help-17

[frail dew]

yeah but theyre saying sub it into 5.

[glossy maple]

They are saying sub (5) into (4)

[frail dew]

oh lmfao i can read i swear

but that is what i said here anyway so still same issue

[glossy maple]

Show the rest of the solution

Including what's before it

[glossy maple]

They are substituting into f_x and equating to 0?

[frail dew]

the rest is just finding partial derivatives

[glossy maple]

Like, into (3)

[half imp]

They should be plugging 5 into 3

And it looks like they are

[glossy maple]

Yeah but they wrote (4) which is a typo so rip

[half imp]

Yeah

[frail dew]

oh

okay

MFW 23 thumbs up. thanks guys

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[glossy maple]

👍

[half imp]

👍

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[echo verge]

Problem from discrete math:

f(n,k) is a function that counts all permutation from S_n that have exactly k fixed points

[echo verge]

so the problem doesn't require me to find the explicit solution to this but still i was wondering if this is correct

ok nvm im wrong

ok how would i get the number of permutation with NO fixed points?

then if I had that i could just do $f(n,k) = \binom{n}{k}*(permutations with no fixed points)$

[twin meteor]

OHHELLNAH

[river minnow]

Note that when with k fixed points, the rest of the permutation can be presented as a permutation of (n - k) elements

[echo verge]

yeah but if i just do $\binom{n}{k} * (n-k)!$ i could get more than k fixed points

[twin meteor]

OHHELLNAH

[echo verge]

i need exactly k fixed

[river minnow]

Yeah, that should be it

Why do you think this is overcounting?

Ah

[echo verge]

yes

[river minnow]

Do you want at least k fixed points or exactly k fixed points?

[echo verge]

exactly

thats why i wanted to find the number of permutations with no fixed points

and my idea was to to find permutations with Atleast one fixed and then subtract that from all permutations but i ust 0 for some reason:

$n! - \binom{n}{1}(n-1)!$

[twin meteor]

OHHELLNAH

[echo verge]

i just learned there is no formula for this

its called Dearangments and the solution is n!/e rounded to nearest integer

[river minnow]

I remember the problem about counting permutations with k non-fixed points

There is no closed form to the solution, it's presented as a sum

However you can derive the recursive formula by picking some non-fixed element x and considering two cases on phi(phi(x)) where phi is the permutation (the cases are phi(phi(x)) = x or phi(phi(x)) ≠ x)

[river minnow]

Yeah, exactly

[river minnow]

There is, it's just that the formula isn't a closed form

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[gleaming eagle]

not sure where the 1/4 came from ? It was never brought up in other problems

[echo verge]

for example f(x) could be defined like f(x) = 1/2 * x^{1/2}

[gleaming eagle]

oh no, I'm asking where the 1/4 slope came from

[echo verge]

wdym where it come from?

Its just that the derivative of the function f in a is 1/4

the slope of the linear function that aproximates f in a is 1/4

[gleaming eagle]

well the original problem only have us the 1/2 and a=1, this is Quizlet and it kinda just gave me the number

gave**

[echo verge]

well they introduced the value of the derivative in this step

[gleaming eagle]

do you want to see the first two?

[echo verge]

Ok but whats the problem

[gleaming eagle]

like this?

parts a and b

[echo verge]

just a sec

[gleaming eagle]

no problem

[echo verge]

ok im back

[gleaming eagle]

wb!

[shrewd flax]

You differntiate the function

plug in 1

then u get 1/4

[gleaming eagle]

the one in step 2 3b?

3a****

[shrewd flax]

The function given

[gleaming eagle]

oh.

.25 as in 1/4

I

am an idiot

I need sleep, lmao

ty guys xD

[shrewd flax]

💀

[echo verge]

👍

lol

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[vast shale]

Can anyone help with #2

[vast shale]

<@&286206848099549185>

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[vast shale]

<@&286206848099549185>

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[dim sinew]

step by step please

[dim sinew]

or just explanation

plleeasseee

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[wild meteor]

how do you get to that last step from the 2nd last

i havent done this in a while and i forgot

[mellow oyster]

you can apply tan^-1 or arctan to both sides

to isolate F

so if you put a calculator in degree mode and type tan^-1(0.82759), you should get that answer

[wild meteor]

how do you apply ir

it

i asumed

but idk how you enter that into your calc

like tan-1*.82

oro what

[hidden tide]

there should be a tan^-1 ( on your calculator

[mellow oyster]

it might even be the same key as where tan is, you might have to press 2nd or alpha or something to select it

[wild meteor]

ik

but how do i use it

do i multiply

the numer by it

or

[hidden tide]

u put the right side in the brackets

[wild meteor]

can you type in the chat

what ytou mean

Tan^-1

i got that in my calc

how do i tell it

my number

[hidden tide]

so it would be tan^-1 ( opp/adj ) to get the angle

[wild meteor]

oh ok

ty

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[proper dock]

arc tan

[wild meteor]

.reopen

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✅

[wild meteor]

what all work would i need to show

just tan^-15.8/4.8

or would i need to show

tan deg = 5.8/4.8 first

[hidden tide]

sorry let me reword

[hidden tide]

so you can solve into the decimal first (as the worked soloution does) and do arctan of the answer, however i sometimes find it easier to just do it in one equation

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[silk warren]

???

[gaunt willow]

do you remember formula for average rate of change

[silk warren]

y2-y1/x2-x1

[gaunt willow]

yeah

the slope of the secant line between the two points

[silk warren]

ohh

[gaunt willow]

so does the slope decrease or increase

[silk warren]

ohh it becomes less negative

[mellow oyster]

maybe you realized the magnitude of the slope was decreasing

but yeah being less negative means it's technically increasing

even though the changes are getting smaller

[silk warren]

i was thinking of this

decreasing at a decreasing rate

i'll keep that in note

thanks guys

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[low lintel]

I need help with question 4

I forgot how to find these 4 things

[jagged sequoia]

Vertex is the max or the min of the parabola

Axis of symetry is the y axis of the parabola

Stretch or compression is determined by the value

A value

As it is 1-1a 2-2a

Like one space right one space down if it is inverted

Direction of opening determines on negative a value

Either upwards or downwards

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[indigo prism]

can someone solve this

<@&286206848099549185>

[solar barn]

I won't solve it ( I mean give tyou the answer). Think, or recall, what is a perpendicular bisector? Does this construction seem like one?

[indigo prism]

alr can u do another problem then?

[indigo prism]

Its no right

because the lines are not straight

so you cant make a right angle

[solar barn]

Fair enough. But, you meant the lines are not perpendicular since they do not form a right angle? Because I see all these lines are straight.

[indigo prism]

Yeah

thanks

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[rain hollow]

In a chess tournament 2 guys played each of them playd 3 games and then left. In the whole tournament 84 games where played how many people where at the tournament and did the original two guys play each other

[orchid wren]

Not clear

Can you send original question

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[dreamy cliff]

.close

[distant saffron]

Hey I need help

This is my work so far

[mellow oyster]

good job, now theta is the angle forming the arc

so assuming the gondolas are equally spaced out, and you have 360 degrees in an entire circle, how many degrees separates each gondola?

[distant saffron]

8.6 degrees?

@mellow oyster

[mellow oyster]

yes

360 / 42

so that's your theta

[mellow oyster]

though wait

theta is in radians

[distant saffron]

ohhhhh

[mellow oyster]

like the formula S = r * theta only works if theta is in radians

so you'd have to convert your angle into radians to plug it in

the other way i'd think about this is that the circumference is 2pi * r, and you're chopping the circumference into 42 different arc lengths

so you'd just divide 2pi * r / 42

[distant saffron]

Yeah I tried it that way, but it doesn’t follow the s=rtheta formula they want us to use

[mellow oyster]

oh okay if they insist you use the formula then that works

just instead of 360 / 42, it'd be 2pi / 42

[distant saffron]

Is this good?

[mellow oyster]

yeah that looks right

[distant saffron]

Could you check my Answrrs I don’t have an answer key

So this s=rtheta formula, theta MUST be in rad?

[urban snow]

Yea, if you want it in degrees it would be (theta/180)(pi)(r)

[distant saffron]

Could u check if my answers r correct?

[urban snow]

Yea everything seems alright

[distant saffron]

Could u explain this question?

@urban snow

[urban snow]

So we know that the unit circle has a radius of 1, where the right triangle formed using the radius as its hypotenuse has side lengths sinx and cosx

[distant saffron]

yes because soh cah toa?

And because tan45 = 1?

So since cos 45/sin45 = 1 my coordinates r sin 45 and cos45?

[urban snow]

Wow you got that quick but yea essentially that's what it's asking for

[distant saffron]

Okay I see

Is there any other variation of this question ?

[urban snow]

But while the answer is correct, i just noticed that you wrote tan45 = 1, what exactly do you mean by that

[distant saffron]

I was referring to the unit circle

Tan 45 is 1

On the unit circle

The chart associated with the unit circle

Ykwim

[urban snow]

Yea it is, but i dont really see how tan 45 relates to the question

[distant saffron]

Because P(45) is tan?

That’s why opp/adj = hyp

?

If thst makes sense

[urban snow]

Ok i think we should go back and revise a little
So the unit circle has a radius of 1
So sinx = opp/hyp, but because hyp is 1, sinx = opp
Same thing for cosx = adj/hyp, so cosx = adj
So the side lengths of the triangle are sinx and cosx
Meaning that p(45) = cos45,sin45

[distant saffron]

Okay I see

Hey what time is it for you right now?

[urban snow]

3 o clock i think

[distant saffron]

Am?

[urban snow]

Pm

[distant saffron]

Oh

R u busy?

[urban snow]

Nah

[distant saffron]

It’s 12 am for me and I have a big final test in the morning right, it’s on trig, I still need to learn how to find equations from graphs, how to graph when given equations, and how to write equations from words

Do you think you could potentially stick with me for a couple hours maybe and teach me these things?

Let me send you pictures

[urban snow]

Yea i guess so why not

[distant saffron]

I need to learn all this

[distant saffron]

Thank you so much, you’re a life saver!!!!

Should I just read through all the material quickly and return in 15 mins?

[urban snow]

Sure

[distant saffron]

Ok I’ll be back in 15, thank you so much ur a life saver!!

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@distant saffron Has your question been resolved?

[distant saffron]

@urban snow hi bro check your messages (dms)

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@distant saffron Has your question been resolved?

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[vast shale]

what happened to -1?

[hybrid flicker]

(-1)^2 is not -1

it's 1

[vast shale]

...

ah

[hybrid flicker]

negative x negative = positive

[vast shale]

okok got it now thanks

lol

[kind iron]

I figured all this out when I was 14 playing around with desmos

[kind iron]

can't solve it lmao

[vast shale]

WHAT

[kind iron]

trust me bro its all piss-easy if you can add and multiply

[vast shale]

im still learning pre algebruh

so maybe soon

[vast shale]

how you got into math?

[kind iron]

school

[vast shale]

aww man

i dont go to school

[kind iron]

💀 then y do you have to learn this

[vast shale]

getting a job

a dream job ig

:3

[kind iron]

oh shite

what required basic trig?

[vast shale]

graphics programming?

[kind iron]

AHHHH i see

I envy you ahahhaha

[vast shale]

oh..

i envy you first ngl

knowing so much math requires alot of time

but soon i will learn i guess, fundamentals are needed

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[still summit]

Need help on how to solve these optimal price questions

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[gaunt willow]

try writing the total revenue for the hall as a relationship between price and the amount of people

[still summit]

This good?

First bracket being 30 dollars per person and -1.5

And second being for the amount of people

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[subtle helm]

if you add those two expressions and simplified by combining like terms, what would you get

[vast shale]

-4x^2?

[subtle helm]

yes

[vast shale]

Aught thanks!

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[vast shale]

How do I convert the slope to parrarel

[snow hedge]

Try plotting the graph of y=-x+3

[vast shale]

Sorry but how?

The y intercept is 3 i only know that point

[snow hedge]

You only need two points to plot a line
So take any 2 random values of x and calculate y for them
Plot the points and join them

[vast shale]

you know what parallel means?

I can just equate anything to x?

[vast shale]

Yes

The other side or opposite

what do 2 parallel lines have in common

The same line

Or slope

so which line has the same slope

C?

the slope is -1

I just wanna know how to convert it to thr pararrel one

So it’s decreasin

yes

the -x means the slope is -1

Yep

so what line is the one parallel

D?

yes

Ohh

Aight thanks

no

np

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[nocturne glacier]

Can someone help with a

[river kettle]

dr du

[nocturne glacier]

Mhm

Ext 2

[river kettle]

lmao nice

[nocturne glacier]

Yes but my brain malfunctions when it sees proofs

[red bear]

instead of factorng a^2, b^2 and c^2, factor a, b, and c, so that you have terms of the form a^2+b^2

[nocturne glacier]

OH

Thanks

[red bear]

np

[fervent wasp]

!done

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[nocturne glacier]

.close

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[lethal quartz]

I’m supposed to show that 0 =< Un =< Un+1 =< 1

With Un+1 = f(Un)

U0 = 0,1

f(x) = 2xe^-x

[lethal quartz]

And i dunno if it’s good

[solar barn]

I recommend to divide it into cases, it can be challenging otherwise.\

• Prove that $U_m \geq 0$\
• Prove that $U_{m+1} \leq 1$\
• Prove that $U_m\leq U_{m+1}$

[twin meteor]

Crystopher

[solar barn]

You probably can use induction and/or some reasoning involving properties of real valued functions.

[vocal sleet]

@lethal quartz Has your question been resolved?

[lethal quartz]

What is this?

[solar barn]

Like I told you, you can divide the inequality into pieces.

You don't have to do as you did in the image you sent.

[lethal quartz]

Wdym?

I’ve never did another method since

[solar barn]

If go into the logic behind it, if you write $$a<b<c$$ it is the same as $$a<b \land b<c$$
So proving that $$a<b<c$$ is equivalent to proving $$a<b \land b<c$$ being true, which in turn holds if $a<b$ and $b<c$ are true by themselves (as per the significance of the $\land$ operator.

[twin meteor]

Crystopher

[solar barn]

If you still are not convinced/don't like the idea of splitting up the inequality, well I find it hard to help since I myself wouldn't know how to do it as you do.

[lethal quartz]

???😭

[solar barn]

Wait, you don't see this?

[lethal quartz]

Ah yes

I didnt understood

[solar barn]

In short, it is valid to split up the inequality, if you don't wanna, well, I can't do much about it, can I?

[lethal quartz]

Okay thx for your help

I finally solved it

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[flint topaz]

Im not understanding a)

[obtuse umbra]

here

[past thicket]

!spoonfeed

[obtuse umbra]

ohh well i am still explaining

[past thicket]

not really

[obtuse umbra]

hmm wdym?

I was tryna say that you could replace the value of d this way and solve the equations to reach this result

[flint topaz]

So thats the answer

Cuz this shit confusing cuz i havent had it in a long time

[obtuse umbra]

they are just tryna give u the value which u had to substitute

[golden yew]

Please delete this

[flint topaz]

So i just write what u wrote

[golden yew]

Let's not give people answers

[obtuse umbra]

okk

[flint topaz]

But basically thats it tho

[obtuse umbra]

no

replace the d with a+c in the equation and simplify

[obtuse umbra]

well it is

[obtuse umbra]

but technically you cant explain it that way in this server so here you go

[past thicket]

!nosols

[vocal sleet]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[past thicket]

you can say maybe, "find a way to write the right hand side in terms of (a + c)^2"

[flint topaz]

Thanks guys

How do i close the ticket again?

[desert hornet]

.close

[obtuse umbra]

ohh i see okk

[vocal sleet]

Channel closed

Closed by @desert hornet

Use .reopen if this was a mistake.

[desert hornet]

.reopen

[vocal sleet]

✅

[desert hornet]

i didnt mean to close it lol

.close

[vocal sleet]

Channel closed

Closed by @desert hornet

Use .reopen if this was a mistake.

[desert hornet]

.reopen

[vocal sleet]

✅

[desert hornet]

jeez

u get the point

[obtuse umbra]

yup

[vocal sleet]

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[manic lance]

x=2^[2^(y)]

can I log_2() both sides?

[glad python]

yes

[manic lance]

ty mate

[glad python]

both lhs and rhs are nonzero

so you can

[manic lance]

I see, ty

[manic lance]

do you know what I can do when I have:
log_2[log_2(x)]=y?

[glad python]

well what do you want to do what x and y

[manic lance]

I want to simplify the equation for y on one side alone

[glad python]

you did it

[main copper]

I need help😭

[manic lance]

but can't I simply further?

[glad python]

no

[manic lance]

oh

[fervent wasp]

Open your own channel

[main copper]

How?

[fervent wasp]

!help

[vocal sleet]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[manic lance]

ty mate

.close

[vocal sleet]

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[vocal sleet]

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as this action is irreversible, and this channel may abruptly lock.

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[high finch]

https://www.youtube.com/watch?v=M64HUIJFTZM

np

[vocal sleet]

@vast shale Has your question been resolved?

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[crimson flower]

Hey, my homework kinda randomly includes using Excel to approximate the sum of an infinite series, but I genuinely have no idea how I'm supposed to do this.

[crimson flower]

this pic is an example of them doing something similar in the textbook

I've got the solutions to my homework but they kinda just show the equation with the sum from n=1 to n=infinity and then go "we put the first 144 terms into excel and get the answer"

[pale widget]

Just make a column in excel where each row is one of the terms in the series and use =SUM to add them up.

[vocal sleet]

@crimson flower Has your question been resolved?

[crimson flower]

I really hope this is right lol

[pale widget]

You’ll have an easier time parsing it if you make columns for different parts of summand and put it together in the last column

[crimson flower]

Well I kinda tried but I've been working on this assignment for over 12 hours straight and my brain barely works at this point lol

[vocal sleet]

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[crimson flower]

.reopen

[vocal sleet]

✅

[crimson flower]

I'll try to look up excel formatting tips or something to figure out how to make this work for 144 terms

[vocal sleet]

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[dim gate]

1+1

[subtle helm]

pee

[full hatch]

11 and please stop trolling

[dim gate]

Is 2

[full hatch]

No, it is 11

.close

[vocal sleet]

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[dim gate]

Im not that type of person that keeps trolling

[vocal sleet]

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[glossy escarp]

am i on the right track?

[glossy escarp]

would it be t(1)?

for week 1 since u cant get the ln of 0

[glossy escarp]

https://tenor.com/view/crying-cry-spongebob-squarepants-gif-13191010122088454973

<@&286206848099549185>

.close

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[slate island]

having trouble figuring out how this works with powers

[jagged cargo]

the former

[slate island]

so their both squared?

[jagged cargo]

yes

[slate island]

thx

.close

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[vocal sleet]

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[wary tulip]

Here's my question; but aren't only 105 and 106 within the range?

[mild flower]

this question is atrocious

[wary tulip]

I should've given context: It's supposed to be about using SRS to gather opinions on racial profiling of the Hispanic Population of Denver

Also I'm legitimately terrible at math; I need to pass this course for the degree requirement, so any help would be appreciated

<@&286206848099549185>

[vast shale]

oh okay so begin with what you do understand about the problem

[wary tulip]

So, certain sets of the random digits listed equate to a housing label, which should be between 0001 and 9552

With that being said, doesn't that mean only 105 and 106 are within that range? I'm not sure if I'm misunderstanding, or this has been worded badly or whatnot

[vast shale]

in my perspective of which I don't really know much about this topic, the simple random sample of 200 mailing addresses, include line 105 and 106 in being part of the mailing addresses or am I wrong?

[wary tulip]

yes, that's what I'm saying

But then, how can I know what the fourth, let alone third household equates to when those are the only options that fit the range?

[vast shale]

will this be the answer or am I just being dumb? 😭

[wary tulip]

I don't think so

[vast shale]

yeah it would be too straight forward lol

[vast shale]

would the fourth household be 0004? orr

[wary tulip]

Don't think so

No offense, but anyone else?

[vast shale]

wait so 105 and 106 lines is the simple random sample that are fixed numbers. it's a random digit table. there's a total of 9552. All of your possibility of choices can actually have the same chance of getting picked so if the last digits are 4 digits, then the first digits will be 4 digits too.

[vast shale]

no offense taken but I GOT THISSS

[wary tulip]

sorry, walk me through this? so, it can be anything above 106 then?

am I getting that right?

[vast shale]

106 is the line with randomized digits

[vast shale]

the range is 0001-9552

[wary tulip]

yes, but I meant within that range it can be anything above 106?

[vast shale]

chose whatever line number that the problem told you for the sample to be a randomized which is 105 in this case

[vast shale]

in the range of 0001-9552, yes. but the number 106 doesn't matter because it's suppose to be randomized with the amount of random number 105 LINE gave us

it said to start with line 105

[wary tulip]

still trying to figure this out

[vast shale]

they already numbered it for us by 0001-9552. number 200. 9559 doesn't work because it isn't in the range of 0001-9552. but 2940 does work because it's in range 0001-9552. do you understand that?

[wary tulip]

yeeeeessssss

okay

Now it makes sense

thanks, and sorry for doubting you haha

[vast shale]

this is the first line but we also have to do this for the second line 106

[vast shale]

do you think you can do 106 on your own?

[wary tulip]

Yep, I think so

[vast shale]

nah it's totally alright, I doubt myself too sometimes 😭

line 106

Line 105
2940, 0769, 1481, 6077, 9537, 9117, 2975, 9335,

Line 106
6841, 7350, 1313, 5297, 2765, 8508, 6750, 2114, 7487

[vast shale]

@wary tulip these were the answers I got, were they any different to the ones you got?

[vocal sleet]

@wary tulip Has your question been resolved?

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[vast shale]

The Hispanic resident's households is labeled from 0001-9552 (total of 9552 homes), to get a simple random sample of 200 mailing addresses (200 randomly selected mailing addresses having an equal probability of being chosen). Our numbers are grouped 0001-9552 so we're going to group the numbers from the random digit table in four digits too. Anything above 9552 is not included in the numbers and any repeats are skipped (not any in this case). The simple random sample are the numbers selected above, the fourth one in line 105 being 6077. In my cognitive opinion, I believe this is the fourth household in this SRS sample. but idk tho

[fallen hawk]

Lowk feels racist

[vast shale]

it's about this bro 😭

[vast shale]

I was giving my answer to the person that needed in this channel before but this channel got closed after I was able to give my complete answer on the problem 😭 😭

[vocal sleet]

@vast shale Has your question been resolved?

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[bitter oxide]

hello

can anyone help me with this?

[half imp]

apply the distributive property

$a(b+c) = ab + ac$

[twin meteor]

thewizardofOU

[bitter oxide]

tq

.close

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[clear badger]

Heyy

im stuck on 11c

[vocal sleet]

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[undone aurora]

for this binomial distribution question.

p = 0.3
n = 10
x = 9, 10
1-p = 0.7

and I solve for

P(9 <= X <= 10) right?

[solar barn]

Seems alright to me.

[orchid wren]

thee re c'rrect brot'r

[vocal sleet]

@undone aurora Has your question been resolved?

[undone aurora]

ty

.close

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[charred lava]

literally cannot tell if this if x at -3 is removable idek

[outer warren]

!occupied

[vocal sleet]

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).

[outer warren]

try first looking up
removeable discontinuity
jump discontinuity
infinite discontinuity

[charred lava]

i watched the khan academy video lol but im not really sure because for the removable one at the point where it dne, it has another point somewhere else but for this one it doesnt

would x=-3 be considered jump discontinuity? oh wait maybe because it doesnt have a two sided limit

so would it just be x=-6?

[outer warren]

yes

[charred lava]

ohh damn i did not think about that thanks anyways

.close

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[onyx marsh]

$\sin^2(2x+\frac{\pi}{6})-\cos^2(x+\frac{\pi}{3})=0\implies (\sin(2x+\frac{\pi}{6})+\cos(x+\frac{\pi}{3})(\sin(2x+\frac{\pi}{6})-\cos(x+\frac{\pi}{3}))=0\implies\sin(2x+\frac{\pi}{6}+\frac{\pi}{2}-\frac{\pi}{2})+\cos(x+\frac{\pi}{3})=0$ or $\sin(2x+\frac{\pi}{6}+\frac{\pi}{2}-\frac{\pi}{2})-\cos(x+\frac{\pi}{3})=0$

[twin meteor]

pirateking0723

[solar barn]

Maybe use this to write the sum of sine and cosine as a sine function, which should be easier to solve I reckon.

[onyx marsh]

i havent seen this before

so i cant use them

[solar barn]

hmm, otherwise expanding the terms of sin and cosine will leave a mess I think.

[onyx marsh]

so from here $\cos(2x-\frac{\pi}{3})=\cos(x+\frac{\pi}{3})$ or $\cos(2x-\frac{\pi}{3})=\cos(\pi+x+\frac{\pi}{3})$

[onyx marsh]

no need to do messy expansions

this form here is reached without messy expansions as above

[twin meteor]

pirateking0723

[onyx marsh]

now $x\in [-\frac{\pi}{2},\frac{\pi}{2}]$ so $2x-\frac{\pi}{3}=x+\frac{\pi}{3}+2k\pi$ or $2x-\frac{\pi}{3}=-x-\frac{\pi}{3}+2k\pi$ or $2x-\frac{\pi}{3}=x+\frac{4\pi}{3}+2k\pi$ or $2x-\frac{\pi}{3}=-x-\frac{4\pi}{3}+2k\pi$ where $k\in\mathbb{Z}$

each must have a +2kπ where k is an integer

[solar barn]

A problem tho, this is false

[onyx marsh]

not a problem

i am trying to find the x that make this true

i am not saying that they are always equal

[onyx marsh]

just like here plot the RHS and the LHS they wont be the same

because they are not

they just intersect at some points (infinitely many points in fact)

[twin meteor]

pirateking0723

[solar barn]

okay, a bit doubtful of your approach but I'll let you cook until you are done.

[onyx marsh]

tysm

[onyx marsh]

now from here we get $x=\frac{2\pi}{3}+2k\pi$ or $x=\frac{2k\pi}{3}$ or $x=\frac{5\pi}{3}+2k\pi$ or $x=\frac{(2k-1)\pi}{3}$

[twin meteor]

pirateking0723

[onyx marsh]

i am not sure if i have some calculation mistakes at the end

but i am pretty sure about my steps

[onyx marsh]

i am done cooking if i have any mistakes or if you have any doubts please tell me to correct or explain myself

ah wait before that

i am not done yet i still need to follow the restriction on x

[onyx marsh]

$-\frac{\pi}{2}\leq\frac{2\pi}{3}+2k\pi\leq\frac{\pi}{2}\implies -\frac{7}{6}\leq k\leq-\frac{1}{3}$

[twin meteor]

pirateking0723

[onyx marsh]

so the 1st gives no k now for the 2nd one will give $k=0$ the 3rd will not give a k and the 4th will give $k=0$ pr $k=1$ so $x=0$ or $x=-\frac{\pi}{3}$ or $x=\frac{\pi}{3}$

[twin meteor]

pirateking0723

[onyx marsh]

is my work correct

@solar barn

[vocal sleet]

@onyx marsh Has your question been resolved?

[onyx marsh]

-pi/3 isnt the solution because it is outside [-pi/2;pi/2]

i am missing one more solution i have a mistake up there

[solar barn]

no, it is correct

[onyx marsh]

probably a calculation mistake

[solar barn]

The problem is I see x=0 is a solution but I don't see it in your calculations.

[onyx marsh]

i did x=0 here

i am missing x=pi/3

[solar barn]

Yes, i see now. Probably missed a $\pm$ somewhere

[twin meteor]

Crystopher

[onyx marsh]

i think ik where i went wrong

i have to add kpi not 2kpi

because originally cos and sin are ^2

so adding a pi wont affect it

because the negative sign will vanish due to the ^2

[solar barn]

From $x=\frac{(2k-1)\pi}{3}$ you should've gotten
$-\frac{\pi}{2}\leq\frac{(2k-1)\pi}{3}\leq \frac{\pi}{2} $\
$-\frac{1}{2}\leq\frac{(2k-1)}{3}\leq \frac{1}{2} $\
$-\frac{3}{2}\leq (2k-1)\leq \frac{3}{2} $\
$-\frac{1}{2}\leq 2k\leq \frac{5}{2} $\
$-\frac{1}{4}\leq k\leq \frac{5}{4} $\
Where $k=1$ was included.

[twin meteor]

Crystopher

[onyx marsh]

ohh yes

[onyx marsh]

i have a question

since x is restricted to lie in [-pi/2;pi/2] then sinx is bijective here right

[solar barn]

Yes, there are no two different $x$ values in the interval that give the same $y$ since it is monotone in that interval, so it is injective. Then the codomain is $[-1,1]$ and the function is surjective since for each value $y$ in codomain there is at least one $x$ such that $\sin x = y$. This can be proven using the intermediate value theorem.

[twin meteor]

Crystopher

[vocal sleet]

@onyx marsh Has your question been resolved?

[onyx marsh]

but that doesnt mean that $\sin^2(2x+\frac{\pi}{6})$ is bijective in fact it isnt right

thats why we couldnt just use arcsin here

or arccos

[twin meteor]

pirateking0723

[onyx marsh]

tysm

.close

[vocal sleet]

Channel closed

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[onyx marsh]

.reopen

[vocal sleet]

✅

[onyx marsh]

if i take x to be any real number

then wolfram gives this solution

[onyx marsh]

are they different from these

or is there any solution that i added

also why npi not 2npi

[oak magnet]

cos = sin is pi periodic so it's npi

[onyx marsh]

why does it have a period pi

[oak magnet]

have you solved it or you still searching ?

[onyx marsh]

i reached these

if i replace 2kpi by kpi will they be correct

[onyx marsh]

and why is the this the period of that

i thought that kpi will be the period because of cos^2 and sin^2

because adding kpi may lead to a negative sign but with the ^2 it will vanish

[oak magnet]

i'll try the problem

[onyx marsh]

ok

if you want check the steps above to save yourself the trouble of solving it again

and if you want solve it and compare

[oak magnet]

so from this

[oak magnet]

the period for cos^2=sin^2 is pi/2

[onyx marsh]

how do you know

also wdym by period of sin^2=cos^2

i mean this is not a function

this is an equation

[oak magnet]

talking about the solution

[onyx marsh]

although wolfram doesnt add kpi/2 but adds kpi instead

[oak magnet]

and i'm trying to figure out why with you, but i'm clueless for now

[onyx marsh]

how did you reach this

this will give the answer

[oak magnet]

cos²(x) = sin²x
cos²x - sin²x = 0
cos(2x) = 0
x = pi/4 + kpi/2

[onyx marsh]

no

[oak magnet]

i did mistake ?

[onyx marsh]

x=pi/4+kpi

[oak magnet]

nah kpi/2

[onyx marsh]

because then 2x=pi/2+2kpi

[oak magnet]

no it's 2x = pi/2 + kpi

[onyx marsh]

so that cos(2x)=cos(pi/2+2kpi)=0

[onyx marsh]

ah yes because cos(pi/2)=cos(-pi/2)=0

[oak magnet]

exaclty

so this is the answer you were looking for ?

[onyx marsh]

we still didnt reach our destiny

[onyx marsh]

we didnt explain the npi here

ah wait

i found where you did the mistake

[onyx marsh]

what you did here works only when the input of sin and cos is the same

but our problem doesnt have this

the only way to explain the npi is that cos and sin are ^2 so that adding odd multiples of pi to the input of cos which makes it -cos the input will also be cos(input)^2

[oak magnet]

same for sin ig

[onyx marsh]

there is a slight difference

because sin(pi+x)=-sinx while sin(pi-x)=sinx while cos(pi+x)=cos(pi-x)=-cosx

[onyx marsh]

so these are correct if i replace each 2kpi by kpi ?

[oak magnet]

idk, maybe someone else could find the trick, sorry

[onyx marsh]

oh ok tysm np

<@&286206848099549185> can anybody check if my answers are the same as wolfram's

[hot tree]

Okay

[onyx marsh]

so what did you find

[hot tree]

Let me check

Yes these are correct

[onyx marsh]

how are they equivalent to wolfram's results

[onyx marsh]

can you explain to me to know how

[vocal sleet]

@onyx marsh Has your question been resolved?

[onyx marsh]

<@&286206848099549185>

[solar barn]

Seems a lot happened since I was last around here. What's the issue?

[onyx marsh]

i want to check how my solution is the same as wolfram's

[onyx marsh]

so how are they the same

thats the issue

[solar barn]

I mean, you can get your solutions out of them.

$x=\pi n$ gives $x = 0$ for $n=0$

[twin meteor]

Crystopher

[onyx marsh]

and 2kpi/3 gives 0 for k=0

but they arent the same

[solar barn]

You mean these solutions?

[onyx marsh]

yes

if these are the same as wolfram's or no

if they are not then i have some mistakes

[solar barn]

we can see if they both contain the same elements in a group of arithmetic sequences

[onyx marsh]

if they are then how are they

[onyx marsh]

can we equate them to each other

and then find n in terms of k

and take the natural n from that

if n turns out to be any integer then they are the same

other wise it is not

like if we get n=k

then they will be equivalent

[solar barn]

for instance, all of wolphrams solutions entails all $x = \frac{\pi}{3}n$, $n \in \mathbb Z$

[twin meteor]

Crystopher

[onyx marsh]

i am not sure if this is efficient

[onyx marsh]

although npi doesnt

[solar barn]

We can mark the solutions in the unit circle and see where they land

I meant together

[onyx marsh]

because n is an integer

[onyx marsh]

ohh

[onyx marsh]

how can we check algebraically

without resorting to graphs and so on

is that possible somehow

[solar barn]

I think I have tried to do that before, but my reasoning was half-baked, it would be hard for me at least. Like we need to prove that the union of the three arithmetic sequences given by Wolframs solutions is the same as the union of your solutions.

[onyx marsh]

$x=\frac{2\pi}{3}+k\pi$ or $x=\frac{k\pi}{3}$ or $x=\frac{5\pi}{3}+k\pi$ or $x=\frac{(k-1)\pi}{3}$

[twin meteor]

pirateking0723

[onyx marsh]

these are the correct solutions

i changed each 2kpi to kpi

because of cos^2 and sin^2

[onyx marsh]

so i am trying to check if these are equivalent to wolfram's

[solar barn]

One thing I can say about your solutions is that $x = \frac{2\pi}{3} + k\pi$ is redundant since $x = \frac{k\pi}{3}$ contains all of the solutions given by the aforementioned.

[twin meteor]

Crystopher

[onyx marsh]

also (k-1)pi/3 is

even 5pi/3+kpi

[solar barn]

yes

[onyx marsh]

because you can make a common denominator

[solar barn]

so you only have one solution

[onyx marsh]

and then everything will be a constant multiplied by pi

but also notice something

[solar barn]

now we need to prove that the union of Wolframs solutions is the same.

[onyx marsh]

the same goes for these

all of these are just kpi/3

but that doesnt prove it right

[solar barn]

exactly, I don't know how strict do you have to be in your solution.

[onyx marsh]

hmmm

[solar barn]

you could assume that every solution from wolfram can be written as $\frac{\pi}{3}n$ and prove it, but then the problem becomes that you also need to prove that there are no more solutions than that.

[twin meteor]

Crystopher

[onyx marsh]

i can take it as something granted

given that wolfram produced the result

but the real problem is to show that my and wolfram's solutions are exactly the same

that there no solutions i included which arent in wolfram's and vice versa

idk how to do this

[solar barn]

I think I have something. We can 'scale' the sequences given by Wolfram. Let's say that we expect a sequence $x = \frac{\pi}{3}n$ from all of Wolframs solutions together. Then, from the sequence given by $\frac{3x}{\pi}$ we can expect all integers, so we expect this new sequence to contain $\mathbb Z$. We can do the same scaling to Wolfram's solutions. For $x = \pi n$ we can have $\frac{3x}{\pi} = 3n$, for $x = \pi n - \frac{2\pi}{3}$ we have $\frac{3x}{\pi} = 3n-2$ and for $x = \pi n - \frac{4\pi}{3}$ we have $\frac{3x}{\pi} = 3n-4$. We notice that each of these are the sequences of modular classes modulo 3, which means that they together form $\mathbb Z$. So they would be the same, we can then 'convert back' if you want and we should see that they are the same.

[twin meteor]

Crystopher

[onyx marsh]

or we can find the 4 solutions that i got modulo 3 and it will turn out that they will together form Z right

[onyx marsh]

also this shows that they are exactly Z not just that they contain Z

because every n is in Z and 3x/pi for example generates only all of these n

not anything more

[solar barn]

Ye probably, I was thinking that the sequence contains all integers, so they contain $\mathbb Z$, I don't know really if a sequence is really a set.

[twin meteor]

Crystopher

[solar barn]

on the other hand you could construct them as a set I think, so maybe they can be considered sets.

[onyx marsh]

i have a question which is unrelated

[onyx marsh]

is there something as non integers modulo integers

[mellow oyster]

you might find the top answer here useful:

https://math.stackexchange.com/questions/586595/finding-modular-of-a-fraction

though it seems to be only for rationals

[solar barn]

yes, I find it hard, or rather unnecesary to define them on reals.

[mellow oyster]

there might be a notion of repeatedly adding or subtracting

like pi mod 2 might be pi - 2 = 1.14159...

but that seems too hand-wavey

[solar barn]

ye, I don't know how useful that would be

Like the closest to a modular function to the reals I can think of is wave functions like sine

[onyx marsh]

this talks about fractions

but not about irrational numbers

[mellow oyster]

^

[onyx marsh]

oh so there is nothing like that for irrationals

[onyx marsh]

so you are still unsure if they are equivalent

but ig it is a set because you can define the sequence x_n=npi/3 where n is an integer

[solar barn]

If we look at Wikipedia (maybe not the best source ik), they are not the same

[onyx marsh]

cant you define the sequence from integers ?

[solar barn]

I think a key difference is that in a set order does not matter while in a sequence it does.

[dim gate]

432 x 3

[vocal sleet]

@onyx marsh Has your question been resolved?

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[languid spruce]

find the number of solutions to this equsion:
x1 + x2 + ... + xn = k, such the the even xi's are even natural numbers, and the odd xi's are odd natural numbers

[languid spruce]

(even natural numbers include 0)

so how do i start?

so we need n/2 even numbers and n/2 odd numbers that sum to k

[young vortex]

if i had to approach this i would probably start by splitting k into two numbers, a and b, where a + b = k and a is the sum of all the even xi numbers and b is the sum of all odd xi numbers

[languid spruce]

that would require a summation no?

for every possibility of a

[young vortex]

yeah

[languid spruce]

is there a way without a summation?

well actually fuck that i dont have time

so

if were doing this approach

to get to a from even numbers the number of ways is s(n/2, a/2)

right?

now we need n/2 odd number than sum up to k-a

[young vortex]

yeah

[languid spruce]

that would be

s(n/2, -(n/2) + (k-a)/2)

is that right?

[young vortex]

i think so ?

[languid spruce]

and then the sum from a=0 to k?

yeah

the thing is the question says to assume (k-n/2) is even

so im supposed to use that somewhere

but i didnt in this solution

[young vortex]

thats a weird condition

[languid spruce]

not exactly a condition

more of a hint

[young vortex]

oh

it makes sense

that just means that if n/2 is odd then there must be an odd amount of odd numbers being added up meaning that k will also be odd

and if n/2 is even there is an even amount of odd numbers being added up so k will be even

[languid spruce]

right yeah

but how is that useful in finding a solution

[young vortex]

well its just an assumption you have to make otherwise it wouldnt make any sense

if you think you need to use that somewhere in your solution you can just state why it is a necessary condition for the problem to make sense

[languid spruce]

but why is it an assumption?

i mean like

that must be true

why did they state it

like you cant not assume that

thats like saying given an even number n blabla assume that n is divisible by 2

[past thicket]

stars and bars

[languid spruce]

wat

[past thicket]

lets start with a simpler version of the problem: how many solutions are there when xn must be a natural number (no constraint about even or odd)? you can think of this as having k "stars" and distributing n - 1 "bars" to divide the stars. for example 1 + 2 + 3 = 6 corresponds to the picture i gave below

then the number of solutions to x1 + x2 + x3 = 6 just becomes the number of ways to arrange the stars and bars

[languid spruce]

n+k-1 choose k

yeah

[past thicket]

exactly. so you apple the same logic

[languid spruce]

but we have constraints here

[languid spruce]

well the problem is the constraints themselved

its that there are 2 kinds of constraints

in the same equation

thats why i split into cases and did the summation

[past thicket]

the summation shouldnt be too bad, i think

[languid spruce]

yeah i did that already

i was wondering how to do it without a summation

thats what i got

[past thicket]

maybe you can think of it as dropping stars into boxes two at a time

[languid spruce]

even numbers that add up to 2i, odd numbers that add up to k-2i

[languid spruce]

yeah thats what i did with s(n/2, i)

[past thicket]

mhm im not sure how youd do it without a summation

[languid spruce]

what am i missing lol

oh

well then

thanks

both of you

[languid spruce]

and nice bio my friend

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[fresh tide]

So for this integration by partial fractions, for problems that basically need us to create systems of equations how can we tell which system of equation to use to find our B?

[fresh tide]

so for this picture for example, I created a system of equations where A+B=0, 4A-B-C=3, and 4A-2B-C=9

when I solve for A+B=0, I get B=2/3, when I solve for 4A-B-C=3, I get B=32/3, and when I solve for 4A-2B-C=9, I get 2/3

A+B=0 and 4A-2B-C=9 both give me 2/3 which is the right answer when I look at the answer key that my professor provides but 4A-B-C=3 doesnt

how do I know if I should use A+B=0, 4A-B-C=3, or 4A-2B-C=9 to get B?

<@&286206848099549185> ?

[vocal sleet]

@fresh tide Has your question been resolved?

[fresh tide]

<@&286206848099549185>

[orchid grove]

ChatGpt 4 Wolfram Alpha plugin

[fresh tide]

.close

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[vagrant sable]

X^(n)=y

[vagrant sable]

So x is positive and y is negative

And are not equal to zero

[vagrant sable]

Is there n that satisfies this equation

?

Uhm- i can help you

[bitter pilot]

is n an integer?

[vagrant sable]

Anything

Integer irrational whatever

[steep finch]

n must be complex

[orchid grove]

yep

[vagrant sable]

Ooooh okay very interesting

Just a sec

[steep finch]

$a^z = -b^2 => z\ln(a) = \ln(-b^2) => z = \frac{\ln(b^2) + \ln(-1)}{\ln(a)}$

[twin meteor]

Obotron

[steep finch]

$\ln$ denotes the complex logarithm.

[twin meteor]

Obotron

[vagrant sable]

Kk ty

[steep finch]

$\ln(-1) = \ln(e^{i\pi \pm 2ki\pi}) = i\pi \pm 2ki\pi$

[twin meteor]

Obotron

[vagrant sable]

1/(3+sqrt(5-x))

Whats the domain of this

[steep finch]

x <= 5

[vagrant sable]

Let me tell u what i did

Ok wow that was quick

Impressive actually

[steep finch]

it's just looking where the sqrt function is defined

i know the denominator is never 0 since sqrt is always positive

so i just look at the inside of the sqrt function and analyse where it is defined

essentially im solving: 5 - x >= 0

[vagrant sable]

(Brb)

[vagrant sable]

X is equal to or less than 5

But like

The denominator

Cannot be 0

Why didnt u consider tbat

[steep finch]

this

[vagrant sable]

Oh

Omg right

Gj! Ty

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[carmine prairie]

Help !

[carmine prairie]

I need help on an exercise about SIGMA ALGEBRA

[crisp ermine]

sure thing

what's the question

[carmine prairie]

If T1, T2 are sigma algebra such that T1 U T2 is sigma algebra

then T1 subset T2 or T2 subset T1

mhh

i suppose that T1 U T2 is sigma algebra and there is no inclusion between T1, T2

then there exists a set x such that x in T1 but not in T2

@crisp ermine are you there ?

[crisp ermine]

yeah sorry i cant help with this i misunderstood what you meant

not a number theory person 😭