Mathematics

12:19 AM

So that question got answered, but there's the weird thing..... the answer is really simple bu goes against the hint on the question.

Alizter

12:36 AM

@PeterTamaroff **Warning**: The following uses of $i$ are noting to do with the imaginary unit.

Nestation operator notation
$\newcommand{\CapPsi}{\mathop{\vphantom{\sum}\mathchoice{\vcenter{\Huge\Psi}}{\vcenter{\Large \Psi}}{\Psi}{\Psi}}}$

$$\CapPsi^n_{i=0}f_i(x)=f_0(f_1(\dots f_n(x)\dots))$$
$$\CapPsi^n_{i=1}\sum^i_{k=0}x=\sum^1_{k_1=0}\sum^2_{k_2=0}\dots\sum^n_{k_n=0}x=x\cdot n!$$
$$\CapPsi^n_{i=1}\prod^i_{k=0}x_k=\prod^1_{k_1=0}\prod^2_{k_2=0}\dots\prod^n_{k_n=0}x_{k_n}=x^{n!}$$

---
More examples

Daniel Rust

@Alizter why would anyone think i was the sqrt of 1 here? :P

Alizter

@DanielRust At some point somebody will try and be a smartass so I'm clearing the road now ;)

Karl Kronenfeld

@DanielRust Well, it is going to take the value of one of the sqrts of $1$ for most $n$.

Daniel Rust

oh -1 ofc

but yes, no one familiar with summation notation would think it's anything but an index

Alizter

The notation was originally designed for the sigma but I realised it's nestation potential and developed it further.

People say it is useless

the same was said about prime numbers.

point proven.

Yay peter is here

Daniel Rust

12:48 AM

@Alizter that's far from 'point proven'. Maybe 'converse of point not proven'.

Alizter

Yeah but it was an opinion based point so meh

Peter Tamaroff

@Alizter How do you define $\Psi$ inductively?

Like $\sigma_{i=0}^0 a_i=a_0$, $\sigma_{i=0}^n a_i=a_n+\sigma_{i=0}^{n-1}a_i$.

Alizter

@PeterTamaroff Do you mean recursive definition?

Daniel Rust

@Alizter they are the same thing

Alizter

@DanielRust I wasn't sure sorry.

Peter Tamaroff

12:52 AM

@DanielRust Actually, no.

I should have said "recursively" =)

Daniel Rust

@PeterTamaroff from wikipedia: "In mathematical logic and computer science, a recursive definition (or inductive definition) is used to define an object in terms of itself"

Peter Tamaroff

@DanielRust Most people don't like "inductively".

Gustavo Bandeira

@DanielRust I've had AGA in my city.

Karl Kronenfeld

First, let $f_1,f_2,\dots$ be functions such that the compositions $f_i\circ f_{i+1}$ are always possible. Then $\Psi^n_{i=1}f_i=f_n^\ast\circ\Psi^{n-1}_{i=1}f_i$.

Alizter

@KarlKronenfeld nope

Karl Kronenfeld

12:55 AM

Do you know what I mean by $f_n^\ast$?

Alizter

$i = 0$ becomes $i =1$

and $n$ stays

and @KarlKronenfeld no i do not

and it is $f_0$ not $f_n$

you go it backwards :P

Karl Kronenfeld

If $f$ and $g$ are composable as $g\circ f$, then $f^\ast g$ is defined to equal $g\circ f$.

$f^\ast$ is a function assigning functions to functions. Weird, eh?

Alizter

ohh

Now can we treat sigma as a function?

Karl Kronenfeld

Yes, it is a function on sequences.

Hm, no.

That's not what we want.

Alizter

what I got was

$\newcommand{\CapPsi}{\mathop{\vphantom{\sum}\mathchoice{\vcenter{\Huge\Psi}}{\vcenter{\Large \Psi}}{\Psi}{\Psi}}}$
$$\CapPsi^n_{i=0}f_i(x)=f_0\left(\CapPsi^n_{i=1}f_i\right)$$

Karl Kronenfeld

1:01 AM

Yes, that is fine as well.

Alizter

@PeterTamaroff Hows that?

Peter Tamaroff

@Alizter Eh?

Alizter

Above

Karl Kronenfeld

You end up with a schema of $\Psi^n$ for various $n$, where each $\Psi^n$ is defined recursively on the starting index. It's a little more cumbersome than mine, but it works.

@Alizter As far as I can tell $\Psi_{i=0}^n\sum$ is an abuse of notation.

I am confusing myself, actually.

Alizter

I am not sure if it abuse or just not as well defined as for a function

Karl Kronenfeld

1:06 AM

Well, it is a function $\mathbb R^{\mathbb N}\to\mathbb R$.

when the upper index $n$ is finite.

Daniel Rust

it maps infinite sequences to real numbers?

Karl Kronenfeld

You're only summing the first few of the members of the sequence.

You can let it end in zeros if you want.

Daniel Rust

you're summing functions though, not sequences of real numbers

and you're outputting a function

Karl Kronenfeld

Are we talking about the same thing?

I am defining $\sum^n_{i=1}:\mathbb R^\mathbb N\to\mathbb R$.

Daniel Rust

yes, which tends to have terms like $(x^i)$ as an input

Alizter

1:08 AM

I never though for one second if n was unbound

Karl Kronenfeld

Yes.

Daniel Rust

which is properly seen as a function

Karl Kronenfeld

Sure.

Daniel Rust

put it also outputs a function

Karl Kronenfeld

What also outputs a function?

Alizter

1:09 AM

No it doesn't

Daniel Rust

not a real number

Alizter

You have seen my definition for it right?

psi$x^i=x^{n!}$

Karl Kronenfeld

Let $f:\mathbb N\to\mathbb R$ be given. Define $\sum^n_{i=1}(f)=f(n)+\sum^{n-1}_{i=1}(f)$ and $\sum^1_{i=1}(f)=f(1)$.

Daniel Rust

$\sum_{i=1}^n x=nx$ which is a function $\colon \mathbb{R}\rightarrow\mathbb{R}$

ok i see our confusion

Alizter

Karl is right it maps from naturual

Karl Kronenfeld

1:12 AM

That would be the composite $\mathbb R\to\mathbb R^\mathbb N\to\mathbb R$ where the first is the diagonal function.

Daniel Rust

i was considering $f\colon\mathbb{N}\times\mathbb{R}\rightarrow\mathbb{R}$

Alizter

Hmm i am now lost

Karl Kronenfeld

@DanielRust Ah, that is an example of what $\Psi$ operates on though.

And it indeed outputs functions.

Alizter

but $\Psi$ can take a function or a sequence. And can output a Real or a function

Karl Kronenfeld

How does it work for sequences?

As far as I can tell, it is the same as taking the value of the output function at a specific real number.

Alizter

1:16 AM

Here and uses

ignore the first LHS sigma it is a typo

Karl Kronenfeld

I stick to my last comment. I still have no idea how you define $\sum$ or $\prod$ as a function to be applied to $\Psi$

Alizter

let $\sum^n_{i=0}x=f_n(x)$

$i=0$ is a property $\forall f_n$

Daniel Rust

Consider the set of all sequences of functions from $\mathbb{R}$ to $\mathbb{R}$ which is $A=(\mathbb{R}^\mathbb{R})^{\mathbb{N}}$. $\Sigma$ takes as input, an element of $A$ and an integer (it's upper index - although we could generalise to arbitrary elements of the powerset of $\mathbb{N}$) and ouputs a function $\mathbb{R}\rightarrow\mathbb{R}$. So $\Sigma\colon A\times\mathbb{N}\rightarrow\mathbb{R}^{\mathbb{R}}$ is defined by $\Sigma((f_i(x))_{i\in\mathbb{N}},n)=\sum_{i=0}^n f(x,i)$.

Peter Tamaroff

Has anyone played Dark Souls here?

Alizter

$\newcommand{\CapPsi}{\mathop{\vphantom{\sum}\mathchoice{\vcenter{\Huge\Psi}}{\vcenter{\Large \Psi}}{\Psi}{\Psi}}}$
So if I $$\CapPsi^m_{n=1}\sum^n_{i=0}x=\CapPsi^m_{n=1}f_n(x)=f_1\left(\CapPsi^m_{n=2}f_n(x)\right)$$

but $\forall f_n$ they have the property $i=0$. This property is set at psi.

Karl Kronenfeld

1:27 AM

@DanielRust But you still don't get the compositions $\sum^n\circ \sum^{n+1}$.

Daniel Rust

@KarlKronenfeld all I was doing was showing that $\Sigma$ can be given as a function.

Alizter

@PeterTamaroff I go this after searching dark souls and it says my account is terminated.

I never played dark souls before

It looks clean but I do not play games

Daniel Rust

@KarlKronenfeld Using currying, we can consider $\Sigma$ to be a function $A\rightarrow A$.

Karl Kronenfeld

@DanielRust Sure, its input sequences can be members of any group.

Daniel Rust

so we can iterate $\Sigma$

Karl Kronenfeld

1:31 AM

@DanielRust Could you be a little more specific?

(I know what currying is though)

It is possible to define $\Sigma\circ\Sigma$, but the result is undesirable, I think, in this context.

Daniel Rust

There's so many sets flying about i'm confusing myself now

but i think the curried version of the $\Sigma$ i defined about gives the 'usual' definition of nested summations

Peter Tamaroff

@KarlKronenfeld @DanielRust LOL; look at what you got into. This is priceless.

Daniel Rust

that is $(\Sigma\circ\Sigma (f_j(x)))(m)=\sum_{i=0}^m\sum_{j=0}^i(f_j(x))$

Peter Tamaroff

This Dark Souls game looks prone to being terrifying. I have to grow some balls.

Karl Kronenfeld

@DanielRust Ah, that somehow helped me see why it really is nesting.

Now Alitzer wants something different, unfortunately.

Daniel Rust

1:42 AM

haha

@KarlKronenfeld aint it always the way? ;)

eXtremiity

2:52 AM

Hi, can I get assistance in solving the complex equation sin(z)=2 ?

Peter Tamaroff

@eXtremiity OK.

Note that means $e^{iz}-e^{-iz}=4i$.

eXtremiity

Yes, correct.

Peter Tamaroff

And in turn that means $$e^{2iz}-1=4ie^{iz}$$

eXtremiity

Next I create a quadratic in e^z

Peter Tamaroff

Yes, that is alright, I guess.

Careful with squareroots and such things here, though.

eXtremiity

2:54 AM

Wait, why 3iz?

Peter Tamaroff

@eXtremiity Sorry. =)

eXtremiity

Ok, so I get e^z = i(2+-\sqrt(3))

Peter Tamaroff

@eXtremiity Shouldn't you be getting $\exp iz =$ blah?

eXtremiity

YES !!

That's it !

Its e^iz not e^z.

My problem.

Now when we do log(e^iz) we get = x+i(y+2kpi) , right?

Alec Teal

3:19 AM

@eXtremiity can you use Latex please, it's MUCH nicer :)

eXtremiity

Ok, so this is where I'm stuck:
I have $iln(2\pm \sqrt{3})$. The answers have $\pm iln(2+\sqrt{3})$

Is there a relationship?

Alec Teal

$\frac{1}{2+\sqrt{3}} = \frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}$

$=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$

Ethan

$$\frac{1}{\pi}=\frac{1}{3}-8\sum_{n=1}^\infty e^{-2\pi n^2}n\coth(\pi n)-2\sum_{n=1}^\infty e^{-2\pi n^2}\text{csch}(\pi n)^2$$

Alec Teal

Remember $-\ln(x)=\ln(\frac{1}{x})$

Peter Tamaroff

@Ethan Dude, patent them bitches.

Ethan

3:26 AM

lol

eXtremiity

Yes, yes I recall. Excellent. So I'm doing these questions correctly :). Thanks A LOT @AlecTeal.

Ethan

youtube.com/watch?v=mEpfyBOdKxU .

Alec Teal

@PeterTamaroff you cannot patent mathematical things. (THANK GOD)

Ethan

$$(a+b+c)^5+(a-b-c)^5-(a+b-c)^5-(a-b+c)^5=80abc(a^2+b^2+c^2)$$

Alec Teal

Also in the EU, software patents are illegal, as are they now in Austrailia too.

@Ethan press up to edit your most recent message (only)

Ethan

3:34 AM

$$1+2\sum_{n=1}^\infty\frac{1}{e^{\pi n}-1}=\frac{\pi^{1/4}}{\Gamma(3/4)}+4\sum_{n=1}^\infty\frac{1}{e^{\pi n^2}(e^{\pi n}-1)}$$

skullpatrol

3:53 AM

(removed)

Alec Teal

4:51 AM

So who's here?

eXtremiity

5:35 AM

What's the PV of $\sqrt{cos(z)}$ ?

Or should I ask - how does one find it?

skullpatrol

6:27 AM

62

Q: History of Math.StackExchange

This thread is used to record significant events in the life of Math.StackExchange. (There are also anniversary posts written from personal perspectives of contributing users.) Where this idea comes from. Long ago, on a distant website, a user asked "What's the story behind MathOverflow?". Mari...

mauna

8:20 AM

I have a doubt on an answer to this question: math.stackexchange.com/questions/472061/…

I don't understand why the following is true

The $\epsilon$-goodness of $f$ on $[a,y]$ implies its $\epsilon$-goodness on any subinterval of $[a,y]$.

Please help me to explain why this must be true

eXtremiity

@mauna . Not sure if anyone here at the moment can.

Chris's sis

Greetings noble souls!

mauna

@eXtremiity it's alright, I will stick around until someone can

eXtremiity

@mauna . Good luck !

mauna

do you have any hints?

eXtremiity

8:27 AM

No, not at all. Yet to complete a Real Analysis course.

mauna

okay

user87637

9:40 AM

@mauna I have posted an answer. It's not hard, just apply the definitions.

what'sup

9:51 AM

guys this is very unfair : i posted perfect answers and another guys got the right answer

http://math.stackexchange.com/questions/480923/evaluate-int-x-cos-6x-mathrm-dx/480929#480929

http://math.stackexchange.com/questions/480733/nonlinear-ordinary-differential-equation-elsgolts/480780#480780

mauna

10:04 AM

@Jasper thanks. I was wondering what was wrong with my proof?

Daniel Rust

10:33 AM

@what'sup Sometimes people prefer a nudge to a full answer. It happens.

user87637

10:50 AM

@what'sup Voting on SE never made much sense.

eXtremiity

12:33 PM

How does one find where Log(iz) is analytic?

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