@robjohn using the same steps, we can make the generalization for $n$ variables.

what was @robjohn steps ? for any question ?

@what'sup the triple series I posted above. Do you know an easy way to prove the convergence?

Did he post an answer ? @Chris'ssis

@what'sup no. I finally managed to handle with that. I was just curious about other ways of proving the convergence.

where is @robjohn ?

brb

ok

@DanielFischer How do we get: $∣Γ(1+i)∣^2=Γ(1+i)Γ(1−i)$

@Alizter $\Gamma(t) \in \mathbb{R}$ for $t \in \mathbb{R}$, so $\Gamma(\overline{z}) = \overline{\Gamma(z)}$.

Choose $z = 1 + i$.

What does the bar indicate?

Complex conjugation.

Hmm i see but I do not get how Gamma being real proves your first statement

Identity theorem. $z \mapsto \Gamma(z)$ and $z \mapsto \overline{\Gamma(\overline{z})}$ are two meromorphic functions that coincide on $\mathbb{R}$.

brb

math.stackexchange.com/questions/496725/… I'm sad this closed, I wanted to write a joke answer about how it follows from the Yoneda lemma :(

How does it follow from the Yoneda lemma?

so, $\mathbb R$ is a poset, the hom-functors look like $\mathbb R(a,-) : \mathbb R \to 2$, which you can think of as indicator functions, and they're actually indicators for the sets of things $\le a$ (or $\ge a$, whatever)

Yoneda says that if $\mathbb R(a,-) \cong \mathbb R(b,-)$ then $a \cong b$ and in this context that means $a = b$

so if the set of things less than or equal to $a$ is equal to the set of things less than or equal to $b$, then $a = b$

i'm back

@BenMillwood: Nice!

the question is closed for very good reasons

certainly

my reasons for wanting it open are entirely silly

but I wanted to share them with someone anyway

@BenMillwood: It follows from the uniqueness of colimits too, haha.

:P true

modulo isomorphism

:-)

isomorphism is equality in posets!

where is @Chris'ssis ?

@Alizter Gamma is analytic and the coefficients must all be real, so when you take the conjugate of the argument, you get the conjugate of the function

since conjugation fixes the coefficients

Hi, I have sphere with radius R, The radius r at any height y is given by $$r = \sqrt{R^2 − (R − y)^2}$$

can anyone explain why this is true? or how it works?

@Link the radius of what at height $y$?

The circle perpendicular to the $y$-axis?

a cross section of the sphere, yes

The radius of that circle is $\sqrt{R^2-y^2}$

why???

The sphere is centered at the origin, right?

The base is at the origin

You need to say that in your question. People will assume that you mean the standard sphere centered at the origin. Then your formula is correct

Sorry, but can you still explain why?

How this equation works?

@Link You have a cone with its base being this circle and its apex the center of the sphere.

Huh?

Do you see it?

Hmm, I think I get it.

@Link This with the top part removed.

So the pointy part of the cone is at the center of the sphere

Anyway, you can measure the height of the cone and the distance along the side of the cone.

and its base is the circle perpeincduclar to y

Use the pythagorean theorem to get the radius of the base.

Oh, so its a pythogan therom?

oh, okay

that makes so much sense

thanks!

Glad I could help. :)

How do I say a statement is correct because it is a form of an identity?

@Link Use the Pythagorean Theorem

Do I say that it must be true because of so and so

how would I actually visualize this

is this like a cone or something?

@Link The right triangle with the hypotenuse of $R$ and the leg $y-R$ will have the other leg $\sqrt{R^2-(y-R)^2}$

rob john, i got that

This is seperate

That was what you asked me to explain

Yup, thanks for that, but how would I draw this second thing out?

Just a second, I have to go back and look at what that might be. I was away from chat making the drawing

Okay then, thank you

@Link what is this second thing you want to draw out?

The part that is confusing is the cross sections

@Link So it has an elliptical ridge along the $x$-$z$ axis

@robjohn, what do you mean?

I still don't get it, do I need an x-y-z space?

sorry gotta go now byeeeeeeeeeee

@Link The tops of the triangles will form a ridge which is like a semicircle scaled twice as high

@robjohn, sorry I still don't get it XD

@Link If the base is in the plane so that it can be $x^2+y^2=1$, I'd say you would probably need 3 dimensions

Yea, In the end I need the volume, but I don't get how the cross sections and stuff would work

@Link Ah, to get the volume you need to know the cross section area at each $y$

@robjohn, Okay, but I still can't visualize it, which is what the problem I'm having is

Since the base of the triangle is $2\sqrt{1-y^2}$ and its height is the same, its area would be $\frac12bh=2(1-y^2)$

Picture a circle on the floor

OKay

draw a lot of evenly spaced lines across it

Which way?

on each of those chords (the lines across the circle), put an isosceles triangle whose height is the same as its base

whoa, whoa, whoa

which way do the lines go?

It doesnt really matter, North-south

Okay then, they're going north south

So, when I put the triangle

it comes out towards me

right?

they are up, away from the floor

Okay then

Now what?

How do you indicate an operation of simplification of both sides of an equation without writing "dividing both sides by x" is there some form of notation?

so you have a weird looking tent if you cover it with a sheet

@Alizter, just strike out the x on both sides

@Alizter just say "divide both sides by 2" or something

Such as $$\begin{align}2x&=ax\\2&=a\end{align}$$However much less trivial.

just strike out the x's as they cancel?

@Link Nono that was an example

Then just write it out I guess

The equation that I am simplifying requires a simplification that is complex and subtle.

So @robjohn, I have these's triangles going across, getting higher and wider when moving to the center of the circle as they're base length?

@Alizter Don't disturb my circles...

@cyberskull Sorry... what?

@Link Yes. As you make the lines closer together, that is the shape that you are trying to measure

Okay, i get that then

@Link good

what else is bothering you about this?

so, now, if I have that original equation

then does that mean that $y = \sqrt{1-x^2}$ is the length of half the base?

and also half the height?

@Link Yes, then the cross section at any point at distance $y$ from the point at the far west of the circle would be $2\sqrt{1-(1-y)^2}$

so, if I have the area of a triangle, it's equal to $2(1-x^2)$

and if I integrate them from -1 to 1

I get the volume of the solid?

@Link yes

Okay, then, that makes sense

thanks!

you're welcome

Hi there! Don't know if it counts as advertising, but please consider (read) give an star to:

indiegogo.com/projects/…

happy birthday Léon Foucault

Why is it that it seems people who put bounties on their own questions seem not to actually give the bounties out? They still lose the full amount of the bounty, but the recipient only gets half.

It says that the length of the semi major axis

@Link volume = $\frac13bh$ where $b$ is the area of the base

Yes, I know that, but it gives me the following

Area of the base is $24\pi$

the semimajor axises length is given by 0.5*(12-y)

and for semiminor it says 0.333(12-y)

Why are these equations true?

They don't seem to be from the pythogran therom

The cross sections decrease linearly in both directions to the vertex

Yes

The ellispe is getting smaller

okay, so why is it 12-y?

each cross section is similar to the base, and shrunk by $\frac{12-y}{12}$

But, why is it 12-y? I think that creates the ellispes of infitesmimally small heights?

@Link as they get to the vertex of the cone at $y=12$

Huh? I mean if the cone is 12 high, then we have say y = 7

12 - 7 = 5

what is the 5?

it seems to be the height of the small cone at that cross section till the vertex?

Wait, I think I get the concept here

@Link everything about the base is scaled $\frac5{12}$ at height $7$

@robjohn how can i find greatest hits page ?

@what'sup Go to your profile and click on the "StackExchange" at the top of the page. choose "Hot Questions"

ok

@what'sup Do you mean network wide or just for this site?

just for this site

Choose hot, week, or month

OK thanks @cyberskull @robjohn

np

Hey, when you're doing the washer method about a line that is not an axis

how do you set up the equations?

equation R is further away from the rotational line

equation r is closer

but we subtract both from the distance of that rotation point and the axis right?

I answered a question, and the OP accepted my answer, then submitted his own answer based on my answer, unaccepted my answer and accepted his own :|

Sounds like cheating to me @Bitrex. Don't answer that OP again.

@Link, yes, but notice they don't cancel.

@TedShifrin Yeah. He didn't even upvote my answer!

@TedShifrin Hey Ted, what's the oldest student you've seen in one of your classes? I'm thinking about going back to school this year, if I'm able.