help-36

[twin pivot]

what did I do wrong?

[shell condor]

What did you do right?

[proper raptor]

ask a real question or sybau

[desert mantle]

sinx is not a product of the factors s, i, n and x

same for all the other things

[twin pivot]

So what?

Why was I wrong?

[latent dragon]

That’s the trolling part?

[old quarry]

LOL

[latent dragon]

wow.

pathetic. People are really doing this?

[old quarry]

bro cancelled the letters

[twin pivot]

Yes

[desert mantle]

how have none of you seen this joke before

same as 64/16=4/1 by cancelling the 6's

[shell condor]

I have, just expected better here

[latent dragon]

Still a little bit new to helping. I don’t see this often.

[desert mantle]

<@&268886789983436800>

anyway

[twin pivot]

Sorry

.close

[final saddle]

Channel closed

Closed by @twin pivot

Use .reopen if this was a mistake.

[delicate falcon]

please only use the help channels for actual questions

[twin pivot]

👍

[final saddle]

Available help channel!

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[fluid blaze]

idk what i shud do to the log

[final saddle]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[latent dragon]

welp

[shell condor]

Whoops

[fluid blaze]

wha

do i still have to close this

.close

[shell condor]

No no it's dealt with

[fluid blaze]

what

oh

[shell condor]

If you still have to ask then type .reopen

Otherwise it's closed already

[fluid blaze]

.reopen

[shell condor]

Well

I was wrong then

Maybe get a new channel

[fluid blaze]

i cant cloe

close

.close

[shell condor]

It's already closed, it'd go away in some time

[fluid blaze]

o

[final saddle]

Available help channel!

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Remember:
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[soft pewter]

i have a rational function

[soft pewter]

A(x) = 72/x

and i wanted to know what was A(x) when x = 0

but when i do A(x) = 72/0 it says math error

[shell condor]

Well

any number divided by 0 is not defined

[craggy plank]

It’s just a hole on the graph

[soft pewter]

so i cant find the value of y when x is 0 guys?

[shell condor]

It'd be not defined, or undefined

[soft pewter]

wait so it means that the rational function isn't touching the y axis

[tired walrus]

indeed it does

the y-axis, a.k.a. the line x=0, is a vertical asymptote

[shell condor]

Basically it touches it at infinity

[soft pewter]

ah

[soft pewter]

i see

so for my table icant use x = 0

[shell condor]

Yes

[soft pewter]

alright ill try another

whn its 12

when y is 12 x is 6

[boreal lichen]

Yes?

[tired walrus]

so what exactly are you trying to do rn

[soft pewter]

it says graph the function

so im getting points to graph it prety much

[boreal lichen]

What's the shape of the curve?

[soft pewter]

uhh its a curve

got my points

[late rose]

label your origin and also choose a scale such that your line takes up at least half the grid horizontally and half the grid vertically

[soft pewter]

what would be the domaine of this function if it was restricted at your realistic interest rates

[soft pewter]

ty for helping again but wait wheres my origin

[late rose]

origin = (0, 0)

[soft pewter]

sorry

im bCk

alright

so label it like u want me to put a point there

[soft pewter]

oh i get it

you want me to choose points where the curve shows in the middle of the graph

[late rose]

sorry im doing smth rn

[soft pewter]

alr

[late rose]

you might want to ask help from someone else

[soft pewter]

alright thanks tho

.close

[final saddle]

Channel closed

Closed by @soft pewter

Use .reopen if this was a mistake.

[late rose]

you can ping helpers if you want

[soft pewter]

<@&286206848099549185>

[late rose]

reopen the channel

[soft pewter]

.reopen

[final saddle]

✅ Original question: #help-36 message

[soft pewter]

what would be the domaine of this function if it was restricted at your realistic interest rates?

[pale steeple]

Est-ce que tu parles en français?

[shell condor]

Sorry I don't get what you mean by "realistic interest rates"

[soft pewter]

oui

[soft pewter]

ill show u the whole image actually

[pale steeple]

mon français c'est pas bien mais je sais français

[soft pewter]

interest rates is taux d’interet which is x

[soft pewter]

pas grave

ta compris?

ill just make it simpler

whats the domaine of the function

[shell condor]

R - {0}

[soft pewter]

XE [6,infinite[

?

[shell condor]

XE?

[soft pewter]

x belongs to

probably written another way in english

[shell condor]

Oh yeah yeah

It just looked odd

So 6 to infinity

[soft pewter]

ah sorry

[shell condor]

But why can't it be less than 6?

[soft pewter]

i got a dot at 6

but im not sure

[shell condor]

Well it shouldn't be TBH

f(x) = 72/x

[jagged urchin]

yo

[soft pewter]

wait

[shell condor]

it's always real as long as x is real and not 0

[soft pewter]

its not acrually

6 isn't the minimum

[jagged urchin]

65_blinking

[soft pewter]

wait

so i write R

X [R]

[shell condor]

Just R, no brackets

also R - {0}

[jagged urchin]

8828joinvc

[deft terrace]

turn it to lnx

[shell condor]

Have you got anything to contribute here @jagged urchin ?

[soft pewter]

what is happening

[shell condor]

God knows

[jagged urchin]

HOP_AlphabetN HOP_AlphabetO

[shell condor]

Then can you please move to #discussion , #serious-discussion or #chill

This is a help channel

[soft pewter]

lmao

alright anyways

it can be less than 6

[shell condor]

<@&268886789983436800>

[soft pewter]

first time seeing this

[shell condor]

Bloody hell, this is worse than scams

[soft pewter]

crazy

[latent dragon]

Holy hell.

[shell condor]

It can also be negative

[shell condor]

But that's the domain for the function, all real numbers aside from 0

I'm not sure about the interest stuff

[jagged urchin]

joobihappy

[soft pewter]

pretty much x represents the interest rate

[jagged urchin]

joobicool

[soft pewter]

like in c(x) = 72/x

[jagged urchin]

fax~1

[soft pewter]

x is interest rate

[jagged urchin]

joobiflushed

[vital crag]

stop shitposting

[shell condor]

Dude once more, can you please stop disturbing

[soft pewter]

okay so the domaine

so the (0.0) is the asymptote

[shell condor]

Not (0,0), that'd be the origin

[soft pewter]

oh

[shell condor]

x = 0

[soft pewter]

alright but the domaine is bit confusing

[shell condor]

Yeah because it's the domain of the function

But it changes in interest

Like I'm not sure if interests can be negative

[soft pewter]

yes

they can be negative

[shell condor]

Oh sorry then

bloody hell

<@&268886789983436800>

[soft pewter]

damn

[shell condor]

Too many bloody scammers today

[soft pewter]

i hope they getting banned and not just muted

[shell condor]

They likely do

Okay so

[soft pewter]

perfect

[cyan kayak]

we ban them on sight, yes

[shell condor]

Now I'm not versed in financial maths, so I don't think I'll be able to help more

[soft pewter]

alright no worries

but

lets just ignore the interest rate part

whas the domaine

of just this

R?

[shell condor]

No no

R - {0}

All real numbers but 0

As at x = 0, the function is undefined

[soft pewter]

i see

one last question

[shell condor]

Yes mate?

[soft pewter]

what is the equation(rule) that helps determine how many years left before the capital is not double if 2 years pass since the capital has been invested?

the equation is in A(x) = ?/x format

we replace y by 2

because y = amount of years

[shell condor]

Yeah sorry mate, that's beyond me

[soft pewter]

ah alright all good

thanks tho dude

.close

[final saddle]

Channel closed

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[final saddle]

Available help channel!

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[whole halo]

how do I prove $S(n):=\sum_{k=1}^{n-1}\csc\left(\frac{\pi k}n\right)^2=\frac{n^2-1}3$

[soft zealot]

mtt

[whole halo]

the best I was able to do was that S(2n) = 1 + 4 S(n)

[atomic moon]

What about complex number

[whole halo]

good idea, can you simplify this then?

whoops theres a - there

it should be +infinity here

I originally got to that above by swapping these sums then expanding a whole lot, not a good idea
if anyone is good with chebyshev polynomials, maybe theres something simple here

[atomic moon]

Ok so complex number not good

[whole halo]

other ideas dont work out so good either

I dont even know how to approach this way of viewing it

[final saddle]

@whole halo Has your question been resolved?

[final saddle]

Channel closed

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[soft pewter]

guys need help with thiso

so c(x) = a/x-h +k

how do i find

h,k

which is the half between the curve from the other curve

if ur wondering what the 1000 and 3500 are

i converted dam^2 to m^2

[final saddle]

@soft pewter Has your question been resolved?

[soft pewter]

.close

[final saddle]

Channel closed

Closed by @soft pewter

Use .reopen if this was a mistake.

[final saddle]

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[slate topaz]

is this correct? Asks me to calculate the derivative

[loud sundial]

you can check this yourself in #bots

,w derivative of 2x^3 - 3x^2 + 2x with respect to x

[soft zealot]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by civil_service_pigeon

[slate topaz]

,w derivative of 2x^3 - 3x^2 + 2x with respect to x

[soft zealot]

Results provided by WolframAlpha

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

Requested by jukeit

[slate topaz]

ah okay

[loud sundial]

it is good practice to put parentheses if you're differentiating more than one term though

[slate topaz]

thanks

.close

[loud sundial]

[final saddle]

Channel closed

Closed by @slate topaz

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[final saddle]

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[wary copper]

my attempt is on the concurrent case

its not exactly obvious to me why (1) = (2) immediately unless its just algebraic hell and i am just meant to keep expanding everything

[sonic crystal]

Damn

[wary copper]

o sorry minor error in (2)

wait nvm its fine

omg

im so stupid

(AB)^T = B^T A^T

hahahahahhaha

literally forgor basic matrix

.close

[final saddle]

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[dense wedge]

The question is asking me to find the volume of the solid but idk how to start it off. This deals with triple integration (calc 3)

[barren hound]

it should be double in this case

[dense wedge]

Why is it double ?

[barren hound]

ultimately what you want is something like $\int_{y=?}^{y=?} \int_{x=?}^{x=?} upper - lower \dd{x} \dd{y}$

[soft zealot]

〈 kitten | teacup 〉

[barren hound]

because you have a 2d shadow that you're integrating over

[dense wedge]

ik that when you find the first bounds for z, the function inside is just 1dz is that why?

[barren hound]

you could also do triple and integrate 1 sure

[dense wedge]

Why can’t you equal both z’s together to give you the function you’re trying to integrate why is its upper - lower

[barren hound]

it's just like in regular integration when you were trying to find the area of a shape

and you did upper curve - lower curve

$\int_{x=?}^{x=?} \int_{y=\blue{?}}^{y=\blue{?}} upper - lower \dd{y} \dd{x}$

[soft zealot]

〈 kitten | teacup 〉

[barren hound]

we need to figure out what those y bounds are

actually i guess it gives them to you in the problem

[dense wedge]

let me see if i can write it out give me a second

so it would be like this ?

[celest crane]

2y^2

[final saddle]

@dense wedge Has your question been resolved?

[dense wedge]

let me try to solve it and then show my answer to confirm

so is this it?

ignore the blue

[celest crane]

,wolf double integral (2 -2x^2 - 2y^2 dy dx) from y=0 to y=(1-x) from x=0 to x=1

[soft zealot]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by kookiemon

[celest crane]

Looks like you got the answer correct.

[dense wedge]

okay thank you

.close

[final saddle]

Channel closed

Closed by @dense wedge

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[final saddle]

Available help channel!

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Remember:
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[whole halo]

need hints for showing that $S(n):=\sum_{k=1}^{n-1}\csc\left(\frac{\pi k}n\right)^2$ simplifies to $\frac{n^2-1}3$
I only managed to find that $S(2n)=1+4S(n)$

[soft zealot]

mtt

[robust mulch]

the hell

first off what is the square operating on

[whole halo]

by default, $\csc(x)^2=\csc(x)\cdot\csc(x)$, similar to $f(x)^2=f(x)\cdot f(x)$

[soft zealot]

mtt

[whole halo]

I dont recommend the other way of placing the square, $\csc^2(x)$, since that can be confused with function composition notation, and the notation is only ever standard for trig functions so its not generalizable

[soft zealot]

mtt

[robust mulch]

a little ambiguous, but fair

[whole halo]

I would argue it isnt at all ambiguous, since f(x) as a notation is so standard that g(f(x)) by default takes f(x) first regardless of the operation g appears to be
also, its commonly accepted that f(x^2) is valid notation, so f(x)^2 is not f(x^2) because they do not look the same

additionally if you had f(3)^2 mean f(9), wouldnt that be f9 instead of f(9)? the parentheses would require a special rule of adding an additional () around the input to simplify them

[final saddle]

@whole halo Has your question been resolved?

[final saddle]

@whole halo Has your question been resolved?

[plucky rover]

I'm wondering if induction would help here

[whole halo]

maybe for S(3n) based on S(n), S(5n) based on S(n), etc.

[plucky rover]

That's not what I was getting at

I was thinking doing S(1) and then S(n+1) using S(n)

That feels like it would be much easier no?

[whole halo]

I just dont see how thats a good idea, n and n + 1 are coprime

that means none of the numbers you use before appear after

[plucky rover]

Oh right I misread

S(n+1) isn't S(n) + (one term)

Lmao

[loud sundial]

$Q(x)=\sum^{n-1}{k=0} x^k \implies \frac{Q'(x)}{Q(x)}=\sum^{n-1}{k=1} \frac{1}{x-e^{2k \pi i/n}}$

[soft zealot]

Civil Service Pigeon

[loud sundial]

now ||differentiate this with respect to x||

[whole halo]

what in the goddamn

lets try that out when Im done helping

[whole halo]

how did you even come up with this?

[loud sundial]

complex numbers came to mind and I knew that $\left|1-e^{2\pi i k}{n} \right|=2 \sin \frac{\pi k}{n}$

[soft zealot]

Civil Service Pigeon

[loud sundial]

and thus $\sum \csc^2 \frac{\pi k}{n}$ is given by $\sum \frac{1}{|1-e^{\frac{2\pi i k}{n}|^2}$ barring constant terms

[soft zealot]

Civil Service Pigeon
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[loud sundial]

fuck it

you can read latex

anyway

[whole halo]

absolutely wild option

[loud sundial]

summing moduli is weird

so $z\bar{z}=|z|^2$

[soft zealot]

Civil Service Pigeon

[loud sundial]

actually

let $\omega_k=e^{\frac{2\pi i k}{n}}$

[soft zealot]

Civil Service Pigeon

[whole halo]

I havent fully worked it out yet to be honest

[loud sundial]

oh

well I won't spoil too much then lol

[whole halo]

I shouldve disclosed that then

[whole halo]

this is a wild spoilers in and of itself

I never wouldve figured on that

[loud sundial]

I love circuitously helping

[whole halo]

I had to derive that like several times each time I see it

dont believe that its real no matter how many times its useful

now here it is again

[loud sundial]

ngl I didn't know the entire result either

I was like

uh

it's got a sin

and a half

probably works

I also derive it each time lol

[whole halo]

going from the chord thing to Q(x) though, Ill have to use that

[loud sundial]

well it becomes more obvious if you work it out in full

like

it makes sense once you see it

[whole halo]

Im going to start from there, yea

every time man

r = cos(theta) being a circle, unbelievable

[whole halo]

to be honest, this shouldve been the hint

but there wasnt any way you couldve known that

[loud sundial]

yeah I made the mistake of giving you the start of a solution that was written forwards

even though the process was NOT forwards at all

[old quarry]

use pythagorus theorem 🫠

[whole halo]

get this nonsense outta here, you cannot match the taste of the forbidden apple

literal thaleure

[loud sundial]

[shadow marlin]

Actually not a bad idea since I've seen you can do that + ||tchebychev polynomials||

[whole halo]

chebyshev polynomials was on the backburner for a while and didnt go anywhere

its a shame too, I shouldve looked into them more

[final saddle]

@whole halo Has your question been resolved?

[whole halo]

pretty much stuck still, I cant really get around what to do after |z|^2 = (z)(z bar)

[whole halo]

to be honest I didnt look on this channel for the past 15 minutes because somehow further magic is happening

man why did the D3 group have to show up

now to clean up my answer

[whole halo]

|1 - e^(2πi k/n)|^2 = |1 - cos(2πi k/n) - i sin(2πi k/n)|^2 = (1 - cos(2π k/n))^2 + sin(2π k/n)^2
= cos(2π k/n)^2 - 2 cos(2π k/n) + 1 + sin(2π k/n)^2
= 2 - 2 cos(2π k/n) = 4 (1 - cos(2π k/n))/2 = 4 sin(πk/n)^2
(on the original r = cos(theta), I used thales theorem to show that two line segments were the same which would justify the circular shape)

by default, let ∑ be from k=1 to n-1

∑ csc(πk/n)^2
= ∑ 1/sin(πk/n)^2
= ∑ 4/(4 sin(πk/n)^2)
= ∑ 4/|1 - e^(2πi k/n)|^2
= ∑ 4/((1 - e^(2πi k/n))(1 - e^(-2πi k/n))) (|z|^2 = (z)(z bar))

Q(x) = ∑_0 x^k = (x^n - 1) / (x - 1) = ∏_0 (x - e^(2πi k/n)) / (x - 1)
ln(Q(x)) = ∑_0 ln(x - e^(2πi k/n)) - ln(x - 1) = ∑ ln(x - e^(2πi k/n))
Q'(x) / Q(x) = (ln(Q(x)))' = ∑ 1/(x - e^(2πi k/n))
(Q'(x) / Q(x))' = ∑ -1/(x - e^(2πi k/n))^2
(Q''(x) Q(x) - Q'(x) Q'(x)) / Q(x)^2 = ∑ -1/(x - e^(2πi k/n))^2

Q(1) = ∑_0 1 = n
Q'(1) = ∑_0 k = n(n - 1)/2
Q''(1) = ∑_0 k(k - 1) = ∑ k^2 - k
= (n - 1)(n - 1 + 1)(2(n - 1) + 1)/6 - n(n - 1)/2
= (n - 1)n(2n - 1)/6 - n(n - 1)/2
= (n - 1)n/2 ((2n - 1)/3 - 1)
= (n - 1)n/2 ((2n - 4)/3)
= (n - 1)n(n - 2)/3 (not a coincidence)

Q'(1) / Q(1) = ∑ 1/(1 - e^(2πi k/n))
= n(n - 1)/2 / n = n/2 - 1/2

(Q''(1) Q(1) - Q'(1) Q'(1)) / Q(1)^2 = ∑ -1/(1 - e^(2πi k/n))^2
= (n(n-1)(n-2)/3 (n) - (n(n-1)/2) (n(n-1)/2)) / n^2
= (n-1)(n-2)/3 - (n-1)/2 (n-1)/2
= (n^2 - 3n + 2)/3 - (n^2 - 2n + 1)/4
= n^2/3 - n + 2/3 - n^2/4 + n/2 - 1/4
= n^2/12 - n/2 - 5/12

4(n^2/12 - n/2 - 5/12) + 4(n/2 - 1/2) = 1/3 n^2 - 5/3 + 2 = 1/3 n^2 - 1/3 (wow!)
= ∑ -4/(1 - e^(2πi k/n))^2 + ∑ 4/(1 - e^(2πi k/n))
= -4 (∑ 1/(1 - e^(2πi k/n))^2 - ∑ 1/(1 - e^(2πi k/n)))
= 4 ∑ 1/(1 - e^(2πi k/n)) - 1/(1 - e^(2πi k/n))^2
= 4 ∑ 1/(1 - e^(2πi k/n)) (1 - 1/(1 - e^(2πi k/n)))
= 4 ∑ 1/(1 - e^(2πi k/n)) (-1/(e^(-2πi k/n) - 1)) (1 - 1/(1 - x) = -1/(1/x - 1))
= 4 ∑ 1/(1 - e^(2πi k/n)) (1/(1 - e^(-2πi k/n)))
which from earlier is the same as ∑ csc(πk/n)^2

.close @loud sundial thanks and thats crazy that you also figured to use partial fractions on the e^(2πi k/n) instead of on the 1 (which lead nowhere)

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[empty plume]

suppose there are 6 different colored balls ina bag,such that there are 7 valls of the first colour, 8 balls of the second colour,9 balls of the third colour and so on. what is the least number of balls that must be picked from the bag without looking to ensure the balls of all colours are picked?

my anwser is 52 but chatgpt and the answer sheet both says 51. my logic is that if ur rly unluck ur get every posisble ball beside one more so u wont be able to complete the set so its 6+7+8+9+10+11 for it and once u done that the very next one will complete the set no matter what. then it equals to 51 then add 1 cus u still needa pick out one ball. c How am i wrong?

[tranquil pine]

@soft pewter

[deft terrace]

from the info you have given
1st colour - 7
2nd - 8
.
.
.
6th - 12 right

so if we take the worst case scenario

we get 8+9+10+11+12 + 1

thats 51

[empty plume]

wait isnt the question saying like u just need like all the balls from 1 bag to be picked?

[deft terrace]

[empty plume]

wait am i stupid

uhhhh

[deft terrace]

no i think it meant to get every color

[empty plume]

ohhhh

[deft terrace]

happens to the best of us

[empty plume]

bro i reread this question like 20 times

😭

and somehow still misunderstood it]

lmao

[deft terrace]

drink some coffee or get some sleep

[empty plume]

ty dude

[deft terrace]

makes you forget the trauma

[empty plume]

no like 20 times thruout the day lol

[deft terrace]

wlcm

[empty plume]

anyways ty, bye

.close

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[jagged urchin]

|1 - e^(2πi k/n)|^2 = |1 - cos(2πi k/n) - i sin(2πi k/n)|^2 = (1 - cos(2π k/n))^2 + sin(2π k/n)^2
= cos(2π k/n)^2 - 2 cos(2π k/n) + 1 + sin(2π k/n)^2
= 2 - 2 cos(2π k/n) = 4 (1 - cos(2π k/n))/2 = 4 sin(πk/n)^2
(on the original r = cos(theta), I used thales theorem to show that two line segments were the same which would justify the circular shape)

by default, let ∑ be from k=1 to n-1

∑ csc(πk/n)^2
= ∑ 1/sin(πk/n)^2
= ∑ 4/(4 sin(πk/n)^2)
= ∑ 4/|1 - e^(2πi k/n)|^2
= ∑ 4/((1 - e^(2πi k/n))(1 - e^(-2πi k/n))) (|z|^2 = (z)(z bar))

Q(x) = ∑_0 x^k = (x^n - 1) / (x - 1) = ∏_0 (x - e^(2πi k/n)) / (x - 1)
ln(Q(x)) = ∑_0 ln(x - e^(2πi k/n)) - ln(x - 1) = ∑ ln(x - e^(2πi k/n))
Q'(x) / Q(x) = (ln(Q(x)))' = ∑ 1/(x - e^(2πi k/n))
(Q'(x) / Q(x))' = ∑ -1/(x - e^(2πi k/n))^2
(Q''(x) Q(x) - Q'(x) Q'(x)) / Q(x)^2 = ∑ -1/(x - e^(2πi k/n))^2

Q(1) = ∑_0 1 = n
Q'(1) = ∑_0 k = n(n - 1)/2
Q''(1) = ∑_0 k(k - 1) = ∑ k^2 - k
= (n - 1)(n - 1 + 1)(2(n - 1) + 1)/6 - n(n - 1)/2
= (n - 1)n(2n - 1)/6 - n(n - 1)/2
= (n - 1)n/2 ((2n - 1)/3 - 1)
= (n - 1)n/2 ((2n - 4)/3)
= (n - 1)n(n - 2)/3 (not a coincidence)

Q'(1) / Q(1) = ∑ 1/(1 - e^(2πi k/n))
= n(n - 1)/2 / n = n/2 - 1/2

(Q''(1) Q(1) - Q'(1) Q'(1)) / Q(1)^2 = ∑ -1/(1 - e^(2πi k/n))^2
= (n(n-1)(n-2)/3 (n) - (n(n-1)/2) (n(n-1)/2)) / n^2
= (n-1)(n-2)/3 - (n-1)/2 (n-1)/2
= (n^2 - 3n + 2)/3 - (n^2 - 2n + 1)/4
= n^2/3 - n + 2/3 - n^2/4 + n/2 - 1/4
= n^2/12 - n/2 - 5/12

4(n^2/12 - n/2 - 5/12) + 4(n/2 - 1/2) = 1/3 n^2 - 5/3 + 2 = 1/3 n^2 - 1/3 (wow!)
= ∑ -4/(1 - e^(2πi k/n))^2 + ∑ 4/(1 - e^(2πi k/n))
= -4 (∑ 1/(1 - e^(2πi k/n))^2 - ∑ 1/(1 - e^(2πi k/n)))
= 4 ∑ 1/(1 - e^(2πi k/n)) - 1/(1 - e^(2πi k/n))^2
= 4 ∑ 1/(1 - e^(2πi k/n)) (1 - 1/(1 - e^(2πi k/n)))
= 4 ∑ 1/(1 - e^(2πi k/n)) (-1/(e^(-2πi k/n) - 1)) (1 - 1/(1 - x) = -1/(1/x - 1))
= 4 ∑ 1/(1 - e^(2πi k/n)) (1/(1 - e^(-2πi k/n)))
which from earlier is the same as ∑ csc(πk/n)^2
mtt — 1:34 PM

[jagged urchin]

solve it guys @everyone

[tired walrus]

pinging everyone in a server with 200k+ people

[worldly mesa]

Does it actually ping?

[tired walrus]

no

[worldly mesa]

The ping is suppressed no?

[tired walrus]

but the entitlement

[latent dragon]

Imagine going into a server with hundreds of thousands of people and attempting to ping every single one of them.

[vital crag]

they definitely are gonna get everyone to ignore them

[sick oriole]

can we talk about the equation

because

[vital crag]

no one's stopping you talking about the equation

[sick oriole]

Really????!

[vital crag]

try it and see

you'll be surprised

[final saddle]

@jagged urchin Has your question been resolved?

[craggy plank]

Nice troll

[candid pulsar]

sigma male grindset

[latent dragon]

Yeah no.

[final saddle]

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[empty plume]

there are exactly n whole number cubes between 7^6 and 5^9 find n

[empty plume]

my initial idea was n is between 125 and 49

but somehow its wrong?

answer is 77

[worldly mesa]

why do you think its 125 or 49

[empty plume]

because since they want cubes

and its 7^3)

(7^3)^2

which makes it 49^3

[worldly mesa]

read the question carefully

[opal plinth]

n is the number of cubes

[empty plume]

and 5^9 = (5^3)^3

125^3

[worldly mesa]

they want the number of cubes between those numbers

[empty plume]

yea so wouldnt it be 125 - 49\

then minus 2

[worldly mesa]

which is very close to 77

[opal plinth]

Why minus 2?

[empty plume]

cus they dont wanna add 125 and 49

[opal plinth]

They do

[empty plume]

nvm my answer is now 74 lol

[opal plinth]

It's 77 including 49 and 125

[empty plume]

uhh i forgot to mention

they dont wanna add 49 and 125

[deft terrace]

minus 1 you already excluded 49 you just gotta exclude 125

[opal plinth]

... then it's 75

[deft terrace]

i think they are including 125 and 49's cubes as well

[empty plume]

yea but question says

[opal plinth]

Post a picture

[empty plume]

dont include 5^9 and 7^6 to ur count

i cant im on pc lmao

[opal plinth]

Where is the question from?

[empty plume]

AMO

[opal plinth]

Do you have a date / question number, anything?

[empty plume]

yea 2017 and its question 14

[jagged urchin]

solved? guys

[empty plume]

nop

[opal plinth]

This?

[empty plume]

yea

\exactly that

[jagged urchin]

@empty plume hey, which class do you study in?

[empty plume]

wdym

[jagged urchin]

Pls tell me

[opal plinth]

Yeah pretty sure they messed up

[jagged urchin]

@opal plinth hey, which class do you study in?

[opal plinth]

125 - 49 - 1 = 75, not 77

[opal plinth]

I don't

[jagged urchin]

?

job?

[jagged urchin]

job?

[empty plume]

wait did u get this by online or did u create this urself

cus if u got it online pls tell me where

[opal plinth]

You want my credit card info as well?

[empty plume]

wait huh 125 - 49 - 1 = 77???

[jagged urchin]

no

[opal plinth]

https://www.scribd.com/document/498950560/AMO-8-2017
Unfortunately they put ads in there but it's still somewhat usable

[jagged urchin]

Guys , I am just doing fun. I read in class 9\

[opal plinth]

Page 24

[opal plinth]

,calc 125 - 49 - 1

[soft zealot]

Result:

75

[empty plume]

tysm, i dont mind ads cus im literally up at 2 am rn trying to understand the concept of these question

[empty plume]

💀 the anwser sheet is wrong

[opal plinth]

Ye

[empty plume]

alright tysj

[jagged urchin]

I don't know about this math!!!!!!!!!!!!!!!!

[empty plume]

u saved me so much time lmao

[jagged urchin]

@tulip coyote

helllooooo

[candid pulsar]

ping of dominance

[jagged urchin]

huh?

[tulip coyote]

...really

[jagged urchin]

yes

Which class do you study in? @tulip coyote

[candid pulsar]

oh nah

[jagged urchin]

what

[tulip coyote]

The class of timing people out for trolling

[candid pulsar]

king

[tulip coyote]

If you wanna be less trolly, come back tomorrow

Anyways, sorry @empty plume, did you get your question answered?

[empty plume]

yes

.close

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[swift apex]

i am confused between option A and C. i dont understand the relation of velocity with respect to displacement

[opal plinth]

Uniform deceleration would mean the velocity decreases linearly over time, right?

[swift apex]

yeah but the graphs are with respect to displacement

[opal plinth]

Yes, I'm just making sure

So v(t) decreases linearly, meaning d(t) decreases quadratically

[swift apex]

yeah that makes sense. 2as=v^2-u^2 but again A and C are the confusion

[opal plinth]

What would be dv/dt and dd/dt ?

(like what kind of function)

[swift apex]

uh dv/dt rate of change of velocity so acceleration and dd/dt would be rate of change of displacement so instanteous velocity

[opal plinth]

By what kind of function I mean is it constant, linear, quadratic, is it positive, negative, is it going up, down, ... ?

[swift apex]

im guessing quadratic since v^2

first constant velocity so straight line and then uniform deceleration

[opal plinth]

Which one are you talking about, dv/dt or dd/dt ?

Also, only consider the uniform deceleration part

[swift apex]

ok i think i get it why its not C, the distance travelled increases as velocity is decreasing. so only option left is A

[opal plinth]

Right

[swift apex]

.close

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[fair socket]

can you help me to find number of paths that connecting the vertex A to the vertex B in the directed graph

[south whale]

4?

nvm

[fair socket]

hhhhhh

[south whale]

9?

[fair socket]

no

it's clearly that it's greater than 9

[south whale]

its frying me

everytime i recheck i find more ways

[fair socket]

actually it's complicated

because you have soo manyy probabilities

[south whale]

19

pls amirite this time?

[south whale]

its not that many

[fair socket]

noo it's manyy you have a directed graph

[south whale]

what is the answer

?

yea im wrong i found more ways

[final saddle]

@fair socket Has your question been resolved?

[bold zenith]

Write the number 1 next to the letter A, then find another point where every arrow going into it has a number written on the point it came from write the sum of the numbers going into that point and repeat until you write a number into B, that is the number of paths

[fair socket]

i find it thanks

@final saddle

[bold zenith]

What did you get for the answer

[bold zenith]

Yeah

.solved

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[rugged hollow]

i have a question on my workbook being cos(2x)+2sin(x)-4sin^3(x) = 0. ive gotten the solutions pi/4, 3pi/4, 5pi/4, 7pi/4, 7pi/6 and 11pi/6 but im not sure if im correct.

my logic was to convert cos(2x) -> 1-2sin^2(x) through the double-angle identity to make the equation have like terms. i then factored by grouping and ended up with (2sin^2(x)-1)(2sin(x)+1)=0 and then went to solving from there. im basically here to ask if

a: im correct or not
b: if theres a better way to do this

[south whale]

how did u convert sin2x to sinx?

[rugged hollow]

where did i say that?

[south whale]

u dint say that but

there is a sin2x in the beginning

so how did u write it in terms of sinx

[rugged hollow]

oh im sorry its supposed to be 2sin(x) it was a typo

[south whale]

oh

thats what i was confused about

then the process looks correct

[rugged hollow]

alright, and would you say my method is efficient or is there a faster or better way

[south whale]

ur method is pretty efficient

did u take sinx common?

because (2sin^2(x)-1)(2sin(x)+1)=0 looks incorrect to me

did u use middle term splitting?

[rugged hollow]

well i had 4sin^3(x)+2sin^2(x)-2sin(x)-1=0, grouped first two by taking out 2sin^2(x) and last two by taking out -1, and both remaining terms were equal (2sinx+1)

[south whale]

nvm i was tripping u are right

[rugged hollow]

alright

.close

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[fervent panther]

Hey Please Help, How to factorise '1/a+1/b+1/x=1/a+b+x

[vital python]

]

[final saddle]

@fervent panther Has your question been resolved?

[bold zenith]

$\frac{1}{a}+\frac{1}{b} + \frac{1}{x}=\frac{1}{a+b+x}?$

[soft zealot]

BBMaths

[bold zenith]

You can’t factorise an equation

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[queen trench]

Just need the answer real quick

[obtuse grail]

Not allowed here

[latent dragon]

!noans

[final saddle]

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.

[obtuse grail]

What have you tried

[queen trench]

.close

[final saddle]

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[latent dragon]

wow. Only for the answer?

[honest carbon]

Lol, probably doing it on test

[latent dragon]

no proof though. God the urge to ping modulators…

hope they keep this in mind though. For future reference so that mr brainrot over here doesn’t do this again.

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[sleek belfry]

i have an exercise:

determine all posible 4-digit numbers that can be created from digits of the number 238832

i have no idea how to start, could i get a hint? we should be using permutations with repetitions

[opal plinth]

Well you can either have 2 or 3 distinct digits

@sleek belfry

[sleek belfry]

i see

so for 2 distinct it would be 4!/(2!2!)
and for 3 distinct it would be 4!/(1!1!2!)

[opal plinth]

Not quite

[sleek belfry]

not even close?

do i have to multiply them by something

or its completely wrong

[opal plinth]

Yes

4!/(2!2!) would be the number of 4-digit numbers you can make with the digits 2233, for example

[sleek belfry]

thought so

[opal plinth]

(it's 6: 2233, 2323, 2332, 3223, 3232, 3322)

[sleek belfry]

i guess 3! ?

but i dont see why

[opal plinth]

No

[sleek belfry]

mmm sec

[final saddle]

@sleek belfry Has your question been resolved?

[sleek belfry]

i feel like i need to find all posible ways we can pick 4 items where 2 are repeating 2 times

but that would be 4!(2!2!) again

[opal plinth]

Which digits can you use?

[sleek belfry]

oh

is it 6!/2!2!2!

i can use 2 twice, 3 twice and 8 twice

[opal plinth]

No

[opal plinth]

Ok but, in each case...?

[sleek belfry]

3*2

we choose pairs where on first place we have 3 options and on second 2 options

[opal plinth]

Do you want my solution? I feel like this is too basic to give a hint for

[sleek belfry]

ye ok i guess

[opal plinth]

So in case A, we have 2 distinct digits. We can choose them from 3 distinct digits. Then, we can arrange them by choosing 2 spots out of 4 for one pair of identical digits, the other pair just takes the other available spots. In total, that's (3 choose 2) * (4 choose 2) = 3 * 4!/(2!2!)

In case B, we have 3 distinct digits. We can choose which one will be present twice, out of 3 distinct digits. Then, we arrange them as if they were all distinct, and finally we divide by 2! for the duplicate. In total, that's (3 choose 1) * 4!/2! = 3 * 4!/2

(note that in case A, you could also choose the digit that is not present in the number, which yields (3 choose 1) instead of (3 choose 2), but that's equivalent)

(same in case B, you could choose the digits that are not duplicated)

Hopefully that explanation is good enough for you, I'm gonna have to go

[sleek belfry]

yes thanks a lot, i dont understand it 100% but its close, i will go over it couple more times

.close

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[silk fjord]

There are 270 vehicles in a lot, 120 of which are cars, 150 are scooters. Random vehicles leave the lot one at a time until there are only scooters remaining (no cars). One of those scooters is mine. What is the probability my scooter is still in the lot?

[silk fjord]

So my initial though is it would be like number of scooters left over 150

and that number is anywhere from 1 to 150

so like 1/150 or 2/150 ... 150/150

but I don't think thats the right way to think about it

maybe need to focus on the probability of which vehicle leaves

like 1/2 chance a car leaves

[mint orbit]

It's not a trick is it

[silk fjord]

prob not? wdym?

[mint orbit]

Like, it's my scooter, and I haven't left

[silk fjord]

ohhh no

like one of the 150 scooters is mine

i could've left

sorry i worded it badly, im typing this question from memory

[proper dagger]

do you have the original wording of the question then?

[silk fjord]

no sorry

wait maybe i do hold on

no i cant access it anymore

[vestal crest]

I have a feeling that the number of scooters isn't important for this problem.

The leaving stops once all cars leave, regardless of whether some scooters have left before or after you, so all that seems to matter is that your scooter outlasts all of the cars.

[mint orbit]

i think its important

because it tells you how many different ways the same thing could happen right

if that makes sense

it might be (all the cars) (your scooter) (the rest of the scooters)

or (all the cars, and one scooter) (your scooter) (the rest of the scooters minus 1)

oh, wait, i see what you mean

so wait then

[mint orbit]

i still feel like we have to fix the wording in the problem though

or maybe, just say 'one of the scooters is mine' in a different place

[tiny kraken]

Does the last vehicle have to be a scooter

[mint orbit]

i think all theyre saying is that we are excluding any ordering of them that has a car last

which only excludes

uhh

man i wish the numbers were smaller

i may have to play with this one

i think its literally just counting, you don't need anything fancy

just consider all possible orderings, remove the ones that end with a car, and then find the mean number of scooters left after the last car leaves of the remaining orderings

that mean divided by the total number of scooters is the answer, right?

[vestal crest]

I was thinking using a simplified version of the problem.
Let's say we have just one car and two scooters. Call the other scooter S2 and our scooter S1.
At a given moment, our scooter has a 1/3 chance of leaving, but so does the other scooter.
The possible orders are:
C1 S1 S2
C1 S2 S1
S1 C1 S2
S2 C1 S1
S1 S2 C1
S2 S1 C1

Note that if the car leaves first, the order of the remaining scooters doesn't matter, because the game ends after the car leaves.
But if a scooter leaves first, it's a 50/50 chance between our scooter and the other one.
If both scooters leave first, there's no chance for our scooter (and order doesn't matter).
I feel like the other scooters are just padding here.

[mint orbit]

This seems like a great problem for bayesian techniques actually

[vestal crest]

OP seems to be in high school (judging from the role), though. Do high schoolers nowadays learn about Bayesian statistics?
Otherwise I'm not sure if bringing out the heavy machinery will help OP here.

[mint orbit]

Ah, yea, i'm htinking more of just a way to get an answer against which to compare

[vestal crest]

I see, sorry for the assumption.

[mint orbit]

But youre right

I think you're probably closer than me, if they do come back

[silk fjord]

this is from a competition problem

[vestal crest]

It's still relatively high-school sounding. Bayesian statistics is usually taught in undergraduate math, I believe?

I don't think you're expected to know how to set up priors and posteriors for a simple counting problem...?

[silk fjord]

lol i have no clue what those are haha

so yeah, i doubt it. the problems are designed for high schoolers

[vestal crest]

Then I believe that we only need to consider your scooter and the number of cars.

[silk fjord]

i see what you mean, but i dont understand how to apply it to the larger population

[vestal crest]

Essentially, all that matters is that your scooter goes after all cars.

Whether other scooters go before or after you doesn't matter.

Can you continue from here?

[steep hatch]

isn't this wrong?
Thought experiment: If there's no other scooters the probability your scooter is left over given the cars leave before all the scooters leave is 1, this isn't true anymore for 2 scooters

[vestal crest]

But this turns into a conditional probability problem. I don't believe the original question intends for any condition, unless I'm misreading the problem (if so, I apologize).

[steep hatch]

The original question has the condition there is at least one scooter remaining right

the cars leave before all the scooters leave

[vestal crest]

Not necessarily. It could be that all the scooters leave before the cars do (unless OP missed something out).

The only stopping condition is that all the cars are gone. The question seems to make no distinction whether the scooters are allowed to go before the cars, so I assume they are.

[steep hatch]

if thats not what the question means then sure what you did is correct, I just interpreted it as saying there was at least one scooter left over

[vestal crest]

I see. I may have misread the problem, but just for a sanity check, I'll requote the problem, with the interesting part bolded.
"There are 270 vehicles in a lot, 120 of which are cars, 150 are scooters. Random vehicles leave the lot one at a time until there are only scooters remaining (no cars). One of those scooters is mine. What is the probability my scooter is still in the lot?"
You may be right on this reread, so I think I'll concede unless OP can further clarify.

Main question to OP: are all scooters allowed to leave before the cars do?

[final saddle]

@silk fjord Has your question been resolved?

[tiny kraken]

Why do I feel like the number of car doesn't matter. Let say we have 3 cars and 4 scooters. we have two states for each which is in the lot and not
Ofc all three car must not be in the lot
That being said, scooters can be outside or in the lot
Each scooter has 1/2 chance of being in / outside the lot and we have to choose 1 scooter, 2 scooters and 3 scooters to be outside the lot as our 3 cases

Maybe I missed smth?

[vestal crest]

I have somewhat of a concern with this argument, but it depends on OP's answer to my question.

[final saddle]

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[lone fog]

i dont get part a

how do u go from 0.3 to 0.304...

[tired walrus]

calculator

[odd seal]

sine of some angle is 0.3

[tired walrus]

ain't much else to it -- you're not expected to calculate sin^-1 at any point except those few "nice" ones by hand

[odd seal]

that angle itself is some different value

they just looks similar coz they are close to 0, and thats how sine behaves near 0

[lone fog]

oh i took sin-1 and forgot to put it into radians

[torpid pond]

hello

[final saddle]

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[lament jolt]

For a linear space equipped with a metric making it a linear topological space, prove the metric's translation invariance.

[lament jolt]

Actually, I want to utilize the continuity of addition to prove that the metric is translation-invariant.

$(d(x_n,x)\to 0 and d(y_n,y)\to 0 \Rightarrow d(x_n+y_n,x+y) \to 0)\Rightarrow d(x+z,y+z)=d(x,y)$

[soft zealot]

Coker2233

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[tranquil pine]

can i get som eurgent help

[tranquil pine]

can someone explain to me where it says "the inductive step"

when they rewrote c_k+1 and b_k+1

[worldly mesa]

What do you not understand?

[tranquil pine]

this part

[proper dagger]

it's the same thing as our discussion yesterday, by the way

just highlighting this for you

[tranquil pine]

yes i understand whats happening now but im confusing on the part where things are expressed in terms of the hypothesis

[proper dagger]

I don't get it. are you confused about why we're using the hypothesis, or something else?

[tranquil pine]

for example see how they rewrote the inductive formula in terms of the inductive hypothesis?

[proper dagger]

uhuh

[tranquil pine]

are they doing the same thing?

[proper dagger]

same thing as what?

as our discussion yesterday?

[tranquil pine]

not exactly

like do you see how theyre expressing the sum up to k+1 as the original sum + the value up to k+1

and the original sum is the induction hypothesis

[proper dagger]

uhuh, yes

[tranquil pine]

yeah so our goal is to express our indunction formula in terms of the induction hypothesis right

in order to prove that k+1 comes from k and then plus 1

you get what i mean?

[proper dagger]

I sort of get your idea, okay...?

and?

[tranquil pine]

like i get what they did for the sum problem

im asking if what they did in the recursive problem is similar

or is the same concept basically

when they expressed both b_k+1 in example b and c_k+1 in example a

[proper dagger]

yes, the concepts are identical, both between this and your original two questions, as well as all three questions here with the one we discussed yesterday

nothing has changed other than the formulas themselves

[tranquil pine]

i see how it worked for the sum problem but not for the recursion examples

idk how they did it

[proper dagger]

it's the same thing, my friend.

let's be reminded of what exactly we're doing ^

[tranquil pine]

i get its the same concept but i cant spot where in the recursive problem that "+1" comes from

for example in the sum problem it came from the fact that the sum having 1 added to it is (k+1)^2

here the + 1 comes from where

for the recursive examples

[tranquil pine]

yes so they just added 1?'

[proper dagger]

it's in the definition of the formula

[tranquil pine]

but shouldnt that be 2(b_k + 1)

[proper dagger]

no? the original formula didn't have that bracket

why did you add it, and where did it come from?

[tranquil pine]

like

[proper dagger]

the original recursion was $b_n = 2b_{n-1} + 1$, not $b_n = 2(b_{n-1} + 1)$

[soft zealot]

fox(x, y); ∂(fox)/∂x (Flower)

[tranquil pine]

ohhhhh

i get it now

its the whole thing + 1 because

its the whole thing + 1

right

[proper dagger]

yea. let's not go adding some random invisible brackets, shall we?

[tranquil pine]

thats what happens when youre on 3 hours of sleep

[proper dagger]

and hence why you should subsist on sufficient sleep before doing math

[tranquil pine]

true, one thing i noticed though is if i remove the one i should have the original b_k on its own right?

or something equivelant to it

[proper dagger]

why are we doing this?

[tranquil pine]

because in the sum problem it was the original thing + 1

[proper dagger]

but why do we want b_k on its own? what's the goal here?

and from which formula are you rearranging for b_k?

[tranquil pine]

idk how to explain it properly

okay so we wanna express b_k+1 in terms of b_k + 1 right

[proper dagger]

in terms of just b_k.

[tranquil pine]

well but plus 1 right

[proper dagger]

which we have already done via the recursive formula

the + 1 is part of the formula

don't think of it as separate

[tranquil pine]

so in the sum problem, the + 1 was part of what b_k+1 is and not seperate ?

[tranquil pine]

where? thats what im confused on idk how

[proper dagger]

it was never separate to begin with! look at the recursive formula again

[tranquil pine]

wait but we arent proving that one right

we are proving b_n = 2^(2+1) - 1

[proper dagger]

we're not, but we are using it to define b_(k+1)

[tranquil pine]

why

[proper dagger]

you really forgot everything I said yesterday?

[tranquil pine]

well we can just plug b_k

cause i thought the goal was to express stuff in terms of b_k

[proper dagger]

and this is doing exactly that!

[tranquil pine]

or the induction hypothesis in general

but b_k is unrelated to the original recursive formula

[proper dagger]

...like I said, you have two choices to define b_k from that point

either you use the recursive formula on b_k again, turning it into a nightmare of an infinite loop... or you use the hypothesis

[tranquil pine]

ooooh

am i trippin

[proper dagger]

yes

I'm not gonna sugarcoat or lie about it

[tranquil pine]

this is what i see

what am i doing wrong

[proper dagger]

I can forgive conceptual errors and even some minor algebraic ones... but let's not copy the wrong sign down, please?

and of course you copied down the wrong sign for the first two lines of the inductive step as well

[tranquil pine]

im plugging in that one

[proper dagger]

oh let's see, my bad

[tranquil pine]

all good

[proper dagger]

this one

the first step in the induction step should always be from the recursive formula

not the general one

you used the general formula for b_(k+1) to prove itself

[tranquil pine]

but i thought whenever you are proving a formula you are plugging in for that formula itself

[proper dagger]

you plug the general b_k into the recursive b_(k+1)

for the third time, I'm gonna send this

[tranquil pine]

but thats what i did

[proper dagger]

no you did not.

your induction step's first line should have had a b_k if you did

but there's none!

instead you had a 2^(k+2) doing... something there

[proper dagger]

(btw, please turn off pings on replying. I'll be lurking anyway)

[tranquil pine]

alright

[proper dagger]

you start with the recursive formula, not substitute it in afterwards

[tranquil pine]

see how they evaluated k+1 then subbed k in afterwards

oops forgot to turn off pings

[proper dagger]

the summation is kinda like the recursive formula here.
a more appropriate example would be the exercise from yesterday.

also, where did they eval k+1?

this?

[tranquil pine]

well they didnt

but logically

i mean thats what chat gpt told me

💀

[proper dagger]

of course....

look. ChatGPT is good at math, sure. better with every passing day even. but if you don't already understand the concept a bit, or are super confused, ChatGPT will not help you.

[tranquil pine]

alright

well then seems like i got this all wrong

so you dont evaluate b_k+1 then plug in the induction hypothesis?

[proper dagger]

AI can only help bolster your skills, in my opinion. If you are not familiar with how GPT words stuff, you are going to be more confused than ever.
which is why I recommend against using GPT if you are confused

[proper dagger]

you state b_(k+1) using the recursive formula, which should involve your IH (induction hyp) term.
then you slot in your IH formula for that term and prove that it is equal to the general formula for b_(k+1)

[tranquil pine]

like that?

bruh i just realized pings arent automatic they reset to on

my bad

[proper dagger]

correct!

but your IH this time is wrong

your IH should be the general formula

otherwise we're gonna do the infinite samba dance

cuz b(k) depends on b(k-1), which depends on b(k-2), which depends on b(k-3)...

[tranquil pine]

but when they evaluated b_k they did it on the other recursive formula

[proper dagger]

I assume you mean the general formula

the one with 2^(n+1)?

that's not a recursive formula, that's the general formula.

[proper dagger]

that is the general formula!

[tranquil pine]

thats not the induction hypothesis???

[proper dagger]

...that IS the induction hypothesis, which is done using the general formula!
more than one thing can be true at once, you know

[tranquil pine]

well yeah thats what i did originally .. then i was good until b_k+1

[proper dagger]

your IH is based on the general formula.
your inductive step, b(k+1), is based on the recursive formula. this recursive formula **should contain the term that the IH is based on (b(k)) **, which allows you to substitute the IH into the inductive step

[tranquil pine]

wait whatttt

why is it not based off of the general formula

i know you probably repeated that 10 times

[proper dagger]

yes I have

[tranquil pine]

🤣

[proper dagger]

this is the fourth time I've shown this

fifth including the original message itself

[tranquil pine]

i think im not asking my quesiton the right way

what i mean is whats the intuition behind it just forget the loop thing

why does it make sense in general

[proper dagger]

why does induction make sense in general, or why this step makes sense in general?

[tranquil pine]

this step make sense in general ,

[proper dagger]

explained in this message.

[tranquil pine]

im gonna be honest, i just dont get it

[proper dagger]

sure. if you say it doesn't make sense, then you're gonna have to tell me which part of that message makes no sense, and why.

also, reminder: no pings

[tranquil pine]

heres the thing, in other examples what ive seen them do is this

as you can see they find the induction hyp from the general formula, then they go on to prove k+1, when they prove k+1 they plug in the induction hyp into k+1

so k+1 is expressed in terms of the induction hyp

oops

bruh i keep forgetting lol

[proper dagger]

if that's your argument, this formula is also eventually expressed in terms of the IH!

you just don't start with the general formula

[proper dagger]

in fact, here the first step uses the "recursive formula" too!

[tranquil pine]

they are plugging the hyp into the recursive formula? see this is where the confusion happens

[proper dagger]

right here

that's the "recursive" formula

[proper dagger]

yes!

[tranquil pine]

😭

but like all i see is them plugging in for K+1

[proper dagger]

ok, since you are this confused, I'm going to be super strict here

[proper dagger]

what do you mean "plugging in for k+1"? plugging WHAT in for k+1? or plugging k+1 into WHAT?

if you have a confusion from now on, I want you to be as unambiguous as possible

[tranquil pine]

i cant unsee that

am i looking at it correctly like that?

[proper dagger]

I don't know what the point of this diagram is.

all you have shown is that a summation of j^2 from 1 to k+1 contains the summation of j^2 from 1 to k

which I'm not arguing with - that's literally the first step of the inductive step

but where your problem is, I'm afraid, is that your IH is missing the actual expression!

this isn't the IH.

THIS is the IH.

yes, together with the equal sign and the RHS (right hand side)

[tranquil pine]

okay look at this example

IH went into k+1

[proper dagger]

you wrote it yourself. what about the text's solution?

[tranquil pine]

they did the same steps

[proper dagger]

I don't get this whole thing

[proper dagger]

show.

[tranquil pine]

"by algebra"

[proper dagger]

so 3 | 2^(2n) - 1 (3 divides [2^(2n) - 1]) is the formula, and the IH here is that 2^(2k) = 3m + 1.
the first three steps make complete sense. you want to expose the 2^(2k) so that you can substitute the IH in.

I don't get what's the issue here.

yes, you expose the 2^(2k) by algebra, because otherwise there's no 2^(2k) for you to sub the IH into!

[tranquil pine]

yes this is the problem

here they say by algebra

the recursion example they say "by definition"

the hell does that mean

[proper dagger]

by algebra means that you have to do some work to expose the IH term
by definition means that you don't have to. you're given the formula that contains the IH term

that's really all

[final saddle]

@tranquil pine Has your question been resolved?

[final saddle]

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[short wraith]

Residue at π/2 of function sinz/(zcosz)

[worldly mesa]

Isn't the function analytic there?

There is no singularity

[short wraith]

Sry

Check now

@worldly mesa

[worldly mesa]

Well what did you try

[short wraith]

I tried with expansion?

(z+z^3/3+..)/z

[worldly mesa]

Yeah expansion is good

[short wraith]

We have no negative terms so??

[worldly mesa]

You forgot about the cos

[short wraith]

I tried with tan

Ohh we are finding 1/z-π/2 cofficients in the expansion

[worldly mesa]

Also note that you have a pole here

So there is one more way you can find the residue

[soft zealot]

Couldn't find an attached image in the last 10 messages.

Oct 15, 2025, 9:01 AM

[final saddle]

@short wraith Has your question been resolved?

[worldly mesa]

Since f(z)=sinz/zcosz has a simple pole at z=pi/2, you have that the residue is the limit of (z-pi/2)f(z) as z-->pi/2

[final saddle]

@short wraith Has your question been resolved?

[short wraith]

Ohh nice

[final saddle]

@short wraith Has your question been resolved?

[final saddle]

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[fallow compass]

p,q,r,s are non-zero numbers. r and s are solutions of x²+px+q=0, p and q are solutions of x²+rx+s=0. What is the value of p+q+r+s?

[fallow compass]

By vietta, we can see that
p = r+s
q = rs
r = p+q
s = pq

We can see that p=r+s, then p-s=r
p-s = p+q, so q=-s

Then
q = rs
-s = rs
-1 = r

Also,
s = pq
s = -ps
p = -1

So p+q+r+s = -2

I don't believe in myself so I need a validation here, is this correct?

[silver cargo]

hm, I got the same final answer as you, but don't you want to at least fill in s and q?