help-27

[devout snow]

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[uncut crow]

<@&268886789983436800>

[devout snow]

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[pastel rose]

Would it be sufficient here to prove that this now tends to 0?

[runic prawn]

mhm

[north roost]

are we

proving

${\sum_{n=0}^{\infty} \frac{a^n}{n!}}$ converges?

[woven radish]

k

[runic prawn]

good question

[uncut crow]

based on the term “sequence ratio test” probably not?

[north roost]

hmm

[pastel rose]

is this where i do scratich work to choose N?

[north roost]

what r u trying to prove

[pastel rose]

not sum

[runic prawn]

i like how we have to speculate on the question while the op doesn't respond

[north roost]

the sequence?

[pastel rose]

sorryy, i was attempoting it 😭

[pastel rose]

yes

[north roost]

isnt the ratio test for series not sequence

[pastel rose]

this is what i was given

[north roost]

oh damn

well if u have that

ye

maybe write

[pastel rose]

I always get confused behind the intuition of where i choose N

like

I have this rn

[north roost]

[ \lim_{n \to \infty} \frac{a^{n+1}}{(n+1)!} \cdot \frac{n!}{a^n} = \lim_{n \to \infty} \frac{a}{n+1} = 0]

[pastel rose]

Is that even correct approach

[woven radish]

k

[pastel rose]

Yep that was the idea

And im at the part where I’m proving that it’s 0

Does that work?

Ignore scratchwork

[misty crest]

your choice of N might not even be in N though?

[pastel rose]

why not?

oh if a is negative?

[misty crest]

a = -1, eps = 1 for example

just take absolute values

[pastel rose]

ohhh

i can do that right?

well tbf yh

nvm that q

[misty crest]

yea showing |a|/(n + 1) --> 0 is sufficient

and its easier to then do |a|/n

since |a|/(n + 1) < |a|/n

[misty crest]

the choice for N is easier here

[pseudo basin]

wait are we actually trying to write an epsilon proof of this

and like deliberately refusing to acknowledge a/(n+1)->0 as obvious

[misty crest]

yes this is kind of the point of an intro analysis course

you can also just do 1/(n + 1) --> 0

[pastel rose]

💀 let me explain, im on my term 1 of analysis

[pseudo basin]

i mean yeah but also it's so ass to have to prove the analysis equivalent of 2+2=4 each and every single motherfucking time

[misty crest]

🤔

sure but he is struggling with it so its obviously worth doing

[pseudo basin]

what, are we also gonna be disallowed from writing a+b+c because we have to make explicit written reference to (+Assoc) every time we rebracket?

[misty crest]

you only get to stop doing that when you actually can prove it easily or if he can prove more general statements

[pastel rose]

why are u geting rage baited 😭

[misty crest]

what was rage bait?

[pastel rose]

yh i lowkey just need to practice it more

me doing epsilon proofs im guessing

[misty crest]

this is an analysis course

the whole point is for him to learn how to do these

😭

[pseudo basin]

yeah and they have to do it PROPERLY and assuming NOTHING and doing EVERYTHING BY THE BOOK AND PROPERLY AND WITH THE PROPER (I.E. ENORMOUS) AMOUNT OF BUREAUCRACY

[misty crest]

are you ok?

if he was experienced enough then he wouldn't have to prove this

[pastel rose]

is there anything wrong in doing this? I have an exam coming up, i want to make sure i do my best in these type of qs

[pseudo basin]

nah i am exaggerating for comedic effect atp i shut up now

[misty crest]

you should become familiar with this sort of thing

bounding it by something that we can do the eps-N proof easier for

which usually means making the denominator less complicated

[pastel rose]

oh yea bounding them i see

ah yea i was thinking that, lol i decided to make N = ceiling(|a|/n+1)+ 1 to try to bypass this

but yea ill bound it

[misty crest]

see like

why introduce floors and ceilings for no reason

[tender wharf]

We thought you were choking

[pastel rose]

yea i see wym

Is this logical?

[misty crest]

choose N = |a|/n makes no sense

you're choosing N in terms of epsilon

we want |a|/n < eps

so n > |a|/eps

(if a = 0 the fact that it converges to 0 is trivial)

but still needs to be mentioned

[pastel rose]

Oh wait, I don’t need the N defined

Yea cause it’s for all N>=n

Ah yeaa 😭 mb

[misty crest]

wait nevermind i flipped it wrong

😭

[pastel rose]

Wdym

[misty crest]

nothing

you don't need to split into cases and mention the possibility of a = 0

anyways

[misty crest]

you see how i got this?

[pastel rose]

yes

[misty crest]

ok so our choice of N is clear

we take N = |a|/eps

i guess a = 0 is only necessary if you consider N = {1, 2, ...}

[pastel rose]

yea i can inclue that

but that should be fine

[pastel rose]

yep and then we have from |a|/n+1 < |a|/ n < eps

[misty crest]

also why are you saying by archimedian property there exists n thing?

if anything you'd be saying that for N

not n

the n part should go like

thus if n > N then n > |a|/eps and manipulate this to get the inequality we want

[pastel rose]

oh the archimedian is for the N

[misty crest]

yes

the n part is really "if n > N" or "hence for all n > N"

[pastel rose]

oh i see

ah so its always the eps and N we deal with

[misty crest]

wdym

[pastel rose]

yea fairs that works

[misty crest]

like N is a function of epsilon?

[pastel rose]

no no

[pastel rose]

nah nvm, this makes sense

alright so we can choose an N > |a|/eps then manipulate it so that |a|/n < eps and hence |a|/n+1 < eps

[misty crest]

perfect

[pastel rose]

alrighty

tysm

[misty crest]

just mention that "if n > N" after you choose the N

[pastel rose]

Yea thats fine

sorry for the confusion 😭, ill probably review these again more often

[misty crest]

good idea

[pastel rose]

.close

[devout snow]

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[devout snow]

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[sand quarry]

If I have a conditionally convergent series of vectors $\sum v_n$, does the set of all possible rearrangement sums have to be a linear subspace?

[lunar harbor]

Idts - take $\left(\frac{(-1)^{n+1}}{n}, \frac{1}{2^n} \right)$.

[woven radish]

Civil Service Pigeon

[lunar harbor]

The second coordinate is absolutely convergent, so its sum is $1$ under any rearrangement. But the series $\sum \frac{(-1)^{n+1}}{n}$ can be rearranged to converge to any real number.

[woven radish]

Civil Service Pigeon

[lunar harbor]

no origin -> not a subspace

take this with a grain of salt though

not perfectly fluent in this content

[pseudo basin]

nah looks solid

[lunar harbor]

[sand dove]

Though I believe it will be an affine subspace

[sand quarry]

I get that but like

my worry is about what happens when the "conditionalness" isn't confined to the first few slots if you can put it that way

[sand dove]

So you want every coordinate to be exactly conditionally convergent

And none are absolutely convergent

Fine, take ((-1)^(n+1)/n, (-1)^(n+1)/n + 1/2^n)

[sand quarry]

​I actually just looked it up earlier and apparently Kadet found a series in L^2 where the set of rearrangement sums consists of exactly two distinct points {x, y}

That completely breaks the affine intuition, right

[sand quarry]

Hmmm

[pseudo basin]

can you share \

[sand quarry]

That still gives you a line though riht

[sand quarry]

One second im trying to find the paper online i just snuffed this from stack exchange thing

Also why the opencry anywyas?

[violet wind]

L^2 is euclidean right

[sand quarry]

Oh wtf ok nvm

Im dumb 💀

I totally misread what I was meant to be reading 💀

[sand quarry]

Yeah

[violet wind]

I'm confused, if you look at one component, either it's absolutely convergent, or it can take on any real value by rearrangement

So if there's 2 possible points of convergence, the component in the direction of their difference is not absolutely convergent

[fossil flower]

I have a stupid question if I had a polyhedron with infinite sides and faces would it be considered a sphere or nah and why?

[violet wind]

! occupied

[devout snow]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[fossil flower]

ok sry

[sand quarry]

Yeah i think I got it. I was just under a misapprehension I think

Thanks

.close

[devout snow]

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[violet wind]

Np

[devout snow]

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[idle dune]

hii

[idle dune]

isn't it s/2 = t?

[winter torrent]

for which, x?

[idle dune]

yea

for x

[winter torrent]

i mean. those are the same equation

but you have 1t+0 on one side and 2t on the other (which we relabel as s)

[idle dune]

wait im a bit confused

i don't even know how he got here I'll be honest can u explain?

[winter torrent]

do you know what a skew line is? can you explain it to me?

[idle dune]

when two lines are not intersecting and never parallel

[winter torrent]

okay, great

so we want to find out whether they're intersecting

we have l1 (t) and l2 (t)

we want to figure out if we can feed numbers into each of these to give us the same point

but they could be different numbers as input so it's confusing that we use t for both

so instead we use s as input for l2

so l2(s) = (2s)i - sj - sk

[idle dune]

ooohh

i gotchu

[winter torrent]

and l1(t) = ti + tj + (t+1)k

[idle dune]

(t+1)k?

not (tk + 1)

[winter torrent]

correct; tk + 1 doesn't make sense

remember, k is a vector

[idle dune]

oh yeah

[winter torrent]

the ijk thing is basically just another way of writing <a, b, c>

[idle dune]

thank you so much

!close

.solved

[devout snow]

Channel closed

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[idle dune]

❤️

[devout snow]

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[solid osprey]

find the least positive integer $n$ such that $\frac{1}{\sin(45^\circ)\sin(46^\circ)}+\frac{1}{\sin(47^\circ)\sin(48^\circ)}+\dots+\frac{1}{\sin(133^\circ)\sin(134^\circ)}=\frac{1}{\sin(n^\circ)}$

[woven radish]

ihave<skissue>

[solid osprey]

yeah... legit no idea where to start

actually wait

product to sum gets

$\frac{2}{\cos(2^\circ)}\left(\frac{1}{\cos(91^\circ)}+\frac{1}{\cos(95^\circ)}+\dots+\frac{1}{\cos(267^\circ)}\right)$

fuck

[ionic harness]

Extra bracket with the first cos

[solid osprey]

ty

[woven radish]

ihave<skissue>

[solid osprey]

🤔

smells like a telescopic

ohh wait

pair cos(x) with cos(180+x) or smth?

wait noo

[tropic sinew]

it just flips signs and won’t collapse the sum

[solid osprey]

[tropic sinew]

try rewriting 1/(sin x · sin(x+1°)) as a difference of cotangents because sums of successive differences naturally telescope causing almost all intermediate terms to cancel

or you can rewrite as difference of cosines which would be harder i think

[solid osprey]

sorry something came up, lemme try

naturally im guessing like cot(x)-cot(x+1)

so like (cos(x)sin(x+1)-sin(x)cos(x+1))/(sin(x)sin(x+1))

top is sin((x+1)-x)=sin(1)?

[tropic sinew]

yes that's correct

[solid osprey]

so
$$\frac{1}{\sin(1)}(\cot(45)-\cot(46)+\cot(47)-\cot(48)+\dots+\cot(133)-\cot(134))$$

[tropic sinew]

shouldn't it be 1/sin 1

[solid osprey]

oh yea

[woven radish]

ihave<skissue>

[solid osprey]

cot(180-x)=-cot(x) right?

so pair 46 with 134, 47 with 133, ect

[tropic sinew]

yes exactly

[solid osprey]

so whats left is (cot(45)+cot(90))/sin(1)

the first is 1, the second is 0, so were left with 1/sin(1), so n=1?

[tropic sinew]

wait the uncancelled terms are cot 45 and cot 134

[solid osprey]

46 cancels 134 no?

[tropic sinew]

nvm sorry

[solid osprey]

uhh yikes (its in degree mode)

did i do something wrong?

or is desmos innacurate

[steel sage]

You got it

Anflo, me, both confirm it

[solid osprey]

oh wait

[solid osprey]

ok ic i summed it wrong

okay then, thank you guys!

.solved

[devout snow]

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[ionic harness]

Ideas on approaches for this

[tropic sinew]

can we try a recursive decomposition by initial patterns?

ythe combinatorics becomes clearer once the first few bits are fixed

[ionic harness]

A what 🙂

[tropic sinew]

oh i just mean split the strings into cases based on how they start (like 00 01 10 11) and see how each case reduces to a shorter problem

[ionic harness]

Ah

As in

If 00, all strings of length 2^n-2 work

If 01, all a_n-2 work

If 10, all strings of length 2^n-3

if 11, same as 01

[steel sage]

Wasn’t this solved by that person who posted it originally himself

Their thought was dividing it into two cases 0x and 1x, and divide the first 0x case into two sub cases 00x and 01x

[ionic harness]

So you'd get as a recurrence a_n = 2^n-2 + 2a_n-2 + a_n-1?

I think that's different from what they got

[tropic sinew]

yeahh that's much cleaner

[ionic harness]

Ah

[steel sage]

So theirs was correct

[ionic harness]

Hm

Am I double counting something then

[tropic sinew]

yes

[ionic harness]

How did mine end up with a different answer

AH

No I don't see

[tropic sinew]

strings starting with 11 are counted both in the 11 case and inside the broader 1x case

[ionic harness]

Oh I see what happened

By separating them I create a problem

Okay

So I would get a_n = 2^n-2 + 2a_n-2 + 2^n-3

Which is the same as theirs with one term expanded

Wait no it's not

Dangit

I feel like my cases shouldn't have any overlap anymore

So why aren't they equivalent to their cases

[tropic sinew]

your cases still overlap in a hidden way because one of your counts ignores the “must contain 00” condition

[ionic harness]

Back

Hm

I have

00 + any string of length n-2

01 + any string containing 00 of length n-2

10 + 0 + any string of length n-3

Oh I guess with 10 the 0 need not go there

And then 11

So easier to just hav

00
01
1

Okay, makes sense

.close

[devout snow]

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[devout snow]

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[lost laurel]

So to start $X_1+X_2+X_3 \sim N(0,3)$

[woven radish]

wai

[lost laurel]

And I was looking at solved examples . I don't get how this is equivalent to
$\frac{X_1+X_2+X_3}{\sqrt{3}} \sim N(0,1)$

[woven radish]

wai

[ionic harness]

I feel like that's a theorem somewhere

[lost laurel]

oh wiat

*wait

[fossil locust]

well, $Var(X_1 + X_2 + X_3) = Var(X_1) + Var(X_2) + Var(X_3) = 1 + 1 + 1 = 3$

[lost laurel]

I think I got it

[woven radish]

south

[tropic sinew]

that’s just variance normalization of a normal random variable

[fossil locust]

cool

[lost laurel]

Thanks

Now from this we get $X_4^2+X_5^2+X_6^2 \sim \chi^2(3)$

[woven radish]

wai

[lost laurel]

Now, I suppose this is a t-dist?

[tropic sinew]

exactl that theorem applies verbatim here, with v=3

[lost laurel]

So I have $\frac{X_1+X_2+X_3}{\sqrt{ \chi^2(3)}} = \frac{W}{\sqrt{3}} \sim t(3)$

[woven radish]

wai

[tropic sinew]

almost

the denominator needs to be sqrt(U/3), not just sqrt(U) Once written that way, it’s exactly a t3

[lost laurel]

hmm?

I accounted for that I think

[tropic sinew]

[
W = \frac{(X_1+X_2+X_3)/\sqrt{3}}{\sqrt{(X_4^2+X_5^2+X_6^2)/3}}
]

[woven radish]

anflo

[lost laurel]

ooos

oops

missed the √3

[tropic sinew]

[
\frac{X_1+X_2+X_3}{\sqrt{X_4^2+X_5^2+X_6^2}}
\sim t_3
]

[woven radish]

anflo

[lost laurel]

Thanks so much!

[tropic sinew]

mhm you're welcome

[devout snow]

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[loud urchin]

Can anybody explain how to do this?

[loud urchin]

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. (Do this on paper. Your instructor may ask you to turn in this graph.)

y = 2x+3, y = 15-x^2,x = -1, x = 2

[tender wharf]

Plot the graphs

[loud urchin]

Got this

Im pretty sure its top minus bottom

[loud urchin]

He then asks for area s of the region, which im confused about.

.close

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[severe iris]

Find the value of x from the following proportion. Help me.

[sand pumice]

what have you tried

[severe iris]

I applied the properties of proportions: (x-1)(x+3)=(x-2)(x+2)

[rapid wigeon]

!status @severe iris

[devout snow]

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

[rapid wigeon]

Does the proportionality keep on going in the question?

?

Eh why is there a sudden silence

[severe iris]

I'm not sure. I just applied the property $(a/b = c/d \Rightarrow ad = bc)$ to simplify the equation. Should I proceed differently?

[woven radish]

YUSUBA

[severe iris]

my english so bad

[rapid wigeon]

Nah np

[rapid wigeon]

Do you know componendo dividendo?

[remote mantle]

Hii do you know how to expand those brakcets?

[winter torrent]

that sounds good, now solve for x

[severe iris]

I've heard of it, but I prefer using cross-multiplication for this one as it feels more straightforward to me.

[severe iris]

Parentheses?

[rapid wigeon]

No but it is easier

[prisma latch]

À

[rapid wigeon]

But if ur getting the answer like that then you can apply the old method?

[prisma latch]

E cứ nhân nó vào là đc e

[ebon coyote]

( and ) [In the US specifically, for some reason, these symbols are referred to as parentheses]

[remote mantle]

Yess the foil method

Or otherwise

[rapid wigeon]

Like what is the doubt then? @severe iris

[severe iris]

Vâng , để em thử

x= -0,5 ?

[deep abyss]

,w (x-1)(x+3)=(x-2)(x+2)

[woven radish]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by ._ch3rry_.7

[severe iris]

thanks.

[devout snow]

@severe iris Has your question been resolved?

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[idle coral]

Why, in the definition of the line integral, is the vector field defined from A ⊂ Rn to Rn, and not in a space of a different dimension?

[pseudo basin]

you need to be able to dot F(gamma(t)) with gamma'(t)

gamma'(t) lives in R^n

F(gamma(t)) lives in the output space of F

so the output space of F also needs to be R^n in order for the dot product to be a thing

[devout snow]

@idle coral Has your question been resolved?

[idle coral]

thankss

[devout snow]

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[gritty saddle]

Ive trouble with this exercise

[rain summit]

!original

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[gritty saddle]

Its in greek

[rain summit]

then try to translate

[gritty saddle]

Yep

Let R³(x, y, z) and the change of vars to (u, v, w) with x=uv, y=exp(v), z=w. Where is this change okay and how can i express the basis (∂u ∂v ∂w) on the basis (∂x ∂y ∂z) with coefficients with respect to (x, y, z)

[steel sage]

The other way round is easy, so do the other direction and take inverse?

[gritty saddle]

For the first qs i don't know which F to use, like the (u, v, w) to (x, y, z) or the inverse. Theres no problem with the det whichever i use, i know, but whats the right option

[steel sage]

Like the inverse is (1/v, -u/ve^v, 0; 0, 1/e^v, 0; 0,0,1). You just rewrite every entries in terms of x,y,z

Like v=ln(y)

u=x/v=x/ln(y)

w=z

[gritty saddle]

Well yes, like we let v≠0 so thath det≠0

Which is equivalent to y≠1

For the second question

[steel sage]

Yeah

[gritty saddle]

(∂u ∂v ∂w)=J(f(id)^(-1)))(∂x ∂y ∂z)

Bec (R³,id) is the chart for coord (x, y, z)

And f(u, v, w)=(x, y, z) as above

Is this right?

[steel sage]

Yeah

[gritty saddle]

Okay

Me problem is

This holds for every p on R³ like (∂u|p ∂v|p ∂w|p)

So p now is on (x, y, z) or (u, v, w)

How i express it?

Because we have Also (∂x|p ∂y|p ∂z|p)

[steel sage]

Oh… I have no idea when p=(u,0,w)….

[gritty saddle]

What do umean

[steel sage]

I mean we solved it when v isn’t 0. My reasoning fails when v=0 so no idea how to proceed at this case

[gritty saddle]

Lets take that v≠0 equiv y≠1

Now, p is like (x, y, z) or (u, v, w)

Its a point in R³ but how i express it

[steel sage]

No idea… Probably you can reply to that 1/0=infty guy

[gritty saddle]

Ok, thanks for your help

[gritty saddle]

Let R³(x, y, z) and the change of vars to (u, v, w) with x=uv, y=exp(v), z=w. Where is this change okay and how can i express the basis (∂u ∂v ∂w) on the basis (∂x ∂y ∂z) with coefficients with respect to (x, y, z)

[devout snow]

@gritty saddle Has your question been resolved?

[devout snow]

@gritty saddle Has your question been resolved?

[devout snow]

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[wanton axle]

can someone tell me whether my proof of transitivity is formally correct? I think it should be because I don't see any mistakes in my implications, but the master solution did it with more case distinctions which seems unnecessary to me. But then again the professor would probably find the better solution than me, so maybe i did sth. wrong.

[wanton axle]

Is my question inappropriate for this server or why the emojis? are these kind of questions not intended for this?

.close

[devout snow]

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[devout snow]

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[open cargo]

I need to solve this equality but I cant get any further than making one polynominal and then moving the 3 to the other side and idk if there is any other way tbh

[astral cargo]

factor the x^2 - 2x -8

[open cargo]

x(x-2)-8?

[astral cargo]

wait im bumb

[urban harbor]

yes, moving the 3 to the other side and then trying to factor the resulting polynomial sounds like the way, what was the polynomial you got after doing that?

[open cargo]

x^3 + 4x^2 - 20x - 51

if i got it right

[astral cargo]

rational root theorem

[open cargo]

already substracted the 3

[urban harbor]

hm not sure, there is a formula for solving general cubics but im sure thats not whats called for here 🤔

[astral cargo]

try rational root theorem

u cant do this w/o calculator

[urban harbor]

how did this problem come up?

it doesnt seem to be something with "nice solutions"

when looking at a graph

[fast garden]

and if this was the problem to begin with, what class was it for?

[urban harbor]

well...you could use the cubic formula but...

thats not the most fun or productive way lol

[open cargo]

well that's actually what made me doubt so much, didnt expect to get answers other than full 'nice' numbers

[astral cargo]

just memorize this and ur good @open cargo

[fast garden]

it's easier to just depress / reduce and then work out in terms of roots of unity

but i think that is not something you'll learn before taking something like galois theory, so it is a bit dubious

despite being elementary

[open cargo]

oh my days ion even think i learned that so might as well just leave this aside (?)

except if thats the only solution tho

[astral cargo]

hes joking

nobody is memorizing that formula

[open cargo]

thanks the lord

[fast garden]

not with that attitude,

parv what was the original problem

[open cargo]

this is the only thing I got

and got told to solve this

[fast garden]

oh i see

[open cargo]

😅

[kind tiger]

by who/from where ?

[fast garden]

yoooo hi acman

[kind tiger]

o/

[fast garden]

been years...

[urban harbor]

yeah it's not a formula thats NOT taught in schools, maybe mentioned in some "history of math" or esoteric algebra college courses but no one is gonna memorize that 😛

[open cargo]

basically a bunch of exercises from uni (given by my teacher)

[urban harbor]

the history of the discovery of that formula is actually kinda interesting and dramatic though

it was a closely guarded secret used for "mathematical contests" for some while or some such

[open cargo]

wow, might wanna learn about it once I'm back on tracks with the basics

okay but if none of y'all got any "nice" solutions then I might have been right the whole time and been doubting for no reasons hahaha

thanks for the help guys!

[urban harbor]

but anyway the point is in general cubic equations are difficult to solve

unless there's some special factoring trick or special case

[open cargo]

.close

[devout snow]

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[north glen]

Leibnitz's mistake

[north glen]

I don't understand why can't we count (6, 6) and (6, 6) as two different solutions

Like, number wise, it's the same.

But if (5, 6) != (6, 5), then (6, 6) != (6, 6)

idk now sounds silly

[misty crest]

oh wait

let me read the problem first

[misty crest]

because these both say the same thing

[uncut crow]

i didn’t know leibniz was this foolish

i usually like to think in terms of a first and second die roll

it’s the same as rolling them “at the same time”

[final drift]

huh whats going on here

[north glen]

Yeah idk what I was thinking. I mean I do, but I now understand why that wouldnt be true

[uncut crow]

i know that too

i should go gambling

[north glen]

me too

[devout snow]

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[devout snow]

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[warm cloud]

help pls do i have to do another integration by parts or do i do something to simplify

[stone knoll]

I mean if you want to there is a real easy way to get around it

[warm cloud]

pls

[stone knoll]

you know about taylor expansions?

[warm cloud]

no

[stone knoll]

darn okay

[warm cloud]

just by parts

[stone knoll]

no I mean for like functions

like do you know about like series expansions of functions?

[warm cloud]

we havent learned taylor yet just standrd integration and integration by parts

[warm cloud]

like binomial expaniosians?

[stone knoll]

no

like

what's e^{i5x}

[warm cloud]

oh like polar cordinates

we havent done that yet

[stone knoll]

not really polor

so the trick to get around by parts is this

[warm cloud]

modulus?

[stone knoll]

nope

[warm cloud]

yeah i dont know i was told to only use by parts for this

explicitly

and i dont rlly know how'

[stone knoll]

oh

[warm cloud]

i think my teach wants to us to suffer

[stone knoll]

well if you have to do by parts just pick dv=e^{5x]dx

[warm cloud]

i did

but using uv -(integralsign) vdu the vdu is like aids

its like 1/5e^5x*-5sin5x

[stone knoll]

that's fine

you just have to do it again

pick the same thing

[warm cloud]

new u and dv?

[stone knoll]

also the 5's cancel

[warm cloud]

how

[stone knoll]

1/5 *5

is one

[warm cloud]

oh yeah lol

and the negative cancels out the - from the uv - equation

from the 1/5 and -5 cancelling out

right?

[stone knoll]

$$\int e^{5x}\cos(5x)dx=\frac{1}{5}\cos(5x)e^{5x}+\int e^{5x}\sin(5x)dx$$
$$\int e^{5x}\cos(5x)dx=\frac{1}{5}\cos(5x)e^{5x}+\frac{1}{5}\sin(5x)dx-\int e^{5x}\cos(5x)dx$$
$$2\int e^{5x}\cos(5x)dx=\frac{1}{5}\cos(5x)e^{5x}+\frac{1}{5}\sin(5x)dx$$
$$\int e^{5x}\cos(5x)dx=\frac{1}{10}\cos(5x)e^{5x}+\frac{1}{10}\sin(5x)dx+C$$

[warm cloud]

and becokes postivve

[woven radish]

Calisto

[stone knoll]

yee

but the short cut is this

$$e^{i5x}=cos(5x)+isin(5x)$$
$$e^{-i5x}=cos(5x)-isin(5x)$$
$$e^{i5x}+e^{-i5x}=2\cos(5x)$$

$$\int e^{5x(1+i)}+e^{5x(1-i)}dx=\frac{1+i}{10} e^{5x(1+i)}+\frac{1-i}{10} e^{5x(1-i)}$$

[warm cloud]

where is 1/5 sin coming from

[stone knoll]

the second integration by parts

[woven radish]

Calisto

[warm cloud]

ok ty

[stone knoll]

yee

As a way to check your understand of this integral see if you can do this one:

$$\int e^{ax}x^2dx$$

[woven radish]

Calisto

[stone knoll]

this one uses the same concepts

but after that one see if you can find a solution to this one

$$\int e^{ax}x^ndx$$

[woven radish]

Calisto

[stone knoll]

what's the equation?

oh.... oh boy okay let's see here

you definitely want to start off by getting rid of the radicals but that's gonna be tricky

[devout snow]

@warm cloud Has your question been resolved?

[lyric hornet]

!occupied

[devout snow]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[stone knoll]

I am 95% sure that person was a troll

[lyric hornet]

[tender wharf]

Yes they were muted

[lyric hornet]

,w cbrt(n+sqrt(n^2+8))+cbrt(n-sqrt(n^2+8))=8

[woven radish]

Results provided by WolframAlpha

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Requested by xxmrfancyu2xx

[lyric hornet]

not a terrible solution

[devout snow]

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[devout snow]

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[covert umbra]

I dont know how to start for Question A

[lost laurel]

,rccw

[sand quarry]

Do you know what a standard matrix is

[covert umbra]

Yes

In this case its 1.0.0 0.1.0 and 0.0.1 i think

[devout snow]

@covert umbra Has your question been resolved?

[lunar harbor]

Yeah the columns are $T(e_1)$, $T(e_2)$, and $T(e_3)$ (in that order)

[woven radish]

Civil Service Pigeon

[lunar harbor]

You’re given $T(e_1)$ so that’s easy

[woven radish]

Civil Service Pigeon

[lunar harbor]

To find $T(e_2)$ and $T(e_3)$, we should use the properties of $T$ since it’s (presumably) a linear transformation

[woven radish]

Civil Service Pigeon

[lunar harbor]

Do you remember the definition of a linear transformation?

[covert umbra]

Yes i do i have some progress

[lunar harbor]

Oh ok that’s good so far

Do you know how to do the second part

[devout snow]

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[lost laurel]

Really sorry but very lost here

[lost laurel]

I suppose I use this?

S is the sample variance

X bar is the sample mean

[vagrant skiff]

Why do you need this?

X and Y are normal

[lost laurel]

yes

[vagrant skiff]

so their sample means are exactly normal

[lost laurel]

so Xbar and Y bar are nomal

fine

[vagrant skiff]

We can work directly with normal dist Y bar - X bar

[lost laurel]

hmm,how do I do that

[vagrant skiff]

What do you mean how

What are the distributions of the sample means

figure that out first

[lost laurel]

normal, right

I'll continute this in a bit, brain is nto braining

thanks

.close

[devout snow]

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[gray oriole]

Use the pigeonhole principle to show that sum of any two elements from any of the six-element subsets of the set ${1,2,3,4,5,6,7,8,9}$ will be greater than 10.

[woven radish]

Sean

[gray oriole]

i dont remember the question correctly guys, this is the closest recollection that i can recall

[celest frigate]

With the Pigeonhole Principle (PHP), we first choose boxes, what do you think would be best

[gray oriole]

idk man

how to identify the boxes?

[celest frigate]

okay what is the least greatest number in a set

[gray oriole]

5

[celest frigate]

hmmm

not exactly

[gray oriole]

wat is it then

[celest frigate]

6

[gray oriole]

oh its a 6 element

mb

alr now ?

[celest frigate]

list out pairs of numbers that make sum 11

starting from 5,6 which is the first pair

[gray oriole]

5,6
7,4
8,3
9,2

done with what

how

we hv to apply php

[ebon coyote]

[You're using the pigeonhole principle to prove that you can't construct a subset that does not have the desired property]

You've now got boxes [1], [2,9], [3,8], [4,7] and [5,6]

[gray oriole]

how does this work

[gray oriole]

sure

how do i proceed

why is 1 a box btw?

[ebon coyote]

You can go 5 boxes, 6 pigeons

[celest frigate]

ooh mb

6 elements. I was having 5 in my mind. my bad. Thanks

[ebon coyote]

because we can freely pick 1 without problem

[gray oriole]

why is that?

[ebon coyote]

It (the number 1) doesn't change whether we can make an 11-sum <----------

[gray oriole]

what do these boxes represent, are they subsets?

[ebon coyote]

they're your literal pigeonholes

[celest frigate]

if you choose two elements from one of the pigeonhole, then the sum becomes 11 - this means our condition is true

[ebon coyote]

Meaning maximally you can pick ONE element from each pigeonhole

But there're only 5 pigeonholes; and you need 6 pigeons (elements)

[gray oriole]

Guys I'm sorry but I am not getting it

I mean I know that the php is about n+1 pigeons and n pigeons

But I can't understand how this applies here

[ebon coyote]

Prerequisite - do you know how the pigeonhole principle works?

[gray oriole]

Well yeah I did a bunch of problems about picking colored objects and stuff

But this one is kinda different

[ebon coyote]

Right - in this case we need to make the boxes ourselves

[gray oriole]

I did some problems with ceil(N/k)

[ebon coyote]

The thing is, when we make boxes, we need to make "being in the same box" have some specific property

[celest frigate]

Ok so if you pick two elements from one of the boxes, the sum exceeds 10, you agree? (dont mind the sloppy arrangement)

[gray oriole]

Okay, but why do we choose the pairs of numbers having sum = 11 as the boxes?

[celest frigate]

because 11>10, which is the question

[ebon coyote]

Because, see here, this is the property we want the boxes to have

[gray oriole]

why not 12

[celest frigate]

It can be 11, but not 10 and may or may not be 12. 1,2,3,4,5,6 for example

[ebon coyote]

Because it suffices to show that there must be two numbers that sum to 11

[gray oriole]

okay so we are using php to show that any two elements from six elem subset sum to 11?

[ebon coyote]

yes

[gray oriole]

alright so now I can pick like the 'two' elements as the boxes

[ebon coyote]

So your elements are the pigeons

[celest frigate]

you have to pick atleast one box where both elements are included

[gray oriole]

i can do 2,9 and 3,8 and 4,7 and 5,6

[ebon coyote]

These boxes are the pigeonholes

You can fill each box with exactly one pigeon

But then you still have an extra pigeon

[gray oriole]

So I need to prove that when we pick the pigeons at least one of the box will have two pigeons

[celest frigate]

yes

[gray oriole]

which is the equivalent of having two numbers summing to 11

[celest frigate]

yes

[ebon coyote]

Right

[gray oriole]

But i want to have only these boxes

1 is kinda unintuitive

[ebon coyote]

You've got an extra box, though

Because when picking the fifth element, you can freely allow it to be 1

You can't make 11 with that 1, because you don't have a 10 to match it with

[gray oriole]

If we don't have 1 as a box then we do not have any place to accomodate when we choose 1 from the set

ig that is why i need a box for [1]

[celest frigate]

1 is a box with only one possible element. You either pick it or not

[ebon coyote]

Okay so

You need to box all the elements beforehand, first

[gray oriole]

alright well i got 5 boxes now

there are 6 elements

ceil(6/5) = 1

[celest frigate]

You dont need ceiling function tbh

[gray oriole]

mb its 2

[celest frigate]

6>5 suffices

[ebon coyote]

Try and learn this principle first

[gray oriole]

but we do it that way

[celest frigate]

Okay then go on

[gray oriole]

what if i was asked sum > 11

[ebon coyote]

That's not true

[gray oriole]

What

[ebon coyote]

Counterexample: take the set of 1 through 6 inclusive

[gray oriole]

alright lets do 7 element subset

[celest frigate]

That works

[ebon coyote]

Then yes

[gray oriole]

1 to 7

7+5

[celest frigate]

yeah

[gray oriole]

alr ima do 7,5
8,4
9,3
6,6

is that all

i cant have 6 two times can i

[ebon coyote]

"6,6"?

[gray oriole]

it is a set

[ebon coyote]

yh

[gray oriole]

ok well then what should i do?

[ebon coyote]

it's going to be a one-element box

[gray oriole]

1 and 6, two one-element boxes?

[1], [2], [6], [5,7], [8,4], [9,3]

These are my boxes

And I have 7 pigeons

Somethings missing

alr

so now we have ceil(7/6) = 2

[celest frigate]

yeah. So this means atleast one pigeonhole has 2 pigeons

[gray oriole]

Can you explain the boxes thing a bit more

[celest frigate]

what part

[gray oriole]

I kinda get it but I don't get it fully

like how do I decide which boxes to make

[celest frigate]

It depends on the question really

Its a skill learnt best by practice

[gray oriole]

well the sum questions would all have the same pattern

do you remember any other type of question

so i could practice

makin boxes yk

[celest frigate]

I can give you some examples, basic geometry is fine?

[gray oriole]

sure sure

Won't hurt to take a look

[ebon coyote]

google

pigeonhole principle questions

Here are some results: ...

Like... ^^^^

[celest frigate]

lets say you have a square of length 3 * 3, and 10 points, we need to show that atleast two points have distance less than square root 2

the points lie inside the square. try this. And for more just google

[ebon coyote]

shit that's a good question tho

[gray oriole]

im ded

[celest frigate]

Its a slightly hard example, its fine if we fine skip it actually

[gray oriole]

10 points at random?

[celest frigate]

just google PHP questions

[celest frigate]

yes

inside the square

[gray oriole]

so i have to find 9 pairs of points with distance < sqrt(2) ?

[celest frigate]

one pair of points

[ebon coyote]

The trick is to determine what boxes to draw

(and hint - "box" is very literal in this example)

[celest frigate]

yeah once you get the boxes, its a baby problem

[gray oriole]

why do i need just a single pair?

[celest frigate]

atleast two points

[gray oriole]

i mean if i am choosing 10 points then i gotta have 4 pairs at least

[junior mauve]

I have a sample question that's along the same lines

[ebon coyote]

You're misunderstanding the question

[gray oriole]

or maybe 6 pairs

[junior mauve]

let me know after this if you're down

[gray oriole]

so we have more than 10 points

[celest frigate]

Sometimes you wont know the boxes directly

[ebon coyote]

If you pick 10 points randomly, there ALWAYS exists A pair of points that is less than root(2) apart; the question is to prove THIS

[celest frigate]

like the previous question was direct

[gray oriole]

Send it here, I will take a look thanks

[gray oriole]

goddamn

Where do I even start

[celest frigate]

think along a box

[gray oriole]

I have to pick 10 points or smth

[celest frigate]

hint, you have to make the boxes/pigeonholes yourself and then place the points in them

[junior mauve]

you are given seven line segments of random lengths. you know that they all are between 1 and 10 inches long, inclusive. show that there exists at least one set of three lines that you can use to form a triangle.

[gray oriole]

Woud drawing a graph help

[celest frigate]

its a square, drawing a square would help

I wouldnt call it a graph though

[gray oriole]

,w plot x=3 and x=-3 and y=3 and y=-3

lmao

[woven radish]

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[gray oriole]

anyway uhh now how to check that root 2 thing

[celest frigate]

I want you to think

[gray oriole]

lets say you have a square of length 3 * 3, and 10 points, we need to show that atleast two points have distance less than square root 2

[celest frigate]

Spend some drawing different kinds of boxes

[gray oriole]

we are gonna pick 10 points so we have 10 pigeons

[celest frigate]

what are ways to divide a square into at most 9 parts - that helps proving the question

[gray oriole]

where did 9 come from

[celest frigate]

well with PHP, the number of pigeonholes is less than pigeons

[gray oriole]

alright I will try it and tell you if I managed to complete it lmao

dont wanna keep u here for long

[celest frigate]

Whats special about root 2 and a square @gray oriole

Try connecting root 2 to something first

[gray oriole]

what

well

root 2 is a number and square is a shape

lmao

[celest frigate]

uhhhh

[ebon coyote]

think of a unit square

What about it should scream sqrt(2) to you?

[gray oriole]

unit square has diagonal root 2

[celest frigate]

yeee

what is special about the diagonal in terms of distances

[gray oriole]

its got the jelly

the points we need

[ebon coyote]

And following on from that:
If two points were in the same unit square, what could you argue about the distance between them?

[gray oriole]

shi idk

i mean

[celest frigate]

draw a unit square

on paper

[tender wharf]

think in terms of max and min

[gray oriole]

max dist is root 2 ig

[celest frigate]

yay

now think about what boxes/pigeonholes we should choose

i think you got this

[gray oriole]

do i have to draw like unit squares in that big square

there will be 9 unit squares that fit into tat

[ebon coyote]

Not a difficult feat

yh

[gray oriole]

oh WOW

[celest frigate]

and 10 pigeons

so what do you conclude?

[gray oriole]

well uhh if i choose 10 points then there will be at least two points that are in the same square

(same bucket)

[celest frigate]

yez

[gray oriole]

cuz there are 9 squares

alr well y'all arent gonna be in the exam

i didnt do anything in this

[celest frigate]

now you can search for more examples on google/firefox/whatever

[ebon coyote]

Combinatorics questions are all about practice

[celest frigate]

you did it almost by yourself

we just gave hints

[junior mauve]

if you wanna try mine, it's kinda along the same lines

[ebon coyote]

You need exposure to a variety of problems, is all

[gray oriole]

sure

[gray oriole]

How do I know three lines make atriangle

I forgot

[junior mauve]

triangle inequality

[gray oriole]

a+b > c

alr

Okay I have 7 lines

We need to pick 3 lines out of 7?

[celest frigate]

yes

but remember with PHP: first you make boxes/pigeonholes

[gray oriole]

Well if we are picking three lines then do we need two boxes?

[celest frigate]

not quite

look at the last two examples

specefically the last one

[gray oriole]

you said one less

boxes are one less

than objects

[celest frigate]

look at the last example. We had to choose 2 points only

[gray oriole]

well we had 10 points and we had 9 squares

[celest frigate]

yes

[gray oriole]

a square has infinite points

[celest frigate]

Now how many pigeons do we have?

[gray oriole]

in the prev problem?

[celest frigate]

this problem

[gray oriole]

7?

[celest frigate]

yeah and how many do we want to choose

[gray oriole]

3

[celest frigate]

yes now think

[gray oriole]

no but why are we not picking 3 lines

i mean we are given 7, and out of those we gotta pick 3 such taht everytime the three make a triangle

if this is untrue then i might be misunderstanding smth here

[celest frigate]

yes

so some pigeonhole must contain 3 points.

how many pigeonholes can we have now?

[gray oriole]

well if we have 7 pigeons

[tender wharf]

also check these out : https://artofproblemsolving.com/wiki/index.php?title=Pigeonhole_Principle

[gray oriole]

do i have to solve ceil(7/k) = 3

[celest frigate]

Hint: 1 is useless

@gray oriole

[devout snow]

@gray oriole Has your question been resolved?

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[gray oriole]

Prove that among any ten points located inside a circle with diameter 5, there exist at least two at a distance less than 2 from each other.

[gray oriole]

To apply the PHP, I need to draw 9 regions within the circle such that the distance of any two points in that region is < 2. But I am not sure how to draw those regions.

I think I might have to draw a circle with radius 2

but what about the other 8 regions

[pseudo basin]

maybe place that circle of radius 2 at the center

how can you place the other 8 around it in a nice way? maybe that way will work

[gray oriole]

yes

,w plot x^2 + y^2 = 25 and x^2 + y^2 = 4

[woven radish]

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[gray oriole]

There's no space for more circles

ion get it

[pseudo basin]

the outer circle has radius 2.5 btw

[untold crow]

Well if you make a rule that says two points cannot be less than 2 units apart, then every point gets their own “circle of protection”. That circle of protection has a radius 2.

So I would look at how many circles can fit inside the 5, then look how many more points you’re left over (that would otherwise be forced to break the rule if placed).

[gray oriole]

root 5

no what

oh it says diameter 5

[gray oriole]

If i end with even one extra section then how will i apply php

rn im tryna go for 9 sections, ceil(10/9) = 2

if i end up with 10 i will have to do ceil(10/10) = 1

[untold crow]

Hmm. Ratio maybe?
Area of big circle
Area of small circle.
But that doesn’t account for points that are near the circumference and a portion of the small circle is outside the big circle, so here I’m a little unsure.

Maybe create another regional circle within this 5 circle with radius 3, then ask yourself how many semicircles fit between the 3 and 5.

I’m just imagining but not applying php. In terms of php, I assume it’s a play of ratios. My input is arguably amateur unfortunately

[gray oriole]

I reluctantly checked the solution

[tender wharf]

welp

[gray oriole]

alright i'll just accept this solution

ty guys

.close

[devout snow]

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[rare kernel]

i used kings rule and stuff and got it to

f(t) = pi integral dx/(1- cos^2 t sin^2x) from 0 to pi

how to proceed from here

[grand edge]

divide by sin²(x)

then sub cot(x) because you have csc²(x) in the numerator

[rare kernel]

ohk

cosec^2 x/(cosec^2x - cos^2 t)

cot x = t
-cosec^2x dx = dt

[grand edge]

?
sin²/sin² = 1

[rare kernel]

oh shit lol

[grand edge]

This kind of trick is hard to see if you don't have enough experience, so you just need to be aware of it

Here's another example that is less obvious to spot $$\int\frac{x\sin(x)+1+\cos(x)}{(x+\sin(x))^2}dx$$

[rare kernel]

cosec^2 x/(1+cot^2 - cos^2t)
cot x = y
-cosec^2 x dx = dy
-dy/(1+y^2 - cos^2t)

[woven radish]

Roy

[grand edge]

Ayo, our variable t is already taken

You don't wanna be mixing it up

[rare kernel]

oh wth here also we have to do the same thing?

[grand edge]

Dividing by cos²(x) gets the job done quickly

Very hard to come up with if you don't know the trick already

[rare kernel]

i have done this before but i usually convert to sec instead of cosec

[grand edge]

So just be aware of it

[grand edge]

yeah that happens when you divide cos

[rare kernel]

👍 ohk got it

[rare kernel]

huh but in our limit we have cot 0 and cot pi

which are both undefined

ohk we can make it 2pi integral (-dy)/(1+y^2-cos^2 t)

from infinity to 0

2pi integral dy/(1+y^2-cos^2t) from 0 to infinity

@grand edge is this fine?

[grand edge]

it should be, although I haven't been working out the integral since I'm preparing to go to sleep

[rare kernel]

oh ok what to do after this

pls stay awake 1 more min

[grand edge]

treat t as a constant

hence it's just an arctan integral

[rare kernel]

ohh alr alr then

ty

.close

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[devout snow]

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[north glen]

Hi, I've a statsd question

[north glen]

I first tried to calculate the probability of NOT having 2 6's. All the possible outcomes are ${61\choose 12}$. Since 72 minus de 11 6's equals 61. Then we choose 12 dice.

[woven radish]

Chuti | Argentina

[north glen]

Then, the possible combinations we can get without restrictions, are ${72\choose 12}$. So the end probability is $1-\frac{{61\choose 12}}{{72\choose 12}}$

[woven radish]

Chuti | Argentina

[north glen]

But this isnt the solution given in the book and I don't understand why.

[uncut crow]

you sample space should be 6^12 and there is no other reasonable one to use honestly

is your sample space not 6^12?

[north glen]

Looks like it is. Just draw the tree of possibilities and it's a tree of height 12, where each node has 6 edges.

So I guess it it.

So then the probability of NOT having a single 6, is 5^12, then we want to calculate the probability of having only one 6, since that's the maximum we can have?

[uncut crow]

that’s what the book does

the (72 choose 12) size sample space thing doesn’t work because…

[north glen]

so what's the probability of having at least one 6?

I thought these kind of problems were easy lol

idk why I'm having such a hard time.

[uncut crow]

well let’s imagine 2 dice for simplicity. and we ask what’s the probability of no 6s. now with the 12 choose 2 sample space perspective it’s like we have a bag with 1,2,3,4,5,6,1,2,3,4,5,6 in it and we just choose 2 of them. but that’s different than rolling two dice. our first choice should not influence the second

but it does here when we choose 2 out of those 12