Mathematics - 2014-04-17 (page 4 of 4)

Mike

19:00

@Shisui The answer you don't understand is not in terms of elementary functions.

Shisui

@Mike I see. So elementary functions include polynomials, trigonometric functions, hyperbolic functions, inv. trig, inv. hyp, exponential and logarithmic functions?

Mike

Something like that. But it includes the dilogarithm, which definitely isn't anything anyone would call elementary:

$$\text{Li}_2(z) = -\int_0^z \frac{\ln(1-u)}{u}du$$

Gabriel R.

@user127001 this would go way beyond real analysis

Daniel Fischer

@PedroTamaroff What was the cause of that assessment?

Chris's sis

Back.

Pedro Tamaroff

19:04

The axiom of choice stuff

Shisui

@Mike That sounds pretty interesting! I've not heard of the dilogarithm before :)

robjohn

@MattN. It comes on at 4-11 PM here. I will be watching and my wife will join in when she gets home.

@Mike definitely a special function

Anthony

Hello again.

Quick question-for space filling curves, they're describe as a limit of curves, how do you know the limit exists…?

robjohn

@Chris'ssis: I can finally recognize you in the avatar bar :-)

Chris's sis

@robjohn :D

Mike

19:09

@Shisui Personally, I find it dreadfully uninteresting.

robjohn

@Anthony because the sequence of curves converge uniformly

Chris's sis

@robjohn I think I'm about to produce something like Ramanujan, honestly ...

Anthony

@robjohn To show it converges uniformly, don't we need to know what it converges to…?

robjohn

@Chris'ssis how are you going to produce a Ramanujan-like entity?

is that like having a cow?

Chris's sis

@robjohn :-))) I'm going to use a family of identities I just came across ...

I'm working on that right now.

Anthony

19:13

I don't see why the Peano curve is so counter intuitive… Can someone explain?

robjohn

@Anthony we do... given a point in $[0,1]$ we can give the point in $\mathbb{R}^n$ to an arbitrary precision.

mirgee

Hi guys, how do you show that $\sum_{n=1}^{\infty}{(-1)^n}{(1+2\cos {\pi n \over 4}) \over 3^n\ln n}$ diverges?

Sorry, meant $\sum_{n=1}^{\infty}{(-1)^n}{(1+2\cos {\pi n \over 4})^n \over 3^n\ln n}$

Chris's sis

@mirgee from $n=1$?

mirgee

@Chris'ssis $\sum_{n=2}^{\infty}{(-1)^n}{(1+2\cos {\pi n \over 4})^n \over 3^n\ln n}$ :)

Anthony

@robjohn What do you mean the point? How would I check to see it converges uniformly?

Oh this site uses a Cauchy sequence...

robjohn

19:16

@Anthony actually, the differences can also be shown to converge absolutely with uniform bounds.

Anthony

I'm not sure what you mean. The differences between…?

robjohn

@Anthony $\sum|\gamma_{n-1}(t)-\gamma_n(t)|\lt B$ for some $B$ and all $t\in[0,1]$

@Anthony that gives uniformly Cauchy

Anthony

Ah.

@robjohn But why is the result so counterintuitive?

Enjoys Math

0

Q: Do I need to handle recursion in this new kind of machine?

(Question at the very bottom) Def 1. Let $F = \Bbb{Z}_p$ be a finite field. Then an $F^k$-machine is a machine with $k$ input / output memory slots. All computations are done in the field $F$ and in a problem-specific amount of arithmetic registers. Thus algorithms on a $F^k$-machine can be ...

robjohn

@Anthony because the curve is fractal. It is Hölder-$1/n$ at all points.

Anthony

19:20

@robjohn I'm afraid I don't know what that means.

Chris's sis

@mirgee split the sum and then you're done. Analyze the behaviour of $(1+2\cos {\pi n \over 4})^n$ and see it yields $3$ sometimes. You have a finite sum + $\displaystyle \sum_{n=2}^{\infty} \frac{1}{\log(n)}$

robjohn

@Anthony it is very kinky

Anthony

The stuff I've read just says it's surprising because you're filling a square with a line in finite time

Enjoys Math

I've shown that $F^k$-machine algorithms have same computational complexity class as their counterparts on standard machines

robjohn

@Anthony well, the length of the curve is infinite of course.

Anthony

19:22

But like, doesn't it make sense that you can stretch the domain infinitely far to cover it?

Mike

@Anthony Any curve I can normally imagine doesn't cover the square.

Anthony

@Mike But why?

Mike

Draw a curve.

Anthony

Haha.

But like doesn't infinite stretching seem to imply the possibility?

Mike

No really.

Chris's sis

19:23

@mirgee I'M WRONG THERE! It's a sum like that, but smaller. Even so, it goes to $\infty$.

robjohn

@Anthony I guess that depends on what your preconceptions of curves are. Most curves we think of are differentiable or at least rectifiable. These are not.

Mike

"infinite stretching"?

Anthony

Well I mean that's what we're doing right?

I guess it doesn't make sense. I don't know.

I have a very twisted sense of math.

I'll be back, gotta go take a test. Thanks for the help.

Mike

You need to be precise with your thoughts.

This will help - a lot.

Enjoys Math

Are there polynomials in $\Bbb{Z}_p[x_1, \dots, x_k]$ that cannot be computed in time polynomial in $p, k$?

Mike

19:26

what do you mean computed

Enjoys Math

I mean on a standard machine mon

your home desktop

in $O(t(p,k))$ time

where $t$ is poly

Mike

what are you computing

Enjoys Math

I don't know if I stated that correctly though

I'm doing an existence proof:

0

Q: Do I need to handle recursion in this new kind of machine?

(Question at the very bottom) Def 1. Let $F = \Bbb{Z}_p$ be a finite field. Then an $F^k$-machine is a machine with $k$ input / output memory slots. All computations are done in the field $F$ and in a problem-specific amount of arithmetic registers. Thus algorithms on a $F^k$-machine can be ...

computer-science finite-fields computational-complexity formal-languages

Chris's sis

@mirgee It's $$\displaystyle \sum_{n=8,16,24 ... }^{\infty} \frac{1}{\log(n)} \rightarrow \infty$$ Q.E.D.

Mike

what are you computing that you want to compute in polynomial time

robjohn

19:27

@mirgee all the terms except those for which $n\equiv0\pmod{8}$ converge absolutely. Those for which $n\equiv0\pmod{8}$ form the series $\sum\limits_{n=1}^\infty\frac1{\log(8n)}$ which diverges by the integral test

So the series is the sum of 7 convergent series and one divergent series

Enjoys Math

@Mike, I want to see whether there exists a polynomial time algorithm for all $F^k$-compatible problems

Chris's sis

@mirgee that was a funny question! :-)

robjohn

@mirgee so you could replace the $\log(n)$ by $n$ and the series would still diverge.

@Chris'ssis did you mention that the other 7 equivalence classes mod 8 give series that are convergent?

Chris's sis

@robjohn I said above the other series are all convergent. (in other words)

robjohn

@Chris'ssis okay, I didn't see that, and it is an important point that all of the others are convergent.

mirgee

19:33

@robjohn And why do they converge? By comparison test with geometric series?

robjohn

@mirgee comparison with geometric, yes

Chris's sis

@robjohn Sure, that is clear.

mirgee

@robjohn @Chris'ssis Thank you guys :)

Have test tomorrow... So nervous! :)

robjohn

@DanielFischer them's fightin' words

Chris's sis

@mirgee Welcome!

robjohn

19:37

@mirgee anytime... good luck tomorrow!

Mike

@EnjoysMath That's fine, I'm not too interested in that. But you asked a specific question: "Are there polynomials in $\Bbb{Z}_p[x_1, \dots, x_k]$ that cannot be computed in time polynomial in $p, k$?" And I don't know what you are trying to compute. Are you trying to evaluate the polynomial?

Enjoys Math

@Mike do there exist polynomial expressions for $f \in \Bbb{Z}_p[x_1, \dots, x_k]$ such that, directly using the expression : e.g. $(x + y)^8$ takes $1 + \ln(8)$ ops

such that they cannot be Evaluated

in easy time

Mike

okay

Enjoys Math

Or, wait...

For each $f \in \Bbb{Z}_p[x_1, \dots, x_k]$ does there exist an expression for $f(x)$ that can directly be evaluated in a certain time. Hard to define here... have to think

Will Hunting

19:56

@EnjoysMath Hey are you doing plant medicine again?

Enjoys Math

Let $m(f)$ be a minimal expression for the polynomial $f \in F[x_1, \dots, x_k]$, i.e. number of operations for computing $f(x)$ at any $x$ directly using the expression spec of $m(f)$ is minimal.

Chris's sis

20:14

$$\frac{\sin(1) \cos(1)^1}{1^2}+\frac{\sin(2) \cos(1)^2}{2^2}+\frac{\sin(3) \cos(1)^3}{3^2}+\cdots=\int_0^{\tan(1)} \frac{\arctan(t)}{t} \ dt+\frac{\pi}{2}\log(\cos(1))-\log(\sin(1))$$

Fernando Martin

Hey guys, I'm having a discussion with a friend about this:

Algebraic number theory is:

(Algebraic number) theory

Shisui

@Chris'ssis Is that a variation of the earlier series you mentioned?

Fernando Martin

or algebraic (number theory)?

Chris's sis

@Shisui not really. (although they might look similar to a certain extent)

Pedro Tamaroff

@FernandoMartin (Algebraic) number theory.

IMO.

Daniel Fischer

20:29

@FernandoMartin Algebra(ic number theory)

Fernando Martin

@Pedro: please join the IRC channel

user127001

20:42

@FernandoMartin what is the IRC channel

Enjoys Math

0

Q: $\iff$ conditions for $\rm P \neq NP$ maybe?

Please review the $\rm P \neq NP$ problem here. I'm working on an algebraic approach to this problem, and all my notes are currently here. Conjecture 1 If there exists no polynomial $f \in F[x_1, \dots, x_k]$ such that $f$ has a minimal expression $e(f)$, e.g. $e(x^2 + y^2 + xy + xy) = (x + y)^...

abstract-algebra polynomials computer-science computational-complexity

Shisui

Is it just me, or is this integral an algebraic slog at best? $$ \int \dfrac{x^2 + x - 1}{(x+1)(x^2 + x + 1)} \text{ d}x $$

Chris's sis

@Shisui this is elementary.

Gabriel R.

Just a quick off-topic question: do you know how executives at SE pay for server-related expenses ? I haven't seen any donation button, or any ads that may generate money.

@Chris'ssis you always say that :P

Chris's sis

@GabrielR. I always like to tell the truth. ;)

Shisui

20:56

@Chris'ssis I decided to split the integral up as follows: $$ \begin{aligned} \int \dfrac{x^2 + x - 1}{(x+1)(x^2 + x +1)} \text{ d}x & = \int \left( \dfrac{1}{x+1} - \dfrac{2}{(x+1)(x^2 + x + 1)} \right) \text{ d}x \\ & \overset{\text{Partial frac}}= \int \left( \dfrac{2x+1}{x^2+x+1} - \dfrac{1}{x+1} - \dfrac{1}{x^2 + x +1} \right) \text{ d}x \end{aligned} $$ Would you have followed a similar route?

Chris's sis

@Shisui Yeah, I'd follows such a way.

Shisui

@Chris'ssis Thank goodness. I had two pages of working to pull out the correct answer and thought there might've been a shorter and neater way that I may have overlooked.

Chris's sis

@Shisui This is the way to follow.

Gabriel R.

@Chris'ssis What's your favorite equation/identity ?

Enjoys Math

@DanielFischer what are your thoughts on my post above ^^ if you have time

Shisui

21:01

@Chris'ssis There's an error in my post above. It should read $2x$ as opposed to $2$ in the numerator of the second part of my first integral on the RHS.

Daniel Fischer

@EnjoysMath The $P\neq NP$ thing? Haven't looked yet.

Chris's sis

@GabrielR. This question reminds of job interview questions where I cannot answer such questions and annoy many people there. :-))) I love many things, and when I say "many" that really means "many".

Gabriel R.

@Chris'ssis yeah, that is to say more than one right ? ;)

Chris's sis

@GabrielR. Exactly.

Gabriel R.

@Chris'ssis I have another question to ask: if you had only one thing to take with you on a desert island, what would it be ? ;)

Shisui

21:06

@GabrielR. One of my all time favourites is for a triangle, $ \alpha + \beta + \gamma = 180^{\circ} $, where the greek letters are the angles of a triangle.

Gabriel R.

@Shisui Good one. It has a beautiful proof

Shisui

Could you link me to it? @GabrielR.

Gabriel R.

@Shisui Hold on, let me draw it

Chris's sis

@GabrielR. The point is that life is not a static thing, but a dynamic one, things change permanently, and all it depends on time you go there. If I were to go there now, I'd probably take Ramanujan's notebook and keep reading it there. Almost each identity there really impress me.

@GabrielR. After some years it might seem boring to me, but not now.

Gabriel R.

@Shisui sketchtoy.com/60367460

Pedro Tamaroff

21:14

@GabrielR. DAT BE COOL

Shisui

@GabrielR. isn't it the other way around for the final two angles you wrote down?

I thought that the z-angles (alternate angles I think they are called) were supposed to be equal.

Gabriel R.

@Shisui right

Will Hunting

@GabrielR. I changed to black square like you

Daniel Fischer

As for triangle identities, I like $$A = \frac{1}{4}\left(\pi - \alpha - \beta - \gamma\right).$$

Shisui

@GabrielR. sketchtoy.com/60367681 right?

Gabriel R.

21:23

@Shisui yes exactly, I apologize for this embarrassing mistake

@DanielFischer is $A$ the area ?

Daniel Fischer

@GabrielR. Yes. And $\alpha,\beta,\gamma$ are the angles.

Gabriel R.

@DanielFischer so if I scale a triangle by 2 the area remains the same ?

Daniel Fischer

@GabrielR. What does "scale by $2$" mean for hyperbolic triangles?

Gabriel R.

@DanielFischer scaling lengths by $2$

Daniel Fischer

@GabrielR. Changes the angles.

(If at all possible)

Pedro Tamaroff

21:29

@Mike

Suppose $M$ is a module over a PID.

Gabriel R.

@DanielFischer but what about similar triangles ?

Daniel Fischer

@GabrielR. All similar triangles are congruent in hyperbolic geometry.

Gabriel R.

@DanielFischer Did you just kill Euclid ?

Anthony

I'm baaaaaaaack

Daniel Fischer

@GabrielR. Nah, Gauss and Bolyai did in the $19^{\text{th}}$ century.

Gabriel R.

21:30

@DanielFischer Euclide se retourne dans sa tombe

@DanielFischer what exactly are the axioms of hyperbolic geometry ?

Daniel Fischer

@GabrielR. Euclidean minus parallel postulate, plus there are infinitely many parallels through each point not on the line, iirc.

Alyosha

$$\sum_{k \ge 0}\left \lfloor\frac{n+2^k}{2^{k+1}} \right \rfloor, n \in \mathbb{N}$$

Gabriel R.

@DanielFischer is there an intuitive approach ? I cannot picture it

Chris's sis

@Alyosha This seems trivial too. Try Hermite's identity.

Mike

@DanielFischer By the end of next week I'll have proven the Gel'fand-Naimark theorem.

Then C* algebras will really by operator algebras.

Daniel Fischer

21:39

@GabrielR. Take the unit disk model. Points are points, straight lines are arcs of orthocircles, that is, circles intersecting the unit circle at right angles.

Chris's sis

@Alyosha You finally get a telescoping sum and you're done.

Alyosha

@Chris'ssis Thanks.

I'll try to post less trivial ones in future.

Chris's sis

@Alyosha It's interesting and nice. Some would really be in trouble with this one.

Alyosha

@Chris'ssis As, admittedly, was I (which is why I posted).

Gabriel R.

@DanielFischer this makes sense. How do you define areas though without integration ?

Chris's sis

21:45

$$\sum_{k \ge 0}\left \lfloor\frac{n+2^k}{2^{k+1}} \right \rfloor=n$$

Daniel Fischer

@GabrielR. Not, we have integration. The area element is $$\frac{dx\,dy}{(1-\lvert z\rvert^2)^2}.$$

Alyosha

@DanielFischer $z$ is the curvature?

Daniel Fischer

@Alyosha No, $z = x+iy$ is the point.

Anthony

Can someone explain to me once more why the Peano curve is counterintuitive?
The reason I feel it isn't is because we can distort the line as much as we want. Why doesn't it make sense you can fill the square just like you can color in a square with a pencil?

N3buchadnezzar

22:01

@Anthony I think some of the problem lie in the "width" of the line

Anthony

But don't we construct areas from integrals with infinitely thin rectangles?

N3buchadnezzar

When drawing functions, mathematically a line consists of points each having width zero'

So by filling a rectangle with lines each of width zero, in some sense one should expect that no matter how many lines one use, you should never be able to fill the square.

Anthony

Don't we do that with integrals?

Mike

no

skullpatrol

There are "infinitely many" rectangles...

Anthony

22:03

I think I found my problem….

N3buchadnezzar

@Anthony No ? en.wikipedia.org/wiki/Space-filling_curve

Anthony

What I'm saying is I though that integrals were the sum of infinitely many infinitely thin rectangles.

But I guess infinitely thin doesn't mean a line?

Mike

be precise with your thoughts

if you think you can fill in a square that way, then describe a way you think it can be done

N3buchadnezzar

$$ \int_0^\pi x \cos^2n(x)\,\mathrm{d}x = \frac{\pi^2}{2^{2n+1}}\binom{2n}{n} $$

I "discovered" this =)

Mike

what is it

scam site?

Anthony

22:06

My thoughts are that if integrals construct areas from arbitrarily small rectangles, which would approach a line, then we should have no trouble constructing a square from lines, albeit a lot of them.

@Ethan lol

Mike

lol

N3buchadnezzar

@Anthony Well with integrals you start of with rectangles and make them smaller and smaller. Each of them actually have a width, even though no matter how small you make them

Anthony

@N3buchadnezzar I see.

N3buchadnezzar

@Anthony In the line case you are going the other way, in some sense. You are trying to add something togheter wich has no width, into something that has width.

If that made sense..

Mike

@Anthony No. A curve is a function from $[0,1]$ to your target space. You're describing something vaguely and noncommittally.

Anthony

22:08

@N3buchadnezzar I see, I see. Is the finite time part also significant?

Mike

How do you do what you want to do so that it's a curve?

Anthony

Mike, I suppose I'd have to make a limit…?

Mike

Skull all you're going to do is muddy the water for someone who's already a little confused.

Anthony

A lot...

skullpatrol

sorry

Mike

22:09

@anthony Desceibe the limit you want to do. Put it on paper.

N3buchadnezzar

@Anthony I do not know. I have not taken many deep analysis courses yet, but soon..

Mike

@skull no problem, man, just chat it up when someone's not already coddled about that stuff :)

confused*

N3buchadnezzar

$$\frac{1}{2}\left(\frac{\pi}{2}\right)^2\binom{2n}{n}$$

Anthony

Perhaps something like sin(n(x)) is what I was thinking?

Mike

@Anthony Anyway, you should write down your idea on paper and see why it doesn't work out.

Anthony

22:11

Where you just start scribbling things tighter together.

Mike

Well, what's the limit of that as $n$ goes to infinity, for fixed $x$?

skullpatrol

1, 2, 3, 4,...

Anthony

@Mike Yeah, there's no limit. But you still fill in more and more of the space.

But we're looking for a function?

Mike

@Anthony that's the definition of a curve

A continuous function $[0,1] \rightarrow X$

Anthony

Indeed.

So what about going across the square with some squiggle, like sine.

Then going back over some gaps.

Repeatedly? I don't know how I would write that.

Mike

22:15

Try it. Be precise.

You can write your functions piecewise, if you want.

Anthony

Agh so hard. Go over it with sin(x) for the first part, then double back with sin(2x), etc?

For a square of side length $a$ for $t \in$ the $i$th partition of $[0,1]$, we have $sin(\frac{2it\pi}{a}-a)$ if i is even and $sin(\frac{2it\pi}{a})$ if i is odd? I think that's continuous…. @Mike

Mike

How are you partitioning the interval?

Anthony

Evenly into $i$ pieces.

Mike

Well, that's not going to cover the square.

So now you're going to want to take the limit of these?

Anthony

Yeah!

Mike

22:27

You're going to fall into the same trap as before :)

Anthony

:(

Mike

I guess I shouldn't be smiling.

Anthony

Why is that the same trap?

Shouldn't every point eventually be hit?

Mike

Fix $x$. If $x$ is rational... the limit of this is gonna be zero!

Anthony

Wait, what I meant to be saying it to wave through the square once, you've hit some points. Next time, do that twice as fast, and then wave back through at a higher frequency, hitting more points. Etc.

Mike

22:30

Or $t$ I guess.

@Well, you want to do this in "unit time" (a map from the interval). So you should be doing that at 2x speed, then 4x speed, then 8x speed... right?

Anthony

Yeah.

Mike

Two problem.

1) I doubt that will cover the square. It will probably cover a dense subset of the square. But not the whole thing. 2) What's $f(1)$ going to be? You won't be able to find an $f(1)$ that makes that continuous.

Anthony

I don't get 2), and how do I know it doesn't fill the square? Why is it less likely, intuitively, than the pean curve filling the space?

Mike

The trouble for continuity at $f(1)$ is that even as time goes on, you keep traveling huge chunks of the square. (To be precise, with the thing you describe, I can find a sequence $x_n$ converging to $1$, with $f(x_n)=f(0)$ for all $n$; but I can find another sequence that converges to $1$ but which always has $f(y_n)$ equal to the other corner!)

Anthony

So again, the curve has to approach the function.

I see.

Mike

22:36

So if you want it continuous, you'll need to make sure you only cover chunks of the square; your first half of the unit interval should map to, say, the left half of the square; your first quarter should map to the bottom left quarter of the square...

Anthony

And then you get to the Peano curve.

Ish.

I think I understand more of my troubles.

Mike

This line of thought... eventually brings you to thinking about Peano-like curves.

Anthony

Thank you Mike.

Mike

No prob. BTW, my guess as to why it wouldn't cover the whole interval is that sine is too nice. I can't prove it off the top of my head and don't want to.

Anthony

I see.

The curve can be piecewise continuous, right?

Or differentiable!

Mike

22:38

Well, the curve has to be continuous.

Anthony

Didn't mean that! Sorry.

Mike

And... no. It can't be differentiable. :D

Anthony

I meant differentiable.

Not piecewise?

I can't have one little part that is?

Mike

Not continuously.

Or maybe.

I don't know too much about that stuff. The derivative would fail to exist at infinitely many points, at least.

skullpatrol

@Anthony Have you tried to put together a question and asking on main?

usukidoll

22:49

http://math.stackexchange.com/questions/758391/please-give-feedback-to-my-answers-sets

duplicate!!!!

Anthony

@skullpatrol No...

skullpatrol

Give it some thought.

Anthony

I mean I've asked a question.

Just not for this.

Also why do they say "Show a limit exists and is equal to L"? I feel like showing existence and finding the limit are one in the same.

MickLH

not necessarily

Anthony

Is it possible to find a limit without going through the work of showing it exists?

MickLH

22:54

You could find that the limit of a series going to infinity exists by the ratio test, without finding the limiting value

Anthony

^Point taken. But so can you do the opposite?

Somehow find a limit for something that doesn't have one?

skullpatrol

You can't find something that doesn't exist first, right?

Anthony

I mean, straightforwardly no… I guess.

MickLH

@Anthony can you find me my car keys real quick?

Anthony

It just seems weird to not just say find the limit.

@MickLH :p

Mike

22:56

@Anthony It's easy to find out that the limit of the following exists.

$a_n1+1/4+1/9+1/16+...+1/n^2$

Anthony

I guess I just didn't like that it said show the limit exists and is equal to something, as opposed to just saying find the limit.

Mike

Finding what it is is harder.

Anthony

Indeed.

Mike

meant $a_n=$

Anthony

Gracias all.

skullpatrol

22:58

thank mike, he's the best :-)

Anthony

He is indeed.

skullpatrol

I hope I didn't confuse you...

...if I did I apologize.

saadtaame

23:20

Are these equal: $$(1+2+3+\dots)=(1+2+2^2+2^3+\dots)(1+3+3^2+\dots)(1+5+5^2+\dots)\dots$$ Where the RHS has a series for each prime. Looks like they are the same series by the fundamental theorem of arithmetic.

r9m

23:35

Dr. Who !!! seriously ???

Alyosha

23:49

@saadtaame In a sense, though neither side converges.

$$\left( 1+ 2^{-s}+ 3^{-s}+ \cdots \right)=\left( 1+ 2^{-s}+ 2^{-2s}+ \cdots \right)\left( 1+ 3^{-s}+ 3^{-2s}+ \cdots \right)\left( 1+ 5^{-s}+ 5^{-2s}+ \cdots \right)\cdots$$

Both sides converge for $s>1$, and LHS$=$RHS.

@saadtaame See here for some related stuff.

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$Mathematics$ Mathematics