Hey all

Hi

@Semiclassical

Suppose $f(x)=g(x)=m$, for some $x$

Let $a\le x\le b \implies f(a)\le f(x)\le f(b)$ and $-g(a)\le -g(x)\le -g(b)$

fricking boggers :/

$\implies f(a)-g(a)\le f(x)-g(x)\le f(b)-g(b)$

@Null Any ideas ?

what do you want to prove?

If I can prove that $f(a)-g(a)\le 0$ and $f(b)-g(b)\ge 0$

arithmetic mean

?

oh sorry

archimedian mean i think

do you want to show that they are then at one point f(x)-g(x)=0?

Did you see the question ?

not now

If f and g are functions, surjections on R, and are increasing and decreasing respectively on R, how can we prove that f(x)=g(x) has at most one solution ?

@Mahmoud Prove that if it has two, you contradict either $f$ increasing or $g$ decreasing.

oh!

okok

@Fargle But this won't implies that it has a solution.

that we have a solution is clear

only not that we have exactly one

Will my approach work if I prove those f(a)−g(a)≤0 and f(b)−g(b)≥0

?

@Mahmoud Yes, assuming you already have the intermediate value theorem.

Which states .. @Fargle

@Fargle was continuity not a prerequisite for that? or can that be assumed?

$\mathcal{L}$

Ah no wait, @Null is right.

I don't have it t be honest;

The power

The absolute power

$\delta \mathcal{L}$

@Fargle yeah, but can't it assumed?

because the domain is R

@Null Oh wait, yes.

Increasing surjections on $\Bbb R$ are continuous, same for decreasing.

@Fargle I still didn't take Calculus yet, so it can be negligible for me.

then prove the intermediate value theorem

not hard really

What is it ?

you can use what you prove, always

@Mahmoud The intermediate value theorem states that if a continuous function on some interval (a,b) takes on values both less than $c$ and greater than $c$, then $f(x_0) = c$ for some $a < x_0 < b$.

@Fargle neat formulation ;)

Still don't quite know how to use it

f(a)-g(a)=c

f(b)-g(b)=d

wlog, assume c<d

maybe draw two functions first to understand it, it's really intuiitive

@Mahmoud You know that if $f(a) - g(a) \leq 0$ and $f(b) - g(b) \geq 0$, then you know that $(f-g)(x_0) = 0$ for some $x_0$ between $a$ and $b$. Thus, $f(x_0) - g(x_0) = 0 \Rightarrow f(x_0) = g(x_0)$, so $x_0$ is a solution.

(and their difference)

@Fargle So my reversed proof is wrong ?

Not sure, didn't see it.

I got to the point where you started your proof,

black,blue = f,g. red=difference

now, f,g can be obviously other functions than lines, but the argument is the same for all decreasing/increasing functions

and the difference can be something else then a str8 line, but it will be too increasing or decreasing

so if the difference at some point is positive

and at some point negative

If f(a)−g(a)≤0 and f(b)−g(b)≥0 both holds, then $\exists p\le 0$ and $\exists q\ge 0$ so that $p\le 0\le q$ which is always right.

then it has to be 0 at some point

(note that this only holds because everything is continuous)

However showing f(a)−g(a)≤0 and f(b)−g(b)≥0 isn't all obvious.

what are f,g?

only incr/decr?

Defined on R, and are surjections to R

suppose $f(a)\not=g(a)$ for some a, then the difference is either >0 or <0. agree?

@Null Simply that ?

Yes I agree, and since $a\le b$ as I assumed.

if $a\not=b$ then a<b, or b>a

in R

Since $f$ and $g$ are inc./dec. surjections respectively, for every real number $a$, there are $r_1$ and $r_2$ real such that when $r > r_1$, $f(r) > a$ and when $r > r_2$, $g(r) < a$.

If you choose some $p > \mathrm{max}(r_1,r_2)$ then $f(p) > g(p)$, so that $f(p) - g(p) > 0$.

@Fargle do we need strictly decreasing/increasing for that?

@Null It's enough that they're surjections.

$g$ is decreasing, and hits everything, so it's eventually less than any number you name. Similarly for $f$.

but if they are not strictly increasing, then they might intersect at more than one point, my thought

@Null Well yes, for the uniqueness of the intersection you need strictness.

But if a function is a strictly increasing or decreasing surjection, it's a bijection.

@Fargle In my text they are strictly incr and decr

ok, then that makes things possible

Then using the argument above, you can find a solution by noting that that the difference is eventually positive to the right, and eventually negative to the left, so in between, the difference is zero.

@Fargle How can I prove it ? The differences,

@Mahmoud I just proved that the differences are eventually positive.

By reversing some of the inequalities, you prove the differences are eventually negative to the left.

Yes, I can't use the $\operatorname{max}(a,b)$

why?

Still didn't take it,

My teacher found me using $\sqrt[n]{x}$, he didn't like it, and told me that he can't accept such over-killed tools :P

@Mahmoud the methods to invent such tools rigorously are such a pain that it evens out

either way

why you want to use max(a,b)?

@Null To use Fargle's proof ?

I couldn't eliminate any of them.

meh, i butt out, im groggy

You mean you feel sleepy ?

no, i just am unorganized and dont think i help you

wait, why wouldn't he like $\sqrt[n]{x}$?

dhmo style^^

@Semiclassical I'm not supposed to know it until next year.

@Mahmoud why restrict yourself?

@Null I don't, I hate doing that, but he will correct my proof and notice some foreign notations and concepts.

@Mahmoud only make sure, YOU understand the stuff you use

@Null That is a must.

@Mahmoud so if he said "you can use bla, without proving it" you could? seems really cringy

@Null What do you mean ?

@Mahmoud did he, in any point required you to use something without a proof? Like "you can use that the root of 2 is irrational without proof"

@Null Yes, when the proof involves tools we didn't study yet.

@Mahmoud mmh, arbitary restrictions then, from your point of view at least

@Null That's why I'm here :D

screw your prof, and use whatever you can proof yourself

that excludes still some stuff

I do, but not in his exams.

because some definitions are only made later

@Null I officially give up (for now).

Sorry to interrupt, but I recently came across this question which asks why different RREFs imply different row spaces. I thought this was a standard proof that I had forgotten how to do, but after Googling it, I could not find any solutions and looking in my Linear Algebra textbook, I also could not find any solution. I eventually figured the proof out myself, but why is there no answer to this online?

According to Wikipedia, it's part of the proof to saying row equivalence is logically equivalent to same row space, so I was really surprised when I couldn't find a proof for it since it seems important.

@TedShifrin Hi! Good holidays; hope all's going well. Just FYI, I'm attempting to order myself to a self-study schedule. The dabbling so far in Spivak and yours has been pretty messy and unproductive—and definitely not intensive/frequent enough. Think I'll be doing the first four parts of Spivak first (functions to series) and then continue with the multivariate stuff. You'll probably think this a sensible (and overdue) change :)

@Fargle can you help me proving an inequality?

@Null Maybe?

@Fargle @Null LOL, math.stackexchange.com/questions/2077883/…

Anyone got any ideas for using Mobius transformations to show that for $\Bbb{R} \ni r \neq 1$, $|z-z_1| = r|z-z_2|$ determines a circle?

$\sum_{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq 1$ for n>1 First: is there a way to show that the function with points (n,sum) is strictly increasing? I never did such a thing. Because then the statement is trivial..

the sum is 1 for n=2 btw

I tried to square and use complex conjugates to show that $(1-r^2)|z|^2 = r^2 \left( |z_2|^2 - 2\Re(z \ \overline{z_2} ) \right) - |z_1|^2 + 2\Re(z \ \overline{z_1} )$.

i plugged in n=100 for curiousity, its ~2.8 something

I'm really bad with series. But, that function is clearly strictly increasing because each term is non-zero and positive.

mmh

i mean

the function (n,sum), but the sum is a very different for each n. (1,1), (2,something)

f(n) if you wish

@Null That's exactly the sequence of partial sums.

$f(n) = S_n$

mmh

i mean

sum for n=1, is 1

n=2 are some fractions

n=100, will be 2.8 something

sorry

there is n in the denominator!

I don't know what you're trying to get at. The function you're defining is literally the exact same as the partial sums.

so each sum is different

Yes, each sum is different. The function is strictly increasing because the partial sums are strictly increasing.

$f(n)$ literally means "plug n in and evaluate the sum", right?

maybe a better example:

ah yes

i thought you wanted to only add the nth term always ;)

just wanted to make sure we talk about the same

Yep.

why is it increasing?

because with bigger n, the fractions we add get smaller

albeit more fractions

@Null For any $n$, $f(n+1) = f(n) + $ some positive number.

The sum gets larger.

@Fargle that I don't understand quite

i even doubt it IN GENERAL

Oh now I see what you're saying.

Yeah, I don't know how to tackle this one.

Let $a_k > 0$ for all $k$. Do you agree that this is true, @Null? $$ \sum_{k=1}^{n+1} a_k = \left( \sum_{k=1}^{n} a_k \right) + a_{n+1} \geq \sum_{k=1}^{n} a_k $$

@Kari That's not the problem. The top value in the summation is also involved inside the summation.

@Kari but n is in the denominator of the terms

@Fargle precise pointing out the problem hehe

I can see the problem now, haha. Sorry for being so dense.

It only took me about 10 minutes >_>

Any ideas for the posted question above ?

Maybe induction, @Null@Fargle?

i couldn't say it, albeit I had seen our miscommunication^^

so im dense too haha

Maybe induction ? @ me or not ?

Oh srr

I'm off to go play some Magic.

No prob

Hi @ Robjohn

@Kari well, with induction we get $$\sum_{k=1}^{n}\frac{1}{\sqrt{k(n+1-k)}}\geq 1$$ with $$\sum_{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\geq 1$$ as our assumption

By entire, you mean complex differentiable on all of $\Bbb{C}$, @mick?

Then you're done, no @Null?

no, because $n\not=n+1$

Wait what, why the $n+1$ in the denominator?

Well the additional question is about being analytic @Kari

@Kari The original sum was defined up to $n - 1$.

The bottom inequality is his inductive hypothesis; the top is what is sought.

But yes that is the defintion of entire from the title and FIRST question @Kari

Right, I was being my own closure for a bit there, @Fargle.

Im looking for answers to the additional question(s) @Kari

@Fargle at least i'm not the only dumb here haha. Have fun at magic!

I'll see myself out before I confuse myself further.

Bye @Null

There are no dumb people here @Null

@mick true dat

The problem is we are not dumb enough to have no questions , but not smart enough to answer them ;) @Null @all

@mick woah, nice saying

Mankinds history in a nutshell ;)

can you read my mind haha

A brief history of time

Yes :p

Looking for answer to the " additional question(s) " here math.stackexchange.com/questions/289325/…

Now i tell myself to stop spamming and hope ppl have seen it :)

@robjohn hi

Hey guys! I was wondering if \mathbb{F}_{2}^{2} is a vector space over \mathbb{R}. Does anyone know?

@John11 $\mathbb F_2$ is the field with 2 elements?

Yes!

How could it be...?

ok, so if you have a vector space $V$ over a field $k$ and $x\in V$, $r\in k$, $x$ not $0$, do you know why $r \cdot x$ is different from $x$ if $r\neq 1$?

that probably isn't the best way to see it, but the point is that every $\mathbb R$ vector space that is not the zero space has infinitely many elements

but $\mathbb F_2^2$ has $4$ elements

Hey guys, I'm trying to know if it's possible to find the smallest triplet (a,b,c) of integers verifying a^2+b^2+c^2 = x, a^3+b^3+c^3=y and a^4+b^4+c^4 = z if (x,y,z) are known. By smallest I mean lexicographically.

short answer would be closure under scalar multiplication

We assume that a solution exists for that triplet x,y,z

@s.harp I'm not sure I understand. The problem is I don't see how to define the scalar multiplication. Whatever I try, it always violates the distributive property for example as I reach a contradiction

@Null Is it closed under scalar mutliplication or is it not?

@John11 every $\mathbb R$ vector space must have a scalar multiplication with elements from $\mathbb R$. You cannot define a multiplication on $\mathbb R\times \mathbb F_2^2\to\mathbb F_2^2$ that is distributive and associative.

$\pi\cdot 1=??$