help-17

[sly mountain]

what

okk

I mean in order to solve that you need to write down 3 equations since you have 3 informations regarding that curve

1. Amplitutde : 4 = (y_max-y_min)/2

so you need to find y_max and y_min

2. period of pi/3, so basically : sin(cx + cpi/2) = sin(cx). So cpi/2 = ?

3. 2 = a + b sin(x*pi/12)

does it help ?

[heady atlas]

I don’t get it help 😭😭

I will js do it with my physics tutor tmr

Thanks for helping me

.close

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[sly mountain]

bruh just tell me what you don't understand lil bro

[heady atlas]

ok so wait

how do you know y_max and y_min

[sly mountain]

Generally speaking, through derivative. You get y' and then you see where y' = 0. This means you have an extremum. But in this case, an easy way to see that is to see that :
y = a+ bsin(cx)

But the highest sin(cx) can get is 1 and the lowest is -1.
So the highest and lowest bsin(cx) can go is b and - b
So the highest and lowest bsin(x)+a can go is a+b and a-b

[heady atlas]

i see

thanks Lebrown

[sly mountain]

np dude

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[merry belfry]

I’m trying to find my second angle using law of sines but I’m getting the same exact answer for my second equation

[halcyon ice]

I feel like you might have gotten some of the sides that youve inputted into the trig function messed up cuz of the drawing thats DEFINITELY not to scale

youre basing your input into the trig functions on the triangle youve drawn which has B as the right angle despite you having C as the right angle in your assumptions

so imo you should redo the drawing to make C a right angle, then redo the angle calculations

[merry belfry]

Ok I’ll do that thank you

My teacher draws the angle like that so I just copied it

[halcyon ice]

yw

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[spiral needle]

Anyone help me solve these problem

[atomic jasper]

!show

[vocal sleet]

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

[spiral needle]

Let first do 2nd problem I try to use sin( theta + 45) formula

sin(theta + 45) = sqrt(2)

[inner ibex]

do you know calculus?

[spiral needle]

ya but trying to do it without it

but let's explore it

[inner ibex]

explore the calculus or no?

[atomic jasper]

try squaring

[spiral needle]

No Noo I know Calculus

[spiral needle]

okay

sin^2( theta + 45) = 2 ?

[atomic jasper]

what?

$(\sin (\theta)+\cos (\theta))^2=2$

[twin meteor]

Bonk

[spiral needle]

I try to take squre on both side

ohh

sin^2(theta) + cos^2(theta) + 2sin(theta)cos(theta) = 2

sin^2(theta) + cos^2(theta) = 1

So, 1 + 2sin(theta)cos(theta) = 2
2sin(theta)cos(theta) = 2 - 1
2sin(theta)cos(theta) = 1

And 2sin(theta)cos(theta) = sin2(theta)

This is correct right?

[atomic jasper]

yeah i think so

i always forget the double angle stuff

,tex .double angle

[twin meteor]

Bonk

[atomic jasper]

its sin(2theta)

[spiral needle]

yup

My answer is theta = 0

[atomic jasper]

thats incorrect

[spiral needle]

ya got it

[atomic jasper]

$\sin (2\theta)=1\implies 2\theta = \frac{\pi}2+2k\pi$

[spiral needle]

pie / 4

[atomic jasper]

uhm

no

pi/2

[twin meteor]

Bonk

[atomic jasper]

,w sin(pi/2)

[twin meteor]

Results provided by WolframAlpha

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Requested by .bonk__

[spiral needle]

but we also got 2theta so when we shift 2 to other side to find theta the answer become pie/4

right?

[atomic jasper]

yeah well it becomes $\theta = \frac{\pi}4+k\pi$

[twin meteor]

Bonk

[spiral needle]

yup got it thanks

What about first one any idea

[atomic jasper]

rewrite to sin(theta)=stuff

[spiral needle]

sin^2(theta) = 1/4 then sqrt[ sin^2(theta) ] = sqrt(1/4)
sin(theta) = 1/2

[atomic jasper]

*$\pm\frac12$

[spiral needle]

theta = sin^-1(1/2)
theta = pie/6

[twin meteor]

Bonk

[spiral needle]

ya sorry

then what is 2npie in the option any idea

[atomic jasper]

$\theta =\pm\frac{\pi}6+2k\pi$

[twin meteor]

Bonk

[atomic jasper]

oh wait, thats just the answer lol

ig i kinda spoiled it now

[spiral needle]

I got the answer but I do not under the purpose of "2kpie" Is it a general term

[atomic jasper]

sin and cos are periodic

sin(pi/6)=sin(pi/6+2pi)=sin(pi/6+4pi)=sin(pi/6+6pi)=sin(pi/6-2pi) etc

[spiral needle]

got it

[vocal sleet]

@spiral needle Has your question been resolved?

[spiral needle]

Thanks for your time

Bye Have a great day

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[green kraken]

can someone help me solve this without a calculator

[finite bay]

you basically assume x = 0.4141...
then multiply 10 both sides

and subtract the original equation from the new equation

[green kraken]

i dont get it

[prisma rain]

is that 0.414141... or 0.41111111...

[finite bay]

lets walk through the steps

what you can do here is
x = 0.411111
then you can make two equations

1. 10x = 4.1111
   and also
2. 100x = 41.1111

then do 2) - 1)

100x - 10x = 41.111... - 4.111...

is this more clear

[green kraken]

after subtracting we just find for x?

[finite bay]

yup

x will be the simplest form, the reccuring decimals all cancel to 0

[green kraken]

uhhh let me check could you give me a min

[finite bay]

sure take your time

[green kraken]

I can't tell if I'm doing this correctly

[finite bay]

close your 90x is correct but check the right hand side
41.11 - 4.111

you have to do 41 - 4 and .111 - .111
you subtract at the same point, you have to line up the point

[green kraken]

Got ittt :D

So for the recurring I make sure that it cancels out?

[finite bay]

yesss

[finite bay]

amazing!

[green kraken]

thanks for helping me out 🫶

[finite bay]

awww welcomee

[green kraken]

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[gusty heron]

How many number I can make from 2 numbers that are greater than 10 and consist of 0,2,3,4

[gusty heron]

Please show me the step

[quiet echo]

!original

[vocal sleet]

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.

[gusty heron]

A 2-digit different number greater than 10 is formed from the numbers 0, 2, 5, 7, 8. How many numbers can be formed?

[reef agate]

Are you allowed to reuse the digits?

(ex, does 22 work?)

[bitter pilot]

ig just make sure your first digit is not 0 and you should be good

[gusty heron]

Sorry it was translated

[gusty heron]

No

No

[reef agate]

Ok

So we have the digits {0, 2, 5, 7, 8}

For the first of the two digits, what are our possible options

[gusty heron]

2,5,7,8

Cuz zero < 10

4 × 3 then?

[reef agate]

No

4 for the first digit

But for the second digit, you can include 0

[gusty heron]

Ahhh I see

4 × 4

[reef agate]

But ||you exclude the digit you selected for the first digit||

[gusty heron]

So that was easy

Thanks guys ♥️

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[green tree]

How do I implicitly differentiate sin(x+y)=1?

[heavy yoke]

take derivative of both sides and use chain rule

[rugged orchid]

What have you tried

[hearty jungle]

!status

[vocal sleet]

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

[green tree]

One sec, ill take a pic of my process.

[quiet echo]

Did you watch https://discord.com/channels/268882317391429632/903486480075354182 ? What did you not understand?

[green tree]

Your explanation

Sorry to be so blunt about it but I'd like two perspectives in order to better grasp the concept.

[quiet echo]

Ok

[green tree]

Whatever’s in the box is my current method. How can i fix this or is there another way?

[heavy yoke]

to be honest i don't see why you really need to use the double angle identity for this problem

[ember verge]

i know

what you did

you took u'v' + u'v'

instead of u'v + uv'

[heavy yoke]

it would have been easier just to leave it as sin(x+y)

[green tree]

Ohhh thank you @heavy yoke. That clarified it a lot.

[ember verge]

$f(x)g(x)$ deriv is $f'(x)g(x) + f(x)g'(x)$

[twin meteor]

gamer75431

[green tree]

Can you stop flooding my help chat, kind stranger?

[ember verge]

💔

[green tree]

I bow down to your intelligence monsieur. What expertise you display; astounding. 👏

[ember verge]

no need for this rude behaviour @green tree

[green tree]

I am simply displaying my affection towards this helpful and generously talented sire. How heinous of you to presume my innocent behavior has an ulterior motive.

[vocal sleet]

@green tree Has your question been resolved?

[quiet echo]

I quite literally think that's what I said.

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[brave remnant]

I made this incorrect proof to show that cos 330 = cos 60.

[brave remnant]

I dont know why it's inccorect

but I know it's definetly wrong

[pallid zenith]

maybe you mean 300?

[brave remnant]

I meant it to be 330 because 330 = 270 + 60

[pallid zenith]

cos300 cong cos60 or cos330 cong cos30

[brave remnant]

cos 330 = cos 60

[pallid zenith]

i dont see any reason that would be true

cos is even about 0

[deft oar]

Do you mean -sin?

[brave remnant]

nope

I know it's not equal so

[brave remnant]

I made this one

[deft oar]

💀

Why’d you make it 🗿🤨🤨

[pallid zenith]

maybe you try to track out

[brave remnant]

I dont know trig so I tried to use what I understand to solve them

[pallid zenith]

cos(270+60) = cos270 cos60 - sin270 sin60

[deft oar]

🤩

[brave remnant]

it's a formular

[pallid zenith]

this is cos330= sin60

[deft oar]

Regrettably, you do have to memorize some of the angles,

[pallid zenith]

not cos60

[brave remnant]

I know the basics angles like 30 60 46 90 and 0

[deft oar]

But there is a sort of pattern so it’s not too bad

[brave remnant]

alright but

[pallid zenith]

all i used is sum and difference and unit circle

[deft oar]

46?🤨🤨🤨

[brave remnant]

45 * sorry

[pallid zenith]

??

[deft oar]

🤨🤨🤨

[brave remnant]

take a look at it first dont it make sense

two triangles are congruent

[deft oar]

No

[brave remnant]

aw

why is it wrong?

[pallid zenith]

you have just made a mistake man

[deft oar]

You were actually measuring sin

I’m assuming

[brave remnant]

wait really?

[pallid zenith]

you put the angle in the wrong place

in the lower

[deft oar]

Probably this actually

[brave remnant]

how should I draw angle 330?

[deft oar]

Yeah, definitely this

[pallid zenith]

its not 330 you want

draw both angles as coterminal

you will get 300 and 60

then the answer will be right

[deft oar]

Actually wait, it wasn’t your angle being placed wrong

[brave remnant]

one moment i'm going to draw iit

[deft oar]

You see how you set x= -y

[pallid zenith]

it is absolutely not true that cos(x+pi/2) = cosx in general or really at all

[deft oar]

-y isn’t cosine of 330 but sin of 330

You had the wrong orientation with your cos and sins

[brave remnant]

this is the incorrect drawing?

I should instead draw

[deft oar]

Also this 👆👆👆

[brave remnant]

this is the correct one?

[deft oar]

Right drawing wrong angle

[deft oar]

But also this

But also 300 instead of 330

[brave remnant]

can you maybe draw the picture for me plz ? I can't imagine it

[deft oar]

Both angle and orientation is wrong

[pallid zenith]

you gotta do it @deft oar

im on a plane lol

[brave remnant]

a plane??

[deft oar]

Soo much work 😟

[pallid zenith]

a plane

[brave remnant]

I'm sorry 😟

[deft oar]

Level infinity helper

[brave remnant]

for real

[pallid zenith]

i paid for wifi to be here lmao

[pallid zenith]

draw two triangle that share the 0 degree line

one goes up, one goes down

[brave remnant]

ok

[deft oar]

I think you’re drawing it right, you just got the wrong orientation

[pallid zenith]

youll find that to make cosx of both equal, the angles need to be the same magnitude

just going opposite directions

[twin meteor]

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Requested by janniku

[pallid zenith]

this makes sense cosx looks the same going from 0 left or right

[pallid zenith]

yea

[pallid zenith]

so cosx is the same here right for both

its the same line

[brave remnant]

yes

[pallid zenith]

but that only makes sense if the red angles are the same

[deft oar]

Remember cos only look at x axis, sin only look at y axis, unless your doing your angles wonky

[brave remnant]

yes

[pallid zenith]

so if you go 60 up, you must go 60 down

[brave remnant]

alright

[pallid zenith]

thats 60 and 360-60=300

if it were 330 thered be two different lines at the 0 angle mark, not the same cosine

[deft oar]

You do angles wonky quite a lot, so don’t make this a habit, just use it to get through a test if your struggling. The real way is to remember that cosine is adjacent to the angle and sin is opposite to the angle

SOHCAHTOA for the win

[pallid zenith]

ye

[brave remnant]

Okay 👍

[pallid zenith]

and sum and difference

and unit circle

[brave remnant]

alright

thanks

[pallid zenith]

np

[deft oar]

Yeah, but he measured up from 270, so I think it was a bit wonky anyways

[brave remnant]

I'll just remember that I can't do that

[deft oar]

Like it was right

Just that you got confused on your orientation

[brave remnant]

you mean it was suppose to change from angle 60 to angle 30?

[deft oar]

You totally can measure up from 270 or any angle you wish, it just means you gotta see the orientation correctly

[brave remnant]

okay

[deft oar]

No idea what this is referring to

[brave remnant]

ike this

[deft oar]

This is right

[brave remnant]

triangle with angle 60 is the incorrect orientation

[deft oar]

The angle 60 is correct

When I say orientation, I’m not referring to your angles

[brave remnant]

oh

[deft oar]

The angles you measured out were right

It’s just that the associative relationship between the angles and the axises, since you were using the 60 up from 270 you have to remember that you’re cos is actually the cosine of 330 down to the x axis and your sin is from 330 to the y axis, at this point you’ve associated that cos of 60 with the cos of 330 even thought that cos of 60 is actually referring to the y axis because of the orientation shift

Actually let me rephrase this a bit less confusingly

You know how you associated the similar triangles measuring the angle 60 up from 0 degrees and up from 270 degrees

When you measure up from 270, the cosine function measures the side adjacent to the angle that’s not the hypotenuse

Which would be measuring the y-axis

[brave remnant]

this is incorrect?

[deft oar]

Therefore, when you switched from thinking about the angle going up from 60 to all the way around 330, you made the mistake of thinking cos(330) still measured the y-axis, when the new adjacent side would’ve been the x-axis

[brave remnant]

just to confirm

[deft oar]

Yes

But then also sin(330) would equal -y

And cos(330) would equal cos(30)

Typo my bad

So cos(60) = -sin(330)

[brave remnant]

I understand a little bit. I will try to draw the correct way

[deft oar]

You’re not drawing it wrong

Drawing is not the problem

You have the right angles at the right places

The real problem is that you don’t yet understand what sin and cos really mean, and you’re making a mistake on orientation

I highly recommend you search up “SOHCAHTOA” if you’re teacher didn’t teach this method to you

[brave remnant]

okay

let me search right now

[deft oar]

It gives a much easier way to understand sin and cos that allows you to do orientation switches like this more comfortably

[brave remnant]

this one?

[deft oar]

🤩 🤩 🤩 🤩

My precious

I highly recommend watching a video explaining it ontop of reviewing the diagrams

Red is opposite, green is hypotenuse, and blue is adjacent

[brave remnant]

oh I think I alreaned it

[deft oar]

Then review it! Use it whenever you see a sin or cos

It will make your life very easy🤩🤩🤩

[brave remnant]

alright

okay

wait so I did sin cos tan wrong?

[deft oar]

Yes, you didn’t see which sides were adjacent and opposite for the angle you were looking at

Look again to the diagram you drew

[brave remnant]

okay let me redo

[deft oar]

For the angle up from 270 and the angle down from 0, ie 60 up from 270 and 330, do they have the same adjacent and opposite sides🤨🤨🤨

[brave remnant]

nope

[deft oar]

Exactly! Now then why would cosine(330)=cos(60) when their adjacent sides are of clearly different length

[brave remnant]

wait

[deft oar]

And their hypotenuse is of the same length and equal to 1

[brave remnant]

do they? beacuse they ar econgruent they must have equal length?

[deft oar]

a/1 == b/1 if a==b 👈👈👈👈👈

But a does not equal b

So a/1 does not equal b/1

[brave remnant]

like this

[deft oar]

Sure, but which side is cos(330) measuring, is it the same side cos(60) is measuring 🤨🤨🤨🤨🤨🤨🤨

[brave remnant]

cos of 330 is the y axis and cos of 60 is the x axis

[deft oar]

Nuh uh uh,

Which side is adjacent 🤨🤨🤨

[brave remnant]

a and b?

maybe?...

[deft oar]

a is adjacent to 60

You agree

[brave remnant]

yes

[deft oar]

But, you’re not using 60, you were using 330

Draw out 330

And see again which side is adjacent

[brave remnant]

I see two ways of drawing 330

one triangle has angle 330 and another has -30

[deft oar]

Neither way is really drawing out 330

[brave remnant]

o

[deft oar]

And both ways you’ll get different adjacent sides yes?

[brave remnant]

yes

how do I draw 330?

[deft oar]

Draw all the way arround from 0

The long way

[deft oar]

Aha

Now

Which side do you think is adjacent

[brave remnant]

I thought now we wont have a right triangle

[deft oar]

It’s a bit abstracted away from triangles now

[brave remnant]

oh

[deft oar]

But, from the hypotenuse there’s still a value that lands on x and a value that lands on y

Cos will land on x

Sin will land in y

Because we aren’t measuring our angles funkily but using the og long way

[brave remnant]

like this?

[deft oar]

So, cosine of 330 would be the x axis length, yes??

[brave remnant]

ohhhh

yes

[deft oar]

🤩 🤩 🤩 🤩 🤩

[brave remnant]

that is true

I just realized what you meant by I'm not drawing 330 angle

[deft oar]

Not really, what I mean is that you’re not getting the orientation of cos and sin for the angles you drew

Now draw the -30 angle again

You’ll see that for -30 and 330, they both share the same x projection

[brave remnant]

yes

[deft oar]

🤩

[brave remnant]

therefore -30 = 330

[deft oar]

No!

Cos(-30) = Cos(330)

[brave remnant]

okay

[deft oar]

And sin(-30) = sin(330)

[brave remnant]

I cleared my understanding

[deft oar]

But cos(60) actually also equals -sin(330)

[brave remnant]

oh

[deft oar]

I highly recommend you memorize your unit circle

[brave remnant]

okay

[deft oar]

Even if it’s not intuitive now, by memorizing the values it’ll be easier for you to develop an intuitive understanding

[brave remnant]

alright. I thought that everything is inside of a right triangle so I try to put them in right tirangle

but I understand now

thanks for taking your time

[deft oar]

There’s is a right angle there

It’s hard to explain without drawings

But watch a few videos on how the sine and cosine functions actually work

[brave remnant]

okay

[deft oar]

Having animations really makes it easier

[hearty jungle]

circle has triangle

[deft oar]

🗿

[brave remnant]

sometimes I just have to remmeber it first

alright

tysm

.close

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[jagged cargo]

what have you tried?

[shrewd stone]

nothing

[deft oar]

😭

[shrewd stone]

i forgot how to solve polynomial diophantine equations

[lean iris]

x³ = -21y² - 5

Maybe you can replace it and then move on.

[shrewd stone]

trial calculations?

[lean iris]

Just giving you advice, ask the experts instead.

Hey, kungfu Panda @hearty jungle

[jagged cargo]

forget everything i said

[lean iris]

Try hard method, u are a man

[lean iris]

.

[deft oar]

Same

[gentle sleet]

could try reducing mod 7

[cedar bone]

Then ||considering the cubes of the residues mod 7 should get you an answer||

[shrewd stone]

yeah i figured trhat out

.close

[vocal sleet]

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[hearty jungle]

oh

nevermind

[vocal sleet]

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[sudden marten]

doin integration by parts and this one just seems to go in circles, heres what i got so far

[heavy yoke]

you didn't apply the integration by parts formula correctly

[sudden marten]

oh?

whered i go wrong

[heavy yoke]

,, \int u \dl v = uv - \int v \dl u

[twin meteor]

cloud

[sudden marten]

i do not see how thats different from what i did 😟

[heavy yoke]

double check the integral part specifically

[deft oar]

that emoji was a mistake, my bad

[deft oar]

hint: integral part on the righthand side

[sudden marten]

did i fix it

[deft oar]

My bad, misread

[deft oar]

yep you fixed it

[deft oar]

lowkey, i'm tripping all the time

[sudden marten]

i still do not know how to go from here, if i IBP again it just goes in circles

[deft oar]

did you teacher teach you add to other side trick?

well, I think you gotta uv - int(vdu) one more time

but, it should be pretty easy after that

[deft oar]

don't qoute me on this, or anythin ever

[sudden marten]

potentionally

lemme try

im getting 0 = 0

it just goes in circles i dont know where to go

[deft oar]

can you show me your work?

[deft oar]

where did the sine come from

[sudden marten]

which gave me this

[deft oar]

in e^(-x)sin(x) on the second uv

[sudden marten]

thats uv

[deft oar]

okay, let me solve this real quick

I'll check to see what's up

Maybe your problem is that you chose e^(-x) as you U

[sudden marten]

holy shit

think i got it

yeah

idk why that messed everything up but when i chose the other thing

it just worked

[deft oar]

My teach taught me ILATE cuase exponents suck

you should get something like (-cos(x)e^(-x)+sin(x)e^(-x))/2 as your indefinite integral

sorry for all the parenthesis, never got arround to learning latex

[deft oar]

actually, I made a mistake with this one

[deft oar]

had the wrong sign somewhere and now I'm too lazy to fix it

[vocal sleet]

@sudden marten Has your question been resolved?

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[arctic knot]

I just want to know if my math is correct right now

[hoary apex]

Impossible algebra equation 2 🥶

[arctic knot]

I’m tryna stump my teacher

[hoary apex]

Its seem correct to me

[arctic knot]

It’s not all the way done yet, but maybe I should leave it like that and make him figure the rest out?

[hoary apex]

Ok?

[vocal sleet]

@arctic knot Has your question been resolved?

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[sudden compass]

Yo

[sudden compass]

So i was doing some integration today (no special functions:\ )

And I had to use riemann sums and stuff

[reef agate]

So what’s your question

[sudden compass]

So i use the definition we were taught (formally called the left rule)

But I remembered there being other rules like darboux upper and lower sums and trapezoidal rule

Is it possible to prove that in the limit of all these sums (ie as n goes to infinity) they all converge to the same thing

[reef agate]

No, because there’s a countereample

Take upper/lower sum for y = x for example

[sudden compass]

o

[reef agate]

One will converge to slightly more than y = x

Other converges to slightly less

And I don’t mean slightly = 1/infty

[sudden compass]

Don't they both converge to the integral of x dx in the limit as n goes to infinity?

[reef agate]

What is n

[sudden compass]

Like if I compute them individually? Wait don't answer this

[reef agate]

Upper bound of integration?

[sudden compass]

The upper bound of the riemann sums?

I assume that's standard notation?

[reef agate]

Then the upper sum for y = x will be significantly more the lower sum

[desert hornet]

Reimann and darboux integrals are equivalent, meaning that a function is reimann integrable if and only if it is darboux integrable, and their values are always the same

[desert hornet]

You want the mesh to go to zero

[sudden compass]

How do I go about proving that? That is the one i have the biggest problem with because I still can't get infinums and supremums

[desert hornet]

https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Riemann_and_Darboux_Integrals

[sudden compass]

Yes that's another way of putting it (we don't use mesh to define it)

[desert hornet]

You have to, to use the epsilon-delta definition

For all epsilon > 0 there exists delta > 0 such that for every tagged partition of [a, b] with mesh < delta, |sum - L| < epsilon

[sudden compass]

i don't think we have used that with integrals before, but I believe I have seen it

[sudden compass]

Yeah yeah this is similar to what's on wikipedia i remember reading it

[desert hornet]

An integral is a limit

So it can be defined by epsilon-delta

[sudden compass]

Will read this when I get home 👍

[sudden compass]

I agree with that just never used it

Anyhow, what about proving this for say, the trapezoidal rule and left rule?

So formally speaking I want to prove that $$\lim_{n \to \infty} \sum_{i=1}^n f\left(\frac{x_i+x_{i+1}}{2}\right)\Delta x_i = \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x_i$$

[twin meteor]

rak³en

[paper depot]

doesn't the general definition of a riemann integral involve a limit taken along all possible subdivisions of [a,b] with all possible choices of points to evaluate f at within each piece

(with the biggest subinterval length taken as the thing that approaches 0)

[sudden compass]

I don't think I get what you mean..?

[paper depot]

all of those other rules are special cases of this

https://en.wikipedia.org/wiki/Riemann_integral?wprov=sfla1

[sudden compass]

How could you choose all possible choices

[desert hornet]

Yeah you need to prove it for all tagged partitions

[heavy yoke]

(which is why the darboux integral is easier to work with, since you at least only have 2 choices of evaluation point)

[sudden compass]

Ohh wait so you mean that since the formula is true for any choice of $t_i$ all the rules are equivalent

[twin meteor]

rak³en

[sudden compass]

Is there a source for that?

[vocal sleet]

@sudden compass Has your question been resolved?

[vocal sleet]

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[turbid snow]

let's say my radius of a sphere goes from value a to value b (it is a varying radius), how would the parametrization of the sphere look like?

would the parametrization in spherical coordinates look like (r, theta, phi)? how would i get ru cross rv?

[paper depot]

well it would not be a sphere for starters

what shape is your thing, an ellipsoid?

!xy

[vocal sleet]

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.

[reef agate]

Consider using $\frac 43 \pi r_x r_y r_z$

[twin meteor]

King Leo [Ping For Help]

[vocal sleet]

@turbid snow Has your question been resolved?

[turbid snow]

what do i use as ru and rv

[turbid snow]

it's a sphere with a varying radius

it goes from r = a to r = b

[paper depot]

there is no such thing as "a sphere with a varying radius"

either this phrase describes a class of shapes way too broad to tell you anything about it,

or it does not describe anything at all

[turbid snow]

a sphere that gets bigger and bigger in size

[paper depot]

ok look

are you doing a math problem right now

yes or no

[turbid snow]

nope it's a physics problem

[paper depot]

ok a physics problem

did you get it from a textbook or did you make it up yourself

[atomic summit]

Give your sphere parameters r,θ,φ

But parameterise r in u

[turbid snow]

wdym

[atomic summit]

Then you’ll have spherical coordinates

x(u,v) = r(u) sin(v)cos(u)

So your radius will be a function of the angle theta

[turbid snow]

they want to find the electric field of a sphere of a radius a concentric with a spherical shell that exists from radius b to radius c. I want to find the electric field between r = a and r = b, the radius is changing, so when we draw the gaussian surface wouldn't its radius change as well?

[atomic summit]

Then u will vary through theta angles 0:2π

[paper depot]

show the problem exactly as it is originally stated

[atomic summit]

For a sphere with a radius varying linearly from a to b define r(u) = a+(b-a)(u/2π)

[paper depot]

right now you are not making any sense at all

[paper depot]

wait, so are you saying OP meant a spherical SHELL?

a sphere of a radius a concentric with a spherical shell that exists from radius b to radius c

ok right

that makes A LOT more sense thank you

no idea what "gaussian surface" is supposed to mean but at least we got the geometry of the problem sorted out?

[atomic summit]

Yes he’s describing a shell

Gauss’s law states that:
$\oint_{\mathcal{S}} \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0}$

[twin meteor]

TLB

[atomic summit]

So you’re going to analyse the sphere in three different regions

Inside the inner radius

[turbid snow]

i want it between r = a and r = b

[atomic summit]

Inside the inner shell

Do you know if the charge is uniformly distributed inside the inner shell?

[turbid snow]

it is

[atomic summit]

Then there’s a formula for it

[turbid snow]

it's uniform with charge Q

[atomic summit]

But you want a<r<b right?

[turbid snow]

yes, the electric field is kQ/r^2 but how do we derive it in spherical coordinates

[atomic summit]

Are you just looking for how you parameterise this s.t. You get yep okay

[turbid snow]

yes

the radius is changing

right?

r = (r, theta, phi)

like this?

but we need two variables to change to get the vector dS

[atomic summit]

Yeah so do u,v

[turbid snow]

what's u and what's v

[atomic summit]

And let the radius be a function of the inner angle

Using r is a little confusing so let’s call it something else

I like p as parameterise.

[turbid snow]

ok

what's p equal to

[atomic summit]

p(u,v) = <r(u) sin (v) cos u, r(u) sin (v) sin (u), r(u) cos (v)>

Then you give r(u) as I did above

[atomic summit]

Here

Yes the derivations will be shitty hahaha

Sort of

[turbid snow]

the gaussian surface is a spherical shell

u is what

which angle

[atomic summit]

Theta

U ranges from 0 to 2π

Is that all you need bro?

[turbid snow]

im trying to calculate it

i wanna get dS to be psin(phi) * pdphidthetaphat

[atomic summit]

Is that what the cross product is supposed to be?

[turbid snow]

ru cross rv should be phat

or maybe not

[atomic summit]

Makes sense

I haven’t done this in ages honestly can’t remember the exact definitions

[turbid snow]

im getting rphi to be <0, 1, 0>, and rtheta to be <pi(b-a)/2, 0, 1>

[atomic summit]

Sounds right, sorry I’m out and can’t write it down and look at it

Trying to visualise the x product in my head and failing spectacularly

[turbid snow]

so n = ap -pi(b-a)/2 az

and then E is E_0 ap (constant)

when we dot E with n, the z component cancels

[turbid snow]

right?

[atomic summit]

Yes

[turbid snow]

how did you get this, what would r(u) be for r>b and r<a

[atomic summit]

It behaves like a point charge in that case so radius would suffice

[turbid snow]

The inner sphere has a uniform charge distribution.

The Gaussian spherical shell has a varying radius r<a

How would I parametrize it

What would be r(u)

You can’t treat it as a point charge for r<a

[vocal sleet]

@turbid snow Has your question been resolved?

[turbid snow]

how did u know that p depends on the angle theta btw? is there a quick way to know it?

[vocal sleet]

@turbid snow Has your question been resolved?

[vocal sleet]

@turbid snow Has your question been resolved?

[vocal sleet]

@turbid snow Has your question been resolved?

[vocal sleet]

@turbid snow Has your question been resolved?

[vocal sleet]

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[dim ivy]

im getting a different answer than the one in the textbook

[dim ivy]

so im confused on what im doing wrong

its a quadratic equation*

[ebon rapids]

Show your work

[dim ivy]

$2x(x-2) = (2x - 5)(2x + 3)$

[twin meteor]

sircule

[ebon rapids]

Ah so uhh

Don’t cross-multiply

[dim ivy]

why?

[ebon rapids]

Because x-2 and 2x+3 could be zero and we can’t have that

[dim ivy]

so then how do i go about this?

[ebon rapids]

Or well hmm

You could keep in mind x-2 and 2x+3 must be nonzero

If you want to preserve that

[dim ivy]

alright...

[ebon rapids]

Just show the rest of the sol

Show the whole sol and I’ll try to spot which line you fibbed

[dim ivy]

that was the first step

if thats wrong

[ebon rapids]

It’s not wrong if you just keep the technicality in mind

[dim ivy]

ah

$2x^{2}-4x = 4x^{2}-4x-15$

[twin meteor]

sircule

[dim ivy]

then -4x cross out on both sides

so then $2x^{2}=4x^{2}-15$

[twin meteor]

sircule

[dim ivy]

$0 = 2x^{2}-15$

[twin meteor]

sircule

[dim ivy]

the factorised form of that would be $(\sqrt{2}x+\sqrt{15})(\sqrt{2}x-\sqrt{15})$

right?

[ebon rapids]

ok stop

Why is the x in the square root?

[twin meteor]

sircule

[dim ivy]

oh mb

[ebon rapids]

Anyway yeah this works

[dim ivy]

so

$x = \frac{\sqrt{15}}{\sqr{2}}$

[twin meteor]

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

[dim ivy]

oop

$x = \frac{\sqrt{15} {\sqr{2}}$

$x = \frac{\sqrt{15}}{{\sqr{2}}$

[twin meteor]

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

[dim ivy]

$x = \frac{\sqrt{15} {\sqrt{2}}$

[ebon rapids]

sqrt

[twin meteor]

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

[ebon rapids]

Need an extra brace

[dim ivy]

$x = \frac{\sqrt{15}{{\sqrt{2}}$

what

$x = \frac{\sqrt{15}} {{\sqrt{2}}

i-

[ebon rapids]

$x=\pm\frac{\sqrt{15}}{\sqrt{2}}$

[dim ivy]

yes

[ebon rapids]

Is that what you mean?

[dim ivy]

yes

[ebon rapids]

You forgot the negative

[dim ivy]

its +-

anyways

[twin meteor]

CST (please ping when replying)

[ebon rapids]

Plug back in, does it work?

[dim ivy]

and i've realised what mistake i've done

[ebon rapids]

Oh, by the way, this rationalizes to $\pm\frac{\sqrt{30}}{2}$

[dim ivy]

the denominator isn't supposed to me sqrt

yea

[twin meteor]

CST (please ping when replying)

[dim ivy]

yea

thats correct

now

next question

i dont know how to put this into equation

the furtherst i've gotten is

$-(x^{2} + 13 -12) = 0$

but i doubt thats correct

Extended version: $(6 * 7) - (x^{2} + (7x - x^{2}) + (6x-x^{2})) = 30$

[twin meteor]

sircule

sircule

[dim ivy]

<@&286206848099549185>

[vocal sleet]

@dim ivy Has your question been resolved?

[dim ivy]

<@&286206848099549185>

[hidden turret]

break it into 3 parts

[dim ivy]

thats what i did here

[hidden turret]

oh yeah

[dim ivy]

which would simply to

$42 - (x^{2} + 13x - 2x^{2}) = 30$

[twin meteor]

sircule

[dim ivy]

$-(-x^{2} + 13x) + 12 = 0$

[twin meteor]

sircule

[dim ivy]

$x^{2}-13x+12=0$

[twin meteor]

sircule

[dim ivy]

$(x - 1) (x - 12) = 0$

[twin meteor]

sircule

[dim ivy]

$x = 1, 12$

[twin meteor]

sircule

[dim ivy]

which is incorrect for some reason

and i can't figure that part out

[sly mountain]

Who needs help

[dim ivy]

🙋‍♂️

[sly mountain]

Where is your question

[dim ivy]

.

[sly mountain]

Draw a line to get two rectangles. One has an area of 6 times x. The second one has an area of x times (7-x). The sum of these two equals 30

[dim ivy]

You mean this?

Oh

[sly mountain]

Yeah -x^2 + 13x = 30

[dim ivy]

so

-(x^2 - 13x + 30) = 0

[sly mountain]

Yep

[dim ivy]

(x - 10)(x - 3)

= 0

x = 10 or 3

[sly mountain]

Yeah

But x=10 makes no sense since x<7

So ?

[dim ivy]

The answer says it's only 3 tho

[sly mountain]

Yeah

[dim ivy]

Oh

Also where does it say that x<7?

[sly mountain]

Like mathematically speaking, two possibilities here. But physically speaking there is only one that works

[dim ivy]

Ohhh

Got it

[sly mountain]

Nice

[dim ivy]

Well that's methods out of the way

Thanks

.close

[vocal sleet]

Channel closed

Closed by @dim ivy

Use .reopen if this was a mistake.

[vocal sleet]

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[coarse wedge]

Can someone find x pls

[coarse wedge]

<@&286206848099549185> answer this asap pls

[vocal sleet]

@coarse wedge Has your question been resolved?

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[wild notch]

@sly mountain

[sly mountain]

Yeah

[wild notch]

Help me please

[sly mountain]

Sure

[wild notch]

So the first way is to give everyone one cake

The second way is to give someone nothing, someone 2 and the rest have just one

Are you here @sly mountain

[sly mountain]

Yeah just one moment lol

[wild notch]

Ok

<@&286206848099549185>

@sly mountain

<@&286206848099549185>

Bro @sly mountain where are you

!close

.close

[vocal sleet]

Channel closed

Closed by @wild notch

Use .reopen if this was a mistake.

[sudden cloud]

is it not n choose n

💀

oh

[wild notch]

What

Sorry you want to help me?

[sudden cloud]

No de

Dw

[wild notch]

What?

[sudden cloud]

You can’t really determine the number of ways if there are no numbers to start with

But you can create an expression

[wild notch]

How

.reopen

[vocal sleet]

✅

@wild notch Has your question been resolved?

[sly mountain]

Yo bro my bad i was doing something. Yeah when it comes to your question I think it should be 2^(n-1)

[vast shale]

@wild notch did u get it

?

[wild notch]

No

@vast shale

[vast shale]

K listen

Whenever you face such problems

Start with a lower number

Let's say 3 pastries and. 3 ppl

Think of the 3 ppl as u and ur friends

It doesnot matter who gets the pastry first

It only matters whether everyone got one or not

So the possibility when p1,P2,p3 gets one pastry is only (1)
One of them getting a 3 and other two getting zero
It can be p1,P2,p3 ,anyone can get 3 pastries
So the possibilities are 3C1=(3)
And the last way is one of them gets 2 pastries and another one gets one
I.e. 3C1×2C1=(6)
This sums up to the total 10 possibilities

And the formula for this is n+k-1Ck-1
Where n is pastries and k is number of ppl

Sorry for the crazy yap but its hard to explain this chapter shortly(atleast for me)

[atomic pine]

one way to think about this is the following: there are two things that can happen when a person buys 1 pastry

1. buy and don't leave the line
2. buy and leave the line (you've fed your family), and next person comes up
   each different choice between 1 and 2, at each stage, yields a totally different outcome. Thus doing the operation n times will give you 2^{n-1} choices (it is not 2^n since the last choice that is made doesn't affect how much pastry they are getting)

for example 11121 would mean the first person buy 4 pastries and the second buys 1

111121111212 would mean the first person buys 5 pastries, the second buys 5 pastries and the third buys 2.

with this intuition, it should not be hard to formalize it: show that every sequence of 1 and 2's of length n-1 gives you a unique way that the people buy the pastries, and every possible way that the people buy the pastries corresponds to a sequence of 1 and 2's of length n-1

[wild notch]

How did he buy 4?

[atomic pine]

111 means buy 3, the 2 that folllows means they buy another one and leaves.

[wild notch]

Ah that s a code

1 and 2 are cases

[atomic pine]

so you can think of person buying pastries as a sequence of 1 and 2's

buy buy buy buy and im outta here.

[wild notch]

Yes

[atomic pine]

and the second person comes up and buy

something like that

[wild notch]

What if one single person buys all

[atomic pine]

that corresponds to 11111111111...

[wild notch]

n-1 times

And then 2

[atomic pine]

yeah, and the last number doesn't matter since it don't matter if they leave or not, pastries are out

[atomic pine]

this works too

so the n^th digit is always a 2

[wild notch]

Than why lenght n-1?

[atomic pine]

because the n^th operation doesn't matter

they are gonna leave anyways (since they are out of pastries)

[wild notch]

So if the lenght is n-1

The number of combinations of 1 and 2

Is 2^(n-1)

?

[atomic pine]

yeah

so the answer should be 2^{n-1}

[wild notch]

Should be?

[atomic pine]

yeah

i mean you'd have to show that how the people buy the pastries indeed correspond to a string of n-1 digits

[wild notch]

Yeah man i understood

Thanks so much

[atomic pine]

no worries

[wild notch]

!done

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[wild notch]

.close

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[woven sail]

so i have like 2 "questions"(or 2 needed clarifications), first what is an indeterminate forms (our school teacher said it's like a value that can change from exercise to exercise like +infinity -infinity) in the exemple of N and N-{0}, it would be 0, but that's not true in every case).
Second, still working with N (for an easier understanding), it comports an infinity of numbers , N= {0,1,2,3,4,5,6,7,8,9,...}, and like 2N = {0,2,4,6,8,10,12,14,...} (multiplied every element by 2), so 0N= {0,0,0,0,0,0,...} = {0}, which means that for N 0N = {0}, and this is true for any other set, so why isn't +infinity*0 = {0}

[rain meadow]

1. Do you mean indeterminate forms?
2. The operation inf*0 does not multiply a set by 0. The notion of multiplying a set by a number is also not used so much, and especially not with 0, you would never even encounter something like 0N

[atomic jasper]

infinity*0 is not defined

and how are you going from infinity*0={0}??

[rain meadow]

Bonus answer: in some contexts, inf*0 is actually defined to be 0

[woven sail]

define not defined better please

[atomic jasper]

consider f(x)=1/x, now f(0) is not defined is implied, but it is defined for R/{0}

[woven sail]

I think that since multiplying any number by 0 gives us 0, if we were to multiply an infinity of numbers, it would still be a bunch of 0

[atomic jasper]

you can define it to be like that, sure

but unless stated, its undefined

by convention

[woven sail]

an exeption, maybe ?

like

if for any set or number

it = 0

why would it be undefined

so it must be an exeption to what I just said (I'm just too stupid to think about it)

[atomic jasper]

my real analysis book did this

but here its explicitly defined

generally, these are undefined

[woven sail]

tbh , I didn't understand how it went from 0*inf to a/inf

[atomic jasper]

there is nothing to understand

theyre merely defining it to be like this

but for brevity, since they define them all to be 0, they write it in the same line

[woven sail]

ohhhhhhh, i understood what u mean

[atomic jasper]

like $5\cdot 0=2\cdot 0=0$

[twin meteor]

Bonk

[atomic jasper]

here you dont go from 5 to 2

its just that theyre both 0

[woven sail]

but 0 is not undefined

[atomic jasper]

so instead of writing $\5\cdot 0=0\2\cdot 0=0\$ its in one line like so: $\5\cdot 0=2\cdot 0=0$

[twin meteor]

Bonk

[woven sail]

i understand what u mean, but if it's 0 then why is it not defined

[atomic jasper]

wdym?

[woven sail]

inf0 = 0, from ur book , however 0 is not undefined so inf0 is also not

[hybrid flicker]

that one is HARD to justify outside of lebesgue integration

[atomic jasper]

yeah its real analysis lol

[atomic jasper]

i still dont know what you mean

0 is just 0

[woven sail]

wtf is a lebesgue integration

[woven sail]

yes

[atomic jasper]

dw about it

[hybrid flicker]

if you forget about this line for a second

maybe this being ok

[woven sail]

and it is not indefined (a constant) so inf*0 is also a constant ?

(0)

[atomic jasper]

a constant is defined?

[woven sail]

yes i think

[hybrid flicker]

inf * 0 is actually very hard to be assigned a value in the realm of limits, much to your dismay

if you have lim(f(x) * g(x))

with f(x) going to infinity

and g(x) going to 0

[atomic jasper]

whats done here is basically $\lim_{x\to \infty} x\cdot 0= \lim_{x\to\infty} 0=0$

[twin meteor]

Bonk

[hybrid flicker]

but that's so arbitrary

[hybrid flicker]

and outside of the context I gave

[atomic jasper]

yeah but thats the intuition of it

0 times something that gets progressively larger (limiting to infinity) is 0

[hybrid flicker]

if you're ready to define 0 * infinity = 0, be ready to get rid of distribution property

which is why 0 * infinity is so not worth being defined at all

[woven sail]

is there a situation in which inf *0 != 0

?

[hybrid flicker]

in limits yes

x * 1/x

make x -> infinity

[atomic jasper]

the problem here is that the question is ambiguous

[hybrid flicker]

the indeterminate form is inf * 0

[atomic jasper]

the screenshot of definitions i gave you make sense in the given context, because in the book youll come across those types of problems

[hybrid flicker]

and when resolved will not equal 0

because x * 1/x = 1 for all x

[atomic jasper]

and its handy to give a list of definitions that youre going to use in the upcoming sections without having to specify at every step the definition

[woven sail]

another writing of the exemple pls

[hybrid flicker]

?

$\lim_{x\to \infty}x \cdot \frac 1x$

[twin meteor]

rafilou is not not born in 2003

[atomic jasper]

either way, at your current level, its unnecessary for now

[hybrid flicker]

also in your original question

you're asking why "+ infinity * 0 isn't {0}"

the reason is we never said that

we said 0N = {0}

because every natural integer

when multiplied by 0

is 0

+infinity is not in N

so it's out of the question

[woven sail]

what tf is even infinity (more of a concept ?)?

[hybrid flicker]

it's a quantity bigger than any number

[woven sail]

my reasonig was that it was true for any set of numbers, so it must also be trur to inf

[woven sail]

*might

[hybrid flicker]

the problem is you're jumping from a conclusion made for every finite number to a conclusion using an infinite quantity

[rain meadow]

a symbol that we define to have certain properties

[hybrid flicker]

infinity is not a number

[woven sail]

so more of a concept ?

[hybrid flicker]

so you can't manipulate it like other numbers

[rain meadow]

yea

[woven sail]

i know

[hybrid flicker]

and if something is true for numbers

it might not be true for infinite quantities

[woven sail]

I think I understand better now

[woven sail]

tk for the help

.close

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[vast shale]

hi i have doubt in physics numerical question which uses math knowledge , sorry for being dumb

[sly mountain]

you have got to tbe kidding right

what do you need help with exactly, cause you only sent numbers so far

[vast shale]

i am trying to solve this but i can't

[finite mist]

Put that in the calculator buds

[sly mountain]

for real

[vast shale]

the answer is coming wrong by book