There is a full solution available online (for those interested) so I shan't repeat it - as it is also very long.
However, the key insight to solve that seemingly monstrous product is to use one of the trig identities - but in an unusual way - so that within just 2 or 3 terms the mind can see that the infinite product simplifies to just two terms.
Such problems are very much borderline A Level Further Mathematics - scanning the MIT paper, I think at least half of the problems could be done at school. Nowadays, a student would have to be studying Cambrdge's new Pre-U Level Mathematics as they have brought back all the topics that they have slowly removed from the (dumbed down) A Level.
I think that that's sort of the appeal of these sorts of problems. In and of themselves, they're not particularly difficult, but they require you to think outside of the box in order to get the solution. Once you see that light-bulb, the solution is obvious and elegant.
On your second point, yes, most of the problems could be done at school back in the day. I don't know the state of British mathematics education, but American maths education is in a sorry state. Most people can't wrap their head around the fact that 1/3 is in fact larger than 1/4.
Oh, I think we're in the realm of 1 in 10,000 who won't freak out just looking at that integral! There is a small counter-current holding on to mathematics as a rigorous subject; I mean, philosophy, for example, has almost totally disappeared from schools. Thankfully the maths competitions and Olympiads are away from the dead hand of government/corporate "mastery (obedience) learning". I teach a few gifted kids who haven't had their lights dimmed as yet, so that's fun!
I see that cos(x/2^1) = cos(2*x/2^2) = 1 - 2sin^2(x/2^2).
When k = 1, the product is cos(x/2^1).
When k = 2, the product is (1 - 2sin^2(x/2^2))*cos(x/2^2).
When k = 3, the product is (1 - 4sin^2(x/2^3)cos^2(x/2^3))*(1 - sin^2(x/2^3)*cos(x/2^3).
Does this lead to finding any useful pattern?
The key step is using
sin(2x) = 2.sinx.cosx
so that
cos(x) = sin(2x)/2sin(x)
and by extension
cos(nx) = sin(2nx)/2sin(nx)
when plugged into the integral product, all the sines cancel out apart from the first and last terms.
There is a full solution available online (for those interested) so I shan't repeat it - as it is also very long.
However, the key insight to solve that seemingly monstrous product is to use one of the trig identities - but in an unusual way - so that within just 2 or 3 terms the mind can see that the infinite product simplifies to just two terms.
Such problems are very much borderline A Level Further Mathematics - scanning the MIT paper, I think at least half of the problems could be done at school. Nowadays, a student would have to be studying Cambrdge's new Pre-U Level Mathematics as they have brought back all the topics that they have slowly removed from the (dumbed down) A Level.
I think that that's sort of the appeal of these sorts of problems. In and of themselves, they're not particularly difficult, but they require you to think outside of the box in order to get the solution. Once you see that light-bulb, the solution is obvious and elegant.
On your second point, yes, most of the problems could be done at school back in the day. I don't know the state of British mathematics education, but American maths education is in a sorry state. Most people can't wrap their head around the fact that 1/3 is in fact larger than 1/4.
Oh, I think we're in the realm of 1 in 10,000 who won't freak out just looking at that integral! There is a small counter-current holding on to mathematics as a rigorous subject; I mean, philosophy, for example, has almost totally disappeared from schools. Thankfully the maths competitions and Olympiads are away from the dead hand of government/corporate "mastery (obedience) learning". I teach a few gifted kids who haven't had their lights dimmed as yet, so that's fun!
I see that cos(x/2^1) = cos(2*x/2^2) = 1 - 2sin^2(x/2^2).
When k = 1, the product is cos(x/2^1).
When k = 2, the product is (1 - 2sin^2(x/2^2))*cos(x/2^2).
When k = 3, the product is (1 - 4sin^2(x/2^3)cos^2(x/2^3))*(1 - sin^2(x/2^3)*cos(x/2^3).
Does this lead to finding any useful pattern?
The key step is using
sin(2x) = 2.sinx.cosx
so that
cos(x) = sin(2x)/2sin(x)
and by extension
cos(nx) = sin(2nx)/2sin(nx)
when plugged into the integral product, all the sines cancel out apart from the first and last terms.
Beautiful