Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    0
    ReputationRep:
    (Original post by TheMagicMan)
    True, but feel sorry for the americans: international students are way more likely to get in
    That's very interesting. Could you explain why/ have any sources for that info? Also, is taking SAT next year/ this year for entry to US for (let's say) after a gap year feasible? Or are you severely disadvantaged when applying with SATs taken after finishing school?
    Offline

    19
    ReputationRep:
    (Original post by hassi94)
    Haha I had exactly the same thought process - took the longer one to make it more worth it!
    Aha, yhupp! Time spent wandering, is time not spent on maths.

    Good answer to the question, I didn't think of that!
    Offline

    3
    ReputationRep:
    (Original post by hassi94)
    I thought i) was extremely easy.

    Natural log both sides to get

    ln(a^lnb) = ln(b^lna)
    ln(b)ln(a) = ln(a)ln(b)

    That's it isn't it!?
    It is. I actually went for the very similar a^lnb=(e^lna)^lnb=(e^lnb)^lna=b^ lna.

    Call it personal taste but I don't like meeting in the middle when I do identities
    Offline

    2
    ReputationRep:
    (Original post by TheMagicMan)
    Let's do some step qs. Ok here's the question.
    Which of the following statements are true and which are false? Justify your answers.
    (i) a^ln b = b^ln a for all a, b > 0.
    (ii) cos(sin µ) = sin(cos µ) for all real µ.
    (iii) There exists a polynomial P such that |P(µ) − cos µ|=< 6^10−6 for all real µ.
    (iv) x^4 + 3 + x^−4 > 5 for all x > 0.

    Nice and easy A gentle start

    I'm not sure that this was a good q to choose. It is extremely easy
    Spoiler:
    Show

    1.True, take logs of both sides
    2.False. mu=0 obvious counter example
    3. Hardest of them all to explicitly prove. The crucial fact is that it is a polynomial. i.e finite length+integer powers. as mu tends to either very big +ve/-ve numbers the polynomial will have to grow to +/-infty Where the inequality certainly wouldn't hold (a bit iffy)
    4. Take the five over to the LHS and turn it into one big ass fraction. Show denominator+numerator are +ve, your done.
    Offline

    9
    ReputationRep:
    (Original post by Aristotle's' Disciple)
    Hello! Dude, why's there no Warwick? :P Interesting UCAS! Your grades look solid, good luck on Oxford, hope to see you participating in our keen mathematical discussions.
    Cheers, why no Warwick though you ask? Well, I'm not too keen on the area to be honest when I did visit. Though I have heard it has a brilliant reputation. I think I preferred Durham and Bath over Warwick and I need some unis that will give lower offers just in case (I know I already have the A*, just being cautious about physics).
    It was an option though.

    Anyway, I look forward to the mathematical discussions and I shall lead out with my all time favourite maths joke:

    Why did the tree fall over?
    It's discriminant was less than zero
    Offline

    3
    ReputationRep:
    (Original post by ben-smith)
    Spoiler:
    Show

    1.True, take logs of both sides
    2.False. mu=0 obvious counter example
    3. Hardest of them all to explicitly prove. The crucial fact is that it is a polynomial. i.e finite length+integer powers. as mu tends to either very big +ve/-ve numbers the polynomial will have to grow to +/-infty Where the inequality certainly wouldn't hold (a bit iffy)
    4. Take the five over to the LHS and turn it into one big ass fraction. Show denominator+numerator are +ve, your done.
    Yep. Hopefully there are some more challenging ones down the road. I think the best way to do 3 is to consider the power series of cos(x) and that P(x) must be finite
    Offline

    19
    ReputationRep:
    (Original post by TheMagicMan)
    Ironically i) is the most difficult
    I'll give you a clue: out of ii),iii) and iv) one can be solved by considering large numbers, one can be solved by counterexample, and one can be solved by a useful substitution

    My tip for i) is that it can be solved by replacing b (or a ) with something involving e
    Well here iv) x^4 + x^-4 > 2 x > 0 however if x = 1 then 1^4 + 1^-4 = 2 which isn't > 2. So it's false. Is that okay? I'm terrible at proving stuff, haven't got the hang of it yet lol. :P
    Offline

    3
    ReputationRep:
    (Original post by Miller693)
    Cheers, why no Warwick though you ask? Well, I'm not too keen on the area to be honest when I did visit. Though I have heard it has a brilliant reputation. I think I preferred Durham and Bath over Warwick and I need some unis that will give lower offers just in case (I know I already have the A*, just being cautious about physics).
    It was an option though.

    Anyway, I look forward to the mathematical discussions and I shall lead out with my all time favourite maths joke:

    Why did the tree fall over?
    It's discriminant was less than zero
    Considering your picture I would think your favourite joke would be:

    Which cat fell off the roof first?

    The one with the lower mu

    EDIT: Does anyone else have that UKMT mug with the jokes on it?
    Offline

    19
    ReputationRep:
    (Original post by Miller693)
    Cheers, why no Warwick though you ask? Well, I'm not too keen on the area to be honest when I did visit. Though I have heard it has a brilliant reputation. I think I preferred Durham and Bath over Warwick and I need some unis that will give lower offers just in case (I know I already have the A*, just being cautious about physics).
    It was an option though.

    Anyway, I look forward to the mathematical discussions and I shall lead out with my all time favourite maths joke:

    Why did the tree fall over?
    It's discriminant was less than zero
    LOL! That's actually pretty good. :P
    Offline

    3
    ReputationRep:
    (Original post by Aristotle's' Disciple)
    Well here iv) x^4 + x^-4 > 2 x > 0 however if x = 1 then 1^4 + 1^-4 = 2 which isn't > 2. So it's false. Is that okay? I'm terrible at proving stuff, haven't got the hang of it yet lol. :P
    I might have missed out an equals sign
    Offline

    19
    ReputationRep:
    (Original post by TheMagicMan)
    I might have missed out an equals sign
    Damn, no worries, quote me with the proper one hehe.
    Offline

    2
    ReputationRep:
    (Original post by TheMagicMan)
    Yep. Hopefully there are some more challenging ones down the road. I think the best way to do 3 is to consider the power series of cos(x) and that P(x) must be finite
    My thinking was along those lines but I decided not to use power series as I figured it was overkill for STEP I.

    BTW High five for Maths+Phys!
    Offline

    14
    ReputationRep:
    (Original post by Miller693)
    Cheers, why no Warwick though you ask? Well, I'm not too keen on the area to be honest when I did visit. Though I have heard it has a brilliant reputation. I think I preferred Durham and Bath over Warwick and I need some unis that will give lower offers just in case (I know I already have the A*, just being cautious about physics).
    It was an option though.

    Anyway, I look forward to the mathematical discussions and I shall lead out with my all time favourite maths joke:

    Why did the tree fall over?
    It's discriminant was less than zero
    Can imaginary roots not hold it up?
    Offline

    3
    ReputationRep:
    (Original post by Aristotle's' Disciple)
    Damn, no worries, quote me with the proper one hehe.
    (iv) x^4 + 3 + x^−4 >= 5 for all x > 0.

    That's what it should be
    Offline

    7
    ReputationRep:
    (Original post by ben-smith)
    Spoiler:
    Show

    1.True, take logs of both sides
    2.False. mu=0 obvious counter example
    3. Hardest of them all to explicitly prove. The crucial fact is that it is a polynomial. i.e finite length+integer powers. as mu tends to either very big +ve/-ve numbers the polynomial will have to grow to +/-infty Where the inequality certainly wouldn't hold (a bit iffy)
    4. Take the five over to the LHS and turn it into one big ass fraction. Show denominator+numerator are +ve, your done.
    Spoiler:
    Show
    for 3 you've pretty much nailed it, as any polynomial degree 1 or higher can't work for the reason you state. However they'd expect you to consider a polynomial of degree zero (ie a constant) too, just to complete your argument. It's just spotting the little things like that which will help you pick up high scoring solutions

    For 4 you'll do better to try and produce an algebraic expression which is a square, purely because it's a slightly neater solution (and you don't risk multiplying by an expression containing an unknown in your proof (this can lead to invalid inequalities)
    Offline

    9
    ReputationRep:
    (Original post by TheMagicMan)
    Considering your picture I would think your favourite joke would be:

    Which cat fell off the roof first?

    The one with the lower mu

    EDIT: Does anyone else have that UKMT mug with the jokes on it?
    That was one my teacher alluded to in year 9 but I had no idea what it meant. So glad I do now though, the geeky maths jokes are always the best followed by groan-worthy puns. I haven't heard of this mug though, it sounds exciting to be honest!

    (Original post by Aristotle's' Disciple)
    LOL! That's actually pretty good. :P
    Cheers, that is one of the better ones I know,my other jokes are awful and there is only ever one person that laughs. That's ok though


    Anyway I should stop distracting people with jokes and get on with some real maths
    Offline

    3
    ReputationRep:
    (Original post by ben-smith)
    My thinking was along those lines but I decided not to use power series as I figured it was overkill for STEP I.

    BTW High five for Maths+Phys!
    Overkill is good. Make the examiner sweat

    Are you doing Phys as well? I figure that I can get away without doing numbers and sets as you can go to the lectures even if you do Phys and you can change easily in the first few weeks
    Offline

    9
    ReputationRep:
    (Original post by hassi94)
    Can imaginary roots not hold it up?

    Maybe in dream land. Though if the tree's discriminant was equal to zero I don't think it would do a lot of good; one root wouldn't be an awful lot of help
    Offline

    2
    ReputationRep:
    (Original post by jonnyboy1993)
    Spoiler:
    Show
    for 3 you've pretty much nailed it, as any polynomial degree 1 or higher can't work for the reason you state. However they'd expect you to consider a polynomial of degree zero (ie a constant) too, just to complete your argument. It's just spotting the little things like that which will help you pick up high scoring solutions

    For 4 you'll do better to try and produce an algebraic expression which is a square, purely because it's a slightly neater solution (and you don't risk multiplying by an expression containing an unknown in your proof (this can lead to invalid inequalities)
    As you can tell from my post, what I wrote was not what any sane person would actually write down in an exam so you'll forgive me if my solutions are less than rigorous.
    For 4, I am pretty much saying what you said. To make it all one fraction you only have to multiply and divide by x^4 which is strictly positive so I don't think you have to worry about such things.
    I haven't actually done the question, just saying how I would roughly do it.
    Offline

    19
    ReputationRep:
    (Original post by ben-smith)
    As you can tell from my post, what I wrote was not what any sane person would actually write down in an exam so you'll forgive me if my solutions are less than rigorous.
    For 4, I am pretty much saying what you said. To make it all one fraction you only have to multiply and divide by x^4 which is strictly positive so I don't think you have to worry about such things.
    I have actually done the question, just saying how I would roughly do it.
    Can you use Latex and show me iv) please? I was trying to do it by counter example, just finding a value that contradicted the statement. :/
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?

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.