help-23

[low vector]

but you can also just fill it in

but its gonna be a bit more writing

[tiny gorge]

ah ok

[low vector]

if we have the area and replace x with what you wrote

what do we get

[tiny gorge]

$A = \frac{2r(18-r(\pi+2)) + \pi*r^2}{2}$

[flat frigate]

marco.py

[low vector]

write it out a bit further

all the way so that you dont have any more brackets

[tiny gorge]

ok

$A = \frac{36r - 2r^2*\pi+4r^2+\pi*r^2}{2}$

[flat frigate]

marco.py

[low vector]

combine like terms

such as r and r^2

otherwise we still have a quite ugly function

[tiny gorge]

should i factorize r out

[low vector]

its not necessary, but you could if you like

[tiny gorge]

ok

so i got

$A = \frac{36r+ (-\pi+4)r^2}{2}$

[flat frigate]

marco.py

[tiny gorge]

now we can differentiate easily

[low vector]

exactlty

we get the maximum

by taking the derivative

[tiny gorge]

so

$A' = 18 + (-\pi + 4)r$

[flat frigate]

marco.py

[low vector]

what does this need to be equal to

[tiny gorge]

0 or DNE

but 0 here

[low vector]

exactly

so set A' equal to 0

and solve for r

[tiny gorge]

but should we determine the range for r yet?

im getting r to be a negative number

[low vector]

not yet

then you did something wrong in a step before

i see it

its when you simplify A

you made a mistake

look at it again

[tiny gorge]

ah ok

$A = \frac{36r + (-\pi - 4)r^2}{2}$

[flat frigate]

marco.py

[tiny gorge]

yeah

so $A' = 18 + (-\pi - 4)r$

[flat frigate]

marco.py

[low vector]

wait actually

one second

[low vector]

this is the correct one

so continue from there

[tiny gorge]

yeah

so i set A' to 0

found r = 2.52

[low vector]

now we have an r

then we check

if this satisifes

so

with r=2.52

does our diagram make sense?

[tiny gorge]

yes i think its a plausible value

[low vector]

then are we done?

[tiny gorge]

yes but we dont need to check the range for r?

like $0 <= r <= something$

[low vector]

what doy ou mean by range?

[flat frigate]

marco.py

[low vector]

the range can be infinite

but we are stopped by the constraint of the perimeter

[tiny gorge]

yes

[low vector]

which is what we used

to write x(r)

[tiny gorge]

yes

so it doesnt matter

for this case

for some cases it matters

cuz like sometimes we have quadratic

and we have to eliminate some values based on range

ok that makes sense

Thank you very much!!

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[limber pine]

inequality unit

[limber pine]

what do i do here?

[grand kraken]

Graph the function on Desmos and try understanding what the graph is showing first

[limber pine]

i already put it on desmos

oh wait

but i wont have desmos on a test. so how do i solve this algebraically?

[grand kraken]

I just want to make sure you understand what's going on first, so you can tell what the answer is approximately, right?

[limber pine]

yes @grand kraken

[grand kraken]

Ok, so now formalize what you can see: set the equation <= 8 and see what positive x-values satisfy that

since the graph starts at 8, and then decreases, you just need to find the second time it reaches 8 to find the full range of values

[limber pine]

can you elaborate on that?

so i should find the y value when x = 8?

[grand kraken]

opposite, x value when y = 8

so set c(t) = 8 and solve

[safe radish]

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[twin prawn]

how do you expand something like $$4^{\frac{1}{4}}$$

[flat frigate]

Maladroit

[plucky elk]

a * 1/b = a/b

[twin prawn]

so 4/4 here?

[plucky elk]

Oh you edited

No

[twin prawn]

here

oh

[plucky elk]

2^2 = 4

[twin prawn]

tes

yes

[twin prawn]

yo

[safe radish]

@twin prawn Has your question been resolved?

[plucky elk]

Did you use that in your expression

[twin prawn]

wdym

[safe radish]

@twin prawn Has your question been resolved?

[plucky elk]

You have a couple 4s here

[plucky elk]

^

[safe radish]

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[void onyx]

Circle x^2+y^2=25
There is a triangle PQR jn it where corrdinates of Q(3,4) and R(-4,3) then find angle QPR will be?

[grand kraken]

What do you mean by "find QPR will be?"

[void onyx]

Angle*

[quiet plume]

What's P? If you don't have more information there's infinitely many triangles with vertices Q and R on the circle.

[timber moss]

^

[void onyx]

Yes

It said triangle is inscribed in circle

[quiet plume]

Doesn't help

There are infinitely many triangles still

P could be any point on the circumference that aren't Q and R.

[safe radish]

@void onyx Has your question been resolved?

[void onyx]

Then we can draw any triangle and use cos angle formula no?

[grand kraken]

Post a picture of the original question

[void onyx]

It is not in English

[grand kraken]

it's ok

[void onyx]

Do you understand Hindi?

[grand kraken]

nope, but maybe it will make what P is more clear

[safe radish]

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[winter path]

got a weird calculus question that i've been struggling with for a while, I can't make any progress cause I dont know where to substitute t into the equation

[safe radish]

@winter path Has your question been resolved?

[winter path]

<@&286206848099549185>

[low vector]

what do you mean substitute t?

t=0 is just your initial value

because wheny ou have a 1st degree differential equation youre gonna get 1 unknown, thus giving you infinite solutions

but having one initial condition means you only have 1 solution

[winter path]

so if i dont need to substitute t anywhere, does that mean i just group the variables then integrate the V?

[atomic drum]

use variable separation to solve the differential equation

then use the data that v=0 at t=0 to find the constant of integration

[winter path]

Hoping im good so far

I believe the first integral should become -ln(V-E)

But the 2nd one I'm not sure

Maybe t/CR ?

[atomic drum]

-ln(E-V)

[atomic drum]

yess

t/cr +k(constant of integration)

[winter path]

they dont swap?

[atomic drum]

no

[winter path]

cause when i looked up the integral of 1/(1-x) it said it's -ln(x-1)

[atomic drum]

no

it will be -ln(1-x)

-ln|x-1| is better

in this case, V will be less than E, hence to satisfy domain of log, we will accept -ln(E-V)

find the value of k using v(0)=0

[winter path]

how do we find k when E, C and R are still unknown?

or do i just sub in the values from the particular solution?

i suppose that would make sense since k is constant

[atomic drum]

yes

find k in terms of E c and r then substitute the values when finding v(2)

either way you'll get the answer

[winter path]

ah ok that makes more sense

[winter path]

alright finally got it done

thanks for everything

.close

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[whole wasp]

Help please

Tried pairing up the terms
Could figure out a suitable substitution for cos^2

[robust river]

How is there an 11

[whole wasp]

To annoy us

[robust river]

I don't think it's even possible

[whole wasp]

We have tried many over past few days
You know brother it became possible somehow

[whole wasp]

N in answer is 22
So it must be n/2
Now how to relate that is the issue

[robust river]

Oh wait I messed up 1 calculation in my head

[final bay]

what the hell is this question

[robust river]

Yeah okay

There might be an answer then

[final bay]

i would love to know the solution to this

[robust river]

The product can be simplified

Then solve for n

The main task is to simplify the product

[whole wasp]

Yeah that is so

Is there some product splitting idea

?

[robust river]

I'm currently trying to take advantage of cos(x)^2 = 1/2 (cos(2x) + 1)

I liked that 1/2^10 appeared after that

[whole wasp]

Oh that's interesting

I converted to sin product

But then the number of terms in numerator are big and even
And denominator is big and odd

[robust river]

I found a fun observation

[haughty oasis]

I don't know if this is helpful but if r/n is an integer it'll equal 0

wait no

But is something like that

It'd be 1

Like cos^2(rpi/n)=1 is r/n is an integer

and r is always an integer

[robust river]

So after this substitution we have $\Pi_{r=1}^{10} (\cos{\frac{2\pi r}{n}} + 1) = \frac{11}{2^{10}}$
And then I thought what if we want to group first factor with the last factor? For example, if $cos{\frac{2\pi}{n}} = -cos{\frac{20\pi}{n}}$, then we will get $(1 - (\cos{\frac{2\pi}{n}})^2)$ womewhere in the product. Funny thing is that n=22 will let us do that

[flat frigate]

EQUENOS

[robust river]

So checking this equality for n=22 is reasonable

We will get $\Pi_{r=1}^5 ( 1 - (\cos{\frac{\pi r}{11}})^2 )$

[flat frigate]

EQUENOS

[safe radish]

@whole wasp Has your question been resolved?

[robust river]

Sorry I messed up my tex code

I meant cos^2

Oh wait maybe we can use Chebyshev polynomials and Vieta formulae

[safe radish]

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[whole wasp]

.close

.reopen

[safe radish]

✅

[whole wasp]

Went to dinner

[whole wasp]

How can we do that?

[whole wasp]

That's some good hit there
But it ain't solving right?

[robust river]

right

[robust river]

Unfortunately this doesn't seem to work

[tiny wraith]

n = 22 seems to work according to desmos, I tried applying the $cos^2$ identity again btw and got this $$\frac{11}{2^{5}}=\prod_{r=1}^{5}\left(1-\cos\left(\frac{2r\pi}{11}\right)\right)$$

[flat frigate]

Jelle

[tiny wraith]

this needs to be proven though

[robust river]

My idea was that Chebyshev polynomials have roots equal to cos(pi (2k+1) / (2n)) for k=0, ..., n-1
And its derivative has roots at cos(pi k / n) for k=1 ,..., n-1
On the other hand, by Vieta formula we know that the product of roots is something like (-1)^n a_0 / a_n

Where a_n is the coefficient before x^n and a_0 is the coefficient before x^0

But unfortunately we can't apply this knowledge in our case

Or does it?

Oh wait it does

[whole wasp]

Howwww

I am unable to catch up honestly

[plucky elk]

You really don't need chebyshev polynomials

[robust river]

Let $T_{22}(x)$ be a Chebyshev polynomial of degree 22.
Then $T_{22}'(x)$ will have the following roots: $\cos{\frac{\pi k}{22}}$, where k=1, ..., 21
Note that one of the roots is zero. We don't want that, so we divide $T_{22}'(x)$ by $x$. The resulting polynomial has the same non-zero roots and they're all symmetrical. So the square of their product is just the product of all positive roots to the 4th power, because all negative roots are just a mirror image of the positive roots.
On the other hand by Vieta formula we know that the squared product of all roots is $\frac{a_1^2}{n^2 a_n^2}$

[flat frigate]

EQUENOS

[whole wasp]

There should be a simple way
These look slightly advanced

[robust river]

All positive roots are cos(pi / 22), ..., cos(10 pi / 22)

[robust river]

Yeah

Sorry I saw Chebyshev polynomials and couldn't unsee them after that

[whole wasp]

What do you suggest?

[whole wasp]

Should I open a new channel xD

[plucky elk]

This should work

[robust river]

This also worked

Here $a_1 = T_{22}''(0) = (\cos(n \arccos{x}))''$ at 0

[flat frigate]

EQUENOS

[robust river]

And a_22 = 2^21

[whole wasp]

That we see

But I don't understand

[robust river]

okay then

a bit of trigonometry will help

[safe radish]

@whole wasp Has your question been resolved?

[whole wasp]

.close

[safe radish]

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[whole wasp]

.reopen

[safe radish]

✅

[whole wasp]

@plucky elk

[whole wasp]

Unable to use the trigo here

How to go ahead as
Terms are multiplied

Used cos2x in terms of tan
Then denominator becomes tan square multiplied

[safe radish]

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[safe radish]

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[lean otter]

how do i find a normalvector created by 3 points?

[lean otter]

what is the formula

[plucky elk]

Normal to what?

3 points can make a plane

[lean otter]

) Find a normalvector to the plane that these points punkter (1, −2, 1),
(2, 1, 3) og (0, 1, 5), and find an equation for this plane

[jolly shard]

Construct a plane using the two vectors coming out of one point, then use the cross product

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

how

i thought i had to take the difference and some dot product stuff

[safe radish]

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[carmine panther]

are my invariant points the coordinates that intercept with the original function ?

[hard crest]

yea

[carmine panther]

so my whole f(-x) is an invariant function ..

.close

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[zinc flume]

Please could someone recommend a good way of proving the convergence or divergence of this series?

[glacial cairn]

List the values cos(n*pi) can take

[zinc flume]

-1 and 1

[glacial cairn]

Right, for which n?

[zinc flume]

It simplifies to (-1)^n

[glacial cairn]

Not quite

[zinc flume]

I'm sorry but I'm pretty sure it does

,calc
f(n) = cos(n * pi)
g(n) = (-1)^n
f(0)
g(0)
f(1)
g(1)
f(2)
g(2)
f(3)
g(3)

[flat frigate]

Results:

f(n)
g(n)
1
1
-1
-1
1
1
-1
-1

[glacial cairn]

Fair enough, that's not how I understood what you said

But if you know how it simplifies, why don't you rewrite the original?

[zinc flume]

$\sum_{n = 0}^{\infty} \frac{(-1)^n}{\sqrt{n + 1} + (-1)^n}$

[flat frigate]

Jacks0n

[zinc flume]

Simply because I didnt know if it was necessary depending on the solution

I should probably note I've solved this already but I dont like my method

[glacial cairn]

Then show it and say what you don't like about it

[zinc flume]

But I just want to see how someone else solves it 😦

[safe radish]

@zinc flume Has your question been resolved?

[zinc flume]

<@&286206848099549185>

[safe radish]

@zinc flume Has your question been resolved?

[safe radish]

@zinc flume Has your question been resolved?

[zinc flume]

😠

[lean mountain]

wouldn't you be able to just use the divergence test on this since you already know it doesn't converge

i havent done calc 2 tho so maybe there's a better test

[zinc flume]

Why do you say we know it doesnt converge?

Wouldn't knowing it doesnt converge mean it diverges by direct implication

[lean mountain]

this shows the alternating pattern

i cant remember for certain but i dont think you can use the alternating series test on this, i dont think the conditions are met

but regardless there's no harm in checking to see if the divergence test works for it; it's a test that even if it fails, some other tests you may use later will use its results anyways

use its results as in evaluating at a limit* not using its convergence/divergence

my memory could be shart tho as i said, i just think it's a good first step here because the values you got implies divergence

[zinc flume]

What is "the divergence test"?

[lean mountain]

i was always taught this was called the divergence test, if u know it under a different name im sry for confusing you

[zinc flume]

Ah I see

[lean mountain]

there are 8 tests total if i remember right

[zinc flume]

Nah you're all good. The limit of the summands colloquially known as the divergence test

[lean mountain]

this one is the easiest tho (normally) so i prefer it when i can/when i think itll work

[zinc flume]

This is easily shown to be 0 with the divergence test

[lean mountain]

ah

i didnt try

i just saw alternating function lol

damn idk what i was thinking lol it is obvious

[zinc flume]

Its a weird one imo

But it might be easy for some people hence I want to see how someone else solves it

My solution ends up with the first term being undefined

So I had to shift the series

[lean mountain]

alternating series test might work, or u could try absolute convergence

[zinc flume]

This?

[lean mountain]

yea

you should be able to just cross multiply over the inequality for the 1st condition

and u alrdy verified condition 2 with divergence test

[zinc flume]

This doesn't decrease monotonically unfortunately

[lean mountain]

damn

[zinc flume]

Well

I'm gonna just put a spoiler out there

[lean mountain]

in that case im not the best then

[zinc flume]

This series diverges

[lean mountain]

i never was but now its confirmed i learned nothing from calc 2

a credit is a credit tho

ye but it was clear it diverges alrdy

[zinc flume]

Thank you for taking the time at least ❤️

[lean mountain]

it was just a matter of proving im just dumb and forget the tests

and am studying for a diff exam and this is just to take my mind off things

sry i couldnt put more effort to helping u i wish u luck tho

[zinc flume]

Why do you say that?

[lean mountain]

1 1 -1 -1 1 1 -1 -1...

[wind shoal]

cant you do Alembert

[zinc flume]

Pretty sure the limit is 1

[lean mountain]

this doesn't tell you 100% but it's a good indication

[wind shoal]

ah alr

[zinc flume]

We'd end up with something like

sqrt(n + 1)/(sqrt(n + 2) + c)

Which intuitively looks like the limit is 1

,w limit sqrt(n + 1)/(sqrt(n + 2) + c) as n approaches infinity

[flat frigate]

Results provided by WolframAlpha

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

Requested by jacks0n1130

[zinc flume]

And the same goes for the reciprocal if my division sucks booty

[lean mountain]

its not super specific to this but i remember when i was in calc 2 i found it rly valuable to use a flowchart for series tests

so u might find value in that and just go down the list until u get it

ik its a pretty painful method initially but the upside to it is as u practice doing these tests u get a rly good awareness for them and can eventually see some series and almost instintually know the test to use for it

idk how much time u have for that tho, if this is in prep for a final u have soon then best of luck

[zinc flume]

I did mention earlier I already proved this but I didn't like my method all that much so I was hoping to see someone else's method of proving it

[lean mountain]

what was u method

like what test did u end up using

[wind shoal]

Comparison maybe

would be good

[zinc flume]

I used the alternating series test after a bunch of rearranging and series shifting

[lean mountain]

wdym series shifting

[zinc flume]

Shifting the series so we started at the second term instead of the first

Which I rationalised as valid since we can just take the finite value of the first term out but I wasn't sure

[lean mountain]

at n=1 instead of n=0?

[zinc flume]

Yeah

[lean mountain]

that's normal iirc

i cant remember textbook exactly

i think that's valid for some tests though

[zinc flume]

The issue I had was my rearrangement of the series had division by 0

[lean mountain]

ah

[zinc flume]

And you know when you're solving an inequality and everything goes to poop because you end up with division by 0

[lean mountain]

u ended up with division by zero u mean?

[zinc flume]

I didnt trust myself enough to guarantee I was right XD

[lean mountain]

like u series adjusted and then got div by 0?

[zinc flume]

So the series ended up being

(Blabla)/n

So the first term of the series at n = 0 was division by 0

[lean mountain]

oh

yeah review ur textbook

im 99% sure u can do n-1

n=1*

i dont remember ever actually doing n=0 for most things, i just cant remember the explanation as to why

[zinc flume]

Yeah I see a lot of series starting either at 0 or 1

Maybe they do it because of the same issue

[safe radish]

@zinc flume Has your question been resolved?

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[wise schooner]

not sure how to do this

[granite idol]

well you know function composition, I assume?

and you should know the properties of a linear transformation.

so what is T([1 1])

[wise schooner]

1, -1

[granite idol]

yes. so you want to find S([1 -1])

how can you use the properties of linearity to do so

[wise schooner]

can you add s(1,1) and s(0,-2) ?

[granite idol]

yes

[wise schooner]

oh so the answer is just e then?

[granite idol]

yes

[wise schooner]

ohhh makes sense

tysm

.close

[safe radish]

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[bold torrent]

Solve for each equation for 0pi<= x <= 2pi

sin(x)(cos(x)+1)=0

cos(x)(tan(x)-1)=0

[west hedge]

Help is on the way!

[marsh walrus]

what have you tried?

[bold torrent]

Ummm

Nothing to be honest. I was out sick for a while and kind of forgot how to even begin solving this material

[marsh walrus]

use this

if $a b = 0$ then $a=0$ or $b=0$

[flat frigate]

jan Niku

[bold torrent]

So 1 of them has to equal 0?

[marsh walrus]

or both, yea

[bold torrent]

Sorry I’m a bit confused, that doesn’t really help me if I’m being honest

[marsh walrus]

sure it does

it turns your weird equality into two

[bold torrent]

I understand that it helps to solve the problem, but I don’t know how to use it

[marsh walrus]

!nosols

[safe radish]

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.

[marsh walrus]

o maybe its not a sols

[west hedge]

No, this is inspiration

[marsh walrus]

OH when you said help is on the way you meant you

sorry i jumped in on ur channel

[bold torrent]

Wait i have a question

[bold torrent]

so i can see it as sin(x) * cos((x)+1) = 0?

[marsh walrus]

yes

[marsh walrus]

this is cool tho

[west hedge]

Thank you

[marsh walrus]

this is the naive way

but i think it works, too

if you had no idea what to do, either way would work out

naive is a little more straightforward

[bold torrent]

Yes but what do i do from here?

[marsh walrus]

so $\sin x (\cos x + 1 ) = 0$

[flat frigate]

jan Niku

[marsh walrus]

what does that mean

[bold torrent]

How did u go from the original question to sinx cosx + sinx = 0

[bold torrent]

One of those (or both) = 0

Can I distribute?

[marsh walrus]

its easier without distributing i think

[bold torrent]

What do i do after sinx cosx + sinx =0

[marsh walrus]

.

[hollow lake]

hey guys i need help on my math hw

so like

what is the answer to this question

like solved by yourself

not google

2- -5^2 - -9x-4

[marsh walrus]

i mean okay lets be clever

well

actually no i dont want to be clever

you need to understand how to solve these in the basic way

$\sin x (\cos x + 1) = 0$ means $\sin x =0$ or $\cos x + 1 = 0$

[flat frigate]

jan Niku

[marsh walrus]

solve these equalities

[hollow lake]

anyone?

[marsh walrus]

#❓how-to-get-help

[hollow lake]

im in 4th grade so im kinda stupid

oh ok

[bold torrent]

i’m gonna be honest i’m like really fucking stupid and i don’t know what to do.

do i set one of them equal to 0?

[marsh walrus]

ive given you two equalities

you need to solve them

start with $\sin x = 0$

[flat frigate]

jan Niku

[marsh walrus]

whats x

you have a range for x, right?

$0 \leq x \leq 2\pi$

[flat frigate]

jan Niku

[marsh walrus]

so which values solve the equation?

don't just say im stupid try something

[bold torrent]

0?

[marsh walrus]

,calc sin(0)

[flat frigate]

Result:

0

[marsh walrus]

okay we found one

are there more

[bold torrent]

not sure

[marsh walrus]

have you ever used a unit circle before

[bold torrent]

Yes

[marsh walrus]

do you know what that is

[bold torrent]

Yes

[marsh walrus]

,tex .unit circle

[flat frigate]

jan Niku

jan Niku

[bold torrent]

0, pi, 2pi(?)

[marsh walrus]

okay, now solve $\cos x + 1 =0$

[flat frigate]

jan Niku

[bold torrent]

pi

?

[marsh walrus]

okay, assemble your answers

whats the final collection

[bold torrent]

x = 0, pi, 2pi?

[marsh walrus]

,w Plot[Sin[x] (Cos[x]+1) , {x,0,2*Pi}]

[flat frigate]

Results provided by WolframAlpha

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

Requested by janniku

[marsh walrus]

graph look good or did we miss some

[bold torrent]

Ummm

The graph looks amazing

[marsh walrus]

[bold torrent]

What does this tell us?

[marsh walrus]

i plotted your original function over the given domain

[marsh walrus]

this is a plot of f(x) on 0 <= x <= 2pi

[bold torrent]

Ok that makes sense

[marsh walrus]

you were trying to solve f(x) = 0 and found x=0,pi,2pi

[bold torrent]

So the graph is correct

[marsh walrus]

the answer is right

[bold torrent]

Well my answer was correct ?

[marsh walrus]

yea

so, theres the basic process

can you apply it again?

you can do more clever things but they will not always work

[bold torrent]

Not if the question says tan

[marsh walrus]

huh?

[bold torrent]

If the equation is = 0, then what can I do if it has tangent

cos/sin = tan

[marsh walrus]

im sure you can solve $\tan x = \frac{\sin x}{\cos x} = 0$

[flat frigate]

jan Niku

[bold torrent]

Okay

[marsh walrus]

lets turn it into a problem similar to the one you just did

[bold torrent]

Well i have one—

[marsh walrus]

$\frac{\sin x}{\cos x} = \sin x \qty( \frac{1}{\cos x} ) = 0$

[flat frigate]

jan Niku

[bold torrent]

Wait can we do mine

[marsh walrus]

okay

[bold torrent]

cos(x)(tan(x)-1)=0

cosx * tan(x)-1) = 0 so one/both will be zero so pi must be an answer

[marsh walrus]

youre skipping steps

but i like your initiative

[marsh walrus]

go ahead and write the two equations

1. cosx = 0
2. tanx - 1 = 0

here

[flat frigate]

jan Niku

[bold torrent]

Ohh so it’s just pi/2 and 3pi/2 for cos

[marsh walrus]

and equation 2?

[bold torrent]

Tanx = 1?

[marsh walrus]

this is where you were concerned right

you werent sure how to handle the tangent

you have some options

first off say you had $\frac ab = 1$

[flat frigate]

jan Niku

[marsh walrus]

can you say anything about a and b?

i mean the relationship between them

[bold torrent]

a=b

Maybe not

Idk

[marsh walrus]

yea

and i think further that neither are 0 but it doesnt matter here

so if we have $\tan x = \frac{\sin x}{\cos x} = 1$

[flat frigate]

jan Niku

[marsh walrus]

then try to apply what you said

[marsh walrus]

where do we see that here?

[bold torrent]

So it basically is just pi/4 and 5pi / 4

[marsh walrus]

yea

[bold torrent]

Ok ok

[marsh walrus]

wonder if i can make the bot plot in terms of pi

one second

ugh its really annoying

lets just use solve

[bold torrent]

I don’t think

You need to make one

[marsh walrus]

ah

well you can just use desmos too

you should check your answers somehow

graphs are just an easy way to do it

[bold torrent]

Ok well, I think that solves all my questions for this area. Thanks for helping me, jan niku and being patient w me. I’m going to close the ticket now if that’s ok

[marsh walrus]

,w Plot[Cos[x] ( Tan[x] - 1 ), {x, 0, 2 Pi}, Ticks -> {Table[{n Pi/4, n Pi/4}, {n, 0, 8}], Automatic}]

[bold torrent]

Oh

[marsh walrus]

no problem

[flat frigate]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by janniku

[marsh walrus]

[bold torrent]

I’ll wait

[marsh walrus]

ah thanks stupid bot

just use desmos

lmao

[bold torrent]

OK bai thanks again!

.close

[safe radish]

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[safe radish]

@keen minnow Has your question been resolved?

[safe radish]

@keen minnow Has your question been resolved?

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[viral elbow]

for part B, would there not also be a relative max at x = 5?

[solar hazel]

yes?

what is the concern here

[viral elbow]

i meant 5

[solar hazel]

oh, then no

[viral elbow]

y

[solar hazel]

,w plot x^3

[flat frigate]

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Requested by slayyla

[solar hazel]

f will have a shape like this there

the derivative is positive everywhere around 5 (and 0 at 5)

which means f is increasing there

[viral elbow]

ahhh, it has to cross the x axis for it to be a max/min?

and 5 only touched

[solar hazel]

sure

[viral elbow]

no like

true or flase

[fluid token]

min/max occur when the derivative changes from positive to negative or vice versa

[viral elbow]

right, so when f'(x) crosses x axis

basically

[solar hazel]

when f is differentiable yes. functions that aren’t differentiable everywhere can still have mins and maxes though

but that’s probably not something that would be thrown at you (actually on second thought yes it could be)

[viral elbow]

im scared now. im in calc 1 only

[solar hazel]

take e.g.

,w plot |x|

[flat frigate]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by slayyla

[solar hazel]

this has a local min at 0

ignoring 0, it’s derivative looks like this though

so what i’m trying to say is points where the function isn’t differentiable are also candidates for mins/maxes

[safe radish]

@viral elbow Has your question been resolved?

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[marsh walrus]

im confused what the question is here

havey ou measured the drawing?

[plucky elk]

How old are you

[lament creek]

.close

[safe radish]

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[safe radish]

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[drifting heath]

x²-y² = 2024 find the number of possible values for 'x' and 'y'.

[drifting heath]

.close

[safe radish]

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[mild bridge]

can anyone help with this question?

[safe radish]

Please don't occupy multiple help channels.

[raven heart]

yea?

[mild bridge]

oh heyyy

whats up

yeah so basically

I figured out the common denominator between the two numbers using the euclidean thing

and Idk what to do from here

I need to find values x and y that satisfy this equation

[mild bridge]

is it readable?

[raven heart]

your dots are damn small but I manage

well now you do the euclidean algorithm in reverse

we know the gcd is 4003 right

[mild bridge]

yup

[raven heart]

let's look at that second-to-last line

[mild bridge]

okk

[raven heart]

52039 = 1 * 48036 + 4003

with some algebra magic, we get that 4003 = 52039 - 1* 48036

so at least that's a good start, we have 4003 = something

[mild bridge]

yeah ok

[raven heart]

and now the goal is to use all these divisions to write 4003 as combinations of bigger and bigger numbers

[mild bridge]

which divisions?

[raven heart]

the ones you did for the euclidean algorithm

which other ones would I be talking about

[mild bridge]

ok ok

[raven heart]

that 3rd to last line tells you 48036 = 256192 - 4*52039

[mild bridge]

oh ok

so now we take the 48o36 and replace it?

[raven heart]

yes

that gets you 4003 as a combo of 52039 and 256192

[mild bridge]

ohh

and you keep going up

[raven heart]

and then you replace 52039 with the 4th-to-last line

etc...

yes

[mild bridge]

until you reach the starting numbers

oh okk

cool cool

[raven heart]

my teacher called it "climbing up the remainders"

cause you're replacing the remainders of the division each time

[mild bridge]

makes

sense

im gonna remember it cuz of the name now lol

and for part b

what is a ring?

[raven heart]

it's a set of objects you can "add" and "multiply" with each other essentially (with the addition and multiplication satisfying some rules we're used to)

one very basic example of a ring is the integers

[mild bridge]

ok

[raven heart]

it's an abstract algebra thing (rings)

[mild bridge]

so the numbers are the objects

and since we can add them together and multiply them with each other

it is also a ring

and cuz doing so follows the rules that we have

wait

for part b

do we basically have to the same thing

[raven heart]

yes

[mild bridge]

except with variables instead of hard numbers

ohh ok

ok im gonna give it a go

and see how far i get

[raven heart]

aight

[safe radish]

@mild bridge Has your question been resolved?

[raven heart]

@mild bridge what's up

[mild bridge]

I's still working on it

on part a still

takes a while

I made a mistak
goddammit

[mild bridge]

yo

I'm confused

i got 441 for x and 470 for y

And I did every step

but the answer doesn't turn out to be 4003

[raven heart]

rip

show your work

[mild bridge]

it took soo long

wait lemme

the dots are bigger here

[raven heart]

yeah that's cause you forgot a digit on one of the final numbers

2264497__1__

[mild bridge]

🤦‍♂️

thanks

for pointing it out

ok

im gonna do b now

[safe radish]

@mild bridge Has your question been resolved?

[mild bridge]

,rotate

[mild bridge]

u still there by any chance?

[raven heart]

yea

what

[mild bridge]

so basically

when i get the gcd

[raven heart]

you can't have 1/2t^2, you don't even have a polynomial anymore

wtf

[mild bridge]

wasn't i supposed to do a long division?

[raven heart]

yeah

how is that long division

[mild bridge]

so i just have 1 as remainder

[raven heart]

yeah

[mild bridge]

bro i looked it up online to learn it

i thought that'S how it wentr

[raven heart]

t^4 + 1 = (1/2 t^2)*2t^2 + 1

yeah so the gcd is 1

[mild bridge]

ah ok

[mild bridge]

I GOT ITTT

lets gooo

thanks bro

[timber violet]

can someone help me about this question

Find the sum of all positive integers with 5 1's and 5 0's in binary notation.

<@&286206848099549185>

[raven heart]

@mild bridge are we finished or no ?

[raven heart]

you should get yourself your own channel, see #❓how-to-get-help

[safe radish]

@mild bridge Has your question been resolved?

Channel closed

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[mild bridge]

sorry bro I didn't see your message

yeah I'm done with that question. i'm working through another one rn.

[safe radish]

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[sour violet]

Okay irl inches to mm problem

[sour violet]

I made a part in inches by accident using a 3D modeling software and there is no normal fix right

But I can scale it down

So I thought I could just use 1/25.4 to scale it down

But the number is wrong

Basically is there a way I can make it so I can scale 100 inches to 100mm

Okay so 1270/25.4= 50

I’m trying to find what scale I’d need to I put I need to I put to get 1270 to 50

.close

[safe radish]

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[vast shard]

i need help with this question, how do i solve for this?

[rough storm]

!status

[safe radish]

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

[rough storm]

!showwork

[safe radish]

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

[rough storm]

,tex .exp rules

[flat frigate]

🫎 A Certain User(Moosey) 🫎

[rough storm]

,tex .log rules

[flat frigate]

🫎 A Certain User(Moosey) 🫎

[vast shard]

so i actually dont know hwo to start T-T what i did was, i moved the 2 to the explonent so it becomes log(x+11)^2=(1/2)^x, im not sure if it helps and idk how to proceed

[drowsy karma]

treat both sides of a given equation as functions of the variable x and plot their graphs in one system, you will see that they intersect and it is not difficult to guess the solution. You cannot solve this equation using algebraic methods.

[vast shard]

oh.. my teacher told us it is solvable, someone asked today in class tho

[drowsy karma]

i guessed solution within 3 seconds

[vast shard]

howd you guess?

[drowsy karma]

i asked myself what x = , an then i imaigne graph

what else

🙂

[vast shard]

oh so id have to memorize the shape of the graphs 😭

[drowsy karma]

yes they are very easy

or

use desmos website

they are easy

[vast shard]

alr, tyty

[drowsy karma]

yw

[vast shard]

.close

[safe radish]

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[split crest]

how do you solve this

i have 2 more i just need to learn how

[safe radish]

@split crest Has your question been resolved?

[safe radish]

@split crest Has your question been resolved?

[safe radish]

@split crest Has your question been resolved?

[safe radish]

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[safe radish]

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[brazen helm]

If I take a complete graph with n edges and color each edge red or blue what is the probability that all the edges are blue

[brazen helm]

I mean intuitively Id have to say its $(1/2)^{n}$

[flat frigate]

bigpufik

[brazen helm]

but the textbook says its (1/2)^{n-1}

and i dont get why

[quasi bison]

!original

[safe radish]

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.

[quasi bison]

sounds like some kind of miscommunication

[quasi bison]

xyproblem

they wanted all edges the SAME COLOR, not necessarily all blue.

it's either all blue or all red

[brazen helm]

OHH so we have to remove one 1/2?

[quasi bison]

yes

[brazen helm]

would that mean we would need to add a 1/2?

since we flip a coin to choose if its blue or red

[quasi bison]

.........

??

[brazen helm]

well like how is it less to worry about

[halcyon carbon]

Show the whole question

[halcyon carbon]

Where does the coin come from

[brazen helm]

Im guessing something is wrong here

[halcyon carbon]

The colour of the first edge doesn't matter, you just need the subsequent edges to be of the same colour.

[brazen helm]

OH RIGHTT

[halcyon carbon]

So you only need to consider (\binom k2 -1)

yeah

[flat frigate]

Pure

[brazen helm]

Ok but how are we happy witht eh expected value being smaller then one?

didnt we show that there never exists a value like that

I mean a collection like that

[safe radish]

@brazen helm Has your question been resolved?

[safe radish]

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[atomic sonnet]

can someone help me find the running time

[safe radish]

@atomic sonnet Has your question been resolved?

[fiery merlin]

Well, what values does s take on?

For example, it starts as 0, then what does it become?

[atomic sonnet]

1

then 1+2

1+2+3

[fiery merlin]

Right, and what do the first n natural numbers sum to?

[atomic sonnet]

idk

some value

[fiery merlin]

Have you seen n(n + 1)/2 before?

[atomic sonnet]

yea but how is that n(n+1)/2

[fiery merlin]

OK, so let's say you have the first 10 numbers.

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10.

We can pair up the numbers from the outside in:

(1 + 10) + (2 + 9) + (3 + 8) + (4 + 7) + (5 + 6)

Those all become 11.

11 + 11 + 11 + 11 + 11

So, there are five 11s.

[fiery merlin]

5 is half of 10, which was the count of numbers we had.

And then each pair adds to n + 1 (here, 10 + 1).

Does that make sense so far?

[atomic sonnet]

i guess so

[fiery merlin]

That becomes n/2 times n + 1.

So, n(n + 1)/2.

[atomic sonnet]

i think you misunderstood my question

i am asking how is that n(n+1)/2

shouldnt it be bigger than that

i = 1+2+3+4+5
but s= s+i

[fiery merlin]

No, i goes up by 1 each time.

s goes up by i each time.

So, i will be 1, 2, 3, 4, 5.

s will be 1, 3, 6, 10, 15.

s will be n(n + 1)/2.

You can try it by making a program that does that algorithm, then printing out s each iteration.

[atomic sonnet]

yea ok

i guess what is confusing is that s = 1+3+6+10

which is not 1+2+3+4+5

[fiery merlin]

No, s is 1, then 3, then 6, then 10.

It's not 1 + 3 + 6 + 10.

[atomic sonnet]

oo

😅

well

my bad

[fiery merlin]

OK, so let's use another variable instead of n, because n is already used in your program.

Let's say t(t + 1)/2, where t is the number of iterations of the loop.

We need t(t + 1)/2 to be greater than or equal to n to finish the loop.

So, we can solve t(t + 1)/2 = n.

Does that make sense?

[atomic sonnet]

yea

can you say t(t+1)/2 aprox t^2?

[fiery merlin]

You can. That would be a shorter way.

[atomic sonnet]

then t^2>n => t > sqrt(n)?

[fiery merlin]

Right, so the number of iterations is O(sqrt(n)).

[atomic sonnet]

we can say theta right?

[fiery merlin]

Yes, that's right.

[atomic sonnet]

hell yea

[fiery merlin]

There's only one case, so the best and worst cases are the same.

[atomic sonnet]

ok thanks mr t rex 🙂

[fiery merlin]

You're welcome.

[atomic sonnet]

.close

[safe radish]

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[carmine panther]

would $y=3\sqrt{16-(-x)^2}$

[flat frigate]

marulk

[carmine panther]

not result in a y axis reflection

i thought you had to replace x with a -x to get it

[lone arch]

Yes, but if f is already symmetric to the y-axis, then that makes no difference

[carmine panther]

so if i did (-x) im just undoing the base function of 16-x^2 and not getting a y reflection ?

[quasi bison]

-(-x)^2 is still -x^2 btw not x^2

[carmine panther]

whats the politically correct way to write it '

.close

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[bold ferry]

?

[lean otter]

uh

<@&268886789983436800>

[quasi bison]

<@&268886789983436800> advertising

[safe radish]

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[hard crest]

strange behavior

[safe radish]

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[proud vessel]

Any idea for this xd?

[exotic panther]

What are u trying to do

[proud vessel]

Find the absolute degree of the sixth term in the development of:

<@&286206848099549185> 😔

[runic summit]

what do u need to fill out for m

[hybrid plover]

what's the question? also is the picture supposed to mean $\frac{n^8+m^8}{n+m}$ ?

[flat frigate]

moriaritie

[arctic burrow]

change the top one into 2^3 * n + 2^3 + m

maybe

idk

im bad

[proud vessel]

Well, the question is in Spanish hehe

[hybrid plover]

translation plz

[proud vessel]

I've made a variable change

Find the absolute degree of the sixth term in the development of:

[hybrid plover]

development means expansion?

[proud vessel]

Yeah

[hybrid plover]

Make another variable change n/m=z

then use polynomial division

[proud vessel]

Alright I am gonna try It

[hybrid plover]

good luck

[proud vessel]

No idea how to follow this xd

I got residual

[hybrid plover]

oops my bad, i forgot the m doesn't cancel. but its still doable by polynomial division (that var change was a simplification)

[proud vessel]

What do you mean?

[hybrid plover]

wait

the exponent is 8, which is not odd so there's gonna be a remainder after you divide

[proud vessel]

Yeah

[hybrid plover]

so after division you get (2 m^8)/(m + n) - m^7 + m^6 n - m^5 n^2 + m^4 n^3 - m^3 n^4 + m^2 n^5 - m n^6 + n^7
[Result via WA]

the thing thats not clear is what the 6th term here is

[proud vessel]

I'm also not sure😅

[hybrid plover]

but it seems to me its m^5n^2

from the end its 6th

5×2+2×3=16

[proud vessel]

But in your solution there is still a division

[hybrid plover]

yeah its not reducible further ig

[proud vessel]

Hmm thanks anyway

I'll be searching information. If I get a way to do It i'll share It here

[hybrid plover]

well the fact that m^8/(m+n) is irreducible is certain

[proud vessel]

Maybe the exercise is wrong

I've found a lot of bad formuled problems in peruvian books xd

[safe radish]

@proud vessel Has your question been resolved?

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[supple bramble]

Can someone help me finding the sum of this series

[keen coral]

Rearrange terms and you can find a geometric series

[supple bramble]

yeah i dont know how they are they are doing that, like i dont know what to do with the 2n

[keen coral]

Rewrite 2^(2n) as 4^n

[supple bramble]

how is that 4n though

because isn't it (2)^n^2

[empty gyro]

$a^{b\cdot c}=(a^b)^c$

[flat frigate]

SWR

[supple bramble]

or (2^n)^2

yes but wouldn't there still be a n^2 left