This is the Realm of Simply Beautiful Art

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Simply Beautiful Art

00:00

Hm, btw, my real name is... Jack [last name denied].

RSerrao

Jack Sparrow :P

Let me think

That is a nice way of putting the problem

I just can't seem to guess that function

The only things I thought of were the sum of the first n terms of an arithmetic series,

(n+1)(n/2)

but that doesn't seem like it

and the derivative of the taylor expansion of $1/(1+x)$ evaluated at $x = 1$

sorry, I meant $1/(1-x)$

but the derivative of it is $-1/(1-x)^2$

and at $x=1$ that gives $-\infty$

Simply Beautiful Art

00:46

@RSerrao so you need to "regularize" it in a different way.

1 hour later…

Simply Beautiful Art

02:15

room topic changed to Simply Beautiful Art's room: Room for totally random people to hang [recreational-mathematics]

7 hours later…

Secret

09:01

in Mathematics, 12 mins ago, by Secret

$$I=\int x^2 \sin x e^x dx$$
Using Feymann's integration trick (Special case of Lebniz rule) and $\sin x= \mathcal{I}(e^{ix})$
\begin{align}
I & =\left.\frac{d^2}{dt^2}\int -\sin (tx) e^x dx\right|_{t=1}\\
& = \left.\frac{d^2}{dt^2}\int \mathcal{I}(-e^{itx}) e^x dx\right|_{t=1}\\
& = \left.-\frac{d^2}{dt^2}\mathcal{I}\left(\int e^{(it+1)x} dx\right|_{t=1}\right)\\
& = \left.-\frac{d^2}{dt^2}\mathcal{I}\left(\frac{1}{it+1}e^{(it+1)x}\right|_{t=1}\right)\\
& = \left.-\mathcal{I}\left(\left(\frac{-2}{(it+1)^3}+\frac{-2i}{it+1}+(it+1)\right)e^{(it+1)x}\right|_{t=1}\right)\\

4 hours later…

Simply Beautiful Art

13:00

@Secret boy, that was a strange one huh?

If you want another problem, I would like you to rigorously evaluate the following divergent series

$$S=1-2+3-4+\dots$$

By rigorously, I mean whatever you try to do to this must be something you can do to a convergent series without changing it.

For example,

$$\sum_{n=1}^\infty a_n=\lim_{x\to1}\sum_{n=1}^\infty a_nx^n$$

Secret

There are many possibilities:
1. Either I made a careless mistake in one of the following (conversion to polar forms, the double differentiation etc.) (I have rechecked it many times and my brain get locked up trying to hunt for the mistake)
2. Or that for reasons I don't know, one cannot interchange the imaginary operator with the $\frac{d^2}{dt^2}$

Otherwise it should work and this solving pathway is the only one I knew that does not directly involve IBP

Simply Beautiful Art

13:17

@Secret it's good enough. Looking at what's supposed to be the solution, it's only human that you might've made a mistake

Ramanujan

@SimplyBeautifulArt good morning :-)

Simply Beautiful Art

@Ramanujan and goodnight for you?

Ramanujan

After few hours, today is it Sunday for you?

Simply Beautiful Art

13:35

Not yet. :-) and I'll be gone most of today sadly.

Secret

13:49

$$S=\sum_{k=1}^{\infty}(-1)^{k+1} k=\lim_{x\to 1}\sum_{k=1}^{\infty}(-1)^{k+1} \frac{k}{x^{k+1}}=\lim_{x\to 1}\sum_{k=1}^{\infty} \frac{S_k}{x^{k+1}}$$

where $S_k$ is the kth partial sum. The latter is a convergent series by p series test (except at $x=1$)

And then I am suck at series, thus I am currently discussing about it in main chat

Secret

14:03

As divergent series do not have well defined values (and I suspect by the rigor you mean the solution will not involve the analytic continuation summing method that is made popular by 1+2+3+...) after some discussion with the group, DHMO, Tobias Kindletoft and I have arrived at the limiting value that a power series with coefficients matching that of $S$ will be 1/2.

DHMO also found that S as a power series is actually $\frac{1}{(1+x)^2}$ which has a radius of convergence of 1 in the complex plane

(Disclaimer: Actually I don't really understand that analytic continuation method, other than there are so many guys from reddit to youtube (including some mathematicians) saying that is not very rigorous)

2 hours later…

DHMO

16:15

Let $\displaystyle I(a,b) = \int x^a e^{bx} \ \mathrm dx$.

Secret

Simpleart: Here's a generalisation of your integral problem. It was found that the symmetry of the integrand under IBP allows the integration to be expressed in terms of differentiation. Theoretically, the feymann trick + imaginary part that is used in the challenge also work for this general case. However I am too lazy to compute the nth derivative and converting between polar forms, thus it will be dealt with later chat.stackexchange.com/transcript/message/34926783#34926783

DHMO

Note that $\left({x^a e^{bx}}\right)' = a x^{a-1} e^{bx} + b x^a e^{bx}$.

Therefore, $x^a e^{bx} = a I(a-1,b) + b I(a,b)$.

Manipulation gives us the reduction formula: $I(a,b) = \dfrac 1 b x^a e^{bx} - \dfrac a b I(a-1,b)$.

Now, $\sin x = \dfrac {e^{ix} - e^{-ix}} 2$.

Therefore, $\displaystyle \int x^2\sin(x)e^x\ \mathrm dx = \dfrac 1 2 I(2,1+i) + \dfrac 1 2 I(2,1-i)$.

$\displaystyle = \frac 1 2 \left(\frac 1 2 x^a e^{(1+i)x} - \frac 1 2 I(1,1+i)\right) - \frac 1 2 \left(\frac 1 2 x^a e^{(1-i)x} - \frac 1 2 I(1,1-i)\right)$

(I spot a mistake on the line above the line above, but I cannot edit it.)

$\displaystyle = \frac 1 4 \left(x^a e^x (e^{ix}-e^{-ix}) - \left(I(1,1+i) - I(1,1-i)\right)\right)$

Now I spot even more mistakes

I shall continue this next time.

5 hours later…

Simply Beautiful Art

21:19

@DHMO Oh heyo!

I totally forgot to invite you man! :D

@Secret Think you made a mistake. If you plug $x=1$ in, you should get $1/2^2=1/4$, not one half.

@DHMO Well, now that you know about this place, welcome, and I usually post some random posts around and stuff :-) In case your interested.

2 hours later…

Simply Beautiful Art

23:21

@projectilemotion How you been mate?

Simply Beautiful Art

23:31

:P Welcome to my private chat room @Simple

Feel free to do anything math

Simple

I am doing my differential equation homework now

Simply Beautiful Art

I don't even have that class

Simple

and English isn't my first language

Simply Beautiful Art

Oh, ok. What language?

Simple

I speak Chinese

Simply Beautiful Art

23:33

Wo hui shuo yi diar zhong wen

Simple

(⊙０⊙)

Simply Beautiful Art

:-)

Akiva Weinberger

Are there any Chinese characters that are more than one syllable

Simply Beautiful Art

@AkivaWeinberger Uh...

I don't know...

Simple

yes

Simply Beautiful Art

23:38

@AkivaWeinberger Well, um, welcome to Simply Beautiful Art's room for random people to hang

Where I usually post interesting questions, but I think you'd figure all my secrets out too fast

@AkivaWeinberger An interesting question if you haven't already seen: evaluate the following integral without integration by parts:

$$\int x^2\sin(x)e^x\ dx$$

Akiva Weinberger

I saw it on the starboard in the main room, sorry

Simply Beautiful Art

@AkivaWeinberger Darn XD

'Twas a fun little puzzle

Akiva Weinberger

You evaluated it as the second derivative wrt $t$ of $\int\sin(xt)e^xdx$, which is the imaginary part of $\int e^{x+ixt}dx$ or $\int e^{x(1+it)}dx$, which is easily evaluated

Simply Beautiful Art

@AkivaWeinberger I personally did it with the $t$ in the $e^x$, but it doesn't really matter :-)

Akiva Weinberger

Oh. I see, that was Secret's solution I was looking at.

Simply Beautiful Art

23:44

Yes. I usually don't post my solutions

Akiva Weinberger

Personally, I would do it with linear algebra, actually.

Simply Beautiful Art

I haven't learned linear algebra. What's it about?

Akiva Weinberger

Here's a beautiful series on it. It's about linear maps from $\Bbb R^n$ to $\Bbb R^m$, essentially

Alternatively, it's about vector spaces

Simply Beautiful Art

Ah, ok

Akiva Weinberger

but I recommend you watch the series to get a better understanding.

Simply Beautiful Art

23:47

I will learn then

Where do I learn about vector spaces?

Akiva Weinberger

In any case, I would first write the thing as the imaginary part of $\int x^2e^{(1+i)x}dx$.

Simply Beautiful Art

Yes, ofc

Akiva Weinberger

Then, I would find the derivatives of $x^2e^{(1+i)x}$, $xe^{(1+i)x}$, and $e^{(1+i)x}$.

And I would try to find what linear combination of them has $x^2e^{(1+i)x}$ as a derivative.

Their derivatives are: $e^{(1+i)x}\left((1+i)x^2+2x\right)$, $e^{(1+i)x}\left((1+i)\vphantom{{}^2}x+1\right)$, and $e^{(1+i)x}(1+i)$ @SimplyBeautifulArt

So, clearly, I'd need $\frac1{1+i}$ times the first one, plus something times the second one plus something times the first one

(to get us the $e^{(1+i)x}x^2$ term)

And I'd proceed from there.

Simply Beautiful Art

Hm, sounds pretty interesting and useful

@AkivaWeinberger You don't answer questions do you?

At least, not much

Akiva Weinberger

Yeah, not very often

I like the chat more than the main site.

I know you're the record-holder.

Simply Beautiful Art

23:54

Mhm...

Yeah

I seriously want to place first in all sites, just for a day

but StackOverflow gets too many questions compared to MSE

-1

A: Do nested integrals exist?

By the Cauchy's repeated integral formula: $$\int_a^{x_1}\int_a^{x_2}\int_a^{x_3}\dots\int_a^{x_n}f(t)\ dt\ dx_n\dots\ dx_3\ dx_2\ dx_1=\frac1{(n-1)!}\int_a^{x_1}f(t)(x_1-t)^{n-1}\ dt$$ which I find analogous to the Cauchy integral formula from complex analysis for $n$th derivatives (instead of...