the no-normie zone - 2017-11-29

Anonymous

4:14 PM

@BalarkaSen There's a question asking "By looking at the surface of $z=x^2-y^2$, determine whether the equation can be solved for $y$ as a function of $x$ in a neighbourhood of the indicated point $(0,0)$"

Anonymous

So, I write $y$ as

Anonymous

$\pm \sqrt{x^2-z}$

Anonymous

I mean can't it always be solved for $y$?

Balarka Sen

@Blue No, you have to be careful. You need to write $y = \phi(x)$

$\pm \sqrt{x}$ is not a function of $x$

It takes two outputs for one input

Anonymous

That is true

Anonymous

4:21 PM

Yes

Anonymous

But I'm not sure what condition to check for

Anonymous

Looking at the graph it seems it can always be written as either $\sqrt{x}$ or $-\sqrt{x}$

Anonymous

Uh, wait

Anonymous

Even $z$ is present

Anonymous

We need only $\phi(x)$

Balarka Sen

4:24 PM

It's not entirely clear to me what it means to solve $y$ as a function of $x$, no.

If it was "solve $y$ as a function of $x$ and $z$" that would make sense

Anonymous

If $z=0$ is the best linear approximation at $(0,0)$

Anonymous

They could be referring to that

Anonymous

So we wouldn't have to worry about $z$ locally

Balarka Sen

I have no idea. I can't interpret vaguely stated questions

Anonymous

Okay, even I'm confused. You can have a look at the question paper:

Anonymous

4:28 PM

Anonymous

12 (ii) @BalarkaSen

Balarka Sen

You didn't explain what the equation was. It clearly writes $f(x, y) = 0$

$f(x, y) = x^2 - y^2$

Anonymous

Oh, oops. I'm sorry

Anonymous

I don't know how I missed that

Balarka Sen

Yeah, so that's much better

Anonymous

4:31 PM

Anyhow, we are looking for $x^2-y^2=0$

Anonymous

And trying to solve for $y$

Anonymous

So it's a pair of lines

Balarka Sen

Mhm

Anonymous

So for one $x$ we get two values of $y$

Anonymous

Doesn't look like a function

Balarka Sen

4:34 PM

Mhm

Quite correct

Anonymous

Only at $(0,0)$

Anonymous

It gives an unique value. Anyhow, I guess the answer will be just, NO?

Anonymous

(To the original question)

Balarka Sen

Correct

5 marks for it though...

Anonymous

@BalarkaSen Yeah, this is a bit much :P

Anonymous

4:35 PM

I saw 5 marks and thought it must be complicated...lol

Anonymous

I was thinking of weird things like differentiability...till I noticed $f(x,y)=0$

Anonymous

XD

Balarka Sen

Maybe you have to ramble a lot. Like, if $f'(a, b) \neq 0$ then $f(x, y) = 0$ can be locally parametrized as $y = \phi(x)$ by implicit function theorem

But in this case $f'(0, 0) = 0$ etc so that doesn't work etc

Ah, I see. So you're supposed to compute $f'(0, 0) = 0$ from the graph $z = f(x, y)$

Because the tangent space at $(x, y) = (0, 0)$ is completely horizontal

Anonymous

@BalarkaSen Huh, I don't know that theorem...

Balarka Sen

Oh it's a beautiful theorem. Learn it!

Shall I explain?

Anonymous

4:38 PM

@BalarkaSen Okay, if you're free!

Balarka Sen

Or are you too busy with preparation for exams

Anonymous

Nope, free today

Balarka Sen

Oh nice

OK

So let me first explain the inverse function theorem

Anonymous

okay

Balarka Sen

Suppose $f : \Bbb R^n \to \Bbb R^n$ is $C^1$ (continuously differentiable) and assume the Jacobian $Df(x_0)$ is an invertible $n \times n$ matrix. Then there is a neighborhood $U$ of $x_0$ and $V$ of $y_0 = f(x_0)$ such that there is an inverse $g : V \to U$ of $f$ such that $g$ is also $C^1$.

Anonymous

4:45 PM

@BalarkaSen So far so good :)

Balarka Sen

That's pretty much the statement of the inverse function theorem

So

The reason behind the Jacobian condition is this

Consider $f : \Bbb R \to \Bbb R$, $f(x) = x^2$

$f'(0) = 0$, so it's NOT invertible as a $1 \times 1$ matrix

That's why $f$ has no $C^1$ inverse near $0$

Makes sense?

Anonymous

Oh, right

Anonymous

Got so far

Balarka Sen

Notice that $\pm \sqrt{x}$ is not a valid inverse: it's not even a function. If you take $+\sqrt{x}$, that's still not defined on a neighborhood of $0$

Because any neighborhood of $0$ is of the form $(-\epsilon, \epsilon)$, and $+\sqrt{x}$ is only defined on the positive half

Anonymous

Right. But we were dealing with $\sqrt{x^2}$, isn't it?

Anonymous

4:52 PM

and $-\sqrt{x^2}$

Anonymous

That's $|x|$

Anonymous

and $-|x|$

Anonymous

So for a particular branch, it is invertible

Balarka Sen

Huh? What is your candidate for $g$?

$g(x) = ???$

Anonymous

$g(x)=x$, in case $f(x)=x$ ?

Anonymous

4:56 PM

(Neglecting the other branch)

Balarka Sen

? $f(x) = x^2$. $g$ is the inverse of $f$

I have no idea what you're talking about

We started with $f(x) = x^2$

Anonymous

We're talking about different functions here. I was talking about the original question $y^2-x^2=0$

Anonymous

Sorry for the confusion

Balarka Sen

Where did that come from

I was taking an example to help you understand the inverse function theorem

Anonymous

Understood your example, yes

Anonymous

5:00 PM

23 mins ago, by Balarka Sen

Maybe you have to ramble a lot. Like, if $f'(a, b) \neq 0$ then $f(x, y) = 0$ can be locally parametrized as $y = \phi(x)$ by implicit function theorem

Anonymous

After this I was having trouble understanding this

Anonymous

23 mins ago, by Balarka Sen

Ah, I see. So you're supposed to compute $f'(0, 0) = 0$ from the graph $z = f(x, y)$

Balarka Sen

I didn't even get to explaining the implicit function theorem

Slow down man

Anonymous

Alright, alright :P

Anonymous

Go on

Balarka Sen

5:01 PM

So do you understand why $f(x) = x^2$ has no ($C^1$) inverse near any neighborhood of $0$? Explain

Anonymous

@BalarkaSen Yes, because $f'(x)=0$

Anonymous

So jacobian not invertible matrix

Balarka Sen

I'd give that 1/2 out of 5 if I were grading your answer

Anonymous

21 mins ago, by Balarka Sen

Suppose $f : \Bbb R^n \to \Bbb R^n$ is $C^1$ (continuously differentiable) and assume the Jacobian $Df(x_0)$ is an invertible $n \times n$ matrix. Then there is a neighborhood $U$ of $x_0$ and $V$ of $y_0 = f(x_0)$ such that there is an inverse $g : V \to U$ of $f$ such that $g$ is also $C^1$.

Anonymous

So, basically here $f:\Bbb R\to \Bbb R$

Anonymous

5:04 PM

It is $C^{1}$

Anonymous

And the Jacobian matrix is not invertible as $f'(0)=0$

Anonymous

Also,

Anonymous

17 mins ago, by Balarka Sen

Notice that $\pm \sqrt{x}$ is not a valid inverse: it's not even a function. If you take $+\sqrt{x}$, that's still not defined on a neighborhood of $0$

Balarka Sen

No, you mean $f'(0) = 0$.

Anonymous

I meant at 0, but yes

Anonymous

5:08 PM

So an inverse $g$, doesn't exist in the neighbourhood of $0$

Balarka Sen

So you understand why naively declaring $g(x) = \sqrt{x}$ is not valid?

Anonymous

Yep, got it

Balarka Sen

Well, why? :) (Don't quote me, explain it yourself. This is a rather important point)

Anonymous

Firstly, it's not defined for negative $x$

Balarka Sen

That's it

Anonymous

5:11 PM

Yes!

Balarka Sen

Any neighborhood of $0$ will contain negatives

@Blue But, is $f(x) = x^2$ invertible at $x = 1$?

Anonymous

There you get two functions for the inverse. No?

Anonymous

$\sqrt{x}, -\sqrt{x}$

Balarka Sen

Sure. But that doesn't answer my question

Anonymous

$f'(1)=2$

Anonymous

5:15 PM

And it is $C^{1}$

Balarka Sen

So, is it a yes or is it a no? :)

Anonymous

Okay, I'm confused

Anonymous

According to the theorem it should be invertible

Balarka Sen

Right.

Anonymous

But, how?

Anonymous

5:18 PM

We get two inverse functions

Balarka Sen

Let's take baby steps. What does it mean to say "$f(x) = x^2$ is invertible at $x = 1$"?

Anonymous

First of all the condition for inverse to exist is that there should be a one one and onto mapping

Balarka Sen

Ok, but all it means is that an inverse exists. Now, what does the condition "at $x = 1$" mean?

How do you incorporate that information?

Anonymous

@BalarkaSen There must be a $g(x)$ which maps the $f(1)$ to $1$ ?

Balarka Sen

Close. Let's write that down rigorously

It means: There is an open interval/neighborhood $I_1$ around $1$ and an open interval/neighborhood $I_2$ around $f(1)$ such that there exists an inverse $g : I_2 \to I_1$ to $f$. That is, $f(g(p)) = p$ for all $p \in I_2$ and $g(f(q)) = q$ for all $q \in I_1$.

That is, $f : I_1 \to I_2$ and $g : I_2 \to I_1$ are inverse to each other.

In this case, $f(1) = 1$ of course.

Anonymous

5:24 PM

Right

Anonymous

So was the previous definition having some loophole?

Balarka Sen

@Blue So, what is your candidate for $g$?

$g(x) = ???$

Anonymous

@BalarkaSen $\sqrt{x}$ ?

Balarka Sen

Excellent, yes.

$g(x) = \sqrt{x}$ does the trick when you take $I_1$ and $I_2$ to be small intervals around $1$.

Anonymous

Oh, presence of $-\sqrt{x}$ doesn't matter, then

Anonymous

5:26 PM

I see

Balarka Sen

Yep. $g(x) = -\sqrt{x}$ maps $g(f(1)) = g(1) = -1$.

Not an inverse of $f$ in $I_1$.

You need to have $g(f(1)) = 1$

Anonymous

Alright. Got it!

Balarka Sen

@Blue But! What if I ask if $f(x) = x^2$ is invertible at $x = -1$?

What would be the candidate for inverse then?

Anonymous

Then we can choose $g(x)=-\sqrt{|x|}$ ?

Balarka Sen

@Blue Very good. But you don't need $|x|$

Ah, no, no, $g(x) = -\sqrt{x}$ suffices.

$g$ does not take negative values.

Recall what the domain of $g$ is.

Anonymous

5:31 PM

Oh, silly mistake

Anonymous

Yes

Anonymous

Domain is positive

Balarka Sen

It's okay, this takes some time to get used to. But yes, exactly, domain of $g$ is a subset of the codomain of $f$, where everything is positive.

In particular, domain of $g$ is a small interval around $f(-1) = (-1)^2 = 1$

Anonymous

Right!

Balarka Sen

So that's quite interesting, isn't it? If you take local inverse of $f(x) = x^2$ for some positive $x$, that's one function $g(x) = \sqrt{x}$, but if you take local inverse of the same $f$ for negative $x$, that's another function $g(x) = -\sqrt{x}$

Anonymous

5:34 PM

@BalarkaSen mmhmm!

Balarka Sen

That's the reason you don't have inverse at $x = 0$ actually.

One one side of $x = 0$ you have one inverse, and on another side you have another

You can't patch them up

Anonymous

What does "patch up", mean?

Balarka Sen

Actually, uh, ignore my last statements

Those are meaningless

Sorry

Anonymous

no prob

Balarka Sen

@Blue It doesn't mean anything rigorous, just something at the intuitive level

This phenomenon actually becomes more interesting in complex analysis

It's called "monodromy phenomenon"

But never mind that

Anonymous

5:39 PM

Do you just mean the inverse functions are different on either side of 0, so we can't patch them up?

Balarka Sen

Yeah

It's a useful intuition to keep in mind

Anonymous

gotcha!

Balarka Sen

So the implicit function theorem, now?

Anonymous

yep!

Balarka Sen

So say $f : \Bbb R^n \to \Bbb R^m$ is a $C^1$ function where $n > m$

Anonymous

5:44 PM

okay

Balarka Sen

Write each vector in $\Bbb R^n$ as $(x, y) \in \Bbb R^{n-m} \times \Bbb R^m$, OK?

Anonymous

Okay so far

Balarka Sen

Suppose that the $m \times m$ submatrix of $Df(x_0, y_0)$ of partials $\partial f_i/\partial y_j$ is invertible. (This is a technical condition)

For some $(x_0, y_0) \in \Bbb R^n$ such that $f(x_0, y_0) = \mathbf{0}$

Or, rather, better to say that $(x_0, y_0)$ is a point in the level set $S: f(x, y) = \mathbf{0}$ inside $\Bbb R^n$

Is this OK?

Anonymous

@BalarkaSen A bit of problem understanding the definition of the submatrix

Balarka Sen

$Df(x_0, y_0)$ consists of partials $\partial f_i/\partial x_j(x_0, y_0)$ and $\partial f_i/\partial y_j(x_0, y_0)$, right?

By definition of Jacobian

Anonymous

5:53 PM

Oh, right. Got it

Anonymous

$m$ components of $y$

Anonymous

$y_j$

Anonymous

j=1,2,3,...,m

Balarka Sen

Right. Those form a smaller matrix inside $Df(x_0, y_0)$

Assume that is invertible

Anonymous

Alright

Balarka Sen

5:55 PM

OK, now so you have the level set $S : f(x, y) = \mathbf{0}$ in $\Bbb R^n = \Bbb R^{n-m} \times \Bbb R^m$ and a point $(x_0, y_0)$ on $S$

Anonymous

Yes

Anonymous

Ok

Balarka Sen

The theorem says the following: there exists neighborhoods $U$ of $x_0$ in $\Bbb R^{n-m}$ and $V$ of $y_0$ in $\Bbb R^m$ and a function $\phi : U \to V$ such for all $f(x, y) = 0$ with $(x, y) \in U \times V$, $y = \phi(x)$

That is, $S$ can be parameterized by $y = \phi(x)$ over $U$

In other words, you can locally write the level set $S$ as a graph near $(x_0, y_0)$

Anonymous

I see

Anonymous

Okay, got it

Balarka Sen

6:00 PM

Here's a nice example. Consider $f(x, y) = x^2 + y^2 - 1$ from $\Bbb R^2 \to \Bbb R$

Anonymous

$m=1$ and $n=2$

Anonymous

Here

Balarka Sen

Correct

Anonymous

I need $\frac{\partial f}{\partial x}$

Anonymous

Which is $2x$

Anonymous

6:02 PM

Our level set is $x^2+y^2-1=0$

Balarka Sen

Wait, you need $\partial f/\partial y$

Anonymous

@BalarkaSen Doesn't matter. It's symmetric in x and y

Anonymous

And it's a matter of choice which one you take as $m$

Anonymous

and which one as $n-m$

Balarka Sen

Haha. Fair enough

Then you'd be looking at $x = \phi(y)$

Anonymous

6:04 PM

So, let's take a point $(1,0)$ on the level set

Anonymous

It's $2$ over there

Anonymous

The partial

Anonymous

So, it's invertible around $1$ ?

Balarka Sen

Yup

Anonymous

Great

Anonymous

6:08 PM

So that was it?

Balarka Sen

Well, not "invertible"

Anonymous

Ah, parameterizable

Balarka Sen

You can write $x^2 + y^2 = 1$ (given by $f(x, y) = 0$) implicitly near $(1, 0)$ as $x = \phi(y)$, is the point. Indeed, $x = \sqrt{1 - y^2}$ works

@Blue Right

There's a nice picture of this in wikipedia

Hm, not really

But here it is

Read the description

Anonymous

Ah. At B half of it lies in the negative quarter

Anonymous

And other half in the positive

Balarka Sen

6:15 PM

Yep!

Anonymous

Makes sense

Balarka Sen

It's the same thing as inverse of $f(x) = x^2$ at $x = 0$

Anonymous

Right. Got it!

Anonymous

BTW I was having some trouble with Lagrange multipliers

Balarka Sen

Ah yes

Shoot

Anonymous

6:21 PM

I have a plane $x+y+z=2a$ and I have to find shortest distance to $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$

Anonymous

So, should I consider a general point $(x,y,z)$ on ellipse and write a distance function

Anonymous

$D=\frac{|x+y+z-2a|}{\sqrt{3}}$

Anonymous

And then $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ is the constraint

Balarka Sen

Yep

Anonymous

Then I can write $D_x=K(2x/a^2)$

Anonymous

6:23 PM

$D_y=K(2y/b^2)$

Balarka Sen

Use $D^2$ instead, because that's differentiable.

Anonymous

@BalarkaSen Okay, I was thinking of that

Anonymous

Lemme try:

Anonymous

$D^2_x=2(x+y+z-2a)(1)$, right?

Anonymous

That is equal to $\lambda(2x/a^2)$, say

Anonymous

6:24 PM

Similarly, $D^2_y=2(x+y+z-2a)(1)=\lambda (2y/b^2)$

Balarka Sen

Right

Anonymous

And $D^2_z=2(x+y+z-2a)(1)=\lambda (2z/c^2)$

Anonymous

Now I need to solve for $x,y,z$ I guess

Balarka Sen

Right

Anonymous

Maybe adding them all up can help

Anonymous

6:26 PM

Nah, doesn't help much

Anonymous

Solving these linear systems are a pain

Balarka Sen

How did you add up?

You have $1/a^2, 1/b^2, 1/c^2$ on the RHS

Anonymous

Duh

Anonymous

$6(x+y+z-2a)=2\lambda(x/a^2+y/b^2+z/c^2)$

Anonymous

Ok?

Balarka Sen

6:28 PM

yeah...hm

Maybe $x/a^2 = y/b^2 = z/c^2$ is useful

yeah, write $y$ and $z$ in terms of $x$ from that

and the plug all of them in $x^2/a^2 + y^2/b^2 + z^2/c^2 = 1$

that's it

Anonymous

Oh, awesome

Anonymous

Done then!

Balarka Sen

Yep

Anonymous

I'm doing a few more of these Lagrange exercises...will ask if stuck :P

Anonymous

I need to get hang of the tricks

Balarka Sen

6:30 PM

For sure

Anonymous

"Determine the maximum value of $OP$, $O$ being the origin of coordinates where $P$ describes the curve: $x^2+y^2+2z^2=5,x+2y+z=5$". I'm trying to maximize $D^2$ i.e. $x^2+y^2+z^2$. The equations are: 1) $2x=a(1)+b(2x)$ 2) $2y=a(2)+b(2y)$ 3) $2z=a(1)+b(4z)$

Anonymous

Any ideas to proceed after this?

Anonymous

@BalarkaSen

Balarka Sen

You want to solve (1), (2) and (3)?

Wait. How did you get $a$ and $b$?

Anonymous

Adding gives $2(x+y+z)=a(1+2+1)+b(2x+2y+4z)$

Anonymous

6:42 PM

@BalarkaSen Two constraints

Anonymous

You need two scaling factors in case two constraints are present

Balarka Sen

Oh, I see, I missed that. Yes, you need $\nabla f = \vec{\lambda} \cdot \nabla g$

Anonymous

@BalarkaSen Yep

Balarka Sen

OK

Anonymous

But I can't figure out a way to solve the equations

Anonymous

6:46 PM

i.e. extract x,y,z

Anonymous

Or $x^2,y^2,z^2$

Balarka Sen

So $2x = a/(1 - b)$

Anonymous

Yes!

Anonymous

$2a/(1-b)=2y$

Anonymous

and $2z=a/(1-2b)$

Anonymous

6:52 PM

Ah. Got an idea. Multiply 1) by x, 2) by y and 3) by z

Anonymous

And add them up

Balarka Sen

That sounds much better

Anonymous

1) $2x^2=a(x)+b(2x^2)$ 2) $2y^2=a(2y)+b(2y^2)$ 3) $2z^2=a(z)+b(4z^2)$

Anonymous

$2x^2+2y^2+2z^2=a(x+2y+z)+b(10)$

Anonymous

$2r^2=a(x+2y+z)+10b$

Anonymous

6:56 PM

$2r^2=5a+10b$

Balarka Sen

So $5a + 10b$

Anonymous

mhm

Anonymous

So, any ideas after this?

Balarka Sen

Not sure. I just think you should do the calculation in the manner I explained

Anonymous

@BalarkaSen Which manner?

Balarka Sen

7:05 PM

Write $x, y, z$ in terms of $a, b, c$

and plug it in the two constraints

It should be a short calculation

Anonymous

Oh, okhay. Trying!

Anonymous

7:49 PM

@BalarkaSen One more! Find the maximum and minimum values of $ax^2+by^2+cz^2+2hxy+2gzx+2fyz$, subject to the conditions $lx+my+nz=0$ and $x^2+y^2+z^2=1$. I get $2ax+2hy+2gz=\lambda (l) + \mu (2x)$, $2by+2fz+2hx=\lambda m + \mu (2y)$ and $2cz+2gx+2fy=\lambda n + \mu (2z)$. Multiplying the first equations by $x$, second by $y$ and third by $z$ and adding I get $2u=2\mu$, where $u$ is the function to be maximized or minimized.

Anonymous

Basically $u=\mu$

Anonymous

In the textbook they are getting a strange condition:

Anonymous

Check number (16)'s answer @BalarkaSen. No idea how they got that determinant

Anonymous

Looks like solving a system of equations using determinant method

Anonymous

7:53 PM

hmm

Anonymous

If I consider $\lambda$ to be another variable

Anonymous

Or rather $\lambda/2$ as another variable

Anonymous

They are using the condition $D=0$ for solutions to exist for the homogenous system

Anonymous

Does that look right?

Anonymous

:/

Balarka Sen

7:59 PM

Sorry, let's see.

SE was breaking down so I was busy talking to people about that

Anonymous

breaking down? O_o

Balarka Sen

Yeah someone raised 20 flags or so in the Portuguese SE chat and that made the chat break

It had effects on the math chat too

Anonymous

lol

Anonymous

crazy

Balarka Sen

Hmm let' see

Ah OK I am seeing it.

It's what you say

Anonymous

8:03 PM

Yeah, I think they're just doing $D=0$

Anonymous

And here $x,y,z,\lambda$ are reals

Anonymous

So no other weird problems

Anonymous

I guess

Balarka Sen

Write down the matrix $A$, you'd end up with $A[x, y, z, \lambda] = 0$

That has a solution iff $\det A = 0$

Anonymous

Yup, that, thanks! And another problem I was facing is: I'm not sure how to parameterize a rectangular parallelopiped. The question is to find the maximum volume of a rectangular parallelopiped within $x^2/a^2+y^2/b^2+z^2/c^2=1$ (probably based on Lagrange multipliers)

Anonymous

8:05 PM

Should I use angles to parameterize?

Anonymous

I think cylindrical coordinates might help

Anonymous

(BTW I hope I'm not disturbing you...I find these optimization questions a bit tricky)

Anonymous

Say $(0,0,0)$ is the centre

Balarka Sen

Oh I am not disturbed, it's totally fine

Anonymous

I consider $r=\sqrt{a^2+b^2+c^2}$

Anonymous

8:09 PM

(Okaies :))

Balarka Sen

I don't think you need to parametrize

Anonymous

@BalarkaSen Okay, then?

Balarka Sen

The rectangular parallelopiped is a symmetric thing. If you have the center of ellipsoid at origin, then corner points of the cuboid are $(\pm x, \pm y, \pm z)$

Anonymous

Uh, that does make sense. So the volume of cuboid would be just $8xyz$?

Balarka Sen

Right

Anonymous

8:14 PM

Oh, awesome. I don't know how I missed that. :P

Anonymous

There's another similar question based on geometry. It asks us to find a point such that the sum of squares from four faces of tetrahedron will be minimum.

Anonymous

There must be some shortcut for this too

Anonymous

It would be too lengthy otherwise

Anonymous

To write the equations of plane faces and find distances individually, or no?

Anonymous

Let's assume that the point is $(x,y,z)$

Anonymous

8:18 PM

0

Q: Sum of the square distances from a point to the sides/faces of a regular polygon/polyhedra

This is a variant of the result discussed in this link: On a constant associated to equilateral triangle and its generalization. Consider any regular polygon and an arbitrary point, $P$, on an arbitrary circle with center at the centroid of the polygon. The sum of the square distances from $P$ t...

geometry

Anonymous

Huh, look at this theorem

Anonymous

There must be some easier method for this... (using Lagrange Multipliers)

Balarka Sen

8:30 PM

@Blue Yeah you'd have to write down the equations for the faces of the tetrahedron

I see no sneaky way to do this

This is all calculations

Anonymous

Huh, too bad

Balarka Sen

I don't want to do this :P

Anonymous

Me too XD

Balarka Sen

lol

fuget about it

Anonymous

I'm skipping it and hoping with all my heart that it doesn't appear in the exam

Balarka Sen

8:31 PM

is this for yourexams?

lolol

@Blue If I'm in a good mood I might try it some time tomorrow

Anonymous

Sure!

Balarka Sen

I think the idea is to use the tetrahedron with vertices $(1, 1, 1), (1, -1, -1), (-1, 1, -1), (-1, -1, 1)$

That's the simplest one I know

Then find the plane passing through three points each, write down the distance function, etc

Anonymous

Yes, that might work. Btw did you see the theorem in that Math SE post? It looks quite elegant

Anonymous

0

Q: Sum of the square distances from a point to the sides/faces of a regular polygon/polyhedra

This is a variant of the result discussed in this link: On a constant associated to equilateral triangle and its generalization. Consider any regular polygon and an arbitrary point, $P$, on an arbitrary circle with center at the centroid of the polygon. The sum of the square distances from $P$ t...

geometry

Balarka Sen

Yeah

It's quite nice

Anonymous

8:40 PM

That however is slightly different from this problem

Anonymous

Yup

Anonymous

Okay, I'll try with your method, thanks

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♦ $Mathematics$ the no-normie zone

Transcript for

♦ the no-normie zone

♦ $Mathematics$ the no-normie zone