Mathematics - 2022-01-30

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LearningCHelpMeV2

12:13 AM

@leslietownes Thanks for the advice

0

Q: Growth rate of a sequence that generalizes the primes?

Let a strictly increasing sequence $A_2$ be defined as starting with $a_1 = 2$ and containing all integers larger than the ones already in the sequence unless $$ x = a_{i_1}*a_{i_2} + a_{i_3}*a_{i_4} $$ for some $i_1,i_2,i_3,i_4$ Or said differently $$ a_{n+1} = a_n + 1 $$ unless $$ a_n + 1 = a_{...

sequences-and-series number-theory asymptotics recursive-algorithms sieve-theory

normally i say hi , but i got my first downvote this year :p

12:28 AM

goodnight

monoidaltransform

12:43 AM

in a spherically symmetric object with radius $r$, if a mass moves around the object

does that mean $dr/dt$ is constant?

12:54 AM

@LearningCHelpMeV2 Make an outline for yourself of the key ingredients/steps of the proof.

@monoidaltransform Your question makes no sense. Edit?

What does $r$ have to with the moving mass?

1:21 AM

why aren't there shovelers at the duck pond?

I give up.

1:50 AM

Hello please I need some x such that [nx]-n[x]=n-1

Where [nx] is the floor function of nx

With $n\geq 1$

try n = 2, x = 3/2

2:16 AM

Ah right it works

Thank you very much

Can we find an example of x writing by n ?

try n = n, x = (n+1)/n

there are whole intervals of x-values that will work for a given n and that example is the starting point of one of them

No it gives me 1

Not n-1

[n+1/n]=1

I don't find an example with x writing in terms of n

2:33 AM

i'm not sure i understand the goal

probably i don't

The question is to find max of this set $A_n=\{[nx]-n[x], x\in \mathbb{R}\}$

So to say that is n-1

I have to find $x$ such that [nx]-n[x] =n-1

Leslie, keep quiet. Vrouvrou, think about trying different values of $x$. Even less than $1$.

Work on it with your brain.

thankfully it's too late to edit my spoiler

A spoilt spoiler gathers no moss.

there's two schools of thought about the spoiler. in one the spoiler is a magic End This Now card. in the other the spoiler is a french fry for the seagulls at the duck pond.

i think the first school of thought is wrong.

2:45 AM

Birds do not need french fries or bread.

of course they don't. it was more, one school of thought is the spoiler ends the conversation, the other is, it's the beginning of a conversation with 50 more birds demanding the french fry.

Send them leeks.

we try not to go to the duck pond too late in the day because that's when people feed the birds, it's annoying.

there's one guy who has really figured out how to feed the herons, though. he gets all of the herons in the park eating his stuff, and none of the other birds seem interested.

he shouldn't be doing that, but i'd love to know how he is doing that

Found a new non-floral gin!

say more..

2:47 AM

In all my choices the n disappear

Four Pillars Olive Leaf

@Vrouvrou make better choices

try 1 minus a little bit.

ted: is it available at any of the usual suspects, or do i need to know a guy.

I got it at Total Wine just opened a mile from me.

@leslietownes You talk too much. This person gets us to do everything .

bevmo near me has a four pillars dry but not the olive leaf. dry sounds suboptimal because i tend not to mix it and that's usually what dry is for.

ted: you cannot spoil the unspoiled.

Garbage.

Go for Olive Leaf. That was my point.

2:53 AM

i wonder if whole foods has it.

I don’t do liquor there.

i need to go to my office in the next month. i bet they have it in costa mesa.

Haven’t checked.

i haven't been that far south since june '21.

I think that x=n-1/n works ?

2:57 AM

Nope, Hi Time has no Four Pillars.

I suggested $x<1$. Why?

[X]=0

Yes

But n disappear

Come on. Work.

X=n-1/n

3:05 AM

Parentheses?

(n-1)/n

OK. So?

[n x]=n-1 and n[x]=0

Perfect.

Thank you professor

3:07 AM

Pas de quoi.

2 hours later…

PenAndPaperMathematics

5:27 AM

1

A: Encoding Turing machine-like behavior using families of sequences of vector spaces and modules. P=NP related

While loops with increasing variables are enough for Turing completeness. For instance, it is enough to implement cellular automata, and cellular automata are Turing-complete. While loops with increasing variables are also enough to implement a Turing machine: you have an outer while loop that ...

5 hours later…

10:22 AM

$\lim_{n\to \infty} \sum_{r=0}^n \binom{2n}{r}\frac 1{4^n}$

how to evaluate this limit?

Is there any approximation to binomials like Sterling's formula?

That's almost half of $\frac{1}{4^n}\sum_{r=0}^{2n} {2n\choose r}$ isn't it?

@Koro the limit is $\frac12$

Actually, the expression equals $\frac 1{4^n}.(1+1)^{2n}-\sum_{r=n+1}^{2n}\binom{2n}{r}\frac 1{4^n}$, the second term is giving me problem.

@Koro The sum is $\frac12+\frac{\binom{2n}{n}}{2\cdot4^n}$

@robjohn right :).

Professor Rob, how?

10:29 AM

$\sum\limits_{r=0}^{2n}\binom{2n}{r}=4^n$

${n\choose k} = {n\choose n-k}$

yes

and $\binom{2n}{r}=\binom{2n}{2n-r}$

$\sum\limits_{r=0}^n\binom{2n}{r}+\sum\limits_{r=0}^n\binom{2n}{2n-r}=4^n+\binom{2n}{n}$

both sums are the same

the extra $\binom{2n}{n}$ is because it is counted twice in the two sums

$\sum\limits_{r=0}^n\binom{2n}{r}=\frac12\left(4^n+\binom{2n}{n}\right)$

divide by $4^n$

Then you can use this answer if you need more

But we have $\frac 1{4^n} \sum_{k=0}^{n}\binom{2n}r=\frac 1{4^n}(\sum_{r=0}^{2n}\binom{2n}{r}-\color{red}{\sum_{r=n+1}^{2n}\binom{2n}{r}})=\frac 1{4^n}((1+1)^{2n}-\color{red}{\sum_{r=n+1}^{2n}\binom{2n}{r}})$

@robjohn looking

@Koro why is this a problem?

10:37 AM

The red colored term is problematic and $\binom {2n}{r}=\binom{2n}{2n-r}$ doesn't seem to simply that either.

$\sum\limits_{r=n+1}^{2n}\binom{2n}{r}=\sum\limits_{r=0}^{n-1}\binom{2n}{r}=\sum\limits_{r=0}^{n}\binom{2n}{r}-\binom{2n}{n}$

@robjohn yes, of course. Thanks a lot :).

I'm getting dumber day by day.

it feels as if you are thinking too hard. you need to relax a bit. Don't focus so narrowly.

yeah, thanks a lot professor Robjohn :).

I messed up because I did: $r=n+1+m$ so when $r=n+1, m=0$ and when $r=2n, m=n-1$. $\sum_{r=n+1}^{2n}\binom{2n}{r}=\sum_{m=0}^{m=n-1}\binom {2n}{n+1+m}$.

This seems wrong but I'm not able to pinpoint mistake in it.

I got it :).

11:54 AM

@robjohn you have the title of a professor?

2 hours later…

2:00 PM

Hello.

Understandable have a great day.

How do I prove that composite numbers are product of prime?

monoidaltransform

@WilliamJohn Fundamental theorem of arithmetic

2:19 PM

@monoidaltransform Thanks.

It seems that Internet lack resource... or I don't have enough knowledge to understand the proof from Wikipedia...

2:41 PM

Ah it seems that this is proof by contradiction.

If composite number is not product of prime then it will be composite number and it will contradict the fact that there isn't smallest composite number.

Is it true that in any Integral domain ideal generated by an irreducible element is maximal ? I know that in a PID ideals are maximal iff they are generated by irreducible element.

Sorry to include recursively.

3:00 PM

@RandomVariable: I have checked out the fix that Davide Cervone suggests for the Responses and Votes tabs, and they work. Hopefully, the devs will implement that fix soon, as well.

1 hour later…

4:29 PM

@cabmetric take x in k[x,y]

@WilliamJohn the key idea behind induction is that every nonempty set of natural numbers has a smallest element

so I can take the set of composite numbers that canNOT be expressed as a product of prime numbers

if we assume for contradiction that some composite number is not a product of prime numbers, then the set I just described would be nonempty

so we can take the smallest composite number $n$ not expressible as product of primes

but then it would have to be the product of two smaller composite numbers, i.e. $n = ab$

since the number we took is the smallest one, these two composite numbers $a$ and $b$ must be expressible as product of primes

but if $a = p_1 \cdots p_r$ and $b = q_1 \cdots q_s$ then $n = p_1 \cdots p_r q_1 \cdots q_s$ is a product of primes

which is the contradiction

@WilliamJohn saying that this is a proof by contradiction is not wrong, but the contradiction isn't the main idea; the induction (i.e. taking the smallest such number) is the key idea

5:02 PM

@LeakyNun Sorry for late reply I will read it tomorrow. Thank you very much for taking effort to reply :).

@LeakyNun Ah this is same idea I thought but may be I couldn't describe it properly mathematically.

Mine thought process was: If $N$ is composite number then the there exist smallest divisor $o$ of $N$ such that it is a prime number. If not then there exist $l$ such that $1<l<o$ and $l$ is divisor of $O$ wich implies it is also divisor of $N$ which contradicts the fact that $o$ is smallest divisor.

5:27 PM

If H and K are subgroups of G of finite index and G can be infinite also. It is given that [G:H]=m, [G:K]=n, then how to show that $lcm (m,n)\le [G:H\cap K]$?

koro: can you show that both [G:H] and [G:K] divide [G:H cap K] ?

G can be infinite also is what scares me.

I wanted to use Lagrange's theorem.

@leslietownes trying

OK. the same set of ideas may be floating around the proof that H cap K is also a finite index subgroup of G (which turns out to be true under your hypotheses)

anyway, once you know that, it's just integer arithmetic. if m and n divide c then lcm(m,n) is not only <= c but a divisor of c.

"just integer arithmetic," he says, as if integer arithmetic isn't difficult

5:51 PM

Okay. So $[G:K\cap H]=[G:H][H: H\cap K]$ and also $[G:K\cap H]=[G:K][K:H\cap K]$

yes! and everything in those formulas is finite under your hypotheses even if G is not

It's called tower law for subgroups. I don't understand why $[H: H\cap K]$ is finite.

Is subgroup of a finite index subgroup H also of finite index?

that proof does not explain why [H:K] is finite. it says it's true "by hypothesis" but it isn't a hypothesis

it's true, and it's basically true by the proof, if the reasoning were made more explicit

i'd find a proof that didn't format the whole thing as a series of set equalities

the "set-theoretic version" at groupprops.subwiki.org/wiki/Index_is_multiplicative at least gives it a little more conceptually although note the use of G/H for the coset space (not necessarily quotient group)

although it does not include a proof

It's been assumed in hypothesis that $[H:K]$ is finite.

Leslie, in your link as well.

But in my case, that's not a problem :).

oh, well, the proof idea works.

6:06 PM

Because if we take any coset of $H\cap K$, say $x(H\cap K)$ then for any $y\in x(H\cap K)$ there exists an $m\in H\cap K$ such that $y=xm$ so $y\in xH$ and $y\in xK$. It follows that $y\in xH\cap xK$, which means that $x(H\cap K)\subset xH\cap xK$.So number of cosets of H cap K are at most number of cosets of H.

the "set theoretic version" gives you the idea and the bijection behind the set arithmetic in the other proof.

a good algebra textbook would present this more clearly than a sequence of web pages, although a lot of the ones i can think of would not have considered the case of infinite G but subgroups of finite index.

even if the proof they wrote down would work in that case.

Leslie: That bijection method looks very confusing. It feels odd to write G/H, when H is not normal in G.

But I understand that they mean set of all left cosets of H in G, by G/H.

write LeftCosetSpace(G,H) then.

i mean, maybe it's goofy notation, but i think the idea embedded in that thought is more valuable than elementwise arithmetic with coset representatives.

at least in terms of generalization and applicability to other situations. maybe for this problem is it overkill

i find it easier to remember stuff like that than elementwise computation even if it is all ultimately the same. maybe just a personal choice

6:21 PM

$$ \left(
\begin{array}{ccccc}
0 & 1 & 0 & 0 & 0 \\
\frac{1}{12} & \frac{11}{12} & 0 & 0 & 0 \\
\frac{1}{3} & 0 & \frac{1}{3} & \frac{1}{3} & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 \\
\end{array}
\right)
$$

Hi everyone! This is a transition matrix of a markov chain. I'd like to calculate the asymptotic probability of transition from the state 3 to the state 1, so I need to obtain the limiting distribution, but it seems that this markov chain is not irreducible. Is there some way around it?

6:34 PM

is it clear that there must be a limiting distribution? is there a way of computing 'asymptotic probability' in the absence of one? i understand the notions but am not up on this terminology.

i note that the vectors (0,0,1/5,2/5,2/5) and (2/5,2/5,1/5,0,0) both have nonnegative entries summing to 1 and are both fixed by that matrix.

Good evening, any tricks to find the rank of a matrix other than counting the pivots in rref form.

6:53 PM

powers of that matrix limit to alternating between $\left(
\begin{array}{ccccc}
\frac{1}{13} & \frac{12}{13} & 0 & 0 & 0 \\
\frac{1}{13} & \frac{12}{13} & 0 & 0 & 0 \\
\frac{1}{26} & \frac{6}{13} & 0 & \frac{3}{8} & \frac{1}{8} \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 \\
\end{array}
\right)$ and $\left(
\begin{array}{ccccc}
\frac{1}{13} & \frac{12}{13} & 0 & 0 & 0 \\
\frac{1}{13} & \frac{12}{13} & 0 & 0 & 0 \\
\frac{1}{26} & \frac{6}{13} & 0 & \frac{1}{8} & \frac{3}{8} \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\

cro: if you are thinking about hand computation, once you get to ref, you know the number of pivots (so if that is all you care about, no need to proceed to rref).

cro: there may be other things that work for specific types of matrices but in general that is all i can think of.

The results of states 1, 2, and 3 seem stable

hmm, that's true $P^1 = P^4$

cro: in theoretical stuff often the rank of a matrix is reflective of some other 'thing' of interest, which can sometimes avoid matrix calculation entirely. but if someone randomly gave me a bag of numbers and put them into a matrix, i'd start row reducing and stop when i knew where the pivots were going to be (e.g. no need to normalize pivots to be 1)

even though for the 3rd, 4th power it's not the case, but for the large n it is indeed! Does it mean that for $n \to \infty$ the $p^n_{31} = 1/3$?

7:00 PM

@WojciechKulma I'd say that if you start in state 3, there is a $\frac1{26}$ chance you'll end in state 1.

is that what you're trying to compute?

yes, exactly, but I just calculated that $P^n = P$ for n very big in mathematica

that's why I would assume, $n \to \infty$ the $p^n_{31} = 1/3$

@WojciechKulma why is that?

3rd row, 1st column, no?

it's $\frac1{26}$ in the limiting matrices I posted.

how did you calculate it?

7:06 PM

@WojciechKulma Do you know the Jordan Decomposition?

Mhm I think I knew it, let me check

but it's strange that for n=50000000000000 mathematica says $P^n = P$

$\left(
\begin{array}{ccccc}
0 & 1 & 0 & 0 & 0 \\
\frac{1}{12} & \frac{11}{12} & 0 & 0 & 0 \\
\frac{1}{3} & 0 & \frac{1}{3} & \frac{1}{3} & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 \\
\end{array}
\right)=P\left(
\begin{array}{ccccc}
-1 & 0 & 0 & 0 & 0 \\
0 & -\frac{1}{12} & 0 & 0 & 0 \\
0 & 0 & \frac{1}{3} & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
\end{array}
\right)P^{-1}$

are these the eigenvalues?

on the diagonal?

They are

now, raise that to a high power

ah it's singular value decomposition!

I know it thanks :)

Akiva Weinberger

7:11 PM

If you apply a circular blur to a complex differentiable function, must it become constant?

$P\left(
\begin{array}{ccccc}
(-1)^n & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
\end{array}
\right)P^{-1}$

which gives the matrices I wrote above

Akiva Weinberger

Like I'm imagining spinning the function really fast and looking at the motion blur… formally this is $f(z)\mapsto\int_0^{2\pi}f(e^{i\theta}z)d\theta$

many thanks @robjohn I was just really confused by the result from the mathematica!!

akiva: well, you'll get something that's constant on circles. you'll get something like the average value over the circle (maybe not normalized).

Akiva Weinberger

Oh yeah I should divide by $2\pi$. I think it should be constant everywhere if $f$ is holomorphic

and my reasoning is, this operation should preserve the complex-differentiability

7:15 PM

oh, now you tell me.

Akiva Weinberger

and if you're constant on circles, and your real and imaginary parts are harmonic (every point is the average of its neighbors), you gotta be constant everywhere

yeah, i didn't see 'complex differentiable' above.

@robjohn sorry, just to make sure, are your eigenvalues in the right order, should it be $\left\{-1,1,1,\frac{1}{3},-\frac{1}{12}\right\}$ ?

Akiva Weinberger

So this basically reconstructs the theorem that the value at a point is the average of the values on a circle centered on that point

through motion blur

which is nice

@WojciechKulma how do you specify an order to the eigenvalues?

7:18 PM

uh no ofc it was a nonsense! I'm really sorry, I confused it with the order of eigenvectors mathing th ecorresponding eigenvectors

it all makes sense now, many thanks once more!

@WojciechKulma: my $P$ is not what you are calling $P$. My $P$ is a matrix whose columns are eigenvectors.

I figured it out, thanks :)

@AkivaWeinberger the integral over each circle is $2\pi$ times $f(0)$, so yes, the values on each circle are the same.

this is your P:

$$\left(
\begin{array}{ccccc}
0 & -1 & 2 & 0 & -\frac{5}{4} \\
0 & -1 & 2 & 0 & \frac{5}{48} \\
\frac{1}{4} & 0 & 1 & 1 & 1 \\
-1 & 1 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 & 0 \\
\end{array}
\right)$$

$P=\left(
\begin{array}{ccccc}
0 & -\frac{5}{4} & 0 & -1 & 2 \\
0 & \frac{5}{48} & 0 & -1 & 2 \\
\frac{1}{4} & 1 & 1 & 0 & 1 \\
-1 & 0 & 0 & 1 & 0 \\
1 & 0 & 0 & 1 & 0 \\
\end{array}
\right)$ and $P^{-1}=\left(
\begin{array}{ccccc}
0 & 0 & 0 & -\frac{1}{2} & \frac{1}{2} \\
-\frac{48}{65} & \frac{48}{65} & 0 & 0 & 0 \\
\frac{7}{10} & -\frac{6}{5} & 1 & -\frac{1}{8} & -\frac{3}{8} \\
0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{26} & \frac{6}{13} & 0 & \frac{1}{4} & \frac{1}{4} \\
\end{array}

You have the correct columns, just out of order

7:28 PM

ok, and how did you choose the order of the eigenvectors?

it is due to the order of the eigenvectors.

it's from the larges eigenvalue to the smallest, right?

it looks like from smallest to largest...

21 mins ago, by robjohn

$\left(
\begin{array}{ccccc}
0 & 1 & 0 & 0 & 0 \\
\frac{1}{12} & \frac{11}{12} & 0 & 0 & 0 \\
\frac{1}{3} & 0 & \frac{1}{3} & \frac{1}{3} & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 \\
\end{array}
\right)=P\left(
\begin{array}{ccccc}
-1 & 0 & 0 & 0 & 0 \\
0 & -\frac{1}{12} & 0 & 0 & 0 \\
0 & 0 & \frac{1}{3} & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
\end{array}
\right)P^{-1}$

hmm, is see, but the order matters for the end result, so I'm just trying to figure out how to pick it

maths researcher

8:08 PM

the resolvent of the operator $D:C^1[a,b]\rightarrow C[a,b]$, $D(f)=f'$ is empty?

hello world

@robjohn, sorry and thank you once more, I learned how to properly do the jordan decomposition

my last 2 questions need more love

8:31 PM

maths: seems right? if you fiddle with the domain of D maybe a different result. but D - kI is never injective (e.g. f_k(x) = e^(kx) is nonzero in C^1[a,b] and (D - kI) f_k = 0)

8:50 PM

0

Q: Growth rate of a sequence that generalizes the primes?

Let a strictly increasing sequence $A_2$ be defined as starting with $a_1 = 2$ and containing all integers larger than the ones already in the sequence unless $$ x = a_{i_1}*a_{i_2} + a_{i_3}*a_{i_4} $$ for some $i_1,i_2,i_3,i_4$ Or said differently $$ a_{n+1} = a_n + 1 $$ unless $$ a_n + 1 = a_{...

sequences-and-series number-theory asymptotics recursive-algorithms sieve-theory

-1

Q: asymptotics of $g(z,s) =\sum_{n=1}^{\infty} (-1)^{n+1} \dfrac{z^n}{p_n^s}$ ??

Let $p_1 = 1, p_2 = 3, p_3 = 5,...$ where the sequence $p_n$ is $1$ and the odd primes ordered by size. Let $Re(s),Re(z) > 0$. define $g(z,s) =\sum_{n=1}^{\infty} (-1)^{n+1} \dfrac{z^n}{p_n^s}$ Notice this infinite sum diverges for $Re(z) > 1$ but we use analytic continuation to define it there. ...

calculus sequences-and-series prime-numbers asymptotics analytic-continuation

9:14 PM

I wanna integrate $\int_{\gamma} \frac{e^{\pi z}}{z-i} dz$ where $\gamma$ is a circle of radius $1/2$ centred at $i$.

Now I know the integrand is not holomorphic on this region

so I figured I should use the deformation theorem, so I decided to then integrate on the circle of radius $1/4$ centred at $i$. Since the integrand is holomorphic on the region in between these two circles if I find out this integral I will have the integral i need

But when I try to parameterise it by letting $z-i = e^{i \theta}$ for $0 \le \theta \le 2\pi$ my integral is too difficult to solve

Is my method right?

the integrand isn't holomorphic on the region, but the numerator of the integrand is. do you know a form of cauchy's integral formula?

Ohhhh

Yeah cheers, you just reminded me

:)

yeah its $-2\pi i$ tysm

2 hours later…

maths researcher

11:04 PM

is $\{f'in C^1[0,1] : f(0)=0\}$ dense in $C[0,1]$?

some latex there. is that ' supposed to be an \in, or a ' \in

in the latter case i have more questions

Grill away.

well, just what is intended by the notation

{f' : f in C^1, f(0) = 0} or something more nuanced

maths researcher

I meant $\{f' \in C[0,1] : f(0)=0, f \in C^1[0,1]\}$

ted, you'll like this. munchkin responded to naptime today by telling my wife to go downstairs, the cat to go downstairs, and for me to go to my office, and to just leave her alone.

maths researcher

11:13 PM

yeah, what you wrote

maths: doesn't the FTC imply that that set is C[0,1]?

ted: she then said "nobody's in control" (a theme she kept repeating and i don't know where she heard it) and that she had "fired" us, which was accompanied by hand gestures indicating casting a magic spell that shot flames at us. i'm assuming this comes from a show she watched at her grandparents house.

Good for munchkin. Be gone, all of you.

She could really get into being Tabatha.

i'm going to pardon everyone

then it's dense in $C[0, 1]$, problem solved

The kid I just interviewed for MIT finished by asking what one super-power I would like to have.

$f(0)=0$ makes it difficult to be dense.

11:15 PM

interesting approach.

she must have been primed to ask that

ted: it's the derivatives of those f, not those f.

Ah. OK.

all my derivatives are zero and i am feeling a bit dense

i want koro's super power. dispelling foxes and rainbows.

11:16 PM

So the question is no longer redundant.

@copper Perhaps. His questions fit with his overall personality and the tenor of our discussion, though.

copper: there's a pill available for the first part of that

i would settle for my 25yo powers back

ted: so what is it?

i like when kids are not obsequious

ignore this, just trying to figure out where the typo is

"One way to approach the problem without solving either equation directly is to consider what it would require to be true: namely, it requires $$r_0 e^{-\alpha t} = \alpha \ln (K/N) \implies \ln N=\ln K +\frac{\alpha}{r_0} e^{\alpha t}$$ That gives a trial solution which can be tested for validity."

11:19 PM

I said to get rid of all hate, suffering, and guns.

maths researcher

@leslietownes no, I don't think so. Not every continuous function is the derivative of a C^1 function

found it

i am afraid trump will be the serb trigger of ww1

@maths really?

$g(t) = \int_0^t f(x)dx$.

just a random comment

11:20 PM

The same one I almost typed, but chose "really?" instead.

maths: so is copper's thing not differentiable, or is its derivative not continuous? is it not 0 at 0?

:-). just chatted with my sister, always puts me in a good mood

i agree that there are many ways of permuting the given input data so that things get very subtle, but this permutation is the nice one.

maths researcher

ah right.

for once, things actually work out.

don't get used to this. :D

nobody's in control

11:24 PM

LOL, @Semiclassic. It's quite a typo.

Negation turned into reciprocal.

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Jan '2230

$Mathematics$ Mathematics

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