Hmm....

Hmm. I think this is a little off-track.

Since the summation is equal to $b-a$

Now what you do is use the fact that $M_{i} = m_{i+1}$

@JayeshBadwaik Oh, right.

@JayeshBadwaik Closed intervals behave so much nicer than open ones...

@PeterTamaroff Yeah they do.

@JayeshBadwaik I could only prove $M_i\leq m_{i+1}$. Maybe I can try showing $M_i<m_{i+1}$ is impossible?

@PeterTamaroff Why? $M_{i} = f(t_{i})$ right?

@JayeshBadwaik DERP.

Is it possible to differentiate $\sin$ without having to use limits?

I know that one can differentiate quadratics without limits

I will explain:

$f(x) = ax^2 + bx +c$

differentiation tacitly invokes limits, so I wonder what you mean by "using" limits. Ultimately, you have to go through the limit definition for the derivative of anything at least once (this can be mitigated for the most part by applying it to product, quotient, and chain rules).

We want to find a line such that

$y=f' x+n$

where

i.e. $f' x+n=ax^2+bx+c$ at $x_0$

Now, the line passes through $f$ once, so

where the line equals the parabola?

@anon At some point

Sorry

@JasperLoy Yes

Using $\Delta$

@JasperLoy I can find the slope of the tangent, however, I do not need limits to find it

(a) how, precisely? (b) but the slope of the tangent is the limit, so you just used geometry to find a limit...

differentiating is a special case of evaluating limits

Hello guys! For simplicity let $(X_i)$ be a countable sequence of Hilbert spaces. How would could we give meaning to $$\bigoplus_i \bigotimes_{k = 1}^i X_k$$ such that it will be a complete space?

M.

I will "differentiate" $f(x) = x^2$ (find the tangent, if you will)

@JonasTeuwen \bigoplus?

at $x_0 = 1$

@JasperLoy Dude, give him a rest, let him explain.

@JasperLoy I will call it "finding a slope"

@anon Yah, was looking it up.

$f(x)=x^2$

$y=mx+n$

$m+n=1^2=1$

@JasperLoy Fermat found minima and maxima of parabolas without limits, but with the $h$ argument we usually use.

I guess it is easy. Just consider the collection of all finite direct sums and complete it.

Ah, I see. If you aren't careful you could allow infinite sums of pure tensors with arbitrarily-many-vectors instead of finitely many, which would be an element out of the space.

$x^2-mx-n =0$

One solution exists; thus

@anon Yep.

$\Delta = m^2+4n=0$

@JonasTeuwen If you complete it, it would include things that are not in the space you wrote down, right?

So, I'd say. The infinite sum is the completion of the space of all finite ones.

@anon Depends on where you complete it.

Like, you first need a norm.

$m^2+4(1-m)=0$

$m^2-4m+4=0$

$\therefore m=2$

@anon If you would take the "infinite sum of norms", that would exclude those thingies right?

@JasperLoy In your face!

Additionally, you can take an uncountable direct sum too.

As it will basically reduce to the countable case otherwise all stuff gets kicked out.

@JasperLoy =D

@JasperLoy Right

Mm. Actually that already will be a complete space me thinks.

If you do it that way.

I was wondering if tricks for trig functions exist

If you take the completion wrt to this norm of all the "finite ones". Then it would be the same.

You would like to have each $X_i$ represent a state space of a system. Then the product would be the space of "interactions between those spaces".

what is the norm on a tensor product anyway? that you are considering, anyway.

The whole thing would represent all possible states and interactions.

@anon That is hard for Banach spaces. For Hilbert spaces take the product of inner products.

That is why I said "for simplicity".

@Argon You need a concept of limit indirectly anyway I suppose, for the following statement.

You could define it this way: $\langle \phi_1 \otimes \psi_1, \phi_2 \otimes \psi_2 \rangle = \langle \phi_1, \phi_2 \rangle \langle \psi_1, \psi_2 \rangle$. Then extend linearly and complete.

okay, cool

It is a kinda explicit construction except the completion step which is like... kinda explicit too.

@JayeshBadwaik To prove it, I guess. It's pretty intuitive,

Anyway, it is something you care little about it is like "I want this vectors in my space. Please add all the nasty shit needed to be able to mess with Cauchy sequences!".

This method can also be used to "differentiate" reciprocal functions, square roots, and possibly others as well

@OldJohn Hi.

@Argon Well, you need some kind of monotonicity, though.

@JonasTeuwen Jonas.

@JonasTeuwen Suppose we have an infinite sum $$\sum_{n=1}^\infty \alpha_n(\psi_1^n\otimes\cdots\otimes \psi_n^n)\tag{$\star$}$$ where $\langle\psi_{i}^n,\psi_{j}^m\rangle=\delta_{ij}\delta_{mn}$ (i.e. the $\psi$'s are orthonormal). Can we not make the $\alpha_n$'s shrink fast enough that the partial sums of $(\star)$ are Cauchy?

@PeterTamaroff "Monotonicity"?

@Argon Monotonous.

@Argon Wait, I think I have caught you somewhere. Because, there are is another line intersection only in one point with the qudractic equation.

The tangent is not the only one.

The vertical line is another. x=1.

@JayeshBadwaik He assumes $y=mx+n$...

@anon I'd say yes.

@PeterTamaroff Yes?

@PeterTamaroff What is the qudratic equation was somewhat rotated in $x-y$ plane. then?

@JonasTeuwen Could you help me prove that if a function is integrable on $[a,b]$ then it is continuous on infinitely many points of $[a,b]$?

Say by $\pi/4$ degrees.

@JayeshBadwaik He also assumes $y=c+bx+ax^2$

@PeterTamaroff The set of discontinuities is of measure zero!

@PeterTamaroff Hmm.

!!

It is not really measure theory.

@JonasTeuwen Heheh OK.

@JonasTeuwen Spivak provides a sketch. I'll try and do it, but maybe I ask for assitance.

But first: coffee!

@anon But the convergence would be like in the sum norm.

@PeterTamaroff are you talking about Problem 31?

@JayeshBadwaik Yup. Not $31.$ alone, $^*31.$!

That means it is tougher.

@PeterTamaroff Yup.

@anon That would be like a space of countably many particles where everything interacts with everything... like Hell.

@JayeshBadwaik Do you think it is hard?

@JonasTeuwen So, then universe is hell! But we can ignore some of the interactions and call it heaven.

@JonasTeuwen Johnny.

@PeterTamaroff Like this: Let $\epsilon > 0$ be arbitrary. Let $A$ be the set of discontinuities. If we can find a cover of $A$ by open intervals $(a_i, b_i)$ such that $\sum_i b_i - a_i < \epsilon$ the stuff would be a null-set.

@PeterTamaroff Yes. But, I think you should be able to get it.

Is it hard to prove that if a function is Riemann integrable then the set of discontinuities is measure zero.

And say you can do integration on $[0, 1]$ (makes sense), this would be like never under epsilon and so you would have infinitely many points left.

@PeterTamaroff Yes, cause it is false.

@JonasTeuwen Sorry!

So.

Let us see, what does it mean for a function to be Riemann integrable?

@PeterTamaroff It is not NP-hard at least. :P (Bad humor I know)

(think partitions)

@JayeshBadwaik If I were a programmer, maybe I'd have laughed.

(use those to get your null set)

(and you're done)

I didn't laugh, and I program. But I am not a programmer.

@JonasTeuwen That for each $\epsilon>0$ there is a partition $P$ for which $U(f,P)-L(f,P)<\epsilon$

And for all refinements.

@JonasTeuwen Pardon?

Use the nets bro. Use the nets.

If you add more points it should be not bigger.

All refinements of the partition also have that property.

@JayeshBadwaik Not necessarily.

Otherwise you could avoid ugly stuff.

Oh, that is new to me.

Okay, I quit! 8-).

@JonasTeuwen ??

I am unable to help somebody with things I don't understand!

So I will go back to my tensors.

Consider $f(x)=0$ for $x\in [0,1]-\{a\}$ and $f(x)=1$ for $x=a$

Yea. So?

What if you refine taking $t_k=a$?

Then you took a sucky refinement.

It is like taking the $N$ in sequence convergement not large enough but it might for for that $N$.

But you should take it large enough such that bad stuff stops happening.

@JonasTeuwen When you say refinement of $P$ is just $P'$ for which $P\subseteq P'$, right? (Maybe you mean sthing else)

Basically it says, from some partition on and all refinements you should have that.

@JonasTeuwen Did I say something wrong? I think I am correct. I am quoting Rudin Theorem 6.7 page 125 (though that does not show the validity.)

There might exist one which does satisfy the property but not the second.

@JayeshBadwaik No, Peter. But I thought I misunderstood integration so I could not help.

@JonasTeuwen What is a null set?

@PeterTamaroff It is like you would say the sequence $a_n = n$ for all $n$ except $a_5 = 0$ converges to $0$ because $N = 5$ "works".

@JonasTeuwen Hmm. Okay. I am still on somewhat shaky ground with misconcepts from my "continuous function" dominant calculus course, so I might utter one or two missteps here or there.

@PeterTamaroff I defined it.

You can take it way further.

@JonasTeuwen I didn't say that... I just said that "and for all its refinements" is not correct.

Spivak doesn't even mention that in the theorem, or the proof.

Meh, whatever. I do something else.

Yes, probably. Doesn't matter that he does not mention.

@JonasTeuwen Don't be so stubborn! =(

It is true.

A null set is just that you can give any covering of the set by intervals arbitrary small total length.

@anon I think it is more fun in reproducing spaces!

@JonasTeuwen So a null set $A$ is a set such that for every $\epsilon >0$ there is a finite open cover of $A$ $\bigcup (a_i,b_i)$ with $\sum_i b_i-a_i<\epsilon$ for each $\epsilon$?

Yea.

@JonasTeuwen Ain't that compactness wizardry, bro?

That's useful cause your partitions are like concatenated intervals.

No.

Compact does not care about size.

@JonasTeuwen (Well, but finite covers....)

It roughly says that you cannot go to far.

The cover is not finite.

The closure is compact.

But the set will be open.

@JonasTeuwen Oh, since you wrote $\sum_i$ I assumed there was some end to it,

If it would be closed that would mean there is like an interval where you thing is not integrable.

Kinda.

@JonasTeuwen OK.

Okay, let us say the thing is discontinuous on at least two distinct points and they live happily together with other points in some interval where the integral is not well-defined.

Where the function is continuous.

Also say it is not a null set.

Now we have to fit in a closed set and get a contradiction! 8-).

The real proof is easy.

Let us make it harder.

Okay, say... say. The set of discontinuities is a closed set. Hence it is complete as a metric space!

@JonasTeuwen So the theorem is. Let $f$ be integrable on $[a,b]$. Let ${\mathscr C}=\{x:x\in[a,b]\text{ and }$f$\text{ is continuous at }x=a\}$. Then $|\mathscr C|\geq |\Bbb N|$.

Hm.

Say the set of discontinuities is closed.

This would mean the thing is of second category.

@JonasTeuwen OK.

@JonasTeuwen OK... whatever that means.

So for any rewriting as an countable union of open sets there is like a non-trivial interval in there somewhere.

With like fixed length.

And hence not the closed one cannot be a null set.

Now what.

So it is like discontinuous on all points in some interval, cannot be integrable!

Okay! SO!

Hmm. Err. Whatever!

It can also be neither open nor closed.

So either open or neither open nor closed.

@JayeshBadwaik I think there's a typo on my book.

@PeterTamaroff why?

That doesn't help as most such sets are. Lol. Whatever.

@JayeshBadwaik It says in $^*31.$ "$[0,1]$" while we are talking about $[a,b]$ all the time...

@JonasTeuwen XD

wow...back

@PeterTamaroff Hmm, yup, did not notice that. But it did get me thinking about why the $M_{i} - m_{i} < 1$.

It doesn;t matter much I think though.

@JayeshBadwaik Suppose $M_i-m_i\geq1$ for all $i$!

or does it?

I did not read the discussion above.

@JayeshBadwaik ?¿

Why is $[0, 1]$ any different from $[a, b]$?

@PeterTamaroff I mean, the discussion between you and Jonas.

@JonasTeuwen Exactly my point.

Do you think analysis in $[a, b]$ is different...?

It is a linear transformation to take the first to the second.

Yes.

Blow and shove.

Or shove and blow!

@JonasTeuwen Yes, still, it is a silly typo, IMO.

@JasperLoy Seems like... obvious?

Why would it depend on what $a$ and $b$ are?

@JasperLoy sorry to ping you without saying anything... it's like was callling you...i do that,just say your name...to see if you are there...

@JasperLoy you can ask me anything you desire

Yes. Today I talked to somebody that had some issues with distribution theory!

But quite basic ones really.

Nope.

This is dude is getting rep amazingly fast

@JasperLoy the inverted V?

Top 0.77 in 32 days.

"I did study math and had a knack for it, but I am sooo out of that business now ..." I hope I never say that!

@JasperLoy Isn't that just divide and conquer?

@PeterTamaroff imagine if was into the business

@Charlie ?¿

I have shown that the complex numbers of modulus one with multiplication form a group. I am now asked to "find a symmetric object which has this group as a a group of, not necessarily all, its symmetries"...Anyone any idea what this object is? Is it just a circle?

@PeterTamaroff I could say that now :)

@OldJohn Oh noes!

I never really understood any maths 8-).

@PeterTamaroff 'fraid so - I have very little maths knowledge now - although I used to be a bit better

@PeterTamaroff if he's out the business and is getting rep that fast,imagine if he was working with it...got it?

@Charlie Nope, if he was in business, he wouldn't waste time here!

@PeterTamaroff who knows

@JasperLoy why?

(I wouldn't, I hope!)

@OldJohn so is it true, that as you grow old, your ability to do math really decreases? My prof (not really a prof, someone I used to know and used to go to ask difficulties) used to tell me that (he was 79).

@JasperLoy even at my best, I would never have won any sort of medal - unless irt was made of chocolate :)

@JasperLoy I think Christian did it quite well. I'd give him full marks! 8-).

@JayeshBadwaik Yes, it is true - I can still do some useful stuff sometimes, but my ability to concentrate hard om something for a long time is less than it used to be

You too.

@OldJohn My ability to concentrate on something for more than like 2 minutes straight is like ill-defined!

@JonasTeuwen :)))

@JayeshBadwaik I already did $(a)$.

Now I have to do $(b)$ and $(c)$.

What is the "trick" Spivak talks about in $(a)$?

@OldJohn I thought that would like reduce my aspects of being a mathematician as I know many people back then which were way better and could focus really well. But now I turn out to be better than them!

@PeterTamaroff "trick"? he mentions no such thing in my version.

oh no..not again

@JayeshBadwaik He says "One can choose $[a_1,b_1]=[t_{i-1},t_i]$ from part $(a)$ except for $i=1,$ or $n$. In this case a simple artifice solves the issue."

My best period of mathematical expertise was when I didn't drink alcohol and I practised yoga concentration exercises regularly :)

@JasperLoy me too! - and it was only useful for impressing classrooms full of kids :(

@JasperLoy i heard that in my uni there's a guy,who drunk knows thousand digits.

@JasperLoy define "leftie" :)

@JasperLoy damn it,jloy,ask anything!

@PeterTamaroff Hmm, I can not think right now what he means.

no - sorry - similarity has its limits it seems

@JasperLoy yes

@JayeshBadwaik But you do have the book with you, right?

@PeterTamaroff yup.

I get now what you meant.

You said in $(a)$ so I was a little confused.

@OldJohn Nah, he is like on the other side of the planet. So left too!

@JasperLoy In my younger days I was very much into yoga and spent a whole summer reading Bhuddist literature :)

And all severely disturbed.

@JasperLoy my son is a leftie - if that counts

@OldJohn Are you sure he is yours...? 8-).

@JonasTeuwen hahah

@JonasTeuwen Oh yes - I was there at the time :)))

@JasperLoy Oh, just didn't think about voting.

@OldJohn Mm. There can be quite some uncertainty eh 8-).

@JonasTeuwen :))

I would request a DNA test.

(is joke)

@JasperLoy go thinking

I spent the afternoon today with a "finacial advisor" - it seems I can afford another holiday :))

"LaTeX for the rest of us" - what the heck. Who is "the rest of us"?

@OldJohn That's like a holiday advisor or what?

@JasperLoy wtf?

@JonasTeuwen well - he told me I was better off than I thought - my conclusion was "more holidays" :)))

Great.

I need somebody to tell me that too.

ah - but I don't understand money - he does

There is such a button!

Uh. Yeah.

Do faithful actions extended bases of polynomials rings?

@JasperLoy Don't say such things...

please

Like, I have the Weyl group (or algebra) $\mathbb Z[t, \partial_t]$. Also, I have the polynomial ring $\mathbb Z[t]$. So $t$ stuff like $t$ increases the grade and $\partial_t$ decrease. These are like faithful actions right?

It kinda embeds the polynomials in the symmetry group of the Weyl algebra.

Why do I not know anything about algebra?

@JonasTeuwen Are you using "like" on purpose?

Nope.

@JonasTeuwen to an analyst, algebra is on the dark side - don't go there - don't even talk to people who go there ...

I am on all sides. An opportunist if you wish.

@JonasTeuwen Kudos.

- but I am converting myself from an analyst (of sorts) to an algebraist :P:P:P

the rot set in when I discovered Banach algebras :)

Yes.

When did you discover those? I use them like all the time.

about 40 years ago - but the rot has spread slowly

Ah!

@OldJohn So. If I understand like intuitively correctly a polynomial ring is like "okay here is this field! I give you like this brother called $X$. Let him join in the fun of your fieldness! They are like "Oh man, now we're no field anymore. Let us make the best of it." And "the best" would be the polynomial ring. Right?

I recall being seriously impressed by the spectral radius thingy - connecting analysis and algebra in some way that i cannot remember at all now :(

It is like you have children play. Good friends. Then you bring in the guy with no friends to have some fun too. You know.

@JonasTeuwen you have a seriously weird way of looking at mathematical objects :)

Spectral radius of $A$ is just the radius of the disk such that for any smaller disc contained (radius $r$) the operator $A - r$ is invertible.

Why is that weird?

@JonasTeuwen it would be nice if everyone thought like that...and teach...

@JonasTeuwen not weird - I just remember being impressed with some mystical connection between algebra and analysis - I forget the details

@OldJohn Hm. I think you need to go to like spectral measures for that.

@Charlie Why?

@JonasTeuwen because it's really cool!

But does it not make any sense?

@JonasTeuwen like like like!