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14:07
@amWhy @SBM Hi there!
SBM
SBM
Hello ‮@Nilknarf
SBM
SBM
14:25
@Nilknarf How are you?
Pretty well. Just getting ready to start a new blog post. :)
And you?
SBM
SBM
@Nilknarf I'm good too; thanks.
14:43
Hello @Nilknarf @SBM
Hi! How are you doing?
@Nilknarf what's this next one about?
Non-telescoping sums
Basically, other methods of evaluating sums that don't telescope
Like with differentiation and integration and such
14:44
Ah, okay
Which reminds me - I was going to ask you how to evaluate sums like
$$\sum_{x=a}^b \sin(x) $$
Something about Euler's "beautiful" formula...?
$$\sum_{n=a}^b\sin(n)=\Im\left[ \sum_{n=a}^be^{in}\right]$$
So you get a geometric series, then take the imaginary part
SBM
SBM
um
what is telescoping series?
14:47
Okay, suppose I did that:
$$\sum_{x=1}^{10} \sin(x)=I\Bigg[\sum_{x=1}^{10} (e^i)^x\Bigg]=I\Bigg[\frac{1-e^{11i}}{1-e^i}\Bigg]$$
@SBM Like this:
$$\sum_{n=1}^{10}\frac1n-\frac1{n+1}$$
Ah, I see. Then I have to split the $e^{in}$ up into $\cos(n)+i\sin(n)$?
$$=\frac11-\frac12+\frac12-\frac13+ \frac13-\dots$$
$$=1-\frac1{11}$$
@SBM That is how a telescoping series works.
@SBM Here are some examples:
http://nilknarf.langlab.org/entries/2017-6-9.html
@Nilknarf Yup. Rationalize the denominator and stuff like that
SBM
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14:50
that looks likes everything cancels almost
Yeah, that's pretty much what a telescoping series is, except a telescoping series has infinitely many terms
SBM
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how did the other part become -1/11
There is one negative terms left over at the end
In a telescoping series, the first $n$ and last $n$ terms don't cancel, where $n$ is how far the negative terms are shifted off from the positive terms.
SBM
SBM
óh
I see.
15:00
@Nilknarf That's a Weierstrass function. If you take the upper bound to infinity, you get a continuous but nowhere differentiable function.
Jeez, how do you know this?
But thanks!
Yeah, makes sense that it would nowhere be differentiable.
...
but what about this one:
$$\sum_{n=1}^k 2^{-n}\sin(2^{-n}x)$$
That one seems to stay smooth
Yes... as you can see in the Wikipedia, you'll want $ab>1+\frac{3\pi}2$ for it to get messy
A fun challenge I have for you guys is to construct a function that is nowhere continuous @Nilknarf @SBM
SBM
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oh
15:04
Goodness gracious
Nowhere?
Can I use a random variable? XD
Must be discontinuous at every point
Nope, must be constructive
SBM
SBM
I remember somebody talking about Dirichlet functions here
but I don't understand them enough
How about something like $f(x)=1$ if $x$ is rational and $0$ if it is irrational?
SBM
SBM
15:07
Discontinuous at every point is hard
@Nilknarf Aw pooey xD
Is that it?
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yes
A dirichlet function is basically $1$ if $x$ is rational, $0$ if $x$ is irrational.
SBM
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the same thing
This function never seem to be continuous.
15:08
One can construct a dirichlet function with the following:
$$f(x)=\lim_{n\to\infty, \ n\in\mathbb N}[\cos(n\pi x)]^{2n}$$
SBM
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Oh
But can't my example be continuous when you have some irrationals that differ by infinitesimal amounts? How do you prove that it is always discontinuous?
Which basically takes advantage of the fact that $|\cos(\pi n)|=1$
SBM
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yes
@Nilknarf there's always a rational number between any irrational number
15:10
Jesus, how do you prove that?
I believe it depends on your definition of real numbers, rational numbers, and irrational numbers
Well, okay
@SimplyBeautifulArt Now I have a problem for you.
Also, that's the Dirichlet function. Just take $a\to\infty$.
Cool!
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@SimplyBeautifulArt looks mindboggling
Suppose you have a regular $n$-gon with side length $l$ "resting" on the $x$-axis with one of its vertices at the origin. Then you roll it and trace the point that started on the origin (basically a cycloid, but not with a circle) until the tracing point is touching the $x$-axis again. What formula gives the length of the path traced by the point in terms of $l$ and $n$?
Heh heh heh
I love this problem
Hm, okay, I'll think about it.
Heh heh heh heh
The beautiful part is, when you take the limit of the formula as the polygon "approaches" a circle, you get the length of the cycloid of a circle.
SBM
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What's a cycloid?
I love it when math works out perfectly like that...
That's a cycloid
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15:19
@Nilknarf oh
Hm, you can split the problem into multiple parts... each of which are just part of a circle, so their lengths shouldn't be hard to find
Yeah, basically it's a summation problem
However, when you want to find the area under it...
You have to be able to notice the "cool thing" about the area under the path it traces. I'm not going to give it away, though.
Wouldn't the area just be another summation problem? Still basically adding up circles
But the sectors don't cover all of the area underneath...
that's okay. It's just parts of circles sitting on rectangles then
15:30
On rectangles?
Perhaps you and I are talking about different circles.
Perhaps
welp, I'm too lazy to do such a tedious problem without figuring out the secret trick here x.x
I'm talking about the circle sectors that are "swept out" when the polygon is "pivoting" on one of its vertices. Thus these sectors would have radius lengths equal to the lengths of the diagonals of the polygon from the tracing point to each other vertex.
I think that's what I'm talking about. But I'm thinking they draw parts of a circle and go downwards, which leaves something I can do with quick calculus + a rectangle
Jeez, all the questions on SE this morning are either lousy or reference things that I don't know yet. :P
15:33
But integrating takes longer.
Ohhhh
Jeeeez
@Nilknarf Not if you're me and already know all the formulas
Go for it
The question, I mean
Still deciding if $n\in\mathbb N$ includes $0$ or not.
15:34
Ha
And some other stuff :P
I hate it when an OP just posts a question and then leaves, and makes everybody wait to find out who will get upvoted. :P
That's happened to me twice this morning
D:<
And I hate it when I'm 10 hours late to the question and the top answer already has 30 upvotes (mine included xD)
SBM
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15:38
oh
SBM
SBM
what is an orthonormal base?
0_0
Yeah, what is an orthonormal base?
15:39
I've never heard that term before
SBM
SBM
I guess somebody meant orthonormal basis.
This is the best I can do:
https://en.wikipedia.org/wiki/Orthonormal_basis
:P
Oh, linear algebra. Not something I know of.
24
Q: Explain non-convergent sums to a bright high schooler.

Jerry GuernI know the higher-math answer to this question, but I'm asking how on Earth to explain it to a bright high school student. Here's the question, paraphrased: “I can see that $\left(1+x+x^{2}+x^{3}+x^{4}+....\right)\left(1-x\right)=1$ because the terms with non-zero powers of $x$ all cancel out. ...

@SimplyBeautifulArt @SBM Have either of you guys ever heard of "Great Courses"?
https://www.thegreatcourses.com
SBM
SBM
@Nilknarf no
15:44
@Nilknarf no
Oh
It's basically like these recorded college courses
That you can watch
@SimplyBeautifulArt For example, that ordinals video you posted was related to Great Courses I think
Yeah
I'm watching this logic/philosophy one and it's kind of mind-blowing
SBM
SBM
It's expensive I guess
Nah
Not really
I mean, if you actually buy the videos it's a little expensive. But you can get a membership thingy and they have an app that you can watch them on.
SBM
SBM
15:48
I don't have a phone
Neither do I :P
I watch them on the laptop or TV
$$\int_1^\infty\frac{z}{x^{1+\alpha}(x-z)}~\mathrm dx$$
How should I estimate this integral?
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it's wrt x?
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is z independent of x?
Whoa
@SimplyBeautifulArt Is asking for help? Whaaat?
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I tried but I'm running out of ideas
16:17
I just went ahead and ended it there, using the conclusion that:
$$\left|\int_b^\infty\frac{z}{x^{1+\alpha}(x-z)}~\mathrm dx\right|<\frac{|z|}{(1+\alpha)b^{1+\alpha}}$$
Assuming $\alpha\in\mathbb R$
 
1 hour later…
17:31
@SimplyBeautifulArt If or when you're around, one more close vote needed here: math.stackexchange.com/questions/2334930/…
18:23
-6
Q: I need help with these 3 equations.

YoankI need help with these 3 equations. 1) $(1+y^2)dx+xydy=0$ 2) $y''-2y+2y=0$, $y(0)=4$, $y'(0)=2$ 3) $y'-xy=x^3\cdot y^3$

delete
@amWhy Hi!
Looks like you already got that question closed. :)
Yes, If I could delete that the first request, I would. The second is an invitation for delete votes.
Wow... she changed her name?
One of the commenters?
Indeed she has. Most likely same account, but change in username.
Goodness gracious, what a mess.
18:36
@Nilknarf Indeed. The main problem is with the questions...too many in one post, no effort shown by the asker.
I can't fathom why people reward questions like that. There isn't even any kind of reputation incentive to do so!
2
...though it's not like it would be any more acceptable if there was reputation awarded for comments.
18:51
@amWhy lol
19:22
@amWhy MariePierredeLeTetou strikes again:
https://math.stackexchange.com/questions/2335031/value-of-an-exponential-function/2335041#2335041
@Nilknarf OMG!! That is really weird?!
Yeah. I told her to just leave an answer but she hasn't responded yet.
@amWhy That is really weird! Why isn't she even responding?
She's not even a new user! She has almost $5000$ rep!
@Nilknarf that's shocking
Yeah it's really weird
The really strange thing is that she would actually get reputation if she would just leave an answer!
Earlier today, she was AnnaTaurogenireva. Note this question and comments (names used to respond to here prior username.) math.stackexchange.com/questions/2334930/…
19:33
Yeah, I saw that one. That's how I recognized her as a notorious long-comment leaver. :P
@amWhy Perhaps we should try and call her into the chat to ask about it?
"She/He" Who knows, has been around here for more than 5 years. S/he has likely rotated through countless usernames, and probably fancies him/her self as a "super hero"... this site is very full of such characters.
How does leaving ridiculously long comments rather than answers make one a superhero?
Delusional Super Hero:
Oops! Yes... I didn't see it had posted the first time...
@amWhy there's math superhero's when you say superhero I think of this
19:45
XD
20:12
@amWhy you busy today I have some freetime :>)
 
1 hour later…
21:13
0
Q: Proving a specifc case of $\frac{1}{\lim \sup_{k \rightarrow \infty}|a_{k}|^{\frac{1}{k}}}$

ZophikelIn the text "Function Theory of One Complex Variable" Third Edition by Robert E.Greene and Steven G.Krantz. I'm having trouble proving the following case of the root test in $(1.)$ $(1.)$ $\text{Lemma 3.2.6 (The root test)}:$ The radius of convergence of the power series $\sum_{} a_{k}(z-P)^{...

0
Q: Formally showing $K_{n} \star f(\theta) \rightarrow f(\theta)$ uniformally in $(\theta)$ as $n \rightarrow \infty$

ZophikelIn the text "Real and Complex Integration Problems" I'm having trouble verifying my approach to proving following conjecture in $(2.).$ $(0.)$ $(0.1)$ $$K_{n}(\theta) = \sum(1 - \frac{|j|}{n+1})$$ $(1.)$ Show that for any continuous $2 \pi$ periodic function $f$ on $\mathbb{R}$ one has: $$ K...

Two questions in one day :), getting ready for the negative reviews :(
@Zophikel I hear OPM season 2 comes out in November.
@Simply :-)
@Simply the one punch man fitness test describes what my IQ test was like
21:46
X-D
22:04
@ylpmiS what shows have you watched
 
1 hour later…
23:11
0
Q: Formally showing $K_{n} \star f(\theta) \rightarrow f(\theta)$ uniformly in $(\theta)$ as $n \rightarrow \infty$

ZophikelIn the text "Real and Complex Integration Problems" I'm having trouble verifying my approach in proving the following conjecture in $(2.).$ $(0.)$ $(0.1)$ $$K_{n}(\theta) = \sum(1 - \frac{|j|}{n+1})$$ $(1.)$ Show that for any continuous $2 \pi$ periodic function $f$ on $\mathbb{R}$ one has: ...

@Simply how does this look ^
0
Q: Proving a specifc case of $\frac{1}{\lim \sup_{k \rightarrow \infty}|a_{k}|^{\frac{1}{k}}}$

ZophikelIn the text "Function Theory of One Complex Variable" Third Edition by Robert E.Greene and Steven G.Krantz. I'm having trouble proving the following case of the root test in $(1.)$ $(1.)$ $\text{Lemma 3.2.6 (The root test)}:$ The radius of convergence of the power series $\sum_{} a_{k}(z-P)^{...

and this ^
@amWhy are you free :>)
23:56
@mbe Hello and welcome to my realm!
@SBM @Nilknarf @Zophikel say, don't know if I've linked this to you guys, but this recent answer of mine may be helpful in understanding the nature of analytic continuation and what a divergent series means. Mainly see the last half of my answer.

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