Calculus and analysis

I'm cool with whatever you fancy telling us ... continued fractions ... the non trivial zeros of the zeta function ... homotopy groups of spheres ... whatever Chief !

Simply Beautiful Art

lol

I was just going to go over boring old "how can you approximate $\sqrt2$"?

Donald Splutterwit

Pell's equation ... I could find that intresting at any time :P

Maybe start with roots of rationals & then go on ... to ?

Mithrandir24601

@DonaldSplutterwit cse.psu.edu/~sjh26/pell.pdf

^ Now that's an application! :P

Simply Beautiful Art

lol

Donald Splutterwit

There are some cool continued fractions for $e^x$

I think I saw a good on for the gamma function recently ... let me see

Incomplete gamma function

In mathematics, the upper incomplete gamma function and lower incomplete gamma function are types of special functions, which arise as solutions to various mathematical problems such as certain integrals. Their respective names stem from their integral definitions, which are defined similarly to the gamma function, another type of special function, but with different or "incomplete" integral limits. The gamma function is defined as an integral from zero to infinity. This contrasts with the lower incomplete gamma function, which is defined as an integral from zero to a variable upper limit. Similarly...

Yeah explain Gauss's continued fraction to us SBA

Simply Beautiful Art

7:06 PM

Lol

It just requires proving a functional equation and iterating over it

Donald Splutterwit

Oh well ... that explains that ... cool ... you have about 2 hours to find something else then ... :-)

Simply Beautiful Art

lol

Well I'm mostly interested in finding something I think heather will be able to understand

@DonaldSplutterwit @Mithrandir24601 How would you go about computing $\sqrt2$?

Donald Splutterwit

Whatever you think ... Me & @Mithrandir24601 will be cool with whatever ?

Simply Beautiful Art

I'm just interested in whether or not you two know any ways to compute things like this

Donald Splutterwit

I would type SQRT 2 into my OX-240 and read off 1.4142...

Simply Beautiful Art

7:15 PM

lol

No, I mean how would you compute it using a calculator that only has addition, multiplication, subtraction, and division

4 function calculator

Mithrandir24601

@SimplyBeautifulArt To be honest, I don't... Normally, I'd just use a calculator or the computer. If necessary, I'd probably just use a series expansion

@SimplyBeautifulArt There's an algorithm for this :)

Simply Beautiful Art

@Mithrandir24601 there's many algorithms for this :P

Donald Splutterwit

I would calculate the first few convergents of the continued fraction ... probably see 7/5 & say I can do that in my head ... 1.4 (to 2 sig fig)

& check on my OX-240 !!

Simply Beautiful Art

Continued fractions are generally very slow though

$$\sqrt2=1.4+\frac1{70}-\frac1{4900}\pm\frac{3}{343000}$$

Where $\pm$ denotes maximum error

I binomial expanded $\sqrt x$ at $a=1.96=1.4^2$

Donald Splutterwit

cool ...

Simply Beautiful Art

7:23 PM

Yeah I'm bored :-/

Donald Splutterwit

I was about to ask ... & then realised ... binomial

Simply Beautiful Art

Well

I've got not much better to do, so here's a challenge if you two want to try it @DonaldSplutterwit @Mithrandir24601

Numerically approximate this integral to 5 significant places

$$\int_0^\infty\frac{\sin(\pi x)}{\ln(x)}~{\rm d}x$$

Mithrandir24601

@SimplyBeautifulArt I've started thinking about the most efficient way to do this on a quantum chip...

Simply Beautiful Art

@Mithrandir24601 Newton's method is usually the preferred approach

Mithrandir24601

That's what I would do classically... But is there a more efficient quantum way?

Donald Splutterwit

7:27 PM

there is a singularity at $1$ ... and interval is quite large (infinte !)

Simply Beautiful Art

@Mithrandir24601 I don't even know how 'quantum chips' affect the problem

Donald Splutterwit

ooops ... there is not a singularity at 1

Simply Beautiful Art

@DonaldSplutterwit Nah, its a removable singularity

Don't worry about the magnitude of the interval

Mithrandir24601

@SimplyBeautifulArt For some problems, the best quantum algorithms are 'better' (quite what 'bett' is, is a bit of a question) than classical ones

Simply Beautiful Art

7:40 PM

@DonaldSplutterwit :P Having fun with that horribly slow converging integral?

I should have the integral evaluated out almost 10 places

Oh, not 10

But definitely 5

Mithrandir24601

@SimplyBeautifulArt My choices here really are: doing it in my head (definitely not even going to attempt that), writing a program or using mathematica...

Simply Beautiful Art

@Mithrandir24601 Bet you can't get a program to approximate that integral to 5 places

:D

Mithrandir24601

@SimplyBeautifulArt Oh no... You just had to say that... :P

Simply Beautiful Art

Okay, maybe not 5 places

Bet you can't do 10 places

Lmao

WA says it diverges

Mithrandir24601

... Although I've just noticed that I've already got a program that calculates square roots, so that's what I'd use to get $\sqrt 2$

Hmm... It actually uses 3 different ways of doing so :P

Simply Beautiful Art

7:50 PM

@Mithrandir24601 :P

Mithrandir24601

@SimplyBeautifulArt Numerically integrating it works though :)

(NIntegrate on Mathematica)

Simply Beautiful Art

@Mithrandir24601 Doubt you'll get anywhere near 5 places, let alone 10 places

Mithrandir24601

-3.21919 apparently

or in more digits: -3.219189999895627

Simply Beautiful Art

@Mithrandir24601 Wrong

Mithrandir24601

@SimplyBeautifulArt :o Bad Mathematica!

Simply Beautiful Art

7:56 PM

Half-decent approximation though

You did numerical integration?

Mithrandir24601

@SimplyBeautifulArt Yeah

Simply Beautiful Art

Dang, that's a lot of places accurate for a numerical integration

Mithrandir24601

@SimplyBeautifulArt What's it accurate to?

Simply Beautiful Art

Error is between $10^{-6}$ and $10^{-5}$

Believe I've got 12 places out right now

Give up yet?

Mithrandir24601

@SimplyBeautifulArt Well, I never got round to trying to write a program as I got side-tracked by Mathematica :P

Simply Beautiful Art

8:07 PM

:P

Mithrandir24601

I am trying various different methods in mathematica and they do have a tendency to give up :P

Ooh, how about -3.219189637131119?

Simply Beautiful Art

Meh, not really better than the last one

Mithrandir24601

Aww :/

Heh. Adaptive Quasi Monte Carlo gave me -79901.4...

Simply Beautiful Art

wtf?

Mithrandir24601

To be fair, it did estimate an error almost as big as the value :P

Simply Beautiful Art

8:14 PM

xD

Mithrandir24601

This is an interesting one: arxiv.org/pdf/1510.04452.pdf

Simply Beautiful Art

8:29 PM

Huh

I've got 13 places out on my integral

Simply Beautiful Art

8:49 PM

\begin{align}\int_0^\infty\frac{\sin(\pi x)}{\ln(x)}~{\rm d}x&\approx\int_0^{50}\frac{\sin(\pi x)}{\ln(x)}~{\rm d}x+\sum_{n=0}^{19}\frac1{2^{n+1}}\sum_{k=0}^n\binom nk(-1)^k\int_0^1\frac{\sin(\pi t)}{\ln(t+k+50)}~{\rm d}t\\&\approx\underline{-3.219190038646}743\end{align}

@Mithrandir24601 @DonaldSplutterwit that's my approximation

Donald Splutterwit

9:47 PM

Sorry Chaps... been busy expressing permutations in terms of $4$-cycles math.stackexchange.com/questions/2472431/…

3-2-1 .. I am back in the room ... right calculus @heather @SimplyBeautifulArt @Mithrandir24601 ... ready when you are ?

Simply Beautiful Art

idk where they are

Donald Splutterwit

They have probably followed my link ... & are trying to express permutations in terms of $4$-cycles ... they will be busy for a while ! ... lol ... @heather @Mithrandir24601

Simply Beautiful Art

lol

Well, how would you approximate the solution to $$x=-2^x$ @DonaldSplutterwit

Donald Splutterwit

That Integral was too bonkers for me ... seemed quite arbitrary to break at 19 & 50

Simply Beautiful Art

Lol

Yeah, the choice of 50 and 19 were somewhat arbitrary

Donald Splutterwit

9:57 PM

$x=-2^x$ ... can this be solved exactly using the Lambert function ? ... let me think

Simply Beautiful Art

...

Donald Splutterwit

desmos.com/calculator/5mp0risqyx ... cheat ? $x=0.641$

But how would I get the approximation using pen & paper ?

Mithrandir24601

@SimplyBeautifulArt I'm here, just making tea :)

Simply Beautiful Art

@DonaldSplutterwit -0.641

Donald Splutterwit

So we know -16/25 is a good approx ... but how would we magic that ?

Simply Beautiful Art

10:06 PM

@DonaldSplutterwit I'll allow you to use a 5 function calculator. +,-,x,÷,2^

Donald Splutterwit

yo .... @Mithrandir24601 Thank goodness ... SBA is asking hard questions ! ... help ?

Simply Beautiful Art

xD

Provide an algorithm that approximates the solution to $x=-2^x$ to arbitrary accuracy, and compute the solution to 10 places.

Mithrandir24601

@DonaldSplutterwit First thing - how does x scale with increasing x (yes, really) and how does $2^x$ scale with increasing x?

Simply Beautiful Art

I'll be back in about 20 min

Mithrandir24601

Also: @heather !!

(this has nothing to do with getting an algorithm, just how you'd guess the value)

Donald Splutterwit

10:12 PM

OK @Mithrandir24601 What do you mean by scale ? ... we have 20 mins, how can we take a good approx & algoritmically get better ones ?

Mithrandir24601

@DonaldSplutterwit so, you wanted to know how we 'magic' -16/25

Donald Splutterwit

Yes ... I cheated ... see the desmos link

but now we know -16/25 ... how do we get better ?

Mithrandir24601

@DonaldSplutterwit OK, assuming that, it's an algorithm, so the first thing you want is an iterative approach, right?

Donald Splutterwit

OK

Mithrandir24601

@DonaldSplutterwit So, I don't know about this specific equation, but the first thing I'd generally do is known as the Newton-Raphson method

Have you ever heard of it?

Donald Splutterwit

10:21 PM

Just tried iterating $-2^{x_n} $ ... glacially slow convergence ... SBA would be sad ? :-(

Mithrandir24601

Only 21 minutes late, @heather !

heather

i know, i'm so sorry.

Donald Splutterwit

... the NR iteration method ... yes I used to teach it 15 years ago ...

heather

my mom dragged me along on errands.

Mithrandir24601

@heather it's fine :P I'm just trying out one of many methods that I only recently learned to try to get people to do things like appear on time :P

@DonaldSplutterwit Well... That makes things slightly easier in terms of explanations... :P Does it work?

Donald Splutterwit

10:24 PM

@heather cool ... you are here ... SBA will be back shortly ... we must aprrox $x=-2^x$

Mithrandir24601

(In this specific case)

Donald Splutterwit

We must have an approx to 87 sig fig before he returns !

heather

hoo boy

Mithrandir24601

@heather Have you done any of these types of things before?

heather

nope. i have an idea (probably wrong) for how to approach it, though.

Mithrandir24601

10:27 PM

@heather What's your idea?

Donald Splutterwit

@Mithrandir24601 can you remind us how the Newton-Raphson method works ?

heather

if you take the derivative of both sides, you get $1 = -2^x\log 2$, or $\frac{1}{\log 2} = -2^x$ or, let me think

$-\frac{1}{\log 2} = 2^x$

or

Mithrandir24601

@heather This looks right, but where would you go from there?

heather

$\log_2( -\frac{1}{\log 2}) = x$

i think

Simply Beautiful Art

@DonaldSplutterwit yeah I'd be sad

heather

10:33 PM

which google says is $1.73202085 + 4.53236014 i$?

Mithrandir24601

@heather Which is unfortunately wrong... Why?

Simply Beautiful Art

@DonaldSplutterwit Yeah I'd be sad if that were all you could think of

heather

I'm not sure why it has $i$ in it.

but i don't have much experience with logarithms, so i don't know.

Mithrandir24601

@heather might be because of the - sign

Simply Beautiful Art

@heather Try log_2(-1/ln(2))

heather

10:35 PM

it says

$0.528766373 + 4.53236014 i$

Donald Splutterwit

@SimplyBeautifulArt we have nearly nailed the Newton-Rhapson method ... we will have an approx to 87 sig fig shortly ?

Simply Beautiful Art

Oh right, we're stupid lol

@heather We're not solving for $x$, we're solving for $dy/dx$

@DonaldSplutterwit lol, okay

I asked for 10 figs... but 87 is good too

heather

@SimplyBeautifulArt oh, so then the derivative of that

Simply Beautiful Art

Gotta go eat

heather

erm, no, nvm

Simply Beautiful Art

10:37 PM

See if you can tell me how fast it converges if you iterate -2^x

heather

but i did take the derivative, just to solve for $x$

Simply Beautiful Art

How fast does it converge if you use Newton's?

Mithrandir24601

@heather So what's $\log\left(2^x\right)$?

Donald Splutterwit

$x_{n+1}=x_n -\frac{f(x_n)}{f'(x_n)}$ where $f(x)=x+2^x$

Simply Beautiful Art

Something like $x-x_n=\mathcal O(10^{-n})$

Mithrandir24601

10:37 PM

@DonaldSplutterwit yeah, this

heather

@Mithrandir24601 google's giving me a graph.

Mithrandir24601

@heather Yeah, so what's happened (I believe) is that you've found the stationary point of the equation

So, I'll explain Newton-Raphson

@DonaldSplutterwit OK - take a function $f(x)$ and Taylor expand about $x_0$ to get $f(a) = f\left(x_0\right) + f'\left(x_0\right)\left(a-x_0\right) + O\left(x_0^2\right)$

Donald Splutterwit

taylor expand to the first order ...

Mithrandir24601

@DonaldSplutterwit yep. Then, at the point where $f(a) = 0$ i.e. $a$ is the number we're looking for, $f\left(x_0\right) + f'\left(x_0\right)\left(a-x_0\right) \approx 0$

which gives $$a \approx x_0 - \frac{f\left(x_0\right)}{f'\left(x_0\right)}$$

Donald Splutterwit

$x_{n+1}=x_n-\lfrac{x+2^x}{1+ \ln(2) 2^x}$ ... rapidly gives ?

Simply Beautiful Art

10:51 PM

O.o

well, how are we doing?

Mithrandir24601

Only now, $a$ is our next iteration, i.e. $a=x_1$ and we repeat the process to get $$x_{n+1} = x_n - \frac{f\left(x_n\right)}{f'\left(x_n\right)}$$

@SimplyBeautifulArt Slowly, but we're hopefully getting there...

Is everyone with me so far?

Simply Beautiful Art

Stupid gifs

It's a root finding algorithm

You take a point

draw the tangent against it

Find where the tangent crosses the x-axis

That's your new point, rinse and repeat

:D

Donald Splutterwit

-0.641185744 ... to 9/10 sig fig ... best my OX-240 will do ?

Simply Beautiful Art

Here's the solution I have

-0.641185744505

Donald Splutterwit

By the way 641 is special ... it is 2^4+5^4 and is the first time a fermat is composite

Simply Beautiful Art

10:58 PM

:|

:I

:P

Well

Are we still having confusions with Newton's method? @DonaldSplutterwit @heather

heather

i'm still a bit confused, yeah.

let me read through it again.

Donald Splutterwit

Fermat conjectured that 2^{2^n}+1 was prime for all n ... it is wrong ... for n=5 ... 641 is a factor

Simply Beautiful Art

@SimplyBeautifulArt what language is that key in?

Donald Splutterwit

Sorry guys ... never meant to kill the thread with a number theory observation ...

Simply Beautiful Art

@SimplyBeautifulArt No idea man

Mithrandir24601

11:06 PM

@DonaldSplutterwit Nah, I was just dosing myself with paracetamol and antibiotics :/

Which my body naturally tried to cough up on me :(

Donald Splutterwit

I am worried about you guys ? ... SBA is talking to himself & you are self medicaticting ?

Simply Beautiful Art

Don't worry, it's fine @DonaldSplutterwit

Mithrandir24601

@DonaldSplutterwit All on the Doctor's orders, don't worry :P

Simply Beautiful Art

:-/

*::Twiddles his thumbs::*

Mithrandir24601

breathes in

...

breathes out

coughs

Donald Splutterwit

11:10 PM

Let's do some continued fractions ? ... that will cheer everyone up ?

Simply Beautiful Art

Sure...

I actually wanted to wrap up the iteration stuff for today

@Mithrandir24601 @DonaldSplutterwit You know how to find the rate at which it converges to our -0.641185744505?

Let's suppose we want to do: x_0 = 0 and x_n+1 = Newton(x_n)

What's the asymptotic behavior of (x - x_n) as n → ∞

Donald Splutterwit

@SimplyBeautifulArt ... good place to end ... we have all gone slightly bonkers ?

Simply Beautiful Art

:P

Mithrandir24601

@DonaldSplutterwit No, I'm actually ill :(

Simply Beautiful Art

@Mithrandir24601 Get better then

Mithrandir24601

11:14 PM

Although I probably went bonkers years ago :P

@SimplyBeautifulArt I'm trying! :P

Simply Beautiful Art

@Mithrandir24601 :D

@Mithrandir24601 trying isn't good enough!

Donald Splutterwit

Oh I am sorry to hear that ... GWS ?

Simply Beautiful Art

Well

Did you guys know that Newton's method usually converges quadratically?

And do you guys know what this means!?

Mithrandir24601

@DonaldSplutterwit Thanks. Hopefully will. If it's not better by Monday, I don't know what the GP will do with me :/

@SimplyBeautifulArt Rings a bell, yeah

Donald Splutterwit

I know it converges quickly ... would need to think about quantifying the speed ?

Simply Beautiful Art

11:17 PM

Well, all it means is that:
(x - x_n+1) = a * (x - x_n)^2 + O((x - x_n)^3)

For some constant a

And if (x - x_n) ≈ 10^-k, then (x - x_n+1) ≈ a * 10^-2k

That is, the amount of accurate digits approximately doubles per iteration

@DonaldSplutterwit D: Cuz that's what an analytic person should do? :P

Transcript for

$Mathematics$ Calculus and analysis