How would one show that e^pi is greater than pi^e without using a calculator?

[meghatronic]

I had interviews for Engineering at both Cambridge and Oxford and was asked similar problem solving questions, you're allowed a pen and paper. The interviews aren't "sit on a chair in the middle of a room and be interrogated," you usually sit down at a table with one or two people and go through some questions.

Both? Through UCAS?

 

[Rebelius]

could've been different years?

 

[Chrisss]

Are you not allowed to apply to both or something?

 

[growse]

Are you not allowed to apply to both or something?

Nope, not in the same year unless you're applying for an organ or choral scholarship.

Oh, and DaveF, nice thinking. Can't say I'd get there these days

 

[Egosh]

if i was asked this question in an interview id just get up and leave cos id know im at the wrong interview..

 

[Johnny Girth]

Nope, not in the same year unless you're applying for an organ or choral scholarship.

Oh, and DaveF, nice thinking. Can't say I'd get there these days

That used to be the case but hasn't it changed now so that you can apply to both?

 

[DaveF]

That used to be the case but hasn't it changed now so that you can apply to both?

As far as I'm aware it's still true. Robinson's (Cam) website says you can't apply to Oxford and Cambridge in the same year, at any rate.

 

[A[L]C]

Consider f(x) = ln(x) / x. f'(x) = (1 - ln x) / x^2. So f'(x) < 0 for x > e, and so f(e) > f(pi).

So ln(e)/e > ln(pi) / pi. So pi ln(e) > e ln pi. Apply exp to both sides to get e^pi > pi ^ e.

wha?

 

[D.Roberts]

ok all these e`s have given me a head ache.

ive never been any good at mathes and never will be ive looked at this whole post and still have no idea what the hell you lot are talking about.

 

[Calder]

could've been different years?

Yeah, was in seperate years, one before my YII and one during.

 

[Solari]

e's and pies don't mix. I think you'd probably hallucinate and then throw up

 

[Stiff_Cookie]

Maybe im just at that level but cant you just replace 'e' with any number?

 

[RastaManBob]

Can you not just say e = ~2, pi = ~ 3 (Obviously poor approximations but...)

e^pi = 2^3 = 8
pi^e = 3^2 = 6

Shows e^pi is greater than pi^e...

I guess I'm missing something

 

[Sleepy]

pi^e = 3^2 = 6

Too much beer?

3^2=9

 

[growse]

Can you not just say e = ~2, pi = ~ 3 (Obviously poor approximations but...)

e^pi = 2^3 = 8
pi^e = 3^2 = 6

Shows e^pi is greater than pi^e...

I guess I'm missing something

Pfft, an engineer's answer

This obviously covers the fact that it's completely wrong too...

 

[RastaManBob]

Too much beer?

3^2=9

I wish I had that excuse... Ignore me.

Pfft, an engineer's answer

I'd have been quite proud of that until I realised my stupid mistake (See above)...

EDIT - So nobody got an "easy" solution to do "on the spot"? The more I look at it the more I find it difficult to prove (easily) but I *know* that the statement is true (Having checked it)... Crazy stuff!

 

[Arcade Fire]

Maybe im just at that level but cant you just replace 'e' with any number?

No:

0.5^pi < pi^0.5

EDIT - So nobody got an "easy" solution to do "on the spot"? The more I look at it the more I find it difficult to prove (easily) but I *know* that the statement is true (Having checked it)... Crazy stuff!

DaveF's solution is an easy, on-the-spot solution. The only obvious thing to do is to take logs of both sides, which leaves you to ask whether e*log(pi)/pi is less than or greater than one. Differentiating e*log(x)/x tells you that it it's always less than 1, which tells you that e^pi > pi^e.

 

[Craig321]

A mate said he was asked this at interview for Maths at Oxford. Sorry it's not a typical GD tea-time query, but there are a few mathmos on here that I thought could help.

e^pi > pi^e

Done.

 

[Zom]

As far as I'm aware it's still true. Robinson's (Cam) website says you can't apply to Oxford and Cambridge in the same year, at any rate.

It's still true - you can't apply to both.

 

[jezsoup]

It's still true - you can't apply to both.

Why do they say you cant apply to both?