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12:02 AM
oeis.org/A001065 might be helpful. OEIS often includes varying descriptions of a sequence.
 
@love_sodam if $\mathrm{cl}(B)\subseteq U$, yes
that's the point I've been trying to get to all along
do you see why this implies $(U,U-B)\subseteq(U,U-y)$ inducing an iso on homology
 
@Thorgott Then I would say from the l.e.s of pairs $(U,U-y)$ and $(U,U-B)$ and five lemma tell they are isomorphism
 
@leslietownes oeis.org/A005114
I found it after running the python
 
oh yeah, that works too
 
derivative, math.dartmouth.edu/~carlp/rangeofs6.pdf might have interesting references.
 
12:04 AM
I was thinking LES of the triple, but either approach is fine
 
if carl pomerance has written a paper about it, it may be insanely hard. :)
 
This is an introductory book :( why are authors like this
 
introductory book in analytic number theory. very different from many uses of the term 'introductory' :)
 
@Thorgott So in the diagram chat.stackexchange.com/transcript/message/59204619#59204619 two vertical maps are iso by excision and horizontal maps are iso by what we just said. So adding $\mathrm{cl}(B)\subset U$ makes the proof works
 
yes
 
12:18 AM
@leslietownes Not necessarily.
Depends on the paper and journal. He did a lot of expository stuff.
Analytic number theory is NOT introductory number theory.
Check out Paul Pollack’s Springer book. That was basically his undergraduate honors thesis.
 
 
2 hours later…
1:57 AM
Given a finite dimensional vector space V with a basis $v_1,v_2,…,v_k$, and a subspace U of V, is it possible to reduce the basis to a basis of U?
 
What do you think?
 
It is clear to me that for any non zero subspace U, we have U= span of some $v_i$’s
 
Clear?
How many subspaces does $\Bbb R^2$ have?
 
3 (zero subspace, straight lines through origin and $\mathbb R^2$ )
 
How many lines are there?
 
2:03 AM
Infinitely many, a point on a straight line through origin is $(x,y)= x(1,m)$. So span {(1,m) } gives a straight line through origin for every real m.
 
Reread your question.
 
Hmm, no. Because standard basis $e_1,e_2$ can’t be reduced to a basis of a straight line $y=5x$ 😀
 
Bingo.
 
Yay! Thanks a lot professor Ted. So the original problem (which I have not stated) needs more work. :)
 
The converse of the extension lemma is false:)
2
Don’t forget to think about concrete examples of things.
2
 
2:48 AM
I see that now professor Ted. Thank you very much. :)
 
 
1 hour later…
4:06 AM
Draw pictures.
 
4:36 AM
Let V,W be finite dimensional vector spaces. Let $T\in L(V,W)$. Let basis of W be given as $w_1,...,w_m$. Then it is to be shown that there exists a basis of V such that first row of matrix of T w.r.t. V and W is either a zero row or has first entry equal to 1 and rest all entries zero.
The idea I have is to assume any basis of $V$ say, $v_1,v_2,...,v_n$. Then there exists scalars c's such that $Tv_i=\sum_r c_{ir}w_r$.
and then finding some linear combination of $v$'s such that the new set obtained say $v_1',v_2'...,v_n'$ is linearly independent and is such that $Tv_1'=w_1$ and $Tv_i'$(for all $i\ge 2$) is a linear combination of $w_j$'s $j\ge 2$.
But it seems difficult.
 
are you familiar with elementary row and column operations? one way of looking at this is that you can either zero out the first row or put a 1 in the (1,1) entry and zeros everywhere else using only elementary column operations.
the corresponding fact with row operations is maybe better known but it's the same principle
 
@leslietownes yes, I'm familiar with those.
 
you could do the exercise you were doing the other day about rank one operators this way too, although it's less explicitly constructive and maybe less illuminating than the direct proof you came up with
the idea would be, pick some basis v_1, ..., v_n for V and consider the matrix M of T with respect to the (v_i) and (w_k) basis. the thing about elementary column operations tells you that there's an invertible matrix A for which MA has either a first row of all zeros, or a 1 in the (1,1) entry and zeros everywhere else.
A or A^(-1) (i can never keep this straight and have to draw a diagram) implements a change of basis from the v_i that you started with to a basis v_i' that has the desired property
the effect on the matrix M of changing a basis for V is right-multiplying by an invertible nxn matrix (which you can think of as a product of elementary column operations), and the effect of the matrix M of changing a basis for W is left-multiplying by an invertible mxm matrix (a product of elementary row operations)
if you actually think about your construction from the rank one problem in these terms you'll see how the formulas you came up with can be thought of as row/column operations on what you started with
 
4:58 AM
We have $[Tv_1, Tv_2,..., Tv_n]$, wherein each $Tv_i$ is a column. I'll do column operations to get desired zeros in the first row. (Column operations because row operations will changes basis for W). Each column operation is equivalent to multiplying by some elementary matrix (just as it is in case of row operations).
 
5:50 AM
@leslietownes I think I understand your idea now.
no, I'm still working on it.
 
6:37 AM
Let $\displaystyle w_{1} ,w_{2} ,...,w_{m}$ be the given basis of W. \



Let $\displaystyle u_{1} ,u_{2} ,...,u_{n}$ be a basis of V.

If $\displaystyle Tu_{i} =\sum _{r} c_{ir} w_{r}$

$\displaystyle \begin{bmatrix}
Tu_{1}\\
Tu_{2}\\
Tu_{3}
\end{bmatrix} =\begin{bmatrix}
c_{11} & c_{12} & c_{13}\\
c_{21} & c_{22} & c_{23}\\
c_{31} & c_{32} & c_{33}
\end{bmatrix}\begin{bmatrix}
w_{1}\\
w_{2}\\
w_{3}
\end{bmatrix} =C\begin{bmatrix}
w_{1}\\
w_{2}\\
w_{3}
\end{bmatrix}$



If the first row is a zero row then we are done. Suppose that the first row has one non-zero entry.
I think I understand how to get the basis now @Leslie.
@leslietownes it seems to be a very powerful way. Thanks a lot :)
 
7:16 AM
Here's an interesting question from a graduate school interview : Let $f:S\to\Bbb R$ be a function which is continuous at $0$ where $S = \{\frac{1}{n}\mid n\in\Bbb N\}\cup\{0\}$. If $f$ uniformly continuous?
 
8:11 AM
Is the universal cover figure-eight embeddable in plane?
I mean from weak topology on universal cover onto subspace topology of plane?
 
8:39 AM
I hope I got the answer. It is on page 59 of Hatcher's book. The main point is placing the real lines with some lengths in the right positions so that an arbitrary collection of vertices should not converge to a point on the tree.
Sorry for inconvenience.
 
9:20 AM
code meatrix
I'm working on the subgraph isomorphism search engine now for Abstract Spacecraft
It's an NP-complete problem, but for small enough graphs, you can do it speedily enough
 
 
2 hours later…
11:24 AM
@Thorgott you once recommended a category theory book, I can't, for the love of me, remember what it's called
 
11:58 AM
Jun 14 '20 at 16:00, by Thorgott
must've been Abstract and Concrete Categories - The Joy of Cats by Adámek, Herrlich and Strecker or Handbook of Categorical Algebra by Borceux
 
 
1 hour later…
1:07 PM
ARG! I left my keys on a shelf at home. So I get to sit outside in the cold for an hour until someone comes to let me into my office. :(
But at least I can grade out here!
 
2:03 PM
1
Q: Galois group of $x^n-p\in\Bbb Q[x]$

love_sodam Compute the Galois group of $x^n-p$ over $\Bbb Q$. Here, $n\in\Bbb N$ and $p$ is a prime number. Let $\eta$ be the primitive $n$th root of unity. Then $\Bbb Q(\sqrt[n]{p},\eta)$ is a splitting field of $x^n-p$ over $\Bbb Q$. Let $H$ be a fixed field of $\Bbb Q(\eta)$ under the Galois correspond...

I have a question related to Galois group. I want to know if the s.e.s argument holds.
 
2:22 PM
Yay! I am inside now! My fingers are thawing!
 
@XanderHenderson. Hello, can you please give hint to solve $\sum_{i=1}^{m}i2^i$ ?
 
2:52 PM
give me lesliecoin
 
@Avra Note that: $s_{m} =\sum i2^{i} =\sum _{i=1}^{m}( i+1) 2^{i} -\sum _{i=1}^{m} 2^{i} =\frac{1}{2}\sum ( i+1) 2^{i+1} -\sum _{i=1}^{m} 2^{i}$
 
can we do is 2sum - sum?
I saw a solution did it like that
12
A: Calculate the sum of series $\sum\limits_{i=0}^{n-1} i2^i$

000Discrete Calculus works here. Via Discrete Calculus, we have summation by parts: $$\sum_{m\le k \le n} f_{k}(g_{k+1}-g_k)=f_{n+1}g_{n+1}-f_mg_m-\sum_{m \le k \le n}g_{k+1}(f_{k+1}-f_k), $$ where $f_k$ and $g_k$ are sequences. Let $f_k=k$ and $2^k=g_{k+1}-g_k$. Via observation, we see that $g_k=2...

That 2Sum - sum will work because it will eventually give us Sum!
@Koro. This is the trick I guess please?
 
The idea is to recognise that the first term on RHS is $\frac{s_{m+1}-2}2$.
 
3:41 PM
@Thorgott yup! thanks
 
for $x,y,z \in \mathbb{R}$, I need to find $\max\{z: z = x^2y \land x^2 + y^2 = 1\}$. Any reason why I can't just do a substitution and find $\max\{z:z=y-y^3\}$ and call it a day?
the condition that $z=x^2y \land x^2 + y^2 = 1$ is equivalent to the condition that $z=y-y^3$, right? or is this mistaken?
 
With what domain for $y$?
 
3:56 PM
I'm figuring out whether I need to explicitly define functions, and so a domain for $y$, or if I can think of this as just all $x,y,z \in \mathbb{R}$ which satisfy $z=x^2y$ and $x^2 + y^2 = 1$
 
You are just repeating yourself.
It is not equivalent to considering the function $z=y-y^3$ for $y\in\Bbb R$, NO.
 
oh!
$z=x^2y \land x^2 + y^2 = 1$ implies $z=y-y^3$, but the opposite isn't true?
 
Of course not.
What if you take $y=5$?
 
you get a contradiction, since $x$ has to be complex and $z,y$ have to be real
oh, that doesn't matter because $x$ gets squared
you obtain $x^2 = -24$ and $z=-120$
 
Are you going to criticize this?
 
4:05 PM
the fact $x^2 = -24$ being independent of the proposition $z=y-y^3$
 
Didn't you say that $x^2+y^2=1$ is the unit circle in $\Bbb R^2$?
 
oh! it does imply a contradiction because we stated $x \in \mathbb{R}$
but we're getting a complex $x$ if the conditions are satisfied
 
So, I return to my question. For what values of $y$ is this valid?
 
$y$ has to be equal or smaller than $1$
 
Close :)
 
4:11 PM
oh, it has to be within the closed interval $[-1,1]$
 
Perfect. There you go.
 
thanks a lot!
 
4:46 PM
Hello,
Can you please show above that it does not make any sense when $a=0 || a = 1$?
 
is there a way to maximize $z = x^2y$ subject to $x^2 + y^2 = 1$ with the pure power of reason, without lagrange multipliers?
 
implying that Lagrange multipliers are not reason
but yes, you can do it
 
it feels more like computing and less like style when doing it with lagrange multipliers
 
Hi, how can I see my extended chat history?
 
What is a simple logic paradox?
 
4:57 PM
@SAJW consider the set of all sets that are not members of themselves
 
Ah that set must include itself, but then it is a member of itself.
 
yep, it is called Russel's paradox and to make sure set theory works properly we change the axioms to prevent it
 
So it's impossible? Is every paradox impossible?
 
at Thorgott: what would be a way to solve it without Lagrange multipliers?
@SAJW gotta specify what you mean by impossible
 
@shintuku I can only think of arguing vague, that x must be larger than y, because it where small the product would be even smaller
 
5:04 PM
Can I ask a question here?
 
Why not make $z=(1-y^2)^2y$?
 
see TedShifrin's answers higher up
 
Hi, how do I separate this into partial fractions?
Some context - this is a Z transform of a transfer function. I'd like to separate this Z transform into partial fractions, so I can go from H(z) to h[n]
 
o.9
how's everyone doing here
my tda course is lit af so I'm feeling good :)
 
Suppose you have a binary search tree T, sorting numbers in the range from 1 to 500, and you do a search for the integer 250. Which of the following sequences are possible sequences of numbers that are encountered in the search?

a) (2, 276, 264, 270, 250)
b) (100, 285, 156, 204, 275, 250)
c) (475, 360, 248, 225, 251, 250)
d) (450, 262, 398, 239, 270, 250)
e) (255, 294, 260, 133, 250)

**Answer**: b by making sure BT properties are satisfied, which is clear. Is this the fastest way to do it please?
 
o.9
5:16 PM
the fastest way is to do it in order
realize that b is feasible
and profit
but I would check all the other options are clearly not feasible if I had over 1m per problem
cuz I'm insecure like that
 
in order means to check it number by number and each time check if it's constrained with binray tree property by checking if it's greater than parent in case if it's to the right...etc. This will take a lot of time if the numbers are large, say 10 numbers
I just did it each time I add a number I traverse the tree back and tick each node if it satisfied the tree properties
 
o.9
what do you mean?
each check takes under 3 seconds
 
@shintuku rewrite $x^2+y^2=1$ as $x^2=1-y^2$, so $z=x^2y=(1-y^2)y$ and this is now a single-variable minimization problem
 
o.9
and you have like 5*5 = 25 checks
so its like 1 minute
but if you are really time constrained you can get lucky and see that b is all good
 
Thanks
 
o.9
5:22 PM
what kind of exam is that for?
in any case selection tests for large number of participants are usually trash sadly
 
@o.9. CS exam :/
 
o.9
for school?
 
It's qualyfying examination yes
 
o.9
for a grade
oh
 
Yes. It's for grad school
 
o.9
5:23 PM
good luck :)
 
Thank you :)
 
o.9
I feel like a good chunk of people who visit math meta are people who dont use math se but visit a bunch of meta pages on the so network
 
:/
 
. How is this right.Finding rate is taking limit of deltafx/deltax.
 
For this question please, What is the maximum number of comparisons performed by a binary search
algorithm when searching an array with 1,048,576 elements? If we take the log of elements we will get the height of a tree, which is supposed to the number of comparisions
 
o.9
5:30 PM
This is probably the easiest site to get banned in that I've ever seen
suspended*
 
Since it's binary, we do $\log{1,048,576} = 13$
 
o.9
binary search on an array
 
The answer I have says it's 21, not sure why
@o.9. I got banned before 7 years ago :/
It's very hard to ask at the begining
 
o.9
that log seems wrong tho
maybe you took natural log?
 
I did take base 2
 
o.9
5:33 PM
no
 
Still upper bound is 13!
 
o.9
that isnt the base 2 log of that number
 
base 10 gives around 8
 
o.9
I think you took base e
 
@Thorgott but $z=y-y^3$ has no maximum, right?
 
5:34 PM
@o.9. :/
Wow, I though ln in my calculater is base 2!
 
o.9
been there
 
It's base e!
OMG
 
you need to consider the domain
 
@o.9. Can we do the same question without using calculator though please?
 
oh, restricting $y$ to $[-1,1]$, thanks
 
o.9
5:36 PM
To be honest I think it kind of depends on the implementation of the search
but
with 1 search you can get up to 2 elements
with 2 searches 4
 
@o.9. The easiest way w/o calculater is to take $2^i$
It take less than 10 seconds to realize the numbers
I guess this would be the fastest way
 
o.9
you can use 2^10 = 1024
2^{20} = 1024*1024
 
:
:|
Yes!
 
o.9
and then that is a bit smaller than what you need
so you need to take 21 I guess
thankfully I don't think I need to take a test like that ever again lol
 
Thanks, yes this is the fastest w/o calculator
@o.9. It's painful :/ Math is easier although you have 1000 lemmas and collaries per exam
:-|
@o.9. Seriously please, in a math course as a math student, I guess on average you have to proof around 200 lemmas, collaries, theorems per course!?
 
o.9
5:40 PM
It depends on the college
 
So, 4 courses per semester would equal to around 1000 lemmas, collaries and theorems in total :/
 
o.9
I think where I'm at, the professor is going to prove 1000 lemmas and corollaries and stuff
but he assumes we are all dumb af
so we just gotta learn like 20
 
@o.9. You just got 20 per course!
Seriously!
 
o.9
yeah like 20 things we gotta learn
 
I got 80 in my graph theory course :(
 
o.9
5:41 PM
oh I think the graph theory course is the hardest
 
:((
 
o.9
but I knew everything in that course like 6 years ago
I'm struggling with the topological data analysis course one
but my coleagues are strugling with graph theory I think
 
Can you imagine if the same professor is giving us 2 more courses, as I said, we would need to proof around 300
 
o.9
Idk I talk with all of my coleagues
So I think we are all sort of on the same page
 
Topological is considered the hardest after abstract algebra and functional analysis
 
o.9
5:43 PM
so I don't think we're all gonna get owned
 
after real and complex analysis, math is no more fun as math students told me :(
 
o.9
my professor asumes we all know how to calculate homologies and a bunch of homotopy groups
 
@o.9. Most don't even care about background
 
o.9
He sort of skimmed over the algorithms for voronoi diagrams
 
They just jump in proofing from day 1
 
o.9
5:44 PM
and I think we are now supposed to know how to implement em
I mean, proofs are easier than codes
 
@o.9. Agree proofs are much easier than coding
 
o.9
When you write the code you can't skip steps
 
If a student knew in advance that he/she is going to study topology/abstract math, he would probably consider going for medical degree
 
o.9
math stops being fun when you realize no one is going to hire you because you can prove sylows theorems
like learning how to change a tire fast is 10x more valuable
 
@o.9. How about measure theory ?
 
o.9
5:47 PM
Well knowing how to formalize markov chains like a chad might help you like 0.001% if you want to be a quant
but I don't think it's a dealbraker tbh
 
@o.9. From what I saw in math and applied math departments, stat students are living a live
 
o.9
they are?
 
haha
 
o.9
I mean some people in my uni got internships at facebook, although they were all already rich af ...
 
@o.9. Facts :/
 
6:10 PM
Here, if we have $T(n) = 2(T(\sqrt{n}))^2$, where $T(1) = 4$, how we can get $T(2)$ please? We substitute the value of $2$ in the equation etc? The answer I have says it's 4? We can not get to $T(1)$ from $T(2)$ as $T(\sqrt{n})$ would be a fraction. What do you think please?
 
o.9
so T is supposed to take on which values?
if it's just positive integers that relation would connect 1 to 2 I think
that's kind of weird
I guess it must be a real function
but even then they wouldn't be connected unless you add a continuity condition
something seems off
maybe they are implictly assuming $T$ is increasing?
if it is for computer science
maybe they want to bound some sort of complexity function
oh lol
yeah Dietrich is right
$T(1)$ must be $0$ haha
so what kind of question is that lulz
 
6:37 PM
@XanderHenderson Me thinks you were sweating, not "in the cold!" What lesson did we learn today? (sumting like bringing your keys with you!) ;P
 
o.9
he locked a question 40 minutes ago
who needs keys in rural areas
 
i suggest damaging the doors of where you live. i once lived in a house where the front door actually wouldn't stay closed unless it was deadbolted shut. i never went off without my keys.
 
o.9
Just climb a couple walls
well I think some people actually close off all the entrances in rural areas
but I have it on good authority that isn't normal
 
7:12 PM
Hello, For the following please
Hello, $$
T(n) = 2(T(\sqrt{n}))^2\Rightarrow \log_2(T(n))=1+2\log_2(T(\sqrt{n}))
$$

$$
t(\cdot) \equiv \log_2(T(\cdot))\Rightarrow t(n) = 1 +2t(\sqrt{n})
$$
what do you mean please when we say $$
t\left( \cdot \right) \equiv \log T\left( \cdot \right)
$$
And also, how we got please $\implies 1 +2t \sqrt{n}$ above?
 
7:27 PM
they're defining a new function $t$ such that $t(x)=\log T(x)$ for any $x$
 
@amWhy Usually, I put all of my keys on one ring. However, I don't actually own a car right now, and sometimes drive my father's truck, and sometimes drive my mother's prius (if my mother needs the truck, she comes to town, steals the truck, and leaves me the prius). So I don't want to put my work keys on my keyring. But that means that I can drive away from home without all of my keys. :\
 
log base 2, i should've said
and therefore in particular $t(n)=\log_2(T(n))$, $t(\sqrt{n})=\log_2( T(\sqrt{n}))$
 
All logs are the natural log. There is no other log.
 
tell that to those blasted computer scientists
 
o.9
what a legend
 
7:32 PM
@Semiclassical Have they accepted $\mathrm{e}$ as their Lord and Saviour? or are they doomed to Perdition?
 
they're CS people, so they're already doomed
 
o.9
hopefully they at least dont lock deleted content
 
did you hear the good news? x d/dx (log x) = 1
the identity function, differentation, and the log are three, but together they are also one.
 
where's 3, i see no 3
 
you know, if you can come up with a better joke about natural logs and the trinity, you're welcome to try.
:)
 
7:46 PM
no thx
if i want to make jokes about the trinity, i'll just point to Borromean rings
 
@leslietownes The Trinity is a Christian concept. $\exp$ and $\log$ are a Duality, and together, they are One.
For $\exp \circ \log = \operatorname{id}$.
 
i'll have to think about this. this sounds a little bit like heresy to me.
more than a whiff of cinders and ash about it.
 
@XanderHenderson. :/ I did half math by the way
@Semiclassical. Deriving such equations is simply based on change of variable please?
 
it's just function composition
if $f$ and $g$ is a function, then $f\circ g$ is a function
 
@XanderHenderson. I did half math, it was too difficult for me though I was very excited :/ I ended dropping it
 
 
2 hours later…
9:54 PM
Good evening, do someone happen to know Abel's theorem and whether a power series ∑a_n z^n (radius of convergence = 1) defined on a stolz sector is always uniformly convergent on that sector please? Even if ∑ a_n does not converge?
 
10:18 PM
0
Q: Mixed type surfaces and type changing metrics

geocalc33I skimmed over and am about to read in depth: "Isometric deformations of mixed type surfaces in Lorentz-Minkowski space" by A. Honda. Trying to grasp the concept of a "type-changing metric." From what I understand you take a surface $S$ embedded in $\Bbb L^3$ which may have a positive definite in...

if anyone can give me insight on mixed type surfaces that would be cool
 
10:33 PM
@shintuku is that what deriviative is for? y-y^3-> 1-2y^2?
to look for a extremum
oops, $1-3y^2$ should be it
 
11:33 PM
yeah, noting the domain restrictions
 
11:47 PM
Hello,
Given the following:
How please S(2n/2) becomes R(n/2)?
Let's define R(n) = S(2n). Then we have that

R(n) = S(2n)

≈ (1/2)S(√2n) + (1/2) lg 2n

= (1/2)S(2n/2) + (1/2) n

= (1/2) R(n / 2) + (1/2) n
 
If R(n)=S(2n) for all n, then by definition R(n/2) = S(n)
n/2 is just another input, after all
 
@Semiclassical. Thanks
 

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