help-27

[idle dune]

I'm thinking for r

it looks like

2x = x^12

not sure how to solve for x though

wait nvm

.solved

[devout snow]

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[devout snow]

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[idle dune]

hi

I don't understand the question that well. isn't z just -2?

[scarlet sequoia]

over reals it would be, but here you're supposed to find all solutions (i.e. solve it over complex numbers)

[idle dune]

ok hm

i don't get it mate

[scarlet sequoia]

in complex numbers, an equation has as many solutions as its degree

[idle dune]

is it the magnitude that's supposed to become -128?

[scarlet sequoia]

in this case 7

[idle dune]

ok go on

makes sense so far

is this roots of unity question or what

[scarlet sequoia]

Are you familiar with this

[idle dune]

nope

wait actually

is there a version with exponentials then may i see it

i see i get it now thank you

[scarlet sequoia]

it follows from this, yeah

bring it to polar form and then find the roots according to this formula

[idle dune]

so it's (128)^(1/7) e^j((pi + 2pik)/7)

that's crazy

thank you bro

.solved

[devout snow]

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[idle dune]

.reopen

[devout snow]

✅ Original question: #help-27 message

[idle dune]

hey @scarlet sequoia thanks for helping out on the last question. Think you can help me on this one?

[scarlet sequoia]

Hint: geometric series

[idle dune]

curious, is this possible without a calculator or not

also, when they say sum = 0, how is that possible if it's a complex number

oh wait

[wind mason]

imag and real parts can cancel out

cause negatives and positives

[idle dune]

im guessing it can be something like e^(j(2pi)) + e^(j(pi)) could be a way

very simple way tho

[idle dune]

thanks brodie

[wind mason]

you can get an exact form w/o, yeah

this funnily reduces to a really simple equation

[idle dune]

what is that so?

im a bit confused

[wind mason]

show me your working so far

unless I'm tripping it does

[idle dune]

all i did was take out e^j(.2026pi)

[wind mason]

ok

did you sum the series

[idle dune]

oh

it's just 40 then

wow you need to remember calc 2 that's crazy

[idle dune]

my friend showed me how to do it

[wind mason]

wha

this is alg

[idle dune]

he told me something about geometric series though

[wind mason]

$\sum_{k=0}^Nr^k=\frac{1-r^{N+1}}{1-r}$

[woven radish]

;(

[wind mason]

this is finite, ones covered in calc are infinite

wait do you see an image anywhere

is texit down

[idle dune]

i see it

my friend showedd me that too

he said something about convergence

[wind mason]

that's for infinite as well

[idle dune]

then he did stuff with the exponent

[wind mason]

i can show you a quick proof

[idle dune]

what proof

[wind mason]

of this formula

[idle dune]

the geometric series convergence?

why not

[wind mason]

ok

sorry I'mma go take a nap soon but i'll just show you before that

$S=1+r+\ldots +r^N$\
$rS=r+r^2+\ldots +r^{N+1}$\
$S-rS=1+r+\ldots +r^N-(r+r^2+\ldots +r^{N+1})=1-r^{N+1}$\
$S(1-r)=1-r^{N+1}$\
$S=\frac{1-r^{N+1}}{1-r}$

[woven radish]

;(

[idle dune]

how is s-rs = to that

nvm i see it

wow

that's actually sick ngl

thank u bro

[sharp obsidian]

does anyone know this sorry

[idle dune]

if I had to give an educatedguess

it would be

[sharp obsidian]

thank you

[idle dune]

wait tell me if im right im curious @sharp obsidian

[sharp obsidian]

it dosent say at all it just gives an average

i thought it was the right and middle collum

[idle dune]

from what i see

if you were to put the first 3 squares' red dots in one single square, then that single square will be filled with red dots

wait ngl

this is a bit hard it could be all middle horizontal red dots filled out

no idea tho

[sharp obsidian]

i see where your coming from icl

[idle dune]

naw i don't know now

oh I KNOW

it could be this

[sharp obsidian]

why do you think that

[idle dune]

going back

if you were to put the first 3 squares' red dots in one single square, then that single square will be filled with red dots

wait

actually

it has to be this

wait no

naw man i give up bro

[sharp obsidian]

its hsrd isnt it

[idle dune]

i'll ask everyone on my dorm floor later tho it's a sick question

[sharp obsidian]

thank you bro

lmk im interested

[idle dune]

ok for sure

im intrested too lol

[lunar harbor]

!occupied

[devout snow]

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

[sharp obsidian]

bro what

[supple knot]

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

[idle dune]

thank you guys for the help

[idle dune]

sick proof bro

cya

.solved

[devout snow]

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[sharp obsidian]

oh i get it my bad

[devout snow]

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[thorny rapids]

I understand the formula 911 + 2[911(0.80)+911(0.80)^2....]
And understand this series converges but
the answer is 8199 and I dont understand how to get that?

I did 911 +2 (911/ 1- (0.80))

<@&286206848099549185>

[plush quarry]

Note that the geometric series we are doing is 911(0.80)+911(0.80)^2....

The first term of that series is not exactly 911

[thorny rapids]

So what would it be?

[plush quarry]

The first term would actually be 911*0.80 (since we are concerned with 911(0.80)+911(0.80)^2....)

[thorny rapids]

Indeed

OH yeah

I thought the "First Term" had to be a single isolated phrase" i understand thx

.close

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[charred eagle]

can someone solve this

[supple knot]

you should do it yourself

,tex .exp rules

[woven radish]

riemann

[supple knot]

familiarize yourself with those

[charred eagle]

i did

nto sure if its right tho

[supple knot]

!show

[devout snow]

Show your work, and if possible, explain where you are stuck.

[charred eagle]

ighnore the top left corner

[supple knot]

,w (6cis(2pi/5))^3 / (1/2 * cis(-pi/4))^(-5)

,w 6912cis(9pi/20)

[woven radish]

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[supple knot]

you did pi/4 * 5 = 5pi/2

oh you said ignore that

oh you messed up 1 / [1/2 ^ (-5)]

the magnitude should be 6^3 / 2^5

[charred eagle]

magnitude?

[charred eagle]

oh i see

thx

oww

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[charred eagle]

is this a identity

[charred eagle]

sin theta = cos theta - pi/2

cos theta = -sin theta - pi/2

[wicked turtle]

yes

you can use the cos(a+b) and sin(a+b) identities on the right hand side to confirm

(they're the only trig identities i bother memorizing)

[charred eagle]

on the right hand side?

if possible could u show pls

[wicked turtle]

well i'll do one of them

$$\cos(a-b) = cos(a)\cos(b) + \sin(a)\sin(b)$$

[woven radish]

Bungo

[wicked turtle]

so

$$\cos(\theta - \pi/2) = \cos(\theta) \cos(\pi/2) + \sin(\theta)\sin(\pi/2) = \cos(\theta) \cdot 0 + \sin(\theta) \cdot 1 = \sin(\theta)$$

[woven radish]

Bungo

[charred eagle]

ohhh

ok tysm

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[true lintel]

laplace transform of $ln(1 + x)/x$

[woven radish]

Einstens Skibidy Rizller

[devout snow]

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

[true lintel]

1

that is why im asking

[young spade]

I think you can assume that theres a heaviside there.

Assuming you use the one-sided version

[sand quarry]

If it is the one-sided laplace transform you dont need a heaviside

[young spade]

I think in most cases the time-shift identity is taught while having heaviside present, or thats what my limited experience tells me

You also will prob need the $\mathcal L {\frac1tf(t)}$ identity

[sand quarry]

,,\lm \LL \left{ \4{g(x)}x \right}(s) = \int_s^\infty G(u) \dd u

[young spade]

That

[sand quarry]

Although the resulting integral contains the Exponential Integral

You won't be able to get a closed form solution

[supple knot]

,w Laplace transform log(1+x)/x

[woven radish]

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[pseudo basin]

!xy

[devout snow]

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

[true lintel]

integral is $\int_{0}^{\infty}\sin\left(x\right)\frac{\ln\left(1+x\right)}{x}dx$

[woven radish]

Einstens Skibidy Rizller

[true lintel]

what does it look like

[mystic scarab]

,w integral of sinx ln(1+x)/x

[woven radish]

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[mystic scarab]

Seems there isn't even a closed form

[true lintel]

equivalent form \begin{align*} I&=\int_0^{\infty}\frac{e^{-x}\tan^{-1}(x)}{x} dx \
I &= \int_1^{\infty} \frac{\text{Ci}(x)\sin(x)+\left(\frac{\pi}{2}-\text{Si}(x)\right)\cos(x)}{x}\ dx \
&=-\left(\frac{\pi}{2}-\text{Si}(1)\right)\text{Ci}(1)+2\int_1^{\infty}\frac{\text{Ci}(x)\sin(x)}{x}\ dx \end{align*}

[supple knot]

Pretty close
https://math.stackexchange.com/questions/2733923/evaluate-int-0-infty-frac-ln-x-sin-xxdx

[woven radish]

Einstens Skibidy Rizller

[true lintel]

that one is ilt of ln(x)/x

saw it before

[supple knot]

Ok buddy

[true lintel]

who

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[mellow sequoia]

How can you check if this graph is planar

[sand quarry]

Yk the colouring method?

[mellow sequoia]

Yea but it only works when edges dont cross right?

i also know this

$v-e+f=2$

[woven radish]

arachi

[sand quarry]

Oh yes

[mellow sequoia]

And if there exists a subgraph that is isomorphic to K5 or K3,3 then it is not planar

[sand quarry]

Yeah you can use Kuratowski's theorem. I think that tangle is probably a K_3,3 subdivision

[mellow sequoia]

so I use edge operations?

[sand quarry]

I think try to trace a Hamiltonian cycle and consider the remaining untouched chords

[sand quarry]

What are edge operations to be clear

[mellow sequoia]

When you add 1 vertex and 2 edges or leave out 1 vertex and 2 edges

If you get something that is isomorphic to K_5 or K_3,3 then the graph is not planar

after you do edge operations

[sand quarry]

Yeah

Thats basically what i meant earlier

Your thing is isomorphic to an M4 Möbius ladder if im not mistaken

[mellow sequoia]

what is a mobius ladder

[sand quarry]

https://en.wikipedia.org/wiki/Möbius_ladder

[hazy leaf]

^

This is $M_8$

[woven radish]

ijo Sani

[sand quarry]

That specific tangle is M_4 i think

[hazy leaf]

-# the example in the wikipedia sidebar annotates the image above as M_16, so i gather that the naming convention is total number of vertices

[mellow sequoia]

I found a subgraph isomorphic to K_3,3

dont mind the arrows

This is $K_{3,3}$

[woven radish]

arachi

[mellow sequoia]

because it's bipartite

[hazy leaf]

good! is your question answered / how do you answer your question?

[mellow sequoia]

I just wanna know what I have to do in general

if I get any graph how should I check if it's planar

[hazy leaf]

this is usually the way to go to show it is nonplanar

to show a planar just find an embedding

[mellow sequoia]

Alr

New question

so here they want me to check eulers theorem

this

$v-e+f=2$

[woven radish]

arachi

[mellow sequoia]

I counted 17 vertices, 32 edges and 21 plains

but that doesnt add up to 2

nvm I counted wrong lmao

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[glossy dew]

i understood the telescopic sum for this qn but the answer i calculated was a bit far off

[glossy dew]

this

but

might it be something to do with the range of arctan

[vital edge]

Those two values look to be about π/2 apart

[glossy dew]

indeed

got it thanks

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[glossy dew]

there were in fact two individual telescopic series to be accounted for

man these are weird

[grand edge]

If you expand it out manually the last two terms will not telescope out

so at the end of it all, there are two copies of arctan(∞)

$\lim_{N\to\infty}\left(-\arctan\left(\frac{1}{2}\right)+\left(\text{telescoping stuff...}\right)+\arctan\left(\left(\frac{N+1}{\sqrt{2}}\right)^{2}\right)-\arctan\left(\left(\frac{N}{\sqrt{2}}\right)^{2}\right)\right)$

[woven radish]

Roy

[grand edge]

etc

[devout snow]

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[charred eagle]

how do i even

factorise this

[grand edge]

Factor z² out of the first two terms

[grand edge]

Can you see a common factor here that may be factored out?

[charred eagle]

oh yeah i do

wow

[charred eagle]

how do u even know to start this

is it just intuition

[grand edge]

Because this idea occurs a lot in factorization and my pattern recognition spotted it

[charred eagle]

ohh

okay thx

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[charred eagle]

hi, does anyone know where i went wrong

[sand pumice]

4*9=13? 🤔

[charred eagle]

oh my gohs

thanks

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[charred eagle]

can someone take me through b)? I'm not quite sure how to do it

[grand edge]

There should be worked examples in the textbook which demonstrate how to do these kinds of questions. Either way, have you started by writing RHS in polar form?

[charred eagle]

yeah, it does go through it but im like lowkey lost

[charred eagle]

Yeah

[grand edge]

That looks good, do you know how to continue?

[charred eagle]

no i dont unfortunately

[grand edge]

Equate the moduli

[charred eagle]

sorry wdym?

oh

oh i see

uhh ok

r = 3

[grand edge]

yup

so now you have three solutions, one for each angle

[charred eagle]

wait what

[grand edge]

should be no problem to write down

[charred eagle]

3cis(pi/6)
3cis(5pi/6)
3cis(-3pi/6)?

[grand edge]

yeah

and you can simplify the last one a bit

[charred eagle]

oh wait yea

[grand edge]

it doesn't really request your answer be in polar or cartesian so leaving it in polar should be fine

[charred eagle]

wait thats the answer?

[charred eagle]

z = this

[grand edge]

yep

[charred eagle]

ohh

[grand edge]

you just need to plot them on the Argand diagram now

[charred eagle]

do i have to

link those

to the center

or draw a circle??

[charred eagle]

like this is mine vs the answrers

[grand edge]

it's fine I guess

in the exam you'll have it predrawn out for you

[charred eagle]

ohh ok

tyty

[charred eagle]

the circle?

[grand edge]

yes

[charred eagle]

ohh

wait doesnt that mean we just know where it is

[grand edge]

it'll look like this

sometimes you'll need to plot a ray or a circle as well

[devout snow]

@charred eagle Has your question been resolved?

[devout snow]

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[ionic harness]

Does this look good?

[ionic harness]

n is taken from 1 to infty (the textbook omits in bracket notation)

[sand pumice]

looks right

[devout grotto]

looks fine but remember you are given epsilon in these problems

the convergence bit is that GIVEN any epsilon > 0 you can find an n

also nitpicky but $N=\frac{1}{\epsilon} + 1$ does not always mean $N \in \mathbb{Z}$

[woven radish]

Mirror

[ionic harness]

Yeah but who said N needs to be in Z

Oh it says index N

[devout grotto]

index N

no your argument is exactly right

[ionic harness]

that's bs 😒 I guess I should have done ceil(1/eps) then

[devout grotto]

that works too!

again your argument is SPOT ON you are just going to get dinged on a technicality because you are given an arbitrary e > 0

[ionic harness]

Originally I was gonna say take eps in R+ but then I was like I guess I can just skip to N = (expression in terms of epsilon), I'll edit to include it 👍

[devout grotto]

ye i mean "given epsilon in R+, set N=ceil(1/epsilon)"

[ionic harness]

Right

[devout grotto]

is basically the same thing but a prof looking for something to mark will get you every time

[ionic harness]

almost done rewriting

[void fox]

usually you would just say pick N > 1/ε

[ionic harness]

Oh you can just let the reader choose their favorite N?

[void fox]

also nice handwriting grandson

[devout grotto]

^^ indeed nice handwriting

ye you can let the reader choose their favourite N because all you have to do is show that at least one N exists

[void fox]

you can, the only thing required in the argument is that N > 1/ε anyway

[devout grotto]

(nb: N > 1/e has infinitely many solutions for all nonzero e)

[ionic harness]

Well that makes my life easier than this floor(1/e + 1) nonsense

This I can do

So we have nonconstructive existence of such N for every eps

Does it look good :)

[devout grotto]

perfect

also again your handwriting is so nice

[ionic harness]

oop didn't get a notification

[ionic harness]

Thank you :))

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[fair wigeon]

hi

[wraith anchor]

hello

[fair wigeon]

I need help on how to rotate an object

[devout snow]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[wraith anchor]

there is a matrix

used to rotate objects

is that what you are looking for?

[fair wigeon]

no not really

[pseudo basin]

!xy

[devout snow]

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

[wraith anchor]

yes we need more context

[fair wigeon]

we were given this question;
find the coordinates of the vertices of the image a parallelogram who's vertices are A(3,5), B (7,5), C (5,0), D (1,0) when rotated about the origin through -90°
with no guide and I have already plotted the points but I do not understand how to rotate the image

[pseudo basin]

try rotating just one vertex at a time

think about how to do that with individual pts

the amount of rotation is -90° so it's going to result in nice and clean values

(it means 90° clockwise btw)

[fair wigeon]

thats the thing, our teacher didnt teach us anything about rotation, so basically I dont know how to rotate

how am I supposed to rotate the vertex?

[pseudo basin]

https://www.youtube.com/watch?v=GqHWdTLL8Qw

[fair wigeon]

tysmm

[devout snow]

@fair wigeon Has your question been resolved?

[devout snow]

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[visual stone]

f(x) = 3(2)^(x+2) + 2 and g(x) = -2x + 3

[visual stone]

Determine

composite

[wicked turtle]

$f \circ g(x)$

[woven radish]

Bungo

[wicked turtle]

thoughts about how to start?

[willow helm]

this looks like $f^{\circ}$

[woven radish]

USS-Enterprise

[willow helm]

This is correct

[visual stone]

i forgot how to do this one

[visual stone]

yeah

[wicked turtle]

recall what composition means

[visual stone]

you had to like

[wicked turtle]

$f \circ g(x) = f(g(x))$

[woven radish]

Bungo

[visual stone]

substitute g into the function f

[visual stone]

yeah

[wicked turtle]

so what do you get if you do that?

[visual stone]

no idea

i dont think this is for exponential right

[stone stump]

substitute the expression of g into all the x's which appear in the definition of f

(aka just one x in this case)

just as if you would substitute 7's into those x's if you compute f(7)

[willow helm]

Doesn't matter

[willow helm]

This is the definition of fog(x)

When it itself is defined is a different story though

[visual stone]

nah i just realised cause i have to learn exponential and log

that wouldnt be in an exam for me

[mystic scarab]

And?

[willow helm]

Yes, could be of use

[visual stone]

oh ok

in what situation

[willow helm]

When you have exponents and logarithms

Or when their use is required

[willow helm]

I mean I don't know what your exam is going to cover

If you haven't learned logs and exponents then no you don't need to know this obviously

[visual stone]

what do u mean by if i have log and exponents

like in a problem and theres an equation of a log and one equation of an exponent?

[mystic scarab]

Huh? Wdym

[visual stone]

f(x) is exponential and g(x) is a log right

[willow helm]

Well in this case f(x) is an exponential function

[mystic scarab]

Such as f(x) = e^x, and g(x) = ln(x)

You should know what g(f(x)) is

[mystic scarab]

Well..... Of course!! 🙈

[visual stone]

yeah

[willow helm]

In this case you need to know exponential functions

[visual stone]

gotcha so like what will doing the composites do

[willow helm]

To be able to work with this

[visual stone]

find x

i forgot what composites do

[mystic scarab]

Haven't you been taught those?

[visual stone]

like 3 months ago

[mystic scarab]

Then revise your notes...

[willow helm]

In short, they combine two processes into one

[visual stone]

like you have f(x) and g(x) and u substitute the equation into one right

[willow helm]

In this case, instead of first computing g(x) for an x, and then f(that output) you can just compute f(g(x)) for that x

[mystic scarab]

https://cdn.discordapp.com/attachments/903486480075354182/1465420131512356954/761155030513352705.png?ex=69790a59&is=6977b8d9&hm=ab6a4d1ca975cb76c3c4ad36555518599a6cc0d31bb916dd15c10ca26e829a32& @visual stone

[visual stone]

ohh

[willow helm]

^ this is all there is to it

[visual stone]

i figured

.close

[devout snow]

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[devout snow]

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[robust bobcat]

for 7 part b I write f'(x) = (x-alpha)^r g(x). Are you allowed to suppose that g(x) = 1 and then you have that f(x) has r+1 zeros. Does assuming that g(x) = 1 make it less general?

[thin fern]

yes that's a big assumption

do you know if the zeros here are allowed to be non-real

[robust bobcat]

how could you then integrate some general function. You can't know if the integral exists and he hasn't shown how to integrate a product of functions

[robust bobcat]

idk

oh wait

[thin fern]

My idea was the fundamental theorem of algebra idk if you've heard of it

a polynomial of degree d has d complex roots

[robust bobcat]

I am going to assume real since that is how he was talking about the mulltiplicity of zeroes

[thin fern]

ok the fundamental theorem of algebra would not work then

hmm

[robust bobcat]

also for this section he talked about rolle, mean value and cauchy mean value but I dont' see how any of those apply

[wraith anchor]

was part (a) also assigned?

[robust bobcat]

ye

[wraith anchor]

i dont have anything concrete as im figuring this problem out alongside you, but i would suggest examining polynomials multiplicities and how taking the derivative affects them

cause that's what im doing

[robust bobcat]

so it could also be true if we had j(x) = r+1g(x) + (x-alpha)g'(x)

but again that limits what our function can be

[wraith anchor]

note that if you look at (x+3)^3 the root shows up in the first and second derivative

this isn't exactly profound

but just something to notice

f^(k)(x) is of the form (x-a_1)(x-a_2)(x-a_3)...(x-a_r)

where a_i and a_k are not necessarily distinct

[robust bobcat]

what is f^(k)(x)? do you mean f(x)^k?

[wraith anchor]

the kth derivative of f

[robust bobcat]

ok

[wraith anchor]

part b is going the opposite direction of part a right?

so if lose a zero every time you take a derivative what happens when you take the integral?

[devout snow]

@robust bobcat Has your question been resolved?

[robust bobcat]

add one zero but that doesn't seem rigorous enough

[devout snow]

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[somber jetty]

is this unimodal, bimodal, or trimodal

[somber jetty]

genuinely curious to the answer. prof is saying bimodal, others online are saying unimodal

conversation had:

me:

also the sampling distribution of the median is not bimodal -- technically it could be described as unimodal with two noticeable bar heights. one had a very noticeable difference in height than the other, and im pretty sure it was like 2:1 ratio in frequency. but i think in retrospect, it would have been better to just have called it a discrete distribution concentrated at two values.

op of the campuswire:

TA wrote bimodal on the key for takehome but I also thought that it was weird

me:

yeah just saw it haha, they didn't give a reasoning though -- @prof? curious about this now because then it comes to what defines a peak mathematically. like is it just "it showed up visually on the graph and also there's like nothing next to the values" or is there a more mathematical way to define bimodal

the OpenIntro Statistics textbook defines bimodal as "any distribution with two prominent peaks" but what if someone interpreted the graph like this

with this attached

prof:

The point here is that these are not histograms of a sample drawn from a population. IF you treat the graph shown above as the population distribution, then it is bimodal with precisely two peaks. Or you could define it as discrete with precisely two values. The sketches superimposed imply the possibility that there are values that we didn't see but could happen. But if this is not the case, then those sketches aren't correct.

so then i wrote:

then by this logic, is my sampling distribution trimodal? it had values at 540 after all

are we defining peaks as bars we can see on the graph or just values which don't really have anything next to them

the qqplot again for reference

he then writes:

yes, it would be trimodal. But you are right....if you push this logic far, than a discrete uniform distribution on the integers 1:1000 would be millenial-modal or something like that.

and that doesn't seem right.

[uncut crow]

[somber jetty]

hello queen

is my prof ragebaiting me

im not sure -- others are saying that it's unimodal online and others say bimodal

[uncut crow]

hm lol

[somber jetty]

wahhhh

its okay

i actually dont gaf i cant really see a reason why id have to use a median sampling distribution in my field

in this specific context

just a little annoyed to get docked off points when other people in industry would call this bimodal/unimodal

with a pretty even split

[uncut crow]

i am trying to find precise definitions of multimodal distribution on the internet and i can't really find any

so yes let's go with your prof is ragebaiting you

[somber jetty]

i too was trying to find one queen

one such rigorous definition is literally if two values have the highest frequency its bimodal

but by the professor’s logic i can splice a normal distribution in a spot that isn’t at the peak and suddenly i have two peaks

guh

[uncut crow]

[somber jetty]

maybe i should have been a mathematics major

👹👹👹

[uncut crow]

amen

let's ask @vagrant skiff he knows everything

[devout snow]

@somber jetty Has your question been resolved?

[somber jetty]

i love you

you are an amazing bot

[nimble solstice]

so delighted!

[devout snow]

@somber jetty Has your question been resolved?

[somber jetty]

love no it hasnt

you're so beautiful though

i love you bot

what a beautiful mind

you have

[devout snow]

@somber jetty Has your question been resolved?

[vagrant skiff]

from what I know, one usually defines define modes as distinct concentrations of probability mass... that is, peaks that are distinct from their surroundings.. since the bar at 660 is separated from the bar at 600 by a valley of zeros, it counts as a second mode... I believe that in practice people reserve trimodal for cases with three comparably visible peaks rather than one major pair plus a trace

[uncut crow]

thank you father

[vagrant skiff]

amen layla

[uncut crow]

[vagrant skiff]

[somber jetty]

amen twin

[uncut crow]

amen mommy

[vagrant skiff]

what the blud

[somber jetty]

.solved

[devout snow]

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[devout snow]

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[green crypt]

in this question the overall velocity is constant, which is vo

why in this question is the overall velocity not just 10 ms^-1, rather its just the x-component of the velocity of peg a

[glossy dew]

i think the slotted link moves at 10 m/s (the huge brown coat hanger shaped one)

and the pegs are connected to the slotted link to be in the same vertical line

so the horizontal component of the pegs are forced to be the same as the slotted link to move along with it

[green crypt]

hmm ig ki kind of get it

.close

[devout snow]

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[brisk cosmos]

anyone can tell me that is this relativity

[brisk cosmos]

I dont understand that weird e

[proud perch]

That's an ε

Epsilon

Transmittivity of vacuum, I believe

[devout snow]

@brisk cosmos Has your question been resolved?

[brisk cosmos]

ok

[devout snow]

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[rapid merlin]

What have you tried

[compact spruce]

hm i had a question in this, do i need to simplify it after multiplying 2x with the other terms

do you want a picture

[rapid merlin]

There is nothing to simplify after multiplying 2x I think

[compact spruce]

but why is it

the answer i got =
14x y^3 + 6x^2 -22 x^2 y^2

[pseudo basin]

there will be a bit of simplification to do on each term, but it looks like you already did it

[compact spruce]

hey again,
but my questions still goes since yesterday that why cant it simplify it

[pseudo basin]

??

[wanton sluice]

Because there is nothing to simplify

[compact spruce]

hmm

ok

[wanton sluice]

All of them are simplified just multiply the two that's all

[pseudo basin]

you've expanded it

so your expression is in expended form already and there's nothing else to expand

[compact spruce]

ohh, so the question only asks my to expand but not anything else if i am not wrong
but if it was not the question then could i have added or subtracted them?

[pseudo basin]

no, bc they are unlike terms anyway

[compact spruce]

ok, but if we had other operations like multiplication then could we have done it?

[wanton sluice]

Yes for multiplication you usually operate them

[compact spruce]

ok, not addition or subtraction, right?

[wanton sluice]

For addition it's only for like terms

[compact spruce]

ok

thanks

.close

[devout snow]

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[wanton sluice]

Np

[devout snow]

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[viral kernel]

$\frac{\ln(1+x)-x+\frac{x^2}{2}}{x^3}$

[woven radish]

Goofy Joe

[viral kernel]

lim x->0

I've redone the calculations I don't know how many times, random numbers come up

[round raptor]

what have you tried

[devout snow]

Show your work, and if possible, explain where you are stuck.

[viral kernel]

I tried Taylor/Hopital but then I get lost in the math

I threw away the papers

[round raptor]

what did you get in taylor ?

[pseudo basin]

$\lim_{x\to 0} \frac{\ln(1+x)-x+\frac{x^2}{2}}{x^3}$

[woven radish]

Ann

[viral kernel]

?

the result?

[round raptor]

yh after simplifying

[viral kernel]

I redid the calculation over and over again. Everything superfluous was erased, the useless terms disappeared... in the end, only one thing remained on the sheet.

but it was wrong

like 7/25 or something

[brittle coral]

...... what did you get with l'hopital

it should be quite straightforward with l'hopital

[viral kernel]

i dont remember

[brittle coral]

try it again

[viral kernel]

ok

[round raptor]

hm..
$\frac{\ln(1+x) - x + \frac{x^2}{2}}{x^3}

\frac{(x - \frac{x^2}{2} + \frac{x^3}{3}) - x + \frac{x^2}{2}}{x^3}$

[meager phoenix]

I have a better method

[woven radish]

scoob

[meager phoenix]

Yeah this explains it

[round raptor]

this simplifies to a good value

[viral kernel]

i got 0

[meager phoenix]

How?

Bruh

[round raptor]

if i see lhopital more than once skip

[brittle coral]

its just one l'hopital step

[meager phoenix]

X and x^2/2 gets cancelled out leaving x^3/3 remaining

[viral kernel]

i get like 1-1/6x=0

[brittle coral]

then the top and bottom cancel out nicely

[round raptor]

ic

[viral kernel]

I get as far as the derivatives, but then I get some strange stuff and it doesn't match anything... so I don't find it.

[twilit field]

it helps to show those strange stuff, I believe.

[brittle coral]

we have

{ln(1+x) - x + x^2/2 } / {x^3}

can you write/type out the derivative of the numerator, and the derivative of the denominator?

[round raptor]

yh lhopital is better here

[round raptor]

he threw his work 🥀

[viral kernel]

(1/(1+x)-1+x/3x^2)

[brittle coral]

now simplify

we dont like having fractions in the numerator itself

if you multiply numerator and denomiator by the same thing, the value doesnt change

[viral kernel]

1/3(x+1)

[brittle coral]

there you go

when x reaches 0

[brittle coral]

what does this become?

[viral kernel]

1/3

[round raptor]

https://tenor.com/view/qawe-asd-gif-26050335

[brittle coral]

see? not that complicated

[viral kernel]

why am I at the first derivation?

since there is x^3 shouldn't I do 3 times?

[brittle coral]

when you do 1 step of l'hopital,

you differentiate the numerator 1 time

you differentiate the denominator 1 time

[viral kernel]

i know

[brittle coral]

so you differentiate the denominator, x^3, only 1 time

[viral kernel]

yes

[twilit field]

are you asking why you only needed one round of l'hop in this question?

[round raptor]

or why does lhopital work?

[viral kernel]

yes

[brittle coral]

call it a coincidence that the question was nicely set

[viral kernel]

but here we have to consider Δ x?

[brittle coral]

you mean the

lim x --> 0

thing?

[viral kernel]

Δ x=x since x->0

[brittle coral]

what is ∆ x?

there is no ∆x in the original qn

[round raptor]

infinitesmal x i supose

[viral kernel]

yes

since x->0

[brittle coral]

for l'hopital we dont need to consider infinitesimal

if you're curious on how it works

or why it works

[viral kernel]

so how do you see the probability that the limit was <1?

without calculation

[brittle coral]

the big idea is that the infinitesimal ∆x on numerator and denominator when differentiating cancel out geometrically

Source: YouTube https://share.google/iZzv3EkKwMdI1Ts1t

[viral kernel]

ok

[brittle coral]

uhhhh not sure what you mean

[viral kernel]

how i know already that the result is <1

[brittle coral]

thats the neat part, you dont

you could compute using calculator/otherwise with tiny values

but at first glance of the entire qn, if nothing is telling, you have to attack it first

and after some manipulation/calculation then you can get an idea

[viral kernel]

but aren't there some kind of demonstration where it is shown that the limit is a certain value?

[brittle coral]

by demonstration do you mean like a graph? or testing values that get closer and closer to 0?

[viral kernel]

with the definition of limit

[brittle coral]

you want to use epsilon delta definition to show that the limit is 1/3?

[viral kernel]

that is <1

[brittle coral]

oh, you want to deduce that the limit is <1 without actually solving the limit?

[viral kernel]

no

[twilit field]

I believe that by the nature of epsilon-delta proofs, you are going to find the limit directly rather than deduce a plausible range for the limit.

if that is what the OP meant.

[viral kernel]

ok

[brittle coral]

using epsilon delta to show that the limit is <1.... ?

[viral kernel]

yes

[brittle coral]

oh god

[viral kernel]

?

[twilit field]

I'm not sure if that kind of question is what an epsilon-delta proof is meant to solve, no?

[brittle coral]

^

[viral kernel]

ok

[brittle coral]

i meeeean you could try

but it wont be a very productive use of epsilon-delta

[viral kernel]

no

[brittle coral]

and its going to be much more tedious than it should be

[viral kernel]

it was just a question

i was interested in resolution

thanks all

.close

[devout snow]

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[devout snow]

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[waxen forum]

Hi!

im having trouble rigorously finding the value of K in
$\int_{a}^{b} \delta(x)dx = K$ for $ b > a $ ,$b < 0$ or $ a > 0$

[woven radish]

Saggitarius

[sand quarry]

rigorously?

what can you use

[waxen forum]

anything i think

its just for an assignment

i believe that it will equal 0 but i have no clue how to show it

[sand quarry]

,, \lm\int_a^bf(x)\delta(x)\dd x =\e{dcases*}{f(0), &if $0 \in (a,b)$ \ 0, &if $0\notin [a,b]$}

[sand dove]

for a < b

[sand quarry]

this is all you need to use, pretty much

[waxen forum]

thinking...

yeah thats true but why i gues

[sand dove]

Don't you have a definition

[sand quarry]

by definition of the dirac delta distribution

[sand dove]

for dirac distribution

[waxen forum]

oh nice then we are good

thanks!

.close

[devout snow]

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[azure parrot]

Find the last digit of $\floor{(2+\sqrt{3})^{20}}$

[woven radish]

lower! <- dum dum

[azure parrot]

This looks like some binominal type shiii, but the expansion looks nasty

[twilit field]

I think I have a trick involving the conjugate.

[azure parrot]

[twilit field]

try summing the expression with its conjugate.

[sand quarry]

you need to establish a recurrence relation

but do what nicole said first

[twilit field]

the recurrence will probably be established through what I said, so consider both Lex's point and my own.

[azure parrot]

Nvm I figured it out

[steel sage]

Use the fact that $a_{n}=(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}$ is always an integer, and you will obtain a recursion of $a_{n}$ thus a recursion of $a_{n}$ mod 10

[woven radish]

Cogwheels of the mind

[azure parrot]

Yeah I see that

One more question b4 I sleep

This question is from CSP bruhhh I haven't figured it out

$x^2-7x+1=0$ Let $\alpha$ be a solution of the equation, find the last 2 digits of $\alpha^{2026}+\frac{1}{\alpha^{2026}}$

[woven radish]

lower! <- dum dum

[azure parrot]

Okay so, I establish a recurrence relation

$a_n=7a_{n-1}-a_{n-2}$ with $a_1=7$ and $a_2=47$

[woven radish]

lower! <- dum dum

[azure parrot]

I managed to figured out the last digit, but the second last

I couldn't find the pattern for the second last digit

[steel sage]

You need a_n mod 100

[azure parrot]

It doesn't seem to follow any pattern to me idk

[twilit field]

given that you are working mod 100, if there is a pattern it would have a maximum length of 100 steps.

[azure parrot]

a1=7
a2=47
a3=322
a4=2207

[azure parrot]

Yeah if it is 100 steps there's no way

I can find that

[twilit field]

working mod 100 means to keep only the last two digits.

[azure parrot]

Ik

[twilit field]

a_3 should therefore be 22.

and a_4 would be 7.

[azure parrot]

Yeah ik

[steel sage]

I found period being 15

[twilit field]

you could probably code something quick to help with finding out a pattern... but damn if I'm skilled enough to do so.

[steel sage]

So done

Any calculator in a phone can do.

[azure parrot]

Not what I have bruhh

It only show 10 first numbers , from the left

[steel sage]

You take mod 100

a_16=a_1=7, a_17=a_2=47 thus a_n period 15 (a_n mod 100 I meant)

[azure parrot]

No way , there has to be some other ways to do the problem

[glossy dew]

yeah getting this remainder stream

i mean its not that arduous so i guess its something

[twilit field]

if it's period 15, it's not terribly bad, esp. working mod 100.

[azure parrot]

No I meant CSP never give a question that require a calculator

[steel sage]

This is doable by hand. Just unnecessary

[azure parrot]

And also a_16 is damn large bruh

[steel sage]

Why you don’t mod 100

[azure parrot]

I can, ofc, but I still have to find a1-a16,17

[round raptor]

CSP shit

[steel sage]

f(x_1,…,x_n) mod m equals f(x_1 mod m, … , x_n mod m) for polynomial f

[twilit field]

you don't need a calculator!

if you ever get to a point where you are using a number that's 3 digits or more, you know something has gone wrong.

[steel sage]

Yeah. Straightforward calculation, don’t seem to be hard to obtain at all.

[azure parrot]

Damn

[twilit field]

for example, consider your a_4.

[azure parrot]

I meant this's bad, so much calculation, I didn't expect CSP to give me this question

[steel sage]

Why you kept saying so much calculation

[azure parrot]

Cuz that's too much for me bruh

[twilit field]

your a_4 has four digits, and your a_3 has three digits.
your a_3 should not have three digits after modding by 100, so a_4 should not exceed 999.

[steel sage]

7x-y for two digits x and y and read its last two digits aren’t much calculation at all

[polar chasm]

You could do it a bit faster by doing mod 4 and mod 25 ig

[twilit field]

though I would understand the slight inconvenience of having to hammer 16 terms.

[steel sage]

7x-y, for two two-digits numbers x and y, and as soon as you have calculated the last two digits you can stop.

[azure parrot]

okay this problem sucks, for me at least thanks y'all

[azure parrot]

To me it's really inconvenient, because I always avoid doing calculations, I meant I usually only do one ore twice/ problem

Idk ig it's just me

I'm a bit bad at doing calculation now tbh

[steel sage]

We all are

[twilit field]

welcome to computational hell.

[steel sage]

Better at math worse at calculation

[azure parrot]

I meant doing them by hand would take me 6-10 min

[steel sage]

But we all have to do some calculations at some point

[azure parrot]

That sucks

Okay thanks guy

.close

[devout snow]

Channel closed

Closed by @azure parrot

Use .reopen if this was a mistake.

[azure parrot]

Oh that's why I hate cases work in Combinatorics 🥀

[round raptor]

https://tenor.com/view/cat-lazy-yawn-bored-gif-11081809427946648824

[polar chasm]

Mod 25 the period is much better btw (only 5)

so doing mod 25 and mod 4 is prolly quicker (then CRT to combine)

[devout snow]

Available help channel!

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[fresh bane]

Just dont get this topic

[devout snow]

Please don't occupy multiple help channels.

[supple knot]

close one of your channels

[fresh bane]

Can u help?

[round raptor]

tf is this question

[fresh bane]

I dont get it

[round raptor]

same

what does the video say

[neon folio]

That aint even a grammatically correct sentence

Or am i trippin

[fresh bane]

Idk

[fresh bane]

Idk

[round raptor]

play it?

[fresh bane]

Later

What abt this

[neon folio]

Im guessing the video explains how to do it?

[fresh bane]

No its not clear to me tbh

[twilit field]

heads-up: you aren't showing the full diagram.

[fresh bane]

Oh mb

[round raptor]

also how are you supposed to use ruler on screen

[twilit field]

I presume OP is to construct it on paper and send an image of their work.

[fresh bane]

K will do in a sec

[fresh bane]

This one I dont get it

[fresh bane]

Im on that

[twilit field]

please do not repeatedly ping me.

[fresh bane]

Mb ill turn off my ping

Didnt mean to do that

<@&286206848099549185>

[round raptor]

?

[fresh bane]

I didnt get it

[round raptor]

nor me

watch the video

[fresh bane]

Oh

How do u find a locus?

[round raptor]

what do you think , point equidistant from B and A are at equal distances from both

[fresh bane]

Is it 3,4

Sorry the other way around

?

[round raptor]

that is the midpoint you need all points that are equidistant,
not only one point

[fresh bane]

Ahh

Mb

[round raptor]

do you know distance formula

[fresh bane]

Time*speed?

[round raptor]

this

[fresh bane]

But isnt the question singular?

[round raptor]

distance b/w two points

[fresh bane]

Im in year 9 so I dont really get it tbh

[round raptor]

it asks to find locus of points

[devout snow]

@fresh bane Has your question been resolved?

[round raptor]

hm..

have you learnt the distance formula

https://youtu.be/YDvUy5VYm9E?si=GjKgU-Rl8mppXgjX

watch this

this is used to find distance b/w twpo points A(x1,y1) and B(x2,y2)

you know coordinates of A and B
let Point P satisfy condition it is equidistant from A and B
P =(x,y)
find distance b/w P and A, then Pand B and equate them you will have the locus

[fresh bane]

Sure will do

Thnx

.close

[devout snow]

Channel closed

Closed by @fresh bane

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sage meteor]

You take x = 0 in this case

[sage meteor]

@round raptor can you use midpoint theorem here?

Can we use ?

.close

[devout snow]

Channel closed

Closed by @sage meteor

Use .reopen if this was a mistake.

[fresh bane]

So its 0,3?

[sage meteor]

yea its on y axis

[fresh bane]

Oh ok

[fresh bane]

Its saying no idk y

[sage meteor]

Hol up

[sage meteor]

I think you calculated it wrong

[fresh bane]

Come in #help-14

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wet haven]

Tried a lot of different stuff nothing works I am lost give me a general direction please

[wet haven]

,rccw

[scarlet sequoia]

What have you tried?

[wet haven]

,rccw

Sorry for the mess but thats the easiest I can show (only 3b)

[scarlet sequoia]

Mhm, it looks like multiplying by denominator conjugate doesn't really help

Let's try to multiply by numerator conjugate instead

i.e.

[wet haven]

I kinda tried that as well with a-b = (a^2-b^2)/(a+b)

[woven radish]

Modus

[wet haven]

Oh the whole thing

The whole thing I haven’t tried

[scarlet sequoia]

And I see it's enough to expand the top part

[wet haven]

Now i am getting somewhere

[pulsar sand]

You should just do the conjugate of both the top and the bottom. If you only do the top you are not going to be able to remove the root from the bottom.

[wet haven]

But also i don’t get the denominator part

If you multiply sqrt by sqrt it’s just gonna be 2x^2 + 1x inside?

[scarlet sequoia]

Did you get 2x-8 at the top?

[wet haven]

Ahh no

[scarlet sequoia]

It will be possible, but it's a little bit tricky in this case

[wet haven]

I changed the sign to +

It was to the power of 2

[scarlet sequoia]

What's more, sqrt(2x+1)^2 isn't 2x^2 + 1

[wet haven]

The sign wasn’t supposed to be included?

[scarlet sequoia]

(a-b)(a+b) = a^2 - b^2, right?

[wet haven]

No it’s still in the sqrt like there but with sqrt still

[scarlet sequoia]

When you square a square root, what remains is the number under the square root

So just 2x+1

[wet haven]

Yes my bad

[scarlet sequoia]

And hence, it becomes 2x+1-9, you see that?

[wet haven]

Yes

[scarlet sequoia]

= 2x - 8

[woven radish]

Modus

[wet haven]

I am talking about the denominator where sqrtx * sqrt(2x+1)

[scarlet sequoia]

Don't touch it yet, believe me

[wet haven]

Oh alr

[scarlet sequoia]

Now, observe that 2x-8 = 2(x-4), I hope you can notice something right now

[wet haven]

I see it

Thats tough to see

Man I wish I had that kind of vision

Maybe I am tired though, thanks for your guidance

[scarlet sequoia]

Perfect

[wet haven]

Math is fun when I can get it done

.close

[devout snow]

Channel closed

Closed by @wet haven

Use .reopen if this was a mistake.

[wet haven]

.reopen

[devout snow]

✅ Original question: #help-27 message

[wet haven]

I don’t know how to change the original question

I didn’t wanna ask again but I know I won’t do it otherwise

,rcww

[willow helm]

About the previous one

[wet haven]

So I know how the module part has to be done the 3 to the power is what gets me

[willow helm]

Is L'Hôpital not allowed?

[wet haven]

Well I can’t say for sure I know it but I am basically retaking this class

But I think we haven’t learned it yet

So no

[willow helm]

I see

[wet haven]

But you could probably do it with L’Hopital

[willow helm]

You definitely can

In 5% of the time

😂

[wet haven]

Wdym

[willow helm]

In the sense it takes 20x less time to do it with L'Hôpital

Exaggerating

But it is much faster

[wet haven]

I would have trouble with it even so

[willow helm]

You just find the derivative of the numerator and denominator and then plug 4 in

[wet haven]

Sqrt with a-b and such would get me

[willow helm]

But yeah this isn't the point if you haven't learned it yet so nevermind