help-33

[fickle shell]

Im not really sure how to get this?

I have literally no idea what to even do

Im guessing its just like a rule or something that im supposed to know?

[amber birch]

yes, it follows from the unit circle

that $\cos(x) = \sin(\pi/2 - x)$

[elfin berry]

southlander!

[fickle shell]

How does it follow from the unit circle sorry?

Hi chartbit ❤️

[amber birch]

the base of the triangle when reflected across y = x becomes the new height

now if you remember that base = x = cos theta

height = y = sin theta

it follows

[fickle shell]

Sorry I still dont really understand, where does the pi/2 come into it?

[amber birch]

cause the angle after the reflection is pi/2 - theta

for the height

[fickle shell]

Im really sorry im still a bit lost

[amber birch]

wait let me edit

[still temple]

Sorry to interrupt, but couldn’t we compare the sine and cosine functions on a graph and see where they intersect? They intersect at a 45-degree angle, which is pi/4 radians. So we know the angle is pi/4 radians, but we need to express it in terms of "a". So wouldn’t the answer be pi/2 - a?

[amber birch]

ah yes that also works

yes if you use the graphs sin and cos are reflections of each other across x = pi/4

[still temple]

yes

[amber birch]

so that means 0 and pi/2, or like k and pi/2 - k work

you can check that |pi/4 - k| = |(pi/2 - k) - pi/4| so same horizontal distance

[fickle shell]

1 moment im reareading everything 100 times trying to understand 😓

I think I get it but im not sure... im going to take a little break and then come back to this one and try more exercises of it haha

Thank you everyone!!

❤️

.close

[marsh citrus]

Channel closed

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[marsh citrus]

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[still temple]

Prove that xy + yz + zx <= x^2 + y^2 + z^2

[rustic field]

Hint one:By cauchy shwartx

[still temple]

I only know AM-GM

[tight meteor]

maybe it can be done with its extension

QM >= AM >= GM >= HM

[rustic field]

try AM-GM

[junior copper]

QM and HM thats new

[grand radish]

root mean square - QM

[still temple]

idk

[junior copper]

oh its also called the quadratic mean

[still temple]

Yes

[grand radish]

HM is harmonic mean, you might have heard about harmonic progression

[rustic field]

x+y+z>=3*cuberoot(xyz)

[hushed egret]

xy <= (x^2 + y^2)/2

[still temple]

I can get, ||nowhere||.

[hushed egret]

write the analogous inequalities for the other 2 terms

[still temple]

no way

I got it

thanks yal

[junior copper]

W

[still temple]

.close

[marsh citrus]

Channel closed

Closed by @fresh rose

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[junior copper]

desmos cube

[still temple]

What's this for?

[junior copper]

idk thats just what your inequality looks like as a cube

[marsh citrus]

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[fallow furnace]

how to show that this limit is 1

[hollow glen]

recall the definition of cosh and sinh

[marsh citrus]

Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem and is confusing for helpers, making it more difficult to help you. Please be patient, even if your channel has not received much attention.

[fallow furnace]

wtf i didnt do that

[fallow furnace]

thx

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[rocky lark]

$\sum_{n=1}^{\infty} (\frac{\pi}{2}-\arctan n^p)$

[elfin berry]

Scruffy

[rocky lark]

I have to see the convergence

[lucid zenith]

what have you tried so far?

[rocky lark]

I don't know where to start

[lucid zenith]

what exactly is p?

do you need to find a range for p where this converges/diverges?

[rocky lark]

Yes

[lucid zenith]

what's with the sob react

[lucid zenith]

do you know any identities involving arctan

[rocky lark]

No

[wide citrus]

Blud has a shaggy rogers pfp

[lucid zenith]

okay well you can't do much without knowing the identities

[rocky lark]

Ggs

[lucid zenith]

$\arctan(x)+\arctan\left(\frac1{x}\right)=\frac{\pi}{2}$ for $x>0$

[elfin berry]

kheerii

[rocky lark]

.

[wide citrus]

https://tenor.com/view/scooby-doo-gif-6984564859876364855

[main idol]

!topic

[marsh citrus]

Please read the channel description before posting, and stay on topic.

[wide citrus]

[lucid zenith]

I just told you the relevant identity..

[rocky lark]

I need a demonstration

pls

[main idol]

You can prove it yourself if you plug in 1/x into the integral formula of arctan

[rocky lark]

What?

[main idol]

Which part of that is confusing you

[rocky lark]

In my exercise there is no integral

[main idol]

I was answering your confusion about this identity

[rocky lark]

I don't know the identity

I'm still stuck

Thanks

@main idol I think the solution is p>1

[lucid zenith]

you're right

[wide citrus]

Nice @rocky lark

[rocky lark]

Thanks

.close

[marsh citrus]

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[plush ember]

what.

[still temple]

Well

[plush ember]

tried using the 45/45/90 rule but idk

all the answers seem... possible?

[still temple]

Let the two equal sides be x and x.

[plush ember]

yeah i did that

[still temple]

Then what's the length of the hypotenuse?

[plush ember]

xsqrt2

[still temple]

Yup. Now the perimeter is x + x + x root 2

Which is equal to 38 + 38 root 2

[plush ember]

yeah

[still temple]

I think you should be able to solve it naow

[plush ember]

x(sqrt2+2)= 38+38sqrt2

[still temple]

yes

[tawdry rampart]

!nosols

[marsh citrus]

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.

[still temple]

Lol

[plush ember]

bro she literally just divided it

[still temple]

You pop immediately

[tawdry rampart]

lmao

i was actually waiting

[plush ember]

so one leg means x

i was thinking the hypotenuse

.close

[marsh citrus]

Channel closed

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[still temple]

Yes

[marsh citrus]

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[atomic pike]

How to calculate the size of the marked angles?

[atomic pike]

any help?

[rancid geode]

is AB and CD parallel?

[mighty bramble]

i just answered the same thing in another channel, stop making more channels

if AB and CD are parallel, 8x + 5 = 10x - 7
you can solve for x

[atomic pike]

yes but How to calculate the marked angles?

[mighty bramble]

by solving that equation?

if you're not sure how to solve linear equations then ask abt that

but yeah solving that equation will give you your answer

[crystal lagoon]

You solve for x then plug x into the angle equation

[marsh citrus]

@atomic pike Has your question been resolved?

[marsh citrus]

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[still temple]

hello how do I solve this??

[still temple]

should I find the average velocity between t=3 and t=1 and then make it positive to make it average speed then use the law average speed = distance/time to get the distance?

[stark sandal]

Oh heeeeeeeey

[still temple]

Hiiii

[stark sandal]

funny seeing you here idk if you recognise me

[still temple]

the name is familiarr

[stark sandal]

yeah samee with the name and pfp lmao

prolly talked to you on the other server

[still temple]

yess the atheism server

[stark sandal]

i'd just integrate

[still temple]

ohh

itd be velocity

[stark sandal]

i dont know if you can use average velocity it feels erroneous

[stark sandal]

uh no, the distance travelled, $x$, is equal to $x(3)-x(1)$ yeah?

[elfin berry]

Percy

[stark sandal]

hold on can't you just subtract lmfao.

[still temple]

thats displacement tho

[stark sandal]

oh distance

sorry

[still temple]

yesyes its oki

omg

I feel so silly

[stark sandal]

I feel sillier dw

[still temple]

isnt d=v*t

xD

[stark sandal]

that's displacement though so

[still temple]

wouldnt it be average velocity/time

[loud linden]

If you take the derivative and then integrate like $\int_1 ^3 |v(t)| dt$ (?)

[elfin berry]

luna-maia

[stark sandal]

aren't you still getting the displacement-

[loud linden]

Absolute value

[stark sandal]

oh oh right

would that work?

[still temple]

I dont remember what that symbol means tbh

[loud linden]

I might be wrong though, hopefully someone else can come over

[stark sandal]

the fact that i cant think of a definitely easy way is silly as all hell

[still temple]

the sum one

[loud linden]

$\int$?

[elfin berry]

luna-maia

[still temple]

yep

[stark sandal]

integration.

[still temple]

yea but the numbers on it

nevermind I vaguely remember

[stark sandal]

upper and lower limits because this is definite integration

[still temple]

we didnt take any of that in the course tho

[stark sandal]

$\int^m_nf(x)dx=F(m)-F(n)$, fundamental theorem of calc

[elfin berry]

Percy

[still temple]

theres gotta be an easier way right??

[stark sandal]

well fundamental theorem 2

[loud linden]

$\int _1 ^2 2x dx$
$x^2|_1 ^2$
$((2)^2)-((1)^2)$

[elfin berry]

luna-maia

[loud linden]

oh.

oops

it did not go how i wanted it to

idk how to make newlines on this thing

[stark sandal]

$\int_1 ^2 2x dx\$
$x^2|_1 ^2\$
$((2)^2)-((1)^2)\$

[elfin berry]

Percy

[still temple]

no worries no worries

ohh right!

I remember that

[stark sandal]

i still prefer the fundamental theorem of calc version lmao

[still temple]

+c

[stark sandal]

no +c.

+c is for indefinite

[loud linden]

<@&286206848099549185>, is this correct for Meow's question(#help-33 message)?

[still temple]

oh

[stark sandal]

here we have limits so we don't need this

[still temple]

just resending cause lunary pinged helpers

[still temple]

we never needed those in this course thi

[stark sandal]

it feels kind of overkill?

i mean i guess you can just

look at the function, see where it gets negative

then add the $x(f)-x(i)$ separately.

[elfin berry]

Percy

[stark sandal]

yeah the mod one is more convenient if it works but they don't know calculus yet.

honestly a bit weird doing physics without calc

[still temple]

i wonder if we can just find the displacement for each second between t=1 till t=3 and just add them up

[stark sandal]

do you have precalc or calc1 or smth

[still temple]

we didnt use any advanced calc in this course

[stark sandal]

that's basically what integration does just for time intervals much smaller than a second

also no that's still displacement

you'd need all signs to be positive

[still temple]

[loud linden]

displacement might work since the function doesnt double back on itself between 1 to 3 right?

[stark sandal]

so the mod

[still temple]

OH

[stark sandal]

it does

[still temple]

so I take every displacement between t=1 and t=3, make them positive, then add them up together?

[stark sandal]

yeah that's what you do when you have calc

when you don't you just look at where it changes signs and do it manually

in this case,

[still temple]

still off sadly

[stark sandal]

,wolf 5+3x-6x^{2}=0 solutions

[elfin berry]

Results provided by WolframAlpha

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

Requested by percykel

[stark sandal]

okay be kind of a bitch

[still temple]

Which is the absolute value of the displacement , by the way

[still temple]

hmmm

[loud linden]

Is this the answer key?

[still temple]

yep!

[loud linden]

if so, thats what i got

[stark sandal]

uh huh.

[still temple]

whaaa but thats the displacement-

heres the same question but asking for the displacement

[stark sandal]

im so confused for literally no reason

it's weird

[still temple]

me and you both brother xD

[still temple]

thats really interesting

[stark sandal]

no it's not sigh

we're just dumb

okay so

for distance and displacement to be different the velocity has to change direction during the motion right?

[rocky lark]

Yo

[loud linden]

if you walk forward 2 steps, then take 1 step back, your displacement is 1 step, but your total distance is 3 steps.

Here, you were walking backwards the entire time, so your displacement is -42 and your total distance travelled is 42

[stark sandal]

we can look at the sign of $f'(x)$ in this case

[elfin berry]

Percy

[still temple]

mmhm! unless its going in a negative direction

then its the absolute value of that

[loud linden]

From x=1 to x=2, the displacement is 0, but the total distance is 2

[stark sandal]

$f'(x)=3-12x$ is negative $\forall x \in [1,3]$.

The velocity doesn't change direction, the distance and displacement are the same. Case closed.

[elfin berry]

Percy

[rocky lark]

Bros are cooking

[still temple]

OH

[stark sandal]

Of course the distance=$|\text{displacement}|$ because you can't have a negative length.

you can have a negative difference though

[elfin berry]

Percy

[still temple]

wow I see I see

[loud linden]

(Only in this case though. For the picture I posted above as an example, that wouldn't be the case)

[still temple]

true xD

[still temple]

how did you know velocity didnt change directions??

[stark sandal]

here you can again see that the derivative first increases then decreases so the distance $\neq$ displacement.

There, we mathed it.

[elfin berry]

Percy

[stark sandal]

do you know how to differentiate yet?

[still temple]

yesyes

[stark sandal]

okay so you can see the sign of the derivative

[still temple]

like negative or positive?

[stark sandal]

or you can just look at the graph and see it doesn't change directions, but you typically won' get a graph n the paper and drawing one can sometimes be a pain

[junior copper]

Is this negative or positive displacement?

[still temple]

negative

[still temple]

yeah thats true, we wont be getting a graph

[loud linden]

Graphically, the graph goes down, but never up within x=1 and x=3

Number-ly, if you have the line $f'(x)=3-12x$, region test it

[elfin berry]

luna-maia

[still temple]

true true

[stark sandal]

here $f'(x)=3-12x$.

This of course has root at $x=\frac14$, and decreases as x increases. (you can see this by noticing that $f''(x)<0 \forall x$)

So it would have the same sign (negative) $\forall x \in [1,3]$

[elfin berry]

Percy

[still temple]

so do I use the derivative to make a graph??

[stark sandal]

you just do this :D

or you can make a graph it's easier lmao.

wavy curve method for life.

[loud linden]

I'm giving you two different ways to figure it out

1st way, if u have a graph

2nd way, using derivative + region testing

[stark sandal]

apart from when it's a pain to factorise ig

[junior copper]

critical points

[still temple]

I see thank you! I dont think we know region testing yet

But I appreciate it

[stark sandal]

please don't use the term critical points here. It's a weird way to refer to 'zeroes of a polynomial.'

A critical point is any value for which the derivative of a function is zero or doesn't exist.

So like totally different a thing.

[still temple]

Oh gosh I see

Yeah that's true I can do that

But we haven't taken that yet and I don't know if this is what they want us to use

[stark sandal]

@still temple if you can understand this, this is basic region testing

also you can always just draw the line for smth that's this easy

[still temple]

It would be really funny if this wasn't even included in our test tomorrow

[stark sandal]

(you should draw graph if it's easy)

[junior copper]

I know what a critical point is...

[still temple]

I could , but I get the feeling that this isn't included , to be honest

I will ask my classmates about them because they know what the teacher gave us

Thanks a lot for the help guys.I appreciate it

<3

[stark sandal]

yeah well we were talking about wavy curve here no? it's gonna get really confusing for someone who hasn't taken calc because they'll think you mean a zero.

[junior copper]

oh well thats a fair point

[stark sandal]

yay good luck

[still temple]

Nice meeting you again , by the way , percy

[stark sandal]

also add me omg

[loud linden]

this thing? (just checking, just in case it goes by different names in different places idk)

[still temple]

Id love to but I only add people I talk to often, if I do id like us to get to know each other more

[loud linden]

if youre wondering why 1/4, its because we know that 3-12x only touches 0 at x=1/4 so thats the only chance it has to change from pos to neg or neg to pos

[still temple]

Oh yeahh, I remember that!! that's not included in our course tho

[loud linden]

I see

[still temple]

that makes sense yesyes

Thank you luna, Sorry for taking a lot of you guys time

I really appreciate the help

[loud linden]

its all good :3

meow :3

[still temple]

meow :3

.close

[marsh citrus]

Channel closed

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[loud linden]

$\omega$

[elfin berry]

luna-maia

[loud linden]

@still temple

:p

[still temple]

woahhh

$ • • $
$\omega$

[elfin berry]

meow

[still temple]

eh

$ . . $
$\omega$

[elfin berry]

meow

[still temple]

$ . . $
$omega$

[elfin berry]

meow
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[still temple]

nevermind xD

.close

[loud linden]

$. .\$
$\omega$

[elfin berry]

luna-maia

[loud linden]

@still temple

[still temple]

YAY

[loud linden]

:3

[loud linden]

btw i recommend #cats

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

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• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

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[stark sandal]

!status

[still temple]

I don’t know how to set this up

[still temple]

!unsolved

I’m not sure if I’m supposed to to that or not

[stark sandal]

lmao

[stark sandal]

!status is supposed to do this

[marsh citrus]

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

[still temple]

Slower trip = 55t

Faster trip = 80(t - 5)

That’s what I did but I don’t think I actually know what’s going on

[still temple]

How do you find T

If you don’t know the distance

The distance travelled is the same in both cases, so you can set them equal to each other.

55t and 80(t - 5) are both distances.

[stark sandal]

what is this

[stark sandal]

assume a variable d for the distance, write s=d/t implies d=st, put values, solve for d

[stark sandal]

missed the 'equate' step whoops

but you get it

[still temple]

That’s how I tried to solve it. I got 16 but I didnt know if that was the answer or if I needed to do more. And the problem is set up wrong

[stark sandal]

wdym the problem is set up wrong

[stark sandal]

16 seems right.

[still temple]

Is that what the set up should look like

[stark sandal]

quick mental math though miiiight be wrong

[stark sandal]

yes but with more words

[still temple]

I’m confused cause with S=d÷t, you don’t know the distance or the time. You only know 2 speeds minus 5 hours of time on the faster speed

[still temple]

So this is correct right?

-25’s on bottom

<@&286206848099549185> can someone just confirm this is how you set it up?

Question is: L.Want
2. Have
3. Connection
AR Unit 13
Ronald is driving to an event that starts at a certain time. He knows that if he drives at 80 mph, he will arrive 5 hours early to the event.
However, if he drives at 55 mph, he will arrive on time. How long is it until the event is set to start?

[stark sandal]

oh no where did the t go

[stark sandal]

doesn't matter if you know the distance, you know it's equal, you can exploit that fact

[stark sandal]

just don't eat your T and it's okay

[still temple]

Are they all T’s? 25 and 400

400 is the distance?

[stark sandal]

.......what?

[still temple]

A better question is to just ask what 400 represents

In this

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

.close

[marsh citrus]

Channel closed

Closed by @heavy onyx

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[tawny herald]

I know how to verify that y is a solution of the differential equation, by why do we need the information y(1)=2?

[tawny herald]

I'm looking at the steps and theyre just plugging it in but idk why

[mystic minnow]

to ensure that y satisfies the initial condition.

you can have different functions that give the same y', for example take y = 2x^3 - 3x + 7, you still get the correct y' but not the correct initial condition

[tawny herald]

I see, so you could have a y that is the solution of the differential equation, but not one to the initial value, right? And the way we check if it matches with the correct initial condition is by plugging it into the original y function and comparing, right?

[mystic minnow]

yes

[tawny herald]

ok thank you

that makes a lot more sense

.close

[marsh citrus]

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[marsh citrus]

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[woeful escarp]

In a Trapezium ABCD where ab is parallel to cd and the daigonals ac and bd intersect at a point 'o' if a line is drawn from a point 'p' on the side ad to a point 'q' on the side cb passing from o and parallel to ab proove that po = qo

[marsh citrus]

@woeful escarp Has your question been resolved?

[marsh citrus]

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[vast umbra]

hey guys this is a 2nd order rlc circuit but i cannot get the transfer function to be 2nd order

[vast umbra]

i get 1/(1+sC/r+C/L) as my H(s)

[marsh citrus]

@vast umbra Has your question been resolved?

[vast umbra]

<@&286206848099549185>

[marsh citrus]

@vast umbra Has your question been resolved?

[marsh citrus]

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[silver plank]

how do i continue with the integration??

nvm i think i got it

.close

[marsh citrus]

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[brave marsh]

Hiya!
I’m getting tired of running into perpetual self questioning loops about this and I can’t find much about a proof like this online.
I’m specifically trying to prove the Drinker’s “paradox” through natural deduction, but for some reason FOL is breaking my brain in a way propositional logic could never.

Here’s the sketch I have so far, hopefully it doesn’t look like a hate crime against logicians.
Could someone just point out any major logic flaws in there? Thanks!

[brave marsh]

I guess I’m a bit unsure about the red bits in there, because I feel like I’m using an assumption I’ve already discharged

[marsh citrus]

@brave marsh Has your question been resolved?

[marsh citrus]

@brave marsh Has your question been resolved?

[marsh citrus]

@brave marsh Has your question been resolved?

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[flint widget]

im quite lost 🥲

[flint widget]

x = 2sin theta
dx = 2cos theta dtheta

2sin(theta)/2sqrt(1-sin^2(theta))2cos(theta)d(theta)

sorry for the messy formatting

[wary kite]

latex

[meager badger]

4sin^2x(theta)/ (2sin(theteta)) * cos(theta) after simplfiying

[wary kite]

but anyways

[meager badger]

2sin(x) * cos(x) dtheta

then use u sub for cos or sin

[flint widget]

hmm

im lost haha

[meager badger]

what part

[wary kite]

$\int \frac{4\sin^2(\theta)}{\sqrt{4-4\sin^2(\theta)}} 2\cos(\theta) d\theta$

[elfin berry]

knief

[flint widget]

yea at that part ^

[wary kite]

well simplify the sqrt

factor out 4

[sacred idol]

simplify

ya

[flint widget]

after taking out the 4 and making 1 -sin^2

[wary kite]

and what’s 1-sin^2

[flint widget]

uhhhhhhhhhhhhh

hmmmm

i forgot lol

[wary kite]

pythagorean identity

sin^2 + cos^2 = 1

so what’s 1-sin^2

[flint widget]

oh

cos^2

[wary kite]

yep

[flint widget]

right right

[wary kite]

now what’s sqrt(4cos^2)

[flint widget]

2cos

[wary kite]

yep

so does anything cancel?

[flint widget]

then dx = 2cos cancels

i see i see

[wary kite]

mhm

then

$\int 4\sin^2(\theta) d\theta$

[elfin berry]

knief

[wary kite]

any ideas

[flint widget]

take out the 4

[wary kite]

sure then what

[flint widget]

then just solve the integral?

[wary kite]

but how

what will you use

how would you integrate sin^2

[flint widget]

hmmmmm

[wary kite]

wouldn’t it be nice if it was ^1

what formula might we use to rewrite sin^2

in terms of some trig function to the first power

[flint widget]

erm

1 - cos^2?

[wary kite]

well the problem with that is now we have another trig function squared

how about one of the angle identities

namely

the double angle identity

do you remember what cos(2x) =

[flint widget]

cos^2(x) - sin^2(x)?

[wary kite]

mhm but let’s rewrite the cos^2 because we want cos(2x) to only be in terms of sin^2

how can we write cos^2 in terms of sin^2

you did it for sin^2 in terms of cos^2

🤷🏼‍♂️

[flint widget]

2sin(x)cos(x)?

[wary kite]

that’s sin(2x)

[flint widget]

hmmmm

[wary kite]

^

[flint widget]

oh

[wary kite]

how do we write cos^2 in terms of sin^2

[flint widget]

1- cos(x)

[wary kite]

no

i think you mean 1-cos^2

but that’s sin^2

[flint widget]

yea

[wary kite]

we want cos^2 =

in terms of sin^2

so we want cos^2 = some function with sin^2 in it

[flint widget]

so we have 4sin^2(x)

[wary kite]

right

[wary kite]

but let’s go back to this

[flint widget]

why do we want cos^2(x) tho

[wary kite]

because of this

$\cos(2x) = \cos^2(x) - \sin^2(x)$

our goal is to write sin^2 in terms of cos(2x)

but the problem is we have cos(2x) in terms of cos^2 and sin^2

we want it only in terms of sin^2

so we can substitute it into our integral

and we can easily integrate something like cos(2x)

it’s a simple u sub

u = 2x

and we know the anti derivative of cos is sin

this is often called power reduction

because we can easily find the anti derivative of trig functions when they are raised to the first power

[flint widget]

doesnt the double angle of cos(2a) have cos^2 and sin^2 which are both not the first power

im lost aaaaa

up to 4sin^2 i get it

[wary kite]

this is true but we can rewrite cos^2 in terms of sin^2

just as you did earlier

[flint widget]

and i know we want to get to the first power

[wary kite]

if sin^2 + cos^2 = 1

you found that 1-sin^2 = cos^2

[flint widget]

right

[wary kite]

so

how about we substitute that in

[flint widget]

sin^2 = 1 +cos^2

[wary kite]

you mean -

[flint widget]

right

yea yea

[wary kite]

ok now let’s put that in our double angle formula

$\cos(2x) = (1-\sin^2(x)) - \sin^2(x)$

[elfin berry]

knief

[wary kite]

now let’s group the sin^2 terms

$\cos(2x) = 1-2\sin^2(x)$

[elfin berry]

knief

[wary kite]

now we have a new formula that relates cos(2x) only to sin^2

so

[wary kite]

let’s solve for sin^2 here

then substitute it into our integral

[flint widget]

one sec lemme just refresh myself on double angle then ill come back

cuz we were cos^2 -1

[wary kite]

wdym

[flint widget]

we switched out sin^2 with this correct?

cos^2 -1

[wary kite]

technically we switched out cos^2

[flint widget]

wwwwwwwwwwwaaaaaaaaat

[wary kite]

$\sin^2(x) + \cos^2(x) = 1 \implies \cos^2(x) = 1-\sin^2(x)$

[elfin berry]

knief

[wary kite]

and

$\cos(2x) = \cos^2(x) - \sin^2(x) \implies \cos(2x) = (1-\sin^2(x)) - \sin^2(x)$

hmm that’s long

i’ll delete the last bit

[elfin berry]

knief

[wary kite]

the last part is what i did here

[wary kite]

we substituted from this

[flint widget]

i see i see, i get this part

but

lol

my brain somehow doesnt understand, how sin^2 is not being swapped with cos^2 -1

[flint widget]

^

[wary kite]

because we don’t want it to be (also sin^2 ≠ cos^2 - 1 unless you meant -sin^2 = cos^2 - 1)

[flint widget]

ah

[flint widget]

i see

gotcha gotcha

[wary kite]

we do this because we’re interested in sin^2

because it’s in our integral

if we were integrating cos^2 we would do the opposite, we’d solve for cos^2

[wary kite]

so we have this

do you follow so far

[flint widget]

yea

[wary kite]

ok can you solve for sin^2 here

like write sin^2 = …

[flint widget]

so we went from 4sin^2(x) to 1 - 2sin^2(x) ;_;

[wary kite]

well

the goal is to write sin^2 in terms of cos(2x) then substitute it into the integral

we didn’t change the 4sin^2 yet

we’re getting there

can you just solve for sin^2 there

$\cos(2x) = 1-2\sin^2(x)$

[elfin berry]

knief

[wary kite]

it’ll make more sense

trust me

[flint widget]

ok so, sin^(x) = 1 - cos(2x)/2

[wary kite]

yes

brilliant

now

$\sin^2(x) = \frac{1-\cos(2x)}{2}$

[elfin berry]

knief

[wary kite]

we will not go back to the integral

$\int 4\sin^2(\theta) d\theta$

[elfin berry]

knief

[wary kite]

but guess what

we’ve just written a formula for sin^2 havent we

[flint widget]

yepppp

[wary kite]

‼️‼️‼️

now let’s substitute that in

$\int 4\left(\frac{1-\cos(2x)}{2}\right) d\theta$

[elfin berry]

knief

[wary kite]

yes?

make sense?

[flint widget]

yep yep

[wary kite]

ok now let’s distribute the 4

can you write what we get

[flint widget]

4 - 4cos(2x)/2

[wary kite]

sure but can’t we cancel a factor of 2

[flint widget]

yea

[wary kite]

so what’s left

[flint widget]

2(1-2cos(2x))

[wary kite]

yea and let’s distribute the 2

[flint widget]

2-2cos(2x)

[wary kite]

hmm not quite

$4\left(\frac{1-\cos(2x)}{2}\right) = 2(1-\cos(2x)) = 2-2\cos(2x)$

[elfin berry]

knief

[wary kite]

there we go

you fixed it

great

so

the integral is now

$\int 2-2\cos(2\theta) d\theta$

[elfin berry]

knief

[wary kite]

do you know how to integrate this

[flint widget]

so then, 2x - 2sin(2x)dx x= theta

[wary kite]

close

check the second part

[flint widget]

• 2sin?

[wary kite]

nope

maybe do a u sub to see why

or take the derivative of what you got to see why

⛓️‍💥

chain rule

[flint widget]

2x - sin(2x) + c?

[wary kite]

brilliant

[flint widget]

😭

[wary kite]

so now convert back to x

$2\theta - \sin(2\theta)) + C$

[elfin berry]

knief

[wary kite]

and we had

x = 2sin(theta)

so what’s theta

[flint widget]

a circle with a dash inside of it 👍

🤯

[wary kite]

🔥🔥🔥

but seriously

do you know how

to find theta

[flint widget]

i have no idea what 1 +1 is anymore tbh

[wary kite]

it’s alright

[wary kite]

we need to isolate theta here

any ideas

[flint widget]

well we could do theta = x/2sin

[wary kite]

no ma’am

we can’t

[flint widget]

[wary kite]

sin is a function

$x = 2\sin(\theta) \implies \frac{x}{2} = \sin(\theta)$

[elfin berry]

knief

[wary kite]

now if sin is that

what is theta

ever heard of inverse functions

[flint widget]

probably

but i dont remember

[wary kite]

well

have you heard of inverse sine?

[flint widget]

yea

arcsin

[wary kite]

as a refresher

suppose f and f^-1 are inverse functions

then

$f(f^{-1}(x)) = x = f^{-1}(f(x))$

[elfin berry]

knief

[wary kite]

with that in mind

we know that we define arcsin(x) to be the inverse function of sinx

hence

$y = \sin(x) \implies \arcsin(y) = \arcsin(\sin(x)) = x$

[elfin berry]

knief

[flint widget]

all of this work for one problem

im cooked hahahahga

[wary kite]

yea that’s how trig sub is

it isn’t much if you commit a few facts to memory

[wary kite]

do you have your final this week

[flint widget]

yep

[wary kite]

lol

[flint widget]

on friday

[wary kite]

good luck

at least you’re studying

[flint widget]

yea hopefully ill pull through

[wary kite]

you will

trust

🤝🏻

[flint widget]

most of the stuff has made sense i just suck with trig stuff

tyty

[wary kite]

yea calculus is only hard if you have weak fundamentals

[flint widget]

[wary kite]

if you’re solid with algebra and trig

[flint widget]

couldnt be me

[wary kite]

it’s trivial

[wary kite]

of course not

anyways

we’re almost there

pull through

[wary kite]

does this make sense?

[flint widget]

is there an alternative way? i remember something about using a right triangle and SOH CAH TOA

[wary kite]

we’re getting there

lol

we will use triangles in a minute

don’t worry

[wary kite]

wait what do you mean?

like for remembering arcsin?

analogous to how sin = O/H and all that

[flint widget]

nah nah nvm

[flint widget]

i honestly dont remember ever seeing arcsin(sin(x) = x in my life tho

[wary kite]

yea high school education is shit

but that’s merely a consequence of them being inverse functions

they probably just taught you to throw it into a calculator

without ever explaining what it really means

[amber birch]

also to jump in, this is not true for all real x

it's only true for the range f arcsin(x), which implies $-\pi/2 \le x \le \pi/2$

[wary kite]

yep but for the purpose of the finding the anti derivative there’s no point in really worrying about all this

[flint widget]

do your classes allow calcs?

[wary kite]

i didn’t for calculus no

also it’s useless

[amber birch]

well like it's also true for -pi/2 <= x <= pi/2

anyways

[wary kite]

a calculator can tell you the function evaluated at some point but it won’t do the algebra for me

like it’s not going to convert our expression in terms of theta to be in terms of x

[elfin berry]

southlander!

[flint widget]

yea

was kinda curious because we haven't been allowed to use a calc for any math classes

[wary kite]

you shouldn’t

what’s the use

math isn’t about crunching numbers

save that shit for physics

[flint widget]

it can be slightly annoying on an exam tho when you can spend like 10 mins in aggregate doing mental math

[wary kite]

you mean dealing with fractions and all that

i mean you should get good at mental math though

good brain exercise

[flint widget]

yea im pretty ok with it

[wary kite]

and you won’t have to do anything too complicated

do you want to talk about arcsin more

or do you understand

[flint widget]

i think im gonna come back to this problem later cuz im quite lost in the sauce

[wary kite]

all good

[flint widget]

the help was much appreciated though

sorry for asking so many questions

[wary kite]

don’t apologize

we were so close

😵‍💫😵‍💫

ping me in a bit if you decide you’re going to try it again

[flint widget]

yea

[marsh citrus]

@flint widget Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[acoustic osprey]

how to solve?

[crystal lagoon]

Post your workings again plz

I didn't see on the other channel

[acoustic osprey]

I'm confused how to get this to a fraction. Because when I do the work it has decimals

[proud arch]

First think about where the curves will bound area

As in where do they intersect

[acoustic osprey]

0 and 1

[proud arch]

Exactly

So between 0 and 1

Which curve comes on top?

[acoustic osprey]

The one with the8

[crystal lagoon]

It's not 2^(3/2) and 8^(9/8)

Those exponents are from the x

Which=1

[acoustic osprey]

It's not?

[proud arch]

Perfect

[acoustic osprey]

Did I do something wrong?

[proud arch]

So to find area between them

[crystal lagoon]

Yes

[proud arch]

You integrate (x)^1/8 - (x)^1/2

[crystal lagoon]

it's 2(1)^(2/3)

[acoustic osprey]

could u explain the steps and what mistake i made?

[crystal lagoon]

Same for the other bounds

[acoustic osprey]

its new material to me and my teacher doesn't explain much

[crystal lagoon]

See the exponents on the x?

[acoustic osprey]

x^1/2

and x^1/8

[crystal lagoon]

no, after you did indefinite integral and haven't put bounds yet

2x^(3/2)/3

[acoustic osprey]

how do i do that

[crystal lagoon]

The power is (3/2) of the X

So if you put the bounds 1,0 you get 2/3

Because the ^3/2 doesn't carry to the 2

[acoustic osprey]

im sorry but i don't really follow much unless u write it out 😦

[crystal lagoon]

I would've written it out but I'm at some restaurant rn

[acoustic osprey]

i see

[proud arch]

What he is saying is the limits are 0 and 1

[crystal lagoon]

Everything you did up till this point is correct

[proud arch]

You put 1

After integrating

[acoustic osprey]

i mean if i shoved in zero doesn't it just subtract 0?

thats why i didn't use the zero as an input

[crystal lagoon]

no I mean when you put the bounds 1

The 0 doesn't matter

Since it cancels out

[proud arch]

1^ (8/9) = 1

[crystal lagoon]

Okay, as an example, let's say I have a function f(x)=2x²

What would f(1) be

[acoustic osprey]

4

wait

w

2

[crystal lagoon]

no (2x)² isn't the same as 2x²

[acoustic osprey]

sorry its 2

[crystal lagoon]

Yeah, so here let's go to your question

[acoustic osprey]

do u got time

[crystal lagoon]

f(x)=2x^(3/2)/3

What's f(1)

[acoustic osprey]

i'll take another shot at the question

but a different one

actually its the same question

heh

[crystal lagoon]

just answer this

You'll understand your mistake

[acoustic osprey]

meaning i did the input wrong

[crystal lagoon]

yeah

[acoustic osprey]

ty

i feel like a dumbass

[crystal lagoon]

It's 2(1)^(3/2)/3

ie 2/3

[acoustic osprey]

ty!

[crystal lagoon]

Happens to the best of us😂

[acoustic osprey]

i love math.....

[crystal lagoon]

I can imagine writing all that out into a screen takes so much time

[acoustic osprey]

yea.

thanks!

.close

[marsh citrus]

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[marsh citrus]

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[fierce crane]

i need help

[fierce crane]

how do i do this

[carmine sandal]

1 significant figure is the first non zero digit in your number

while looking at it from left to right

[fierce crane]

Oh okkk

[marsh citrus]

@fierce crane Has your question been resolved?

[marsh citrus]

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[carmine sandal]

My thought process is to prove by contradiction, trying to construct such plan and show that this is not possible. In order to do so I first picked any arbitrary point

and pick a colour, say blue

then we know that a unit circle around it

the colour of each point is red or white, then by picking some random point on the unit circle

and by just shifting a bit along the circle

this can probably be replicated, and then picking any point on this circle

but then i dont see a way this is contradictory

[undone kite]

example of Linear Equation in two variable in real life

pleasee

<@&286206848099549185>

[carmine sandal]

!occupied

[marsh citrus]

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).

[undone kite]

tyy

[carmine sandal]

bt yes i need help on this qs i suppose

<@&286206848099549185>

[plush radish]

hello, everyone..

I am here to help people with mathematics.

pls send md DM if need help

[carmine sandal]

.close

[marsh citrus]

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[cosmic raptor]

With these propositions, i’ve noticed as part of the proof that they already use property they’re proposing

[knotty trellis]

which part?

[cosmic raptor]

“by the inductive hypothesis n+m=m+n”

How does that usually work?

For proposition 2.2.4

[knotty trellis]

they are doing induction on n

for that, you need to prove that
if n+m = m+n, then (n++) + m = m + (n++)

i.e. if this result holds for some n, then it holds for n++ as well

[cosmic raptor]

I got up to (n+m)++ = (m+n)++ using definitions and lemmas from previous results

But don’t know how we proved n+m = m+n

[knotty trellis]

this gives you a hint that you should be using induction

imagine that we have proved
0 + m = m + 0

[knotty trellis]

then by this, we also know that 1 + m = m + 1

because
1 + m = (0 + m)++ = (m + 0)++ = m + 1

and we also know 2 + m = m + 2....

and to formalize this argument, we use induction on n

do you know how induction works?

[cosmic raptor]

Ohhh it’s already proven when we got (n+m)++ = (m+n)++ right?

[knotty trellis]

I wouldnt say proven, but it's assumed

[cosmic raptor]

Sort of, with discrete maths course i’ve done yeah where we select base case n, and prove case in n+1 still holds

But not this definition

[knotty trellis]

you can think of induction as a theorem, which says that if you have a statement about natural number n, call it P(n) (which in our case is n + m = m + n),
then if prove that

1. P(0)
2. P(n) => P(n++)
   then you can conclude
   P(n) for all n

[knotty trellis]

this is a good way to phrase it

in our case, P(0) is 0 + m = m + 0

which i suppose is already proven

and then you need to prove that if P(n), then P(n+1)

meaning that you need to prove that
If n + m = m + n, then (n++) + m = m + (n++)

and you can prove that by assuming n + m = m + n and then concluding (n++) + m = (n + m)++ = (m + n)++ = m + (n++)

[cosmic raptor]

Oh so P(0) is the base case, then P(n) is the inductive step, then if it is true, then P(n++) must be true as well?

[knotty trellis]

P(n) is called the inductive hypothesis

the inductive step is proving P(n) -> P(n++)

to do the inductive step, you assume P(n) and prove P(n++)

[knotty trellis]

but ye, this is the main idea

and if you manage to prove all that, then the induction principle rewards you by concluding P(n) for all n

[cosmic raptor]

Oh so we have to prove 3 cases? P(0), P(n), and P(n+1)? I would assume n is just a step after 0?

Like maybe n=0+1?

[knotty trellis]

no, not really

we need to prove 2 things:

Firstly, prove P(0) without assuming anything

then prove P(n++) assuming P(n)

(or equivalently, prove "If P(n), then P(n++)")

[marsh citrus]

@cosmic raptor Has your question been resolved?

[cosmic raptor]

Ohhh, makes sense, thanks!!

[marsh citrus]

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[kind lagoon]

does somebody know how i could use cauchy to solve this

[kind lagoon]

i seperatet the row in 1/2^n and the otther part

the 1/2^n part isnt hart but the other

i tried it with partial fracture decomposition
but the extent of the upper part the fraction ist larger then the lower part and now iam stuch

[lucid zenith]

cauchy?

[kind lagoon]

yes

in germany we say Cauchy kriterium

something like thsio

[devout mauve]

nothing in the question says to use the cauchy criterion. why do you want to?

[lucid zenith]

the cauchy criterion is to check for the convergence of a series, not to find its value

[kind lagoon]

ok

then how should i move on

can or should i convert it to a telescope sum#

or something else

[lucid zenith]

I think you will have to change the order of the summations

[kind lagoon]

ok illtry

[marsh citrus]

@kind lagoon Has your question been resolved?

[marsh citrus]

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[drowsy stirrup]

Can someone help with these 2 determinants?

[drowsy stirrup]

please someone

[marsh peak]

For the first matrix, do a cofactor expansion along the first column

For the second matrix, subtract the first column from the other columns, do a cofactor expansion along the first row and for the determinant of the submatrix, you will have common factors along the rows, factor them out of the determinant and you should be able to recursively figure out the rest

Or just look up Vandermonde matrix

[marsh citrus]

@drowsy stirrup Has your question been resolved?

[drowsy stirrup]

one moment lemme try

[drowsy stirrup]

Is that what you mean for the first one?

[marsh peak]

I meant expanding along the first column, you seem to be doing it along the first row in that equation

[drowsy stirrup]

oh yeah mb

[drowsy stirrup]

So this then

[marsh peak]

Yeah, now you can notice that the submatrices A_11 and A_21 are equal and the same as the original matrix except with size (n-1)x(n-1)

So you could define the sequence det(A_n) by det(A_{n+1}) = 2det(A_n) and by induction show that det(A_n) is something times 2^n

The base case being det(A_1) = 1

[drowsy stirrup]

Wait I get the size being n-1 but how are they equal

[drowsy stirrup]

.

[marsh peak]

For the first matrix

[drowsy stirrup]

they dont seem equal to the first matrix

maybe i just dont see it

[marsh peak]

"They" being what?

[drowsy stirrup]

the submatrices a11 and a12

you said "you can notice that the submatrices A_11 and A_12 are equal and the same as the original matrix"

i dont see it tho

[marsh peak]

If you are talking about these A_11 and A_12, I never meant the submatrices of the second matrix anyway

Do a cofactor expansion along the first column in the first matrix, not the second

[marsh peak]

Mb I meant A_21 here

[drowsy stirrup]

omg so sorry im a dumbass i got the mixed up again

[drowsy stirrup]

so det= 2det(A11) or det = 2det(A21)

[marsh peak]

A11 = A21, so both hold

[drowsy stirrup]

is this enough of an answer?