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Mathematics Applicants 2012

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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 :tongue:

A gentle start

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

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.

Reply 1261

TheMagicMan

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 :tongue:

Reply 1262

Miller693

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. :smile:

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 :biggrin:

Reply 1263

TheMagicMan

Original post by ben-smith

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

Reply 1264

Aristotle's' Disciple

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

Reply 1265

TheMagicMan

Original post by Miller693

Considering your picture I would think your favourite joke would be:

Which cat fell off the roof first?

The one with the lower mu :tongue:

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

(edited 12 years ago)

Reply 1266

Aristotle's' Disciple

Original post by Miller693

LOL! That's actually pretty good. :P

Reply 1267

TheMagicMan

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

Reply 1268

Aristotle's' Disciple

Original post by TheMagicMan

I might have missed out an equals sign

Damn, no worries, quote me with the proper one hehe.

Reply 1269

Dirac Spinor

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!

Reply 1270

Intriguing Alias

Original post by Miller693

Can imaginary roots not hold it up? :tongue:

Reply 1271

TheMagicMan

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

Reply 1272

jonnyboy1993

Original post by ben-smith

Spoiler

Reply 1273

Miller693

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 :tongue:

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 :smile:

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

Reply 1274

TheMagicMan

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 :tongue:

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

Reply 1275

Miller693

Original post by hassi94

Can imaginary roots not hold it up? :tongue:

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 :smile:

Reply 1276

Dirac Spinor

Original post by jonnyboy1993

Spoiler

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.

(edited 12 years ago)

Reply 1277

Aristotle's' Disciple

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? :smile:

I was trying to do it by counter example, just finding a value that contradicted the statement. :/

Reply 1278

jonnyboy1993

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.

Yeah, of course, I'd never dare to try and write out a full solution on TSR...

But yeah I'm sure you'll be fine, it's just a case of looking out for the little details and special cases because Cambridge love nothing more than to try and trick you...

Reply 1279

twig

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

When I did that Q originally, I considering the solution to dP/dx=0, which is also a polynomial. So there a finite number of solutions bounded by the order of this derivative polynomial. So after the largest absolute solution, the gradient must remain + or -. And the difference would eventually exceed the limit given. This is just as waffly as you/ Ben's posts, but nonetheless. Nice way to waste extra time on a step exam.

Edit: just noticed some of the above posts, nevermind.