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7:00 AM
i have a question
 
I guess to get there requires modifying the default axioms of ZF such that e.g. one can only count up to three
 
can we write $$5^{log x} = x ^ {log 5}$$
their graphs are same
 
Plug in x = 0
 
thats not in the domain of log
 
7:03 AM
so, they are not equal?
 
No, they are not
 
we can see that by putting x =1 also
nice idea man
why did not i try this at first? silly me
wait a sec
 
Well, 1 does not work
 
What is the best homework help
2
site
2
as paid
2
 
@Lagranian chegg
@orbit-stabilizer yeah! in case of one, they become equal
 
7:07 AM
You can try any other number though, if this is log base 10, then plug in 100. You have 25 != 100^log(5)
easiest is just 0
 
hmm
xactly
 
$\arrow$
 
but, this is log to the base e
 
$5^{\log 1} = 5^0 = 1^{\log 5} = 1$
 
so, might as well put e
 
7:09 AM
A necessary conditions for two functions to be equal is that their domains need to be the same
 
you can restrict the domain of one of them, but then you cant have equality without stating that
 
Not that it helps in your case
 
gotcha
 
7:10 AM
If you want an easier way of showing that they're not equal. Take $log_5$ of both sides
 
ok
then, they are becoing equal to log x
 
$5^{\log x} = x^{\log 5}$

$\log x \ln 5 = \log 5 \ln x$

$\frac{\ln x}{\ln 10} \ln 5 = \frac{\ln 5}{\ln 10} \ln x$
 
@Abhishekstudent My payment is being declined
while I'm trying to
 
i took log on both sides
@Lagranian did u contact them regarding this issue?
 
7:13 AM
@Abhishekstudent No, I didn't.
I'm so confused right now.
 
@Secret huh
 
@Lagranian well, there is another site that you might try
 
Okay wait what's this star spam?
 
@Abhishekstudent What is it?
 
that surprises me
 
7:14 AM
@Lagranian wyzant
 
so we restrict to x>0 and they are equal
 
@orbit-stabilizer yeah, me too, but then ,I recall there's something called natural operations which does commute , so I am not sure if it is related
 
i never should have passed pre-cal
 
@Abhishekstudent but I wanna chegg
 
@Secret how did u get to the first step
 
7:15 AM
@Abhishekstudent take ln both sides
 
@Abhishekstudent I wonder why it is being declined
Would it be billing adress?
 
@Lagranian call their customer care number or write an email
 
if you have an exponent in the thing inside the log, you can bring it out front
 
@Lagranian do u have any other card
 
@Abhishekstudent Currently I don't have
 
7:16 AM
@Lagranian calling them is the best thing you can do right now
 
yeah, who is doing this star spam?
 
@Abhishekstudent However, does it depend on country?
 
@Lagranian absolutely
 
@Secret huh, never heard of natural operations
 
@Abhishekstudent aaaah!
@Abhishekstudent The problem would be it then
 
7:17 AM
@Lagranian why?
 
@Abhishekstudent If it depends on country
I couldn't see my country on adresses
 
@Lagranian ohhhhh
 
@orbit-stabilizer I might misremebered something from wikipedia, let me check quickly...
 
@Abhishekstudent What do you think about it?
 
@Lagranian thats why they are not accepting the payment
your country is currently not supported by them @Lagranian
 
7:18 AM
@Abhishekstudent w8
@Abhishekstudent My all payment info was true
about cc
then the country blocks me
 
@Lagranian absolutely
 
@Abhishekstudent What would you recommend to me atm
Paypal or something?
 
Have you considered doing the problems yourself?
2
 
I'm already doing myself.
 
^that deserves a star
 
7:20 AM
but sometimes I'm stuck
 
Is there support outside of the class?
Teachers, or students in your class with whom you can work with?
 
ok I did misremembered natural operations from ordinals and thinking about logarithms, sorry about that.

But I suspect those two functions should differ when complex domain is involved
 
If that fails, there is youtube, and a plethora of online resources
 
uno why I'm searching a site
2
 
@Secret ah okay. Man this is going to be one of my go to examples for things that shouldn't work, but do
it looks so wrong
x^log(5) = log(5)^x
I doubt this works for other numbers
 
7:23 AM
$x^{\log_b y} = y^{\log_b x}$
 
@Abhishekstudent Isn't there any other site where is Q&A
3
 
checking:
 
5^logx = x^log5
 
$\log_b y \ln x = \log_b x \ln y$
 
oh it does
wait
 
7:24 AM
but what about the complex case (for that I need to revise its conversion formula...)
 
even just in the reals tho, this is suprising to me
anywho, thanks for that @Abhishekstudent
night all
 
$z^{\log_u w} = w^{\log_u z}$
$e^{\log_u w \text{Log} z} = e^{\log_u z \text{Log} w}$
$e^{\frac{\text{Log} w}{\text{Log} u} \text{Log} z} = e^{\frac{\text{Log} z}{\text{Log} u} \text{Log} w}$
 
Turns out that 2 hidden layers are indeed (qualitatively) more powerful than 1, despite the universal approximation theorem.
 
$n\in \Bbb{Z} : \frac{\text{Log} w}{\text{Log} u} \text{Log} z = \frac{\text{Log} z}{\text{Log} u} \text{Log} w + 2n\pi i$
so the complex case seemed to differ only at the branching points by the factor of $2\pi i$
0
Q: Understanding why $x^{\log_b(y)} = y^{\log_b(x)}$

PP1211According to wikipedia, we have that $$ x^{\log_b(y)} = y^{\log_b(x)} $$ because $$ x^{\log_b(y)} = b^{\log_b(x) \log_b(y)} = b^{\log_b(y) \log_b(x)} = y^{\log_b(x)} $$ But what justifies that first leap? $$ x^{\log_b(y)} = b^{\log_b(x) \log_b(y)} $$

Ah I see, it's because the products of the indices of exponentials commute
$a^{bc}=a^{cb}$
So, if we want that to fail, we need to use noncommutative rings such as matrices
i.e. $A^{BC}\neq A^{CB}$ in general
13
Q: Matrix raised to a matrix

kηivesGood evening, I was wondering if there is such a valid operation as raising a matrix to the power of a matrix, e.g. vaguely, if $M$ is a matrix, is $$ M^M $$ valid, or is there at least something similar? Would it be the components of the matrix raised to each component of the matrix it's raise...

[Random]
Matrix tetration:
${}^{c}A = A^{A^{A^{\cdots^A}}} : = e^{ e^{ e^{\cdots e^{A \log A} \log A} \log A} \log A}$
 
7:52 AM
Can we use 'Gluing lemma' for Lipschitz function instead of continuity?
Please someone help
 
 
1 hour later…
9:00 AM
@Daminark you only need your forms to be locally lipschitz
 
Oh really now? Huh
Also hey @Alessandro!
 
In fact if you know it for C1 the proof is like two lines
You might not even need that, there's probably a version that works for horrible things
 
9:17 AM
Hi @Dami
 
@EricSilva sp00k'd
@Alessandro how's it going?
 
9:43 AM
Quite well, I've got more uni work to do than time to do it, but that's pretty common :P
What about you?
 
[Conversation fragment]
... the idea is to let them know first, see if they think it is a good idea, and then on the day it happens, the MO and MSE users will probably head to it ...
... think about it, this is basically graphically speaking, 3 streams of users all converging to one point in spacetime in cyberspace ...
 
@AlessandroCodenotti doing a bio paper, suffering less from a lack of time and more from a lack of motivation
 
Bioinformatics stuff?
 
Not quite, I did do a class which involved that last year but this is about mathematical methods in epidemiology
 
hmm, sounds like you need a lot of population modelling and need to involve solving a lot of dynamical systems...
 
9:52 AM
So the simplest model involved is called SIR, which is about large scale population dynamics
It is very simplistic and makes many assumptions, but what I'll be focusing is more along the lines of the info limitation
Since it just gives a bunch of numbers based on means calculated wrt some population, but it has little resolution on interactions between subpopulations of different natures, and how that's important to know about as well
Ramble on for a bit on graph theory, then give some exposition on a new model
Anyway I'm tired as shit since it's nearly 4 so good night!
 
have a good sleep
 
10:49 AM
@MatheinBoulomenos Your answer was my original approach, but I thought I was wrong. hahaha
 
@ÍgjøgnumMeg your approach does work as well, you just have to phrase it a bit more carefully
I edited
 
@MatheinBoulomenos Ah okay, thanks for the answer
 
glad to help
 
11:05 AM
[Chemistry] hmm... the more I read the data analysis manual, the more I wonder whether instead of trying to crawl through all chapters covered by the basic manual, I should look at my data and then decide on how to slice it first, before returning to the manual...
 
Does the ideal $(2, \sqrt{3}) \subset \Bbb Z [\sqrt{3}]$ contain $1$? I feel like it shouldn't but maybe I'm missing something...
 
It shouldn't contain $1$
 
Yeah, why do I think it does
lol
$(2, \sqrt{3}) = \lbrace 2a + b\sqrt{3} : a, b \in \Bbb Z[\sqrt{3}]\rbrace$ right?
 
So $a = -1$ and $b = \sqrt{3}$ give $1$?
 
11:18 AM
Ah, wait, $a=-4$ and $b=3\sqrt{3}$
 
8
Q: Proving that the set of natural numbers is well-ordered

ErnestConsider the following theorem: "Every non-empty set of positive integers has a minimum element". The proof I usually see is one that uses contradiction, and does not seem like the easiest possible proof. I think there is an easier proof, and I wonder why I never see it. Does it contain an inva...

hmm, I cannot seemed to express this fact into a binary tree...
 
$\Bbb Z[\sqrt{3}]/(2,\sqrt{3}) \cong \Bbb Z[x]/(x^2-3,x,2) \cong \Bbb F_2[x]/(x^2+1,x)$. $(x^2+1,x) = (1)$, so the quotient is zero and the ideal is the whole ring
$1 = x^2+1 - x\cdot x$
this may seem overkill, but it's a very systematic approach
you can always reduce this kind of questions to polynomials over finite fields
 
Nice! Thanks.
 
ah ok, so if the ordering is given by reading from left to right, then it is dense and hence not a well ordering
but if begin from zero and then label each level, then every chain that one pick will terminate in finite number of steps. Hence the naturals are well ordered under that ordering scheme
So looking at the elements generated from this binary tree that enumerates the binary digits of the naturals, for every two elements, there are countably many progressively larger elements
 
11:36 AM
0
Q: $L=\lim_{n\to \infty}\frac{1}{\sqrt[n]{n!}}$

Maneesh Narayanan $L=\lim_{n\to \infty}\frac{1}{\sqrt[n]{n!}}$. Then (1)$L=0$ (2)$L=1$ (3)$0<L<1$ (4) $L=\infty$ Let $x_n=\frac{1}{\sqrt[n]{n!}}$. Taking logarithm on both sides, $\log(x_n)=\frac{1}{n}\sum_{k=1} log(\frac{1}{k})$ Using the Cauchy's first theorem on limits,$\lim_{n\to \...

Answer given by Find_x is wrong .right?
 
By picking some intervals in the resulting number line, then we are selecting countable subsets of the naturals, which because there is a limit on how short each stick can become, the existence of a minimum is guarenteed, hence well ordering is established
So... generalising this ordering to unbounded sequences, it will mean this ordering, it will somehow need to go "way beyond". Before the investigation continues, let's rewrite the well ordering of the naturals as a binary tree properly:
 
@ManeeshNarayanan Why didn't you read the question completely? - It's also their fault for not reading it entirely, but you shouldn't be that picky, given that you have edited because you made a mistake in the first place.
 
Actually in the title only there was a typing error.
Where should I note the edited thing? Please help me. @Mr.Xcoder
 
I know, but why are you so aggressive to the answerer by blaming them for not reading carefully? You could have commented on their answer once, saying that you made a mistake and then asking them to edit in, but you shouldn't accuse them of not being careful readers.
@ManeeshNarayanan I'd say the following: Remove all your current comments, and then write another comment on their answer explaining your changes and apologising for the mistake in the title. That's all I can help you with.
 
ok. I was not aggressive. sorry for my bad tone.
 
12:10 PM
@Mr.Xcoder You may enjoy reading this.
@Secret might as well ^
 
@SimplyBeautifulArt looks like some kind of array function
and takes in cantor normal forms
very similar to what I tried to do in the past by extending veblen all the way down to the lower hyperoperations, except much cleaner
So, I guess one can do OCF on that?
 
Hey chat. Is there any way to show that if two diagonal matrices are similar, the diagonal entries of one are a permutation of the other.
So far I've figured out that one must be obtained from the other by conjugating by a permutation matrix, but I'm stuck at this point.
I've also shown that if one matrix is $P$ and the other $Q$, then for any $i$, $P_{ii} = q_{jj}$ for some $j$.
 
12:32 PM
@Secret Yeah
In fact, that's what I'd call an OCF.
 
@TimTheEnchanter Well, what are the eigenvalues of each matrix.
 
@user104729 The diagonal entries, which are the same for both, but I'm not sure how to show that the number of times a particular entry appears in P is the same as that for Q.
 
@TimTheEnchanter Do you know Jordan Canonical Form?
 
@user104729 I'm afraid not.
 
@TimTheEnchanter Do you know how to show that changing basis preserves eigenvalues, and that change of basis corresponds to conjugation?
 
12:47 PM
@user104729 Yes
I know that because similar matrices have the same minimal polynomial.
 
So you are done.
 
@user104729 What I don't get is why in the case of diagonal matrices where one entry is repeated, they cant differ in the number of repeats. Say if two 3x3 matrices have minimal polynomial (x-1)(x-2), how would I show that a matrix with diagonal entries 1,1,2 cannot be similar to one with entries 1,2,2?
 
@Secret Please have look to the new answer here: math.stackexchange.com/questions/164422/…
 
Your method will be very useful to deal with matrices that has the form isomorphic to the complex numbers. Meanwhile the exp and log case by Qiaochu will handle all matrices that are not in $G_2(\Bbb{R})$ provided the log converges
Now it wonders me, what about matrices that behave like quaternions, an analogous pathway should also be possible
 
1:11 PM
:P
 
And now... I need to figure out how to write 2-adics in a tree, for an illustration of mine need that
 
1:28 PM
Equilibirum sign anybody? (the one of chemistry) I forgot the mathjax code for it...
 
@Abcd $\rightleftharpoons$
@Abcd \rightleftharpoons
 
@ÍgjøgnumMeg Thanks, I saw an easier one too (seconds ago) : $\ce{<=>}$
\ce{<=>}
 
1:51 PM
p-adics are soooooooo weird
(illustration came later)
 
2:07 PM
Hi all; I have a question
 
3
Q: Can I prove the divergence of the series $\sum_{n=1}^{\infty}\frac{1}{n}$ using this technique?

Maneesh NarayananWhen I see the answer to the problem. How to check the convergence of the series whose elements are taken from the set $A$? One question raised in my mind, Can I prove $\sum_{n=1}^{\infty}\frac{1}{n}$ diverges using the following technique? By the Prime decomposition theorem, $n=2^{j_1}3^{j_2}...

Please help me.
 
So basically, the tree of the usual ordering of the natural numbers and the p-adic integers are identical, except the notion of magnitude is flipped around due to the p-adic norm
 
Why can we test if a number is prime by testing all the prime divisors of it under $sqrt(k)$
 
Can you please explain? I don't understand your question.
 
2:17 PM
I'm learning number theory, and one of the ways I learned to test if a number k is prime, is by dividing it under every prime under sqrt(k). I just don't understand why that process works.
 
@DarkRunner what happens if $k$ is not prime?
 
@MatheinBoulomenos If k is not prime, it will be divisible by one of the prime numbers under sqrt(k), and thus be determined composite.
 
What do you not understand, why dividing only by primes or why stopping at sqrt(k)?
 
If $k$ is not prime than there at least two prime factors of $k$ (which may not be distinct) what can you say about their size?
 
I understand we divide by primes because they are, in essence, "the building blocks" of products, by I don't understand why we stop testing at sqrt(k).
@MatheinBoulomenos One prime factor will be less than the other (if distinct)
 
2:24 PM
Because a number k must have a prime factor smaller than sqrt(k) if it's composite. To see why is that suppose that k=ab, with a,b>sqrt(k), do you see what goes wrong?
 
Okay let's call the smaller one $p_1$ and the (maybe) bigger one $p_2$, then $p_1^2 \leq $?
 
Ah, basically if the prime factors are >\sqrt{k}, they "overstretch", in essence, over the number, right?
 
2:37 PM
Corrections:
Other weird things about the p-adic world include:
1. The further away two numbers from each other, the closer they are (because the p-adic norm is the reciorpcal of the largest prime that divides the number)
2. 0 is simultaneously the largest number (because all p adic numbers with more non decimal integers converges to it) and the smallest number (because its p adic norm is 0)
3. There are many numbers that will become closer together the "further away" one walks from the origin (the value of the p-adic norm depends only on the largest prime that divides the number, thus e.g. numbers 31 to 16, under the p-adic norm, are all equal to $\frac{1}{16}$. Since the prime powers become further apart as they increase, it means more numbers will be close together the larger their digits are)
So... a p-adic space seemed to share some similarities with hyperbolic spaces in that it becomes harder and harder to travel close to the origin, but very easy to travel far away from it
(To be refined later when I study them proper, as so far I only read part of the sections of this maths.gla.ac.uk/~ajb/dvi-ps/padicnotes.pdf to prepare the illustration) So when touring such space, one need to be mindful of both their location and proximity to something
In particular, it should in general very hard to meet somebody in such space, because not only it is required for the individuals to do the opposite: to move away from each other, but also it becomes exponentially harder to get close together as the distance in the space need to cover becomes exponentially larger
This is analogous to how in hyperbolic space, there are many coplanar lines that don't intersect
 
2:59 PM
@Secret surely -1 is the largest number
 
@Balarka are you here?
 
1
Q: Primary Ideal in a PID

user193319 Let $R$ be a PID. An ideal $P$ in $R$ is said to be primary if $ab \in P$ and $a \notin P$ implies $b^n \in P$ for some $n \in \Bbb{N}$. Show that $P$ is primary if and only if $P = (p^n)$ for some $n \in \Bbb{N}$ and some prime element $p \in P$. Here is my attempt: Assume that $P = (p...

 
@LeakyNun Yes, you are right, hmm, seems that the negative p-adic integers are in some sense "wraps around" as they have countably many nonzero digits
but then what is the valuation of $-1_p$ since using the definition of p-adic norm, it can be divided by a prime power arbitrarily large, hence suggesting its p -adic norm is close to 0. Meanwhile, 0 is defined to have p -adic norm 0, so which one is larger in the p -adic norm?
 
@Secret I think $\| (-1) \|_p = 1$
 
3:17 PM
hmm interesting, so a p -adic number countably long gets arbitrarily close to 1 as its rightmost digits increases
 
what do you mean?
 
Suppose p=2. Then -1=...111111111, -2=...111111110, -3=...111111101 etc.

Now since $|-a|_p=|a|_p$ it follows that the larger the negative digit in steps of powers of 2, the smaller their corresponding p -adic norm
so wrt the origin, the sequence {...,-3,-2,-1} approaches 1 from the inside of a disk of radius 1
 
3:37 PM
Start at $1_p$ which has $|1|_p=1$
Walk to larger and larger p -adic integers, 1,2,3,4,5,...
For every prime power you passes, you get closer to the origin in steps of $\frac{1}{p^n}$
Note as the prime powers become more spaced out, it becomes harder and harder to approach the origin (light gray contours, the steeper it is, the smaller the effective distance you travel towards the origin under the p -adic norm)
Then some unspecified point later, the digits become countable and thus you end up with e.g. ...aaaaaaaa stuff, where a=p-1. These are the negaitve p adic integers
Now since the p -adic norm is symmetric under negation, it follows the larger the negative digit, the smaller its corresponding p- adic norm
And so, you end up taking a round turn some countable moment later and heading straight back towards the circle with radius 1
...aaaaaaaa(a-2),...aaaaaaaaa(a-1),...aaaaaaaaa
(and then I realised I had a mistake in my diagram)
Edited diagram = ignore all dark gray circles
Now noticed how the trip back starts out not so easy (the negative number is large in the digit sense, or the countable string of digits have a lot of digits not equal to a), but as the journey continues, it becomes easier and easier (because the prime powers needed to make aaaaaaa... becomes less). Eventually you found yourself back at where you start again: $1_p$
Now back at $1_p$ you plan to journey away from the origin. To do so, you travel "towards" it
$1,\frac{1}{p},\frac{1}{p^2},\frac{1}{p^3},...$
Again, the trip becomes easier and easier as you go. Eventually, you are indefinitely far away from the origin in finite amount of time
 
3:57 PM
So that means, the current attempt to establish a well ordering of countable binary strings should not suffer from infinitely decreasing chains by extrapolating from the ordering of the naturals (where e.g. numbers arbitrarily close to 0 become larger and larger instead of smaller).
The only challenge is then, when is the onset of countable sequences occur. It seems clear that using the stick diagram of the naturals as a reference, the resulting number line will have uncountably many elements between any two elements
The observation that a certain undetermined threshold need to be pass to cross from finite strings to countable strings as the strings get more entries of 1s suggest there are elements that act as limits in the structure
More investigation needed in order to determine how it works in detail
The existence of infinite elements (elements that are unreachable from below) thus demonstrate whatever this construction is, it is not embeddable in the reals and vise versa.
Our aim, if possible, is to demonstrate this generalisation of the natural numbers are in bijection with all the ordinals in $\omega_1+1$. In particular, to show that $\sup (n < \omega : \text{n digits of } 1) = \omega_1$
In mathematics, the supernatural numbers, sometimes called generalized natural numbers or Steinitz numbers, are a generalization of the natural numbers. They were used by Ernst Steinitz in 1910 as a part of his work on field theory. A supernatural number ω {\displaystyle \omega } is a formal product: ω = ∏ p p n p , ...
This is the closest thing that I think resembles this construction to be carried out
 
4:14 PM
Hi there!
I have a stupid simple question: I am trying to integrate by substitution (3x)(3x-1)^-3
what i have done:
du/dx=3
therefore du/3=dx
so xu^-3 du
x=(u+1)/3
therefore integrate (u+1)/(3u^3)
 
i need to prove that each borel measure -$\mu$ - on $\Bbb R$ that satisfy :
1) each point has an interval containing that point with finite measure
2) invariant - $\mu(A+x) =\mu (A)$
is lebegue measure up to a multiplicative constant
someone can help ?
 
this is wrong what have i done wrong
 
@audittxl There is nothing you have done wrong. Note the next step you can then simplify it into two fractions 1/(3u^2)+1/(3u^3) which can be easily integrated
Also use tinyurl.com/cfqcvpc LATEX, it is much easier to read
 
4:29 PM
$$x^2$$ testing....
$x^2$
nope... not working
 
(cont.) O wait a sec... I found an infinitely decreasing chain:
$...1111111,...1111110,...11111101,...111111100,...11111111011,...$
The cofinite binary sequences are the infinitely decreasing chains
O well, that one is easy to solve, by virtue that there are only countably many cofinite sequences since there are only countably many finite sequences
throws cofinite sequences into the bin
Now, what do we left, obviously an incomplete set, but can we well order the remaining elements :?
->
 
4:44 PM
If $N(I) = \lvert \mathcal{O}_K/I \rvert$ for some number field $K$ then why should $N(I)(x + I) = I$ for all $x \in \mathcal{O}_K$ without the fact that $N(I) \in I$? The book I'm using makes this step but I can't quite see why it's true
 
Hello mathein
 
Can you give me some ideas on how you view rings ideals
I find stuff that you say are very usefull :D
 
@ÍgjøgnumMeg $N(I)$ is the order of the additive group of $\mathcal{O}_K/I$, so this is just Lagrange
 
Ideals have the property that if a in I and r in R
ar in I
if we compare ideals with normal subgroups
 
4:56 PM
@KasmirKhaan (Right ideals [for sick people])
 
normal means gHg' =H , for g in G
if we look at the structure they seem to share some stuff
@user104729 I just do commutative rings this course
 
yes, ideals are similar to normal subgroups in that they are the kernels of homomorphisms
 
but am aware of right and left arent the same
well the way i see it also
 
they both have some structure of hmm
I dont know how to say it but
 
4:58 PM
Ok, now to check why my proof did not capture this case.
 
we make operation of them with elements outside the set
and we get back to the set
 
You should, even at this point, be careful with your notation. If you don't be careful with notation, you'll be less practiced later on. Most people use left ideal notation though. (In my experience)
 
in normal Groups we have ghg' in H
ideals ar in I
@MatheinBoulomenos please tell me if am making sense or not :D
 

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