help-26

[wind ember]

What's the question?

[neon iron]

You need to solve the right equations, according to the right circumstance

Did you do that?

[wild zephyr]

Idk😭

I don’t know how to factor it (?? Sorry my english is bad

[wind ember]

Pull out 2 from the equation so it's 2(27x^3 + 125y^3)

Notice that 27 and 125 are cube of 3 and 5

Combine it with their x and y then u can use the normal cubic formula

[wild zephyr]

Like this??

[timid junco]

For the questions ur answering u don’t hafta factor

Well for that one u do

[wind ember]

Yes

Btw write it as (3x)^3

So u won't confuse urself

[wild zephyr]

ok!!!

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[sullen scaffold]

I know this isn't strictly maths, but anyone who does hsc physics or has a physics background I would like some assistance pls

[sullen scaffold]

the formula they use is only correct for very small values of theta, but in this case it is very large

and you cannot use d sin(theta) = m lambda

as the slit size is much too small

so how would you properly work this out?

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[faint comet]

for applications of trignometry when should I use sin theta/ tan theta

[fallow heart]

What do you mean?

sin/tan is cos

[fair thorn]

think he means sin t or cos t

[sullen scaffold]

maybe asking for the difference of sin and tan?

[faint comet]

Nooo

I mean

Sin and tan are different and if the question doesn’t say which one to use how should we know that we gotta use tan or sin

[sullen scaffold]

do you have an example?

[restive inlet]

by definition of what the trig functions do

[visual lark]

obviously

[restive inlet]

and what you're given / trying to find

[visual lark]

depends on question.

[restive inlet]

if they're both applicable, you can use either

[faint comet]

K

.close

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[neon iron]

Is this correct

[wary tulip]

[wind ember]

Is this astronomy?

[wary tulip]

i think you mean astrology

[fair thorn]

@neon iron what exactly is your question

and how is it related to mathematics?

[sullen scaffold]

a small something tells me this isn't maths related

lmao

[wind ember]

I mean constellation names doesn't always mean horoscope

[neon iron]

is it correct

[sullen scaffold]

why would the mathematics discord know?

[river thunder]

lmfaooooooooooooo

[neon iron]

coz i remember a mod said that

[sullen scaffold]

what? lmoa

[neon iron]

this can also be used for science

can anyone check?

[sullen scaffold]

is it constelations or astrology

[neon iron]

Im not sure if the websites

[sullen scaffold]

bc if its astrology

[neon iron]

I used is correct

[wary tulip]

ok no need to hound her for the question

[neon iron]

its science

[river thunder]

LMFAOOOOOOO

[sullen scaffold]

astrology or ur question?

bc astrology is not science lol

[neon iron]

in my science class yes it is

[river thunder]

astrology = science 👏 👏 👏

[neon iron]

💔💔

[wary tulip]

it's not astrology i was joking jeez

[sullen scaffold]

💀

[river thunder]

ok wait it looks like astronomy now that i think about it

[neon iron]

Are these correct thp

tho

[river thunder]

like it's saying when the constellations appear

[fair thorn]

was about to say that

and frankly, no idea

[sullen scaffold]

rightttt

[river thunder]

why are you asking here this is math discord not science discord

ask google

[sullen scaffold]

thats chill then lol

[fair thorn]

not the sort of stuff we typically do

on mathcord

[river thunder]

ye

[sullen scaffold]

I apologise

[river thunder]

thats respectable

i also apologise

[sullen scaffold]

But yeah, i think google or a more specific discord would be more appropriate

[wind ember]

I guess u can brute force google it for each constellation but yeah we can't answer ur question since it's not about math

[wary tulip]

or you could just let the question sit here for longer than 2 minutes without spamming the channel

[neon iron]

but is it correct

[wary tulip]

can't answer ur question since it's not about math
lmao

[neon iron]

Idk if the sites i used in google are right

[fair thorn]

by that logic you should never google anything, you'll never know if the sites are factually correct

[wary tulip]

yea that's definitely what that logic says

[neon iron]

ok

.close

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[neon iron]

thankss

[fair thorn]

what?

[wary tulip]

🙄

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[charred moat]

i just wanted to clear a simple/silly doubt

[charred moat]

this is true right?

[fallow heart]

Yep

[charred moat]

only (a^b)^c= a^bc

am i right?

[fallow heart]

Correct

[charred moat]

alright thanks

is there any identity

for that case

[fallow heart]

I don't think so

[charred moat]

ok

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[cunning holly]

there is but nvm ig

[fair thorn]

??

you mean

@cedar wagon ann's back btw, go say hi

[cedar wagon]

its false for two cases, b = c = 2 or a = 1

[thorny flame]

ren

[cedar wagon]

i did say, ask how she was doing but didn't get an answer back xd

[fair thorn]

o

[cedar wagon]

it's fine, she told others she was fine so

[fair thorn]

hlounge is going really fast rn

[cedar wagon]

yeah, many things should be in help channel tho

[fair thorn]

eh

[fading owl]

the last line of ur bio is a quote from koro sensei in the anime 'assassination classroom'

off-topic

[fair thorn]

idt ppl'd appreciate me asking random questions abt metrics and inner products and inf-dim hilbert spaces

[cedar wagon]

never watched but i trust you

[fading owl]

k 😂

sry for disturbing bye

[cedar wagon]

lmao you're not dw

[cedar wagon]

no i mean, theres no prob to ppl asking help questions there ofc, it's just a matter of having maybe more advices in help channel

cuz some aren't helpful

but they have many knowledge

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[frosty iron]

Let S,E,N,D,M,O,R,G,L,Y be distinct digits (0 - 9) and SEND + MORE + GOLD = MONEY, where SEND is a four digit number with S in the thousands, E in the hundreds, and so on. find the value of every variable

[frosty iron]

the only thing i found is that M is no more than 2

not sure what to do next

[jovial pasture]

Can their leading digit be 0?

[frosty iron]

the question didnt say

[jovial pasture]

Okay, so maybe try to sub in M = 0, 1, 2. Honestly just do 1 or 2, cause one can safely assume leading digits are not 0

[frosty iron]

still clueless

😭

[jovial pasture]

alright, try M=2

sorry if this is not helping btw, but usually these type of problems require some creative thinking + noticing bounds

[frosty iron]

its alrihgt

[sage mica]

bruh usually a cryptarithm puzzle only has 2 terms and result

y does urs have 3 😭

[jovial pasture]

actually I think you should show how M=2 does not work

[frosty iron]

quirky type maths

[frosty iron]

oh yeah

the maximum of x + y + 2 is 20 if x = y = 9 but x,y are distinct

so m = 1

[jovial pasture]

yes, assuming leading digits cannot be 0, but like I said before this is a safe assumption

[jovial pasture]

now sub this in

[frosty iron]

i still cant get anywhere😭

[jovial pasture]

so we get SEND + ORE + GOLD = 9000 + ONEY, right?

[frosty iron]

yeah

S + G is 7,8,9

oh

<@&286206848099549185> someone with python brute fforce this pretty please😇😇😇😇

[jovial pasture]

this is a weird cryptogram cause I feel like there might be many solutions

[frosty iron]

huh

[jovial pasture]

idk, now maybe try S+G = 9

[frosty iron]

this was the practice questions in a number theory olympiad book😭

(the practice questions dont have any solutions in the book for some reason)

[barren marten]

Can you send the question?

[sage mica]

Let S,E,N,D,M,O,R,G,L,Y be distinct digits (0 - 9) and SEND + MORE + GOLD = MONEY, where SEND is a four digit number with S in the thousands, E in the hundreds, and so on. find the value of every variable

[prisma mesa]

||{'S': 2,
'E': 8,
'N': 5,
'D': 6,
'M': 1,
'O': 3,
'R': 4,
'G': 9,
'L': 7,
'Y': 0,
'SEND': 2856,
'MORE': 1348,
'GOLD': 9376,
'MONEY': 13580}||

this is python

[frosty iron]

thats crazy

thanks

[jovial pasture]

I think there’s many solutions

[frosty iron]

there probably is idrk

[jovial pasture]

found this online: https://www.pleacher.com/mp/puzzles/mathpuz/mobcry4.html

[prisma mesa]

oh wait

there are more

[frosty iron]

no way

[barren marten]

There could be.
We need to use computational power ig

[frosty iron]

theres definetly a way to use logic for this tho

uhh i gotta go, do i leave it here for you guys to talk about it or?

[prisma mesa]

Assume M = 1.
SEND + 1ORE + GOLD = 1ONEY

[jovial pasture]

funny thing is that SEND + MORE = MONEY I think has only one solution given the constraints. It’s covered by mind your decisions too

[frosty iron]

lmao

[prisma mesa]

python yields like 10 sols now

i gotta verify if it's correct

yeah, idt this is easy to do by hand

you could probably go over all the cases

but its painful

[quartz ingot]

Since SEND and MORE are four-digit numbers, ( S ) and ( M ) cannot be zero.

( M ) must be 1 because MONEY is a five-digit number.

(S + M + G ) Carry over to the ten-thousands place. Since ( M = 1 ), ( S + 1 + G ) must be at least 10.

So, ( S + G ≥ 9 ).

( E + O + O ) Should carry over to the thousands place.

( O ) is a digit, ( E + 2O ) is at least 10.

( N + R + L ) Must carry over to the hundreds place.
So, ( N + R + L ≥ 10 ).

( D + E + D ) Carries over to the tens place.
Meaning ( 2D + E ≥ 10 ).

Solving,
( S = 9 )
( E = 5 )
( N = 6 )
( D = 7 )
( M = 1 )
( O = 0 )
( R = 8 )
( G = 3 )
( L = 4 )
( Y = 2 )

Values are then:
SEND= 9567
MORE = 1085
GOLD= 3647
MONEY = 14299

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[dense plover]

Let $E$ be an elliptic curve. How do we show that the "multiplication by n" morphism is a finite morphism?

[thorny flame]

DavidL1450

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[neon iron]

\begin{align*}
ax^2+bx+c &= a(x-\alpha)(x-\beta)\
\cancel{ax^2}+bx+c &=ax^2-a\alpha x -a \beta x +a\alpha\beta\
\text{comparing coefficients}&\
b=-a(\alpha+\beta) &\text{and}\quad c=a\alpha\beta\
\text{now in equation}& a(x-2)^2-b(x-2)(x-3)+c(x-3)^2=0\
\text{putting values of c and b}&\
&a(x-2)^2+a(\alpha+\beta)(x-2)(x-3)+a\alpha\beta(x-3)^2=0\
&a[(x-2)^2+(\alpha+\beta)(x-2)(x-3)+\alpha\beta(x-3)^2]=0\
\end{align*}

[thorny flame]

praee2k

[neon iron]

thats my progress is but idk how can i solve further

[unique perch]

i was about to claim this channel, my question was also polynomials lmao😭. i'll add what input i can

[neon iron]

ping me when u re replying

[neon iron]

better claim other free channels

[unique perch]

I will, but if i can help out with something i will too

[neon iron]

sure sir thanks

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@neon iron Has your question been resolved?

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@neon iron Has your question been resolved?

[brave coral]

You could just go further and solve this with the quadratic formula but that's a pain

Consider this instead

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[autumn sparrow]

What is the values of sec^2(3x) + sec(3x) = 2 in the domain 6π<x<9π ? Can someone do it so I can check my answers for a test I did

[craggy haven]

why don't you show your work and we'll tell you if you're right

[autumn sparrow]

I can't since it was on a test, but I can try solving it again

[prisma mesa]

,w sec^2(3x) + sec(3x) = 2

[thorny flame]

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[prisma mesa]

if you just need verification, you can use command like this

,w sec^2(3x) + sec(3x) = 2, 6π<x<9π

[thorny flame]

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[autumn sparrow]

How would you do this question though? I just found cos at that domain and then divided it by 3

[prisma mesa]

hmm

I'd substitute u = sec(3x)

u^2 + u - 2 = 0

solve this

and get solutions, say p, q

then solve
sec(3x) = p
sec(3x) = q

this would be my approach

[slender pebble]

damn i keep missing those in your face substitutions

[prisma mesa]

tbf i wouldnt even do the substitution

[slender pebble]

yeah

[prisma mesa]

I'd directly just solve the quadratic in terms of sec

[slender pebble]

yup

[prisma mesa]

but why not ig ¯_(ツ)_/¯

[autumn sparrow]

I did that and got sec as -1/2 and 1

But then the domain kept screwing me up

[prisma mesa]

,w u^2 + u - 2 = 0

[thorny flame]

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[autumn sparrow]

yeah, cos would then be the Inverse of those

[prisma mesa]

mhm

so cos(3x) = 1

and cos(3x) = -1/2

3x, pardon

[autumn sparrow]

and then I found the values of cos at those angles

Or I guess found the angles at those values

[prisma mesa]

this would mean

3x = pi + 2npi
3x = pi/3 + 2npi
3x = -pi/3 + 2npi

and with this you should be able to extract the required sols

[autumn sparrow]

Those are general solutions then, no?

I think the question was asking for the exact values

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[supple socket]

2cos(x)tan(x) = 0
I am thinking I have to subtituta tan(x) for sin/cos as an identity, butn then, can I cancel out the cos at the top and then bottom, but there's the 2 in front of the cos,, so I feel lost

2cos(x)sin(x) / cos(x) = 0

[hallow depot]

$x s.t f(x)g(x)=0\iff x s.t. f(x)=0 or g(x)=0$

[thorny flame]

everg

[acoustic pecan]

in this case you can cancel the cos, yes

[hallow depot]

but here you have a singularity lol

[supple socket]

so then 2sin(x) = 0 ?
and then divide by 2, but it'd still equal 0, so

so sin(x) = 0

this is the domain or interval or whatever it is called

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[sullen herald]

Its sequences and series so ig precalc?

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[thorn surge]

does anyone have any idea on how to do part a and part b?

do i just integrate fx
fy and fz?

[raven sparrow]

That's usually the idea. You integrate fx, that gives you some expression but your constant is a constant in terms of y,z, since any function of y and z is constant with respect to x.

So then you can redifferentiate in y and compare with the y coordinate of F

[thorn surge]

okk

wait what

im confused 😭

[raven sparrow]

When you integrate the x coordinate with respect to x, you'll get like
f(x,y,z) = (whatever you integrate) + g(y,z)

But you need to know g(y,z)

So you differentiate with respect to y, and compare with F's second coordinate

[thorn surge]

i differnetaite what and compare it to 2nd component of F

[raven sparrow]

You want to find $f(x,y,z)$ such that $\vec{F} = \Delta f = (f_x, f_y, f_z)$.

If you integrate the first coordinate, then that should correspond to integrating $f_x$.

You'll get $f(x,y,z) = \int f_x(x,y,z) dx + g(y,z)$ for some unknown function $g(y,z)$.

You need to figure out what $g(y,z)$ is (because we want an expression for $f(x,y,z)$), so you can differentiate that with respect to y (this means you'll also be differentiating g(y,z)).
This gives you an expression for $f_y(x,y,z)$ which you can compare with the second coordinate of $F$ (in particular you can identify what $g_y$ is).
Then it's a matter of doing the whole thing again (integrate with respect to y, you'll be left with an unknown function $h(z)$, differentiating with respect to $z$, identitying $h'(z)$ and finding $h(z)$.

[thorny flame]

Azyrashacorki

[raven sparrow]

This is probably clearer

https://mathinsight.org/conservative_vector_field_find_potential

[thorn surge]

kind of

im confused on steps

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[urban barn]

am i being stupid, or is my answer already right?

1 - p^5 is the probability of X happening at least once

X does not happen at least once is 1 - (1 - p^5) = p^5

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[river thunder]

Could someone help me take a look at d

Ah shoot

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[river thunder]

What the

[sonic belfry]

oop

.close

[river thunder]

Is it broken

[sonic belfry]

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[autumn frigate]

could som 1 help me rq

[autumn frigate]

this is my problem and I was wondering if there are 5 real zeros

its odd negative, I believe there are 5 distinct zeros and 2 relative maximum's or maybe 3 I am not sure if the one farthest to the right rises again

yellow, is anyone there?

[neon iron]

i think there would be 3 relative maxes

bc of the right-most one

since f'(x)=0 at that point

wait no

that would be a min

[autumn frigate]

mhm

ur right

but 5 zeros right?

[neon iron]

yes

[autumn frigate]

alright thanks for your help man

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[last lava]

a) increasing at (-2,0) and (5,∞). decreasing at (-∞,2) and (0,5)
b) increasing at (-∞,-1) and (-3,∞). decreasing at (-1,3)
c) local min at x=-2 and x=5 because they cross the x axis from negative to positive
d) local max at x=0 because it crosses x axis from positive to negative
e) x=-1 and x=3

I need help with part f please

[neon iron]

have you done integrals

[last lava]

no

[neon iron]

ok

what does f(0)=100 mean

in this context

[last lava]

at x=0 y=100?

[neon iron]

yes

so you apply all of the characteristics and move it to 0,100

[last lava]

wdym apply characteristic?

[neon iron]

what you found a-e

[last lava]

is a-e correct? im not sure if i did those correctly

[neon iron]

ya

looks right

we also know local max at x=0, so (0,100) is a ?

[last lava]

abs max?

[neon iron]

at least a local max

idk if we can say its absolute yet

[last lava]

would it be something like this?

[neon iron]

that could work

make sure not to make the curves too pointy bc someone might think its an absolute value

[last lava]

got it, will look better on paper

[neon iron]

other than that looks good

[last lava]

if the graph kept going before -3 is it gonna be be above the x acxis?

[neon iron]

we don't know that

that's why they ask for a sketch

[last lava]

oh ok

thanks

[neon iron]

yup

[last lava]

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[strong sable]

Am I doing correct so far?

[fallow heart]

Why is f' = f² ?

[strong sable]

Uhh it's just a given condition

[fallow heart]

Ah alright

[opal vault]

not sure about f'''

[strong sable]

Yea I doubted that either but idk where's wrong

[opal vault]

should be 6f^2 * f'

since it's 2 * (f f f' + f f' f + f' f f)

or use power rule

f^3 differentiates to 3f' f^2

[strong sable]

Aah finally found the fault

Ty!

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[opal vault]

A trick I often use

PFD

won't need recurrence

partial fraction decomposition

[strong sable]

Ooh but they asked to find the recurrence relationship😭

[strong sable]

I know how to do in this way

But I'm really unfamiliar with recurrence

[opal vault]

I have no idea how to get this recurrence relationship then

Unless you already know what the n-th derivative looks like

[strong sable]

Ok ty!

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[empty fossil]

Let d be a metric on $\Omega$. Then let f be continous on the positive real numbers, concave and strictly increasing also f(0)=0. show that the function $g(x,y)=f \circ d(x,y)$ is a metric for $\Omega$.

[thorny flame]

bigpufik

[empty fossil]

Most of the definitions a metric has a have are obvious except the triangle inequality, I mean

g(d(x,y)) < g(d(x,z)+d(y,z)) is obvious

But this problem would make more sense if f was convex, just jensen of karamata. Any ideas?

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[sleek bluff]

hey guys

4sin(2pix) where x [0,2]

dilation from y axis by factor of 2, translated 3 units right.

What is the new domain and rule? I wanna check my answer

[sleek bluff]

<@&286206848099549185>

[desert peak]

It should be $y'=8 \sin(2\pi(x'-3))$

[thorny flame]

cristorenzo99

[desert peak]

Where the transformation is:
$$\begin{cases}
y'=2y\
x'=x+3
\end{cases}$$

[thorny flame]

cristorenzo99

[sleek bluff]

isnt dilation from the y axis within the brackets

i though dilation in the x axis went in front of sin - effected amplitude

[desert peak]

I wrote a transformation from (x,y) [old variables] to (x',y') [new variables]

[sleek bluff]

the A is affected by the dilation in the x axis

werent we given dilation in the y axis though?

like if its stretched from the y axis wont it affect the period?

so if its still 2pi the period is the same

[desert peak]

Ah, maybe I didn't understand what were you saying

A dilatation from the y axis you mean the y axis remains still and you move the x axis?

[sleek bluff]

yeah like its stretched like this

[desert peak]

Then I swapped the equations

[sleek bluff]

it should say 'dilation FROM y axis'

so how do you do the image translations

[desert peak]

I find the law of the transformation for the new variables (x'), find x from this relation as a function of x' and sub x

[sleek bluff]

is it gonna be something like this

and y stays unchanged?

[desert peak]

I believe so. Let me draw the graph to be sure

[desert peak]

I believe this is the correct answer:

[sleek bluff]

it should be stretched outwards

from the y axis

the pink one is dilated from the x axis isnt it

[desert peak]

The y coordinates of the pink one are the double of the green one

[sleek bluff]

yea i get what u did

but based off of the wording

'dilation from y axis'

i thought it meant u stretch it outwards from the y axis, therefor effecting the x values

[desert peak]

You wrote "in", not "from" here

[sleek bluff]

sorry i meant to write from

i didnt think it made a difference so i didnt edit it

im sorry thats my bad

did that change ur entire interpretation of the question 😂

[desert peak]

1. Dilatation from the y-axis by a factor of 2: $x'=2x$

2. Translate x units by 3: $x'' = x'+3$

Putting all together: $x'' = x'+3 = 2x+3 \Rightarrow x''-3 = 2x \Rightarrow x = \frac{x''-3}{2}$

[thorny flame]

cristorenzo99

[sleek bluff]

so 4sin(pi(x-3))

[desert peak]

From $x'' = 2x+3$ you can find the new domain by subbing $x=0$ and $x=2$ and you can find the new law subbing x in $y=4\sin(2 \pi x)$

[thorny flame]

cristorenzo99

[sleek bluff]

ahhh thats how you find the domain

so [3,7]

thank u so much

sorry it took so long

[desert peak]

You're welcome. It was my fault too because English is not my mother language

[sleek bluff]

All good, have a good one bro

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[neon iron]

I NEED AN ANALYSIS TEACHER ON YOUTUBE PLZ

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[neon iron]

a bag contains 6 red balls, 3 yellow balls, and 8 green balls. what is the probability of drawing a red ball and a yellow ball if two balls are drawn at a time ?

[neon iron]

this is the second step using the combination formula to calculate the number of ways to draw 1 red ball and 1 yellow ball after determining the total number of balls.

my doubt lies in where the 16 comes from

[woeful drift]

it comes from algebra

17! = 17*16*15!

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[long fog]

What to do in cases where neither the limit at LHL,RHL or at the point is defined

[long fog]

like say lim x->0 ln(sin 3x)/ln(-sinx)

[agile fulcrum]

the domain does not include x = 0+

[neon iron]

<@&268886789983436800>

[agile fulcrum]

so we only consider x = 0-

[long fog]

0- makes the numerator ln(0-)

[agile fulcrum]

yes…

Oh wait

so the limit at 0 does not exist

its the same case as when lnx as x tends to -1

[long fog]

the solution given is 1

[agile fulcrum]

oh… hold on a second then..

no…

Are you sure that’s the question?

[long fog]

yes

[karmic dove]

this function has no value at least near 0 so what are we discussing? limit DNE for sure

and moreover this function doesn't not even exists

it does not has any value

[long fog]

Wolfram also says 1

😭

[karmic dove]

aparently i am wrong

[torpid matrix]

,w graph ln(sin(3x))/ln(-sin(x))

[thorny flame]

Results provided by WolframAlpha

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

Requested by tuzis

[torpid matrix]

,w lim x->0 ln(sin(3x))/ln(-sin(x))

[thorny flame]

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Upgrade to WolframAlpha Pro!

Requested by tuzis

[torpid matrix]

huh

[karmic dove]

this guy says that it dne hut aparently there is some values for this function which i wonder how

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[trail jacinth]

does anyone know the word for the result of monoid folding?
is “product” fine?

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[bitter nimbus]

∀xPx → Qc, ∀x(Px → Qc)
would this serve as a counter-model and if so, why?
D={1,2}
a: 1
c: 2
P={}
Q={2}

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[cosmic siren]

Can someone explain this working to me?

[cosmic siren]

How is this part achieved?

[smoky sparrow]

(r - 4)! = (r - 4) * (r - 3)!

just like how 11! = 11 * 10! for instance

4! = 4 * 3!

it's this property used a lot of times at once

[neon iron]

i think you mean (r-3)! on the lhs

[smoky sparrow]

corrected yeah

[cosmic siren]

ohh

I was confused how 10! = 11x10!

[smoky sparrow]

brainfart moment

[wind ember]

Factorial is every number from 1 to n timed together

So 3! Is 3*2*1

Which is also 3*2!

[cosmic siren]

Wait how is b-8 ! = b-8 x b-9 !

[smoky sparrow]

if a variable is too hard to think about, use examples with numbers

so then the pattern is n! = n * (n - 1)!

now just sub n = b - 8

(b - 8)! = (b - 8) * (b - 8 - 1)!

[keen plinth]

i think it will be much easier to observe if it is addition, but b-9 is smaller than b-8

[cosmic siren]

Ohh
I was thinking why is the number bigger

Forgot about the ‘-‘ sign

[wind ember]

[smoky sparrow]

yeah it's confusing with variables

so reducing to a simpler case always helps

[cosmic siren]

Another one, why is x=24 used to explain why the estimate is unreliable? Is it because it is an extrapolation?

And also if level of significance is not given can I still do hypothesis testing?

Or it it purely impossible

And is it possible to find the level of significance by working backwards somehow?

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[uncut bone]

Just need slight help
for $\lim n\to\infty \frac{1}{log(1+\frac{1}{n}) · a^{n+1}}
How do you simplify for this limit

[uncut bone]

$\lim n \to \infty \frac{1}{log(1+\frac{1}{n}) · a^{n+1}}

damn

[outer salmon]

$\lim_{n \to \infty} \frac{1}{log(1+\frac{1}{n}) \cdot a^{n+1}}$

[uncut bone]

not meant to be an u there

but a multiplication

[thorny flame]

FirstNameLastName

[uncut bone]

log being logarithm of base 10

thank you

[uncut bone]

slightly confused as to how to simplify this as i can't just plug in the values for log(1+1/n) but =0 for n->inf

also a is a real number

if you were to plug in for n->inf,
0<a<1, 1/0 · 0 = inf
|a|>1, 1/0 · inf, denominator has an indeterminate form

so how do i fix for this case |a|>1?

(i'm just looking for an appropriate rewrite)

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[zenith night]

I don't even know where to start here. They're asking me to find the limit of the series using P-Test, Integral test, or comparison test

[agile harness]

well do you think you can integrate that

you could also do comparison

[agile harness]

which uses the result of a p series

@zenith night

[zenith night]

Hmm, how would I do a comparison with this?

Usually the comparison questions have more terms than what you are comparing it to

[unreal quest]

Integral test could be a good idea

[agile harness]

$\sum_{n = 1}^{\infty} \frac{1}{n}$

[thorny flame]

knief

[agile harness]

does this diverge

[unreal quest]

Remember the integral is a continuous sum, and, as such, will be greater than or equal to the original series

[unreal quest]

Yes, that's the harmonic series

[agile harness]

i wasn’t asking you but yes

[unreal quest]

(oh, sorry, thought you were the guy of the channel)

[agile harness]

💀💀💀

[unreal quest]

I'll leave it to ya then, and sorry for interfering 💀

[agile harness]

@zenith night

boss

you opened a help channel then dipped

twice

[zenith night]

I'm too late now but yes it divergers

[agile harness]

very good observation

is ln(n)/n > 1/n

take a moment to think about that

[zenith night]

It also diverge bc of p test right

[agile harness]

no

by comparison

1/n diverges by p test

[zenith night]

p>=1?

[agile harness]

first of all this is just wrong

it diverges for p<= 1

and that’s not relevant because the numerator is ln(n) not 1

[zenith night]

Ahh okay

[agile harness]

^^

[zenith night]

Yes??

[agile harness]

great now let me ask you a follow up question

if we know that the harmonic series diverges and we know that the terms in ln(n)/n are greater than the terms in 1/n what do you think the sum of ln(n)/n will do

[zenith night]

Converge?

[agile harness]

no

we’re adding terms that are bigger

and are decreasing less rapidly

than a divergent series

if it’s decreasing less rapidly

what makes you think this one would converge

[zenith night]

You're right about that. I see why it would diverge now. Why did you use 1/n to compare again?

[agile harness]

because we know it diverges and clearly since ln(n) > 1 for n>=3 we know comparison would work great here

alternatively we could’ve used the integral test

since ln(n)/n is positive decreasing and continuous for n>=1

$\int_1^{\infty} \frac{\ln(x)}{x} = \frac{1}{2} (\ln x)^2 \rvert_1^{\infty} \to \infty$

[thorny flame]

knief

[zenith night]

Why is n>=3?

I understand why ln(n)>1 though

[agile harness]

well, because ln(1) = 0, ln(2) = 0.693 and those are less than 1 so 1/n would be greater than ln(n)/n in those cases

but this doesn’t affect the divergence

we’re not concerned with the size of the series

it’s not really about how large it is

[zenith night]

Ohhh I understand now

Thank you. I appreciate it! @agile harness

[agile harness]

you’re welcome sir

have a nice day

[zenith night]

You too

[topaz sinew]

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[kind rose]

I have to differentiate h(x) = ln(1/cos(x) + tan(x)), i dont know how to do it

[agile harness]

hmm this is a famous one

[unique quarry]

Chain rule

[agile harness]

$\ln(\sec(x) + \tan(x))$

[thorny flame]

knief

[kind rose]

what is sec(x)

[agile harness]

1/cosx

[kind rose]

does this stand for sectant

[agile harness]

secant

[kind rose]

Thank you

So all i did was wrong?

[agile harness]

i mean i see what you’re doing but it’s just a bad approach

you should know by memory that

$\frac{d}{dx} \sec(x) = \sec(x)\tan(x)$

[thorny flame]

knief

[agile harness]

and $\frac{d}{dx} \tan(x) = \sec^2(x)$

[thorny flame]

knief

[kind rose]

Do i then do the fraction technique and add it together

[agile harness]

it isn’t sec(tan(x))

it’s sec(x) * tan(x)

and it should be + sec^2

not * sec^2

[kind rose]

But dont i have to replace the x

[agile harness]

wdym

[kind rose]

The cahin rule

[agile harness]

$\frac{d}{dx} \ln(\sec(x) + \tan(x)) = \frac{\frac{d}{dx} \sec(x) + \frac{d}{dx} \tan(x)}{\sec(x) + \tan(x)}$

[thorny flame]

knief

[agile harness]

derivative of sec(x) = sec(x) * tan(x)

derivative of tan(x) = sec^2(x)

adding them together yields

sec(x) * tan(x) + sec^2(x)

what do you notice

🙈🙈

[kind rose]

trigonometric functions arent put together in the x, but plussed on the side?

[agile harness]

wdym

[kind rose]

Like when i do the chain rule, i have f'(g(x))

[agile harness]

yep

your g(x) is sec(x) + tan(x)

f(x) is ln(x)

[kind rose]

oh

[kind rose]

is sec(x) the derivative of 1/cos(x) or the same thing?

[agile harness]

no sec(x) is literally just defined as the reciprocal of cosx

similarly csc(x) is the reciprocal of sinx, 1/sinx

and cot(x) is the reciprocal of tanx, 1/tanx

[kind rose]

I dont understand what a reciprocal is, my teacher did not teach me this

[agile harness]

1/

ex, reciprocal of 2 is 1/2

reciprocal of 3/4 is 4/3

reciprocal of 1/e is e

[kind rose]

Its the opposite

[agile harness]

you can call it whatever you’d like

[kind rose]

Is then the answer

[agile harness]

yes but you can simplify

factor out sec from the chain rule stuff

you should see something cancel

[kind rose]

I dont see the vision bro

Maybe i should make it a fraction?

[agile harness]

$\frac{\sec(x)\tan(x) + \sec^2(x)}{\sec(x) + \tan(x)} = \frac{\sec(x)(\tan(x) + \sec(x))}{\sec(x) + \tan(x)}$

[thorny flame]

knief

[agile harness]

what cancels

🤔

[kind rose]

sec^2 becomes sec

But how is that possible

[agile harness]

i factored out sec(x)

from sec(x)tan(x) and sec^2(x)

[kind rose]

How

It doesnt change except forthat one

[agile harness]

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

[thorny flame]

knief

[agile harness]

pretend a = sec(x)

and b = tan(x)

[kind rose]

Ooooh you make a paranthese

This looks like the opposite of a double product

[kind rose]

@agile harness can it be reduced any further

[topaz sinew]

@kind rose Has your question been resolved?

[agile harness]

yea the sec(x) + tan(x) cancels

and you’re left with sec(x)

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[true dirge]

hii need help w this

[true dirge]

A and then 1/3?

<@&286206848099549185>

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@true dirge Has your question been resolved?

[true dirge]

@helper

<@&286206848099549185>

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[woven geode]

hello

[woven geode]

I'm having trouble finding A as a function of r

I don't know the steps to invert this equation

I know I need to replace r with y and solve for x

or in this case a and r

but i'm not certain where to start

or the precise steps to take

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[red socket]

how do i go about solving for m? already got that n is 6 but i didnt see any guides on that side

[jade cape]

ok, 30-60-90 triangle

that means n = 6, which you already got

[fading owl]

alright

[jade cape]

m = n * root3

ooops lemme add latex

$$ m = n\sqrt{3} $$

[thorny flame]

reaver

[red socket]

i see

6 square root 3

ah it worked!

[fading owl]

we can deduce this because $\sqrt{12^2 - 6^2} = 6\sqrt{3}$

[thorny flame]

Death.cv

[jade cape]

that too

[red socket]

interesting

so if this problem follows the same layout and all, the same formula is applicable right

[fading owl]

yes

try to get x

[red socket]

20 sqrt 3 is what i got

[fading owl]

wait what

[jade cape]

?

[red socket]

system must be bugged or smth

[jade cape]

somethings wrong there i think

[fading owl]

isnt it 20 sqrt 3

[jade cape]

should be?

[fading owl]

yeah

[red socket]

yeah site is def wrong

[fading owl]

fr

[red socket]

ty for your help yall

[fading owl]

np

[jade cape]

np

[red socket]

its bugged on every other one

.close

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[lone smelt]

part c

I tried the rational indicator function where the upper sum is 1 and the lower sum is 0. However these exist as Riemann sums right?

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@lone smelt Has your question been resolved?

[lone smelt]

<@&286206848099549185>

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[lone smelt]

<@&286206848099549185>

[sweet shard]

You're gonna have to define what s and S are for anyone to know what you're talking about

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[barren panther]

How to solve this? Do I have to use "L'Hôpital's rule"?

[loud oasis]

you can use the strategy of multiplying by the conjugate

[barren panther]

Do you mean this:
lim x->0 ((x^2+1)^(1/2)-1)/((x^2+16)^(1/2)-4)* ((x^2+16)^(1/2)+4)/ ((x^2+16)^(1/2)+4)?

At the end it will be 0 in the denominator anyway

[loud oasis]

yes, then repeat the same strategy for the numertor

[barren panther]

Thanks!

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[surreal silo]

i tried that second one so many times, i just dont seem to understand this

[surreal silo]

the home work is "disc method" so im trying to use disc method

[wind ember]

You use the shell method for that

[true coral]

or you would need to invert

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[surreal silo]

@true coral @wind ember gotcha, ty for the help. realized what i was doing wrong :#

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[keen lance]

I want to use the method shown in the first image to finish off the second image but I can't figure out what it happening (with the red arrow)

[keen lance]

I know that in the first problem, i have multiples of 3 as coefficients os I could use that, but how would i do that for the second problem here?

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[neon iron]

.

prove that (a,b) = (a, b +ax)

[neon iron]

Kindly help.

[cursive patrol]

d = (a, b)

[neon iron]

yes

[cursive patrol]

proceed

[neon iron]

?

how

you just labelled the thing

<@&286206848099549185>

[barren marten]

Can you describe the question in detail?

[neon iron]

yes

the question is complete just so you know

well to describe is another thing

So we have to prove that the HCF of the pair (a,b) = the HCF of the pair (a, b + ax)

Where x is an integer

[cursive patrol]

so, d leq (a, b + ax), yes?

[neon iron]

?

[neon iron]

what

[cursive patrol]

leq means less than or equal to

this problem, do by inclusion

Show d leq (a, b + ax). Show (a, b + ax) leq d

[fair thorn]

d < (a, b + ax)

[cursive patrol]

from this, you will be able to conclude d = (a, b + ax)

i have provided the easier direction. try the other

[neon iron]

what da hell

I am not aware of inclusion

So this way is not good for me

[cursive patrol]

ok whatever i call it “inclusion” but really it is very simple

it means, if x is less than or equal to y and y is less than or equal to x then it follows that x = y

[cursive patrol]

refer here

[neon iron]

oh

Can't we do it the following way?

Let K = ax + b
Assume a > b
Where a is the divisor of K and x is the quotient and 0 <= b < a.
Then (K, a) = (a,b)
(ax+b, a) = (a,b)

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[balmy spruce]

i am not sure how to start

[spare smelt]

@balmy spruce let's start with a simple example

[balmy spruce]

ok

[spare smelt]

Let E = {0,1,2}

how many subsets are there of E?

[balmy spruce]

6

[spare smelt]

how do you know?

[balmy spruce]

the number of ways to group these elements into sets is 3!

these sets if considered as elements form the power set of E

[spare smelt]

actually no

it's not 3!

but 2^3

[balmy spruce]

{0},{1},{2},{1,2},{0,1},{0,2}

ohhh wait

i forgot about E and the empty set

[spare smelt]

{}, {0}, {1}, {2}, {0,1}, {0,2}, {1,2}, {0,1,2}

[balmy spruce]

yes exactly

[balmy spruce]

but does this also work in general

[spare smelt]

yes

do you want to know why?

[balmy spruce]

yes

but wait a second

[spare smelt]

the power set of a set with n elements has 2^n elements

[balmy spruce]

if there are n elements you can arrange them in n! sets + the whole set and the empty set

[spare smelt]

n! + 2 is not 2^n

[balmy spruce]

ik

[balmy spruce]

thats why i am trying to find what i am missing here

this question just popped up during this conversation

[spare smelt]

the power set of a set with n elements has 2^n elements
this is because each element in the set corresponds to a lightbulb that is either on or off, and if the lightbulb is on, you include the element in the subset, and if it's off, then you exclude the element from the set

2^n accounts for all combinations of the lightbulbs being on or off

[balmy spruce]

ah wait its clear now since every element has 2 choices and there are n elements

this yields 2^n subsets

now that ik the number of elements of P(E) which is the power set of E what to do next

[spare smelt]

let's look at our equation

put this in english

[balmy spruce]

we need all the elements that are in A and X at the same time to be all elements of B and no other elements

[spare smelt]

good

let's look at P(E) again:
{}, {0}, {1}, {2}, {0,1}, {0,2}, {1,2}, {0,1,2}

give me an example where this equation would hold

[balmy spruce]

A={0,1},X={0,2},B={0}

we need B to be the only subset of both A and X other than the sets themselves and the empty set

i thought about this but i am not sure if this is a necessary and sufficient condition

so is it

[spare smelt]

yes, but let's break this down

you're saying a lot in just one sentence

first off, what do you notice about the size of B?

[balmy spruce]

=< the size of both X and A

in case of equality, A=X=B

[spare smelt]

yes

so here's what we have so far

$B \subseteq A \land B \subseteq X \land B \cap A = B \land B \cap X = B$

[thorny flame]

Mr. Gamer 🇵🇸

[balmy spruce]

isnt $A\land B$ the same as $A\cap B$ then why did you use different notations

[thorny flame]

pirateking0723

[spare smelt]

no

these are 4 different logical statements

joined together with the AND operator

B is a subset of A, AND
B is a subset of X, AND
B intersect A = B, AND
B intersect X = B

[balmy spruce]

oh wait i read it as $B\subseteq (A\land B)\subseteq ...$

[thorny flame]

pirateking0723

[balmy spruce]

thats why i thought about this

[balmy spruce]

i mean this

[spare smelt]

yep actually the first two conditions are sufficient my bad

the last two are redundant

$$(B \subseteq A) \land (B \subseteq X)$$

[thorny flame]

Mr. Gamer 🇵🇸

[balmy spruce]

np dw

ok so thats what we reached till now

these conditions are sufficient but how can we prove that they are necessary

i mean we need to reach logical equivalence not just an implication because thats what the question is asking about

[spare smelt]

yeah so now we just need to figure out the condition on X.

so find an equation that says X =

actually we need one more condition

[spare smelt]

asides from what is contained in B, the elements in X and A do not overlap

[balmy spruce]

yes i said that above

and you agreed on it

[balmy spruce]

you and me just forgot to add it here

but if we add it like this will be just restating the equation right ?

$(B \subseteq A) \land (B \subseteq X) \land((A\cap X)\subseteq B)$

[thorny flame]

pirateking0723

[spare smelt]

(A-B) intersect (X-B) = {}

[balmy spruce]

ohh yes that works

i have the solution of the professor but i didnt see how it relates to the question at all

so i decided to solve it on my own with some help here

after we are done can you check the professor's solution

[spare smelt]

ok now we just need to 'solve for X'

[balmy spruce]

yes

[spare smelt]

$(B \subseteq A) \land (B \subseteq X) \land ((A-B) \cap (X-B) = \emptyset)$

[thorny flame]

Mr. Gamer 🇵🇸

[spare smelt]

now we have our necessary and sufficient condition

[balmy spruce]

as an answer to the question is it valid to just drop these

or should i do some explanations

i mean the questions says give a ...

[spare smelt]

that's up to your professor

[balmy spruce]

so this is probably enough but i will ask the professor about it later

[balmy spruce]

ok so now we have to do this

let me think about it for a sec

AnB is a solution

[balmy spruce]

$A\cap B\subset E$ since $A,B\subset E$ which satisfies the condition that $X\subset E$ and $A\cap (A\cap B)=A\cap A\cap B=A\cap B=B$ since $B\subseteq A$ from here

[thorny flame]

pirateking0723

[balmy spruce]

but i am not sure if this is the only solution

any subset K of E which contains B will lead to X=KnB being a solution am i right

because in this case KnB=B this leaves AnB which is B

[balmy spruce]

but then comes this again

[spare smelt]

huh i'm kinda lost here ngl

[balmy spruce]

if you want forget about what i said for now

how where you going to solve it