help-27

[topaz axle]

fire is the right analogy, it's the burning embers

[dry oak]

if its just dust, couldn't you just, like, wipe it off and be all good?

[topaz axle]

if you remove the embers from the building, the light they gave off is not left behind

yes

if you spill it on the floor, you wipe it off and you;re good

[dry oak]

interesting

thank you

[topaz axle]

then you have to bury the rug

we can't stop it, we wait until it burns out

@dry oak oh also UV from the sun is the same thing right?

so that's confusing

[dry oak]

yeaaa idk about the sun, but ive heard that it damages skin cells and can cause cancer, i tan anyway :P

[topaz axle]

oh ok microwave is the opposite end, i thought it's between uv and gamma

makes sense then

so the left side cooks you a lot, but doesn't touch dna

and the right side is opposite

i guess it's like that

[dry oak]

dang so higher frequency is basically more dangerous

which one here is radiation?

[regal moat]

They're all radiation

[topaz axle]

all of this is radiation, starting from UV it's "dangerous"

starting from X-Ray it's nuclear

[regal moat]

A bit wrong

All rays can be generated without nuclear

Photon is a bit special

They carry energy using a packet

Before Enstein

[dry oak]

huh? dont we get xray in hospitals 👀

[topaz axle]

they use nuclear decay

maybe not always

[regal moat]

Photoelectric effect is unexplainable

People thought photon energy derive from intensity

But its frequency

Its a long paper

[topaz axle]

@dry oakalso nuclear bombs do all of this, and the goal is to make the left side radiation, one that cooks

the "dangerous" side is pointless

just can;t avoid it

[regal moat]

Nope

Pretty sure a bunker build underground from steel (for support) and lead (for shielding) can stop the gamma ray

[dry oak]

water can stop radiation too right? hence the blue light effect

[regal moat]

Not that effective agaisnt gamma

You want dense material like lead and tungsten or depleted uranium

[topaz axle]

can't avoid producing it i meant

i started doubting that it's insignificant in total energy

but seems true, it's less than a third at least

[dire forge]

The X-rays you get in a hospital are called hard X-rays, and they're more on the right of that image, so they're more energetic. Because of how energetic they are, X-rays can give you increased risk for cancer, if they are absorbed by DNA and cause mutations. But for medical use, X-rays are generally very controlled and you are only exposed to them for a very short amount of time. Generally, if you are in need of getting an X-ray, the benefits of getting the X-ray and being able to get a medical diagnosis and treatment outweigh the potential risks. X-rays generally come from electrons, not the nucleus of an atom. This happens when an electron jumps down from a higher energy level to a lower one, releasing the energy in the form of a photon. The nucleus of an atom generally emits gamma rays, which are even more energetic.

For the discovery of X rays, the physicist Röntgen was awarded the first Nobel Prize in Physics

But yes, everything from radio waves to microwaves to the light you see to X-rays operates under the same principle. The only difference is the frequency of the radiation, which changes how much energy each photon has. This form of radiation is called "electromagnetic" radiation because it travels as a wave through the electromagnetic field.

[devout snow]

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[devout snow]

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[finite briar]

could someone verify (numerically) if $$\psi \left(\frac{2n+1}{4} \right) - \psi \left(\frac{2n+3}{4} \right) = \Phi \left(-1, 1 \frac{2n+3}{4} \right)$$? Here $\psi$ is the digamma function and $\Phi$ is the lerch transcendant

[woven radish]

rak³en

[finite briar]

Okay nvm seems false

yeah def false

chat could yall help me with this

I used the fact $\psi(z) = \gamma + \int_0^1 \frac{1-t^z}{1-t}dt$

[woven radish]

rak³en

[finite briar]

so basically that means $$\psi \left(\frac{2n+1}{4} \right) - \psi \left(\frac{2n+3}{4} \right) = \int_{0}^{1} \frac{ t^{\frac{2n+3}{4}} - t^{\frac{2n+1}{4}} }{1-t} dt$

[slender mirage]

At n = 1, LHS is pi - 4, rhs is

[woven radish]

rak³en
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[finite briar]

yes i confirmed that with wolfram just now

[finite briar]

hence this

[finite briar]

wait actually nvm got my blunder

thx

.close

[devout snow]

Channel closed

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[finite briar]

.reopen

[devout snow]

✅

[slender mirage]

blunder returns

Hey

[finite briar]

Hey lol

Tf isn't the imaging going

is this correct?

also assume first line is 100% correct

also i think the mistake is when i switch the sum and the integral

(done by fubinis theorem, but i never check the conditions xd)

pretty sure its simply convergence though, which it should

HMMMMMMMMMMMM

OKAY so I am getting

[woven radish]

rak³en

[finite briar]

Where in reality it is

[woven radish]

rak³en

[finite briar]

<@&286206848099549185>

oops missed a comma in both latex

[slender mirage]

umm

[sage burrow]

hmmm did you miis the change from 1/2 to 1/4?

[slender mirage]

you're geting: 1/2 of left sde = Right side tho

[finite briar]

oh yeah lol right i missed wayy too much

1 sec lemme re latex

got $\psi \left( \frac{2n+3}{4} \right) - \psi \left( \frac{2n+1}{4} \right) = 2\Phi \left(-1,1, \frac{2n+5}{2} \right)$, but supposed to be $\psi \left( \frac{2n+3}{4} \right) - \psi \left( \frac{2n+1}{4} \right) = 2\Phi \left(-1,1, \frac{2n+1}{2} \right)$

[woven radish]

rak³en

[finite briar]

And I can't spot the mistake 😐

[finite briar]

the mistake shouldnt be after this page, because all I did was simply write it as a lerch transcendant on the next page

also I gtg for a bath so yall if someone finds the mistake pls do share it

[devout snow]

@finite briar Has your question been resolved?

[slender mirage]

this is wrong

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[slender mirage]

.reopen

[devout snow]

✅

[slender mirage]

$\psi(z) = - \gamma + \int_0^1 \left(\frac{1-t^{z-1}}{1-t}\right) \dd t$

[slender mirage]

is the right formula. what you're using has {z} in place of {z - 1} in the integral @finite briar

that is shifting your Lerch transcendent's a by 2

[devout snow]

@finite briar Has your question been resolved?

[devout snow]

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[daring barn]

Hi, when doing data normalization, how do we choose between max-normalization or min-max normalization?

[finite briar]

OMFG TYSM!

[devout snow]

@daring barn Has your question been resolved?

[devout snow]

@daring barn Has your question been resolved?

[devout snow]

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[devout snow]

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[viral lynx]

I drew a lil diagram sorry its not the best

The equation of the circle is (x+1)^2 + (y+2)^2 = 9

[fossil locust]

what did you get as the radius of the circle?

oh cool

[viral lynx]

r = 3

Im not really sure how to solve it though

[fossil locust]

a few different approaches exist

[viral lynx]

We dont have either the x or y coordinate of the tangent

[fossil locust]

I would recommend subbing in y = ax + b, expanding and rearranging to get a quadratic in x

then you can use the fact you need discriminant = b^2 - 4ac = 0

since you only want 1 x-value, for 1 intersection -> tangent

and also -2 = a * 11 + b, so you have -2 - 11a = b

so you really just have a quadratic in x, and a as some unknown constant

[viral lynx]

1 moment im rereading a few times to try and understand

[viral lynx]

Ok I think I get it all except this bit

Where do these numbers come from?

[fossil locust]

you sub (x, y) = (11, -2) into y = ax + b

[viral lynx]

Ahh

Alright I will try

(x + 1)^2 + (y + 2)^2 = 9

(x + 1)^2 + ((mx + c) + 2)^2 = 9

x^2 + 2x + 1 + (mx + c + 2)^2 - 9 = 0

(mx + c + 2)^2 is really really messy 😵‍💫

[ebon coyote]

( ... = 0 at the end there)

Yh it'll be a little

Treat it as ( [mx] + [c + 2] )^2

(because it will sort out your x-powers for you)

[viral lynx]

Can I sub in the -2 and 11 now? That would make things easier

Or is it too early?

[viral lynx]

This -2 and 11

[fossil locust]

up to you, so you'd get mx + c + 2 = mx - 2 - 11m + 2

yeah actually that's easier now

[viral lynx]

Ok thank god 😅

x^2 + 2x + 1 + (mx + c + 2)^2 - 9 = 0

x^2 + 2x + 1 + (mx - 2 - 11m + 2)^2 - 9 = 0

x^2 + 2x + 1 + (mx - 11m)^2 - 9 = 0

x^2 + 2x - 8 + (m^2)(x^2) - 22xm^2 + 121m^2 = 0

Actually im not sure

What is a, b and c in b^2 -4ac

Do I get a = 1 from x^2? Or is it a = m^2 because of the (m^2)(x^2)?

[viral lynx]

Im stuck on this part

[fossil locust]

try collecting the terms of x^2, the terms of x, and the constant terms (including 121m^2)

also you mean -22xm^2

[viral lynx]

What do you mean sry

I think thats the part im stuck on

Like is x^2 the only x^2 term or is (m^2)(x^2) also an x^2 term

[fossil locust]

you can group x^2 + m^2 x^2 = (1 + m^2) x^2

[viral lynx]

Ah ok

Ahhh I see what I need to do

x^2 + 2x - 8 + (m^2)(x^2) - 22xm^2 + 121m^2 = 0

(1 + m^2) x^2 + (2 - 22m^2) x + (-8 + 121m^2)

So a = (1 + m^2)
b = (2 - 22m^2)
c = (-8 + 121m^2)

b^2 - 4ac = 0

(2 - 22m^2)^2 - 4(1 + m^2)(-8 + 121m^2) = 0

[fossil locust]

try substituting m^2 = u by the way

makes your life easier

[viral lynx]

4 - 88m^2 + (- 4 - 4m^2)(-8 + 121m^2) = 0

4 - 88u + (- 4 - 4u)(-8 + 121u) = 0

4 - 88u - 484u^2 - 452u + 32 = 0

-484u^2 -540u + 36 = 0

-4(121u^2 + 135u - 9) = 0

[fossil locust]

yeah hold on

(2 - 22m^2)^2 = 4 - 88m^2 + (-22m^2)^2
= 4 - 88u + 484u^2

so you actually just get 4 - 88u - 452u + 32 = 0, everything else is right

[viral lynx]

4 - 88u + 484u^2 - 452u + 32 = 0

[fossil locust]

no, you are missing +484u^2

so if you add it back you get 4 - 88u - 484u^2 - 452u + 32 + 484u^2 = 0

[viral lynx]

Ohhh

I thought you meant my sign was wrong haha

Okok I see

4 - 88u - 484u^2 - 452u + 32 + 484u^2 = 0

36 - 540u = 0

540u = 36

u = 36/540

u = 1/15

m^2 = 1/15

m = +-sqrt(1/15)

Those are the slopes of the tangents

[fossil locust]

yeah

you just need one of the tangents so you can choose say m = sqrt(1/15)

also remember from the very beginning, -2 - 11m = c

[viral lynx]

For a min i forgot what the original question was 😅

So many u's and m^2's

but ok

-2 - 11m = c

-2 - 11(sqrt(1/15)) = c

Cant simplify that

[fossil locust]

yeah just leave it then

[viral lynx]

so c = -11sqrt(1/15) - 2

The final equation is

x/15 -11sqrt(1/15) - 2

[fossil locust]

you mean, x/sqrt(15) - 11sqrt(1/15) - 2

[viral lynx]

Ah yeye

[fossil locust]

wait can I ask you, how far are you in trig?

could you simplify something like tan(arcsin(u)) using a right triangle?

[viral lynx]

🤩

[viral lynx]

I have covered the whole course but I took a looooong break and now I have to relearn / remember everything hehe

[fossil locust]

I have a quicker way if you know that, cause this question is quite specific

the centre happens to lie on the same horizontal line as (11, -2)

[viral lynx]

I dont think so...

[fossil locust]

okay then don't worry about it

[viral lynx]

Is it an identity of some kind?

[fossil locust]

yeah, it's actually not that bad

[viral lynx]

Could you teach me? If you dont mind

[fossil locust]

if theta = arcsin(u), then sin(theta) = u

so you can let opposite = u and hypotenuse = 1
so adjacent = sqrt(1 - u^2)

and tan(theta) = u/sqrt(1 - u^2)

for this question u happens to be 3/12 = 1/4

tan(theta) = sin(theta)/cos(theta) = y/x

so that gives you the slope of a line

[viral lynx]

Ok 1 moment I need to reread that a few times to understand

[fossil locust]

when you're ready

,w 3/sqrt(12^2-3^2)

[woven radish]

Results provided by WolframAlpha

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

Requested by sooouth.

[fossil locust]

,w 1/sqrt(15)

[woven radish]

Results provided by WolframAlpha

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

Requested by sooouth.

[viral lynx]

Ah cool

That is actually a really neat way of doing it

I thiiiiiiiiiink I get it but im gonna do some more examples with this method to make sure I do

[fossil locust]

yeah then the identity follows if you replace 3 with u, and 12 with 1

so that sin theta = opp/adj = u/1 = u

[viral lynx]

Ah yeye I think I get it

My brain is fried after this question so Im gonna take a quick break before I try some more examples hehe

But thank you sooooo much for your help

[fossil locust]

no worries!

[viral lynx]

You are a really good teacher!!

❤️

.close

[devout snow]

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[native stone]

so we got this exercise, dont mind the different language because ill show my solution so that should help

thing is, i came up with a solution, got to the right answer....but the solution has a flaw in it that actually makes me wonder why it works

at the top of the page i put what we know so that its easier to understand

the problem here, is that i amplified with z-|z|

so multiplied and divided

which we dont know if its zero or not

because if it zero, then i couldnt do that

one of the multiple choice actually even sais that z=1 so it was a possibility

anyone got a clue why this works?

[frozen aurora]

if $\overline{z} \neq -|z|$ then $z \neq -|z|$ as well

[woven radish]

artemetra

[frozen aurora]

because |z| is real

[native stone]

why

oh

thats cool, thanks

[slender mirage]

You have to check, for which z, you have Im(z² + 2z|z|) = 0

[native stone]

it sais that z is complex

so for no z

or wait

idk what you wrote there honestly

but it makes sense that z is different that |z|

since one is complex and one is real

[thin inlet]

|z| not equal z conjugate implies z is not purely real

[native stone]

oh

so then how do i show that z-|z| is different than 0

[thin inlet]

or wait

[slender mirage]

The conditions should be Im(z) = 0 or Re(z) + |z| = 0

[native stone]

ill show you a pic of what i got

which is wrong

here

[slender mirage]

No need to over complicate

Since it says w is real, multiply num, denom with (z + |z|)

(z{bar} + |z|)(z + |z|) is now real

So, (z + |z|)² is real

=> z² + 2z|z| is real

Plug z = a + ib, and you have your conditions, 2ab + 2b|z| = 0

So, b = 0 or a + |z| = 0

[native stone]

i dont understand what youre saying by multiply with denom conjugate

like, z conj - |z|?

[slender mirage]

multiply num, denom with (z + |z|)

[native stone]

how do you know that thats different than 0

[slender mirage]

huh? what is different than zero

[native stone]

z+|z|

how do you know its diff zero

[slender mirage]

z{bar} + |z| non zero is given

so its conjugate is also non zero

wdym

[native stone]

wdum ,,its,, conjugate

the conjugate of the sum?

[slender mirage]

YES

[native stone]

whats the conjugate of a sum

[slender mirage]

bruh

conjugate of z1 + z2 = conjugate of z1 + conjugate of z2

[native stone]

wow

oki

[slender mirage]

now which option says Im(z) = 0, or Re(z) + |z| = 0

[native stone]

if im(z)=0 doesnt that just mean that z is real?

just hold on until i write it down, i may have some more questions if i dont get something

what do i do now

@slender mirage

[slender mirage]

BRUH

Don't you think you're over indulgent with your FOILS?

[native stone]

foils?

huh?

whats that?

[slender mirage]

basically, multiplying algebraic expressions and expanding

[native stone]

how would you do it otherwise?

[slender mirage]

It's like I give you (a + b)(b + c)(c + a) and you expand it all the way to get 2abc + ...

=_= I never told you to expand denominator

[native stone]

what are you supposed ot do?

[native stone]

isnt that what you do?

[slender mirage]

NO, not when it's not required

but you will =_=

we multiplied denominator with its conjugate because

For any complex number w, ww{bar} = |w|² is real

So denominator is real even if you do not multiply and expand

Got it?

[native stone]

NO

[slender mirage]

=_=

[native stone]

all this time

[slender mirage]

the denom is (z{bar} + |z|) yes?

[native stone]

you meant to multiply w*w conj?

is that what u wanted?

[slender mirage]

YEAH

wait no

[native stone]

i feel violent

[slender mirage]

no no lmao

[native stone]

good

[slender mirage]

it's not w wconj

jeez

I ran out of variables

is why I used w

[native stone]

brother, what do you want me to do

i dont understand

shouldnt i have multiplied how i just did/

[slender mirage]

Alright let's do from step 1

Okay?

[native stone]

sure

[slender mirage]

denom is (z{bar} + |z|) yes?

[native stone]

damn right

[slender mirage]

(z + |z|) is its conjugate yes?

[native stone]

hell yea

[slender mirage]

so we multiply num and denom with (z + |z|)

alr?

[native stone]

no no

we dont

[slender mirage]

=_=

[native stone]

im kidding

keep going

[slender mirage]

so the denom automatically becomes real, yes?

[native stone]

wow

i didnt see it that way

[slender mirage]

Yes. No?

[native stone]

yeye

i see it now

so now we got the num which we should show it real

[slender mirage]

Yes

And num is (z + |z|)²

[native stone]

damn

[slender mirage]

so we gotta have z² + 2z|z| as real

[native stone]

and |z| is real

wait

oh yea i see

[slender mirage]

plug z = a + ib

[native stone]

i thought that if (z+|z|)^2 is real then z+|z| is also real

[slender mirage]

we need Im(z² + 2z|z|) = 2ab + 2b|z| = 0

[native stone]

i need to check this myself

[slender mirage]

it doesn't mean that

[native stone]

howd did you get that so easy?

im tempted to just literally see what (a+bi)^2 is

[slender mirage]

you only need the imaginary part

[native stone]

and then substitute in z as well

[slender mirage]

Yes that is what I did

picked the imaginary part from the expression after substitution

[native stone]

right right

so you did do the ,,foil,, or whatever

[slender mirage]

did not -_-

[native stone]

so you would get 2b(a+|z|)=0

[slender mirage]

Yes

So either b = 0, or a + |z| = 0 which converts to b = 0, a < 0

[native stone]

why

cant a= -|z|?

[slender mirage]

Yes lol, solve that equation now

[native stone]

oh yea, thats right

[slender mirage]

what is |z|

[native stone]

oh

b=0 in that case too

[slender mirage]

✅

[native stone]

sensational

[slender mirage]

so z is compulsorily real,

not necessarily < 0 tho

[native stone]

this was way harder than i thought

[slender mirage]

which we could've just guessed by putting numbers

[native stone]

thought itd be easy peasy

[slender mirage]

=_= which part of what I showed you was hard

[native stone]

you sound like u just like gambling, man

[native stone]

none i guess, its just some properties i forgot about

im reminding you that im recapping for an exam rn

[slender mirage]

no when you do competitive exams a lot, you need to have a way to quickly deal with stupid problems

[native stone]

so thats why i forgot some things

[slender mirage]

those problems are called no brainers

[native stone]

damn

[slender mirage]

unfortunately there's not many no brainer problem up here

[native stone]

well

you probably dont have multiple choice problems

[devout snow]

@native stone Has your question been resolved?

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[regal berry]

$(-4)^{73} / 2^{144} - 25^{12} / 5^{23}$ idk how to start

[woven radish]

Simon James B

[regal berry]

$(-2^2)^{73} / 2^{144} - (5^2)^{12} / 5^{23}$ like this?

[woven radish]

Simon James B

[regal berry]

but from here i am stuck

[toxic grove]

Try expanding out the brackets

[small jackal]

4^evennumber = (-4)^evennumber

Same with 2

Simplify the right side first

[regal berry]

so it's -(2^2)^73

sorry but i am confused

[toxic grove]

He doesn't have that though, he's got (-4)^oddnumber

[small jackal]

Simplify the right side first before the left. The right side is much easier

[toxic grove]

$(a^b)^c = a^{bc}$

That should be enough

with proper formatting though lol

[small jackal]

.

[toxic grove]

ukwim

It's still -4^oddnumber though

The - does not cancel o ut

[regal berry]

still confused on what to od

[woven radish]

depression

[regal berry]

I know! But is this the easiest way? we are supposed to do to the easiest

[toxic grove]

That is the easiest way

[small jackal]

Okay I just re-read

[toxic grove]

Just simplify it down and the answer will fall out

[small jackal]

-4^73=4^72 x -4

2^144=4^72

[regal berry]

$-2^{146} / 2^{144} - 5^{60} / 5^{23}$

[woven radish]

Simon James B

[toxic grove]

Might wanna check that arithmetic

But yeah that's the right idea

[small jackal]

Just sub in and simplify

[regal berry]

but we do not have the same base how do i make 2 negative

[small jackal]

-(2^146)=-2x2^145

[regal berry]

?

or it does not matter

[small jackal]

Sub this

[toxic grove]

It both matters and doesn't matter

You already solved it without realizing I think

[small jackal]

Do you know how to factor?

[toxic grove]

The minus sign is effectively outside the brackets

So yeah just forget about that

[small jackal]

That's true

[regal berry]

$-2^2 -5^{37}$

[toxic grove]

You still need to double check your arithmetic from back here

[woven radish]

Simon James B

[small jackal]

Check the right side

[regal berry]

ok lemme do it again

$(-2^2)^{73} - (5^2)^{12} / 5^{23}$

[woven radish]

Simon James B

[small jackal]

No just evaluate the right side

[regal berry]

omg-

what is there to evalute

i can't see

[toxic grove]

How did you work out 5^37

[small jackal]

Reduce it to a whole number

[regal berry]

IDKK

[small jackal]

Do this first $(5^2)^{12}$

[woven radish]

David

[regal berry]

5^24

[small jackal]

Yes so 5^24/5^23=...

[regal berry]

5

[small jackal]

Remember the minus sign multiplying it

So what is your final answer

[regal berry]

-5?

[small jackal]

So the right side is equivalent to -5

Now put that together with your answer for the left.

[regal berry]

hold on cuz idk what is the answer from the lest side

[small jackal]

Ping me when you're ready

[regal berry]

I do not understand this problem at all.

• (2)^73 / 2^144 - (5^2)^12 / 5^23
• 2 ^146 / 2^ 144 - 5^24/ 5^23
  -2^2 - 5

-4-5 = -9

[devout snow]

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[small jackal]

Yes

[small jackal]

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[vast violet]

pls...

[vast violet]

i am just annoyed and ready to smash tjis calcukator

[tropic skiff]

Click on "mode"

[vast violet]

ok

[viscid onyx]

That's in radians.
Convert it into degrees.

[tropic skiff]

And in the 4th row, change to DEGREE instead of RADIAN

[vast violet]

Ohh ok thnx

so i should keep it there for a physics class

[tropic skiff]

In my limited experience studying physics, you only need degrees

[vast violet]

Ok thnxs

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[hasty flame]

does a polynomial like f(x) = (x+1)^2 (x-2)^2 have a vertex when graphed?

[distant helm]

From the given function there are definitely 3 vertex

2 lies on x-axis which are zeros of the function and 1 above x-axis

[ancient sluice]

x< -1 fn , x (-1,2) fn , x >2 fn

[hasty flame]

yeah- got the x intercepts

[smoky nimbus]

Did you try graphing it?

[distant helm]

You can expand f(x) and solve f'(x)=0 which instantly gives 3 critical points

[hasty flame]

I did after, I've never seen a graph like this one before so I got a little confused.

[hasty flame]

minimums and maximums right?

[distant helm]

But in this case its unnecessary because you can guess how many vertex

[distant helm]

Ye

[distant helm]

Guess: f(x)=0 has 2 double roots so they act as vertex as well, between them must also have 1 more vertex otherwise the graph doesnt exist

Double roots act as vertex because f(x) doesnt change sign through those points

[hasty flame]

to get min/max i have toooo

expand the function

[distant helm]

Ye, expand, solve f'=0

[hasty flame]

whats that little ' for?

[distant helm]

f'(x)=0, to be more clear

[smoky nimbus]

Notation for derivative

[hasty flame]

ok how's this?

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[stray orbit]

not even sure what the u is here imma be honest

Very lost

[faint gorge]

maybe 2tan(theta)

for x

[stray orbit]

What?

[potent tusk]

how does bro see this

[stray orbit]

😭😭😭

[faint gorge]

or maybe trig sub won't work

nicely

[potent tusk]

why cant you use usub

[stray orbit]

I have not learned of any such trig sub yet

[stray orbit]

I’m supposed to

[faint gorge]

the next thing that came into mind would be u = x² that would cancel an x

and then another linear substitution with u+4

[stray orbit]

Bro what is linear substitution

I just started Calc 2 bro 😖😖😖

I can only use u sub

[faint gorge]

when you substitute something in the form of ax+b

that's also u-sub lol

just a fancy name

[stray orbit]

Oh

[faint gorge]

yea try u = x²

[stray orbit]

I assumed it was some crazy bs idk

[faint gorge]

no dw

and then do z = u+4

step by step should get you to work

it out

[stray orbit]

What am I to do with this Z

[faint gorge]

did you first apply the u-sub

[stray orbit]

Am rn

Ngl bro he did not prepare us for a lot of the questions on this hw

How are they assigning this after 1 lecture

[stray orbit]

@faint gorge

[faint gorge]

you forgot the dx and to change that into du

[woven radish]

anti-algebraist 𝔸dωn𝓲²s

[faint gorge]

And then another "u-sub" let's call it z and that will be z = u+4

and then use power rule

[stray orbit]

Where does the * 1/2x come from

[faint gorge]

differentiate u = x^2

[stray orbit]

So u sub inside u sub?

[faint gorge]

du/dx = 2x

dx = 1/(2x)

[faint gorge]

basically and that's pretty common

you could have done u = x²+4 too but I was afraid that would be too much at first

anyway gtg u will figure it hopefully

[stray orbit]

If you don’t mind would you also tell me how I’d go about this?

[faint gorge]

same

it's the very same procedure

rewrite everything in terms of u and change dx in terms of du

[stray orbit]

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[velvet bear]

can i get some help solving this

[wicked turtle]

!status

[devout snow]

What step are you on?
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2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
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7. None of the above

[velvet bear]

1

[wicked turtle]

what have you tried?

[velvet bear]

i really dont even know where to start so nothing yet

[strange nimbus]

What can you do to the left side to get just s by itself?

[velvet bear]

wait i think i know what to do now thanks for the help

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[strange nimbus]

No problem.

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[devout snow]

What step are you on?
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7. None of the above

[rugged crypt]

1

[winter torrent]

draw a sketch

[rugged crypt]

how?

[winter torrent]

well you could start by getting a pencil and some paper...

I see an equation. is it in a standard form that you recognize?

if not, can you put it into one?

it says ellipse. is there something in your book or notes about an ellipse?

[rugged crypt]

wanted to confirm this is the sketch

[winter torrent]

I mean it's a graph so sure

[rugged crypt]

tbh i dont really know much bout how to solve these types of word problems

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[burnt leaf]

For Calculus 2, how do I get the volume for this ? I'm honestly very confused and don't know how to do it on paper.

[wind mason]

,w plot y=x^2, x=y^2

[woven radish]

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[wind mason]

do you know the disk method

[burnt leaf]

Kinda, it's using the area of the circle as the first step, right?

[wind mason]

yes.

$\int_a^b \pi (r(x))^2dx$ or in terms of $y$, where $r(x)$ or $r(y)$ is the radius function.

[woven radish]

;(

[wind mason]

but here, there is 2 radii.

[burnt leaf]

So for x, it'd be pi (x^2)? then y, pi (y^2)?

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[burnt leaf]

.reopen

[devout snow]

✅

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[hazy cradle]

i made a GP like:
a/r^3, a/r^2, a/r, a, ar, ar^2, ar^3, ar^4

3rd * 5th = a^2 = 49 => a=7

nvm

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[hazy cradle]

idk how to solve

what i tried :
took b as the common diff, a as the first term, thought of it like an AP

but idts im going anywhere with that

[toxic grove]

You should start by calculating m

There's quite a limited number of values for a and b to try so it shouldn't take long to just count the elements of R

[hazy cradle]

wont that waste time, if i try finding it by counting?

[feral agate]

anything beyond n=3 results in 0

so if m >= 3 you only need to go upto n=3

ah wait you still need m though

yeah you just need to count then

[hazy cradle]

lemme try

m=8?

[feral agate]

yes

[hazy cradle]

oo i got it

i hv one more ques

[distant helm]

I think you can use binary here? not sure but give it a try

[hazy cradle]

what is binary?

[distant helm]

yeah no that is too overkill

you can simply determine the row by trial and error
(if you notice that you can determine the first and final number of the row)

for example, to know 8 is in what row, guess that 8 is in-between 7 and 10

the first and final numbers of a row also have a formula that is easy to derive

[solar goblet]

what i would recommend is to write the upper and lower bound of a row as a formula of k

notice that the lowerbound for each row is 1, 2, 4, 7, 11, ...

1 + 0 = 1
1 + 1 = 2
2 + 2 = 4
4 + 3 = 7
7 + 4 = 11

one can see that as the pattern continues, the second term increments. thus this is a quadratic trend

likewise for the upperbound

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[slender mirage]

Now, solving the recursive relation gives L(n) = n(n+1)/2.
Also, the start term of nth row is nothing but L(n - 1) + 1```

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[hazy cradle]

oh well i did it a little differently, i got it tho

[slender mirage]

let's check

,calc 103*104/2

[woven radish]

Result:

5356

[slender mirage]

looks like it does contain 5310. Congratulations

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[hazy cradle]

yeah ggs

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[restive river]

need to write the negation of this

i.e f is not increasing smth

My attempt

f is not increasing if $f(x) \geq f(y)$ for some x,y such that $x<y$

[solar goblet]

[restive river]

hehe

sorry

[distant helm]

equal also works

obviously

[solar goblet]

negation of < is not >

[restive river]

i forgot the difference between increasing and strictly increasing

so wasnt sure if i had to include equals to

[dark tundra]

They’re the same

[solar goblet]

they are the same

[slender mirage]

basically, you're given p -> (q -> r)

can you write it's negation

[dark tundra]

There is a difference between increasing and nondecreasing

[stone stump]

be careful

this very much depends on the author

[steady robin]

i thought increasing was >= strict increasing was >

[stone stump]

for me increasing and strictly increasing are different

[steady robin]

https://math.stackexchange.com/questions/3333418/difference-between-strictly-increasing-and-increasing-functions

[solar goblet]

i guess 1 2 3 3 4 is increasing but not strictly increasing

[steady robin]

https://mathworld.wolfram.com/StrictlyIncreasingFunction.html

[solar goblet]

but it seems like the question writer doesn't bother to make a distinction between the two

[dark tundra]

Well either way the given definition of increasing in this question is f(x) is strictly less than f(y) for x<y

So what is the negation?

[steady robin]

the distinction is important
f(x) < f(y) whenever x < y implies increasing but the converse is not true

[restive river]

👍 ?

[dark tundra]

In this case, yes

[steady robin]

idk who put that i think it was arya but

[slender mirage]

it was OP making changes to their latex

[steady robin]

oh

[slender mirage]

if you're required to entirely negate it, should look like:
||f is an increasing function and there exists x < y such that f(x) ≥ f(y)||

[restive river]

i should have posted the complete question :P

[slender mirage]

however, I think you want something different

[restive river]

ii

[slender mirage]

:p

[restive river]

increasing function ?

[slender mirage]

:p I assumed you wanted to negate the whole statement, turns out it only requires you to negate the increasing property

[restive river]

yeah

arent both the same?

[slender mirage]

No lmao, one is ¬(p -> (q -> r)) other is ¬p -> ¬(q -> r)

they're not the same

[restive river]

damn

ah

[slender mirage]

So basically, what you wish to express is: \ f is not increasing if $\exists x, y,$ s.t. $x < y$ and $f(x) \geq f(y)$

[restive river]

s.t ?

[slender mirage]

such that

it's same anyways, you can use your wordings

[restive river]

thanks

[slender mirage]

^^" next time present original q before interpretation

helps

[restive river]

¬(p -> (q -> r)) is the same as p -> (q -> ¬r)) ?

@slender mirage

[slender mirage]

nope

[restive river]

whats ¬(p -> q) ?

p -> ¬q right ?

[slender mirage]

¬(p -> q) is p and ¬q, p -> ¬q is ¬p or ¬q

are they both same?

[restive river]

i should make a truth table to check that right?

[slender mirage]

or simplify, it's easy to do that

[restive river]

simplify ?

oh

p and (q and ¬r))

[slender mirage]

💀

[restive river]

lol

thats rigged

[slender mirage]

Lmao, you should get back on topic

I need to go too. I'm running late on my Curves n Surfaces assignment

[restive river]

okay

thanks

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[pseudo pollen]

A password contains the digits 0,1,...,9 with no restrictions, except it has to have at least two and at most four digits. Also, the password must contain an 8. How many possible passwords are there?

[topaz axle]

you solve it three times

2 digit, 3-digit, 4-digit

[pseudo pollen]

I did

and I got 2386

is it right?

[topaz axle]

no i get much more

[pseudo pollen]

how much 💀

[topaz axle]

how much did you get

separately

[pseudo pollen]

from 10 to 99 18 numbers

From 100 to 999 252

[topaz axle]

18 is right, 19 with 00

3 digit is 271

[arctic field]

why are you doing 10 to 99?

its a password it doesnt have to be a number

[pseudo pollen]

wdym

[topaz axle]

oh yeah, it's jsut a conicnidence

[pseudo pollen]

wait so it’s much higher

[topaz axle]

like you missed 10 paswrods and found 9 extra

[pseudo pollen]

wait a sec

but wdym I missed 10 passwords?

[topaz axle]

well you started from 10, there are 10 passwords less than 10 i forgot about 8 facepalm

but you still got 18 anyway

the right number is 19

[pseudo pollen]

okay

but for example, from 100 to 199 there are 19 numbers right?

and from 800 to 899 there are 100

and from 200,300,400,500,600,700 and 900 there are also 19 numbers for each of them

so

isn’t it 19*8+100?

or am I missing something

[topaz axle]

now you're missing 19 from 000 to 099

[arctic field]

008 is valid

[pseudo pollen]

o

oh

wait so

there are also numbers like 0008?

[topaz axle]

sure

[arctic field]

its a password, not a number

[pseudo pollen]

oh

[topaz axle]

that's why they said no restrictions

[pseudo pollen]

that makes sense

can I dm you when I’m done?

[topaz axle]

there's like no other restriction they meant

i don't mind

[pseudo pollen]

ohh

[devout snow]

@pseudo pollen Has your question been resolved?

[pseudo pollen]

ok final answer 3729

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[proven canyon]

My book tells me that a function such as the one on the upper part of the image cannot have a primitive function because of darboux's theorem.
However could a function such as the one on the bottom have one?
Thanks in advance

[proven canyon]

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[spring gate]

any idea on how to integrate this

[faint prawn]

Let r2 + x2 = u

2xdx = du

Then you can do it i think

[spring gate]

lemme try

wont it come like this

nah it seems way too complecated

ther's gotta be a better method

ight lemme try soemthing

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[woven radish]

rak³en

[devout snow]

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[finite briar]

BRUH

[devout snow]

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[lost laurel]

Kind of confused as to How I'd sketch this

[finite briar]

jacobian

[winter torrent]

it's pretty simple to see the region 1 ≤ u ≤ 2

the region 1 ≤ uv ≤ 2 is a little harder but still very doable

[lost laurel]

Oh, I thought one of the axes was a hyperbola here 🤦

[lost laurel]

It's the region between two hyperbolas is it not?

[winter torrent]

sure seems good

[lost laurel]

Thanks

.clopse

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[lost laurel]

Evaluate $\int \int_{S} x^2 y^2 dy$ where it's bounded by $y=0, y=x, x+y=1$

[lost laurel]

I intend to use a change of variable

I was thinking $ y= v, 1-x=u$

[woven radish]

What a wonderful world!

[lost laurel]

So like the integral before I change the order would be

$\int_{0}^{0.5} \int_{y}^{1-y} x^2y^2 dx dy$

[cinder bobcat]

I think you mean y to 1-y right?

[lost laurel]

oops

yeah

[cinder bobcat]

and do you have to use a change of variable?

[lost laurel]

Yes

[cinder bobcat]

oof

okay well what's your next step then

[woven radish]

What a wonderful world!

[lost laurel]

so now y=v

and x=u?

[cinder bobcat]

well you said the transformation was y = v, x = 1-u if you move stuff around

[lost laurel]

yeah

oops

that works too

[cinder bobcat]

yeah, best thing to do is graph the transformation

[lost laurel]

I am

[cinder bobcat]

you kinda have to in order to know the new bounds

oh good

oh I gotta go for now sorry, hopefully someone else can help or you can finish it off 🫡

[lost laurel]

Thanks

I feel like the bounds don't chnage at all

<@&286206848099549185>

[faint gorge]

Why would this go from 0 to 0.5

[lost laurel]

what should it be

,w y=0, y=x, x+y=1

[faint gorge]

I think 0 to 1

[woven radish]

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[lost laurel]

y=x intersecst y+x=1 at (0,5,0.5)

[faint gorge]

isn't that the region?

[lost laurel]

oops

my bad

[faint gorge]

no my bad

[lost laurel]

yea

[faint gorge]

you are right

[lost laurel]

I

I'll use horizontal lines to integrate

[faint gorge]

now what's the sub you plan to do

[lost laurel]

so is $\int_{0}^{0.5} \int_{y}^{1-y} x^2y^2 dx dy$ fine?

[woven radish]

What a wonderful world!

[faint gorge]

yea

[lost laurel]

then y = v , x= u

wait no

[faint gorge]

this wont do anything

[lost laurel]

yeah

Any suggestions

Maybe x+y=u,, x-y=v

[faint gorge]

maybe u = x² and v = y²

[lost laurel]

hmm

yeah

that works

[faint gorge]

if think anything else would make it worse than it is

[lost laurel]

So I now have $\int_{0}^{0.25} \int_{\sqrt{v}}^{1- \sqrt{v}} uv du dv$

ooops

[woven radish]

What a wonderful world!

[lost laurel]

Is that right

[faint gorge]

I think the jaboian is missing

[lost laurel]

oh right

I can manage that

is it fine other than that

I want to ensure I know how to change the limits easily

[faint gorge]

the bounds for u might be off

[lost laurel]

Hmm?

[faint gorge]

x = (1-y) and x² = u
So u = (1-y)² = (1-sqrt(v))²
and for lower x = y then x² = y² = v

[lost laurel]

yea

[faint gorge]

I feel this sub made things worse lol

[lost laurel]

Let me check if my book suggests any subs

[faint gorge]

[ \int_{0}^{0.25} \int_{v}^{(1- \sqrt{v})^2} uv \cdot \abs{J} : du dv ]

[woven radish]

anti-algebraist 𝔸dωn𝓲²s

[lost laurel]

It suggets x=u+v, y=u-v

my book

[faint gorge]

ok then let's do that

[lost laurel]

so $u= \frac{x+y}{2}; v= \frac{x-y}{2}$

[woven radish]

What a wonderful world!

[faint gorge]

ok

[lost laurel]

hmm

hmm

[lost laurel]

I don't think this helps much

[faint gorge]

y in [0, 0.5] and x in [y, 1-y]

[lost laurel]

no, no

I mean

How do I do the substitution now

[faint gorge]

x²y² = (u+v)²(u-v)²

then you'd need the Jacobian

[lost laurel]

Yeah, trying to sketch the region though

[faint gorge]

I think you need to get the intersections

[lost laurel]

yeah, trying to figure that out

[faint gorge]

well that's here

and then plug them into your substitution and draw the new intersections in the uv plane

[lost laurel]

wait

so I have x=u+v

and y = u-v

one boundary is x=y

so u+v=u-v

so v=0

another boundary is 1-y=x

so 1-(u-v)= u+v

so 1-u+v=u+v

which gives be u=1/2

which is very sus

[lost laurel]

Any suggestions

[faint gorge]

The new points are (0,0), (0,0.5) and (0.5,0.5)

[lost laurel]

How did you get that

[faint gorge]

the intersections of this plugged into the sub

[lost laurel]

What I did was,

uh

x=y

so u+v=u-v

so v=0

then $u+v=1-(u-v) \implies u+v= v+1-u$

[woven radish]

What a wonderful world!

[lost laurel]

so u=1/2

[faint gorge]

ok

continue

[lost laurel]

and then

uh

u-v=0.5

so the points I',m getting are (0,0), (0.5,0.5), (0.5,0)

[faint gorge]

(u,v) or (v,u)?