help-13

[ebon burrow]

and both the LHS and RHS = 2

[dreamy fern]

oh i did it wrong

[undone sage]

Yes

[dreamy fern]

just a sec

[undone sage]

When u divide by cosx cosy , you have to divide both numerator and denominator

So that the net change is 0

And we achieve our goal

[dreamy fern]

both ??

[undone sage]

Yes

So that there's no change in the expression

If you only do to denominator/numerator then you are changing your expression

[dreamy fern]

let me try

[ebon burrow]

i find my method a lot easier

6 steps

simple

[undone sage]

It is 2 steps

[ebon burrow]

ehhh but this is clearer and the 6 steps are like year 5 maths

[undone sage]

First express as 1/cos(x-y) and expand

2 nd divide

Done

[ebon burrow]

i done this

Made x and y = 30

so

both the LHS and RHS = 2

[undone sage]

But it is not the correct way

[ebon burrow]

works

[undone sage]

But this method won't teach you anything

It's not a test

[patent cape]

just because it works for 1 value doesnt mean it works for all of them

[undone sage]

That you have to solve anyhow and tick the correct option

[ebon burrow]

i mean it taught my whole class to ensure full 3 marks

[ebon burrow]

it did that time

thats why i did

it

if it didnt i dont do it that way

[patent cape]

that proves nothing

[undone sage]

I know but while practicing problems you should avoid such methods

[patent cape]

you have to prove its equal to that you dont know it is

[dreamy fern]

it worked ty

[ebon burrow]

tbf i've always prefered different ways then most people

[patent cape]

youre literally taking a shot in the dark and assuming it does with no basis other than 1 specific example

[undone sage]

Yes correct

[dreamy fern]

close.

[obsidian coral]

Because you teacher expects you to that. It's not the proper method that is suppose to be used

[obsidian coral]

.close

[patent cape]

its .close

[undone sage]

Calm down boy

.close

[cedar kiln]

Channel closed

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[patent cape]

i am calm?

[ebon burrow]

ik when it works or doesnt

[patent cape]

youre supposed to prove when it does and doesnt

[cedar kiln]

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[crimson sedge]

AB = BC
angle CAD = alpha

What is the ratio between area of triangle CDM and ABM

[patent cape]

hmm

is this drawn to scale?

[lusty grotto]

is ABC equilateral

or somrthing

oh AB = BC

[soft owl]

isosceles

[lusty grotto]

and CAD = alpha

[lusty grotto]

yeah i didnt notice that they wrote anything before the pic

[crimson sedge]

Oh as in show it using alpha

[lusty grotto]

angle BAC = angle ACB = angle ADB = angle BDC

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

@lusty grotto well sure I got that but we still need line sizes

(sorry English terms for geometry are complex)

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[scenic ferry]

Why can’t you integrate cos^2 directly using reverse chain rule, instead of substituting in double angle formulas?

[ebon burrow]

You don't because the Chain Rule is for compositions of functions, and consists of just one step, for which you already have a rule of differentiation

[scenic ferry]

So we don’t a rule of differentiation for cos^2?

[crimson delta]

how exactly would you want to apply the reverse chain rule

[scenic ferry]

Think about what cos^2 would integrate to and fix the coefficients?

[upper garnet]

What do u think it will integrate to

[scenic ferry]

That’s where I’m stuck

So I asked if it’s even possible

[crimson delta]

what exactly do you think the reverse chain rule is?

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[humble quartz]

Is there a pattern shown in this residual plot ? (it’s for ap stats and i need to see if there’s a pattern or not to determine if a linear relationship is appropriate for the data)

[humble quartz]

i feel like there isn’t a pattern, but the overlapping y values are kinda throwing me off so im not sure

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[humble quartz]

<@&286206848099549185>

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@humble quartz Has your question been resolved?

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[cosmic steppe]

,eve

,rccw

[cosmic steppe]

Have you proven the x acoordinate is 10/3?

[royal leaf]

no

thats what im stuck on

[cosmic steppe]

You can create an equation of the tangent line

[royal leaf]

i can do everything after that

[cosmic steppe]

It'll be a quadratic though

[royal leaf]

thats the thing

idk how

what !?

quadratic for straight line?

[cosmic steppe]

Well ... the slope is f'(x)

And then it'll be f'(x)x

Eh I guess it won't be a quadratic per se

Like you understand why the tangent line is y = f'(x)x right

[royal leaf]

i understand f'(x) idk where the second x came from

ohhh

no wait

idk

[red pumice]

general equation for a straight line through the origin is y = ax

[long arrow]

equation of the line passes through the origin is ax

[red pumice]

a = f'(x) in this case

[royal leaf]

yes

i got that

y = mx i have

but idk m

big problem

[cosmic steppe]

you do have m

m = f'(x)

The derivative is also the slope of the tangent line

[royal leaf]

but i cant fin d f'(x)

i know

[long arrow]

you can also do it by solving for discriminant of quadratic if you don't understand calculus concept

[royal leaf]

but i cant use 10/3 to prove that

bc i have to prove its 10/3

[cosmic steppe]

You just differentiate sqrt(3x-5)

That's all

[royal leaf]

yes

but thats only the gradient function

i need the gradient at 10/3 without using 10/3

i need to find the grad at 10/3 without knowing its 10/3

how

[red pumice]

you must solve $\sqrt{3x-5} = f'(x)x$ of the point x such that the equation has only 1 solution

[cosmic steppe]

...

^

[royal leaf]

hmmmmm

what does that mean

[cosmic steppe]

Well, would P_x would be x

[red pumice]

actually my bad you can just write

[cosmic steppe]

It should@be f'(x)

And you solve for x

[royal leaf]

yes

[wraith dagger]

Gijs

[cosmic steppe]

^

You can find f'(x)

[royal leaf]

i d9nt get the f'(x)x part

[long arrow]

$\TangentLine$

[wraith dagger]

Modus

[long arrow]

last line

[royal leaf]

what is x0

[cosmic steppe]

Except in our case, x0 = x

[royal leaf]

okay

x isnt a constant tho

[cosmic steppe]

We need to prove x = 10/3

[royal leaf]

and i need the gradient at a point

yes

oh okay

[cosmic steppe]

So technically in this case, yes it is a constant

[royal leaf]

okay

[cosmic steppe]

It's simply just finding the intersection of the tangent line and f(x)

[royal leaf]

how to do this

bc if i make them equal

i have y = mx

i have y = (3x-5)^1/2

mx = (3x-5)^1/2

m is in terms of x

[carmine bronze]

The derivative of y=sqrt(3x-5) will be the slope of your tangent line. You can use the point (0,0) in the Point-Slope formula to find an equation for the tangent line.

[royal leaf]

x isnt a constant here

[royal leaf]

wdym

[long arrow]

Maybe this concept will be easier for you:
y = ax is the tangent line
y = sqrt(3x - 5) is the function
tangent line has exactly one common point with the function
then ax = sqrt(3x-5) need to have exactly one solution
in other words a^2x^2 = 3x - 5 (still one solution, a > 0)
a^2x^2 - 3x + 5 = 0 (still one solution)
When will it have exactly one solution (it's quadratic)?

[royal leaf]

when it touches the x axis

[carmine bronze]

You do not need an exact value for the slope. You can use an equation for the slope in the Point-Slope formula.

[royal leaf]

so i make

y = ax is the tangent line
y = sqrt(3x - 5) is the function

equal each other, rearrange for a and use discriminant?

[long arrow]

which means discriminant is equal to ... ?

yes might be also

[royal leaf]

0

[long arrow]

yes

but in this case you'll need to solve also for x coordinate, which is probably longer way but maybe clearer for you

so what's discriminant (in terms of a)?

[royal leaf]

(-3)^2 - 4(a^2)(5) = 0

[long arrow]

yes, solve for a (a > 0)

[royal leaf]

rearranges for a = sqrt(9/20)

why >

i thought x is supposed to only have one answer

or a

in this cae

[long arrow]

it's not 'x coord' yet

just slope

[royal leaf]

so what is it then

[royal leaf]

whats the gradient there

[carmine bronze]

You don't need to know the exact value.

[royal leaf]

isnt in terms of x

but 10/3 is an exact value

[long arrow]

let the explanation be the fact that tangent line is increasing here

for a > 0 statement

[royal leaf]

ok sure

[long arrow]

so a = sqrt(9/20)

right?

[royal leaf]

eya

yea

[long arrow]

now you know tangent is y = sqrt(9/20)x

[royal leaf]

sure

yea

[long arrow]

you can solve for x coordinate now

[royal leaf]

ok done

got 10/3

buhahahahah

✅

thank u guys

[long arrow]

now you can try with derivatives

[royal leaf]

how

[long arrow]

follow tips above

[royal leaf]

how to know that only denom is m

[carmine bronze]

That underbrace applies to that whole fraction, not just the denominator.

[royal leaf]

oh okay

[carmine bronze]

That should make it more clear.

[royal leaf]

how to find 10/3 from that

thats just another function

[carmine bronze]

You have an equation for y1 and y2.

You know they intersect at one or more points.

[royal leaf]

equal and solve for x?

[carmine bronze]

Yes.

[royal leaf]

and get 10/3

wow

crazy stuffs

[carmine bronze]

As shown by y1 = y2 at the bottom of the image.

[royal leaf]

yes

[carmine bronze]

Some of the steps you need to take in math to solve a problem can get crazy.

[royal leaf]

yes

maths is crazy

🧠

😵

thank u guys

.close

[cedar kiln]

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[cedar kiln]

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[ebon burrow]

so

[violet night]

what is 45 minutes as a fraction of an hour?

[ebon burrow]

ok steal my question

UGH

[violet night]

if there are 60 minutes in 1 hour

[ebon burrow]

okay

if there is 60 minutes in 1 hour

the fraction for 1 minute in in 1 hour would be

so what would 45 be

now

simplify

what is the HCFof 45 and 60

can they be divided by 2

and give a whole number

[obsidian coral]

If you do that, what do you result in?

[ebon burrow]

you can simplify it further

you wanna divided it by the heighst common factor of the 2 numbers

yeah so what is 3/4 as a number

[obsidian coral]

I am trying to demonstrate that to the OP

[ebon burrow]

and you know

1 hour = 170

so how would you find out what .75 hours is

[obsidian coral]

And it is not necessary, it's ideal, but you can keep simplifying if you can still simplify further

[ebon burrow]

thats why i said you wanna

want to and have to are different

yes

good

[grim token]

ok

thanks

[ebon burrow]

npnp

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[still tapir]

hello! so there's this сauchy's convergence test for infinite series. and there's the test for function.
they're all given for finite limits, and we're expected to make version for infinite limits ourselves
so I've stumbled upon a problem: it seems like you can't make a convergence test for series with a limit of just infinity (picrelated) — because it has that modulus sign that you can't remove safely without knowing the sign, and also because if a function changes sign rapidly from + to - infinity it just. doesn't converge
so my question is: do functions like that, that have limits of just infinity, not +-infinity, exist?

[still tapir]

also should I take it to #calculus?

[crimson sedge]

This open?

[still tapir]

the question? yeah

the channel is occupied tho

[cedar kiln]

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[balmy pivot]

.help

[cedar kiln]

Commands:
clopen: .close, .reopen, .solved, .unsolved
consensus: .poll
factoids: .tag
help: .help

Type .help <command name> for more info on a command.

[balmy pivot]

ok so

[balmy pivot]

my question is:

what temperature in Celsius is 10 more than the temperature in Fahrenheit

the equation for both Celsius and Fahrenheit are:

[ebon burrow]

whata are you trying to find

[balmy pivot]

this

[ebon burrow]

so can you word it differentley

[balmy pivot]

what temperature at x°F is equal to 10 more than its degrees in Celsius

[ebon burrow]

so like e.g. 50F and 40F

[balmy pivot]

its like 20F = 30C

thats not a solution its just what i mean

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[quartz gorge]

help

[dire geode]

#help-0 message

[mossy mango]

riemann is correct

[cedar kiln]

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[fossil coral]

I am confused

[fossil coral]

Oh wait

Do i do the exponent first?

Then multiply by 2?

.close

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[crimson sedge]

Help

[crimson sedge]

idk where to start

[latent bloom]

Ooo

Okay so

Here they tell you that the average temperature is 10 degrees warmer than the previous month

And they ask for how much less Jasmin will use this month compared to last month

That means

Let's say that consumption is y

That means y2-y1

They want this difference

Also

You are given x2-x1

y2-y1 and x2-x1?

That feels like a slope/gradient thing right?

@crimson sedge

Welp

I guess mayo either went offline or there is a connection issue

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[woven storm]

How would I solve this using a set of linear equations?

[woven storm]

I found this online

what is R1 my diagram has 2 paths with both a reward

https://cs.stackexchange.com/questions/142128/how-to-setup-the-bellman-equation-as-a-linear-system-of-equation

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@woven storm Has your question been resolved?

[woven storm]

.close

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[high whale]

Just a random question

Is it distributable

Like the formula is x^n+1/n+1 if i remember correctly

But who gets the exponent

Or is it distributable

[upbeat lotus]

You can distribute it

[crimson sedge]

[
\int ax + b \dd{x} = \int ax \dd{x} + \int b \dd{x}
]

[wraith dagger]

♡LexQa♡

[upbeat lotus]

$\int_{2}^{4}{2xdx}+\int_{2}^4{3dx}$

[wraith dagger]

Arnab

[high whale]

Oh you do them separately

[upbeat lotus]

Yeah

[high whale]

Give me a second

Like this?

Then you add them?

[sand meadow]

no

[upbeat lotus]

That 3 part is not right

Integration of a constant @high whale

$3\int_{2}^4{dx}$

[wraith dagger]

Arnab

[high whale]

Theres nothing to integrate then

[upbeat lotus]

This is as same as $\int_2^4{3dx}$

[wraith dagger]

Arnab

[upbeat lotus]

Do you know the integration of dx?

[high whale]

X hatches of and it becomes 3x

[upbeat lotus]

$3\int_2^4{x^0 dx}$

[wraith dagger]

Arnab

[upbeat lotus]

Yes

[high whale]

Why is it 0 instead of 1

So it becomes 1?

[upbeat lotus]

The exponent become 1

$\frac{x^{0+1}}{0+1}$

[wraith dagger]

Arnab

[upbeat lotus]

Something like that

[high whale]

Magic

[upbeat lotus]

Yupp

[high whale]

So then it bcomes x² + 3x in the end?

[upbeat lotus]

Yes but this is indefinite integration

So you have to put the values in first

[high whale]

Yes substitue

4 and 2

And subtract

[upbeat lotus]

Subtract 2

$(4²-2²)+3(4-2)$

[wraith dagger]

Arnab

[upbeat lotus]

This will be your step

[high whale]

But its substitution right

You just substiute the terms

[upbeat lotus]

Not exactly

Completely wrong

You have to
(4²-2²)+3(4-2)

[high whale]

Oh so its distibrution all again

[upbeat lotus]

Yeah

[high whale]

Lets see

(16-4) + (12 - 6)

How did you know it wasnt gonna be minus

12 + 6

[upbeat lotus]

You just subtract the lower limit

[high whale]

18

[upbeat lotus]

Right

[upbeat lotus]

Yupp

[scarlet breach]

16-4+12-6=18

[high whale]

But the answer isnt even on the choices

[upbeat lotus]

What the fuck

What's that 4² thing?

[scarlet breach]

4×4 =4²=/

[high whale]

Its definitely not a radical

Its supposed to be 4 at the top or 2 the bottom

[high whale]

Get wolfgram alpha to integrate 4²(2x+3)

[upbeat lotus]

It is 4^2

[high whale]

Which is 16

[upbeat lotus]

No I m asking you

,w 4²(2x+3)

[high whale]

Well lets try it out

[wraith dagger]

Results provided by WolframAlpha

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

Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[high whale]

We may have read the problem wrong

Not even one of the choices

💀💀💀

[upbeat lotus]

@crimson sedge sorry for tag but what is this 💀

⁴² thing

[crimson sedge]

typing error probably

or the latex didn't load up properly

[upbeat lotus]

Yeah maybe

@high whale load this page again

[high whale]

Rigged ahh website

[upbeat lotus]

Don't know bro this type of integration or this type of notation dosen't exist

[high whale]

Lol

Thanks a lot btw

.close

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[valid nexus]

(a strange question) Why I can't divide 0?

[foggy merlin]

look at the proof 2=1 on youtube
weird shit happen if you do that

[tropic oxide]

you mean divide by zero?

[valid nexus]

yes

[tropic oxide]

well

[snow zodiac]

you will get into trouble if you divide by zero

[tropic oxide]

a/b is the number whose product with b equals a

[tropic oxide]

OP's question is surely about what kind of trouble one would get into, no?

[tropic oxide]

@valid nexus is this point familiar to you?

[valid nexus]

if a=b,a/b=1(do you mean this?

[tropic oxide]

no, that is not what i mean.

a/b is the number which, when multiplied by b, gives a.

does this make sense to you?

[valid nexus]

yes

[tropic oxide]

right

[valid nexus]

you mean a/b xb=a

[tropic oxide]

i mean that (a/b) * b = a, yes.

i was highlighting the fact that this is in fact the definition of division

now, let's try working out what 1/0 could be, based on this

if 1/0 existed, it would be a number that, when multiplied by 0, would give 1.

however, anything multiplied by 0 equals 0.

[valid nexus]

1/0=? 1/0*0=0? but 1=0isn't right

bbbut if I let 0 divide by 0?

[tropic oxide]

1/0=? 1/0*0=0?
(1/0)*0 is 0 no matter what value we assign to (1/0), yes. but it is supposed to be 1.

[dire palm]

1/0 = undefined
2/0 = undefined
does that mean 1 = 2 ?

[tropic oxide]

if you try to examine 0/0, a different issue arises: there is no longer a unique value that can be assigned to it, unlike all other division problems.

after all, 1 * 0 = 0, and 2 * 0 = 0, and 3 * 0 = 0, and so on... so what should 0/0 be? 1? 2? 3? something else entirely?

[valid nexus]

sowwy,I think my problem is stupid:(

[tropic oxide]

no, this isn't stupid.

[valid nexus]

But what if I give this problem some meaning

[tropic oxide]

"Why can't you divide by 0?" is a legitimate question, and what i've been explaining here is its answer.

[valid nexus]

like I have 1 dollar ,I see some products are free

How many things can I get

[dire palm]

well

infinite?

but if u have 2 dollar

its also the same

so like division produces multiple results/

thats kind of undesireable

it shouldnt be multivalued

[valid nexus]

strange,I can get many different answers by dividing by 0

.close

[cedar kiln]

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[crimson sedge]

Are circumcentre, orthocentre and centroid always colinear?

[tropic oxide]

did you list the same point twice by accident or on purpose?

also are we talking about a triangle here

[crimson sedge]

By accident

[crimson sedge]

Mhm

[tropic oxide]

apparently, yes

https://www.quora.com/How-do-I-show-that-the-orthocentre-circumcentre-and-centroid-of-any-triangle-are-always-collinear-with-the-centroid-between-them-Also-how-do-they-coincide-in-case-of-an-equilateral-triangle

there's a proof written out here

didnt check it but it appears legit at first glance

[crimson sedge]

Thanks!

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[crimson sedge]

I have a question

[crimson sedge]

how does 2 = 10 - 2 * (64 - 10 * 6) turn into 2 = -2 * 64 + 13 * 10

[ebon burrow]

i’m not quite sure it does

[crimson sedge]

in the advanced euclidean algorithm it does

[ebon burrow]

oh right then

[crimson sedge]

but I dont get how

it essentially replaced 4 with (64 - 10 * 6)

and then solves

but how do we get to 2 = -2 * 64 + 13 * 10

[ebon burrow]

its perhpahs just showing another way

of solving it

bc both = 2

[crimson sedge]

wdym by that

[ebon burrow]

give the context of question

i’m a bit confused if i’m being honest

[crimson sedge]

<@&286206848099549185> can anyone here help out with some advanced euclidean algorithm confusion

[crimson sedge]

sure thing

[violet night]

can u post the full question

[crimson sedge]

yeah

[violet night]

is it asking for 2 = ab + cd

[crimson sedge]

solve the advanced euclidean algorithm for gcd(74,212)

its just that

[naive ginkgo]

you factor out the 2*, so you get $10-2\cdot 64 + 12\cdot 10$, then you factor the +10 into the multiplier of 10, so it becomes $-2\cdot 64 + 13\cdot 10$

[wraith dagger]

Zabbx

[crimson sedge]

wdym by "factor out"?

[naive ginkgo]

I mean distribute

not factor out

[crimson sedge]

[crimson sedge]

212 = 74 * 2 + 64
74 = 64 * 1 + 10
64 = 10 * 6 + 4
10 = 4 * 2 + 2
4 = 2 * 2 + 0

just normal euclidean algorithm to find gcd

then the advanced one flips it around and starts from 10 = 4 * 2 + 2

turns it around and makes it 2 = 10- 4 * 2

[violet night]

yes , 2 = ab+cd

[crimson sedge]

and yeah just that

[crimson sedge]

not aware of this formula existing

[violet night]

no its not a formula, just form

[naive ginkgo]

but was THIS not your question?

[crimson sedge]

it was but I don't get what you wrote

can you elaborate a bit?

[naive ginkgo]

there isnt more to elaborate?\
looking at right side, you had $10 - 2 \cdot (64 - 10 \cdot 6)$\
which turns out to $10-2\cdot 64 + 12\cdot 10$ after you distribute the 2 over the brackets\
then turns to $-2\cdot 64 + 13\cdot 10$ after you make the +10 as part of the latter multiplication

[wraith dagger]

Zabbx

[violet night]

^^^

[naive ginkgo]

since 12*10+10 = 13*10

[crimson sedge]

how am I expected to shorten this by default

is there something I'm missing here

[violet night]

you dont need to strictly, but it makes it easier going up the ladder to get 2=ab+cd

[crimson sedge]

surely shortening 12 * 10 + 10 to 13 * 10 isn't standard?

[violet night]

you notice how your just going 'backwards' from your algo

makes it easier to plug stuff in.

[crimson sedge]

yea

[crimson sedge]

what does this form represent

the gcd = ax + by?

[violet night]

13 x 10 is easier to manipulate than 12 x 10 + 10

[violet night]

yes,

[crimson sedge]

so what's really happening here in detail is:
10 - 2 * (64 - 10 * 6)
10 - 2 * 64 - 2 * (10 * 6)
10 - 2 * 64 - 2 * 60
10 - 2 * 64 - 120
-2 * 64 - 120 + 10
-2 * 64 - 12 * 10 + 10
-2 * 64 - 13 * 10

[crimson sedge]

ah, but this is no must is it? It's just to make it simpler (even tho it confused me)

[violet night]

absolutely not required however you will realise how painful it is if you dont collect terms and start substituting .

you might even end up going in circles

besides, you need to collect the at the end anyway , so may as well do it as you go along

[crimson sedge]

what are ab+cd again

[violet night]

I wouldnt actually do 10*60=60, just leave it 6 x 10

only change the coefficents of your algorithm values, so dont evaluate your remainders essentially

[crimson sedge]

hm

I don't even know what the values are

a and b are the initial numbers

and c and d are the linear extensions aren't they?

the ones we are trying to get

the ones for which the advanced euclidean algorithm was made

[violet night]

sure or vice versa

just trying to emphasise on the process of keeping it neat and easy to substitute

its hard to work backwards with mess

post once you get an answer and we can compare

[crimson sedge]

this algorithm is so filthy

[crimson sedge]

will do

give me a minute

(or more)

[violet night]

no worries

[violet night]

yeah its a job for computers now

[crimson sedge]

just one last thing before I go

what do you mean by coefficients of the algorithm values

[crimson sedge]

yet our professor still wants us to do it
disadvantages of having a university professor as your school teacher in advanced math class

[violet night]

so like in your example of 2=-2 * 64 + 13 * 10

the coefficents would be -2 and +13

oop typo

[crimson sedge]

could we not say the same about -2 and 10?

seeing as factors are interchangeable

[violet night]

yes but I mean the values you worked out you dont want to mess around with, if your goal is to get back to 2=ab+cd where you are dealing with your original values then you NEED the remainder values to substitute back towards it

[crimson sedge]

what are remainder values?

I'm sorry for asking so many questions but I did it in German

so I have to try and replace all the German words I know with the English ones you guys are using

[violet night]

ahhh ok

[crimson sedge]

one second

check this out

https://docs.google.com/document/d/1iyHV7u-qFow4ZGmFDuk1T9UbWW9nHnq2myNsHP_ROPc/edit?usp=sharing

might have to turn your translator on for this one

[violet night]

im just using words that describe the values you work out, so when you do the algorithm you have

74 = 64 * 1 + 10

which is

p = qs + r

where r is your remainder value and p,q,s are just random letters for demonstration

[crimson sedge]

p probably being the gcd

[violet night]

in this case yes but just in general

if you have like

25

or no

27

you could write 27 = 5*5 + 2

just a way of writing numbers

[crimson sedge]

ah

so the remainder is just what is being added

[violet night]

sure

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

wait up bot

I'm still solving

[crimson sedge]

almost there

😊

damn it

I messed something up

[toxic moat]

twerk

[crimson sedge]

what

@violet night quick question

in 2 = -2 * 64 - 13 * (74 - 64 * 1) could I not just calculate the parenthesis and call it a day?

74-64=10

[toxic moat]

so strange

its just an =

u can just use 2

n call it a day

[crimson sedge]

what

please stop trolling me

[toxic moat]

um

im not?

[crimson sedge]

context: advanced euclidean algorithm

[toxic moat]

o sry

im dumb

[crimson sedge]

no worries

I'm just wondering why I can never solve the parenthesis

but rather always have to multiply by the coefficient

[toxic moat]

hm ive seen this b4 but i kinda forgot o well

sry

[crimson sedge]

oh no worries

maybe another <@&286206848099549185> can help out

or not

screw the euclidean algorithm

@violet night you still there or nah

[toxic moat]

ok im back

isnt that the gcd then

sry not gcd but

its a step

2 = -2 * 64 - 13 * (74 - 64 * 1)
2 = -2 * 64 - 13 * 10

[toxic moat]

can i ask what the prev steps or what were

n what 2 numbers u r trying to find gcd of

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[upbeat lotus]

In some problems of limits at the end of the day we get something divided by 0 and we conclude the solution as infinity

[upbeat lotus]

Like at the end we get

3/0

And we conclude that the limit approaches to infinity

But if we look at the graph of 3/x

,w plot y=3/x

[wraith dagger]

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[upbeat lotus]

We can conclude that when x approaches to 0

[tropic oxide]

we don't

[upbeat lotus]

Oh really

I have seen some of the teachers explains 1/0 as infinity

In physics

Is it possible?

[frosty ocean]

The graph is 'asymptotic', it seems like tending to 0, but it never does.

[tropic oxide]

maybe there was some detail from context that you overlooked

[upbeat lotus]

It was about heisenberg uncertainty principle

[tropic oxide]

and you were in fact interested not in lim[x->0] 1/x but in lim[x->0**+**] 1/x

[tropic oxide]

do you have a video or an exact transcript of what was said

[upbeat lotus]

Yeah he said that
" ∆p.∆e≥h/4π where h is plank's constant and e= error in position and p is error in momentum what if your ∆e is equal to 0 here then what happens let us see
∆e.∆p≥h/4π
=0.∆p≥h/4π
=∆p=h/4π×1/0
Therefore ∆p=infinity cause anything multiplied by infty is infty"

And I was like is this possible

He also said that infinity in quantum world is barely 100m or 1km in our world

[crystal raptor]

They're taking the limit from the right like Ann said

[upbeat lotus]

Ohh

[crystal raptor]

Δe is always positive so you can only approach zero from the right

[upbeat lotus]

But this concept is not true?

[upbeat lotus]

Yeah yeah yeah 😳

[crystal raptor]

This is a bunch of rubbish

Physics waffle

[upbeat lotus]

Yupp infinity is undefined so you can't do that right

Okk thank you btw @crystal raptor @tropic oxide

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[gentle ether]

What changes on calculating sin(x) if the x is expressed in radians or in degrees?
I mean is there any preferred usage?

[inner glacier]

Well

Its comparable to cgs and si system per se

1 radian = 180/ pi degrees

[gentle ether]

What do you mean?

[inner glacier]

It's like saying 5cm or 0.05 m

[timber plover]

Sin x value will be same dosen't matter if you use x in degrees or express it in radians

[inner glacier]

Whether you use 5cm or 0.05 m in a calculation the answer in its core remains the same

[timber plover]

sin60°=sinπ/3=√3/2

[gentle ether]

I don't think that this is the right way to exmplain as both radians and degrees don't have an International system (They are pure numbers)

[inner glacier]

No

Radian is si system

[obsidian coral]

But you can still convert them between each other

[gentle ether]

Why the the preferred use is radians?
In radians definition is told that radians are used to simplify calculations, that's the only difference i can notice...

[obsidian coral]

Because most everything, online calculators and stuff, defaults to radians

[crystal raptor]

Also most calculus results involving trig functions only rly hold with radians

In degrees you get ugly extra stuff

[gentle ether]

However, isn't that a difference?
Because if we convert 60° in radians then it's : rad = 60*pi/180
Where pi is a real number, so sin(rad) would be more "precise" than sin(60) (by not adding approximations)?

[obsidian coral]

Ugly can be beautiful too

[gentle ether]

Yes, that's the only difference i can see

[crystal raptor]

Beauty is in the eye of the π/180

[obsidian coral]

That's like asking if we convert from inches to cm

[gentle ether]

Ok, so the only difference is calculus

Ok, i got it. Thank you guys

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[viral junco]

I feel like this is clearly wrong, but I don't know where I went wrong

[fiery whale]

only if they’re independent

in step 3

you assume A and B are independent events

[viral junco]

When I expand the and?

[fiery whale]

Yeah

[viral junco]

I see, I'll have to brush up on probability lol. Thanks

[fiery whale]

Nw:)

[viral junco]

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[royal leaf]

what to do if an equation im solving has a ln(-ve)

[royal leaf]

do i just assume its 0

what do i do to it

[dire geode]

show the equation

you probably made a mistake

[royal leaf]

i got that by intergrating this

[dire geode]

,rotate

[dire geode]

[dire geode]

do you know what |-1| = ?

[royal leaf]

yes i do1

it is 1

[dire geode]

do you have a log of a negative number?

[royal leaf]

😅

I guess not

thank

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[royal leaf]

how to integrate the ln(2+e^y)

[crimson sedge]

i dont believe it would have an elementary anti derivative

[violet night]

impossible?

[crimson sedge]

i did an integral similar to this

[royal leaf]

idek what that means]

[crimson sedge]

it would result in the logarithmic integral

[royal leaf]

what does that mean

[crimson sedge]

,w Li(x)

[dire geode]

show the original question

[wraith dagger]

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[crimson sedge]

oh lol

[crimson sedge]

,rotate

[dire geode]

nahhhhh

use $e^{x+y} = e^x e^y$

[wraith dagger]

riemann

[dire geode]

always send the original question with your work so you don't make people solve problems you don't need solved.

[royal leaf]

i dont know how to integrate the y side of this either

[dire geode]

,w rotate

,rotate

[dire geode]

int by parts

[royal leaf]

ok i will try that

[cedar kiln]

@royal leaf Has your question been resolved?

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[bright sorrel]

I do not understand at all how tan 55 becomes cot 35

[bright sorrel]

at the denominator

[dull oxide]

90-55

[bold hinge]

cofunction identity

sin(90-x) = cos(x)
same with: csc & sec
tan and cot

[bright sorrel]

oh ok

thank you

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[wraith pumice]

what are the zeros of this function?

[wraith pumice]

f(x) = e^x(x+3)(x-1) = 0

[wraith dagger]

Stephen

[runic garnet]

Set each individual component equal to 0 and solve

[wraith pumice]

okay thanks

[crimson sedge]

$e^x \neq 0$

[wraith dagger]

♡LexQa♡

[crimson sedge]

Like ever

[runic garnet]

Keep that in mind^

[wraith pumice]

ty

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[neon cypress]

i need help!

[neon cypress]

so i have an event that has a 4% chance of success, and i have 12 attempts to succeed

how do i calculate the probability of succeeding in 1 or more attempts out of 12?

[fair geyser]

1 or more is the opposite of never

calculate the never

subtract the result from 1

[neon cypress]

well yeah

how do i calculate the never

[fair geyser]

0.96 × 0.96 ... × 0.96

[neon cypress]

based. appreciate it!

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[idle cedar]

Hey, I have a question regarding homomorphisms on graphs

[idle cedar]

So lets say I have a graph G1, which is not a complete Graph, and I wanna build a Graph G2, such that there is no homomorphism from G2 to G1.
Am I getting it right, that I could simply take G1 and say G2 is the complete graph K_n where n = |V(G1)| ?

[blissful grove]

hi guys

what is 3+4

[hushed spoke]

dont troll in help channels

[idle cedar]

this help channel is occupied by me, pls if thats really your qeustion, open your own, thank you

[blissful grove]

im not

[crimson sedge]

yes, you are correct

[idle cedar]

thank you :D !!

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[mint oriole]

Ermm I don’t get how they get 6144 as a answer in the textbook

[dull oxide]

3*2^11, not 32^11

Their syntax is not great

[cerulean sail]

if there’s any an example

[dire geode]

holy shit that's awful

$a_{12} = 3\cdot 2^{12-1} = 3\cdot 2^{11}$

[wraith dagger]

riemann

[dire geode]

,calc 3 * 2^11

[wraith dagger]

Result:

6144

[mint oriole]

Oh thanks

[dire geode]

where do you first get lost?

okay glad that was the only issue

[rugged eagle]

those exponents are awful too, literally a new line

[cedar kiln]

@mint oriole Has your question been resolved?

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[trim crescent]

Here we go again

[crimson sedge]

I hope "examen" doesn't mean exam

what have you tried?

[earnest socket]

seems like a past exam (2022)

[trim crescent]

Yep it does

In class right now

No i’m joking

It’s from last year

[crimson sedge]

oh yeah, we are in 2023 arent we oops

[trim crescent]

🙂

Well i was suggested to use the exponential form

[dull oxide]

No puedo leer tu papel

[trim crescent]

So here

[trim crescent]

?

[dull oxide]

The problem

I can't read it

[trim crescent]

The question is to prove that there is n roots for P

In the complex world

[crimson sedge]

if you want to do it like that, you have to show the nth roots of unity are distinct

[trim crescent]

Of unity ?

What’s unity ?

[earnest socket]

1

[trim crescent]

Ah

Figured yeah but i have no idea how

[earnest socket]

you're seeking all complex x such that x^n = 1

use the property that multiplication in the complex numbers adds the arguments of the terms in the product i.e. arg(xy) = arg(x) + arg(y) (mod 2pi)

taking the argument mod 2pi of both sides of x^n = 1, you get n * arg(x) = 0 (mod 2pi)

solve the above equation for arg(x)

[trim crescent]

I see

Wtf

Still can’t do it lol @earnest socket

I haven’t worked with complex numbers in a while

Don’t think this will be in tomorrows exam but now I’m itching to know how to solve it lol

[earnest socket]

can you give me one value for arg(x) that satisfies n * arg(x) = 0 (mod 2pi) ?

[trim crescent]

Fuck

Well 1 right ?

[earnest socket]

not 1

n * 1 = n

okay let's suppose that n = 2

2 * arg(x) = 0 (mod 2pi)

i will argue that there are two solutions to this equation

arg(x) = 0 and arg(x) = pi

do you see that?

[trim crescent]

Yeah i think so

Wait no

Oh yes ok i see

My bad

[earnest socket]

2 * 0 = 0 and 2 * pi = 2pi, which is the same as 0 mod 2pi

what about n = 3?

if you provide a general solution to this equation for arbitrary n, then you've solved the problem

[trim crescent]

0 and 2Pi/3 ?

[earnest socket]

there's one more

i should've said you are looking for all solutions in [0, 2pi]

[trim crescent]

-2Pi/3 ?

[earnest socket]

yes

we'll go with 4pi/3

[trim crescent]

?

[earnest socket]

because of this

i just added 2pi to your solution

[trim crescent]

Oh yeah sure ok

[earnest socket]

so you see 3 * 0 = 0, 3 * 2pi/3 = 2pi, 3 * 4pi/3 = 4pi, which are all equivalent to 0

so the key to the this problem is being able to provide n general solutions to this equation for any n

can you do that from what you've learned from these examples?

[trim crescent]

How can there be n roots tho

[earnest socket]

generalize this string of equations

[trim crescent]

Oh woops yep

[earnest socket]

because, for example when n = 3, there are 3 distinct angles that, when added (which is what happens when complex numbers are multiplied), are coterminal to where 1 lies in the complex plane (angle 0)

[trim crescent]

Wait we have arg(x) = Pi as a solution too

[earnest socket]

we don't

for n = 3? 3 * pi = 3pi, which is not coterminal to 0

[trim crescent]

Oh i’m so dumb wtf ok

Yeah i see

So for n we have 0, 2Pi/n and -2P/n

[mint oriole]

So anyone know where the -7come from

😭

[earnest socket]

channel occupied

[dire geode]

!help

[cedar kiln]

Please read #❓how-to-get-help

[earnest socket]

please post in an available channel

[earnest socket]

let's work within [0,2pi]

you're right that 0, 2pi/n are distinct solutions for any n > 1

the question asks for n distinct solutions

for n = 2 we have 0, 2pi/2

for n = 3 we have 0, 2pi/3, 4pi/3

for n = 4 we have 0, 2pi/4, 4pi/4, 6pi/4 (i haven't simplified these so the pattern is more obvious)

maybe you can see the pattern?

for n = 4, the four solutions are simply the angles 0, 90, 180, 270, which, when multiplied by 4 are all equivalent to the angle 0

there's only one more step once you see the pattern

[earnest socket]

you can also note that all of these solutions, when multiplied by 4, equal 0, 2pi, 4pi, 6pi

which are the multiples of 2pi from 0 to 3

[trim crescent]

So 0 + 2Pi/n n times

0 included

[earnest socket]

not quite

[trim crescent]

N-1 times

[earnest socket]

you can try for n = 5 if you're having trouble

oh nvm i see what you're saying

0, 2pi/n, 4pi/n, etc.

[trim crescent]

Yeah my bad haha

The {}

[earnest socket]

stopping at the (n-1)th multiple of 2pi

that's exactly right

[trim crescent]

Damn

[earnest socket]

so you have n solutions for any n

let's parameterize these solutions

for any integer n, you have given me n angles: 0, (1)2pi/n, (2)2pi/n, (3)2pi/n, ... (n-1)2pi/n

and all of these angles, individually when multiplied by n, are coterminal to the angle 0

when complex numbers are multiplied, the angles get added, and the magnitudes (real numbers) get multiplied

so what does x^n = 1 tell you about |x|?

where |x| is the modulus of x

[trim crescent]

I’m scared bro 😂

This is not my strong suit

But 0 ?

No

[earnest socket]

lol, well you know that |xy| = |x||y|, for any two complex numbers

the magnitudes/modulus get multiplied when the complex numbers get multiplied

the modulus of 1 is 1

so if x * x * x ... (n times) = 1, then |x|^n = 1

and |x| is a real number

there's only one real number |x| such that |x|^n = 1 (for all n)

and that's |x| = 1

hopefully that makes sense

and now the problem is solved

a complex number is just an angle and a modulus

by the last discussion you've determined that the modulus of any complex number satisfying x^n = 1 must be 1

by the earlier discussion, you've determined that there are n distinct possible angles for x: 0, (1)2pi/n, (2)2pi/n, (3)2pi/n, ... (n-1)2pi/n

so from these facts you can form n distinct complex numbers such that x^n = 1

[trim crescent]

Damn i see

[earnest socket]

you can express this succinctly if you know euler's formula, which allows you to describe a complex number by it's modulus and argument