I'll be impressed when you've put John Conway's Game of Life in Mathematica Oh yeah that's pretty cool. he says as he nods and pretends to understand what an Euler-Maclaurin sum Formula is

We greatly appreciate the one for this room @robjohn

@AMDG I've actually thought of doing that. I wrote one in C, and in Java.

I'm sure everyone's seen this, right? youtu.be/xP5-iIeKXE8

@skullpatrol That's just a link to MathJax. The one for Mathematica is also just drawing on Mathematica's native LaTeX support.

hello

need help of understanding big O notation

$f=O\!\left(\frac1x\right)$ as $x\to0$ means that $|f(x)|\le C\frac1{|x|}$ as $x\to0$

same thing as $x\to\infty$

For example $\cot(x)=O\!\left(\frac1x\right)$ as $x\to0$

Of course, the $C$ should not change as $x\to0$

how did cot(x) be O(1/x)

I am confused on how e^x = 1+x+(x^2/2) + O(x^3)

so you taking anything after x^3?

$\cot(x)=\frac{\cot(x)}{\sin(x)}$ and $\cos(x)\le1$

@EM4 Okay that is looking at $x\to0$

@EM4 as $x\to0$ the sum of the terms past $x^3$ are smaller than $Cx^3$, so we don't need to worry about them

oh okay

today in class, complex analysis we did derivative of $z^n$

As $x\to0$, $x^4=O\!\left(x^3\right)$ so it wouldn't make sense to write $O\!\left(x^3\right)+O\!\left(x^4\right)$

@EM4 which luckily turns out to be $nz^{n-1}$

@EM4 and you saw that $(z+\delta)^n=z^n+n\delta z^{n-1}+O\!\left(\delta^2\right)$ as $\delta\to0$

assuming that $z$ is a distance away from $0$

so you $O\!\left(\delta^2\right)$ you taking anything past that is smaller, so we don't focus on them.

@EM4 yes. properties of geometric series allow us to say that as $\delta\to0$, the remainders of convergent series can be ignored.

so yes, we can ignore the terms past $\delta^2$

this is what we did.

so he only focus on $delta^2$ then

@EM4 That is essentially what I wrote above

yes, I didn't saw it until now.

He focused on $\delta$, the $\delta^2$ is the remainder term

The sum of the convergent series past the $\delta^2$ terms are bounded by $C\delta^2$ for some fixed $C$.

I need more learning on it then.

You know about radius of convergence for series?

$$\sum_{k=0}^\infty a_kz^k$$ converges for $|z|\lt R$

right, I need to refresh series from calc 3 again haha.

$$\sum_{k=n}^\infty\left|a_kz^k\right|=|z|^n\underbrace{\sum_{k=n}^\infty\left|a_kz^{k-n}\right|}_{\le C\text{ for some constant}}$$

If $|z|\lt R$, then we can find an $R_1$ so that $|z|\lt R_1\lt R$

$$\sum_{k=n}^\infty\left|a_kz^k\right|\le|z|^n\sum_{k=n}^\infty\left|a_kR_1^{k-n}\right|$$

So the constant may get bigger as we get closer to the boundary of $|z|\lt R$, but we are looking at $z\to0$ so we don't worry about that.

it makes sense right about now, if I have more questions I will come again

my only question is why from step 2 and to step 3.

big O changed.

it was $O(\delta^2)$ to $O(\delta)$

can one construct a manifold using decorations on the boundary assumed to be projections from the manifold onto the boundary?

akin to the holographic principle in cosmology

Right, so going back to basics... has the distance from a circle to the circle-like shape created by sine and cosine actually been studied, and do we have a set of functions that gives such a norm? desmos.com/calculator/41f8edgfft

Also, what is it called if it even has a name?

It kind of looks like a square trying to be a circle.