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The Proof is Trivial!

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Reply 40
Avatar for und
und
OP
16
Original post by j.alexanderh
Solution 7:

Spoiler


Haven't seen you around in a while! :hi: How have you been?
Problem 9**

Show that there is no function f:  N→Nf:\;\mathbb{N}\to\mathbb{N} satisfying f(f(n))=n+2013f(f(n))=n+2013
(edited 11 years ago)
Original post by und
Haven't seen you around in a while! :hi: How have you been?


I'm pretty good, thank you. Et toi?

Problem 10*:

Evaluate ∫0π2sin⁡17xsin⁡17x+cos⁡17x dx\displaystyle\int^\frac{\pi}{2}_0 \frac {\sin^{17}x}{ \sin^{17}x + \cos^{17}x}\ dx
Solution 10

x→π2−x⇒∫0π2sin⁡17x dxsin⁡17x+cos⁡17x=∫0π2cos⁡17x dxsin⁡17x+cos⁡17x\displaystyle x\to \frac{\pi}{2}-x\Rightarrow \int_0^{\frac{\pi}{2}} \frac{\sin^{17} x\,dx}{\sin^{17} x+\cos^{17} x}= \int_0^{\frac{\pi}{2}} \frac{\cos^{17} x\,dx}{\sin^{17} x+\cos^{17} x}

Hence I=12∫0π2dx=π4\displaystyle I=\frac{1}{2}\int_0^{\frac{\pi}{2}} dx=\frac{\pi}{4}
(edited 11 years ago)
Original post by Lord of the Flies
Solution 10

x→π2−x⇒∫0π2sin⁡17x dxsin⁡17x+cos⁡17x=∫0π2cos⁡17x dxsin⁡17x+cos⁡17x\displaystyle x\to \frac{\pi}{2}-x\Rightarrow \int_0^{\frac{\pi}{2}} \frac{\sin^{17} x\,dx}{\sin^{17} x+\cos^{17} x}= \int_0^{\frac{\pi}{2}} \frac{\cos^{17} x\,dx}{\sin^{17} x+\cos^{17} x}

Hence I=12∫0π2dx=π4\displaystyle I=\frac{1}{2}\int_0^{\frac{\pi}{2}} dx=\frac{\pi}{4}

Dat substitution :sexface:
Solution 8

Let mm run from 22 to n+1n+1 and pip_i is the ithi^{\text{th}} prime number (i<pi)(i<p_i).

Consider Nm=(p1⋯pn)q+1+mN_m=\big(p_1\cdots p_{n}\big)^{q+1}+m where qq is the highest power occuring in the prime factorisations of all possible values of mm

When mm is not a prime power NmN_m has a factor of the form pi⋯pjp_i\cdots p_j where pi,⋯pjp_i,\cdots p_j are the primes in the prime fact. of m.m. When mm is a prime power we need to check that Nm=pka(P+1)N_m=p_k^a(P+1) is not a power of pkp_k. This is clearly true since PP is a multiple of pkp_k and thus P+1P+1 is not
(edited 11 years ago)
Reply 46
Avatar for Noble.
Noble.
17
Original post by Lord of the Flies
Problem 9**

Show that there is no function f:  N→Nf:\;\mathbb{N}\to\mathbb{N} satisfying f(f(n))=n+2013f(f(n))=n+2013


Solution 9

Looking at the sets S={f(n):n∈N}S = \{f(n) : n \in \mathbb{N}\} and T=f(f(n))={n+2013:n∈N}T = f(f(n)) = \{n+2013 : n \in \mathbb{N}\} clearly T⊂S⊂NT \subset S \subset \mathbb{N}. If such an ff exists, it is injective and is also bijective between the sets N\S\mathbb{N} \backslash S and S\TS \backslash T so (N\S)∪(S\T)=N\T(\mathbb{N} \backslash S) \cup (S \backslash T) = \mathbb{N} \backslash T contains an even number of elements, which is contradicted by the fact the cardinality of N\T\mathbb{N} \backslash T is 20132013
Original post by Mladenov
Let me try again.

Solution 6

Firstly, f(F)=⊔{fn+1(1)∣n∈N∪{0}}\displaystyle f(F) = \sqcup\left\{f^{n+1}(1)|n \in \mathbb{N}\cup\{0\}\right\}, and f(1)≥1f(1)\ge 1 lead to f(F)=Ff(F)=F
Secondly, by induction, we obtain fn(1)≤fn+1(1)\displaystyle f^{n}(1)\le f^{n+1}(1). Consequently, {fn(1)∣n∈N∪{0}}\displaystyle \left\{f^n(1)|n \in \mathbb{N}\cup\{0\}\right\} is a directed complete partial order, and its supremum is obviously FF.


Excellent. :smile: However the steps after this point can also be completed (more cleanly I think) without the use of contradiction.

Consider any N∈NωN \in \mathbb{N}_\omega satisfying f(N)=Nf(N) = N. Since 1≤N1 \leq N we have f(1)≤f(N)=N f(1) \leq f(N) = N. Hence by induction fn(1)≤N f^n(1) \leq N for any natural nn and therefore any NN satisfying f(N)=Nf(N) = N is an upper bound for the CPO {fn(1)∣n∈N∪{0}}\displaystyle \left\{f^n(1)|n \in \mathbb{N}\cup\{0\}\right\}. However out of all of them FF is the supremum of the CPO, so by definition the smallest.

The point of the question was to get you to prove an instance of Kleene's fixed-point theorem.
(edited 11 years ago)
Original post by Lord of the Flies
Solution 10

x→π2−x⇒∫0π2sin⁡17x dxsin⁡17x+cos⁡17x=∫0π2cos⁡17x dxsin⁡17x+cos⁡17x\displaystyle x\to \frac{\pi}{2}-x\Rightarrow \int_0^{\frac{\pi}{2}} \frac{\sin^{17} x\,dx}{\sin^{17} x+\cos^{17} x}= \int_0^{\frac{\pi}{2}} \frac{\cos^{17} x\,dx}{\sin^{17} x+\cos^{17} x}

Hence I=12∫0π2dx=π4\displaystyle I=\frac{1}{2}\int_0^{\frac{\pi}{2}} dx=\frac{\pi}{4}


Concise, as usual. :wink:
Original post by Noble.
...


Nail on the head! :wink:
Reply 50
Avatar for shamika
shamika
16
Am I allowed to bombard this thread with problems?
Problem 11* (if you have never seen substitution before, **)

Find the general solution, by a suitable substitution or otherwise of this differential equation.

dydx=sinx(cosx+y)cosx−y\displaystyle \dfrac{dy}{dx}=\dfrac{sinx(cosx+ y)}{cosx-y}
(edited 11 years ago)
Reply 52
Avatar for und
und
OP
16
Original post by shamika
Am I allowed to bombard this thread with problems?

Yes, go right ahead! :tongue:

Original post by j.alexanderh
I'm pretty good, thank you. Et toi?


I'm good. Applying for maths at university I take it?
Problem 12*

I have a cube and I want to paint each side a different colour. How many different ways can I paint the cube with nn colours, n∈N,n≥6n \in \mathbb{N} , n\geq 6 ?
(edited 11 years ago)
Reply 54
Avatar for und
und
OP
16
Original post by Star-girl
Problem 12*

I have a cube and I want to paint each side a different colour. How many different ways can I paint the cube with nn colours?


Solution 12

First we choose the six colours, so there are n choose 6 possibilities =n!6!(n−6)!=\frac{n!}{6!(n-6)!}. Without considering identical cases we have 6!6! possibilities for colouring the cube, but we divide by 4∗6=244*6=24 where 6 is the number of possible anchor faces and 4 is due to rotation about the anchor face, so we get n!24(n−6)!\frac{n!}{24(n-6)!}
(edited 11 years ago)
Original post by Star-girl
Problem 12*

I have a cube and I want to paint each side a different colour. How many different ways can I paint the cube with nn colours?

A cube has 6 sides and there are 24 ways to orientate a cube (each one of six faces can be placed and each placement can be rotated about 0, 90, 180 or 270 degrees). So the number of distinct cubes that are possible if we consider each side to be different is 6!/24 = 30. Given n colours, we must choose 6 (one for each face). So overall we have 30*nC6

(I assumed that n>6).

Problem 13*

Find the exact value of:
42+42+42+...\sqrt{42+\sqrt{42+\sqrt{42+...}}}
As an extention, try the above with 6 instead of 42. Also try 2 and then try 8. Can you spot a pattern and/or find a way to spot more numbers that "work"?
Solution 13

x=n+n+⋯⇒x=n+x⇒x2−x−n=0x=\sqrt{n+\sqrt{n+\cdots }}\Rightarrow x=\sqrt{n+x}\Rightarrow x^2-x-n=0 which has solutions for all positive n (pick the positive one)

The required solution is an integer when n=k(k−1)n=k(k-1) (the solution is then k).
Problem 14**

Prove that 260−12^{60}-1 is divisible by 61.
Reply 58
Avatar for Noble.
Noble.
17
Original post by Star-girl
Problem 14**

Prove that 260−12^{60}-1 is divisible by 61.


Solution 14

As 61 is prime and 2 is not divisible by 61

260≡1(mod61)2^{60}\equiv 1 \pmod{61} by Fermat's Little Theorem. Hence:

260−1≡0(mod61)2^{60} - 1\equiv 0 \pmod{61}
Problem 15*/**

Evaluate lim⁡n→∞∫1nn1+xn dx\displaystyle \lim_{n\to\infty} \int_1^{n} \frac{n}{1+x^n}\,dx

What if the upper limit in the integral were fixed, say π?\pi?
(edited 11 years ago)

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