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The Proof is Trivial!

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Ach, this thread is overloaded with calculus!

Here is an interesting problem I encountered not too long ago: it is tough in my opinion, but the solution is very short.

Problem 454**/***

Consider ϑ:  R(R)\vartheta:\;\mathbb{R}\to \wp(\mathbb{R}) where ϑ(x)\vartheta (x) finite and x∉ϑ(x)x\not\in \vartheta(x) for all xx. The question is whether there exists SRS\subset\mathbb{R} s.t.:

Sϑ(S)=S\cap \vartheta(S)= \varnothing and S=c|S| =\mathfrak{c}
Reply 2761
Original post by Lord of the Flies
Ach, this thread is overloaded with calculus!

Here is an interesting problem I encountered not too long ago: it is tough in my opinion, but the solution is very short.

Problem 454**/***

Consider ϑ:  R(R)\vartheta:\;\mathbb{R}\to \wp(\mathbb{R}) where ϑ(x)\vartheta (x) finite and x∉ϑ(x)x\not\in \vartheta(x) for all xx. The question is whether there exists SRS\subset\mathbb{R} s.t.:

Sϑ(S)=S\cap \vartheta(S)= \varnothing and S=c|S| =\mathfrak{c}



Well, if you don't need ϑ\vartheta to be injective,
ϑ(x)={{0} if xR{0}[br]{1} if x=0[br]\displaystyle \vartheta(x)=\begin{cases} \{0\} & \text{ if } x \in \mathbb{R} \setminus \{0\} \\ [br]\{1 \} & \text{ if } x= 0[br]\end{cases}

and S=R{0}RS= \mathbb{R} \setminus \{0\}\subset \mathbb{R}. Obviously, ϑ(S)={0}\vartheta(S)= \{0\}, so Sϑ(S)=(R{0}){0}=S \cap \vartheta(S)= ( \mathbb{R} \setminus \{0\} ) \cap \{0\}= \emptyset, and R{0}=1 | \mathbb{R} \setminus \{0\} | = \aleph_1 . Thus there is some ϑ\vartheta for which this property holds.

I guess I've made a mistake or misinterpreted the question.
(edited 10 years ago)
Original post by henpen
...


Indeed you have misunderstood the question: it asks whether for any ϑ\vartheta there exists such an SS, not whether you can find both SS and ϑ\vartheta satisfying the conditions (as an aside: 1\aleph_1 is the successor cardinal of 0\aleph_0 and nothing more. It seems you believe 1=c\aleph_1=\mathfrak{c}; this is an undecidable assertion).
Reply 2763
Since this thread loves integration:

Problem 455 **/***

Let J0(x)=2π0π2cos(xcosθ) dθJ_0(x) = \dfrac{2}{\pi} \displaystyle\int_0^{\frac{\pi}{2}} \cos(x\cos\theta) \ d\theta. Show that

0J0(x)eax dx=11+a2\displaystyle\int_0^{\infty} J_0(x)e^{-ax} \ dx = \dfrac{1}{\sqrt{1+a^2}} , for a>0a > 0


Problem 456 **/***

For b>a>1b > a > 1 find the value of 1xaxblog(x) dx\displaystyle\int_1^{\infty} \dfrac{x^{-a}-x^{-b}}{\log(x)} \ dx


To make it easier, you may assume both the Tonelli and Fubini theorem hold for both.
Original post by Lord of the Flies
Ach, this thread is overloaded with calculus!

Here is an interesting problem I encountered not too long ago: it is tough in my opinion, but the solution is very short.

Problem 454**/***

Consider ϑ:  R(R)\vartheta:\;\mathbb{R}\to \wp(\mathbb{R}) where ϑ(x)\vartheta (x) finite and x∉ϑ(x)x\not\in \vartheta(x) for all xx. The question is whether there exists SRS\subset\mathbb{R} s.t.:

Sϑ(S)=S\cap \vartheta(S)= \varnothing and S=c|S| =\mathfrak{c}



Original post by henpen
Well, if you don't need ϑ\vartheta to be injective,
ϑ(x)={{0} if xR{0}[br]{1} if x=0[br]\displaystyle \vartheta(x)=\begin{cases} \{0\} & \text{ if } x \in \mathbb{R} \setminus \{0\} \\ [br]\{1 \} & \text{ if } x= 0[br]\end{cases}

and S=R{0}RS= \mathbb{R} \setminus \{0\}\subset \mathbb{R}. Obviously, ϑ(S)={0}\vartheta(S)= \{0\}, so Sϑ(S)=(R{0}){0}=S \cap \vartheta(S)= ( \mathbb{R} \setminus \{0\} ) \cap \{0\}= \emptyset, and R{0}=1 | \mathbb{R} \setminus \{0\} | = \aleph_1 . Thus there is some ϑ\vartheta for which this property holds.

I guess I've made a mistake or misinterpreted the question.


Solution 454

As stated in the original problem, the solution is very short.

A very short solution does not allow time to consider different cases.

Threfore the answer is either yes or no for all such functions.

Since we have found a function for which the answer is yes, the answer must be yes for all such functions.
Problem 457**

Find all automorphisms of the real numbers

i.e. find all functions f:RRf:\mathbb{R} \rightarrow \mathbb{R}

such that

f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y)

and

f(xy)=f(x)f(y)f(xy)=f(x)f(y)

Now do the same of the complex numbers***
Original post by james22
Problem 457**

Find all automorphisms of the real numbers

i.e. find all functions f:RRf:\mathbb{R} \rightarrow \mathbb{R}

such that

f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y)

and

f(xy)=f(x)f(y)f(xy)=f(x)f(y)

Now do the same of the complex numbers***


Done now - thanks, LotF!

Spoiler

(edited 10 years ago)
Original post by Smaug123
...


Hint: f(x)0f(x)\geq 0 for x0x\geq 0. Two possible routes now: prove that a solution to Cauchy's eq. satisfying either 1. ff increasing or 2. there exists a real interval on which ff is bounded below [the former case is easier to deal with in my opinion], must be f=axf=ax

Spoiler


Original post by Flauta
Problem 439**

Find 0πdθsinθ+cscθ\displaystyle\int^{\pi}_0 \frac{d \theta}{\sin \theta + \csc \theta}

Nothing particularly ground breaking, just thought the graph looked pretty cool


Solution 439**

I haven't seen the graph yet, but Wolfram agrees with my final result :smile:

Spoiler


Original post by Lord of the Flies
...


Loving your new avatar :wink:
(edited 10 years ago)
Original post by Smaug123
Done now - thanks, LotF!

Spoiler



Nice solution. Although you are wrong about the automorphisms of the complex numbers. It was a bit of a trick question because you have found all the contructablle automorphisms, but the set of automorphisms of C\mathbb{C} is the same size as the power set of the reals.

EDIT: If you assume that the automorphisms must be continuous, you have them all.

EDIT 2: Your mistake with the complex numbers bit is you assume that the restriction of an automorphism of C\mathbb{C} to R\mathbb{R} is an automorphism of R\mathbb{R}, this is not true. In your proof for R\mathbb{R} you assumed that y^2>0 for all y, but this may not be true in C\mathbb{C}.
(edited 10 years ago)
Original post by james22
Nice solution. Although you are wrong about the automorphisms of the complex numbers. It was a bit of a trick question because you have found all the contructablle automorphisms, but the set of automorphisms of C\mathbb{C} is the same size as the power set of the reals.

EDIT: If you assume that the automorphisms must be continuous, you have them all.

EDIT 2: Your mistake with the complex numbers bit is you assume that the restriction of an automorphism of C\mathbb{C} to R\mathbb{R} is an automorphism of R\mathbb{R}, this is not true. In your proof for R\mathbb{R} you assumed that y^2>0 for all y, but this may not be true in C\mathbb{C}.


Ah, that did twinge in my mind at the time, but I dismissed it because I was thinking about increasing-ness :frown: I loved the problem, though!
Original post by Smaug123
Ah, that did twinge in my mind at the time, but I dismissed it because I was thinking about increasing-ness :frown: I loved the problem, though!


Try and find one that is not continuous (hint: you need the axiom of hoice, in particular zorns lemma).
Problem 458***

For u,v>0u, v >0 show that

ucosvxxsinvx(u2+x2)2dx=π2u2euv.\displaystyle \int^{\infty}_{-\infty} \dfrac{u\cos vx - x\sin vx}{(u^2+x^2)^2} dx = \dfrac{\pi}{2u^2e^{uv}}.
(edited 10 years ago)
Original post by Lord of the Flies
Well done :wink: I'll type it out for you (credit to Izzy in the OP):

1nn1+xndx=1nnxn(11+(1/x)n)dx=1nnxn(11xn+1x2n)dx=n1n1xn1x2n+1x3ndx=n[1n112n1+]n[n1nn1n12n2n1+]0\displaystyle\begin{aligned}\int_1^n \frac{n}{1+x^n}\,dx &=\int_{1}^n \frac{n}{x^n}\left(\frac{1 }{1+(1/x)^n}\right)dx\\&=\int_1^n\frac{n}{x^n}\left(1-\frac{1}{x^n}+\frac{1}{x^{2n}}-\cdots\right)dx\\&=n\int_1^n \frac{1}{x^n}-\frac{1}{x^{2n}}+\frac{1}{x^{3n}}-\cdots\,dx\\&=n\left[\frac{1}{n-1}-\frac{1}{2n-1}+\cdots\right]-\underbrace{n\left[\frac{n^{1-n}}{n-1}-\frac{n^{1-2n}}{2n-1}+\cdots\right]}_{\to\,0}\end{aligned}

Hence the limit is ln2\ln 2

Replacing nn with π\pi obviously makes no difference.


how does one evaluate the last line to get ln2?
Original post by Khallil

Spoiler




Solution 439**

I haven't seen the graph yet, but Wolfram agrees with my final result :smile:

Spoiler




Loving your new avatar :wink:


nice solution though i think a more basic one would be to change the sin^2 in the denominator into 1-cos^2 and let u=cosu, then a little partial fractions and were done.
Original post by newblood
nice solution though i think a more basic one would be to change the sin^2 in the denominator into 1-cos^2 and let u=cosu, then a little partial fractions and were done.


I noticed that way but I really wanted to bash some of the algebra out in the integrand with the Weierstrass sub :tongue:
Anyone like stats? Here are a couple of questions i particularly liked that id done recently [not very hard, especially for someone whos done any uni stats or S3/4 even]:

Problem 459**
A fair coin is tossed n times. Let un be the probability that the sequence of tosses never has two consecutive heads. Show that u_n =0.5u_(n-1)+0.25u_(n-2) .Find also u_n

Problem 460*
The radius of a circle has the exponential distribution with parameter lambda. What is the probability density function of the area of the circle.
(edited 10 years ago)
Original post by newblood
How does one evaluate the last line to get ln 2?


nkn11k[]k>0(1)k+1k\displaystyle\frac{n}{kn-1}\to \frac{1}{k}\Rightarrow \big[\cdots \big] \to \sum_{k>0} \frac{(-1)^{k+1}}{k}
Original post by Lord of the Flies
nkn11k[]k>0(1)k+1k\displaystyle\frac{n}{kn-1}\to \frac{1}{k}\Rightarrow \big[\cdots \big] \to \sum_{k>0} \frac{(-1)^{k+1}}{k}


Why are you justified in swapping the order of limits here?
Original post by james22
Why are you justified in swapping the order of limits here?


if youre referring to the original integral, i think he also changed the signs as well to change the limits, just for ease of manipulation id imagine

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