12:06 AM
okay I see !
I think I understand now
@MatheinBoulomenos @LeakyNun @TedShifrin thank you guys! all the methods work
drawing the picture was also good

great

seems like a black hole

you'll see the magical 5 appear out of nowhere

yes
the points cant escape haha
am so used to doing math by a general approch that works
so I find it difficult when they don't tell us that things don't work this way in general
like when you have to find a homomorphism between groups or rings
@LeakyNun in the solution of mathein, do you know how to go from line to line?
seems like alot of steps missing

@Jacksoja do you know the third isomorphism theorem?

12:12 AM
they number them differently
but I know the isomorphism theorems
@LeakyNun I mean from ( x^2+1 , 2+x ) to ( 2+x , 5) ideals
and (2+x , 5) to (5)

(x^2+1, x+2) = (x^2+1-(x+2)(x-2), x+2) = (5, x+2)
it's how you simplify ideals

we never did this

you use a sort of Euclidean algorithm
like row operations to a matrix
add a multiple of a row to another row

okay but last question
from (2+i) why we have ( x^2+1 ,2+x)
am sorry wait not like that
Z[i] / ( 2+i)

Z[i]/(2+i) = (Z[x]/(x^2+1))/((2+x)/(x^2+1)) = Z[x]/(x^2+1,2+x)

12:16 AM
and Z[x] / ( x^2+1 ,2+x)

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?

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

@LeakyNun where did 2+i dissappear?
in the double quotient?

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

@Jacksoja it became 2+x

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

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

1 hour later…
2:46 AM
0

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 ... 3:34 AM On yucky mathematics and mathematical quagmires: Consider the following equation: 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 If a ring isn't a unique factorization domain then it isn't euclidean? Is this true 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 @Secret can you clear the domain thing? 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. 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}$4:00 AM Well, here's an identity for that$\frac{1}{1+e^z}=\frac{e^{-z}}{1+e^{-z}}$@Sonal_sqrt if you can write anything in terms of prime factors in the ring or you have zero divisors, you cannot possibly find a euclidean division algorithm in it and another:$\frac{1}{1+e^z}=\frac{1}{2+\frac{2}{\text{coth}(z/2)-1}}$the second one looks a lot more "yucky" (subjectively) than the original, yet the existence of that identity means the expression on the right cannot have a very small$Y$value, hmmm or$\frac{1}{1+e^z}=\frac{d}{dz}(z-\text{log}(1+e^z))$yeah you are right, just the number of identities alone can increase rapidly, further making$Y$not correspond to what we think as "yucky" 4:06 AM Bingo In fact, your examples showed something more: There are expressions that are written in very complicated and nested way, which can be rewritten into a very simple form because the symmetries are just right for the rewriting rules applied to cascade it into a simple expression Yeah, forms are very often equivalent. I've actually graded differential equations quizzes where two people got seemingly wildly different answers to a problem, but closer inspection revealed that they were actually one and the same. Stuff can catch you off-guard sometimes 4:22 AM Another identity:$\frac{1}{1+e^z}=\frac{1}{4}\int_1^2\int_{-2x}^1\frac{1}{1+e^z}dtdx$that last one is soooooooooo lame, lol (Now, that's just a way to represent multiplication by one, yeah) Sometimes, I wonder whether algebraic identities are really n-tuple operators e.g. in groups, the binary operator$\cdot$takes in two elements and spits out one You might be able to interpret them as such, but you'd have to be very careful in how you define things So something like:$\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, +$in the structure 4:31 AM Ah, I see what you mean. So any function induces a type of algebraic object. You could have the only operation be the$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}$you want associative wrt$f$, or wrt$a,b$? are you thinking about something like$f (f (a,b),c)$or just$f(a,b)*c$? Well,$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))$ah, I don't think there is a simplication of this for general$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$at some abstract line of symmetry thus there will be fixed points there. I forgot the geometric counterpart to associativity, need to check 3 hours later… 7:08 AM Hello 7:26 AM What is complex conjugation on the Riemann sphere? What happens to$\infty$? 8:13 AM I guess it doesn't move? hmm annoying little prick hahaha the fact that it just destroys everything is making finding its pre-image under a transformation annoying So did they trick you into doing complex analysis by selling it as number theory? lol no this is in modular forms, just recapping möbius transformations Ah, makes sense in particular, I have the inverse of the relevant transformation but stuffing$\infty$into it gives the wrong pre-image :( unless I pretend$\infty/\infty = 1$which actually seems to be quite common in this context lol 8:27 AM So,$\infty/\infty+\infty/\infty=2$? sure why the hell not Hmmm, and what about$\frac{\infty}{\infty+\infty}$? that's 1 obviously hehe I would have guessed that$\frac{\infty}{\infty+\infty}=\text{stop}$2 rofl 8:40 AM if you're trying to plug infty into (az+b)/(cz+d) you're really asking for the limit of this as z -> infty and that's made more obvious either by saying lhopitals or rewriting as (a+b/z)/(c+d/z) yeah I realised this and the problem has now stopped being a problem lol thanks :P Anyone familiar enough with graph theory? Trying to prove something with graph theory and induction and Im not sure if my intuition is correct 2 hours later… 10:40 AM apologies if this is trivial: if$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$have? it looks like none.. is that right? 10:59 AM cancel that question please 1 hour later… 12:18 PM @krauser126 “Just ask, don’t ask to ask.” 12:45 PM @MJD Since you're the user who created (or at least started) the MathJax tutorial on meta, maybe you' would have some comments on this recent suggestion on meta: Organizing The MathJax Tutorial. (Basically, the post suggest to add some kind of ToC to the post.) If needed, we can discuss this further in the Math Meta Chat. (This seems to be close on-topic in that room - I used this room simply because it was a place where I was able to ping you.) 1:29 PM$f(a,f(b,a))f(f(a,b),a)$1:51 PM is that notation functional? nah, just a two variable function nested twice Trying to understand what happens if they are equal how are you pal? as usual @MartinSleziak Thanks for the ping. I left a comment in Meta. I think it's a great suggestion. Over the years I have zealously defended the main post against all sorts of additions. But I think this would be an addition worth having. OMG! a talking potato :O 2:02 PM Where?!!? OMG I'M A TALKING POTATO 3 Thanks for the response! @martin: I value your judgement. Do you see any drawbacks that I might have overlooked? I did not interact with the MathJax tutorial on this site too much, so I don't really have strong opinion on this. Since this was mentioned in Math Meta Chat, perhaps you can ask for additional feedback there. Maybe some of the users who visit that room responds. (Although I am not sure that much feedback is needed here.) The only thing that came to my mind was whether the OP is actually allowed to edit the post - considering that they have relatively low reputation. However, for CW-post this should not be on problem. (They would not be able to edit regular posts on meta.) 2:45 PM What are some suitable change of of variables to show that 1/(x+y) is a hyperbolic paraboloid, i.e. it is of the form$z=x^2/2-y^2/2$? 2 hours later… 4:18 PM hello what is the best method to open complement using demorgans law ( in boolean algebra ) inside- out or from outside to inside 1 hour later… 5:30 PM @Noob not sure what you mean by inside -out 5:54 PM @ShineOnYouCrazyDiamond hello :) Hey guys! Will the function$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$. 6:35 PM No. That only works if f(x)=x^2 for nonnegative x. If two functions are different for nonnegative x, then they’ll yield different f_e. Oh, I see. So what is the question asking us to really do here? I am not entirely sure. Think about function composition, maybe. Can you think of a function such that f(x)=x when x>0? 6:53 PM a function such as$f(x) = |x|$since we can split it into two cases? 7:04 PM |x| does seem like a good candidate, yes. Now, the question wants$f_e(x)=f(x)$. How then would |x| play into this? 7:14 PM Wouldn't$|x|$satisfy$f_{e}(x)=f(x)$since it will always produce$x≥0$? hello, a small question please$f: R\setminus\{0\}\to \mathbb{R}\\ x\to \frac{1}{x}$f is a function or a map ? Abwatts: what if$f=x\text{sin}(x)$or some other outlandish function. Then what is$f_e$? Is it |x|? No because$f=xsin(x)$is a bounded function? So, what would$f_e$be? (Also,$x\text{sin}(x)$isn't bounded, but that's a nonsequitur, here.) Remember the initial hint: think about function composition. 7:33 PM Umm then$f_{e}$would be$f_{e}(f(x))$? since$f_{e}$will directly depend on the structure of$f(x)$? Well, you still haven't said what$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$So wouldn't$f_{e}$be$x^2$because it will always spit out positive numbers? But when does$x^2=x\text{sin}(x)$? But wouldn't it equal to$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$? 7:54 PM I came to know about complex numbers of the from x+yj, where j^2=1 In the plane of such complex numbers, what would be the equation of the circle with center at 0? The unit hyperbola is given by ||z||=1 But I couldn't find a formula for circle with center at 0. On average, what's the population density of a car (measured in people per square mile) Well, there is a bit to unpack, there. First,$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)$@Abwatts the point is to define$f_e$for an arbitrary$f$(in terms of the arbitrary$f$) 8:10 PM (Well$f^2$might equal$f$for a specific$f$, but not in general) Oh, I see. This question is fairly tricky. a hint is to think about the graph. specifically, what does a function being even mean to the graph of the function? 8:28 PM it has symmetry with respect to the y axis so can you rephrase the question in terms of this interpretation using graphs? 8:50 PM Yes, I will try doing that. Thanks. 9:15 PM I have another question about increasing functions if you guys don't mind. Is the following conclusion that I made about this statement is true? This is the statement: If functions$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)$. Would this be a valid approach? 9:31 PM 9:53 PM$f(x)=\frac{1}{f(x)}$How do I solve this equation given$f(f(x))=x$nvm 10:23 PM @Ultradark hey @ShineOnYouCrazyDiamond hey @shi worked 10:42 PM @Random-15 No, you're computing$|z|$incorrectly. That is, in fact, the unit circle centered at$0$. Remember that if$z=x+yj$, then$|z|^2 = x^2+y^2 = z\bar z$. 10:54 PM Hello!! Is someone of you familiar with Turing machines? I have the following question: 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\$.

11:56 PM
@TedShifrin hi.

is a block diagonal matrix always nonsingular?
i think yes, as the matrix is still diagonal in some sense