help-23

[somber oriole]

Maybe he couldn't figure out why it works

[feral linden]

He said he got it and closed it. I thought he got it

Anyway, hope you don’t have to ask a third identical one

[nimble vine]

if they ask then there is some confusion with the concept

[feral linden]

Okay. Whenever you see a_k multiplied by some r^k, consider f(r)

Ask for even or odd terms together consider f(r) and f(-r)
(only certain k mod m terms appear consider root of unity of order m)

[safe radish]

@short dawn Has your question been resolved?

[safe radish]

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[midnight sonnet]

Can I disprove transitivity without picking particular values?
Like I can take |x-y| < 1 and |y-z| < 1 as premise then I could add
both which gives |x-z| < 2

[quasi bison]

just because |x-z|<2, does not mean |x-z| couldn't also be <1

[midnight sonnet]

Is there another way

[quasi bison]

another way to do what

[midnight sonnet]

disprove transitivity

[quasi bison]

not really

[midnight sonnet]

oh

i see

.close

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[midnight sonnet]

a.) is a partial order right?

[safe radish]

Please don't occupy multiple help channels.

[midnight sonnet]

.close

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[daring rune]

So this question answer is (1,0) and (2,1) and i understand why but when i continued i faced another question with standard matrix as (2,1) (1,0) that produce the exact same shape so could the 2 answers be correct?

[daring rune]

this is the question from the text book

and this is the 2 giving same points only difference is that the basis 1,0 is the same as 0,1 on the other one

ultimately my question is could these 2 standard matrices be considred an answer to 23b

[rustic goblet]

2 1
1 0

is also perfectly fine for b)

though the determinant would have a sign change :p

[daring rune]

yea but in the end he is just asking for the standard matrix from what happened to the unit square from the picture no?

[rustic goblet]

mhm

there're two possible answers

but the more "standard" one is probably the one they gave

since it doesn't cause the square to "flip over" during the transformation

[daring rune]

could u explain this a bit more the square flipping over please

[rustic goblet]

in the upper transformation, the red arrow stays to the left of the blue arrow and the green square stays the same orientation

in the lower transformation, the red arrow ends up to the right of the blue arrow and the green square is upside down

you can think about what happens when you pass the red arrow through the blue arrow as they swap positions from the upper right diagram to the lower right diagram

as they trade places, the square has to flip over

[daring rune]

okkkk got it now kinda of spin around or flip yea

like if each face of the square had a letter and the original one facing us was lets say N now it wouldnt be N

[rustic goblet]

yeah! it'd be a mirrored N

[daring rune]

yea thanks thanks much apperciated for the help

[rustic goblet]

both transformations are valid, but the one that doesn't mirror is probably more "standard"

so that's what I'd go with for your problem

[rustic goblet]

no problem, I'm happy to help

also, welcome to the mathcord

[daring rune]

happy to be here i hope i really get to the point where i just help people hopefully not too far in the future

[rustic goblet]

you're probably able to do so already :p

there're plenty of foundational problems going around, so feel free to chime in and help others out c:

[daring rune]

Alright thank you again for now i better continue studying my final is tmrw

[rustic goblet]

good luck!! you got this

if you're done here, you may close the channel

[daring rune]

alrighty

.close

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[last moth]

someone do this and solve it for me

[uncut magnet]

what part are you struggling with?

[nimble vine]

!noans

[safe radish]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[last moth]

walk me through the whole question

[uncut magnet]

do you know how to sketch log(x)?

[last moth]

no

[burnt notch]

Haven't you been taught it? 🤔

[last moth]

yeah but i didnt really get it

[burnt notch]

What of it didn't you get?

[last moth]

i wasnt paying attention in class

[burnt notch]

Alright... 😬

[graceful lichen]

You probably then need to patiently review the definition of logarithm and it's graph once you'll be able to get it

[burnt notch]

Then I believe a video will help you more than anything

Look for "how to sketch logarithm function" on YouTube, I'd say

[nimble vine]

https://youtu.be/-nptxS9rZNA?si=hdxRkJ3BAbVDpOsW

Try this

[last moth]

you guys are lazy as hell

you guys dont even know how to do math

[nimble vine]

If you still have a doubt after watching ask

[graceful lichen]

Lol

[last moth]

my test is tomorrow man

[graceful lichen]

That's why I am saying to not hurry and patiently review the stuff

[nimble vine]

Breakdown

[last moth]

its fine ill just skip the test

[nimble vine]

You have to give it time and effort

[last moth]

schools not that important anyways

[uncut magnet]

okay give up

[edgy breach]

Do NOT skip the test thats a terrible idea

[tender sluice]

@last moth did you watch the video?

[last moth]

ye

[tender sluice]

did you get it?

[last moth]

nah

[tender sluice]

[last moth]

bro i need to releave the stress before i fail my test

[tender sluice]

sounds like your stress comes from a lack of effort

[last moth]

FUCK

I HATE MATH

IM ABOUT TO RIP MY PAPER

AHHHHHHHHHHHH

screw this

[edgy breach]

i hate english but i put the effort in to get a reasonable result in it

[graceful lichen]

Everything is different with me I like physics and math and chem but little less and I put efforts in all and get poor scores

[last moth]

where u guys from

canada

?

[eternal breach]

you'll get good at it if you learn to enjoy it

[tender sluice]

wait am I tired?

you know that if $y=ln(x)$, then $x=e^y$ right?

[flat frigate]

elijah

[safe radish]

@last moth Has your question been resolved?

[primal bone]

<@&268886789983436800>

You know they have delete logs?

In any case:

[primal bone]

Sorry, this coming from someone who and I quote "wasn't paying attention in class" is, at best, a bit rich. Oppenheimer did suggest a video to watch because it already has the prerequisite information you need to understand how to draw the functions you're being asked to, and I recommend you do indeed watch it.

If you still have specific questions, feel more than free to ask them here, but don't simply go around insulting people.

[lilac crane]

@last moth you can voice distress and dislike without being overtly NSFW

[cloud hound]

Someone still needs to .close this channel

[median vigil]

.close

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[kindred slate]

i need some help with the 2nd part of this question plz

ie showing the EL equations

i get this but this doesnt match

[safe radish]

@kindred slate Has your question been resolved?

[safe radish]

@kindred slate Has your question been resolved?

[gusty pasture]

Are you used to index notation? It might be better to solve this problem

[kindred slate]

i suck at using index notation i always prefer things in matrix vector form

tbh i have got somewhere

i realised the gradient of A(q) is acc a matrix not a vector right?

so u cant really dot it

[gusty pasture]

Yes, that's basically the problem, because in your notation you can't really distinguish between the expression that is given in the hint

[kindred slate]

hmm

let me send u what i have so far

[gusty pasture]

If you are taking physics, I would suggest you getting used to it, trust me it's much better, I used to think matrix vector form was better too

[kindred slate]

im up to here

[kindred slate]

i will try. tbh this is the only physics module im taking as im doing a maths degree

[gusty pasture]

Oh, I don't know if it will be useful there so don't focus too much on it

[kindred slate]

its the only physics module i do

[iron estuary]

the EL equation naturally produces the antisymmetric part of the jacobian of A

which in $\mathbb{R}^3$ this skew symmetric matrix corresponds to the cross product with $\nabla \times A$

[flat frigate]

anflo

[kindred slate]

think i got it

.close

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[modern fractal]

Hi, guys. I'm studying and finding out this is confusing me. I have no idea how to switch from n to ∞. Can anybody explain it to me? Just a simple answer no problem, something like what theorem/etc I forget about.

[cloud hound]

Fancy

I guess it helps to know that once $k > d$, we define $\binom{d}{k} = 0$

[flat frigate]

Coolempire93

[cloud hound]

So all of the factors $x^k$ with $k > d$ (meaning they have too high a power to exist in the expansion in the first place) get cancelled out

[flat frigate]

Coolempire93

[cloud hound]

So we can let k go to infinity, but this sum will simplify to $\sum_{k=0}^\infty\binom{d}{k}x^k = \sum_{k=0}^d\binom{d}{k}x^k$

[flat frigate]

Coolempire93

[cloud hound]

Which is the original formula given at the top

[modern fractal]

Aaaa I see. Since the sum notation works using integers, so that when we use real instead, we might be facing a situation where k can't be equal to d so we just wrote it to ∞ with the main idea whenever k > d then the combination is equal to 0

Thanks, brother @cloud hound

Fiuh lucky me that fast

.close

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[cloud hound]

No problem!

[cloud hound]

Yes 👍

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[clever gale]

While playing around with random functions I stumbled across $\int_{-\infty}^{\infty}\prod_{n=1}^{\infty}\left(\cos\left(\frac{x}{2^{n}}\right)\right)^{n},\dd x$, which numerically seems to evaluate to $\pi$. How does one show this result?

[flat frigate]

Roy

[clever gale]

which is equivalent to $\int_{-\infty}^{\infty}\prod_{n=0}^{\infty}\frac{\sin\left(x/2^{n}\right)}{x/2^{n}},\dd x$ by the infinite product of sinc

[flat frigate]

Roy

[keen tulip]

Hm

diamond For $a>0$ if $U sim Unif[-a,a] then$,
[
mbb E(e^{i t U}) = frac1{2a} int_{-a}^a e^{itu} dd u = sinc(at).
]

[flat frigate]

クーリー

クーリー

[safe radish]

@clever gale Has your question been resolved?

[clever gale]

Hmm, I haven't learnt fourier stuff yet

Could you perhaps explain it to me, or maybe redirect me to suitable resources that explain this

[keen tulip]

Give me a moment.

[clever gale]

all good, take your time

[slim lion]

for some reason this reminds me of the borwein integrals

[safe radish]

@clever gale Has your question been resolved?

[faint seal]

yeah this is just a borwein integral

[safe radish]

@clever gale Has your question been resolved?

[safe radish]

@clever gale Has your question been resolved?

[slim lion]

wait deadass?

[safe radish]

@clever gale Has your question been resolved?

[buoyant dragon]

it is the "Borwein" Pattern

$$\int_{-\infty}^{\infty} \text{sinc}(x) \prod \text{sinc}(a_k x) , dx = \pi \quad \text{if} \quad \sum |a_k| \le 1$$

[flat frigate]

Shikhar

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[pine island]

Is it possible to determine a family of self referential functions using a specific alphabet set of symbols (e.g. the usual a, b, c, x, y, z, α, β, γ, +, =, () etc.) [using some arbitrary symbols and positioning] which plot their literal written form on the Cartesian plane within the bounds a </= x </= b and c </= y </= d [following a specific rule of symbol placement/size/etc.]?

Why or why not?

[pine island]

wtf

[round mango]

Oh, I think conway did something like this

Let me look

[pine island]

Kinda like quine programs Ig?

I dunno I just know abt Tupper's formula

[round mango]

Oh yeah that's the one I was thinking about

[safe radish]

@pine island Has your question been resolved?

[pine island]

Oh cool

But tupper's formula needs a bitmap seed (k) which is kinda like a parameter you can change willingly

I was talking abt literal symbols and necessary variables for plotting not parameters

[safe radish]

@pine island Has your question been resolved?

[pine island]

So... any updates?

[safe radish]

@pine island Has your question been resolved?

[safe radish]

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[humble sail]

Hello, i have a questión regarding a calculus 1 problem. The problem states that there are 2 functions, f(x) and g(x), which are continuous in the interval [a,b], and that f(a)<g(a) and f(b)>g(b). I have to prove that there is a value c belonging to the interval [a,b] such that f(c)=g(c).

[quiet plume]

Are you sure those inequalities are the right ones?

[opaque fern]

Uh this seems mighty sus

[humble sail]

yea it was wronjg

[cedar widget]

anyway what is your question?

[opaque fern]

Anyways, intermediate value theorem

Just gonna dump this here before anything

[humble sail]

i dont know how to prove it,

[hardy lion]

➖

[fathom adder]

Define an alternative function h that is f-g is how i'd go by first thought

And then i think it fall from definition

[humble sail]

ok i understood the idea, thanks

.close

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[night heath]

is this possible

[split kayak]

As is it doesnt make much sense, no.

You can do dsin(x) tho.

But thats mostly reserved for evaluating definite integrals

[cloud hound]

https://math.stackexchange.com/questions/2359562/what-does-int-sindx-mean

And the process of evaluation
https://youtu.be/xrXejjEkMxk

But I agree personally it's nonsensical

[split kayak]

iirc it relies on the idea that sin(x) = x near 0.

[cloud hound]

Yes

[night heath]

so technically theres no solution or elementary antiderivative?

[split kayak]

not really, its mostly an algebraic manipulation of the expression that hardly matches whatever it should mean.

[night heath]

👍

alr then

that was all

.close

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[cloud hound]

<@&268886789983436800>

[safe radish]

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[clever gale]

How does one show $\mathfrak{Re}\left(\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n+1}}{\left(n+i+1\right)^{2}}\right)=\frac{1-\pi^{2}\coth\left(\pi\right)\operatorname{csch}\left(\pi\right)}{2}$

[flat frigate]

Roy

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[still charm]

Looks like incomplete zeta I think

The left side is kind of a p series

[clever gale]

I thought it looked similar to the trigamma series but it's alternating 🤔

[safe radish]

@clever gale Has your question been resolved?

[tardy mango]

$S=\frac{1}{4} \left(\psi^{(1)} \left(\frac{i+2}{2} \right)-\psi^{(1)} \left(\frac{i+1}{2} \right) \right)$

[flat frigate]

Civil Service Pigeon

[tardy mango]

then try using this?

I feel like that should do smt

[clever gale]

Oh this was it, smart idea

Then just reflection formula

[tardy mango]

where did this come from lol

I feel like I see you with these sums and integrals all the time

[clever gale]

The original was $\int_{0}^{1}\frac{\cos\left(\ln\left(x\right)\right)\ln\left(x\right)}{x+1},\dd x$

[flat frigate]

Roy

[clever gale]

Wrote cos(ln(x)) as Re(x^i) then geometric seriesed the denominator and subbed ln(x) -> -x to turn it into Laplace(x)(s = i + n + 1)

Which got me that series

[tardy mango]

[safe radish]

@clever gale Has your question been resolved?

[clever gale]

writing out a solution

then i'll send it here

[clever gale]

Note that: $$\psi_{1}\left(-\frac{i}{2}\right)=\psi_{1}\left(1-\frac{i}{2}\right)-4$$
$$\psi_{1}\left(1+\frac{i}{2}\right)+\psi_{1}\left(-\frac{i}{2}\right)=-\pi^{2}\operatorname{csch}^{2}\left(\frac{\pi}{2}\right)$$
$$\implies \psi_{1}\left(1+\frac{i}{2}\right)+\psi_{1}\left(1-\frac{i}{2}\right)=-\pi^{2}\operatorname{csch}^{2}\left(\frac{\pi}{2}\right)+4$$
$$\implies \mfk{Re}\left(\psi_{1}\left(1+\frac{i}{2}\right)\right)=-\frac{\pi^{2}}{2}\operatorname{csch}^{2}\left(\frac{\pi}{2}\right)+2$$
Also:
\begin{align*}
\mfk{Re}\left(\psi_{1}\left(\frac{1+i}{2}\right)\right) &=\frac{1}{2}\left(\psi_{1}\left(\frac{1+i}{2}\right)+\psi_{1}\left(1-\frac{1+i}{2}\right)\right) \
&=\frac{\pi^{2}}{2}\csc^{2}\left(\frac{\left(1+i\right)\pi}{2}\right) \
&=\frac{\pi^{2}}{2}\sec^{2}\left(i\frac{\pi}{2}\right)=\frac{\pi^{2}}{2}\operatorname{sech}^{2}\left(\frac{\pi}{2}\right)
\end{align*}
Hence:
\begin{align*}
\mfk{Re}\left(\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n+1}}{\left(n+1+i\right)^{2}}\right)&=\mfk{Re}\left(\sum_{n=0}^{\infty}\left(\frac{1}{\left(2n+2+i\right)^{2}}-\frac{1}{\left(2n+1+i\right)^{2}}\right)\right) \
&= \mfk{Re}\left(\frac{1}{4}\sum_{n=0}^{\infty}\left(\frac{1}{\left(n+\frac{2+i}{2}\right)^{2}}-\frac{1}{\left(n+\frac{1+i}{2}\right)^{2}}\right)\right) \
&= \mfk{Re}\left(\frac{1}{4}\left(\psi_{1}\left(1+\frac{i}{2}\right)-\psi_{1}\left(\frac{1+i}{2}\right)\right)\right) \
&= \frac{1}{4}\left(\left(-\frac{\pi^{2}}{2}\operatorname{csch}^{2}\left(\frac{\pi}{2}\right)+2\right)-\left(\frac{\pi^{2}}{2}\operatorname{sech}^{2}\left(\frac{\pi}{2}\right)\right)\right) \
&= \frac{4-\pi^{2}\operatorname{csch}^{2}\left(\frac{\pi}{2}\right)-\pi^{2}\operatorname{sech}^{2}\left(\frac{\pi}{2}\right)}{8} \
&\underset{\text{double angle biz}}{=} \frac{1-\pi^{2}\coth\left(\pi\right)\operatorname{csch}\left(\pi\right)}{2}
\end{align*}

[flat frigate]

Roy

[clever gale]

is there a shorter way...

also used $\mfk{Re}\left(\psi_{1}\left(z\right)\right)=\frac{1}{2}\left(\psi_{1}\left(z\right)+\psi_{1}\left(\bar{z}\right)\right)$

[flat frigate]

Roy

[night heath]

in what math do you learn $cos(x)+isin(x)=e^(ix)$

[flat frigate]

Cowking

[clever gale]

Uhhh... this channel is occupied. And you learn it in your average course with complex numbers / calc 2

[gusty trench]

it might be easier (not sure though) if you reindex the original sum as the sum from n = 1 to inf of (-1)^n / (n+i)^2, then interchange the real part operator and the infinite sum, you should get that Re( (-1)^n / (n+i)^2 ) = ((n^2 - 1) * cos(nπ) + 2n * sin(nπ)) / (n^2 + 1)^2

and if you sum this over n = 1 to inf, you'll notice that the term with sin(nπ) will always be 0, and cos(nπ) will be (-1)^n, so it comes down to computing the sum from n = 1 to inf of (-1)^n * (n^2 - 1) / (n^2 + 1)^2

that's of course still hard to find but at least it's much more concrete than the original sum and it's purely real

[safe radish]

@clever gale Has your question been resolved?

[clever gale]

That's very nice

I thiink you can use residue theorem to evaluate the series??

[clever gale]

I believe we have $$\sum_{n=-\infty}^{\infty} (-1)^n g(n) = -\sum_{\text{Poles of } g(z)} g(z) \pi \csc(\pi z)$$ which follows from integrating $g(z) , \pi \csc(\pi z)$ over a square contour with side lengths that tend to infinity, and also of course given that $g(z)$ decays quick enough, because then the entire contour integral is 0

[flat frigate]

Roy

[clever gale]

\begin{align*}
I &= \mfk{Re}\left(\sum_{n=0}^{\infty}\frac{\left(-1\right)^{n+1}}{\left(n+i+1\right)^{2}}\right) \
&=\sum_{n=1}^{\infty}\frac{\left(n^{2}-1\right)\cdot\cos\left(n\pi\right)+2n\cdot\sin\left(n\pi\right)}{\left(n^{2}+1\right)^{2}} \
&=\sum_{n=1}^{\infty}\frac{n^{2}-1}{\left(n^{2}+1\right)^{2}}\left(-1\right)^{n} \
&\underset{\text{Symmetry}}{=} \frac{1}{2}\sum_{n=-\infty}^{\infty}\frac{n^{2}-1}{\left(n^{2}+1\right)^{2}}\left(-1\right)^{n}+\frac{1}{2} \
&= \frac{1}{2}\left(-\sum_{z \notin \mbb{N}} \Res\left(\frac{z^{2}-1}{\left(z^{2}+1\right)^{2}}\cdot\pi\csc\left(\pi z\right)\right)\right)+\frac{1}{2} \
&= \frac{1}{2}\left(-\pi^{2}\operatorname{csch}\left(\pi\right)\coth\left(\pi\right)\right)+\frac{1}{2} \
&= \boxed{\frac{1-\pi^{2}\operatorname{csch}\left(\pi\right)\coth\left(\pi\right)}{2}}
\end{align*}

[flat frigate]

Roy

[clever gale]

Thanks @gusty trench for your suggestion, it worked out perfectly

If nobody has any more comments I shall close this

[clever gale]

You can actually differentiate the Mittag-leffler series of $$\pi\operatorname{csch}\left(\pi x\right)=i\sum_{n=-\infty}^{\infty}\frac{\left(-1\right)^{n}}{ix-n}$$ to get: $$-\pi^{2}\operatorname{csch}\left(\pi x\right)\coth\left(\pi x\right)=\sum_{n=-\infty}^{\infty}\frac{\left(-1\right)^{n}\left(ix+n\right)^{2}}{\left(x^{2}+n^{2}\right)^{2}} = \sum_{n=-\infty}^{\infty}\frac{\left(n^{2}-x^{2}\right)}{\left(n^{2}+x^{2}\right)^{2}}\left(-1\right)^{n}$$ and it yields the answer pretty quick

[flat frigate]

Roy

[clever gale]

@gusty trench this is pretty interesting so I thought you should know

[gusty trench]

that is pretty cool actually, thanks

[keen tulip]

Damn I forgot about this

Sorry

Oh it's a new thing

[safe radish]

@clever gale Has your question been resolved?

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[steady yoke]

How do I do the challenge

[delicate socket]

Part a or part b?

[steady yoke]

B

I got a-b for (a)

[mortal sandal]

have you done problems with rationalizing the denominator before

[mossy lotus]

you can start by rationalizing the denominator

[steady yoke]

how do I rationalise the whole thing tho

it’s a sequence

[mossy lotus]

just do one fraction at a time

you dont have to deal with it all at once

[steady yoke]

so I start with 1/root1+root2

I rationalise this

[mossy lotus]

yea

[steady yoke]

okay give me two seconds

[slim lion]

you should be able to notice a pattern after the first one

u may use part (a) to make it faster

[steady yoke]

I got root 2 - root 1

[slim lion]

ok

now try rationalise the 2nd term

[steady yoke]

Root 3 - root 2

[slim lion]

now lay it out

and notice any patterns

you have

sqrt(2) - sqrt(1) + sqrt(3) - sqrt(2) + ...

[steady yoke]

It cancels out

It repeats every 4 terms

[mortal sandal]

I'd write out more until you're sure

[steady yoke]

Can I do it without writing it out becuase I’ve noticed it’s just the second term inversed plus the first term

[mortal sandal]

If you know the answer then yeah

[faint hornet]

well what would your answer be in the end then

[quasi bison]

thats a lot of effort

[faint hornet]

oh wow

[quasi bison]

also you are meant to ADD all of these

[steady yoke]

yes

[quasi bison]

also would you do the same thing if you had to write out 2029 of these

[steady yoke]

I thought i was meant to do this

[slim lion]

u dont need to write it all out

[steady yoke]

No I was thinking that

[steady yoke]

.

[faint hornet]

don’t doubt the perseverance…

[slim lion]

$\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{2}+\ldots+\sqrt{24}-\sqrt{23}+\sqrt{25}-\sqrt{24}$

[flat frigate]

Flatus

[steady yoke]

never doubt my dedication

[faint hornet]

I think he only meant like a couple more

[slim lion]

and add little ticks on them to show you've crossed them out

to show how

they cancel out

[mortal sandal]

Oh yeah I just meant until you know how they cancel

Sorry 😭

[steady yoke]

now we just pray I never get a question that goes up to 8382928

shouldn’t we do this in a way where like

1/(rootn + rootn+1)

[delicate socket]

Have you encountered the sigma notation for sums ?

[steady yoke]

Nope

[mortal sandal]

Well you can use dots

[steady yoke]

well then are we good

[faint hornet]

I think low-key you were fine dw

but perhaps it would in fact to just be easier to write out a few terms and use dots ^

[open wedge]

try rewriting each fraction in a form of $\frac{1}{\sqrt{n} + \sqrt{n+1}}$

[flat frigate]

1 divided by 0 equals Infinity

[steady yoke]

yeah I said that

[steady yoke]

.

I got root (n + (rootn+1))/n-n+1

[open wedge]

simplify the denominator

:)

[steady yoke]

everything over 1

[open wedge]

and what happens if the denominator is 1

[steady yoke]

and hence its n + rootn+1

[steady yoke]

Stays the same

[open wedge]

cool

[open wedge]

then you should notice that each term is like this in the figure you made

[steady yoke]

What

Ididnt get that tho

[open wedge]

denominator wrong calculation

[steady yoke]

Why

[open wedge]

denominator is $n - (n + 1) = n - n - 1 = -1$

[flat frigate]

1 divided by 0 equals Infinity

[steady yoke]

Why would this help tho

[open wedge]

how about instead of multiplying by $\sqrt{n} - \sqrt{n + 1}$

[flat frigate]

1 divided by 0 equals Infinity

[open wedge]

why don't you multiply by $\sqrt{n + 1} - \sqrt{n}$

[flat frigate]

1 divided by 0 equals Infinity

[steady yoke]

multiply what

[open wedge]

multiply numerator and denominator by this

[steady yoke]

I did

[open wedge]

you multiplied by $\sqrt{n} - \sqrt{n + 1}$

[flat frigate]

1 divided by 0 equals Infinity

[limpid lodge]

I feel like you can notice that any series of this form will leave you with $\sqrt{n+1}-\sqrt{1}$ since all middle terms will cancel out.

[flat frigate]

ch3rry

[safe radish]

@steady yoke Has your question been resolved?

[thin bridge]

either conjugate will work
but you appear to have notation issues

[safe radish]

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[worldly yacht]

Hi, I was interested in getting some help. I tried some math books at the library, like calculus, geometry, algebra but they were way too easy to understand and werent too difficult like some problems feel. I know this is probably because its entry level math, but Is this how learning math feels the whole time? Just easy tidbits of learning all throughout the journey or can I make bigger steps that feel more challenging throughout? Thank you.

[worldly yacht]

Im pretty sure this is everyone, but if you gave me a book that had a difficult problem and all the resources required to solve the problem and more of that difficult problem over time you can start to see paterns is there a book like that

[cedar widget]

what is your level

it would depend on that

[worldly yacht]

I just graduated highschool

and Im in college

[lone void]

i have a book for difficult problems..dont there it has the resources required to solve it tho..its a question book

[worldly yacht]

cool cool

[cedar widget]

there are plently of difficult questions in most fields in math

[worldly yacht]

Lol

I mean

Yea

But its more the roadmap too that Im looking for

to solve those problems

[lone void]

ohk then nvm

[worldly yacht]

like a very direct one

that isnt broad

Ok can you teach me for instance a little bit of analytical geometry and how you would describe the creation of a square using numbers

[cedar widget]

describe the creation of a square using numbers
wdym?

[worldly yacht]

like perhaps a physics equation or something

or formula

[cedar widget]

what

[worldly yacht]

that represents the creation of a square

nvm

Is math going to be easy and boring for some people?

[cedar widget]

boring maybe

easy up to some level

but eventually everyone needs to put in work

[worldly yacht]

I mean isnt there a saying that any good teacher can make any hard problem seem very easy or something

[cedar widget]

occupied

[silk cove]

Sorry

[buoyant dragon]

just copying my msg

[cedar widget]

thats very true in a lot of cases

[worldly yacht]

apparently einstein said

“If you can’t explain it simply, you don’t understand it well enough.”

[cedar widget]

but a good teacher is not everything

[worldly yacht]

The thing is I have a hard time trusting people online too

I want to go back to my orginal question which is, is learning math just a bunch of small easy steps?

Or is there a more challenging way to learn math that people are using that could suit me better

[hard crest]

it's a bunch of small hard steps xD

i guess it depends

[cedar widget]

ideally it should be easy steps

but that is not the case

[hard crest]

computational / algorithmic math problems like the ones you'd find in calculus and stuff are important for gaining familiarity and experience

but a lot of people would say "real" math is more about arguments and words, proofs and logic rather than numbers and derivative rules

[austere goblet]

that being said, OP

I'm glad you seem to have at least grown up since the last time I saw you in a help channel and are actively seeking books abt math to read

kudos

[worldly yacht]

hahaha lol

Guys

Riemanns Hypothesis

Alright

And quantum mechanics

You know anything?

And the Evangelic

[austere goblet]

just as I give you the benefit of the doubt

[worldly yacht]

values

Or eulers formula

I read about it

You guys want to talk about it

Honestly I still dont get how that equals 0

[rugged cave]

#math-discussion is that way if you're looking to discuss about math.

[worldly yacht]

.end\

.end

.done

[rugged cave]

.close

[safe radish]

Channel closed

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[worldly yacht]

.reopen

[safe radish]

✅ Original question: #help-23 message

[worldly yacht]

I have a problem... Why should I learn math?

It feels too boring

[obsidian basin]

that is probably not a question suited for the help channels

[worldly yacht]

then where

mb

[obsidian basin]

#math-discussion

or #discussion

[worldly yacht]

It feels like brain rot

in there

everyone it talking too much

[cedar widget]

thats valid xD

[worldly yacht]

I cant focus

Genuinly... why should I learn math

Theres a million reasons why

but not one that strikes a chord

[obsidian basin]

sure, but this isn't exactly a math question per se, and more of an interest question

[worldly yacht]

do you want me to close this

is that what your saying cuz Ill just do that fam

[obsidian basin]

well I'm just telling you that those two highlighted channels are probably a better place for questions like what you're asking

[worldly yacht]

lets see

how long

it takes for a response

[obsidian basin]

if that fails #math-discussion is also open

[worldly yacht]

well

I guess there is my answer

🙁

[obsidian basin]

!done

[safe radish]

If you are done with this channel, please mark your problem as solved by typing .close

[obsidian basin]

.close (OP moved to #math-discussion)

[safe radish]

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[verbal lily]

is this right?

[rigid blaze]

the working is very low visibility

[verbal lily]

i have a very bad phone

[buoyant palm]

Is this algebra or set theory?

[loud wren]

Group theory

[loud wren]

I can't see anything

do you mind typing out the premise of your idea

[verbal lily]

the if side is trivial
the only if side
i supposed that H is not a subset of K then proved that K is a subset of H
supposed x in K then x in H\cup K and since H\cup K is a subgroup of G then it is closed under the operation of G
since H is not a subset of K then there exists some h in H such that h is not in K
and since h in H then it is in H\cup K which implies
xh in H\cup K so xh is either in H or in K, then i considered the cases
if xh in K we have x in K then the inverse of K is in K (since it is a group) then x^{-1}xh\in K and this gives us h in K which is impossible thus we remain with xh in H and using the same reasoning with h we get xhh^{-1} in H , x in H

the whole idea is that i think the closure is the problem so i worked on deriving my result from it

[safe radish]

@verbal lily Has your question been resolved?

[crude bear]

Your proof is correct

[safe radish]

@verbal lily Has your question been resolved?

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[safe radish]

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[proud tree]

can you think of a substitution that would make the inner roots go away?

[lucid raven]

uhh no

[lost jewel]

Complete the square lol,it's common trick

[proud tree]

try x-1=u²

[upper moat]

Easy substitute

[lucid raven]

Yeah i solve this but what is the motivation to use x-1=u^2

[tardy mango]

^^

||both of the radicands are actually perfect squares||

substituting x-1=u^2 is largely pointless imo

[proud tree]

mb 😭

[lucid raven]

oww ok

thanks

.close

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[silver oar]

any algebraic graph theory peeps willing to help me a little with something?

[silver oar]

its to do with graph spectra

[main mural]

just ask directly, although if it's super advanced you might want to ask in #combinatorial-structures or something

[silver oar]

ok bet wait i need to formula my question first

ill ask late

later*

.close

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[steady yoke]

what have I done wrong here

[timid ridge]

(5x + 3)^2 = ?

[steady yoke]

would it not be 5x^2 + 9

If not why not

[limpid lodge]

$(a+b)^2=a^2+b^2+2ab$

[flat frigate]

ch3rry

[steady yoke]

oh yes

sorry

[limpid lodge]

Or if u hvnt learned tht

$(a+b)(a+b)$

[flat frigate]

ch3rry

[limpid lodge]

Just expand this

[steady yoke]

thank you

.close

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[smoky nacelle]

.close

. lalala

[knotty grove]

guys...

[knotty grove]

is pi=0

cuz i just made a few multiplications

[last heath]

[knotty grove]

e^iπ + 1 = 0
e^iπ = -1
(e^iπ)^2 = (-1)^2
which gives:
e^2iπ = 1

so if we take
e^2iπ * e = 1 * e
it becomes:
e^2iπ + 1 = e

which is:
(e^2iπ + 1)^(2iπ + 1) = e

if we use the power rule (a^m)^n = a^(mn)
(e^2iπ + 1)^(2iπ + 1) = e
e^((2iπ + 1)(2iπ + 1)) = e
(2iπ + 1)(2iπ + 1) = (2iπ)^2 + 2iπ + 2iπ + 2iπ + 1
= -4π^2 + 4iπ + 1

(e^2iπ + 1)^(2iπ + 1) = e
e^(-4π^2 + 4iπ + 1) = e

if we use a^(m+n) = a^m * a^n
e^(-4π^2) * e^(4iπ) * e^1 = e

so e = e… but it doesn’t work like that
e^(-4π^2) * e^(4iπ) = 1

using Euler's formula
e^(4iπ) = cos(4π) + isin(4π) = 1 + i0
which means
e^(4iπ) = 1

then we get
e^(-4π^2) = 1

using ln(a^b) = b * ln(a) we do
ln(e^(-4π^2)) = ln(1)
-4π^2 * ln(e) = 0
using ln(e) = 1 we get
-4π^2 = 0
which is π^2 = 0
which means π = 0
because 0^2 = 0, so π equals 0

so does that mean pi is equal to... 0?

[last heath]

please dont troll

,w pi

[knotty grove]

im not trolling

[flat frigate]

Results provided by WolframAlpha

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Requested by kensukeken

[last heath]

whatever u made seems nonsense

[knotty grove]

read it

brother

pi is 0

[last heath]

<@&268886789983436800> troll

[hard crest]

(a^m)^n = a^(mn)
doesn't work in complex numbers

[knotty grove]

ok

.close

[safe radish]

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[small rampart]

A is given an ordered triple (1,1,1). Each step A can perform one of two operation to the current (x,y,z) triple:
i/ (x,y,z) → (y,z,x+z)
ii/ (x,y,z) → (x+z+1, x+y+z+1, x+y+2z+1)
a) Prove that A needs exactly 4 moves to get to (a,b,6)
b) Find the smallest natural number k such that after k steps, A can get to (c,d,129)

[small rampart]

I got the a) part, which is quite simple

The last number of the triple always increases so you only need to check a few cases and see that applying the first operation four times is enough

I'm needing help with b)

(I'll define the functions f and g to be operation 1 and 2)

[iron estuary]

since the target is (c,d,129) first ignore x,y and track only how the third coordinate changes

what does the third coordinate become after one step of f(x,y,z) and after one step of g(x,y,z) ?

[small rampart]

x+z and x+y+2z+1

[iron estuary]

now compare those two expressions carefully

starting from positive integers which operation makes the third coordinate grow much faster?

[small rampart]

g grows much faster, i'm trying to see how many g's it takes to get close then work from there

[iron estuary]

yeah

since g grows much faster it can only be used a few times

[small rampart]

It only takes 3 to get close

g_3 (1,1,1) = (37,54,78)

Btw i'm trying to find a clever way, instead of exhausting the cases

Though that is definitely possible by hand

[iron estuary]

applying g again would overshoot 129 so g can appear at most 3 times

[iron estuary]

ah i see

[small rampart]

Maybe some arguments with the parity or some modular

[iron estuary]

you don’t need full case exhaustion

[iron estuary]

you can avoid parity/mod arguments

once you know g appears exactly 3 times the rest must be only f’s

[small rampart]

But then you're still testing the cases by hand

[iron estuary]

not really 'cases' once g is fixed to appear exactly 3 times the order is essentially forced

under f the third coordinate follows a deterministic fibonacci type recursion

so you’re not branching anymore you just iterate a single recurrence until you hit 129

[small rampart]

Yeah i see how that solves a problem, but this is an omlympiad question, i expected better

I guess it's just not a very good question

Unless there's a cleverer way, ofc

.close

[safe radish]

Channel closed

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[small rampart]

.reopen

[safe radish]

✅ Original question: #help-23 message

[small rampart]

I might be wrong but the g function might be a distraction?

I haven't checked but you can get there only using f

Idk if that's optimal, but if it is then there might be an argument

[iron estuary]

not really a distraction but it’s a filter

[small rampart]

Maybe you could prove that g isn't used

[iron estuary]

the point of g is to show rigidity - it grows so fast that it can only appear a fixed number of time

[small rampart]

Eh...

I wonder how i could code a program for this stuff

I'm new to programming

[iron estuary]

if you want to code it try thinking of this as a shortest-path / BFS problem on states (x,y,z)?

[small rampart]

.close

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[obsidian walrus]

I don't get this

[obsidian walrus]

It's a question from a mock exam

I know normally you'd find the area for the 3 squares, the black part and then subtract the diamond

But I don't know how to find the area of the middle circle

I get it's 153 per circle but then what

[feral linden]

Union of three circles - diamond as you said

[safe radish]

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[safe radish]

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[thorn coral]

A quadrilateral has four vertices A,B,C ,D , we want to color each vertex from 4 colors (c1 ,c 2 , c3, c4) , such that every side of quad and diagonal AC , hve end points with different color

[thorn coral]

Now , if we assign 4colors to A

[devout shale]

🔠

[thorn coral]

we may go to C nad say it's 3 and the rest have 2

(here numbers are choices)

In which case it would be 48

But if we do B or D after A you may get 4x3x3x2

So yea I need to hear ur thoughts

[vocal badger]

no, the cases are different (i believe)

i think you should write out the cases

[thorn coral]

Lol do it

I don't know

What do you want me to do , lol Add the 2 ?

and what are the cases

[vocal badger]

ok lets see

if we go to b then d then a

we fixed a

we have 3 choicess for b

3 choices for d as well but they have different significances

if you choose same as b, we have two choices for c

but if you choose different from b, c is fixed

[thorn coral]

Hmmm

So the 2nd way will ahve to cases huh

Ok Ok

[vocal badger]

alright then

is this closed?

[thorn coral]

wait wait

[vocal badger]

ok

[thorn coral]

so if I got by 2nd way

you want me to add B , D are diff and B , d are same

?

[vocal badger]

yes

like 4 * (3 * [ 2(1)+ 1(2)]

[thorn coral]

Hehehe

ty sir

.close

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[vocal badger]

which part?

[rare crest]

is this correct

anyone

?

[astral glacier]

What exactly does "modify" mean

What are you allowed to do

[rare crest]

it doesnt specify anything else

i just took my liberty to do it any how

[vocal badger]

then i guess clarify it with your prof

this is the 2nd time i told you the same thing lol

[feral linden]

Yeah define modify. This graph certainly can’t be K3,2

[rare crest]

lol

alright

[devout shale]

oh my

more topology

amazing

[safe radish]

@rare crest Has your question been resolved?

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[rare crest]

.reopen

[safe radish]

✅ Original question: #help-23 message

[rare crest]

would it be okay if i tagged you for help? i'll be diving deeper into graph theory, this was just part 1.

[devout shale]

yes please tag me

[rare crest]

thank you!

.close

[safe radish]

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[midnight sonnet]

1. its gonna be a square matrix of order n with entries a_i_j where a_i_j = 0 whenever i = j and a_i_j = 1 otherwise

[stark spruce]

I should point out that this channel may close any moment.

[cedar widget]

.reopen

[safe radish]

✅ Original question: #help-23 message

[midnight sonnet]

what

[stark spruce]

This is not seans channel.

[cedar widget]

Right

.close

[safe radish]

Channel closed

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Use .reopen if this was a mistake.

[magic junco]

Just claim another one

[midnight sonnet]

complete bipartite means the neural network looking thingy?

[vocal badger]

please tag me as well if your doing graph theory

[rare crest]

alright!

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[rare crest]

for question f
answer: v5,v6,v10?
or answer: just v10?

[rare crest]

@vocal badger @devout shale

[vocal badger]

hi

[rare crest]

hi :)

[devout shale]

define descendant vertex

[vocal badger]

iirc its the ones below( in the root)

[devout shale]

but that seems like it is probably right

[rare crest]

understand

[vocal badger]

5,6,10

[vocal badger]

uh

[rare crest]

i understand

[vocal badger]

ohk

[rare crest]

:) thank you!

[vocal badger]

i didnt do anything tho

[rare crest]

its the effort!

thank you!

[safe radish]

@rare crest Has your question been resolved?

[rare crest]

.close

[safe radish]

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[thin lion]

Hi could I please have some help on this

[thin lion]

I don’t rly know where to begin 😭

Other than the definition

Clueless

[iron estuary]

okay

what does it mean to say that f is continuous at a point a∈R ? can you write the ε–δ definition explicitly

[thin lion]

Alright. If f is continuous at a point a in R then for all eps >0 there exists a delta >0 such that for every x in R with |x-a| < delta we have |f(x) -f(a)| < eps

[iron estuary]

right

now fix an arbitrary point $a \in \mathbb{R}$ and fix an arbitrary $\varepsilon > 0$

[flat frigate]

anflo

[iron estuary]

the problem gives the inequality $\lvert f(x) - f(y) \rvert \le L \lvert x - y \rvert$

[flat frigate]

anflo

[iron estuary]

what happens if you choose y = a?

[thin lion]

we have |f(x) - f(a)| <= L|x-a|

[iron estuary]

we want $\lvert f(x) - f(a) \rvert < \varepsilon$ right

[flat frigate]

anflo

[thin lion]

Yep

So we want

[iron estuary]

what condition on $\lvert x - a \rvert$ would be sufficient to guarantee
$\lvert f(x) - f(a) \rvert < \varepsilon$?

[flat frigate]

anflo

[thin lion]

L|x-a| < eps

If we have |x-a| < eps/L

[iron estuary]

exactlyy

so |x−a|<ε/L what should you choose for δ?

[thin lion]

eps/L?

An yes that works

Tysm

[iron estuary]

yes

you understood right?

[thin lion]

Wait so the general idea w these qs to prove continuity is that we have a fixed eps and a fixed a, then we want to choose a delta in terms of eps so that we can have that |f(x) - f(a)| < eps

Alright Ty

[thin lion]

Yea ❤️

.close

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[iron estuary]

one last check : in this problem, what feature of the inequality
$\lvert f(x) - f(y) \rvert \le L \lvert x - y \rvert$
made it especially easy to choose $\delta$ in terms of $\varepsilon$?

[flat frigate]

anflo

[thin lion]

is it the fact that it already gives us |x-y| bounds |f(x) - f(y)|

[iron estuary]

almost? just flip it arounf

[thin lion]

|f(x) - f(y)| bounds |x-y| ?

[iron estuary]

oh wait nvm i read that wrong 😭

[iron estuary]

this is correct

[thin lion]

Oh yea I was wondering 😭

[iron estuary]

yeah sorry

[thin lion]

No worries, thank you

[iron estuary]

no problem!

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[small rampart]

Given positive integer $n$. Polynomial $P_n = x^{3n}-3.4^{n-1}x^n-2^{3n-3}$\
a) Prove $P_n$ has exactly one positive solution (denoted $a_n$)\
b) $b_n = \frac{2-a_n}{n}$ and $c_n = b_1 + b_2 + ... + b_n$. Prove that $(c_n)_{n=1}^{\inf}$ has a finite limit

[flat frigate]

Nerdyasianguy

[small rampart]

For a), i let $t= \frac{x^n}{2^{n-1}}$, which turned $P_n$ into $t^3-3t-1=0$ (after dividing by $2^{3n-3}$) and it was easy from there

[flat frigate]

Nerdyasianguy

[small rampart]

For b), i need help

[muted sapphire]

what have you done so far

[small rampart]

I got the bounds 0<a_n<2

So b_n and c_n > 0 for all positive integer n

But i honestly don't know what to do next

[iron estuary]

hmm.. you can actually reuse your substitution from part (a)

[small rampart]

Sorry, i don't see how yet

I tried using the fact that a_n is a solution of P_n but couldn't

[iron estuary]

don’t go back to P_n directly

[keen tulip]

if t_0 is the solution to t^3-3t-1=0, what do you know about t_0

[small rampart]

Uhm... t_0^3-3t_0-1=0

[small rampart]

[iron estuary]

right.. and the important point is that t_0 is a fixed positive constant, independent of n

[keen tulip]

you should get from the first part that 1<t_0<2?

[iron estuary]

so in particular, $a_n^n = 2^{,n-1} t_0$ all the dependence on n is explicit

[flat frigate]

anflo

[iron estuary]

what does taking the n-th root do to a fixed constant?

😭

[keen tulip]

I'm merely expressing my agreement. You needn't cry.

[iron estuary]

haha fair dw

[iron estuary]

so $a_n$ is essentially controlled by $2^{(n-1)/n}$, and that tells you how fast $a_n \to 2$

[flat frigate]

anflo

[iron estuary]

yes that's all

😂 okay okay i get it

[small rampart]

Yes

[small rampart]

Oh it's fixed, yeah

[small rampart]

Make it tiny?

[iron estuary]

yeah

more precisely for a fixed $t_0 > 1$ we have $t_0^{1/n} \to 1$ very slowly as $n \to \infty$

[small rampart]

(also sorry i took a bit for each response, i turned off the computer to try it myself)

[flat frigate]

anflo

[iron estuary]

so $a_n = 2^{(n-1)/n} t_0^{1/n}$ differs from $2$ by a term of order $1/n$

[flat frigate]

anflo

[iron estuary]

that’s why dividing by n makes b_n summable

[small rampart]

Sorry i don't think i get this

What does "a term of order 1/n" means?

[iron estuary]

of order 1/n just means bounded by a constant times 1/n

[small rampart]

Does that refer to the big O notation?

[iron estuary]

yeahh

[small rampart]

So $a_n=2.\sqrt[n]{c}$ with $c$ being a constant in $(0,1)$

[flat frigate]

Nerdyasianguy

[small rampart]

(because t_0 <2)

I think i could calculate c too, though it doesn't seem necessary

[iron estuary]

yeah that's the right intuition

nah u don't need to

[small rampart]

What do I do now though?

[iron estuary]

all that matters for (b) is that $0 < c < 1$ since then $c^{1/n} = 1 + O(1/n)$

[flat frigate]

anflo

[small rampart]

Okay

[iron estuary]

ithis should be enough to control 2-a_n and conclude convergence of c_n

but do u understand

[small rampart]

Using derivatives and limits and such?

[iron estuary]

yes but very basic ones

wou only need the standard limit $\lim_{n \to \infty} c^{1/n} = 1$ for $0 < c < 1$

[flat frigate]

anflo

[iron estuary]

and the fact that $c^{1/n} = e^{(\ln c)/n} = 1 + O(1/n)$

[flat frigate]

anflo

[small rampart]

Why do we rewrite it with e?

[iron estuary]

because exponentials are easy to approximate when the exponent is small

that’s the cleanest way to see how fast c^(1/n) approaches 1

[small rampart]

ln c is negative too i guess

[iron estuary]

yes exactly

[small rampart]

Idk what to bound c_n with

Do i use inequalities?

Like Bernoulli's i guess

[iron estuary]

mhm inequalities are exactly the right idea

[iron estuary]

you don’t need bernoulli specifically any inequality that gives this is enough

[small rampart]

I think i just did it without inequalities

I'll type it out

[iron estuary]

you can just send the picture yknow..

[small rampart]

,tex
$1-c^(\frac{1}{n})=\frac{1-c}{\sum_{k=0}^{n-1} c^{\frac{k}{n}}} < \frac{1-c}{1+(n-1)c}$\ (Because 0<c<1)\
So $c_n < \sum_{k=1}^n \frac{2(1-c)}{(1-c)n+n^2c < n \frac{2(1-c)}{c} . \sum_{k=1}^n \frac{1}{n^2}.\
But i know that \sum_{k=1}^n \frac{1}{n^2} converges so i'm done.

[flat frigate]

Nerdyasianguy

$1-c^(\frac{1}{n})=\frac{1-c}{\sum_{k=0}^{n-1} c^{\frac{k}{n}}} < \frac{1-c}{1+(n-1)c}$\ (Because 0<c<1)\\
So $c_n < \sum_{k=1}^n \frac{2(1-c)}{(1-c)n+n^2c < n \frac{2(1-c)}{c} . \sum_{k=1}^n \frac{1}{n^2}.\\
But i know that \sum_{k=1}^n \frac{1}{n^2} converges so i'm done.
```Compilation error:```! File ended while scanning use of \frac .
<inserted text> 
                \par 
<*> 934710587122913311.tex
                          
I suspect you have forgotten a `}', causing me
to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.```

[iron estuary]

..

[small rampart]

Damn

[keen tulip]

,tex
$1-c^(\frac{1}{n})=\frac{1-c}{\sum_{k=0}^{n-1} c^{\frac{k}{n}}} < \frac{1-c}{1+(n-1)c}$\ (Because 0<c<1)\
So $c_n < \sum_{k=1}^n \frac{2(1-c)}{(1-c)n+n^2c} < n \frac{2(1-c)}{c} . \sum_{k=1}^n \frac{1}{n^2}.\
But i know that \sum_{k=1}^n \frac{1}{n^2} converges so i'm done.

slop

[flat frigate]

Oléagineux Distillièr IX
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[keen tulip]

this is so sloppy

[iron estuary]

..

[small rampart]

I forgot a }

[small rampart]

Uhm... My bad?

[keen tulip]

you forgot many things

[small rampart]

Sorry i have skill issues then

[flat frigate]

Oléagineux Distillièr IX

[iron estuary]

thiss looks basically right; its the right idea atleast

but i think it would be easier using inequalities

oh you're back again

[keen tulip]

also the indexing is wrong

[keen tulip]

i just fixed his latex

[iron estuary]

indexing aside the core idea is correct

??

[keen tulip]

I agree

[small rampart]

Thanks

.close

[safe radish]

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[small rampart]

a,b,p are odd numbers (p is prime). a+b ⋮ p, a-b ⋮ p-1
Prove a^b+b^a ⋮ 2p

[small rampart]

Here's my approach:
a=2m+1
b=2n+1
Then a+b=2(m+n+1) ⋮ p
=> m+n+1 ⋮ p
And a-b = 2(m-n) ⋮ p-1

Then i look at a^b+b^a and hope to use a^n+b^n=(a+b)(a^(n-1)-...+b^(n-1)) for odd n somehow

But i am unable to continue here

[faint hornet]

what are these 3 dots

divide?

[iron estuary]

yeah right?

[small rampart]

x ⋮ y is like y|x

"is divisible by"

[faint hornet]

oh what bizarre notation

okay

[small rampart]

I'm a bit surprised this notation isn't universal but ok

[stoic saddle]

try to use fermat's little theorem

[faint hornet]

well there is no need if you have |, but it's okay

[small rampart]

Hm...

So i need to get some stuff to the power of p somehow

[stoic saddle]

you can do better

[small rampart]

To the power of p-1?

a-b=k(p-1)

[stoic saddle]

yeah

[small rampart]

Oh

[faint hornet]

you can do it like this, almost

perhaps, think about flt like you said and rewriting your exponents in some way

[small rampart]

Well if i want them to have the same odd exponent, i could just let d=(a,b) i guess

[faint hornet]

?

[small rampart]

What?

[faint hornet]

i don't understand what you are trying to do

[small rampart]

If you suggest i use this approach then a=dx, b=dy (d=(a,b)) would give me a^b+b^a = (a^y)^d+(b^x)^d

That's what i meant

[faint hornet]

i see, i don't think that's useful

[small rampart]

Me too

[faint hornet]

rather, what i was going for is

of you find something that is a^n + b^n, then you are done since a+b is divisble by p and the sum is obviously even, okay

the problem is getting there

perhaps, look at your second thing where you know that p-1 | a-b

and stare at how this is useful keeping in mind flt

[small rampart]

Wait i think this is quite simple

a ≡ -b (mod p)

a^b+b^a ≡ (-b)^b+b^{a-b}.b^b
≡ 0 (mod p)

a^b + b^a is also obviously even, so it's divisible by 2

And (2,p)=0 completes the proof

[stoic saddle]

yeah

[faint hornet]

that's a nice way yeah

if you wanted to use your method from before, what i was getting is that seeing that this is equivalent to a^a + b^b, so if you added them together you could factor out (a+b) = 0 mod p

[small rampart]

.close

[safe radish]

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[rare crest]

im having trouble with the last question

[rare crest]

you may ignore the annotations

[open wedge]

oh gosh

[rare crest]

um?

[woven hound]

which part are you having trouble with?

[rare crest]

"Using your results...."

yelp

[vocal badger]

just calculate the estimated number of defective pieces per machine out of a total of 100 produced( 100 in whole factory)

[rare crest]

hmm, i was wondering if i could do that but then i chatgpt it and then it broke my brain

thanks

.close

[safe radish]

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[languid monolith]

hi. so suppose im on a job, and i make x dollars per hour. if i work over 40 hours a week, every hour above 40 gets 160% more than the regular hourly. if for example, i clocked in 50 hours, that would mean that the 10 hours get 160% the normal rate. if the hourly is x, would i multiply it by 1.6 to find how much more i got, or is it 2x+0.6x? or neither? im so bad at this shit

[opaque fern]

your second guess is correct

[languid monolith]

ooo okeoke also hi lex

[opaque fern]

if the rule is strictly "160% MORE than the regular hourly"

[languid monolith]

i was rooting for the second too

[opaque fern]

hii fijoo

[cloud hound]

Good point

I missed that

[languid monolith]

yeye its a programming exercise so we keep it simple

thank you

.close

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[safe radish]

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[cloud hound]

New one

[cloud hound]

Here for handwriting criticism as usual

[solar hazel]

your handwriting is medium

[cloud hound]

In all seriousness so the mods don't take me down, I'm looking for if the proof seems sound/applicable

[hard crest]

does your procedure work for statements such as \ $\forall x: \red( P_1x \implies \blue(\exists y: P_2(x, y)\blue)\red)$?

[flat frigate]

boughs of hayley 🪷

[neon vapor]

this is more readable than 90% of what i see from my students in university by the way

[cloud hound]

I didn't think about generalizations to n-variable predicates, that's a great point

[feral linden]

Where is this from?

[cloud hound]

I think technically this falls under case 1 and then case 2 (treating it as exists y P2(y) with x bound) but it should have been clearly established

[hard crest]

i think the main thing is that you'll need to make your induction step a little clearer, or maybe i just didn't notice it

[cloud hound]

Rosen discrete I got it from @midnight sonnet

Just doing a quick runthrough

[feral linden]

Thank you

[cloud hound]

Haha tbf I am a 3rd year student and I have just under 2 years of pure math research experience so although it many not be the best writing it's likely far from the worst :))