Mathematics - 2019-10-18

Hey guys! If function $f$ is even and function $g$ is odd, then what would be $(f\circ g)(x)$? I have deduced that $(f\circ g)(x)$ would be even. Is this conclusion correct?

Leaky Nun

@Abwatts f(g(-x)) = f(-g(x)) = f(g(x)) so it is even

Jacksoja

@LeakyNun where did 2+i dissappear?

in the double quotient?

Abwatts

@LeakyNun Great. I just wanted to confirm that. Thanks a lot!

Leaky Nun

@Jacksoja it became 2+x

Jacksoja

12:26 AM

okay I think I need to continue later

am not seeing clearly why this is happning and i dont want to keep bothering you with bad questions

thank you @LeakyNun

Shine On You Crazy Diamond

1:15 AM

Any undergraduate of computer science here want to publish a paper together?

I need help publishing is the reason. The algorithm is already mostly figured.

@Ultradark

Send an email to enjoysmath@gmail.com if interested.

N. Maneesh

2:46 AM

0

Q: Prove that $\sum _{n=1}^{\infty }\frac{g'\left(n\right)}{g\left(n\right)}$ converges iff $lim_{ n→∞}g(x)<∞.$

Let $g,g',\left(g'\right)^2-gg''$ are positive valued functions that exists on the domain $[1,\infty).$ Prove that $\sum _{n=1}^{\infty }\frac{g'\left(n\right)}{g\left(n\right)}$ converges iff $lim_{ n→∞}g(x)<∞.$ My attempt We know from the Hypothesis that $g(x)>0, g'(x)>0, (g')^2>gg'', \forall ...

Secret

3:34 AM

On yucky mathematics and mathematical quagmires:

Consider the following equation:

Secret

3:46 AM

$$\frac{\sin \Bigg(e^{\zeta (z)} + \frac{1}{1+\frac{\cos^5(z+\ln z)e^{e^z}}{z+\tan(z^2-4z)}}\Bigg)}{1 + e^{z-z^8+e^{z^z}}}=f(\zeta (z))$$

This is very yucky. But how yucky:

Define $Y$, the yuckiness to be the number and complexity of possible ways one can rewrite an expression using identities and inferences

Sonal_sqrt

If a ring isn't a unique factorization domain then it isn't euclidean? Is this true

Secret

Then an equation is considered yucky if $Y < 10$ and each of these paths are not simply connected

An equation is considered very yucky if $1< Y < 5$ and is irreducible if $Y = 0$

The mathematical quagmire, is defined to be the set of all yucky equations

What remains to be explored, is how yucky each region of the quagmire is relative to each other (relative yuckiness)

and whether, given a quagmire restricted by some constraints (e.g. all equations involving exponentials and polynomials) whether there exists a unique path between the most yucky equation to the least yucky equation that composed of the same type of letters

Sonal_sqrt

@Secret can you clear the domain thing?

Rithaniel

I don't think your concept of "yuckiness" can be well defined. I can always just invent a new function and corresponding identity and $Y$ increases.

Secret

ah right, I forgot just special functions alone, there are at least countably many to choose from, one for each power series

and I think the case where only elementary functions are involved is already solved by some kind of Galois theory and hence they cannot be very yucky

actually, I am not sure... functions involving reciprocals and sums tends to get yucky very fast

e.g. fractions like $\frac{1}{1+e^z}$

Rithaniel

4:00 AM

Well, here's an identity for that $\frac{1}{1+e^z}=\frac{e^{-z}}{1+e^{-z}}$

Secret

@Sonal_sqrt if you can write anything in terms of prime factors in the ring or you have zero divisors, you cannot possibly find a euclidean division algorithm in it

Rithaniel

and another: $\frac{1}{1+e^z}=\frac{1}{2+\frac{2}{\text{coth}(z/2)-1}}$

Secret

the second one looks a lot more "yucky" (subjectively) than the original, yet the existence of that identity means the expression on the right cannot have a very small $Y$ value, hmmm

Rithaniel

or $\frac{1}{1+e^z}=\frac{d}{dz}(z-\text{log}(1+e^z))$

Secret

yeah you are right, just the number of identities alone can increase rapidly, further making $Y$ not correspond to what we think as "yucky"

Rithaniel

4:06 AM

Bingo

Secret

In fact, your examples showed something more:

There are expressions that are written in very complicated and nested way, which can be rewritten into a very simple form

because the symmetries are just right for the rewriting rules applied to cascade it into a simple expression

Rithaniel

Yeah, forms are very often equivalent. I've actually graded differential equations quizzes where two people got seemingly wildly different answers to a problem, but closer inspection revealed that they were actually one and the same.

Stuff can catch you off-guard sometimes

Rithaniel

4:22 AM

Another identity: $\frac{1}{1+e^z}=\frac{1}{4}\int_1^2\int_{-2x}^1\frac{1}{1+e^z}dtdx$

Secret

that last one is soooooooooo lame, lol

Rithaniel

(Now, that's just a way to represent multiplication by one, yeah)

Secret

Sometimes, I wonder whether algebraic identities are really n-tuple operators

e.g. in groups, the binary operator $\cdot$ takes in two elements and spits out one

Rithaniel

You might be able to interpret them as such, but you'd have to be very careful in how you define things

Secret

So something like: $\frac{d}{dz} (z - \ln (1+e^z))$ is something like $f(a,b,c) = \frac{d}{db} (b - \ln (a+e^b))$

and then you have some kind of algebraic structure $(S, f, \cdot, +)$ where $S$ is the underlying set

not sure if this can really get anywhere though, there are so many such $f$s

The resulting space does not look like function space to me, because we have $\cdot, +$ in the structure

Rithaniel

4:31 AM

Ah, I see what you mean. So any function induces a type of algebraic object. You could have the only operation be the $f$. Then, provided the $f$ is binary, you have a magma

What has to be true about $f$ for the corresponding magma to be associative though? Like, $f(a,b)=a\cdot b+1$ isn't, provided we are pulling $a,b,\cdot,+$ from $\mathbb{Z}$

Secret

you want associative wrt $f$, or wrt $a,b$?

are you thinking about something like $f (f (a,b),c)$ or just $f(a,b)*c$?

Rithaniel

Well, $f$ in this context isn't the element, but the operation, so we want to know whether or not $f(f(a,b),c)=f(a,f(b,c))$

Secret

ah, I don't think there is a simplication of this for general $f$, but it definitely suggests the two arguments are symmetric somehow. Need to think

$f(a,f(b,c)) = f(f(b,c),a)$ is for commutativity, so that suggests $f(b,c) = a$ at some abstract line of symmetry thus there will be fixed points there. I forgot the geometric counterpart to associativity, need to check

krauser126

7:08 AM

Hello

ÍgjøgnumMeg

7:26 AM

What is complex conjugation on the Riemann sphere? What happens to $\infty$?

Alessandro Codenotti

8:13 AM

I guess it doesn't move?

ÍgjøgnumMeg

hmm

annoying little prick

hahaha

the fact that it just destroys everything is making finding its pre-image under a transformation annoying

Alessandro Codenotti

So did they trick you into doing complex analysis by selling it as number theory?

ÍgjøgnumMeg

lol no this is in modular forms, just recapping möbius transformations

Alessandro Codenotti

Ah, makes sense

ÍgjøgnumMeg

in particular, I have the inverse of the relevant transformation but stuffing $\infty$ into it gives the wrong pre-image

:(

unless I pretend $\infty/\infty = 1$

which actually seems to be quite common in this context lol

Rithaniel

8:27 AM

So, $\infty/\infty+\infty/\infty=2$?

ÍgjøgnumMeg

sure why the hell not

Rithaniel

Hmmm, and what about $\frac{\infty}{\infty+\infty}$?

ÍgjøgnumMeg

that's 1 obviously

hehe

Rithaniel

I would have guessed that $\frac{\infty}{\infty+\infty}=\text{stop}$

2

ÍgjøgnumMeg

rofl

Mike Miller

8:40 AM

if you're trying to plug infty into (az+b)/(cz+d) you're really asking for the limit of this as z -> infty

and that's made more obvious either by saying lhopitals or rewriting as (a+b/z)/(c+d/z)

ÍgjøgnumMeg

yeah I realised this and the problem has now stopped being a problem

lol

thanks :P

krauser126

Anyone familiar enough with graph theory?

Trying to prove something with graph theory and induction and Im not sure if my intuition is correct

felipa

10:40 AM

apologies if this is trivial: if $z, \lambda_1, \lambda_2 > 0$, how many solutions for $z$ does $\lambda_1^2 z^{\lambda_2 z}+\lambda_2^2 z^{\lambda_1 z} = 0$ have?

it looks like none.. is that right?

felipa

10:59 AM

cancel that question please

Semiclassical

12:18 PM

@krauser126 “Just ask, don’t ask to ask.”

Martin Sleziak

12:45 PM

@MJD Since you're the user who created (or at least started) the MathJax tutorial on meta, maybe you' would have some comments on this recent suggestion on meta: Organizing The MathJax Tutorial. (Basically, the post suggest to add some kind of ToC to the post.)

If needed, we can discuss this further in the Math Meta Chat. (This seems to be close on-topic in that room - I used this room simply because it was a place where I was able to ping you.)

Secret

1:29 PM

$f(a,f(b,a))$

$f(f(a,b),a)$

skullpetrol

1:51 PM

is that notation functional?

Secret

nah, just a two variable function nested twice

Trying to understand what happens if they are equal

skullpetrol

how are you pal?

Secret

as usual

MJD

@MartinSleziak Thanks for the ping.

I left a comment in Meta. I think it's a great suggestion.

Over the years I have zealously defended the main post against all sorts of additions. But I think this would be an addition worth having.

skullpetrol

OMG! a talking potato :O

MJD

2:02 PM

Where?!!?

OMG I'M A TALKING POTATO

3

Martin Sleziak

Thanks for the response!

MJD

@martin: I value your judgement. Do you see any drawbacks that I might have overlooked?

Martin Sleziak

I did not interact with the MathJax tutorial on this site too much, so I don't really have strong opinion on this.

Since this was mentioned in Math Meta Chat, perhaps you can ask for additional feedback there. Maybe some of the users who visit that room responds. (Although I am not sure that much feedback is needed here.)

The only thing that came to my mind was whether the OP is actually allowed to edit the post - considering that they have relatively low reputation. However, for CW-post this should not be on problem. (They would not be able to edit regular posts on meta.)

schn

2:45 PM

What are some suitable change of of variables to show that 1/(x+y) is a hyperbolic paraboloid, i.e. it is of the form $z=x^2/2-y^2/2$?

Noob

4:18 PM

hello

what is the best method to open complement using demorgans law ( in boolean algebra )

inside- out

or from outside to inside

Shine On You Crazy Diamond

5:30 PM

@Noob not sure what you mean

by inside -out

Ultradark

5:54 PM

@ShineOnYouCrazyDiamond hello :)

Abwatts

Hey guys! Will the function $f_{e}(x) = x^2$ statisfy the following requirement? $$$$ Assume that $f: [0, ∞) → ℝ$. $$$$ Define a function $f_{e}: ℝ → ℝ$ which is even and $f_{e}(x)$ = $f(x)$ for all $x≥0$.

Semiclassical

6:35 PM

No. That only works if f(x)=x^2 for nonnegative x.

If two functions are different for nonnegative x, then they’ll yield different f_e.

Abwatts

Oh, I see. So what is the question asking us to really do here? I am not entirely sure.

Rithaniel

Think about function composition, maybe. Can you think of a function such that f(x)=x when x>0?

Abwatts

6:53 PM

a function such as $f(x) = |x|$ since we can split it into two cases?

Rithaniel

7:04 PM

|x| does seem like a good candidate, yes. Now, the question wants $f_e(x)=f(x)$. How then would |x| play into this?

Abwatts

7:14 PM

Wouldn't $|x|$ satisfy $f_{e}(x)=f(x)$ since it will always produce $x≥0$?

Poline Sandra

hello,

a small question

please

$f: R\setminus\{0\}\to \mathbb{R}\\ x\to \frac{1}{x}$

f is a function or a map ?

Rithaniel

Abwatts: what if $f=x\text{sin}(x)$ or some other outlandish function. Then what is $f_e$?

Is it |x|?

Abwatts

No because $f=xsin(x)$ is a bounded function?

Rithaniel

So, what would $f_e$ be? (Also, $x\text{sin}(x)$ isn't bounded, but that's a nonsequitur, here.)

Remember the initial hint: think about function composition.

Abwatts

7:33 PM

Umm then $f_{e}$ would be $f_{e}(f(x))$?

since $f_{e}$ will directly depend on the structure of $f(x)$?

Rithaniel

Well, you still haven't said what $f_e$ is. Defining a function in terms of itself doesn't tell you anything about it

Though, it does indeed depend directly on $f$, you are correct.

So you should probably use $f$ in some way to define $f_e$

Abwatts

So wouldn't $f_{e}$ be $x^2$ because it will always spit out positive numbers?

Rithaniel

But when does $x^2=x\text{sin}(x)$?

Abwatts

But wouldn't it equal to $xsin(x)$ if we will use the function composition, like so $x^2sin(x^2)$, where $f_e(f(x))$, $f = xsin(x)$ and $f_e=x^2$?

Random-15

7:54 PM

I came to know about complex numbers of the from x+yj, where j^2=1

In the plane of such complex numbers, what would be the equation of the circle with center at 0?

The unit hyperbola is given by ||z||=1

But I couldn't find a formula for circle with center at 0.

Akiva Weinberger

On average, what's the population density of a car (measured in people per square mile)

Rithaniel

Well, there is a bit to unpack, there. First, $x^2\neq f(x)^2$, and your description leads me to believe that you mean to say $f_e=f^2$ that being said, $f^2\neq f$ and so does not satisfy your requirement. Also, $(x\text{sin}(x))^2=x^2\text{sin}^2(x)\neq x^2\text{sin}(x^2)$

Leaky Nun

@Abwatts the point is to define $f_e$ for an arbitrary $f$ (in terms of the arbitrary $f$)

Rithaniel

8:10 PM

(Well $f^2$ might equal $f$ for a specific $f$, but not in general)

Abwatts

Oh, I see. This question is fairly tricky.

Leaky Nun

a hint is to think about the graph. specifically, what does a function being even mean to the graph of the function?

Abwatts

8:28 PM

it has symmetry with respect to the y axis

Leaky Nun

so can you rephrase the question in terms of this interpretation using graphs?

Abwatts

8:50 PM

Yes, I will try doing that. Thanks.

Abwatts

9:15 PM

I have another question about increasing functions if you guys don't mind. Is the following conclusion that I made about this statement is true?$$$$ This is the statement:

If functions $f$ and $g$ are given to be increasing, then $fg$ would also be increasing on the interval $[a,b]$? This what I did to prove this statement is indeed true. $$$$ Let $x_1$ and $x_2$ be elements of the interval $[a,b]$. Using the defintion of increasing functions, I deduced that $f(x_1)$ < $f(x_2)$ and $g(x_1)$ < $g(x_2)$. So this can be rewritten as $(fg)(x_1)$ = $f(x_1) * g(x_1)$ < $f(x_2) * g(x_2)$ = $(f*g)(x_2)$. Would this be a valid approach?

Abwatts

9:31 PM

Ultradark

9:53 PM

$f(x)=\frac{1}{f(x)}$

How do I solve this equation given $f(f(x))=x$

nvm

Shine On You Crazy Diamond

10:23 PM

@Ultradark hey

Ultradark

@ShineOnYouCrazyDiamond hey

Shine On You Crazy Diamond

@shi worked

Ted Shifrin

10:42 PM

@Random-15 No, you're computing $|z|$ incorrectly. That is, in fact, the unit circle centered at $0$. Remember that if $z=x+yj$, then $|z|^2 = x^2+y^2 = z\bar z$.

Mary Star

10:54 PM

Hello!! Is someone of you familiar with Turing machines?

I have the following question:

schn

11:46 PM

When changing variables in a double integral, is there any strategy for finding suitable transformations so that the non-rectangular domain of integration gets mapped to a rectangular domain? Say one is given the domain bounded by $x=y^2,x=3y^2,y=x^2,y=2x^2$.

Aladdin

11:56 PM

@TedShifrin hi.

johnny09

is a block diagonal matrix always nonsingular?

i think yes, as the matrix is still diagonal in some sense

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$Mathematics$ Mathematics