulae at all, but it's quite intuitive.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 18, 2017
Poll
“Yanny” or “Laurel”

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.