Mathematics - 2017-04-15 (page 1 of 3)

Kasmir Khaan

00:57

how to solve sin(z) = i ?

Meow Mix

Use the definition of sine using $e$...

Kasmir Khaan

in the solution that i did not understand , they jumped from (e^iz -e^-iz ) /2i = i

to (e^iz+1)^2 =2

How are those equivalent ?

Nver mind I got it =p

Bilateral

01:16

Why do you guys think Lorentzian geometry is so unpopular

in maths?

being the universe Lorentzian?

0celouvsky

It's very difficult and there's little payoff in terms of "big theorems"

Harry Evans

03:16

Good evening!

Balarka Sen

06:08

Hi @Alessandro.

DHMO

06:30

challenge: $\displaystyle \int \dfrac {2x^5-3} {x^6+x} \mathrm dx$

CowperKettle

I wonder if relative share is a good phrase. This is my translation from Russian.

Another translator translated this as relative percentage - but I'm not sure, because there is no multiplication by 100 in the formula

DHMO

@CowperKettle amount

CowperKettle

06:45

@DHMO "relative amount"? Okay, thank you

DHMO

@CowperKettle Just "amount".

CowperKettle

The Russian original expression is относительная доля, literally relative share. I'm afraid to change it to mere amount

DHMO

you already said "per one mole of protein"

so the idea of relativity is already here

CowperKettle

yes.. well, okay ))

I will put amount there.

0

Q: How to multiply two common fractions in MS Word formula editor?

In MS Word 2007 formula editor, I started creating a formula: I now need to multiply this common fraction by another common fraction, but the editor lets me put the multiplication sign either in the numerator or in the denominator. I want to "leave the first common fraction behind", put th...

microsoft-word-2007

Maybe someone knows the answer to this

DHMO

@CowperKettle I can do that

I don't know why you asked

CowperKettle

06:55

I asked because I don't know

DHMO

just go right until you go beyond the fraction

CowperKettle

aaah

oops

Thank you

DHMO

no problem

32 mins ago, by DHMO

challenge: $\displaystyle \int \dfrac {2x^5-3} {x^6+x} \mathrm dx$

16 hours ago, by DHMO

@Secret Use linear algebra to find the minimal polynomial of $\sqrt 2 + \sqrt 3$

Mary Star

07:12

Hello!!
We have the sets A=\{(x,y)\in \mathbb{R}^2 \mid x<4\}$ and $B=\{(x,y)\in \mathbb{R}^2 \mid y<2x, x>3\}$ and the function $f:A\rightarrow B$, $f(x)=y$ is defined as follows:
$$\begin{matrix}
x & 0 & 1 & 2 & 4 & 7 & 13 & 15\\
\hline
y & 11 & 8 & 6 & 5 & 3 & 0 & -1
\end{matrix}$$

The function is injective since for two different values of x we get different values for f(x), right?

The function is not surjective, since there are elements in $B$ that are not an image of the function. Is this correct?

DHMO

Use $\{1,\sqrt2,\sqrt3,\sqrt6\}$ as basis.

To find the matrix representing multiplication by $\sqrt2+\sqrt3$:

$1 \times (\sqrt2+\sqrt3) = \sqrt2+\sqrt3$
$\sqrt2 \times (\sqrt2+\sqrt3) = 2+\sqrt6$
$\sqrt3 \times (\sqrt2+\sqrt3) = 3+\sqrt6$
$\sqrt6 \times (\sqrt2+\sqrt3) = 3\sqrt2+2\sqrt3$

$(\sqrt2+\sqrt3) v \equiv \begin{pmatrix}0&2&3&0\\1&0&0&3\\1&0&0&2\\0&1&1&0\end{pmatrix} \vec v := A \vec v$

Multiplication by $1$ is represented by the identity matrix

thus, we have to find a polynomial $p$ such that $p(A) = I$

Or, $p(A)-I = 0$

To do this, use the characteristic polynomial of $A$.

$\det \begin{pmatrix}-x&2&3&0\\1&-x&0&3\\1&0&-x&2\\0&1&1&-x\end{pmatrix}$

$= -x \det \begin{pmatrix}-x&0&3\\0&-x&2\\1&1&-x\end{pmatrix} - \det \begin{pmatrix}2&3&0\\0&-x&2\\1&1&-x\end{pmatrix} + \det \begin{pmatrix}2&3&0\\-x&0&3\\1&1&-x\end{pmatrix}$

$=-x[-x^3+3x+2x] - [2x^2+6-4] + [9-6-3x^2]$

$=x^4-5x^2-2x^2-2+3-3x^2$

$=x^4-10x^2+1$

@Semiclassical @Secret ^

@MaryStar can you fix your typos?

Mary Star

Oh yes.

We have the sets $A=\{(x,y)\in \mathbb{R}^2 \mid x<4\}$ and $B=\{(x,y)\in \mathbb{R}^2 \mid y<2x, x>3\}$ and the function $f:A\rightarrow B$, $f(x)=y$ is defined as follows:
$$\begin{matrix}
x & 0 & 1 & 2 & 4 & 7 & 13 & 15\\
\hline
y & 11 & 8 & 6 & 5 & 3 & 0 & -1
\end{matrix}$$

The function is injective since for two different values of x we get different values for f(x), right?

The function is not surjective, since there are elements in $B$ that are not an image of the function. Is this correct?

DHMO

It doesn't make sense

Elements of $A$ are coordinates

So are elements of $B$

and $f$ takes domain $A$ and codomain $B$

so inputs of $f$ are elements of $A$

how are they not coordinates?

@MaryStar are you here?

Mary Star

07:33

Yes I found that also confusing. So, is the exercise statemant wrong? @DHMO

DHMO

Where is this from? Can you take a photo/screenshot?

Mary Star

Yes (it is in german) :

Is it maybe meant two general sets A and B and not the one of the question (a) ? @DHMO

DHMO

(a) and (c) are not related.

Mary Star

Ahh

So, so that f is surjevtic B must be the set {-1, 0, 3, ,,5, ,6, 8, 11}, right?

DHMO

Korreckt

Mary Star

07:38

And the function is strictly monotonically decreasing since for bigger values of x we get smaller values of y, right?

DHMO

richtig

Mary Star

Great! Thank you so much!! :-)

BAYMAX

08:00

0

A: Complex numbers involving roots of unity

Well I tried this,may be of some help- $z + z^2 + ... +z^n = n|z|^n$ differentiating the above series wrt $z$ and as right side is a constant for a fixed $n$ so rhs will be $0$ on differentiating $1 + 2z + 3z^2 +...+nz^{n-1} = 0 $ subtracting the latter from the first equation we get- $1 +...

Is my solution there correct or some flaws?

SBM

Oh, wait.

DHMO

@BAYMAX you can't differentiate both sides of an equation

e.g. let's solve x^2-2x-3=0

differentiate both sides to get 2x-2=0

therefore x=1

Secret

@DHMO I thought the condition for minimal polynomial is $p(A)=0$, not $p(A)=I$?

BAYMAX

Oh yes@DHMO

Why my mind said to do that,I am also thinking now!

DHMO

@Secret you're right, $p(A)=0$. The answer is still correct

Secret

08:05

I see

BAYMAX

Its a nice question though.

SBM

:)

Secret

Meanwhile, besides managing my quantum chemistry stuff, I am thinking about this:

$\{a+b(\frac{p}{q}+\epsilon),a,b,p,q\in \Bbb{Z}, \epsilon > 0, \epsilon \in \Bbb{R}/\Bbb{Q}\}$

Basically, to investigate whether the transition from a discrete set to a desne set is abrupt by using a very small irrational number as a parameter

Expanding, we get: $\frac{aq+bp}{q}+b\epsilon=\frac{aq+bp+bq\epsilon}{q}=\frac{n}{q}+b\epsilon$

DHMO

In Chinese we say 差之毫釐，謬之千里

Secret

indeed

DHMO

08:15

2 hours ago, by DHMO

challenge: $\displaystyle \int \dfrac {2x^5-3} {x^6+x} \mathrm dx$

SBM

@DHMO please explain what that means.

As I do not know Chinese

DHMO

the slightest difference leads to a huge error (idiom)

BAYMAX

kira kira killer!

SBM

It's really astonishing how eight characters spell out that much

Secret

well, for chinese, each character is already a single word

SBM

08:19

How difficult that must be to remember

$$\ln |x^5 + 1| - 3 \ln |x|$$

plus the constant maybe

DHMO

How did you do that?

Secret

Looks like partial fraction expansion to me

Astyx

Hi chat

SBM

Was the answer correct though?

@DMHO

Secret

one can always differentiate and check

DHMO

08:26

@Astyx hi

Secret

yes, it's correct

SBM

I just verified it.

Thank you

DHMO

How did you do it?

Secret

$\frac{2x^5-3}{x^6+x}=\frac{5x^4}{x^5+1}-\frac{3}{x}$. I don't like partial fraction decompositions because I can never compute them fast enough in my brain and must use paper

SBM

Partial fractions.

Secret

08:28

The first term can be integrated by inverse chain rule

SBM

It's clearly visible.

Secret

Partial fraction asks the following question: Given a rational function $\frac{p(x)}{q(x)}$ where $q(x)=r(x)s(x)$, find the numerator of rational function that has the denominator $r(x)$ and $s(x)$ respectively

$\frac{p(x)}{q(x)}=\frac{A(x)}{r(x)}+\frac{B(x)}{s(x)}$

SBM

Yes

I find integration interesting unless

Secret

One can then solve for $A(x)$ and $B(x)$ by comparing coefficients after multiplying both sides by $q(x)$

SBM

I get some non-integrable function

Secret

08:32

Non integrable as in no elementary primitive or it is unbounded?

SBM

Or super difficult ones

DHMO

Challenge: given two square matrices of the same order A and B. Can AB-BA = I? @Secret

Secret

what's a matrix order again, is it the same as its rank?

DHMO

I mean, n x n

SBM

OK

Astyx

08:44

@DHMO Challenge : $$\sum_{k=0}^{\infty}{1\over 16^k}\left({4\over 8k+1}-{2\over 8k+4}-{1\over 8k+5}-{1\over 8k+6}\right)$$

DHMO

@Astyx $\pi$

SBM

I don't know what to try first

Astyx

Can you prove it ?

DHMO

I can't

SBM

What to prove @Astyx?

Astyx

08:45

@SBM computing the sum above

SBM

I've not learnt that skill so sorry

Astyx

@DHMO I'm sure you can figure it out

DHMO

@Astyx how?

SBM

Could you teach me summation

Secret

@Astyx You might wanna invite Then into this challenge. She is really good at infinite sums

Astyx

08:47

What does $4\over 8k+1$ make you thing of ?

DHMO

@Astyx power series?

Secret

and as far I knew, she had not seen exponential terms in a sum before

Astyx

Not quite

DHMO

I mean

SBM

Um

DHMO

08:48

$\displaystyle \int_0^1 4x^{8k} \ \mathrm dx$

Astyx

Exactly

SBM

?

Astyx

And then compute, compute, compute

SBM

How'd you get that @DHMO

DHMO

ok let me try...

$\displaystyle \quad \sum_{k=0}^{\infty}{1\over 16^k}\left({4\over 8k+1}-{2\over 8k+4}-{1\over 8k+5}-{1\over 8k+6}\right)$
$= \displaystyle \int_0^1 \sum_{k=0}^{\infty}{1\over 16^k}\left(4x^{8k} - 2x^{8k+3} - x^{8k+4} - x^{8k+5}\right) \ \mathrm dx$

Astyx

08:49

Why is that ? (getting the integral out)

DHMO

I can't?

SBM

Oh, how?

Astyx

Yes you can

But not without justifying it

DHMO

I'm not sure how to continue

Well because it converges?

SBM

It appears convergent

DHMO

08:51

Because of the $16^k$

Topologicalife

Hi.

Astyx

It converges uniformly

Hi @Topologicalife

Anyway, go on

SBM

Wait, what?

I see some strange pattern

DHMO

I suppose it is a geometrical series

with common ratio $\dfrac{x^8}{16}$

Topologicalife

I have the following problem: Write the positive number a as the product of four poisitve factors such that their sum is a minimum.

I don't understand the problem. do I have to minimize $f(x,y,z,w)=x+y+z+w$ such that $x,y,z,w>0$ with $a=x\cdot y\cdot z\cdot w$?

DHMO

08:53

@Topologicalife yes

Topologicalife

How can I do this?

DHMO

AM-GM

Topologicalife

It is supposed to be done with multivariable calculus.

SBM

...

Alessandro Codenotti

Hi @Balarka I saw you greeted me earlier only now

Topologicalife

08:54

I want to avoid AM-GM :)

Astyx

Hi @Alessandro

DHMO

On the other hand, I want to avoid multivariable calculus

SBM

.?

DHMO

$= \displaystyle \int_0^1 \left(4 - 2x^3 - x^4 - x^5\right) \dfrac{1}{1-\frac{x^8}{16}} \ \mathrm dx$

@Astyx Is this correct?

SBM

¿:?

Topologicalife

08:55

I'm asked to do it with multivariable calculus :D

Astyx

Yes it is @DHMO

DHMO

I'm still not seeing any $\pi$ in it...

SBM

How do you get that?

@DHMO

Astyx

You will soon

DHMO

$= \displaystyle \int_0^1 \dfrac{64 - 32x^3 - 16x^4 - 16x^5}{16-x^8} \ \mathrm dx$

Astyx

08:57

$\pi$ often appears when you get integrals with trigonometric function or their derivatives

DHMO

I'm seeing a $\pi$ in the denominator

SBM

Trig substitution

?

Astyx

I believe you can compute that integral (evil laugh)

DHMO

I believe not

I'm seeing partial fraction

SBM

What I was thinking

Astyx

08:58

Rationnal fractions integrals are always computable

DHMO

@Astyx doesn't mean I can compute them

Astyx

You know fraction polar decomposition right ?

DHMO

I know partial fractions

Astyx

That must be it

DHMO

$\dfrac{64 - 32x^3 - 16x^4 - 16x^5}{16-x^8} \equiv \dfrac {Ax+B}{x^2-2} + \dfrac{Cx+D}{x^2+2} + \dfrac{Ex+F}{x^2+2x+2} + \dfrac{Gx+H}{x^2-2x+2}$

Astyx

09:01

And you also know primitives of such functions

DHMO

Doesn't mean I can compute $A$ to $H$

too tedious

Astyx

Also $x^2-2$ is not as simple as it gets

DHMO

well I might have a few shortcuts

Astyx

Also you forgot something

DHMO

what is it?

Astyx

09:03

The numerator has roots

So you're making it more tedious than it should be

DHMO

$4-2x^3-x^4-x^5=0$

oh, $x-1$ is a root

Astyx

Well 1 is a root, yeah

DHMO

$x^5+x^4+2x^3-4 = (x-1)(x^4+2x^3+4x^2+4x+4)$

but it wouldn't help

unless the root can cancel out the factors in the denominators

$\dfrac1{x^8-16} \equiv \dfrac18 \left[ \dfrac1{x^4-4} - \dfrac1{x^4+4} \right]$
$\dfrac{x^4}{x^8-16} \equiv \dfrac12 \left[ \dfrac1{x^4-4} + \dfrac1{x^4+4} \right]$

Astyx

Try some roots of the denominator then

DHMO

just let me use my method thanks

$16\dfrac{x^5+x^4+2x^3-4}{16-x^8} = 2(2x^3-4)\left[ \dfrac1{x^4-4} - \dfrac1{x^4+4} \right] + 8(x+1)\left[ \dfrac1{x^4-4} + \dfrac1{x^4+4} \right]$

$=\dfrac{4x^3+8x}{x^4-4} - \dfrac{4x^3-8x-16}{x^4+4}$

Now we can see the roots

$=\dfrac{4x}{x^2-2} - \dfrac{4x^3-8x-16}{x^4+4}$

Secret

09:12

@DHMO $A=\begin{pmatrix}a & b \\ c & a\end{pmatrix},B=\begin{pmatrix}0 & f \\ g & 0\end{pmatrix}$ with the condition $bg=2cf$

That's just one of them

DHMO

@Secret show me

Now:
$\dfrac1{x^2-2x+2} + \dfrac1{x^2+2x+2} = \dfrac{2x^2+4}{x^4+4}$
$\dfrac1{x^2-2x+2} - \dfrac1{x^2+2x+2} = \dfrac{4x}{x^4+4}$

$(2x^2+4)(2x-4) = 4x^3-8x^2+8x-16$

$8x^2-16x = (4x)(2x-4)$

Wait what

I must be blind

$4x^3-8x-16 = 4(x^3-2x-4) = 4(x-2)(x^2+2x+2)$

Mary Star

Is the set $f:\{\text{set of all people}\} \rightarrow \{\text{female, male}\}$ surjective?
I think yes, since every male or female is a person of the set of all people.
Is this correct?

DHMO

@MaryStar yes

$=\dfrac{4x}{x^2-2} - \dfrac{4x-8}{x^2-2x+2}$

Secret

I am not good at the properties of matrix commutators, thus I basically brute force it by picking entries from $A=\begin{pmatrix}a & b \\ c & d\end{pmatrix},B=\begin{pmatrix}e & f \\ g & h\end{pmatrix}$. Note how the diagonals give the strongest restriction to the elements as they have to add up to 1, i.e. after expanding $[A,B]=AB-BA$, we get $bg-fc=cf-hc=1$. Now setting $h=0$ to narrow down cases, we get $bg=2cf$.
Now because in the original equations they have to add up to 1, it means $b,g,c,f\neq 0$. Looking at the equations for the off diagonals, we have $af-eb-fd=-(d-a)f-eb=0=ce+dg-ga…

Mary Star

@DHMO Ok! Thanks!

DHMO

09:19

@Astyx am I correct?

@Secret I mean, can you show me what $AB$ and $BA$ are, using the matrices you gave me?

Astyx

Give me the time to reread

Mary Star

Does maybe someone of you have an idea about my question: https://math.stackexchange.com/questions/2234892/do-we-get-an-information-about-the-extrema-from-the-graph

According to the answer the extrema is the intersection point of the contour lines and the constraint. But in this case it doesn't hold, does it?

Topologicalife

if $f:C\setminus\{0\}\to C$ is analytic, has at worst a pole at $0$ and at worst a pole at infinity, must it be of the form $az^{-k} + bz^{-k+1} + ... + dz^{k'}$?

Astyx

It looks fine to me

I might be wrong though

Topologicalife

Oh, yes.

Astyx

09:25

So you're nearly there @DHMO

DHMO

so now
$\quad \displaystyle \int_0^1 \left(\dfrac{4x}{x^2-2} - \dfrac{4x-8}{x^2-2x+2}\right) \ \mathrm dx$
$= \displaystyle \int_0^1 \dfrac{2\mathrm d(x^2-2)}{x^2-2} - \int_0^1 \dfrac{2\mathrm d(x^2-2x+2)}{x^2-2x+2} + \int_0^1 \dfrac{4\ \mathrm dx}{(x-1)^2+1}$

Secret

$AB=\begin{pmatrix}bg & af \\ ag & cf\end{pmatrix},BA=\begin{pmatrix}fc & fa \\ ga & gb\end{pmatrix}$. $[A,B]=AB-BA=\begin{pmatrix}bg-fc & 0 \\ 0 & cf-gb\end{pmatrix}$. Now imposing $cf=1$ and $bg=2cf$, we get...

Uh wait a sec, not cf=-cf again? I thought I have eliminated this contradiction
Ok, I have no idea how to solve this more elegantly

DHMO

I'm surprised that you're not seeing the answer from $A=\begin{pmatrix}a & b \\ c & d\end{pmatrix},B=\begin{pmatrix}e & f \\ g & h\end{pmatrix}$ already, @Secret

Astyx

There is a far less tedious way of doing it @Secret @DHMO

DHMO

@MaryStar are you still here?

Secret

09:30

@DHMO yeah, I just realised, right from the start $bg-fc=-(cf-gb)=1$, thus the contradiction is already there unless the field is $\Bbb{Z}/2$

Mary Star

@DHMO Yes

DHMO

@MaryStar here's a shortcut for you

@Secret but you have only dealt with 2x2 matrices

Mary Star

@DHMO Can we get from that graph an information about the extremas of the function under the constraint?

Secret

@Astyx Does it have something to do with commutators?

Astyx

Not really

Alessandro Codenotti

09:31

No @Secret

DHMO

@MaryStar I have no idea at all about your question. All I've showed you is a shortcut in desmos

Mary Star

Ahh ok!!

Astyx

@DHMO You're one step away now

Secret

@AlessandroCodenotti In that case I have no idea, the only strange thing I saw is that the contributions from the product of diagonal elements always cancel out

DHMO

@Astyx what do you mean?

Astyx

09:33

From getting $\pi$

Alessandro Codenotti

@Secret you're going in the right direction then

DHMO

how is it so?

Astyx

You can integrate these

DHMO

@MaryStar must $x$ and $y$ be positive?

Mary Star

We don't have any information about that. @DHMO

DHMO

09:35

@MaryStar then I can make $f(x,y) = x^2y$ arbitrarily big

by making $x$ arbitrarily small

Astyx

Have you given up @DHMO ?

DHMO

@Astyx well the first two terms cancel out

And the last term evaluates to $\pi$

because I recognize the $\arctan$

Astyx

Good job

DHMO

(I'm borrowing complex numbers in the first term)

Astyx

You don't have to

DHMO

09:40

how?

Astyx

By taking the module of $x^2-2$

DHMO

oh...!!!

Astyx

Instead of itself

DHMO

I almost forgot

Astyx

I gotta go now

See ya

Secret

09:45

2

Q: Consequences when the commutator is a scalar multiple of the identity matrix

I just stumbled over the question below. As to the first, I could easily find out the answer (D) by invoking the commutation relation. But I don't figure out how to solve other two. Could anybody give a hint? Here's the question: The commutator $[B,A] \equiv BA-AB = \lambda I$ of two $n\ti...

linear-algebra matrices

I really need to pay more attention to the bigger picture when solving maths problems...

DHMO

basically tr(AB) = tr(BA)

Mary Star

@DHMO Why is f getting arbitrarily big when x is arbitrarily small?

Secret

56

Q: Trace of a commutator is zero - but what about the commutator of $x$ and $p$?

Operators can be cyclically interchanged inside a trace: $${\rm Tr} (AB)~=~{\rm Tr} (BA).$$ This means the trace of a commutator of any two operators is zero: $${\rm Tr} ([A,B])~=~0.$$ But what about the commutator of the position and momentum operators for a quantum particle? On the one hand: $$...

quantum-mechanics mathematical-physics commutator

And then things get more interesting in infinite dimensions

DHMO

@MaryStar because |x| and |y| are very big

you can try it for yourself

let x=-1000

what is x^2 y?

SBM

10:03

Wait what?

What just happened?

Mary Star

10:16

@DHMO Ahh I see.

Secret

Can We Combine pi & e to Make a Rational Number? | Infinite Series

►

SBM

Yikes

Fawad

11:20

$\mp$

Today I learned^

SBM

11:48

idk

BAYMAX

12:04

nice one secret!

DHMO

12:33

no table is not able to be notable

Will Njundong

TF ლ(ಠ益ಠლ)

https://imgur.com/a/ZVJWy

SBM

What?

Will Njundong

someone please explain this shit to me ಠಠ

I have looked over my work several times, i can understand how they got $2^{36}-2$

where did the 2 come from

SBM

@WillNjundong do you mean finding $$\sum_{k=0}^{35} 2^k$$

Will Njundong

yes

I first chose A, turns out that was wrong

so I justrandomly picked another one because I didnt know how else to get the other answers

SBM

12:40

Use Geometric Progression to find what it is.

DHMO

@SBM no

it is k=1

SBM

Oh

What made me say that

Factor

it out.

What

am I doing?

Will Njundong

$$\frac {1-(1)2^{36}}{1-2}$$ works out to $$\frac {-2^{36}+1}{-1}$$

soooo where does 2 come from in their answer?

DHMO

comes from your unawareness that it starts from k=1 not 0

Will Njundong

ahh, didnt spot that.

SBM

12:55

-1

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$Mathematics$ Mathematics