help-10

[lucid flume]

what will be it's theta then?

the answer suggests the theta in a x b should also be 90 degrees

here's another thing

I found a Pythagorean triplet identity here

3,4,5 isn't it a Pythagorean right triangle identity?

so, angle between A and B should be 90 right?

@vapid hamlet

[vapid hamlet]

The fact that the magnitudes are 3, 4, 5 might be important

But I don't see directly how this (together with a + b + c = 0) implies that the angle between a and b is 90°

[lucid flume]

if the magnitude of c is 5

and (a,b)=(4,3)

so isn't 4^2 + 3^2 = 5^2 ?

So, the angle between A and B is 90 right?

[vapid hamlet]

This already assumes that a and b are orthogonal

So, your reasoning is circular

[lucid flume]

wdym?

[vapid hamlet]

The Pythagorean theorem works if the angle between a and b is 90°

But you don't know that a priori

[vapid hamlet]

Sorry, it's 3 am, you're actually right

[warm shale]

Zanarcane

Zanarcane

[lucid flume]

yeah, I just arrived at the solution

thanks

@vapid hamlet

.close

[obtuse pebble]

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[timid silo]

how do i do c

[forest sinew]

its an identity directly

||look at double angle||

[timid silo]

huh

whats the exact value then

ohhhhh

[obtuse pebble]

@timid silo Has your question been resolved?

[low herald]

$cos(2x)=cos^{2}x-sin^{2}x$

[warm shale]

謝墨離

[obtuse pebble]

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[timid silo]

how do i find the period for

[obtuse pebble]

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[frozen swift]

43

[frozen swift]

How solve? Or forced to check the answer choices

[crimson berry]

there are like 10 questions in this photo

[frozen swift]

Number 43

[short spruce]

is this a test

[frozen swift]

No it’s literally the weekend

I don’t think teachers give out paper tests on weekends as homework

[short spruce]

thank you telling me literally what day it is

why does it say test at the top

[frozen swift]

friend gave it to me

its his binder im using it to practice

[short spruce]

so your friend gave you a blank test that is in his binder to practice on

[crimson berry]

googling the header, it appears to be a practice test for a state competition

[frozen swift]

bro....

am i getting help or getting questioned

.close

[obtuse pebble]

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[native cove]

I’m stuck on part (b)

[obtuse pebble]

@native cove Has your question been resolved?

[native cove]

<@&286206848099549185>

[forest sinew]

i would draw a tree

first branch is each bag

0.5 for each

then the probabilities associated with the colors

[native cove]

Ok, let me try ... I tried previewly just addi g number of balls up, and didn't work

Now let me do this again

I tried the tree, and got 111/176, which wasn't the answer

[forest sinew]

is it 101/144?

[native cove]

Yes

[forest sinew]

[native cove]

How did you get it, can you please teach me?

[forest sinew]

so

whats the probability of selecting the first bag?

[native cove]

0.5

[forest sinew]

and whats the probability of drawing a blue ball out of this bag

[native cove]

5/8

[forest sinew]

yea

so the probability of drawing a blue ball from this bag is the product of these two numbers

think of moving down a tree

[native cove]

10/16

[forest sinew]

no

[native cove]

I men 5/16

[forest sinew]

yea

[native cove]

*mean

[forest sinew]

what do you get for the second bag?

its an identical process

just the numbers change

[native cove]

7/18

[forest sinew]

yup

so we have the prob of drawing a blue ball from the first

and a prob of drawing from the second

what do we do

[native cove]

multiply? or add? This is where I'm confused

[forest sinew]

where things are on the same branch, you multiply

[native cove]

Oh, you just add

[forest sinew]

when things are on different branches, you add

[native cove]

Ok, thanks

For C I know wht to do

[forest sinew]

you can verify this conceptually since all branches have to add to one

(something has to happen)

[native cove]

Ok, thanks

[forest sinew]

np

[native cove]

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[next trellis]

how do i prove that this relation is not transitive ?

[obtuse pebble]

@next trellis Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

Find the value of λ such that lines x+2y=3,3x−y=1 and λx+y=2 can not form a triangle.

[timid silo]

I thought of using the idea that if area of triangle is zero, this means all 3 points are collinear

[vernal wren]

Sounds good

[timid silo]

but here in the question lines are given instead of coordinates

[lucid flame]

think about the graphs

if 2 lines are parallel, can it form a triangle?

[timid silo]

never

[lucid flame]

so 2 lines are parallel iff?

[timid silo]

slopes are equal

[lucid flame]

ok, so put the equations in y-intercept form, and find the slope

but there's one more exeption which i'll tell after you find it

[vernal wren]

So write all equations in the form

ax + by + c = 0
And use this

[timid silo]

the slopes are -1/2, -3 and - lambda

[vernal wren]

Like

1 2 -3
3 -1 -1
λ 1 -2

[lucid flame]

in this case, when the determinant is 0, it will only give the solution where the lines are concurrent

but that's not all the values for lambda

[lucid flame]

so what can lambda be to make at least 2 of the lines parallel?

[timid silo]

-1/2 or 3

[lucid flame]

the signs

think carefully

[timid silo]

sorry 1/2 or 3

[lucid flame]

ok, so thats the case where the lines are parallel

but the other exception is when the lines are concurrent

[lucid flame]

which is basically this

[timid silo]

in this case lamda is coming 6/5

[lucid flame]

yep

[timid silo]

thank you I got all 3

.close

[obtuse pebble]

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[astral mural]

is it possible to find theta in this situation

I have 2 points and an angle with the x axis ( 30deg) and need to find the angle to that new plane axis

[royal basin]

what is that symbol that looks awfully like a crossed zero

[astral mural]

dont even lol

[royal basin]

i'm serious. what is this

[hollow sun]

must be theta

dont do him like that yoo

😔

[astral mural]

i choose math not art lol

[royal basin]

yeah well

the line through theta is supposed to be horizontal

while yours is almost vertical

hence your symbol cannot be read as theta no matter what

[astral mural]

because that was the way my pen happened to be going

[royal basin]

but anyway, to answer your question: no, it is not possible to find theta here. not with the info you've provided

[astral mural]

again lazy

what would I need you reckon

[royal basin]

you would probably need one of the red lengths

[astral mural]

in its context

[royal basin]

so...

is the second red line meant to be horizontal or what

er

[astral mural]

yes

[royal basin]

ok

[astral mural]

its like the x axis

[royal basin]

parallel to the x axis

then you can calculate with relative ease the third (unmarked) angle in your triangle

[astral mural]

if you may, paralell

yes its the "with ease" part I need more depth for

[royal basin]

find the slope of the cyan line and take its arctan

[astral mural]

wouldnt that get me this

oh i see lol

180- (angle - (90-(180-atan(m))) )

[obtuse pebble]

@astral mural Has your question been resolved?

[obtuse pebble]

@astral mural Has your question been resolved?

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[blissful solar]

Calculate all Fourier polynomials of the 2π-periodic function
Also decide whether the corresponding Fourier series converges.

[blissful solar]

Hi I couldnt understand how to apply this to fourier series formula

Fourier series is new for me dont actually know what to do

I guess the function is odd

so should I apply what I have to this?

<@&286206848099549185>

[obtuse pebble]

@blissful solar Has your question been resolved?

[blissful solar]

<@&286206848099549185>

[blissful solar]

.close

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[glossy grove]

For the sequence, 5,7,10,15,22,33,46,63. Is this AP or GP and how do I tell the general formula?

[pine sail]

Neither.

[olive hawk]

It is neither an AP nor a GP but a general formula can be written

[glossy grove]

oh ya neither AP nor GP. How do i find the general formula tho

[pine sail]

You should notice, the difference each time is a prime number.

[glossy grove]

how to get started

[glossy grove]

oh true!

didnt see that

.close

[obtuse pebble]

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[tame berry]

what patterns are there for prime numbers?

[pine sail]

Yes exactly, that's what I expected then to ask, then again they did not and somehow their problem was solved I don't really know why that happened.

[tame berry]

lmao

ok so there is no general formula I assume

.close

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[high shell]

hi

[high shell]

a quick question

i got this as the answer

and the correct answer is 25x² + 40xy +16y² -4

where did + 40xy come from

[scarlet locust]

I don't know why I feel like I've seen this before

anyway
(a+b+c)(d+e+f) = a(d+e+f) + b(d+e+f) + c(d+e+f)

[timid silo]

There are 3 terms in each set of brackets

[scarlet locust]

not ad + be + cf

[high shell]

cant we take it as (a + b + c)(a+b-c)
as the numbers remain the same

[scarlet locust]

yes you can

but you have to expand it correctly

[high shell]

also , isn't it
(d + e - f)

[scarlet locust]

f can be any number so if its negative the equation would become -f

so you don't need to worry about signs

[high shell]

.close

[obtuse pebble]

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[alpine raven]

Don't ping people

[robust sleet]

^

[wary vigil]

which question and what do you not understand

what about the question do you guys not understand

but what about it

but what do you not understand

when you read the question

where are you stuck

that's not how this server works

[robust sleet]

why are u in such hurry

[lucid flame]

what

[robust sleet]

[timid silo]

hey sorry to disturb you mate...I got all 3 answers but am thinking how by concurrency of these 3 lines u deduced their deteminant is zero....I mean in the above screenshot I was told if the points are collinear then deteminant is zero but here u told if lines are concurrent too then it is zero I didn't get how?

[robust sleet]

next time dont do homeworks late

[lucid flame]

in dm's

[lucid flame]

this is someone else's channel

[timid silo]

yeahh got it

[lucid flame]

i told @timid silo , not you

chill out

[keen cave]

<@&268886789983436800>

[robust sleet]

😆

[sterile latch]

hm

.close

[obtuse pebble]

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[sterile latch]

[robust sleet]

fun gone :((

[obtuse pebble]

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[wintry stream]

can someone please explain how to solve this step by step

[novel knoll]

use ratio test?

[hollow sun]

you have to define you sequence first

$a_n:=\frac{(-1)^n 3^{2n+1}}{(2n+1)!}$

[warm shale]

Big xdddd

[hollow sun]

and the ratio test is: $|\frac{a_{n+1}}{a_n}|$

[warm shale]

Big xdddd

[hollow sun]

you just have to do this and simplify as much as possible. your answer is f(n)

[obtuse pebble]

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[timid silo]

hi. can someone verify my solution if I solved my problem correctly? i am trying to find the absolute extrema of a given function in a given interval

[keen cave]

Sure.

[keen cave]

Yep it’s correct.

[hollow sun]

just want to make sure something: here in the first row you wrote f(-1) = x-2tan^-1(x) but you havent insert your -1 into the right side of the equation. That is not formula. (Just want to help you)

[timid silo]

oh just a minor typo error

@hollow sun but is my answer correct?

[keen cave]

Yes it is.

[timid silo]

this might be a very stupid question, but when I use sci cal to compute arctan (-1), arctan (1), and arctan (√3). should I set my sci cal in degree or radian?

[keen cave]

Good question. It should be radian.

Since your doing calculus.

[timid silo]

really? so should I change it since I used degree mode

[keen cave]

Yep.

[timid silo]

okaay quick thing. will it change the outcome?

will have to resolve it later

[keen cave]

Always use radian unless your professor says so. Not sure can you post again with the updated calculations.

Wait I can use wolphram.

,w -1-2arctan(-1).

,w 1-2arctan(1)

[warm shale]

Results provided by WolframAlpha

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Requested by Plegasus#2179

[keen cave]

,w sqrt(3) -2arctan(sqrt(3))

[timid silo]

i think it should be -1-2arctan(-1)

[warm shale]

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Requested by Plegasus#2179

[keen cave]

Yep the result is different. Now you absolute minimum is at x = 1 not sqrt(3).

Always use radian unless your teacher says otherwise.

[timid silo]

yeah so the absolute max should be at x = -1 and the absolute min at x = 1

am i right?

[keen cave]

Wait

[timid silo]

just before i close it

[keen cave]

Nvm

Sorry.

[keen cave]

Yep.

[timid silo]

@keen cave thank you so much. you're the best

[keen cave]

Remember to close both channels. Np.

[timid silo]

.close

[obtuse pebble]

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[hazy sorrel]

The coordinates of the point C are C(5.5), and the coordinates of its image resulting from a rotation of 100° degrees around a certain point are C'(-5, 7.5) , find the coordinates of the center of rotation. Explain your answer.

[obtuse pebble]

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[tranquil kiln]

Jenny is going to attend a sports camp for 7 days. Each day, she will play exactly one
of three sports: hockey, tennis or camogie. The only restriction is that in any period of 4
consecutive days, she must play all three sports.
Find, with proof, the number of possible sports schedules for Jenny’s week.

[tranquil kiln]

how do i even start

[obtuse pebble]

@tranquil kiln Has your question been resolved?

[tranquil kiln]

<@&286206848099549185>

[winter steppe]

bump

i would break it down like this

i'd first find how many possible schedules there could be in 3 days with one sport each, to get the requirement out of the way

then figure out how many possible schedules for the remaining day alone

the remaining day can be either 1st, 2nd, 3rd or 4th. take that into account when calculating everything

[tranquil kiln]

yh i tried these steps but i don't know how to calculate i thought of trying Bernoulli trials to try to calculate the 3 days with one sport and rest but i don't think it works i am bad at possibilities.

[obtuse pebble]

@tranquil kiln Has your question been resolved?

[winter steppe]

it's not bernoulli trials

it's just combinatorics @tranquil kiln

[tranquil kiln]

oh i tried for (3 days with one sport each) = 3C1 = 3 , for one day than x by 3 3x3=9 for the requirement. possible sport for the remining day 4x3= 12 total so (12C3 of the sport )=220 . so total for the week 220+9=229 ??

[winter steppe]

close but not exactly

think of a bookshelf and 3 books; how many possible sequences are there?

that will help you calculate the first 3 days

[timid silo]

<@&286206848099549185>

[tranquil kiln]

ohh 3! right so 6 so 6C3=20 total is 20+220=240?

[winter steppe]

@timid silo this channel is occupied; go to one of the available help channels

why 6c3?

3c3 = 3! = 6

yes

keep that

now how many possible ones do you have for the remaining day?

3 * 1 = 3

now; the last day can be either 1st 2nd 3rd or 4th

which means 4 scenarios

do you get the picture now?

[tranquil kiln]

beacuse each day you play one sport so 3 day is 3 the total scenario is 6 so 6C3

[tranquil kiln]

3x6=18 ? last day 4 scenarios = 4! 24 you chose 3 sport = 2024+18=2042?

[tranquil kiln]

.close

[obtuse pebble]

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[umbral kayak]

heya

[umbral kayak]

the question isn’t done but

im pretty sure I did this wrong

like at the part where I make it 0

what am i actually supposed to do?

[obtuse pebble]

@umbral kayak Has your question been resolved?

[umbral kayak]

<@&286206848099549185>

[obtuse pebble]

@umbral kayak Has your question been resolved?

[timid silo]

@umbral kayak

[umbral kayak]

yes

[timid silo]

Oh you are

What are you trying to do exactly?

[umbral kayak]

i dont know what this is called in english

but i have done the derivatives down so far

now i have to try to get it to 0

[timid silo]

Oh right

The derivative of 12 is 0

So f''''(x)

[umbral kayak]

whaaaaaaaa

[umbral kayak]

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[short spruce]

• After 15 minutes, feel free to ping @Helpers.

[short spruce]

.

even then you didn't wait 15 minutes lmao

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

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@timid silo Has your question been resolved?

[tardy epoch]

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[deep imp]

Looking at review and just don't understand how this equations transforms the way it does, anybody get this?

[warm canopy]

Which step

[deep imp]

all to be honest

where does the whole e part go in the second point

and then why does it come back again lol

nevermind just gonna skip this for now

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[reef pond]

what would be a good way to approach this problem

[reef pond]

I tried splitting it into sin/cos

and then using double angle

but my result was root2-root6/root2+root6

[fierce lagoon]

Half angle or double angle

Personally, I would do half angle

[reef pond]

okay cool one sec

okay got it

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[fossil ether]

Hi, how to prove the image question?
Thanks.

[timid silo]

It says to not prove it

show that it is not true in general

[fossil ether]

Yes but in logarithm we know for the left side power k will come in-front of log so that is true.

[brave bramble]

You're thinking of
log(n^k) = klog(n)

[timid silo]

difference of log(n^k) vs (log(n))^k

[fossil ether]

Yes

[brave bramble]

Note for this question we instead have
(log(n))^k on the left side

[fossil ether]

Ok, then

[obtuse pebble]

@fossil ether Has your question been resolved?

[fossil ether]

Still I'm confused, how it is not true in general?

[forest sinew]

what have u tried

[fossil ether]

Just confused how to prove it

[forest sinew]

you cant prove it, its not true

just give a counter example

[fossil ether]

But in general, log n ^ k is k log n

[forest sinew]

write that again but with parens

then compare to the problem statement

[fossil ether]

(logn)^k = klogn

[forest sinew]

the slightest amount of googling could resolve your confusion

[fossil ether]

Would you like to provide me a reference URL where I can read this issue and clear my confusion?
I think it is related to exponential and parens.

[forest sinew]

oo this one conveniently leaves off the parens

[fossil ether]

I know this mate 🙂 could you prove the problem with rule number 3?

[forest sinew]

no, you cant

because rule 3 is true

but the statement in your problem is not the same as rule 3

[fossil ether]

Hmm

[forest sinew]

your problem is asking you to recognize that $\log \qty (b^k) \neq \qty( \log (b) )^k$

[warm shale]

jan Niku (join us for @pomo)

[fossil ether]

Hmm that is right, log (n)^k is not equal to k log n

[forest sinew]

yup

so think of a counter example

[fossil ether]

So normally, (logn)^k is not equal to log(n)^k?

[fossil ether]

Can we prove it with a real number like say here k = 2 and n = 3

[forest sinew]

👀

im still not convinced you understand what the notation means

your problem is saying to recognize:

$\log (b) \cdots \log (b) \neq \log \qty( b \cdots b )$

[warm shale]

jan Niku (join us for @pomo)

[fossil ether]

So with this case we can say (logn)^k is not equal to logn^k.
So only we can say that if it was logn^k then we can write the right side expression is klogn
So in general it is not true 🙂

[forest sinew]

you usually come up with a specific example

but youre right

its only true in specific circumstances

but in general we dont expect it to be true

[fossil ether]

How to make this channel to resolve?

[forest sinew]

.close

[fossil ether]

.close

[obtuse pebble]

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[stable sedge]

what are rational / irrational periods in functions

[stable sedge]

my guess would be rational or irrational number ? after which it repeats

but i am not sure

also irrational period is further divided into Like and unlike 🤔

[obtuse pebble]

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[coral parcel]

Hey there I was wondering does this require simeltaneous equations to find the a and b value?

and if so how is it done

<@&286206848099549185>

[compact shadow]

The condition tells you that

P(1)=2 and P(-1)=-4

This gives you a system of linear equations of a and b you can solve them now

[coral parcel]

so then i use simeltaneous

?

[compact shadow]

What?

You understand what I said?

P(1)=2 so a+b=4

[coral parcel]

ahhhh yep

[compact shadow]

P(-1)=-4 so a-b=0

Solve a,b

[coral parcel]

gotcha

yep so a=2 and b=2

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[coral parcel]

cheers cogwheels

[obtuse pebble]

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[wanton dagger]

I need to find x here

[slim lake]

What's the formula for the area of a circle?

[wanton dagger]

What I did:

But answer is 18

[lucid flame]

if you're comparing the ratios of 2 areas, then the lengths must be squared

[slim lake]

whats the formula for an area of a circle?

[wanton dagger]

Pi * r^2

[slim lake]

what is r in this case?

So $r = x, \pi x^2 = 324\pi$ then we need to solve for x

[warm shale]

Max..

[wanton dagger]

Ok

[slim lake]

solving that and x = 18

[wanton dagger]

What did I do wrong here though?

Yh I forgot to put the 2 but ans is still same

[wanton dagger]

I get this way but I just don’t understand what’s wrong here

For our sides I used x and 6(the radius)

[slim lake]

Not sure but if the radius is 6 the area of 1 isnt 144pi

Thats why I didnt use the first circle cuz it looks like they labelled the diameter

so radius would be 6 and then ye what u put

but if the radius is 6 then that is not the area

[wanton dagger]

Why?

Isn’t it the ratio of the area of the side length and it’s area

[slim lake]

well $A_1 = \pi r^2, r = 6 \Rightarrow A_1 = \pi \cdot 6^2 \Rightarrow A_1 = 36\pi$

[warm shale]

Max..

[slim lake]

and if it isnt a proper circle then u cant use this method as it's only for similar shapes

[wanton dagger]

Ok

[slim lake]

$\frac{x^2\pi}{12^2\pi} = \frac{324\pi}{144\pi}$ should work

[warm shale]

Max..

[wanton dagger]

But why did we use the 12 (the diameter) and x(the radius)?

This does work but that’s what I don’t understand

[obtuse pebble]

@wanton dagger Has your question been resolved?

[wanton dagger]

Anyone

[cedar lichen]

12 is the radius. The image is a little misleading

[wanton dagger]

Oh that’s why

Right

Got it

Thanks

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[wanton dagger]

For part e) how would I find the range algebraically?

[wanton dagger]

f(g(x))=sqrt(4-x^2)

[lucid flame]

find the domain and then bound f(g(x))

[wanton dagger]

What do you mean by that

I have the domain

[-2,2]

[lucid flame]

so -2 <= x <= 2

0 <= x^2 <= 4

and so on

[wanton dagger]

Hmm.. I feel confused

Someone told me to make y the subject and test different values for it to see which values make it not true, something like that but I can’t exactly remember that method

[novel knoll]

Confused about? How did you find range in the other questions

[wanton dagger]

Completely forgot

I think I tested different values

But then I got confused in this question with the range and I forgot

[novel knoll]

Testing random values doesn’t sound very “algebraically” to me

[lucid flame]

you can't test random values

unless you're finding the domain/range of a relation

[wanton dagger]

Yh then I forgot

[lucid flame]

in that case, you're just testing to see which side of the graph satisfies the relation

[novel knoll]

You were told here how to do it

[wanton dagger]

I don’t understand it

[novel knoll]

If -2<=x<=2 then 0<=x^2<=4, agree?

[wanton dagger]

How?

[novel knoll]

-2<=x<=0 => 0<=x^2<=4, agree?

[wanton dagger]

Ok I got it

Thanks

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[timid silo]

for this question

im getting -1-(0/-2)

idk what im doing wrong

using `x2 = x1- f(x1)/f'(x1)

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

😢

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185> sorry for the double ping, just have not got help

[chilly peak]

is -1 not the answer?

[timid silo]

im stupid, I was typing in 1 for some reason

sorry!

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[vapid oak]

Hello there, am I doing this question right

[obtuse pebble]

@vapid oak Has your question been resolved?

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[cyan pagoda]

what do you need help with

[somber jackal]

how to do these with steps?

part b and c

<@&286206848099549185> i think i can ping? 15min passed

[novel knoll]

Split up in cases if 1st letter is there only once or several times in the word

[obtuse pebble]

@somber jackal Has your question been resolved?

[somber jackal]

how to do the steps?

[novel knoll]

consider the two cases I said

its just application of law of total prob

[obtuse pebble]

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[teal harbor]

Hello how to find area of triangle using trigonometric stuff?

[teal harbor]

İ made smthng like this but idk

[fierce lagoon]

Wdym

It's just

$\frac{1}{2}ab\sin{\left(C\right)}$

[warm shale]

Umbraleviathan

[teal harbor]

@fierce lagoon ok but how do you use sin if theres no 90°

[fierce lagoon]

Don't need to

It's one of the area formulas

And you would use a calculator

[teal harbor]

You just put the value of sin depending on angles value then right?

[fierce lagoon]

Well yeah

[teal harbor]

.close

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[quiet kraken]

.reopen

.open

[fading stirrup]

how to prove "there does not exist a periodic sequence {an},where all terms are +1 or -1 , to every rational number theta"

[fading stirrup]

so i think we should find a theta , to every sequence, the sum is diverge

i denote the period of {an} is k ,and let theta =1/k

so e^(2 pi i n/k) is nth root of 1-x^k=0

and the common period of this sum is k

so i use the taylor expansion to rewrite the sum as

since 1-x^k is 0 here, the expansion is not converge

but how can i prove the former term is not 0?

or are there other methods to prove this one doesn't converge?

[alpine raven]

maybe you can do it with a contradiction proof ? 🤔

[fading stirrup]

how to make a contradiction

[alpine raven]

like the famous proof of sqrt2 is irrational, you say that there exists a periodic sequence where all terms are 1 and -1 and for every rational number theta, sup of the series diverges

I think thats the contradiction 🤔

[fading stirrup]

emmmm i tried

[alpine raven]

and what did you get ?

[fading stirrup]

may be i did wrong and i cant find more through this contradiction

[alpine raven]

well if said periodic sequence exist, lets a_n = (-1)^n = exp(inpi)

[haughty coyote]

I don't think so.
We're proving $\forall (a_n), \exists \theta$ such that that sup is infinite. That would be a direct proof.
A proof by contradiction would assume the existence of a sequence such that for every theta, the sup is finite and have a contradiction, not that the sup always diverges like you said

[warm shale]

themateo713

[alpine raven]

oh yea you're right, my bad

I had contrapositive proof mixed with contradiction proof in my mind

[fading stirrup]

but it is more difficult to prove "every" rational theta

i think

[haughty coyote]

I think contrapositive (and I also think this is weird) would be that if for every theta this sup is finite, one of the a_n is not 1 or -1, or a_n is not periodic

which sounds very useless

[alpine raven]

indeed

contradiction must be the best option

So let's assume the existence of a period sequence {a_n}

and a_n = (-1)^n

[haughty coyote]

you don't just assume the existence of something and then set its value immediately

[fading stirrup]

i try to use my method but honestly, i dont learn any about complex analysis

[haughty coyote]

for any theta since it's rational the sequence of exponentials has a period. The product of 2 periodic sequences is periodic. That sum goes to infinity iff the sum of the product over its period (or any multiple of it) is nonzero. So you go from a series to a finite sum

this is purely algebra, so no analysis should be involved

[fading stirrup]

yeah i also did that

but i don't know how to prove the sum of each nth roots and some are with different signs is nonezero

[fading stirrup]

actually that is what i asked

[haughty coyote]

I feel like it doesn't have to be though. If you take n multiple of 3, then 1, j, j² are included (the 3rd roots of unity). The sum of the n roots is 0, now subtract 2 +2j + 2j². Since 1+j+j² = 0, the sum is still 0 but you have a bunch of +1 and 3 of them are -1

so unless I'm wrong, there should be sums where it is still nonzero

[fading stirrup]

well…but what about nth?

may be some of the nth roots without same signs can still have the sum=0

[haughty coyote]

but you agree that this disproves that ?

[fading stirrup]

wait,but that makes the period becomes 1/3 of what you set

[haughty coyote]

I don't see how it would. Or at least I'm missing something, because to me there are still as many terms

[fading stirrup]

the sum of nth roots means you set the coefficient of each terms with +1

and you subtract 2,2j,2j²

so that makes these three terms with -1

eg.if n=9

what you did is to let the sequence becomes

1 1 -1 1 1 -1 1 1 -1

which.is actually witg period 3

one third of n,where n is initially 9

[haughty coyote]

it is still a sequence of period 9, it is just no longer the smallest period. I don't think we ever stated it had to be minimal

[fading stirrup]

yes so i tried to escape that case and let theta equals to 1/k,where k is the 'true'minimal period of sequence

otherwise that makes this method useless due to the existance of your exanple

[haughty coyote]

I do think your idea works, I'd like to believe in it but I don't know how to prove it rn

<@&286206848099549185> is it true that if you have a strict subset of the nth roots of unity, then their sum is nonzero ?

[tardy epoch]

no

$1 + e^{i\pi/4}$

[warm shale]

riemann

[lean wigeon]

only if n is prime, I think, eg with 6, just take every other one

[haughty coyote]

that is nonzero, but we were wondering if it could ever be 0 ?

[tardy epoch]

i + (-i)

[lean wigeon]

if you take the 1st, third, and 5th 6th roots of unity, they will sum to 0

[haughty coyote]

I feel like this is not going to work

[tardy epoch]

i'm sure there's a better geometric proof

[lean wigeon]

if you are depending on 9, it wont work

[tardy epoch]

or something with groups. groups are the shit

[haughty coyote]

I phrased it poorly, but the idea was to avoid a set that is "all the nth roots of unity for a certain n", but your other example circumvents that

never seen helpers respond that fast as well, wow

[lean wigeon]

I think its still equivalent to the third roots of unity, eithet that or then just shifted by 1. Look up test tube balancing and you can find a better example.

[haughty coyote]

so if you have any idea for the original problem feel free to have a look, personally I feel like I'm a bit out of ideas

[fading stirrup]

but i cant find other method😫

[lean wigeon]

Can you say the problem again? Im not sure I understand how it is said originally

[fading stirrup]

how to prove "there does not exist a periodic sequence {an},where all terms are +1 or -1 , to every rational number theta"

[haughty coyote]

I don't think he wanted you to just copy-paste it

[lean wigeon]

wdym "to every rational number theta"?

[fading stirrup]

yes

[lean wigeon]

I dont understand what you mean by that.

[fading stirrup]

because ¬（∃{an},∀theta,sup of sum < inf)

is the same to

∀{an},∃theta, sup.of sum diverge

and i think.the second statement is easier to prove

[lean wigeon]

What im asking is, what is the purpose of theta? You never define what to do with it.

did you mean "for every rational number theta"?

[tardy epoch]

@fading stirrup just show the original problem rather than your interpretation of it

[fading stirrup]

but the original problem is in chinese……

sorry my english is poor

[fading stirrup]

yes,for every rational number theta,originally.

and i use the contrapositive

[obtuse pebble]

@fading stirrup Has your question been resolved?

[fading stirrup]

thanks for everyone's suggestions and i will keep thinking this question.

[obtuse pebble]

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[austere knoll]

hey guys so I am slightly confused why these two linear non-homogenous SODE questions have different methods

[austere knoll]

This is the first question where we had to use Ae^-3x as the particular solution

I thought it was the same as this which is why in my first attempt at the other question I used xAe^ax as the particular solution

However in the first question you don’t use XAe you just use Ae

Why is this

<@&286206848099549185>

[tardy epoch]

depends on repeated root

https://tutorial.math.lamar.edu/classes/de/RepeatedRoots.aspx

[austere knoll]

okay thanks so much will research that!

[tardy epoch]

but it doesn't look like either of the ODEs have repeated roots, so i'm not sure

[austere knoll]

after asking friends they say its to do with the fact that c1e^2x in the first question woudl have the particular solution Ae^2x as well

which would mean that it would = 0 so you have to use xAe^2x as the particular

but i dont fully understand this concept or what this scenario is called

[obtuse pebble]

@austere knoll Has your question been resolved?

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[obtuse pebble]

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[oak pewter]

Can someone help find the function of f(x) = tan x. this is what I have so far and the question.

[oak pewter]

i am not sure if i am asking a good question. i heave already read the rules, but i am not sure why i dont get a respone

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[obtuse pebble]

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[dreamy gale]

Super confused, is the period 2pi/B or is it |B|?

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[obtuse pebble]

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[warm canopy]

@dreamy gale hopefully you closed this because you noticed how the two equations are slightly different, hence the periods being labelled differently

[dreamy gale]

yes

2pi/(2pi/B) would give us B

[warm canopy]

yep nice one

[obtuse pebble]

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[timid silo]

All I need is the first 4 numbers of the sequence

[warm canopy]

$a_2 = 2a_1$

[warm shale]

iCaird

[warm canopy]

see if you can write similar equations for $a_3$ and $a_4$

[warm shale]

iCaird

[timid silo]

6

18

72

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[obtuse pebble]

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[cinder schooner]

So uh

Idk if I got this one right

All I know is that theta=s/r

<@&286206848099549185>

Anyone?

I guess they all sleep

[fierce lagoon]

Lemme check

No

[fierce lagoon]

Find the circumference of the circle

67.85 radians is a ... it's a concerning number

Since a full circle is 2π radians

67.85 is almost 10 full revolutions lmao

[cinder schooner]

Oh she

@fierce lagoon I got 56.54, as the circumference

Now what

[fierce lagoon]

Well

Keep it in terms of π

[cinder schooner]

k

[fierce lagoon]

This is gonna make things a lot easier

[cinder schooner]

I put the pi symbol next to the number right?

[fierce lagoon]

Usually, yeah

[cinder schooner]

ok

[fierce lagoon]

It doesn't really matter

[cinder schooner]

😅

[fierce lagoon]

But preferably after

[cinder schooner]

Now what do I do?

[fierce lagoon]

Well what's the circumference

[cinder schooner]

56.54

[fierce lagoon]

I meant in terms of pi

Don't give me a decimal

[cinder schooner]

Oh

[cinder schooner]

Hmm idk man

254?

[fierce lagoon]

Well the formula for the circumference of a circle is $2πr$, correct?

[warm shale]

Umbraleviathan

[cinder schooner]

Yessir

[fierce lagoon]

So just treat π is a costanr, plug in 9

You'll get 18π

[cinder schooner]

oh alr

[fierce lagoon]

You understand?

[cinder schooner]

OH

yeah now I get it

[fierce lagoon]

Yea

[cinder schooner]

It’s so easy now

I never thought of it that way

So when it means in terms of pi

I just have to do that?

For every question

[fierce lagoon]

Well yeah for the circumference

But for now, let's focus on this question

The angle of a sector will be proportional to the length of the sector to its circumference

[cinder schooner]

Oka

[fierce lagoon]

But we need to know what the angle of the sector will be proportional to

So what is the radian of a full circle

[cinder schooner]

I haven’t learned much about radians 😥

[fierce lagoon]

Alright

180° = π radians

So what would a full circle be

[cinder schooner]

360

2pi

?*

[fierce lagoon]

Yes!

[cinder schooner]

Ok

[fierce lagoon]

2π is a full circle

[cinder schooner]

Oh

Okay

[fierce lagoon]

That means that $\frac{θ}{2π}=\frac{\left(\frac{12π}{5}\right)}{18π}$

[warm shale]

Umbraleviathan

[cinder schooner]

Bro…

But idk what the angle is tho

[fierce lagoon]

You can solve for the angle

What would you do to both sides to isolate θ?

[cinder schooner]

Cancel the 2pi

Put it on the other side?

Wait but would that make it a negative?

[fierce lagoon]

Well 2π is in the denominator

So what can you do to remove the denominator

[cinder schooner]

Divide both sides?

[fierce lagoon]

Ehh, I mean technically? But then you would be dividing by a fraction, which is just a multiplication

[cinder schooner]

oof

Hmm 🤔

[fierce lagoon]

So multiply both sides by 2π

[cinder schooner]

What about multiplying

[fierce lagoon]

Well yeah

[cinder schooner]

That would cancel out 2pi

[fierce lagoon]

It would

You could also, as you said, divide by 1/(2π) but that's the same as multiplying by 2π

[cinder schooner]

True

[fierce lagoon]

That means that $frac{\left(\frac{12π}{5}\right)}{18π}=\frac{12π}{5}\cdot{1}{18π}=\frac{12}{80}$

[warm shale]

Umbraleviathan

[fierce lagoon]

Now I would keep your answers in terms of π

Let's see what you get

[cinder schooner]

0.83?

[fierce lagoon]

No

If you want a decimal answer, just plug it into a calculator

[cinder schooner]

Ok

[fierce lagoon]

But I'd rather keep it in terms of π for the exact answer

[cinder schooner]

46.93pi?

[fierce lagoon]

No

Let's break it apart

[cinder schooner]

All I got left with is 0.83

[fierce lagoon]

Wait where did my message go

Bro did it void it

Bruhhhh

[cinder schooner]

I think so

[fierce lagoon]

I gotta retype it skfhekfjdkdkf

[cinder schooner]

e

[fierce lagoon]

$frac{\left(\frac{12π}{5}\right)}{18π}=\frac{12π}{5}\cdot{1}{18π}=\frac{12}{80}$

[warm shale]

Umbraleviathan

[fierce lagoon]

What the fuck

[cinder schooner]

Weird

[fierce lagoon]

It's breaking

The bot is breaking

[cinder schooner]

Oh noes

[fierce lagoon]

$\frac{\left(\frac{12π}{5}\right)}{18π}=\frac{12π}{5}\cdot\frac{1}{18π}=\frac{12}{80}$

[warm shale]

Umbraleviathan

[cinder schooner]

What the

How was I supposed to-

[fierce lagoon]

So we go back to fraction division

Using your multiplication-division rules

And then basic/pre algebra stuff

[cinder schooner]

k

[fierce lagoon]

Is there anything you'd want me to clarify?

[cinder schooner]

Idk how you got all that

12/80

I don’t remember my teacher teaching me this

._ .

[fierce lagoon]

Okay so basically what a fraction is, A/B, is A divided by B

[cinder schooner]

okay

[fierce lagoon]

So I'm gonna get A = 12π/5 and B = 18π

[cinder schooner]

Kk

[fierce lagoon]

A/B = A divided by B = A • (1/B)

[cinder schooner]

Oh ok

[fierce lagoon]

Thus, this

So that takes care of the fraction in a fraction

So now

$θ = 2π\cdot\frac{12}{80}$

[warm shale]

Umbraleviathan

[cinder schooner]

Ohhhh

[fierce lagoon]

Which shouldn't be too hard

[cinder schooner]

I got a decimal 0.94

[fierce lagoon]

Yeah

[cinder schooner]

Now wut

[fierce lagoon]

That's your answer

[cinder schooner]

._ .

Finally some sleep~

[fierce lagoon]

0.94 radians, or 0.3π radians

Lmao yes

Get some sleep

[cinder schooner]

Jesus how am I gonna do all this for the test on Friday

[fierce lagoon]

You're gonna need a lotta practice

[cinder schooner]

She only rushed us for 2 weeks instead of the whole month ;-;

Anyways thanks a lot for the help

Gn man

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[obtuse pebble]

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[fickle lantern]

mistake, this should be 12/90

[fickle lantern]

wait what

oops

@cinder schooner not sure if this is important but just wanted to point that out

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[obtuse pebble]

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[obtuse pebble]

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[woeful cove]

The question is from physics but it's based on proportionality can someone help?

[timid portal]

There was a survey on vacation plans of students. Every 3rd student said they're staying home. The remaining 40% said they're going south and 20% north. The rest of the 8 students in the class didn't say anything. Find out how many students there are in the class.

[visual onyx]

make a new channel

don't use others

[timid portal]

Ok

[woeful cove]

@timid portal pls paste this in other channels

We are given 2 equations
P=v²/r

P=i2r

Is p∝r
Or
P∝1/r

[warm shale]

||𝓚𝓪𝓼𝓹𝓮𝓻||

[woeful cove]

No

P≠2ir

[visual onyx]

why?

P=i2r

oh

p=i^2r

p= i squared r

[woeful cove]

Yes

[visual onyx]

,tex $\frac{v^2}{r} = i^2 r$

[warm shale]

||𝓚𝓪𝓼𝓹𝓮𝓻||

[visual onyx]

cause p=p

[woeful cove]

Yes

V=ir

[visual onyx]

but what do you want to find

ohm's law

ic

[woeful cove]

Noo

[visual onyx]

one of the ways to express ohm's law is v-ir

v=ir

[woeful cove]

I wanted to find whether P is directly or inversely proportional to R

Someone told me the answer

[visual onyx]

ic

[woeful cove]

When we take i as constant, p is proportional

[visual onyx]

yes

[woeful cove]

When we take V as constant, p is inversely proportional

I was having trouble with how p is inversely and directly proportional can coexist together

Hahaha

Thank you

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[obtuse pebble]

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[west sierra]

For an object projected with speed u, the horizontal range is two times the maximum height attained by it. The horizontal range of the object is?

[west sierra]

a) 2u^2/3g, b) 3u^2/4g c) 3u^2/2g d) 4u^2/5g

The answer I came with matches neither of these options..

My answer was 2u^2/25

Formula of range is u^2 sin2theta/g and formula for maximum height is u^2 sin^2theta/2g

I equated them by multiplying 2 on RHS and I got the relation 2costheta = sintheta