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12:48 AM
suppose we have the cubic polynomial function $y=f(x)=-0.06 x^3+0.9x^2+3x$ with $x\in[0,5]$ and $y\in[0,30]$
how can we find the inverse function?
I have tried the cardano method and the hyberbolic functions method
but both fail
as the coefficients violate the conditions (eg the discriminant is negative so no real root)
are there any other methods to find the inverse function?
1 hour later…
2:18 AM
Imagine if this was modern game advertising:
Q: Expected time to fuse a $D_{10}$ dragon

TedI play a game called DragonSky merge and idle (or something like this). The basic premise of the early game is that dragons will spawn, and will be fused in pairs. You continue to fuse these until you have a level $10$ dragon. Let me be precise. Let $\{D_1,D_2,\dots, D_{10}\}$ denote the set of ...

Formulate mathematical questions about your game to utilise the great SEO of the stack exchange network
3:34 AM
$$\sum_{n \in \Bbb Z} \operatorname{sinc}(n) = \pi$$
2 hours later…
5:15 AM
$\widehat{\delta_{x_0}}(s) = \exp(-2\pi i x_0 s)$
$\widehat{\delta_{x_0/2\pi}} + \widehat{\delta_{-x_0/2\pi}} = 2 \cos$
$\newcommand{\sgn}{\operatorname{sgn}}\newcommand{\sinc}{\operatorname{sinc}}$$\displaystyle \frac{4}{\pi} \sum_{n=0}^\infty (-1)^n \frac{\cos\left(\left(2n+1\right)x\right)}{2n+1} = \sgn(\sinc(x)\sinc(2x))$
$\sgn(\sinc(x)\sinc(2x)) = \displaystyle \mathcal F\left\{ \frac2\pi \sum_{n=0}^\infty (-1)^n \frac1{2n+1} \left( \delta_{(2n+1)/2\pi} + \delta_{-(2n+1)/2\pi} \right) \right\}$
$\displaystyle \sinc(x) = \dfrac{\sin(x)}{x} = \dfrac{\exp(ix) - \exp(-ix)}{2ix} = \pi \int_{-1/2\pi}^{1/2\pi} \exp(-2 \pi ixt) \ \mathrm dt = \mathcal F \left\{ \pi \chi_{\left[-\frac1{2\pi}, \frac1{2\pi}\right]} \right\}$
$\displaystyle \sinc(2x) = \frac12 2 \sinc(2x) = \frac12 \mathcal F \left\{ \pi \chi_{\left[-\frac1{2\pi}, \frac1{2\pi} \right]}\left( \frac12 t \right) \right\} = \frac12 \mathcal F \left\{ \pi \chi_{\left[-\frac1{\pi}, \frac1{\pi} \right]} \right\}$
5:33 AM
@LeakyNun What is sinc? hyperbolic sin?
$\chi_{[-1/2\pi, 1/2\pi]} \ast \chi_{[-1/\pi, 1/\pi]} = \begin{cases} 1/\pi & |x| \le 1/2\pi \\ 3/2\pi - |x| & 1/2\pi \le |x| \le 3/2\pi \\ 0 & |x| \ge 3/2\pi \end{cases}$
or circular sin?
@AbhasKumarSinha no, $\sinc(x) = \dfrac{\sin(x)}x$
@LeakyNun why?
> The term sinc /ˈsɪŋk/ was introduced by Philip M. Woodward in his 1952 paper "Information theory and inverse probability in telecommunication", in which he said the function "occurs so often in Fourier analysis and its applications that it does seem to merit some notation of its own",[2] and his 1953 book Probability and Information Theory, with Applications to Radar.[3][4]
5:35 AM
@LeakyNun oh okay... :)
@LeakyNun Do you know functional analysis?
not much
@LeakyNun hermitian operators?
there's the spectral theorem
For some operator $\hat f$ $$ \left < \phi_1 , \hat f \phi_2 \right> = \left < \phi_2, \hat f \phi_1 \right> $$
1. Is it true for all $a, b$? $$ \left < a, b \right> = \left < b, a \right> $$
Let $\displaystyle g = \frac2\pi \sum_{n=0}^\infty (-1)^n \frac1{2n+1} \left( \delta_{(2n+1)/2\pi} + \delta_{-(2n+1)/2\pi} \right)$
@AbhasKumarSinha no
so I think you got the definition wrong
5:40 AM
@LeakyNun no inner products are not commutative?
it should be $\langle \phi_1, \hat f \phi_2 \rangle = \langle \hat f \phi_1, \phi_2 \rangle$
they're anti-commutative
$\langle a,b \rangle = \overline{\langle b, a \rangle}$
@LeakyNun But, $$ \int a b dx = \int b a dx $$?
so $|\sinc(x) \sinc(2x)| = \mathcal F \left\{ g \ast \chi_{[-1/2\pi, 1/2\pi]} \ast \chi_{[-1/\pi, 1/\pi]} \right\}$
@AbhasKumarSinha well are you real or complex?
@LeakyNun complex
then the inner product is $\displaystyle \int a \overline{b}$
5:42 AM
@LeakyNun ugh!
correction: let $g = \displaystyle \sum_{n=0}^\infty (-1)^n \frac1{2n+1} \left( \delta_{(2n+1)/2\pi} + \delta_{-(2n+1)/2\pi} \right)$ and $|\sinc(x) \sinc(2x)| = \pi \mathcal F \left\{ g \ast \chi_{[-1/2\pi, 1/2\pi]} \ast \chi_{[-1/\pi, 1/\pi]} \right\}$
so $\displaystyle \int_{-\infty}^\infty |\sinc(x) \sinc(2x)| \ \mathrm dx = \pi (g \ast \chi_{[-1/2\pi, 1/2\pi]} \ast \chi_{[-1/\pi, 1/\pi]})(0) = \pi \left( \frac1\pi + \frac1\pi \right) = 2$
so $\displaystyle \int_0^\infty |\sinc(x) \sinc(2x)| \ \mathrm dx = 1$
@VladimirReshetnikov there you go
@LeakyNun Bro, how are dot products defined in complex numbers?
so on $\Bbb C^n$?
@LeakyNun yes
@LeakyNun ..?
@AbhasKumarSinha $\langle v, w \rangle = \displaystyle \sum_{i=1}^n v_i \overline{w_i}$
5:52 AM
@LeakyNun proof?
it's a definition
@LeakyNun does it works same with vectors?
@LeakyNun see $$ \left< a, b \right> = \int a.b^* \, dx $$?
that's the definition of inner product for functions
5:54 AM
$a, b$ are funcs
@LeakyNun then prove that $$ \int a^* (\hat f b) \, dx = \int b^* (\hat f a) \, dx $$ is always a real valued eigenoperator $\hat f$ for $a(x), b(x)$ as complex valued functions.
your question doesn't make sense
^The def of hermitian operators
@LeakyNun When $\hat f$ operator acts on functions $a, b$ it always gives real values
it doesn't
@LeakyNun strange....
I think I know what you mean to ask, but the way you say it makes no sense
so you misunderstood the statement / question
5:58 AM
@LeakyNun perhaps...
$$ \langle a, b \rangle \stackrel{def}{=} \int a^*.b \, dx $$ Is this true?
@LeakyNun ?
"def" means it is a definition
@LeakyNun yes
definitions are definitions
they are not true or false
@LeakyNun Then $$ \langle b, a \rangle = \int b^*.a \, dx$$?
6:04 AM
@LeakyNun For an operator $\hat f$ to be hermitian, $$ \langle a, \hat f b\rangle = \langle b, \hat f a \rangle $$?
that's an incomplete statement (missing quantifiers)
@LeakyNun what? then please complete it
you need $\langle a \hat f b \rangle = \langle b, \hat f a \rangle$ for all $a$ and $b$
@LeakyNun yes, $a, b$ are functions...
the point is "for all $a$ and $b$"
6:06 AM
@LeakyNun yes
$$ \int a^*. \hat f b \, dx = \int b^*. \hat f a \, dx $$ for all complex valued $a, b$ for an operator $\hat f$?
no, $\hat f$ needs to be hermitian (that's the whole point)
@LeakyNun Yes, sorry forget to mention that..
@LeakyNun $\hat f$ is hermitian...
then that's what you get by expanding the definition
Then, $$ \hat f a = \alpha a $$ for some constant $\alpha$, prove it is real!
that's the right question
6:10 AM
@LeakyNun yeaaaaaaaaah!! $\ddot \smile$
@LeakyNun proof? :P
ah you need $a \ne 0$ also
@LeakyNun yes yes
$\langle a, \hat f a \rangle = \langle \hat f a, a \rangle$ by hermitian
on the other hand, $\langle a, \hat f a \rangle = \overline{\langle \hat f a, a \rangle}$ by property of inner product
6:14 AM
so $\langle \hat f a, a \rangle$ is real
and $\langle \hat f a, a \rangle = \langle \alpha a, a \rangle = \alpha \langle a, a \rangle$
so $\alpha$ is real
@LeakyNun damn... I was integrating whole night for this!! Night Wasted
It's so easy!!?
hehhehe XD
@LeakyNun Can you help me to find some good references for the introduction of innerproduct of vectors and complex numbers on web?
In linear algebra, an inner product space is a vector space with an additional structure called an inner product. This additional structure associates each pair of vectors in the space with a scalar quantity known as the inner product of the vectors. Inner products allow the rigorous introduction of intuitive geometrical notions such as the length of a vector or the angle between two vectors. They also provide the means of defining orthogonality between vectors (zero inner product). Inner product spaces generalize Euclidean spaces (in which the inner product is the dot product, also known as the...
6:32 AM
it's not true
@LeakyNun now?
now it doesn't even make sense
@LeakyNun $\left < a, b \right> \in \mathbb{R}$
that's not true
@LeakyNun Inner products can be complex too?
6:34 AM
@LeakyNun sorry, I forgot what to ask...
@LeakyNun What is the graphical meaning of inner products for 2D complex plane?
so $\Bbb C$
@LeakyNun yes
so that's $\langle z, w \rangle = \overline{z} w$
so you reflect $z$ along the real axis
and then multiply by $w$
@LeakyNun ok
6:49 AM
so the magnitude is the product of the magnitudes (which agrees with the dot product of the two complex numbers viewed as elements of $\Bbb R^2$)
but now you have a complex number
the argument is the difference of the two arguments
@AbhasKumarSinha ^
7:43 AM
Q: equivalence relation over real number

maths student$ xRy$ iff $x^2 -y^2 = x-y$ is relation we have defined over $\mathbb{R} $ I have shown this is equivalence relation. Now we have asked to find equivalence class of 3 which can be found out to be -2 and 3 . Also next we need to find equivalence class of general x which turn out out to be 1-x...

8:18 AM
My calculations indicate that 25.8db is about 5% quieter than 26db. Am I correct?
9:06 AM
This question was posted a few days ago with a theorem I haven't seen before in finite group theory
has anyone come across this result anywhere?
3 hours later…
11:54 AM
$${\{a_j}\}_{j=1..n} \subset \mathbb N$$
$$n\,\varphi\Bigl(\sum^n_{j=1}a_j\Bigr) \leq \gcd({\{a_j}\}_{j=1..n})$$
Has to correct the previous time I put this up to put emphasis that uniqueness for ${\{a_j}\}_{j=1..n}$ is necessary
12:54 PM
What exactly is $\Bbb{Z}[\frac{1}{2}]$? Is it the polynomial ring $\Bbb{Z}[x]$ with $x = \frac{1}{2}$?
$\Bbb Z[x]/(2x - 1)$, yes.
Ring of rational numbers with denominator some power of $2$
Alternatively $\Bbb Z$ localized at the multiplicative subset $\{1, 2, 2^2, \cdots\}$
1:20 PM
Thanks @BalarkaSen
Suppose I have two matrices $A, B \in SL_2(\Bbb{Z}[\frac{1}{2}])$ that satisfy the relationship $BAB^{-1} = A^2$. Can I say that they generate a group isomorphic to the abstract group $\langle a,b \mid bab^{-1} = a^2 \rangle$?
You can't say a-priori that they don't satisfy any other relations. You need something like the ping-pong lemma.
What are your matrices?
$A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \\ \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 1 \\ 0 & 1 \\ \end{pmatrix}$
So showing they satisfy the same relationship isn't sufficient. Okay. That was my gut feeling. How exactly does ping pong help. I thought that concerned free groups.
Hm, I am not sure how to go about this
“Ping-pong lemma”
That’s a great name
From the relation $BA = A^2B$ we see that any word in $A$ and $B$ is actually of the form $A^k B^j$ for some integers $k,j$. So it suffices to check that this is never the identity.
1:34 PM
Meh, nevermind, nonsense.
One should be able to ping-pong this
$P \subset N \subset M$ are modules. If P is a direct summand in N, then is it also direct summand in M?
Why should it be?
Take P = N
Then you're asking if every submodule is a summand
Hm. Let $G = \langle A, B \rangle$. We know $H = \langle A \rangle$ is an infinite cyclic normal subgroup of $G$.
$G/H$ is obviously $\langle B \rangle$, which includes into $G$, giving a section, right?
So $G$ is a semidirect product of $H$ and $G/H$.
Yeah, because $G/H$ is a free - there is always a section
This forces the group to be $\langle a, b | bab^{-1} = a^k \rangle$ for some $k$. You already know the relator $bab^{-1} = a^2$ is in the presentation.
Fun pictures here: BS groups
Ffs, I hate when links break like that
1:47 PM
@MikeMiller's argument is much more down to earth though, but I think what I said works (right?)
Nicely, the first pic there is for B(1,2) which is the group of interest here
I am reading proof of a baby version of Universal Coefficient Theorem, in which $H_{n-1}$ is assumed to be free and we show that $H^n \approx Hom(H_n, \mathbb{Z})$. The isomorphism is given by the map $k([\phi])([c]) = \phi(c)$, where $\phi$ is a cocycle and $c$ a cycle.
Q: Solving a set of simultaneous equation for periodic boundary conditions.

Rajesh DachirajuGiven $$g(x) = e^{-|x-x_0|/\alpha} + Ae^{-x/\alpha} + Be^{x/\alpha}$$ $\alpha >0$, $x_0 \in (0,1)$ Constraints are $$g(0) = g(1)$$ and $$g'(0) = g'(1)$$ Solve for $A$ and $B$. My solution : $$A = \frac{e^{(x_0-1)/\alpha}}{1-e^{(-1/\alpha)}}$$ and $$B = \frac{e^{-x_0/\alpha}}{e^{1/\alpha}-1}$$...

$k$ takes a cohomology class $[\phi]$ and map it to the element in $Hom(H_n, \mathbb{Z})$, given by $[c] \mapsto \phi(c)$. It is well-defined, surjective.
I am trying to show that $k$ has kernel $0$.
Suppose $k([\phi]) = 0$. This means that $\phi(c)$ is zero for all cocycles $c$, ie. $\phi$ vanishes on kernel of $\partial: C^n \to Im \partial$. So, it factors through $Im \partial$.
What I mean is, it induces a map $\tilde \phi : Im \partial \to \mathbb{Z}$.
such that $\tilde\phi \circ \partial = \phi$.
Since $H_{n-1} = \ker\partial / Im \partial$ is free. $Im\partial$ is a summand in $\ker\partial$.
What I'd like to show is that $Im\partial$ is a summand in $C_{n-1}$.
Because then $\tilde\phi$ can be extended to $f : C_{n-1}\to \mathbb{Z}$. Then $\delta f(c) = f(\partial c) = \tilde\phi(\partial c) = \phi(c)$.
so $\phi = \delta f$
or $[\phi] = 0$.
6 mins ago, by feynhat
What I'd like to show is that $Im\partial$ is a summand in $C_{n-1}$.
2:23 PM
$\ker\partial$ is a direct summand in $C_{n-1}$.
because, $C_{n-1}/\ker\partial = Im\partial$ which is free, being a submodule of a free-module.
$Im\partial$ is summand in $\ker\partial$ is summand in $C_{n-1}$.
So, $Im\partial$ is summand in $C_{n-1}$.
Is this good?
Obviously, $C_n$'s are singular chain modules with coefficients in PID, $R$. I don't know why I wrote $\mathbb{Z}$ all over. Please read it as $R$.
3:03 PM
What is a formula for $\begin{pmatrix} 2 & 1 \\ 0 & 1 \\ \end{pmatrix}^i$? I thought I was able to discern a pattern for $i=1,2,3,4$; but after that the (1,2) entries blows up.
3:28 PM
@BalarkaSen Seems fine to me
My argument takes longer
I'd find a formula for the left column of B^k and then a formula for the top-left term of A^j B^k --- if A^j B^k = I, what is k? Then simply find a formula for A^j for all j.
Whence we get that none are the identity except j = k = 0
@feynhat Sorry, this will take too long for me to think about right now
though actually now that I look I do think we have explicitly $B^k = \begin{pmatrix}2^k & 2^k - 1 \\ 0 & 1\end{pmatrix}$ for all $k$
1 hour later…
4:40 PM
Well if I digress enough to reveal my personality I get banned so ill keep booking this up since I get told to only discuss math $${\{a_j}\}_{j=1..n} \subset \mathbb N$$
$$n\,\varphi\Bigl(\sum^n_{j=1}a_j\Bigr) \leq \gcd({\{a_j}\}_{j=1..n})$$
just a counter example and ill leave everyone alone
and that can be spun into a multidimensional argument no doubt which is popular with folk hmmm? …. hmmmm?
and then Iran hikes their petrol prices by 300 percent all the while plastic <---> petroleum
1 hour later…
6:07 PM
So, the powerset of a set equipped with the union operation qualifies as a monoid. However, there don't exist any rings with this monoid as their multiplicative structure, correct?
Hello, please I need help in this simple question, il must use the definition to prove that $\lim_{x\to 1} (x^3+2x^2-2} =1$ I started by
$|x^3-1+2x^2-2|\leq |x^3-1|+2|x^2-1|=|x^3-1|+2 |(x-1)| |(x+1)|=|x-1|(|x^2+x+1|+2|x+1|)$
how to find $\delta>0$ such that $|x-1|<\delta$
@Rithaniel trivial case: the null ring and the powerset of the empty set
Ah, yeah, fair enough. I still forget to rule out trivial cases
6:22 PM
$\mathbb{Z}/2\mathbb{Z}$ works too
someone help me on analysis ?
generally, such a ring would only have one invertible element (corresponding to the empty set)
so it must be of characteristic $2$
I suppose the set of a single element corresponds to the multiplicative structure on $\mathbb{Z}/2\mathbb{Z}$?
@PolineSandra I don't think a $\delta$ such that $|x-1|<\delta$ is what you want. You want a $\delta>0$ such that, if $|x-1|<\delta$, you can bound the RHS with a given $\varepsilon>0$.
So, any set union itself is itself, so every element in this ring is idempotent, so this is a boolean ring
6:28 PM
That checks out, cause $0^2=0$ and $1^2=1$
I'm not sure about examples with more elements
so you have stone's representation theorem
In mathematics, Stone's representation theorem for Boolean algebras states that every Boolean algebra is isomorphic to a certain field of sets. The theorem is fundamental to the deeper understanding of Boolean algebra that emerged in the first half of the 20th century. The theorem was first proved by Marshall H. Stone (1936). Stone was led to it by his study of the spectral theory of operators on a Hilbert space. == Stone spaces == Each Boolean algebra B has an associated topological space, denoted here S(B), called its Stone space. The points in S(B) are the ultrafilters on B, or equivalently...
given any compact Hausdorff space, its clopen sets form a Boolean algebra
and every Boolean algebra arises in this way
I was just reading up on that, though I'm not familiar with the result
But I think that gives you a contradiction with more than two elements
Yeah, if $A\bigcup B=S$ then $A\bigcup(A+B)=A\bigcup A+A\bigcup B=A+S=A$ because $S$ would have to be the "zero" of the ring, so $A+B=A$ and, by a symmetric argument, $A+B=B$
so given a set X, the power set of X can be equipped with a ring structure that makes multiplication the union
by having symmetric difference as addition
and the whole set the zero
the empty set the one
hi @Semiclassical
6:39 PM
@Thorgott yes that is it
how to do please
to prove using the definition that $\lim_{x\to 1} (x^3+2x^2-1)=2$
@Semiclassical what's the eigenvalue of a symbol?
atiyah macdonald!
In part ii) shouldn't it be under the assumptions that \mathfrak q is primary?
Yes it is!
"Let $\mathfrak q$ be a $\mathfrak p$-primary ideal"
6:43 PM
Yes but if q is not primary then there are easy conterexamples in the ring of integers
"Let $\mathfrak q$ be a $\mathfrak p$-primary ideal" means that $\mathfrak q$ is primary
Oh I didn't see that assumption in the book so just to confirm. Thank you;)
it's literally the first sentence
@LeakyNun what
@Semiclassical of an parabolic PDE
6:45 PM
ah. dunno
@PolineSandra try factorizing $x^3+2x^2-3$
7:02 PM
How does one integrate a function in a complicated open set. For instance, I have been given $f(x,y) = \frac{1}{(y+1)^2}$ and have been asked to integrate this in the region $A(x, y ) = \{ (x,y) | x >0 , x^2 < y < 2x^2\}$.

Theoretically I need to find a nested sequence of open sets sets $\{U_n\}$ such that $\int_{U_{n}} f$ is bounded. But how do I find such sets here
Also I need to use fubini's theorem but again I need to give proper limits. And I dont get how to do that for a relatively complicated open set that needs to be taken given my set $A$
7:21 PM
Since $f(x,y)$ doesn't depend on $x$, it seems reasonable to just cut off large values of $x$ to achieve such a sequence
how is $\int_{U_n} f$ defined? this seems a little circular to me
$\int_{\mathbb{R}^2}1_{U_n}f$, id say
@LeakyNun you define the integral in terms of compact rectifiable sets.
@Thorgott Could you elaborate on this?
compact sets can't be open
take $A_n=\{(x,y)|0<x<n,x^2<y<2x^2\}$, then $A_n\uparrow A$
7:28 PM
@LeakyNun Yes but using the compact set definition you can arrive at this one.
@LeakyNun what about the empty set
@Thorgott don't be that guy
@Thorgott I see. Cool. That's an obvious choice
i had to
@Thorgott though if I take the limits to be $0$ and $n$ for $x$ in the integral, you arrive at having to evaluate $1/x$ at $0$ which is a problem.
7:35 PM
Huh? @Sayan
where does that term come from?
Oh never mind. I am making calculation errors
There's a $1+$ in there.
Yeah, I didn't notice that. I should sleep lol
Mathing without sleep isn't good.
7:39 PM
Yeah, especially if its analysis
7:57 PM
This is weird. If I take the same $A_n$ but instead of $x^2 < y <2x^2$ I take $x < y <2x$ (i.e change the region to $\{ x>0 , x<y<2x\}$) the integral $\int_{A_{n}} f$ is bounded for all $n$. Specifically it is $\log(\frac{n+1}{2n+1})$. But the text says that $f$ is not integrable in the changed region.
Well it's not exactly that but the integral is bounded for all $n$
It should be $\log\left(\frac{n+1}{\sqrt{2n+1}}\right)$
That's not bounded
Oh I am messing up big time. Shit. I have to sleep
8:13 PM
Q: How to evaluate $\lim_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)$

didgocksI have to find the limit of following $$\lim_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)$$ I have no idea how to start this one off. How would I do it? Do I just substitute the $0$? It doesn't look that easy and simple. The answer says it's negative infinity. Please show me a solution w...

Do you guys see anything wrong with my answer here?
8:56 PM
Let $A \in \mathbb{Z}+\cup$ $\{$ $0$ $\}$ and $B \in \mathbb{Z}^+$ then there exists $C\in \mathbb{Z}$^+ $\cup$ $\{$ 0$\}$ with $C\leq A$ and $P\in \mathbb{Q} \cup [0,1)$ so that $A=BC+BP$

Is this statement true?
In simpler terms: Suppose $A,B$ are nonnegative integers with $B\neq 0$. Then there exists a nonnegative integer $C\leq A$ such $C\leq A/B<C+1$.
How'd you prove it though?
actually. should your initial statement have had $\mathbb{Q}\cup [0,1)$ or $\mathbb{Q}\cap[0,1)$?
Break it up into cases. If $A<B$ and if $B<A$
9:03 PM
okay. I want to sat that's just the archimedian principle?
(and if $A=B$)
$A=B$ should be very easy, though
Well, I shouldn't say that. All three are easy, if you already know the answer
The version I wrote can be summed up as this: For every nonnegative rational number, there is some smallest integer which is bigger than it.
9:34 PM
@Semiclassic: Why are you saying $C\le A$?
45 mins ago, by topologicalmagician
Let $A \in \mathbb{Z}+\cup$ $\{$ $0$ $\}$ and $B \in \mathbb{Z}^+$ then there exists $C\in \mathbb{Z}$^+ $\cup$ $\{$ 0$\}$ with $C\leq A$ and $P\in \mathbb{Q} \cup [0,1)$ so that $A=BC+BP$

Is this statement true?
@TedShifrin because it was in there
oh ...
@k170 I write such proofs very differently. I would also try to make it all about positive numbers, instead of negative ones, because inequalities are a pain with negatives. In particular, since $|x-1|>3/4$, I think your $5/4$ should be $3/4$ in your final stuff.
10:10 PM
@TedShifrin Thanks for the feedback. I agree that these edits would make it smoother to read.
But I think you did mess up your inequalities, getting the $5/4$.
I didn't take the time to check line by line.
Are you referring to this part?
$$-\frac54\lt x-1\lt -\frac34$$
No, that part is fine. It's when you're bounding $(x-1)/x^2$.
Note that $3/4<|x-1|<5/4$. So $|x-1|>3/4$.
We're doing an > argument, not an < argument.
Ahhh Yes that makes sense
Thanks again. I'll edit my post
If I may offer a more concrete suggestion, I prefer to start by rearranging $|f(x)-\ell|$, then see $|x-a|$ appear and make my preliminary assumption (in your case $|x|<1/4$), and then work out the final requirement for $\delta$ to make $|f(x)-\ell|<\epsilon$ (or, in your case, $>N$).
10:25 PM
why is the content of the chat room considerably less impressive and or helpful compared to responses for questions I post? Like how come only a tiny selection of users use the chat feature to commutate almost entirely exclusively with one another?
1 hour later…
11:34 PM
Can anyone explain how answer key got the arctan value here? Im super confused

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