Mathematics - 2023-04-05 (page 1 of 2)

Ted Shifrin

00:14

@D.C.theIII Integral looks right. I have no idea what your last clause means.

D.C. the III

When I was playing with it before I wasn't taking into account that I want the area in the cube $0 \leq x,y,z \leq 1$, so I was letting my upper limits go all the way to the top plane

Jack Ohara

@TedShifrin Hello Ted

Ted Shifrin

00:41

Hi Jack

@D.C.theIII It helps to notice what vertices of the cube or on that tilted (not top) plane.

D.C. the III

Riiiiiiight....I fell into the trap of just jumping to a conclusion of what the area should be like instead of being meticulous.

Ted Shifrin

00:57

I don’t know to jump to conclusions in problems like this, seriously.

D.C. the III

01:42

Suppose $f: [a,b] \times [c,d] \rightarrow \mathbb{R}$ is continuous and $\frac{\partial f}{\partial x}$ is continuous. Define $F(x) = \int_c^d f(x,y)dy$ prove. Show $F$ is continuous.

Proof attempt:

$\|F(x_1,y) = F(x_2,y\| = \|\int_c^df(x_1, y) dy - \int_c^d f(x_2,y)dy \| = \|\int_c^df(x_1 - x_2, y)dy\| < \int_c^d\|f(x_1 - x_2, y)\| dy$. By uniform continuity I know there exists a $\delta$ s.t $\|f(x_1-x_2,y)\| < \frac{\epsilon}{d-c}$, therefore $\int_c^d\|f(x_1 - x_2, y)\| dy < \int_c^d \frac{\epsilon}{d-c} = \frac{\epsilon}{d-c} \cdot (d-c) = \epsilon$, so $F(x)$ is continuous. I don…

Ted Shifrin

01:57

Regular absolute values. So $f$ is linear?

D.C. the III

sigh...I don't know if $f$ is linear. The regular absolute value because this is only a function of $x$ correct?

$|\int_c^df(x_1, y) - f(x_2,y)dy | < \int_c^d |f(x_1, y) - f(x_2,y)|dy $ and I can still apply uniform continuity to this object.

Ted Shifrin

Guess again.

$\le$, not $<$.

D.C. the III

Yes, I was just typing it quick not to make you wait, but in my proper solution I would have it correct

What am I not taking account of there?

Ted Shifrin

What is the meaning of $\|\cdot \|$ versus $|\cdot|$?

You also need to be more precise about what $\delta$ means.

user2236

Also the act of discovering what you missed is called learning.

D.C. the III

02:09

$\|\cdot\| = $ norm while $| \cdot| = $ absolute value

As for $\delta$ it means that for all $x_1, x_2 \in [a,b]$ when $|x_1 - x_2|< \delta$ then $|f(x_1,y) - f(x_2,y)| < \epsilon$

shintuku

02:53

@XanderHenderson thanks for math.stackexchange.com/a/3110351/755010

3

user 7269591

03:06

+1

👍👍

Silly Goose

I am having a bit of trouble understanding why in (14) the RHS of the inequality is $e$. Is the $\lim_{n \rightarrow \infty} \text{sup} s_n = e$?

I can maybe intuitively see that at the very least $\lim_{n \rightarrow \infty} \text{sup} s_n \leq e$. But I am having trouble formalizing this result

leslie townes

is rudin, where e is defined to be lim s_n?

if a_n <= b_n holds for all n, then the same inequality holds if you slap limsup on both sides

Silly Goose

yes it is

right so we get the relation $\lim_{n \rightarrow \infty} \text{ sup}t_n \leq \lim_{n \rightarrow \infty} \text{ sup}s_n$. But I am having trouble getting to $e$.

robjohn

@SillyGoose This answer might be useful

Silly Goose

is there some result that $\lim_{n \rightarrow \infty} a_n \leq \lim_{n \rightarrow \infty} \text{ sup}a_n$?

robjohn

03:21

what does the limit on the right mean?

what is $\sup a_n$?

Silly Goose

So to my understanding $\{a_n\}$ is a sequence of real numbers. Then, sup$a_n$ is the supremum of the subsequence $\{a_1, a_2, ..., a_n\}$.

Here is exactly what Rudin says, though it's not really explicit

robjohn

So you really mean $\sup\limits_{k\le n}a_k$

Silly Goose

I think so

I am having trouble understanding the notation but that seems reasonable

robjohn

Usually we define $\limsup\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}\sup\limits_{k\ge n}a_k$

$\liminf\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}\inf\limits_{k\ge n}a_k$

both of these exist or are $\pm$ infinite for any sequence

Silly Goose

Can I also write $\lim\limits_{n\to\infty}\sup\limits_{k\ge n}\{a_k\}$? Since we should be taking the supremum of a set

robjohn

03:31

it would be $\lim\limits_{n\to\infty}\overbrace{\sup\{a_k:k\ge n\}}^{\sup\limits_{k\ge n}a_k}$

Silly Goose

I see

Oh wait should the inequality be flipped?

robjohn

Nope

limits deal with things that happen in the tail of the sequence

not in the beginning

Silly Goose

ah

Wait but my understanding was that we are taking the supremum of subsequences starting with the first point and then adding more and more points each iteration, so this is the wrong understanding?

robjohn

well, it depends on what you are doing with those subsequences.

Silly Goose

well so I think $\{s_n\}$ should be a sequence of real numbers. each $s_n$ is the partial sum (up to $n$ terms) of a series.

robjohn

03:42

now you're talking about a series. The sequence is the sequence of partial sums of the series.

Silly Goose

Since taking sup$s_n$ literally would not make sense ($s_n$ is a single real number, not even a set), I am confused

robjohn

13 mins ago, by robjohn

it would be $\lim\limits_{n\to\infty}\overbrace{\sup\{a_k:k\ge n\}}^{\sup\limits_{k\ge n}a_k}$

so $\sup\limits_{k\ge n}a_k$ is the sup over a set

leslie townes

look at theorem 3.19

some of this static may be due to rudin not adopting the more standard definition of lim sup

robjohn

is this baby Rudin?

leslie townes

yes

robjohn

03:46

I do have that, but not right with me

Silly Goose

here we are

robjohn

I used to have RCA, but it got lost when my parents' house was sold.

leslie townes

yes, he's doing that (with the sequences labeled slightly differently)

lim sup s_n in the argument in 3.31 is e, by his definition of e

roughly speaking, you can "lim sup" both sides of an inequality of sequences to get an inequality of lim sups, he's doing that in a situation where when you lim sup the RHS, the lim sup is actually a limit that actually exists and was earlier defined to be e

Silly Goose

how do I see that lim sup s_n is e? I am still confused what lim sup as an operation does, which is why I was struggling to show that lim sup s_n = e

I guess I could try to show that e is the supremum of the set of all subsequential limits, but that kind of beats around the bush

leslie townes

look at definition 3.30

Silly Goose

03:51

$e = \lim_{n \rightarrow \infty} \sum_{k=0}^n \frac{1}{k!}$

leslie townes

and the remarks that follow

if i were to add more words to rudin, it would be "the sequence s_n = sum k=0..n 1/k! is a convergent sequence, and we are defining its limit to be e"

Ted Shifrin

@SillyGoose Careful not to overuse the letter $n$.

Silly Goose

oh whoops

Ted Shifrin

Oh, you changed it.

Silly Goose

per your comment yes

Ted Shifrin

03:52

LOL, OK.

leslie townes

so the fact that "lim s_n" exists is noted/explained somewhat right after definition 3.30, and e is just rudin's name for it

Ted Shifrin

Might be my name for it, too.

robjohn

$e=\lim\limits_{n\to\infty}\left(1+\frac1n\right)^n$ is mine

the series is derived

leslie townes

rudin promoted that to theorem status

robjohn

it is valid either way

leslie townes

03:57

they're just lemmas to me

Ted Shifrin

@robjohn That’s certainly mine when talking to precalc students. Neither is the definition for calc theory students.

Silly Goose

well so we have $\lim_{n \rightarrow \infty} t_n \leq e \implies \lim_{n \rightarrow \infty} \text{ sup }t_n \leq \lim_{n \rightarrow \infty} \text{ sup } e $, (I don't think this is what you were getting at, but this is a new confusion) well this doesn't make sense though :P.

robjohn

@TedShifrin what's the definition for them?

leslie townes

i forget what stewart does. is it the a for which the slope of a^x at 0 is 1?

Ted Shifrin

Down with $e$. It’s just a left-wing Democrat conspiracy.

robjohn

03:58

Ah

leslie townes

silly: at that stage in rudin's argument i don't know if he knows or you know that lim t_n exists yet

Ted Shifrin

For me, it’s $\log ^{-1}(1)$.

leslie townes

lim sup always exists, so you can infer from t_n <= s_n for all n that limsup t_n <= limsup s_n

and limsup s_n is e

robjohn

@TedShifrin Just don't mention it in Tennessee (which has 4 e's)

Ted Shifrin

Ugh.

Silly Goose

04:00

wait but i am not getting why lim sup s_n is e

I must be missing something :P

robjohn

What is your definition of $e$?

Silly Goose

i get that lim s_n = e

wait the one I am using @robjohn?

robjohn

if $\lim\limits_{n\to\infty}s_n$ exists, then it equals $\limsup\limits_{n\to\infty}s_n$

Silly Goose

maybe this result is in the exercises... I have never seen it before :P

leslie townes

silly: the limit existing is equivalent to the liminf and limsup being equal (and in this case their common value is the value of the limit)

Silly Goose

04:04

oh

well this is nice to know XD

leslie townes

this falls right out of rudin's definition. if s_n converges to e, all subsequences of s_n also converge to e, so the set "E" in rudin's definition of lim sup (definition 3.16) is the one-point set {e}

Silly Goose

wait but why do all subsequences converge to the limit?

leslie townes

it's a little surprising he doesn't say much more about this, but it's also not like he's hiding anything too major

hooray for rudin

robjohn

@SillyGoose if a sequence converges to a limit, then all subsequences also converge to that same limit.

leslie townes

silly: that falls out of the definition of convergence (given epsilon, take N that 'works' for the sequence in the sense of the epsilon-N condition, the same N will 'work' for every subsequence)

rudin includes this as a 'remark' after definition 3.5

he's nice enough to say "we leave the details of the proof to the reader," which is ten more words than he usually says about stuff like that

Silly Goose

04:07

ah gosh i must have read over that bit

well this is helpful

Koro

04:48

@DLeftAdjointtoU Is there any solution manual to Ireland and Rosen's number theory book?

shintuku

05:27

how could we give a rigorous description of a space where distance does not preserve volume

e.g. the mandelbox here youtube.com/watch?v=QhMdL4kSnsg

Koro

$(q^{n^2-n-1})= (q^n-q^2)...(q^n-q^{n-1})$

?

nvm

user 726941

06:17

(removed)

Koro

06:39

In the search of solutions to Ireland and Rosen's book, I subscribed to numerade and some of the solutions there turned out to be incomplete (truncated and in some mathjax didn't render properly) and some even looked completely wrong.

Apparently there are no solutions available online after chapter 9 of the book.

The teacher finishes off 1 chapter of the book in one day in class.

user 726941

:O

Koro

In the same way, I can teach even biology students and can 'finish off' five chapters in half an hour and complete their entire syllabus in just a week.

it can be done faster. Just project the book on a screen and turn pages at high speed.

user 726941

Medical intern students have to memorize under sleep deprived conditions.

< 6 hours sleep

Koro

same here.

user 726941

for 6 consecutive days

Koro

06:51

I for the first time life developed a new disease: lack of sleep. I went to the doctor here and I hate to say this he was not helpful at all (as expected). He loaded me with so many medicines. I was worried and thought I may have to take sleeping pills.

leslie townes

just read a few pages of 'real and complex analysis.' ha, ha.

Koro

But then I stopped taking the medicines the doctor prescribed to me and brought some medicines from the outside and I was fine.

I was able to sleep again. That made me so happy. This happened without me taking sleeping pills.

user 726941

Good for you pal 😁

Koro

@user726941 I understand, it is a very difficult and painful situation.

Due to lack of sleep, there is headache and you feel like every cell in your body is demanding sleep but as you go to bed, close your eyes, you can't fall asleep even if the whole body is tired!!

It's a very very painful situation. I wish it never happens to anyone.

user 726941

The movie The Paper Chase shows the same for law students.

Koro

06:57

And then some strange thoughts start to kick in. I was thinking: 'am I going to collapse now?'

because see if I walked away from room, I felt dizzy. That was due to weakness caused by the medicines the doctor here prescribed.

user 726941

Don't push too hard to stay awake, your brain needs time to absorb information.

Psychologists call it "consolidation."

Koro

I asked the doctor here: what is the cause of it? This never happened to me before. And he was browsing something on his device... and I said: here are the medicines etc.

Franklin

Can anyone please help me with this: math.stackexchange.com/questions/4673094/… ?

Koro

Only teachers are not terrible here.

lethargy is very common here.

user 726941

Memory consolidation

Memory consolidation is a category of processes that stabilize a memory trace after its initial acquisition. A memory trace is a change in the nervous system caused by memorizing something. Consolidation is distinguished into two specific processes. The first, synaptic consolidation, which is thought to correspond to late-phase long-term potentiation, occurs on a small scale in the synaptic connections and neural circuits within the first few hours after learning. The second process is systems consolidation, occurring on a much larger scale in the brain, rendering hippocampus-dependent memories...

"within a few hours after learning"

Koro

07:06

yeah . Brain processes what was learnt before sleep.

user 726941

Sleep makes it solid.

user 726941

07:27

Sleep and memory

The relationship between sleep and memory has been studied since at least the early 19th century. Memory, the cognitive process of storing and retrieving past experiences, learning and recognition, is a product of brain plasticity, the structural changes within synapses that create associations between stimuli. Stimuli are encoded within milliseconds; however, the long-term maintenance of memories can take additional minutes, days, or even years to fully consolidate and become a stable memory that is accessible (more resistant to change or interference). Therefore, the formation of a specific memory...

Sourav Ghosh

07:44

@Franklin I have added a solution. math.stackexchange.com/a/4673113/977780

Franklin

08:15

@SouravGhosh I have edited it. Is my complete solution valid, now?

@SouravGhosh A little suggestion(optional) : Please edit your post, so that I can accept it!

@SouravGhosh thanks for your clarifications and validations!

Astyx

08:53

@TedShifrin $\operatorname{nnn}_{n\to n}\operatorname{\Large{N}}^n_{n=n}\frac{1}{n!}$

PNDas

11:18

Hi can you please give references for this theorem: Let $A$ be a bounded linear operator on a Banach space $X$. Then the IVP $u_t=Au, t\in\mathbb R$ and $u(0)=f\in X$ has a unique solution $u\in C^{\infty}(\mathbb R;X)$.

I know that the solution is $u(t)=e^{At}f$.

Franklin

13:20

0

Q: Permutation Combination

Let $C = \{(i, j)|i, j \in \mathbb Z,\; 0 ≤ i, j ≤ 24\}$. How many squares can be formed in the plane all of whose vertices are in $C$ and whose sides are parallel to the X−axis and Y − axis?

I stumbled upon this. Any ideas how to solve this, all the comments and answer seems unclear. 🤔

Xander Henderson

@Franklin Have you attempted to do what was suggested in the comments?

Franklin

I tried doing it for smaller arrangements, say, $i,j\leq 2$ as suggested by a commentor

Xander Henderson

And?

Franklin

@XanderHenderson I am in search of patterns here, to be honest

Xander Henderson

I often get the impression that you approach mathematics from the point of view that the goal is to produce an answer. When I ask you if you have done a thing or not, I am not looking for a "yes" or "no" answer---I am asking you to explain your thinking. The implicit question is "did you try [x]? and, if so, how did it go? what did you actually do?"

Franklin

13:24

Let me play with increasing nos a bit more

@XanderHenderson ok, ok, ok, ok. Let me formalize my approach a bit, I will definitely let you know about my ideas

, once it looks presentable and less convoluted

Xander Henderson

Again, remember that the goal of mathematics is, ultimately, to communicate ideas. Solving a problem is pointless if you can't explain your solution to another person.

user 85795

13:45

In mathematics the art of proposing a question must be held of higher value than solving it.
-Georg Cantor

user 85795

14:05

After watching "Clouds Are Not Spheres," I
now see why Mandelbrot was considered such a "think outside the box" kind of mathematician.

Koro

14:38

D^2 can not retract to S^1.

It's true but hard to believe.

Here is a proof: Let $i:S^1 \hookrightarrow D^2$ be inclusion. The induced homomorphism $i_*$ should be an injection if there existed the said retraction.

But this is not possible as pi_1 of D^2 is 0 and that of S^1 is Z.

D^2 can contract to any point on its boundary.

shintuku

15:04

https://www.youtube.com/watch?v=QhMdL4kSnsg

is there any way to formalize the fact that the volume of this structure, when you're outside of it, is smaller than its volume when you're inside of it? and any time you go deeper into it, the volume sort of expands

geocalc33

Take $X=\Bbb R^{1,1}$ and its conformal compactification $\overline X.$ How do you construct an immersion $f:\overline X \to \Bbb R^{2,1}?$

Xander Henderson

15:22

So I got totally nerd-sniped by math.stackexchange.com/q/784895, which @Franklin cited, above. I have written an answer, which I will undelete once @Franklin gives me a more complete response to my questions. I'll not that my essential idea is to integrate $1$ over the space of possibilities (where, because the space is discrete, the integral is just a sum). This is a different approach than the one suggested in the comment and other answer.

So, @Franklin, please respond to my questions. :D

In fact, I would encourage you to figure out the problem per the given hints, and then post your solution.

user 85795

Have you seen Clouds are not Spheres? @XanderHenderson

Xander Henderson

@user85795 Yes.

Koro

7. Covering Spaces; Lifting Criterion - Pierre Albin

►

I have one confusion here. I think that in the covering space, path a should also be returning to the bottom point.

What is the meaning of this ?
Now let $\tilde X$ be any other graph with four ends of edges at each vertex, as
in X.

what is "graph with four ends of edges at each vertex"?

edge has two ends.

how come 4?

what's going on?

Mr. Feynman

15:43

Hello. I want to prove that $B^n/S^{n-1}$ is homemorphic to $S^n$. By means of stereographic projection I've proved that $B^n\setminus\partial B^n\cong S^n\setminus\{N\}$. Could you help me understand if what I did afterward works? I noticed that $B^n/S^{n-1}$ is basically $B^n\setminus\partial$ plus another point (the boundary identified to a point). At this point I extend the homeomorphism from $B^n\setminus\partial$ to $S^n\setminus\{N\}$ to a map from $B^n/S^{n-1}$ to $S^n$.

which maps the boundary $\partial B^n$ to $N$. Since both of this are closed set, I concluded that this extended map is continuous (and obviously invertible)

Koro

I think the homeomorphism that you proved is wrong. What is $S^n\setminus \{N\}$ ? Are you removing north pole?

Mr. Feynman

Yes

Koro

then it is wrong, I think.

Circle minus a point is homeomorphic to R for example.

Mr. Feynman

And the circle is $S^1$

Koro

Try to show that the quotient is one point compactification of $B^n- S^{n-1}$

Intuitively, the space $B^n -S^{n-1}\cong R^n$, which 1 pt. compactifies to $S^n$

Mr. Feynman

15:48

I used the fact that the $n$-sphere $S^n$ without the north pole is homeomorphic to $\mathbb{R}^n$ and then the fact that the open ball $B^n-\partial B^n$ is homeomorphic to $\mathbb{R}^n$. What's wrong with this?

Thorgott

@Koro and that's the same as $S^n\setminus\{N\}$

Koro

Oh I'm sorry. I thought you have written $B^n/\partial B^n$

Mr. Feynman

Oh, ok

Koro

But you have written $B^n- \partial B^n$

which is as you said $R^n\cong S^n - \{N\}$.

@Thorgott: can you please also look at my question? This is regarding covering space of S^1 wedge S^1.

Mr. Feynman

The part that I am not sure about is if it's ok to extend the homeomorphism as I said above and conclude such extension is a homemorphism

Koro

15:53

but you said that you proved it to be continuous.

maths student

A firm that refurbishes personal computers (PCs) recently received 24 faulty computers
of a similar make and model. 21 PCs have been fixed but an intern mistakenly packaged all the PCs to storage without labeling. The repair engineer will need to test the PCs one after another until he identifies the 3 PCs that are not yet fixed. Find the probability that he tested 15 computers to complete the task

$P(\text{test 15, exactly 3 faulty}) = \frac{{21 \choose 12}{3 \choose 3}}{{24 \choose 15}} = \frac{154}{2,704,156} \approx 0.00006$ Is this correct ?

Koro

The result you are proving is true in more generality :-). X= compact hausdorff, A = a closed subspace of X, then X/A = one pt. compactification of (X-A)

The proof is almost similar to how you are trying to do it (but it doesn't use any stereorgraphic projection)

maths student

@Koro

Mr. Feynman

@Koro I said I concluded it is continuous because in $B^n-\partial B^n$ it is continuous by construction and I basically added a point (closed set in this case) to the domain and its image is another point (still a closed set)

@Koro Sorry, I'm not familiar with topology so I don't know such result :/

maths student

@Mr.Feynman

A firm that refurbishes personal computers (PCs) recently received 24 faulty computers
of a similar make and model. 21 PCs have been fixed but an intern mistakenly packaged all the PCs to storage without labeling. The repair engineer will need to test the PCs one after another until he identifies the 3 PCs that are not yet fixed. Find the probability that he tested 15 computers to complete the task
$P(\text{test 15, exactly 3 faulty}) = \frac{{21 \choose 12}{3 \choose 3}}{{24 \choose 15}} = \frac{154}{2,704,156} \approx 0.00006$ Is this correct ? @XanderHenderson

Xander Henderson

16:02

@mathsstudent Off the top of my head, I have no idea.

What is your thinking?

And why are you pinging me?

I don't have any particular expertise on counting.

Thorgott

@Mr.Feynman that's not quite enough. if $X$ is a space, it is possible to put different topologies on $X\sqcup\{\ast\}$ such that the point $\{ast\}$ is closed and the subspace topology on $X$ is the original topology. for example, you can take the disjoint union topology on $X\sqcup\{\ast\}$, but also the topology whose open sets are the open sets of $X$ and the entire space.

@Koro which question?

Koro

33 mins ago, by Koro

I have one confusion here. I think that in the covering space, path a should also be returning to the bottom point.

The link takes to that diagram.

A016090

@mathsstudent Your numbers for your combinations are incorrect. Furthermore, you need to think of the problem at hand. If the technician finds all 3 faulty computers in the first 3 tries, he will not try the remaining 12 times. What does this imply about the problem?

Mr. Feynman

@Thorgott I'm using the quotient topology in thr quotient set, which apparently satisfies the requirements you mention. I don't understand why (or what) it is not enough.

Koro

@Mr.Feynman Second line: The part 'I noticed that ...'. Question is: Suppose X is a space, and suppose you say X plus a point p, then I think you mean X union {p}. Right? But what topology are you giving to 'X plus p'?

Mr. Feynman

16:25

@Koro Yes, I mean $X\cup\{p\}$. In this case, since I was saying that $X\cup\{p\}$ was basically my quotient set (because the boundary is identified to a point), I'm assuming there is the quotient topology (the original topology was the one subspace topology of the sphere in $\mathbb{R}^n$)

If I wasn't clear enough, I'm saying that $B^n/S^{n-1}=(B^n-\partial B^n)\cup\{P\}$ (where P is the identified boundary) and in the LHS I have the quotient topology

Thorgott

@Koro I'm not sure what the phrase "path should return to the bottom point" is supposed to mean nor how to read the diagrams in the video.

@Mr.Feynman I'm saying just because you have two spaces and the results of removing a point from each is homeomorphic does not quite mean the two spaces themselves are homeomorphic. But yes, using the quotient property to get continuity of the map in one direction is a good step.

Xander Henderson

16:43

Boy... it is getting to be a thing today. I try to interact with folk, and they ignore me entirely...

robjohn

@XanderHenderson except for those with whom you don't try to interact.

Ted Shifrin

@XanderHenderson Are you saying something?

Xander Henderson

No. Nothing at all. :/

Ted Shifrin

That's what I thought.

Xander Henderson

Yeah. Carry on.

robjohn

16:45

It sure is quiet in here...

Ted Shifrin

@Thorgott Too bad I never thought to give this as a test question on a topology exam :)

Franklin

@XanderHenderson Sorry for being late, I was busy with some of my daily coursework problems. To be back again, I think the number of squares in an $n*n$ grid is $1+2^2+3^2+...+n^2$ . No, I have not been able to prove this, but surely this is the pattern it follows

Xander Henderson

@Franklin Why?!

3 hours ago, by Xander Henderson

I often get the impression that you approach mathematics from the point of view that the goal is to produce an answer. When I ask you if you have done a thing or not, I am not looking for a "yes" or "no" answer---I am asking you to explain your thinking. The implicit question is "did you try [x]? and, if so, how did it go? what did you actually do?"

Franklin

@XanderHenderson I did observations :) with $0\leq i,j\leq 2,3,4...$

But I really dont understand if you mean greatest integer function by [x] ?

Xander Henderson

@Franklin No, I mean [x] to mean some fill in the blank.

In the previous conversation, [x] is the statement "to use the hint in the comments? or the answer already given?".

@Franklin This is a start. What observations did you make?

Franklin

16:51

@XanderHenderson number of squares in an $n*n$ grid is $1+2^2+3^2+...+n^2$

Xander Henderson

Presumably, you spotted a pattern. Can you extend that pattern?

@Franklin WHY?!

What is the proof?

Franklin

@XanderHenderson extend to what ?

@XanderHenderson if I had the proof, I wouldh've been done :?)

@XanderHenderson trying to figure out

But I am makinv it too complicated

Xander Henderson

@Franklin You made observations for a small number of cases. Can you extend that observation to larger cases?

Franklin

@XanderHenderson For the proof part, as you mention previously, my idea is: Finding k×k squares possible in an n×n grid, where n\leq k,

I tried to consider squares from origin, then translate k units in x axis, build a k*k square

Xander Henderson

Okay... that might work (I don't know). How does the proof go?

Franklin

16:55

And continue this, until we are exhausted, and move to y=1, line and do the same and continue this.

@XanderHenderson not a desirable outcome, as I only have this idea, and when I try implementing it, it becomes too much convoluted

And confusing as well...

Now, I am out of ideas

Xander Henderson

Well, the suggestion was to start with small cases.

Often, when you start with small cases and work them out, you spot some kind of pattern.

Since the pattern extends over the natural numbers, a natural (ahem) approach might be induction?

Franklin

@XanderHenderson cool! Let me try again with this idea fresh in my mind

Obliv

what if we did math with numbers in prime factorization, would make addition/subtraction harder but multiplication/division easier

would also save space for larger numbers

Koro

@Thorgott there are 3 dots above. I referred to the dot at the bottom in the image.

A016090

@Franklin Not to chime in when you might have already done this, but since you're restricting your grid to be square, don't forget to let rotational symmetry help you!

Koro

17:05

From that dot, one path should leave and one should enter.

Mr. Feynman

@Thorgott Now, the quotient topology grants that the point I excluded constitutes a closed set. If I define a function $f:\{\partial B^n\}\rightarrow\{N\}\quad f(\partial B^n)=N$ that is continuous since the preimage of a closed set is closed (the excluded points are the only two non-trivial sets there)

And then I wanted to consider the union of $f$ and the homeomorphism I had found before

Xander Henderson

17:23

Okay, I've now written up two solutions to this stupid problem. After you figure it out, @Franklin, let me know, and I'll undelete those solutions.

Though I don't know how I managed to get nerd-sniped by this.

Franklin

@XanderHenderson This is gonna be long:

To find the total number of squares in an $n \times n$ grid, we can count the number of squares of each size from $1 \times 1$ up to $n \times n$, and then add up the totals.
There are $n^2$ squares of size $1 \times 1$ in the grid.
There are $(n-1)^2$ squares of size $2 \times 2$ in the grid, since there are $(n-1)$ rows and $(n-1)$ columns in which a $2 \times 2$ square can be placed.
Similarly, there are $(n-2)^2$ squares of size $3 \times 3$, $(n-3)^2$ squares of size $4 \times 4$, and so on, down to $1 \times 1$ squares.

@XanderHenderson I think what I did is trash, but you might take a look,...

Xander Henderson

Okay... so you are fixing a side length $\ell$, then counting the number of squares of that size. That seems totally reasonable.

Franklin

@XanderHenderson So, you're saying, this is valid ?...

Or am I too fast for a conclusion...

Xander Henderson

If I choose a side length $\ell$, I can put a square with lower-left vertex $(i,j)$ at any point where $i+\ell < n$ and $j+\ell < n$. There are $(n-\ell)^2$ such choices.

And then I add up all possibilities with $\ell \in \{1,2,\dotsc, n\}$.

Yes, it seems to work.

I would suggest that you post that as an answer. It is a third approach, distinct from the two I came up with.

Franklin

@XanderHenderson Ohh...you posted a solution! I didn't notice it. No, please don't delete it, I want to know your approaxh as well :)

Xander Henderson

17:30

@Franklin I've posted two solutions, which I will now undelete, as you have given me your argument.

Franklin

@XanderHenderson I remain obsecreted!

Xander Henderson

Again, I would suggest that your approach is also interesting, and merits posting.

Franklin

@XanderHenderson I will post it, once I plug in more details and make it look a bit more formal. But I liked your second solution using induction so much! Indeed, I said this earlier and I say it now, you are gifted. I never ever thought about it like this or even if I thought I could not implement it. I have a long way to go!

Xander Henderson

17:45

@Franklin Gifted? No. I have simply trained a lot longer than you. Experience---working through a lot of hard problems---is the key.

Koro

Suppose we have a path that is not a loop. Can it be homotopic (path) to a constant loop?

Xander Henderson

@Koro What is a "constant loop"?

Is that not just a single point?

Koro

It's a constant map [0,1]-->X

Xander Henderson

@Koro Right, so the image of the "loop" is a point.

Koro

yes.

Xander Henderson

17:55

Won't any non-intersecting map $\gamma : [0,1] \to X$ which is not a loop deformation retract to a point?

Koro

I think the said homotopy can exist.

@XanderHenderson depending upon the space yes and no.

In R^2, yes.

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$Mathematics$ Mathematics