A few problems to have a go at...

Indeterminate

3

Four problems that I hope you'll like. A level knowledge (inc FM where appropriate) is required.

Problem 1

Spoiler

Problem 2

Spoiler

Problem 3

Spoiler

Problem 4

Spoiler

(edited 7 years ago)

Reply 1

DFranklin

18

I have a funny feeling none of these are actually appropriate for "A level students"...

Reply 2

Indeterminate

OP

3

Original post by DFranklin

I have a funny feeling none of these are actually appropriate for "A level students"...

Well, definitely inappropriate for an A-level paper :tongue:

Whenever I've done this in the past I've always received a decent response. We'll see what happens :lol:

Original post by DFranklin

I have a funny feeling none of these are actually appropriate for "A level students"...

Having done A-Level maths + FM just last summer, I can safely say I have never seen anything like this unless I missed something... :lol:

Reply 4

DFranklin

18

Original post by Indeterminate

Well, definitely inappropriate for an A-level paper :tongue:

Whenever I've done this in the past I've always received a decent response. We'll see what happens :lol:

Since there's a set of 10 problems I posted only yesterday with little response (and although some of them were hard, some were reasonably accessible) I'm not expecting a great deal...

Reply 5

Indeterminate

OP

3

Original post by RDKGames

Having done A-Level maths + FM just last summer, I can safely say I have never seen anything like this unless I missed something... :lol:

Well here's a start. For problem 4 consider the quadratic in the denominator and try a substitution.

Original post by DFranklin

Since there's a set of 10 problems I posted only yesterday with little response (and although some of them were hard, some were reasonably accessible) I'm not expecting a great deal...

Oh.

Well, erm, at least we try. :lol:

Reply 6

solC

14

I think I may be able to do 4...
I'll have a crack at it tomorrow :smile:

Edit: Couldn't hack it

(edited 7 years ago)

Reply 7

BobBobson

3

Can you confirm whether or not Problem 2 is just f(x) = x. If so, do you want us to prove it?

Reply 8

A02

10

Original post by BobBobson

Can you confirm whether or not Problem 2 is just f(x) = x. If so, do you want us to prove it?

I think f(x) = 1+x works as well. Plug a,b = 0 in to get f(0)=0 or 1. Then let a=0, and see what you get. Also, with functional equations substitute back in to see if what you get works.

Reply 9

Indeterminate

OP

3

Original post by BobBobson

Can you confirm whether or not Problem 2 is just f(x) = x. If so, do you want us to prove it?

Original post by A02

I think f(x) = 1+x works as well. Plug a,b = 0 in to get f(0)=0 or 1. Then let a=0, and see what you get. Also, with functional equations substitute back in to see if what you get works.

If you think you've found one (or more than one) solution, can't you find any others? Or can you show that there aren't any others?

Reply 10

A02

10

Original post by Indeterminate

If you think you've found one (or more than one) solution, can't you find any others? Or can you show that there aren't any others?

Yes.
I think this is right:

Answer

Reply 11

Zacken

22

Problem 4 is just ugly. Problem 3 isn't so bad, on the other hand (doubt an A-Level student would manage it, though):

Use

x \mapsto (\sin x)^2

to get

Unparseable latex formula:

\displaystyle [br]\begin{align*}\int_0^1 \frac{\log \frac{3+x}{3-x}}{\sqrt{x(1-x)}} \, \mathrm{d}x &=\int_0^{\pi/2} \log\left(\frac{3 + \sin^2 x }{3-\sin^2 x}\right)\frac{2\sin x\cos x \, \mathrm{d}x}{\sqrt{\sin^2 x(1-\sin^2 x)}} \\ &= 2 \int_0^{\pi/2}\log\left(\frac{3 + \sin^2 x}{3-\sin^2 x}\right) \, \mathrm{d}x \end{align*}

Now use

x \mapsto \frac{x}{2}

and

\sin^2 x = \frac{1-\cos \frac{x}{2}}{2}

to get

\displaystyle \int_0^1 \frac{\log \frac{3+x}{3-x}}{\sqrt{x(1-x)}} \, \mathrm{d}x = \int_0^{\pi} \log \frac{7 - \cos x}{5 + \cos x} \, \mathrm{d}x

Now use

\displaystyle \int_0^{\pi} \log(\kappa \pm \cos \theta) \, \mathrm{d}\theta =\pi \text{arccosh}\, \kappa

on our integral (pretty nifty identity, try proving it) to get:

\displaystyle I = \pi (\text{arccosh} \,7 - \text{arccosh} \,5)

(edited 7 years ago)

Reply 12

Zacken

22

"An algebraic integer is a root of some polynomial with leading coefficient (that is the non-zero coefficient of the highest power) 1."

Really? :tongue:

Reply 13

Indeterminate

OP

3

Original post by Zacken

Problem 4 is just ugly. Problem 3 isn't so bad, on the other hand (doubt an A-Level student would manage it, though):

Use

x \mapsto (\sin x)^2

to get

Unparseable latex formula:

\displaystyle [br]\begin{align*}\int_0^1 \frac{\log \frac{3+x}{3-x}}{\sqrt{x(1-x)}} \, \mathrm{d}x &=\int_0^{\pi/2} \log\left(\frac{3 + \sin^2 x }{3-\sin^2 x}\right)\frac{2\sin x\cos x \, \mathrm{d}x}{\sqrt{\sin^2 x(1-\sin^2 x)}} \\ &= 2 \int_0^{\pi/2}\log\left(\frac{3 + \sin^2 x}{3-\sin^2 x}\right) \, \mathrm{d}x \end{align*}

Now use