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12:40 AM
Since the integrand decays like $1/|z|^2$
$$
\oint\pi\cot(\pi z){\small\left(\frac1{(8z+1)^2}-\frac1{(8z+3)^2}-\frac1{(8z+5)^2}+\frac1{(8z+7)^2}\right)}\,\mathrm{d}z=0\tag{1}
$$
There are singularities at each integer and at $\left\{-\frac18,-\frac38,-\frac58,-\frac78\right\}$.

The residue of $\pi\cot(\pi z)$ is $1$ at each integer, so the contribution of the residues at the integers is
$$
2\sum_{k=0}^\infty\left(\frac1{(8k+1)^2}-\frac1{(8k+3)^2}-\frac1{(8k+5)^2}+\frac1{(8k+7)^2}\right)\tag{2}
 
@user1 Yello.
 
@PeterTamaroff Hi, what's up?
 
@user1 Bernstein Polynomials are oozing my chillies.
 
1:19 AM
@PeterTamaroff /me looks at wikipedia, so are they mainly used to approximate stuff?
 
@user1 They are a (very nice) option to prove Weiertrass' Approximation theorem, say.
@user1 (Yes)
 
1:35 AM
@user1 In fact, I am trying to prove that.
Wanna see?
 
2:08 AM
@robjohn
 
@PeterTamaroff yes?
 
@robjohn Suppose $f$ is continuous over $[0,1]$.
For each $x\in [0,1]$ and $n=1,2,\ldots$, define $\varepsilon_n(x)=\max \left|f(x)-f(k/n)\right|$ where the $\max$ is taken over the $k$ such that $|x-k/n|<n^{-1/2}$.
I want to show $\varepsilon_n\to 0$ uniformly over $[0,1]$. I will give it a try first. Just sharing.
 
@PeterTamaroff okay
 
I think uniform continuity of $f$ should take care of it.
 
hi guys
 
2:13 AM
@PeterTamaroff do you need the discreteness of the $k/n$ or is a $\delta-\epsilon$ good enough?
 
Isn't $$x_n=\sup_{|w-y|<n^{-1/2}}|f(w)-f(y)|$$ a vanishing upper bound independent of $x$ as $n\to\infty$?
(By uniform continuity)
 
@PeterTamaroff since there is no $x$ mentioned, it would be independent of $x$
 
@robjohn Yes, I know that! =)
I was simply being detailed in what I wrote.
 
@PeterTamaroff $x_n\to0$ is equivalent to uniform continuity
 
@robjohn Aha.
That was my point.
Actually, it is $n^{-1/4}$, but that is slightly irrelevant.
 
2:20 AM
@PeterTamaroff as long as the function of $n$ tends to $0$.
 
@robjohn So, I guess that is all, yes? $$\varepsilon_n(x)\leq x_n$$ over $[0,1]$.
Since $x_n\to 0$, $\varepsilon_n\to 0$ uniformly over $[0,1]$.
 
looks like it. It appears you are restating the conditions for uniform continuity
 
hi
 
I'm wondering if there is a difference between a statement and a proposition in logic.
It seems that they both mean: a sentence that is true or false (but not both)
but i'm not sure
 
2:36 AM
@Kasper It may depend on the particular text used, but I think, in general, that sounds right.
 
@robjohn ok thanks
 
2:52 AM
@Kasper , in my text , a sentence is a statement which has no variable which occurs free . so the statement may have a variable which occurs free , but the sentence don't have this property . a proposition is used in propositional logic not in predicate logic . this is how it's used in the text i study , i don't know what about yours , you can go back and check the definitions .
 
So... is anyone here in the Coursera class on Linear Algebra? (class.coursera.org/matrix-001)
I'm somewhat questioning their approach to what a vector is...
 
@anorton: What is their approach to what a vector is, and why do you question it?
 
The professor says that a vector is a function from a set $D$ to $|D|$ values in a field $\Bbb{F}$. I'm familiar with vectors as a tuple in $\Bbb{R}^n$, but not in the sense of a function. For example, a valid vector in this class would have
$$D = \{\text{one}, \text{two}, \text{three}\}$$
$$\Bbb{F} = \Bbb{R}$$
So, a particular vector in the vector space $\Bbb{R}^D$ could be represented as a function:
$$\text{one} \to 1 \\ \text{two} \to 2 \\ \text{three} \to 3$$
@dfeuer Sorry for the lag... I wanted to make sure I wasn't misrepresenting what the professor was saying. :)
 
3:09 AM
Well, that's .... a fairly awful way to think about vectors, mathematically. However, it's a great way to think about what computer scientists think of as a vector.
 
That makes sense... the professor is a computer scientist... :)
(But he is trying to teach it as a math class, not a CS class)
 
From a mathematical perspective, a vector space is a field (or a division ring) along with a set of "vectors" and the operations of vector addition and scaling.
The vectors themselves need not have any internal structure.
 
@dfeuer What is a "vector"? (I've googled around for a precise definition, but couldn't find one...) Does a matrix count as a vector? (I think that they can be a part of a vector space...)
Meaning, is a matrix a type of vector in some interpretations? (It seems like this course is introducing matrices as a form of vector...)
 
Well, it tends to depend on perspective.
I'm thinking about how to formulate a good answer.
From the perspective of a mathematician, there is no way, when looking at an individual thing, to say whether it is a vector or not. The "vectorhood" of a vector is not intrinsic.
It has to be considered within the context of a vector space.
There are some very important vector spaces, such as $\Bbb R^n$ over $\Bbb R$ with the "standard" operations, which is the one you appear to be studying right now.
It's sort of like an "element". You can't really say if something "is an element". You can only say if it is an element of a certain set.
@anorton, have you attained enlightenment?
 
4:20 AM
Off Topic: Hi. In probability, what does Prob(X) mean in the context "Prob(X) is always compact"?
 
Hello
 
4:34 AM
Hi.
 
4:51 AM
@mixedmath Hey, why do I seem to be confused here ?
 
5:19 AM
@BenjaLim What does "Spec" denote?
 
6:18 AM
@BenjaLim why would you get $A_i$? The sheaf you have put on $spec(A_i)$ is not the usual one
@PeterTamaroff spec means the set of prime ideals of the ring (with the Zariski topology)
 
7:08 AM
@TobiasKildetoft Hey
 
@BenjaLim hi
 
@TobiasKildetoft Somehow I am still confused
 
@BenjaLim the spec of a ring and of its reduced version are homeomorphic, right?
 
@TobiasKildetoft Because having shown $(U_i, \mathcal{O}_X|_{U_i})$ is isomorphic to $(\text{Spec}(A_i), \mathcal{O}_{\mbox{Spec}(A_i)})$
isn't quite right
@TobiasKildetoft ah of course!
 
so all you change when going to the reduced version is the rings
 
7:11 AM
you also need to change the sheaf no?
 
sure, that's what I meant
but it remains a sheaf on the same space
 
ok.
 
nice term: Quasi-exactly-solvable Problems
whatever it means
 
7:41 AM
hello !!!
 
8:18 AM
@robjohn may I get your opinion on my wording of this property?
If a, b, c, and d are real numbers with b $\neq$ 0 and d $\neq$ 0, then

$$\frac{ac}{bd} = \frac{a}{b} * \frac{c}{d}.$$
"You can write any quotient as a product of quotients."
 
8:53 AM
@robjohn I just noticed your worked above that is somewhat similar to mine. When I saw that series I knew is related to $$\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{z-n}=\frac{\pi}{\sin(\pi z)}$$
Then I performed the differentiation with respect to $z$ and I was done.
 
9:24 AM
@Chris'swisesister Heya
 
9:40 AM
@N3buchadnezzar hello! How are you? :-) I'm a bit busy and that's why I answered you back with delay.
 
@skullpatrol Great one! How is it going? :-)
 
@Chris'swisesister Fine thanks, how are you?
 
9:43 AM
@skullpatrol a bit busy right now, but I'm fine, thanks.
 
Good, good, don't let me disturb you :D
 
hehe, brb
 
@Chris'swisesister Just a quick question do you remember which year this integral showed up in Putnam? $$ \int_0^1 \frac{\log(1+x)}{1+x^2}\,\mathrm{d}x $$?
I think it was $1968$ A5, but I can not seem to find it. I know you asked a question about it, so i thought you might knew.
 
@skullpatrol I see that rather as "a product of quotients is also a quotient"
@Chris'swisesister You'd need to add at least two of those together.
 
@robjohn Yes, but that would be more like a multiplication rule for fractions, right?
What I am trying to put into words is the property of quotients.
2 hours ago, by skullpatrol
If a, b, c, and d are real numbers with b $\neq$ 0 and d $\neq$ 0, then

$$\frac{ac}{bd} = \frac{a}{b} * \frac{c}{d}.$$
"a quotient is also a product of quotients"?
 
10:06 AM
You mean something like "to multiply fractions, multiply numerators and multiply denominators"?
 
@skullpatrol that sounds much too vague
the important thing is that the product of two quotients is the quotient of certain specific products
 
@skullpatrol that sounds like "a number is also a product of numbers". That just sounds odd.
 
@robjohn Yes, that would be your rule for multiplying fractions :-)
 
@skullpatrol I guess I am confused about what you want here.
 
@robjohn Does this wording sound odd?
 
10:12 AM
@skullpatrol Yes.
 
2 hours ago, by skullpatrol
"You can write any quotient as a product of quotients."
Thanks for responding :-)
 
@skullpatrol it sounds odd because it is not really saying much
 
@skullpatrol That sounds as if you are saying that given any quotient, you can split it into a product of two quotients
 
@robjohn Yes, that is what I'm trying to say :D
Any quotient can be split into a product of two quotients.
 
@skullpatrol but why is that an interesting statement?
(since 1 is a quotient)
 
10:16 AM
@skullpatrol so given $\frac34$ there is no canonical product of fractions. I don't know what you are trying to say. Sure, $1=\frac11$ and so we can say $\frac34=\frac34\cdot\frac11$, but who cares?
 
Students trying to learn this stuff for the first time?
@robjohn @TobiasKildetoft
 
@skullpatrol I would think the other direction would be a lot more interesting, that the product of two fractions is also a fraction is a lot more interesting.
 
@N3buchadnezzar it's A5-2005 (12 solutions here to it). I saw that it was added in our country in some books for the students in the last high school year.
They usually recommend you to make us of the trigonometric substitutions.
 
Yeah, but the problem is much older than that!
 
@N3buchadnezzar agree.
 
10:27 AM
I remember seeing a refference to a book on diff.equations from the 1890 ties using the value of the integral for the first time.
Also I remember that the integral appeared in the first edition of Putnam.. But again, I can not find the first issue to check this.
Says on page 2 that it appeared in $1844$
 
@N3buchadnezzar Interesting, but not something unexpected. Major part of the problems we solve now were already solved a long long time ago. :-)
The good part is that their beauty remains the same. :-)
 
10:43 AM
As I learn newer math, arithmetic becomes more beautiful :-)
 
@skullpatrol I was thinking these days here I feel at home, a part of my family, and not because of the site itself, but because I'm surrounded by people that like math at least like me. :-)
 
@Chris'swisesister Well put :D
 
@N3buchadnezzar funny your question here: math.stackexchange.com/questions/441106/…
@skullpatrol :D
 
10:59 AM
hi all
hows this manipulation possible? puu.sh/3AckF.jpg look at the RHS, the max zips through inside.
hows that possible
 
$\max\limits_n\, (a_n+b_n)\le (\max\limits_n a_n)+(\max\limits_n b_n)$ ... now generalize
 
ok generalizing..
 
11:32 AM
@PeterTamaroff Sorry, I am constantly in and out. I would have liked to have seen. :)
@PeterTamaroff I am not sure if anyone answered this question for you. It is the set of prime ideals of a ring, often endowed with a topology.
 
@N3buchadnezzar I answered that question
 
key phrases: "spectrum of a ring," "zariski topology"
 
11:46 AM
It is greater than God and more evil than the devil.
The poor have it, the rich need it and if you eat it you'll die.
What is it?
 
Nothing can stop me from saying it.
 
String or nothing...
2
 
nothing
 
a heart
 
11:52 AM
...and don't forget you can't divide by it :-)
 
@skullpatrol there is nothing you can't divide by?
 
@TobiasKildetoft you can't divide by nothing.
Welcome @nesreen we were just talking about your Master's Thesis :-)
 
hi,@skullpatrol
about what?
 
zero divisor
 
oh
 
12:08 PM
Mathematician Love letter #1

My Dear Omega,

Yesterday, I was passing by your rectangular house in face,conical nose and spherical eyes,standing in your triangular garden. Before seeing you my heart was a null set, but when a vector of magnitude (likeness) from your eyes at a deviation of theta radians made a tangent to my heart, it differentiated. My love for you is a quadratic equation with real roots, which only you can solve by making good binary relation with me.

The cosine of my love for you extends to infinity.
Mathematician Love Letter #2

Dear Sigma,

With reference to the syllabus of my feelings, I want to prove that the locus of my point is directly proportional to your heart. On seeing you, I feel like a triangle with only two angles; the third being you! At times it seems to me that I am a circle without a radius or a matrix without determinants. We are like two simultaneous equations without any solutions.

You never seem to notice that I am a point lying on your vector equation, or in other words, we are collinear. Your smile makes my head rotates 360 degrees anticlockwise and on applying
 
where were you read this??????????????
 
On a kid's math web site.
 
I love it !
 
@N3buchadnezzar how can they know that?
 
12:19 PM
@robjohn I guess they assume it, although it is quite early
 
@skullpatrol are you there
 
Yup
 
are you good in ring theory?
 
@robjohn Yeah, I was just wondering about when it showed up =)
 
@nesreen no, sorry
 
12:22 PM
noooo, it's okay
 
@N3buchadnezzar So I would say at least by 1844, perhaps earlier.
 
=) It is a fun integral family
 
but you were talking about zero divisor.,@skullpatrol
 
@robjohn Do you know if Euclid's Elements is still used in graduate math courses?
 
@skullpatrol I would have no idea.I am sure it is cited.
 
12:28 PM
@nesreen That was just a joke ;-)
 
sorry
 
6 mins ago, by nesreen
noooo, it's okay
:D
 
but my Master's Thesis in graph, zero divisor graph.,@skullpatrol
it' oky
 
12:41 PM
Welcome @TRiG onometry
 
1:19 PM
wop wop
QWOP
 
Hello =)
 
1:33 PM
What kind of algorithm do they use to determine the position and placement of electronic components on a PCB. I just removed my PC cover and this board looks so beautiful.
 
is there a decent polynomial/exponential approximation of binomial coefficients?
can you use stirlings formula for it?
to be more precise, I want to approximate the behaviour of $\max_{k\in\mathbb{N}}{ n\choose k}$
which is $n\choose n/2$ for even n
is there some compact formula to approximate it?
 
@CBenni I think Stirling will give something nice
 
ok ty
 
@skullpatrol Hi. But my name predates any knowledge I have of trig.
> Timothy Richard Green
 
@TobiasKildetoft only thing I fear is that the convergence can fail when dividing stirling by stirling
which is probably necessary
 
1:44 PM
@TRiG np, we all respect the power of trig around here :D
 
@CBenni not sure what you mean by that
how good an approximation it is for given $n$ might be worse, but the convergence should not change
 
ok
I meant asymptotic behaviour
my mistake, sorry
 
since the asymptotic in this case is good (by which I mean we actually know the quotient of the actual thing and the approximation converges and what it converges to), this does not change when we take a quotient
 
ok ty
@TobiasKildetoft I got $\frac{2^{n+1/2}}{\sqrt{n\pi}}$ which is pretty nice (yet bad for me...)
 
why is that bad for you?
 
1:53 PM
because its still $2^n$ :D
but that was to be expected
a $2^n$ algorithm really is bad
but thats the stated problem
 
@dfeuer Sorry for a (super late) response. That helped a lot! :) Thanks!
 
usually the worst case shouldnt happen
if it should ever appear, I can fall back to my $n^2$ solution if I want to
and maybe I can break down the $2^n$ to polinomial with dynamic programming
 
2:14 PM
Are you guys ever surprised at what ends up being your most up-voted answer? I sure am.
 
yes
 
I am not
Hurr durr why isnt 0/0=<insert any number here>
insta 10 upvotes
 
:D
 
next, $0^0$
then $\log_0(0)$
 
2:16 PM
$\int_0^0\infty dt=1$
 
I thought that it was just eh, and that amWhy's answer was the one that would be upvoted most/accepted, but ??
 
lol
srsly
why has that question gotten so much attention
math.se getting overrun by 10y olds
 
I don't know... somehow, it was triggered to be one of the "hot questions" in the notificaitons dropdown.
 
yeah
they should remove that
 
I have the pleasure of saying that my most upvoted answer is a joke and my second most upvoted answer took 30 seconds to write.
 
2:19 PM
it promotes single questions (usually bad ones) to get way more attention and other, more interesting questions to fall down
lol
 
 
2 hours later…
4:05 PM
@anorton I'm surprised at my most up-voted answer; not so with my most up-voted question
 
@Mariano Hey
 
@user1 Helloes?
 
@PeterTamaroff Hello
 
@Arkamis Hello there!
 
@Mariano I seem to be misunderstanding something basic here. Even though I seem to have solved the problem, it does not really seem to be the case...
 
4:14 PM
@BenjaLim A prime ideal is one such that $ab\in P\implies a\in P$ or $b\in P$, yes?
 
Yes that is correct.
 
(And what does the topology have to do with what the prime ideals are?)
 
You put the Zariski topology on the set of all prime ideals
 
@BenjaLim Ah, OK. How is that topology defined?
 
In algebraic geometry, the Zariski topology is a topology chosen for algebraic varieties. It is due to Oscar Zariski and took a place of particular importance in the field around 1950. The more subtle étale topology is a refinement of the Zariski topology discovered by Grothendieck in the 1960s. The classical definition In classical algebraic geometry (that is, the subject prior to the Grothendieck revolution of the late 1950s and 1960s) the Zariski topology was defined in the following way. Just as the subject itself was divided into the study of affine and projective varieties (see ...
 
4:16 PM
@BenjaLim Cannot you explain XD?
I like talking instead of reading Wikipedia =)
 
Peter, have you ever done any topology studies (this question unrelated to current conversation)
 
In commutative algebra we define a closed set in $\text{Spec}(R)$ to be one of the form $V(a)$ for $a \subseteq R$ an ideal. $V(a) = \{ P \in \text{Spec}(R) : P \supseteq a\}$
 
@Arkamis Just a little.
@BenjaLim Oh, OK. Nevermind.
 
for which distributions $\varphi$ is there a distribution $\Phi$ with $\Phi'=\varphi$?
like, is there a integrability condition
 
@PeterTamaroff In classical algebraic geometry you only consider the maximal ideals which correspond to points on a variety by the Nullstellensatz (over $K$ algebraically closed at least).
 
4:20 PM
@PeterTamaroff I'm looking to go through Munkres this summer in preparation for a class -- if you had time/interest, we could do a collaboration.
 
@Arkamis Oh, when would you be starting?
(And what do you mean by "collaboration"?)
 
Well, we could set a rough schedule and some problems, then use mathb.in or so to post solutions, and bounce ideas off each other.
 
Anyways, I am off to buy a ream of paper. I ran out!
Be back in a while.
 
I should have started a few weeks ago, as it is, I'll probably start in a week or so.
hasta luego
 
@Arkamis OK. I have my finals (Algebra I and Analysis I) coming in two weeks, but I can handle it. However, I will have to think about it. Looks like something nice to do.
 
4:24 PM
Alright, I'll let you know. I'm crazy busy, so I'll probably set a limited pace, but if you're interested I'll try and find something that works for both our schedules.
I always find it easier to get things done with I have a target and a deadline, even if those things are completely synthetic.
 
5:05 PM
@PeterTamaroff Lots of helloes.
 
5:16 PM
@user1 $\sup$?
 
peters so smart so is arkamis
 
@PeterTamaroff Supremum of what?
Answer is 42 btw.
 
(outside of math, "sup" is slang for "what's up")
 
@user1 You so silly.
2
@user1 So, I learned to cut up cubes real good, like Jonas says.
 
lol user1 noob
peter are u in advanced math
 
5:28 PM
@FrankEpps Who are you?
 
i'm just a user, what's the matter
 
@anorton Now I know.
 
@user1 :)
 
@user1 Bernstein Polynomials!
 
@PeterTamaroff !
 
5:33 PM
lol peter u must be antisocial, unable to interact with new faces
 
@FrankEpps What does noob mean? \me takes it as a compliment.
 
yes its a compliment, it means ur quite slick
 
@FrankEpps be nice. :)
 
anorton i am
 
@FrankEpps What does slick mean? \me takes it as a compliment.
 
5:35 PM
yes ur suave
 
Urban definition of noob :-)
 
@FrankEpps What does ur mean?
 
i mean ur so cool
 
Ah, recursive definitions. \me likes.
 
ur = you are
 
5:39 PM
oh lol i thought u asked what 'suave' meant
 
@FrankEpps what does lol mean? :)
@user1 I think I may have just caught on to what you're doing in this conversation... :)
 
lol means ur so funny
 
Seriously, though... I've heard lol to mean multiple things. Some people mean it as "lots of luck," others say "lots of laughs," and I'm sure I've seen others.
 
no i've never heard of those definitions...........
 
@user1 Note that we can write $$1=\sum_{k=0}^n\binom{n}{k}x^k(1-x)^{n-k}$$
I choose $x\in [0,1]$.
 
5:43 PM
I lol'ed all over myself probably uses a nonstandard meaning of lol.
@PeterTamaroff Indeed, those things partition unity.
 
@user1 Right. Now, the question is, where is the contribution bigger? That is, for what $k$ is the contribution bigger?
 
Quick--Everyone post a bunch of messages so that @skullpatrol's fit-inducer rolls off screen! :D
 
PETER IS ANTISOCIAL
 
@anorton that will make my COPTER take-off :-)
 
5:50 PM
@skullpatrol :)
Let's see how fast it rises!
 
@PeterTamaroff Those polynomials have local maxima at n/k and nowhere else, so if there is justice we could use that somehow.
 
@FrankEpps how 'bout asocial :D
 
@user1 Now, we look at $$\left| \frac kn-x\right|$$ If we split the sum above into $\sum_1$ and $\sum_2$ where the first contains the $k$ such that $$\left| {\frac{k}{n} - x} \right| < {n^{ - \frac{1}{2}}}$$ and the second those for which $$\left| {\frac{k}{n} - x} \right| \geqslant {n^{ - \frac{1}{2}}}$$ the claim is that $$\sum{}_2<\frac 14n^{-1/2}$$
The proof is easy, once one uses that $$\sum_{k=0}^n \left(\frac kn-x\right)^2\binom{n}{k}x^k(1-x)^{n-k}=\frac{x(1-x)}n$$
For $x\in[0,1]$, $x(1-x)\leq 1/4$, so that is basically it.
 
Right.
 
@user1 And that is essentially all we need to prove that the polynomials $$K_n(x)=\sum_{k=0}^nf\left(\frac kn\right)\binom{n}{k}x^k(1-x)^{n-k}$$ converge uniformly to $f$ on $[0,1]$ for $f$ continuous!
 
6:00 PM
@PeterTamaroff That is actually pretty neat.
 
@user1 It is! =D
 
yeah peters asocial
not receptive to new relationships, what a pity
 
Hello
 
@FrankEpps What's your analysis of Jacobadtr based on his/her one message?
 
6:04 PM
@user1 I mistyped something.
We want to look at $${\left| {x - \frac{k}{n}} \right| < {n^{ - 1/4}}}$$ and at $${\left| {x - \frac{k}{n}} \right| \geq {n^{ - 1/4}}}$$
 
Can I ask a reaaaally basic question?

I can't tell if my book has a mistake or i'm being an idiot
 
Have you ruled out the possibility of both?
I KID, I KID
;)
 
Hahaha
 
@user1 Anyhow, because of uniform continuity we kill off the sum with $$\mathop {\max }\limits_{\left| {x - \frac{k}{n}} \right| < {n^{ - 1/4}}} \left| {f\left( x \right) - f\left( {\frac{k}{n}} \right)} \right|$$ and because of the other estimate we kill the remaining sum.
 
I'm still rubbish at the format of this site, so it won't be pretty

y = nth root of (1 / x^m)

dy/dx = -m/n x ^ -(m-n)/n
 
6:08 PM
$$y=x^{-m/n}$$
 
Yes, no, i'm stupid?
 
Use the rule $(x^a)'=ax^{a-1}$. That's it.
 
sqrt(n, 1 / x^m)
Yeah, i know that, but my book is saying f'x = 1m/n x ^ -(m+n)/n
-m/n x ^ -(m+n)/n
*
 
Well, $$-\frac{m}n-1=-\frac{m+n}n$$
 
Stepdads home, i'll show him the actual book.

Much easier than this format - for some reason your replies aren't coming out nicely presented, i'm seeing the raw 'code'
Thanks for the help anyway
 
6:11 PM
@Jacobadtr Your mistake is the following:
You have -m/n-1
You can write this as -m/n-n/n
 
@Jacobadtr to see the math nicely, you need to follow the instructions in the link to the right
 
Agreed
 
Uppon collecting the -1/n factor, you get -(m+n)/n
 
the chat-LaTeX one
 
Not -(m-n)/n
 
6:13 PM
ohhhhhhhh
 
Thank you!

I knew it was more likely myself where the error lied
And thanks Tobias
Looks much better now
 
@Jacobadtr that often turns out to be the case. But it is still a good idea to ask to make sure (and of course, it does happen that the error lies somewhere else)
 
@TobiasKildetoft What do authors generally mean when they say "trigonometric polynomial"?
 
Definitely - especially when it's been a while since I used much maths :)
 
6:22 PM
@PeterTamaroff no idea
 
@PeterTamaroff According to my fourier series reference (by some russian guy translated by Silverman), they sums of the form $$A+\sum_{k=1}^n\left(a_k\cos\frac{\pi kx}l+b_k\sin\frac{\pi kx}l\right)$$
 
@user1 Ah, OK.
What would qualify as the "degree" of the polynomial? $n$?
 
Yes.
The constant 2l is referred to as the period of the polynomial.
 
@user1 Ah, because I ought to prove that any continuous function on $[0,2\pi]$ can be approximated to any degree of accuracy by trigonometric polynomials.
 
The degree of a polynomial is the greatest of the degrees of its terms after it has been simplified.
 
6:29 PM
@PeterTamaroff ;)
 
@user1 Cannot I use the previous fact on usual polynomials? Because we can always express $\sin nx$ and $\cos nx$ in terms of powers $\sin^k x$ and $\cos^k x$.
 
@PeterTamaroff I am not seeing it. I am also kinda brain dead right now.
 
$$\cos nx+i\sin nx=(\cos x+i\sin x)^n$$
Equate $\Im$ and $\Re$.
 
I see that, lol. I don't see how it applies.
 
DERP.
 
6:35 PM
@PeterTamaroff mistyped: a moped for riding in the fog?
 
@robjohn I don't understand!
 
@PeterTamaroff ;)
misty ped
 
@PeterTamaroff you wrote that you had mistyped... I parsed it as misty-ped.
 
@robjohn LOL!
 
@robjohn Imagine adding dyslexia to the mix.
 
6:37 PM
@user1 ouch
 
Bring back the LOL chopper?
Nah...
:D
 
Skully
Why is it not football season yet
 
6:53 PM
It is in Canada.
 
Tom Brady does not play in the CFL
 
3 weeks till preseason :D
 
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