The condition is equivalent to ∫01(1−xn)m1dx=∫01(1−xm)n1dx. After the substitutions x↦xm and x↦xn, we have to prove that ∫01mxm−1(1−xmn)m1dx=∫01nxn−1(1−xmn)n1dx. Now let x↦(1−x)mn1. Hence ∫01mxm−1(1−xmn)m1dx=∫01n1(1−x)n1−1xm1dx, ∫01nxn−1(1−xmn)n1dx=∫01m1(1−x)n1xm1−1dx. Finally, integrating by parts, for example, ∫01m1(1−x)n1xm1−1dx, we see that the equality holds true.
As expected, the problems I'm interested in are more quaint than the usual Here's one that's strange but rather easy once you get it.
Problem 6**/***
Let Nω=N∪{ω} where we define for all n∈Nω, ω≥n. For any set S⊆Nω we write ⊔S to denote the smallest member of Nω such that ⊔S≥n for all n∈S. (Hence ⊔Nω=ω.)
Let f:Nω→Nω be a function such that:
•
For all m, n, m≤n⟹f(m)≤f(n).
•
For all S={a1,a2,…}⊆Nω, we have f(⊔S)=⊔{f(a1),f(a2),…}.
Show that the value F given by F=⊔{fn(1)∣n∈N∪{0}} satisfies f(F)=F and is the smallest member of Nω to do so.
Just to clarify... Nω is the set of natural numbers up to and including ω, and ⊔S is the greatest element of any subset S of Nω or in case of the empty set it is 1?
I see why that is the case but the problem for me here will be clarity and rigour in how I present the proof.
Just to clarify... Nω is the set of natural numbers up to and including ω, and ⊔S is the greatest element of any subset S of Nω or in case of the empty set it is 1?
I see why that is the case but the problem for me here will be clarity and rigour in how I present the proof.
Yes all you said is correct.
EDIT: Actually just a slight clarification/correction: suppose S is infinite but does not include ω, in this case ⊔S is still clearly ω so your definition that it is the "the greatest element of any subset S of Nω" is not technically right.
Actually just a slight clarification/correction: suppose S is infinite but does not include ω, in this case ⊔S is still clearly ω so your definition that it is the "the greatest element of any subset S of Nω" is not technically right.
Let p be a prime number. Find all triples of positive integers (a,b,p) such that 2a+pb=19a.
@ukdragon37 I am still unable to comprehend your definition of Nω. Can we write Nω={α∣α∈{N∪{ω}},α≤ω}? And if this is the case, then is Nω not finite?
@ukdragon37 I am still unable to comprehend your definition of Nω. Can we write Nω={α∣α∈{N∪{ω}},α≤ω}? And if this is the case, then is Nω not finite?
I'm not sure what you are trying to do. All am saying is Nω is the same as the set of natural numbers except with the addition of an "infinity" ω element that is defined to be geq all the naturals (and trivially geq itself). Nω is definitely infinite.
I can't even begin to comprehend ukD's question so here's joostan's
Objectively the difficulty of actually doing the question I think is lower than a lot of the interesting techniques that I know people here are capable of (and in fact it only relies on A-level knowledge and manipulation on sets). It's just the material seems alien since nobody at high school level, even STEP, emphasises set theory to any depth. It's a trend that I think should change, especially if we want to be able to produce the next generation of theoretical computer scientists.
The first assertion is true since f(F)=⊔{fn+1(1)∣n∈N∪{0}}, and f(1)≥1 I shall edit, if I manage to prove that F is the smallest element such that f(F)=F. By induction, we obtain fn(1)≤fn+1(1). Now suppose that there is F′ such that: F′<F; f(F′)=F′. Then there is m with F′=fm(1)=f(fm(1)). Therefore, for every n>m, we have fn(1)≤fm(1) - contradiction.
Objectively the difficulty of actually doing the question I think is lower than a lot of the interesting techniques that I know people here are capable of (and in fact it only relies on A-level knowledge and manipulation on sets). It's just the material seems alien since nobody at high school level, even STEP, emphasises set theory to any depth. It's a trend that I think should change, especially if we want to be able to produce the next generation of theoretical computer scientists.
Sets does look quite interesting, definitely something to look at over the summer
Objectively the difficulty of actually doing the question I think is lower than a lot of the interesting techniques that I know people here are capable of (and in fact it only relies on A-level knowledge and manipulation on sets). It's just the material seems alien since nobody at high school level, even STEP, emphasises set theory to any depth. It's a trend that I think should change, especially if we want to be able to produce the next generation of theoretical computer scientists.
IMHO it looks like a horrendous mess of notation But still, I agree things like groups, sets and rings seem fascinating and applicable.
IMHO it looks like a horrendous mess of notation But still, I agree things like groups, sets and rings seem fascinating and applicable.
To be fair, I am trying to sneak in a few interesting concepts, which is why I had to expend notation defining them. But this kind of question is very much like what a nasty university examiner (or even interviewer ) would give you - define some new concepts and see what you can make of them.
To be fair, I am trying to sneak in a few interesting concepts, which is why I had to expend notation defining them. But this kind of question is very much like what a nasty university examiner (or even interviewer ) would give you - define some new concepts and see what you can make of them.
Spoiler
I can see you being a totally nasty supervisor to the undergrads at Cornell
I can see you being a totally nasty supervisor to the undergrads at Cornell
They don't have supervisions there, only recitations to medium-sized groups where the grad student demonstrates example questions. I don't get to actually set the questions for undergrads.
They don't have supervisions there, only recitations to medium-sized groups where the grad student demonstrates example questions. I don't get to actually set the questions for undergrads.
Aww You'll have your dictatiorial rise to power some day
The first assertion is true since f(F)=⊔{fn+1(1)∣n∈N∪{0}}, and f(1)≥1 I shall edit, if I manage to prove that F is the smallest element such that f(F)=F. By induction, we obtain fn(1)≤fn+1(1). Now suppose that there is F′ such that: F′<F; f(F′)=F′. Then there is m with F′=fm(1)=f(fm(1)). Therefore, for every n>m, we have fn(1)≤fm(1) - contradiction.
Nearly there, you want to make the last bit clearer. For example it is not a contradiction if f is a constant function. But I'm glad someone is able to look beyond the notation and make quite a bit of headway.
Nearly there, you want to make the last bit clearer. For example it is not a contradiction if f is a constant function. But I'm glad someone is able to look beyond the notation and make quite a bit of headway.
Let me try again.
Solution 6
Firstly, f(F)=⊔{fn+1(1)∣n∈N∪{0}}, and f(1)≥1 lead to f(F)=F Secondly, by induction, we obtain fn(1)≤fn+1(1). Consequently, {fn(1)∣n∈N∪{0}} is a directed complete partial order, and its supremum is obviously F. If f is a constant, let f(n)=α for some α∈Nω. Then obviously the smallest n∈Nω such that f(n)=n is n=α. Furthermore, F=α. Now suppose that f is not a constant function. Suppose also that there is F′ such that: F′<F; f(F′)=F′. Then there is m with F′=fm(1)=f(fm(1)). Therefore, for every n>m, we have fn(1)≤fm(1), which implies that F is not the supremum of {fn(1)∣n∈N∪{0}}.