help-42

[candid pilot]

@polar frost

[polar frost]

What do u mean?

[polar frost]

yes?

[candid pilot]

In the part I circled

Like is there supposed to be another line there or is it right

also how can i find the point of intercepts for f(x) = 6- x^2 and its inverse root 6-x

[polar frost]

No

For every x there should 1 value of f(x)

[candid pilot]

point of intersection i mean

would i use substitution and then solve?

[polar frost]

U dont neef to prove it

Its a fact

Those x value has a corresponding f(x) and they're there

[candid pilot]

Oh that ones for a different question

[polar frost]

Im so confused

Whats the question again

[candid pilot]

Sorry well start a new question

how can i find the point of intersection for f(x) = 6- x^2 and its inverse root 6-x

[polar frost]

Like the intersection of the 2 lines?

[candid pilot]

yeah

[polar frost]

inverse root 6 - x like

[candid pilot]

oh sorry no like

f^1(x) = root 6-x

so originally i found the reciprocal of 6 - x^2

[candid pilot]

and got that

the next step was the point of intersection

[candid pilot]

yeah

[polar frost]

Just set them =

[candid pilot]

i also substituted for y and got x^4 + 12x^2 + x +30 = 0

[polar frost]

6 - x^2 = sqrt(6-x)

[candid pilot]

yeah i wasn't able to factorise it

[polar frost]

square them

[candid pilot]

yeah and then i got x^4 - 12x^2 + x +30 = 0 is what i mean

and wasn't able to factorise it

[polar frost]

Use calculator then

[candid pilot]

is there any like way to do it without calculator?

[polar frost]

its a 4 degree poly

[candid pilot]

yeah so 4 intersections

[polar frost]

Only 2

[candid pilot]

oh 2?

but like for the root equation

[polar frost]

U need to restrict that sqrt

[candid pilot]

would you have a plus minus in front of it

[polar frost]

So x must be positive

discard the negative

[candid pilot]

okok

wait so

im a bit lost

i squared the square root meaning i squared both sides

and then i brought everything to the LHS

and got x^4 - 12x^2 + x +30 = 0

but im not sure how to solve for x so i can then substitute it and find y

unless this process is wrong

@polar frost

[polar frost]

Just use the calculator for this part

then u should get 4 values of x

but u should only accept x > 0

[candid pilot]

Is it not possible without?

A calculator

[calm coral]

@candid pilot Has your question been resolved?

[calm coral]

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[timber timber]

Option 1: 2(lw + lh + wh), l + w + h, lwh Option 2: 3 layers of 10, 7 layers of 21, 3 layers of 21, 7 layers of 3. Option 3: volume, edge lengths, base area Option 4: equal to, less than, greater than
Image

[timber timber]

pls help 🙂

[calm coral]

@timber timber Has your question been resolved?

[timber timber]

<@&286206848099549185>

i give up

.close

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[mortal heath]

could i possibly get some help on this question? i'm just honestly quite confused on where to even start with this

[polar frost]

Set R(x) > 60

Then find x

[brazen elbow]

also be aware of the fact that rainfall cannot be negative

[mortal heath]

yeah i'm just confused on the > thing but should i just treat it as = when solving for x?

[polar frost]

Doesnt matter

[brazen elbow]

kind of yeah, but you also swap the direction when multiply both hand side with negative

[polar frost]

For that function y>60 only return x in positive

[brazen elbow]

the problem is just convenient to give that result

without considering it, the answer is inconplete

[mortal heath]

wait so i should write it out as 60 > 7500x/x^2 + 2500?

[polar frost]

No?

its more than 60

[mortal heath]

ohhhhhhh

right sorry 😭

[mortal heath]

i'm just confused because the final answer is a range

[remote mural]

there is some algebra you have to do

so like, do you understand that [
\4{7500x}{x^2+2500} \ge 60
]
is the inequality you are meant to solve

[mortal heath]

yes

[remote mural]

ok

so a good way to start is to subtract 60 from both sides, and combine everything on the left to a single fraction

What does that get you

[mortal heath]

oh i thought it would be better to multiply each side by x^2 + 2500 to cancel out that fraction

[glass heart]

I would say its easier to multiply both sides by x^2+2500

[remote mural]

i mean sure the denominator goes away either way

[mortal heath]

yeah

[glass heart]

note that x^2+2500 is certainly >0, so you dont have to switch the inequality

[mortal heath]

so then i got that 60x^2 + 150000 < 7500x

yeah

or wait i can just send what i did

i know i should've put < instead of = but yea

i still don't think i got the right answer idk

[polar frost]

Bruh

[mortal heath]

maybe i'm stupid but i don't know what's wrong with that

[polar frost]

You should use the => arrow instead of -> this one

I thought u were substitute that part in

[mortal heath]

oh i didn't know that made a difference it's just kinda how i write stuff down for myself when working out

[polar frost]

😭

[remote mural]

it's not really important its fine, you can always just include the endpoints

[polar frost]

Hmm

[remote mural]

anyways

[polar frost]

Why the answer looks a bit wrong

[remote mural]

im pretty sure you erased a 0 from the 150000 @mortal heath

[polar frost]

Did u remove one 0 from 150000 to 15000

[mortal heath]

oh i did my bad 😭

[remote mural]

also just simplify all of ig

750, 15000, and 6 all divide 6

[polar frost]

Wait no

[mortal heath]

i got it thank you!!

[polar frost]

Did she devide everything by 10

[mortal heath]

i did yeha

[remote mural]

yeah i think its fine

anyways divide by 6 what do u get

[polar frost]

I use my calculator and get 25 100

So miku probably made a mistake when calculate x

[mortal heath]

yeah i did

my mistake was that i calculated with 1500 instead of 15000

i got the right answers fixing that haha

[remote mural]

yeah she erased a 0 like we said lmao

[mortal heath]

thank you! sorry if this work is really simple but i appreciate the help a lot 😭 need to pass this class

[polar frost]

2 values of x mean u need to limit it

[mortal heath]

yeah i set it as

25 < x < 100

[polar frost]

Good

[mortal heath]

thank you once again haha

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[lunar ridge]

Please help

[timid oracle]

you can solve for x since the all the probabilities sum to 1

split that thing into P(B intersect A')/P(C)

[calm coral]

@lunar ridge Has your question been resolved?

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[fringe hamlet]

Can the fraction be reversed when there is a + in the middle?

Like I did

[ancient helm]

where?

[fringe hamlet]

On the penultimate line of the photo

[ancient helm]

this?

[fringe hamlet]

Yes

[ancient helm]

like to go from the line above it to that?

[fringe hamlet]

Yes

[ancient helm]

well in general no

you should combine LHS into a single fraction

then cross multiply

LHS has common denom already

don't simplify the 3^(n+1)/3^(n+1) part

[fringe hamlet]

Ah ok

Thanks !

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[midnight field]

working on triple integrals! what am i doing wrong here?

[warm warren]

Something’s wrong why are you integrating over T but then you say a ball B

[midnight field]

oops

forgot that i wrote t

should be b

nothing else is changed i just had T in my notes and instinctually put it there

im confdent theres something wrong with the setup

[warm warren]

I don’t see it soz I’m very sleepy tho

[midnight field]

all good

im guessing its a dumb mistake somewhere i havent seen

Yup

forgot rho ^2

lmao

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[uncut narwhal]

Hi

[calm coral]

Please don't occupy multiple help channels.

[uncut narwhal]

please help

934.38

i got that

[unkempt drift]

Please show your work in future

[uncut narwhal]

I calculated it.

[unkempt drift]

I know but the steps are also important

It saves us from having to do your work again

[thin gulch]

It is much easier for us to fix your error from your work. It doesn't really help you if I just give you the right answer right?

[calm coral]

@uncut narwhal Has your question been resolved?

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[blazing coyote]

I split this into

$\int_{\frac{1}{3}}^{\frac{1}{2}}e^2dx+\int_{\frac{1}{2}}^1edx+\int_1^3dx$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

would that be right

if right that would be

$\int_{\frac{1}{3}}^{\frac{1}{2}}e^2dx+\int_{\frac{1}{2}}^1edx+\int_1^3dx=e^2\left(\frac{1}{6}\right)+\frac{e}{2}+2$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

just checked the answer key

it;s apparently 5/6

how is that?

[rustic osprey]

Did you remember the ln part

[blazing coyote]

what ln part

oh

so it would be the ln of these exponetial functions ?

[rustic osprey]

,w 2(1/2-1/3)+(1/2)

[potent lotus]

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[blazing coyote]

so $\int_{\frac{1}{3}}^{\frac{1}{2}}\ln\left(e^2\right)dx+\int_{\frac{1}{2}}^1\left(e\right)dx+\int_1^30dx=e^2\left(\frac{1}{6}\right)+\frac{e}{2}+2$

[rustic osprey]

Yeah e^2 should be replaced with 2 and so on

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

thanks!

[rustic osprey]

[blazing coyote]

.close

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[blazing coyote]

.reopen

[calm coral]

✅

[blazing coyote]

so I was thinkng of solving this by. breaking it down into seperate intervals

$\int_0^1xdx+\int_1^22x^2-1dx+\int_2^33x^3-2dx$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

would that be right

[rustic osprey]

Yeah

[blazing coyote]

thanks!

[clever ruin]

another to try if u want

$\int_0^\infty e^{-\lfloor x \rfloor}dx$

[potent lotus]

Max

[blazing coyote]

hmm

let me think

thanks for the nice problem

hmm

so that would be

$e^0+e^{-1}+e^{-2}....$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

which is an infinite GP

which would be

[clever ruin]

yea

[blazing coyote]

$\frac{e}{e-1}..$

[potent lotus]

ƒ(Why am. I here)=I don't know

[clever ruin]

yea

[blazing coyote]

thanks! That was fun

[clever ruin]

if u like integrals

there are Bees such as the MIT int bee and UK Integration Bee (Cambridge)

[blazing coyote]

I do solve them, that's where I get these problems from, Thanks though!

could you link to the cambridge bee

can't find it

[bronze adder]

saw this in the caltech one lol

[blazing coyote]

it is from the caltech one

[clever ruin]

https://integration.soc.srcf.net/

[blazing coyote]

thanks!

I converted this into two forms

form a

[bronze adder]

utilize symmetry about pi/2

[blazing coyote]

$\left(\int_0^{\pi}\frac{\left(\sin\left(\frac{x}{2}\right)-\cos\left(\frac{x}{2}\right)\right)^2}{\left(\sin\left(\frac{x}{2}\right)+\cos\left(\frac{x}{2}\right)^2\right)}dx\right)$

[potent lotus]

ƒ(Why am. I here)=I don't know

[bronze adder]

after some simplifcation it turns into tan^2(x/2) which is easy to integrate

oh lord

[rustic osprey]

If we’re doing integration then have another fun one
Hint: ||binomial theorem||

[blazing coyote]

thanks!

[blazing coyote]

thanks!

[bronze adder]

do you know how to get there

[blazing coyote]

thinking about it

[bronze adder]

alright

[blazing coyote]

this could be converted to $\int_0^{\pi}\left(\tan\left(\frac{\pi}{4}-x\right)\right)^2$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

which is integrable

[bronze adder]

fair

[rustic osprey]

3 more random ones I had lying around, I’ll leave you alone now

👋

[blazing coyote]

thanks a lot

[bronze adder]

for the 19th one ||you can split it as arctan(1 + x) - arctan(x)||

and the 20th is just ||an embellished euler log trig so thats -pi(ln 2)/2||

[blazing coyote]

spoilers !

[bronze adder]

mb

49 ||x --> 1/x and add||

[rustic osprey]

👍 feel free to ping me if you want more

[blazing coyote]

this is $\int_0^{\frac{3\pi}{4}}1-\sec^2\left(x-\frac{\pi}{4}\right)+\int_{\frac{3\pi}{4}}^{\pi\ }1-\sec^2\left(x-\frac{\pi}{4}\right)$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

thanks!

[blazing coyote]

which evaluvates to

$\int_0^{\frac{3\pi}{4}}1-\sec^2\left(x-\frac{\pi}{4}\right)+\int_{\frac{3\pi}{4}}^{\pi\ }1-\sec^2\left(x-\frac{\pi}{4}\right)=\frac{3\pi}{4}-\tan\left(\frac{\pi}{2}\right)+\frac{\pi}{4}-\tan\left(\frac{3\pi}{4}\right)-\tan\left(\frac{\pi}{\left(2\right)}\right)$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

what am I doing wrong here

oo

it should be x/2

$\int_0^{\pi}1-\sec^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

whcih is $\pi-1$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

huh

the answer is 4-pi

,w integrate 1-sec^2(pi/4-x/2) from 0 to pi

[potent lotus]

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[blazing coyote]

must have made a calculation mistake

nvm

thanks

[blazing coyote]

now onto these

$\left(\int_0^{\infty}\tan^{-1}\left(\frac{\left(x+1-x\right)}{1+x\left(x+1\right)}\right)\right)$

[potent lotus]

ƒ(Why am. I here)=I don't know

[bronze adder]

why do you surround the whole thing in parenthesis lol

[blazing coyote]

I'm using approach 0 for formatting

which is

[bronze adder]

dunno what that means but alright

[blazing coyote]

$\int_0^{\infty}\tan^{-1}\left(\left(x+1\right)-\tan^{-1}\left(x\right)\right)$

[potent lotus]

ƒ(Why am. I here)=I don't know

[bronze adder]

whatever your approach 0 is

[blazing coyote]

which can be integrated by parts

[bronze adder]

its misplacing parenthesis willy nilly

[blazing coyote]

https://approach0.xyz/search/

[blazing coyote]

that's my doing

[bronze adder]

lol the whole thing is in an arctan

[blazing coyote]

$\int_0^{\infty}\tan^{-1}\left(x+1\right)-\tan^{-1}\left(x\right)$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

,w $\int_0^{\infty}^{-1$\ln\left(x\right)\arcsin\left(x\right)-\int \frac{\arcsin\left(x\right)}{x}$$\int \frac{\ln\left(1-x\right)}{\sqrt{2x\left(1-x\right)}}$}\left(x+1\right)-\tan^{-1}\left(x\right)$

lol

thanks

[bronze adder]

use arctan

[blazing coyote]

yeah, will in the future, I think in this problem, getting it to this form was the hard part, so I'll skip evaluvating it for now

onto problem 20

$\int_0^1\frac{\ln\left(x\right)}{\sqrt{1-x^2}}$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

let $x=e^t$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

$\int_1^e\frac{te^t}{\sqrt{1-e^{2t}}}$

[potent lotus]

ƒ(Why am. I here)=I don't know

[bronze adder]

-infinity to 0

[blazing coyote]

$\int_1^e\frac{te^t}{\sqrt{1-e^{2t}}}=t\sqrt{1-e^{2t}}=\int \sqrt{1-e^{2t}}$

ah

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

$\int_1^e\frac{te^t}{\sqrt{1-e^{2t}}}=t\sqrt{1-e^{2t}}-\int \sqrt{1-e^{2t}}$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

$\int_{\ -\infty}^0\frac{te^t}{\sqrt{1-e^{2t}}}=t\sqrt{1-e^{2t}}-\int \sqrt{1-e^{2t}}$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

let $e^t=u$ for the second part

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

I think I;ll just be going in circles if I do that

[bronze adder]

the IBP is

[blazing coyote]

let me IBP instead

[bronze adder]

not right

[bronze adder]

here

shouldn't it be arcsin(e^t)

[blazing coyote]

oh yeah

it should

that makes in much harder IMO

[bronze adder]

i mean as it stands for t --> - infinity you don't even get a real answer

[blazing coyote]

$t\arcsin\left(e^t\right)-\int \arcsin\left(e^t\right)$

[potent lotus]

ƒ(Why am. I here)=I don't know

[bronze adder]

you'll have to do some limit fuckery

[blazing coyote]

I think I better IBP the first function instead

[bronze adder]

easy enough though just multiply and divide by e^t i'd think

[blazing coyote]

this

[bronze adder]

sure try it out

[blazing coyote]

$\ln\left(x\right)\arcsin\left(x\right)-\int \frac{\arcsin\left(x\right)}{x}$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

which isn't integrable afaik

hmm

[bronze adder]

,w integrate arcsin(x)/x

[blazing coyote]

oo

[potent lotus]

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[blazing coyote]

king's rule

maybe

[bronze adder]

F

[blazing coyote]

which would be

$\int \frac{\ln\left(1-x\right)}{\sqrt{2x\left(1-x\right)}}$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

nah

that won't work either

hmm, maybe I use a polynomial approximation for bith functions

and integrate that

[bronze adder]

you cant approximate this stuff

[blazing coyote]

hmm

[bronze adder]

the choices are all closed form

[blazing coyote]

let me think some more

,w integrate te^t/\sqrt{1-t^2}

[potent lotus]

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[blazing coyote]

oo

$\int \frac{\ln\left(x\right)}{x\sqrt{\frac{1}{x^2}-1}}$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

let $ln(x)=u$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

so that gives

nvm

[bronze adder]

same thing as before

do you want a hint

[blazing coyote]

yes please

[bronze adder]

trig sub

[blazing coyote]

hmm, x=sin(t)

so dx=cos(t)dt

$\int \frac{\ln\left(\sin\left(t\right)\right)}{\cos\left(t\right)}\cos\left(t\right)dt$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

which is integrable

oof

that was easy

thanks!

[bronze adder]

its not

,w integrate ln(sin x)

[potent lotus]

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[blazing coyote]

oh

[bronze adder]

you need to use more properties and stuff to get it

the definite case is doable

[blazing coyote]

hmm

ok

hmm

$\int_0^{\frac{\pi}{2}}\ln\left(\sin\left(t\right)\right)+\ln\left(\cos\left(t\right)dt\right)$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

this is 2I

is it -pi/2 ln(2)

$\int_0^{\frac{\pi}{2}}\ln\left(\sin\left(t\right)\cos\left(t\right)\right)dt\ $

$\int_0^{\frac{\pi}{2}}\ln\left(\frac{\sin\left(2t\right)}{2}\right)dt\ $

$\int_0^{\frac{\pi}{2}}\ln\left(\frac{\sin\left(2t\right)}{2}\right)dt$

[bronze adder]

no space after the last character

[bronze adder]

yeah

[potent lotus]

ƒ(Why am. I here)=I don't know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[blazing coyote]

thanks

now to try this

[blazing coyote]

the denominator's form looks similar to this

[cursive linden]

oh i see

[bronze adder]

1 + x + x^2 in the denominator always screams a sub

[blazing coyote]

yeah

$x=tan(u)$

[potent lotus]

ƒ(Why am. I here)=I don't know

[bronze adder]

not what I meant but keep going 🫠

[blazing coyote]

hmm

let me think of something else

[cursive linden]

first keep going with what you were thinking

[blazing coyote]

$\int_0^{\frac{\pi}{2}}\frac{\sec^2\left(u\right)}{\sec^2\left(u\right)+\tan\left(u\right)}$

[potent lotus]

ƒ(Why am. I here)=I don't know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[blazing coyote]

$\int_0^{\frac{\pi}{2}}\frac{u\sec^2\left(u\right)}{\sec^2\left(u\right)+\tan\left(u\right)}$

[potent lotus]

ƒ(Why am. I here)=I don't know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[blazing coyote]

nah

I'll try some other sub

$\int_0^{\infty}\frac{\arctan\left(x\right)}{\tan\left(\operatorname{arccot}\left(x+1\right)-\operatorname{arccot}\left(x\right)\right)}$

[potent lotus]

ƒ(Why am. I here)=I don't know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[blazing coyote]

nah, that won't work either

maybe I IBP

that definietly won't work

I guessing I have to do something to the denom

write it in terms of trig functions , is that right?

[cursive linden]

keep trying ig 🤷‍♂️

[blazing coyote]

hmm

[bronze adder]

what have I returned to

[blazing coyote]

$\int_0^{\infty}\arctan\left(x\right)\tan\left(\arctan\left(x+1\right)-\arctan\left(x\right)\right)dx$

[potent lotus]

ƒ(Why am. I here)=I don't know

[bronze adder]

do you wanna know what id do

[blazing coyote]

No

please

[bronze adder]

alright

keep going

[blazing coyote]

I guess IBP is wrong here

[pure breach]

oh no big scary integral

r u even doing that

[blazing coyote]

did that long ago

[pure breach]

ah this is what ur trying

[blazing coyote]

$\int \frac{\arctan\left(x\right)}{\left(1+x\right)^2-x}$

[potent lotus]

ƒ(Why am. I here)=I don't know

[blazing coyote]

I feel like an inverse trig sub may help here

[pure breach]

not really imo but u can try

[blazing coyote]

hmm

ok

in that case I think I have to express the denom as an inverse trig function?

[pure breach]

that seems to be it

[blazing coyote]

so this is what I thought of

which I don't think helps very much

[pure breach]

arctan a - arctan b might? unsure

[blazing coyote]

ah

that's where I went wrong

[pure breach]

wdym?

[blazing coyote]

this is almost in teh same form except for the tan

$\int \arctan\left(x\right)\left(\arctan\left(x+1\right)-\arctan\left(x\right)\right)$

[potent lotus]

ƒ(Why am. I here)=I don't know

[pure breach]

oh 💀

[blazing coyote]

am i right until here

[pure breach]

i have no idea what u subbed

[blazing coyote]

no sub yet

[pure breach]

wth r u doing then 💀

[blazing coyote]

just tried re-writing 1/(1+x+x^2)

that's why I wrote it as $tan\left(\arctan\left(x+1\right)-\arctan\left(x\right)\right)dx$

[potent lotus]

ƒ(Why am. I here)=I don't know

[pure breach]

but thats wrong no?

oh wait mb

am dumb

[pure breach]

so it will indeed be this

[blazing coyote]

but that isn't integrable , is it?

this

[pure breach]

i dont think u can solve this normally

u might need something more higher-end

[blazing coyote]

hmm

maybe eularian trig

[pure breach]

i am trying feynman 🙂

[blazing coyote]

That's way beyond me

[pure breach]

tbh it isnt

as long as uk partial diff

which everyone knows as long as they r good at differentiation

hmm feynman gives something that i might be able to partial frac

[blazing coyote]

found this hint by Neon

oof

that works

[pure breach]

hmm oh wait i forgot the infinity

[blazing coyote]

why am I so slow, lol

[pure breach]

ye why didnt u just that if he gave u that 💀

[blazing coyote]

I forgot about it

thanks

.close

[calm coral]

Channel closed

Closed by @blazing coyote

Use .reopen if this was a mistake.

[pure breach]

well if u want i can show u the feynman way @blazing coyote

its pretty easy that way

[calm coral]

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[remote mural]

How do I prove this obvious statement

[atomic idol]

construct a bijection between the sets

[remote mural]

Can the subsets have the same elements?

[atomic idol]

the subsets don't have all the same elements

[remote mural]

One has n the other does not

[atomic idol]

A and B are different sets, they just have the same cardinality

[remote mural]

Ok

[remote mural]

How do I create a bijecriom

[atomic idol]

for any $A \in \mathcal{A}$ it corresponds to the set $A \cup {n} \in \mathcal{B}$

[remote mural]

What

I don’t understand

[potent lotus]

Awesam

[atomic idol]

I assume you learned how to show that sets have the same cardinality?

what did you learn as the method

[remote mural]

Logic

[atomic idol]

so you're not looking for a formal proof just intuition?

[remote mural]

Ye the statement is obvious

But how do I do a formal proof

Using BP

[atomic idol]

what's BP

[remote mural]

I wasn’t taught how to use bijective principle

(BP)

[atomic idol]

oh

yeah then you do it the way I described

[remote mural]

What even is it

[atomic idol]

do you know what a bijection is

[remote mural]

No

It’s like one to one

[atomic idol]

yes

[remote mural]

When u map it

Each element corresponds to 1 other element

[brisk pelican]

And onto

[remote mural]

Ex: {1,2,3,4,5} and {5,6,7,8,9}

[atomic idol]

yes so if you map a set $A$ to a set $A \cup {n}$ then this is a 1 to 1 correspondance between $\mathcal{A}$ and $\mathcal{B}$

[potent lotus]

Awesam

[atomic idol]

for example if I have the set ${1, 2, 3}$ then it maps to the set ${1, 2, 3, n}$

[potent lotus]

Awesam

[atomic idol]

then every set maps to a different set and every set containing n is accounted for

[remote mural]

1,1 2,2 3,3 it’s one to one and onto

Because n has no element mapped

Surjective

I’ll show u my proof

[remote mural]

And you see if it’s good ok?

[atomic idol]

okay

[ancient helm]

that's not onto

also that's not even what A and B look like

A is the set of all subsets not containing n

B is the set of all subsets containing n

[calm coral]

@remote mural Has your question been resolved?

[remote mural]

So what do I do

[ancient helm]

.

[remote mural]

But I don’t understand that

[ancient helm]

He gave you the correspondence

[remote mural]

?

Which correspondence

[ancient helm]

.

[remote mural]

But I don’t understand what it says tho

[ancient helm]

You map the subset not containing n

To the exact same subset but with n

[remote mural]

Ok

Half contain n half don’t

?

[remote mural]

Yes

Yes but depends what n is

[ancient helm]

No

[remote mural]

There is one to one correspondence

?

[ancient helm]

Yes you're trying to show this mapping is a bijection

So you need to show it's injective and surjective

[remote mural]

But isn’t a bijectiom also mean it’s onto?

[ancient helm]

Yes

Surjective = onto

Injective = one to one

[remote mural]

What does onto mean

One to one means the mapping goes 1 element for each corresponding

But onto?

Sorry I never learnt this lol

[ancient helm]

Let f:A -> B then f is onto if for every b in B, there exists an a in A with f(a)=b

[remote mural]

I see

And one to one means that f(a2) neq to f(a1)

[ancient helm]

if a1 and a2 are distinct

[remote mural]

Yes

So it’s obviously one to one the set

[ancient helm]

How is it obvious

[remote mural]

Because each subset will be different

[ancient helm]

Yes but why wont f map different subsets to the same thing

[remote mural]

The sets which hold the subsets are equal caridinality I mean

[remote mural]

So now, how do I construct the proof

Cuz I’m still lost I can’t visualize anything

@ancient helm can u show me an example?

Let’s say i take the set X={1,2,3}

[ancient helm]

Ok

What is A and what is B

[remote mural]

A = {{1,2},{1},{2},{}}

Since n = 3

[ancient helm]

yes

What is B

[remote mural]

B = {2,3},{3}, {1,3},{1,2,3}

Oh nvm

A and B are equal in cardinality

[ancient helm]

That's not what B is

[remote mural]

Is it that?

[ancient helm]

Youre missing {1,2,3}

And in A you're missing the empty set

[remote mural]

Oh

[ancient helm]

Ok now show me the mapping

[ancient helm]

Yes

Now prove it for general sets X

[remote mural]

Ok

[ancient helm]

You need to show your map is injective and surjective in your proof

[remote mural]

That’s hard there’s so many elements

[ancient helm]

That's why you need to generalize

Not explicitly write out the mapping

[remote mural]

How do I generalize?

[remote mural]

Aren’t all possible subsets 2^n

?

Half contain n, half don’t

So I’ve proved it but not by BP 😂

[remote mural]

Look I’ve never proved something using BP. I know what it is now, but I don’t know how to use it

Ok first how do I prove that it’s is invective (ie at least 1 x connects to exactly 1 y)

https://www.mathsisfun.com/sets/images/function-mapping.svg

<@&286206848099549185> can someone tell me how to prove injectivity?

[amber roost]

Injective is when B can only have 1 A or 0

[potent lotus]

3ndlyb Alabyd

[remote mural]

I see but how do I connect that to the problem I need to solve?

[amber roost]

What's the problem? I am too lazy to scroll up

[calm coral]

@remote mural Has your question been resolved?

[untold compass]

idk what BP is but you can pair up items in A and B to prove it

like {stuff without n} pairs with {same stuff with n added}

and that hits everything

[calm coral]

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[calm coral]

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[proper panther]

im having a tough time understnading logical statements and negations. could someone verify that my process is correct for these problems?

[calm coral]

@proper panther Has your question been resolved?

[proper panther]

.close

[calm coral]

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[calm coral]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[keen grove]

how do I solve this?

[keen grove]

this is what ive done so far

[lethal oak]

you know how log laws work right

[keen grove]

yeah

[lethal oak]

do you think u can apply it here

[keen grove]

so do i do ln((1+4R^2)/1+R^2))? i dont see how that helps tho

[lethal oak]

oh brother it helps

what can you do with limits here

in this situation

given that you have a inside function nested inside an outside function

[keen grove]

idk

what can u do?

[lethal oak]

Can you not push the limit inside if the limit inside exists?

[keen grove]

inside the log?

[lethal oak]

yes

[keen grove]

ahhh ok

[lethal oak]

you see now

[keen grove]

ohh then its 1/2(ln(4))

right i see

[lethal oak]

mhmmm

[keen grove]

thanks for that

[lethal oak]

np

[keen grove]

.close

[calm coral]

Channel closed

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[calm coral]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[mossy wigeon]

Where did the random 2 come from

Also why the bounds of integration was changed

[neon sage]

,w plot 1-x^2

[potent lotus]

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[neon sage]

notice how the function is symmetric about x = 0

so instead of integrating from -1 to 1, we integrate from 0 to 1 and double the value

[mossy wigeon]

so

its the same thing

like

i dont need to change it

but

its for

easiness

[calm coral]

@mossy wigeon Has your question been resolved?

[neon sage]

yes

[mossy wigeon]

ty

.close

[calm coral]

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[calm coral]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[novel jackal]

yo

[novel jackal]

I don't understand why f(x) is a bijection from the natural numbers to the positive rational numbers

in there example, there is never any rational number greater than 1 in there list

Because of that, isn't this a bijection from the natural numbers to all rational numbers between 0 and 1?

So why did they claim its a bijection from the naturals to all positive rationals?

[clear delta]

you're right

[novel jackal]

wait so is the slide wrong

HA

[swift laurel]

an actual example for your reference, from wikipedia (note that the red numbers are duplicates and are discarded)

[calm coral]

@novel jackal Has your question been resolved?

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[remote mural]

I have to prove

$Z_1 \bar{Z_2} + Z_2 \bar{Z_1} = 2 |Z_1| |Z_2| \cos (\theta_1 - \theta_2)$

[clear moat]

great!

[remote mural]

[clear moat]

now, is Z and Z^C complex numbers

[remote mural]

Z1 and Z2 are complex numbers, the $\bar{}$ represents their conjugates and $\theta_1 \ and \ \theta_2$ represent their arguments respectively

[potent lotus]

fredthebread69

[clear moat]

fair enough what did you try

[remote mural]

I have tried putting $Z1 = |Z1|e^{i\theta}$

[potent lotus]

fredthebread69

fredthebread69

[clear moat]

try using the respective form for complex numbers

$$z_1 = r_1 (\cos \theta_1+ i \sin \theta_1)$$

[potent lotus]

ℕaive

[clear moat]

$$z_2 = r_2 (\cos \theta_2+ i \sin \theta_2)$$

[potent lotus]

ℕaive

[remote mural]

Can I not use exponents and try?

[clear moat]

sure if you feel so

[remote mural]

Could you help me out?

[calm coral]

@remote mural Has your question been resolved?

[remote mural]

<@&286206848099549185>

I think I'm doing something wrong

[calm coral]

@remote mural Has your question been resolved?

[green haven]

does the RHS look like some familiar identity/law/rule?

(idk if it's helpful actually i just noticed it immediatly)

[remote mural]

It would help if someone knew what the name of the identity haha

And yes it's an identity

[green haven]

yea is it possible to visualise the LHS as what was usual in that identity?

ir transpose/convert etc

somehow idk

[remote mural]

What?

[green haven]

like if u think about it the cos rule works out the square of the side opposite the angle

now u have angle and the 2 sides

can u make z1__z__2+__z__1z2 into that somehow

[remote mural]

The cosine rule?

[green haven]

yea

[calm coral]

@remote mural Has your question been resolved?

[remote mural]

.close

[calm coral]

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[austere moth]

Let $f(x, y, z) = 14 + x^3y^4e^{x(z+4)}$ and $A = (1, 1, −4)$ \
(a) Specify the tangent plane to the level surface f(x, y, z) = 15 at point A.\
(b) At what rate of change $f′_v (A)$ does f increase or decrease in the direction (−6, 2, −3) away from the point A? \
(c) In what limits does the directional derivative $f_v (A)$ depending on the departure direction v away from A?

[potent lotus]

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

[austere moth]

Anyhow.... i have a formula for calculating the tangent plane

Does this formula work with f(x,y,z)

or is specifically f(x,y,)

[potent lotus]

themadchessplayer

themadchessplayer

[clever loom]

Where v is the direction

[austere moth]

Hold on, do I just input the point A into the formula essentiallly to get the tangentplanes equation?

[clever loom]

Yes

[austere moth]

one sec ! c:

You mean this right?

[clever loom]

I don't understand what you want in (c)

[clever loom]

No. The other one

[austere moth]

Oh

[clever loom]

Because the surface is f(x,y,z)-15 = 0

[austere moth]

oh i see

how do i handle the -15?

I assume i dont?

[clever loom]

Vanishes when you take derivative

[austere moth]

Ah right makes sense

god ur good at this haha

don't you dare leave

[clever loom]

LOL

Please clarify c

If you want an answer for that

[austere moth]

"In what limits varies the directional derivative $f'_v (A)$ depending on the direction of departure v from A"

[potent lotus]

Merineth

[austere moth]

It's written in very poor swedish so a little hard to translate it

[clever loom]

Are you asking for minimum and maximum values of f_v'(A)?

[austere moth]

I honestly have no idea, it's written in very poor swedish to begin with so it's quite unclear

I could ask my teacher for more clarification

[clever loom]

Please do so. There is no point attempting a question which is unclear

[austere moth]

either way, do i need to handle the triangle differently?

Cant remember what it's called

delta?

gradient?

[potent lotus]

themadchessplayer

[clever loom]

Otherwise part (c) does not make sense

[clever loom]

Yes, it is gradient. The symbol is called nabla

[clever loom]

Nah

[austere moth]

Oh ok! And gradient is a derivative vector?

[clever loom]

Transpose of derivative (Jacobian) matrix

[austere moth]

"The gradient of a line is the measure of the steepness of a straight line." seems right?

But i can't just inpout a,b,c into the fornula right?

I have to do something with the gradient?

[clever loom]

Since we are not screwing around with weird bases of R^3.

[austere moth]

$\nabla F(1,1,-4) \cdot (x-1, y-1, z+4) = 0$

[potent lotus]

Merineth

[clever loom]

We can. Just calculate $\nabla f$ and plug (a,b,c) in it. The specific values ensure that the point is actually on the surface $f = 15$

[potent lotus]

themadchessplayer

[austere moth]

the gradient of that point will give a value, right?

[clever loom]

No!

Gradient is a vector

[austere moth]

Oh, i thought gradient was to calculate the slopes incline / decline

Is this the one i should use ?

[clever loom]

Yes

You might have confused with divergence, but that has nothing to do with slope

[austere moth]

Hmm yeah i might have to do some repetition on it

Either way

$\nabla f(x) =$

[potent lotus]

Merineth

[austere moth]

x is in bold, does that mean it's a vector?

I don't have a vector, i have a point

Are you sure i can use that formula to calculate the gradient?

[rare anchor]

hi can any body help with a question for solving a differential equation?

[austere moth]

!occupied

[calm coral]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[rare anchor]

oh

[austere moth]

@upper sparrow

[clever loom]

Don't worry. Boldface is often used for points in R^n.

[austere moth]

So i want to use this part?

[clever loom]

Yes

[austere moth]

Why is f lowercase here but uppercase on the formula for calculating the tangentplane?

[clever loom]

That's the third or fourth time you asked this

[clever loom]

Different places have different notations

For functions

[austere moth]

$\nablda f(1,1,-4) = (\frac{\partial f}{\partial x} (1,1,-4), ... )^T$

[potent lotus]

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

[calm coral]

@austere moth Has your question been resolved?

[clever loom]

Yes do this calculation

[austere moth]

Got a little trouble with the derivatives

$\frac{\partial f}{\partial x} = 14+x^3y^4e^{x(z+4)}$

[potent lotus]

Merineth

[austere moth]

14 is a constant so it disappears

I think i have to use product rule here to derive x^3 and e^(x(z+4))

However i'm not sure what happens with y^4 when i try to apply product rule

$x^3y^4e^{x(z+4)}$

[potent lotus]

Merineth

[austere moth]

Specifically this part

$y^4(x^3 * e^{x(z+4)})$

[potent lotus]

Merineth

[austere moth]

I guess i can apply the product role within the

$\frac{\partial f}{\partial x} = 14+x^3y^4e^{x(z+4)}$

[potent lotus]

Merineth

[austere moth]

<@&286206848099549185> Can someone please help me with deriving this expression with respect to x

I understand that the constant 14 dissapears

and that i can apply the product rule to

$y^4(x^3e^{x(z+4)})$

But i just can't seem to get the right answer...

[potent lotus]

Merineth

[timid oracle]

partial derivative of this wrt to x?

[austere moth]

Yea

[timid oracle]

yeah just use product rule

treat z+4 as a constant

[austere moth]

$y^4(3x^2e^{x(z+4)} + x^3(z+4)e^{x(z+4)})$

[potent lotus]

Merineth

[timid oracle]

Yeah i think

[austere moth]

I'm not getting the same on wolf or derivator-calculator.net

,wolf derive with respect to x (14+x^3y^4e^(x(z+4)))

[potent lotus]

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Requested by merineth

[timid oracle]

thats same

[austere moth]

I tried simplifying but i'm not getting the same afaik?

[timid oracle]

pretty sure it is

[austere moth]

glad to hear

and when i do with respect to y then it's literally the same?

except 4y^3

,w derive with respect to y (14+x^3y^4e^{x(z+4))

[potent lotus]

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Requested by merineth

[austere moth]

,w derive with respect to z (14+x^3y^4e^{x(z+4))

[potent lotus]

Results provided by WolframAlpha

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Requested by merineth

[austere moth]

uh why does x^3 become x^4?

nvm

[austere moth]

I did the calculation and i think i got the right answer

Does it seem right?

I never utilized 15 tho?

So i might have done it wrong?

[austere moth]

<@&286206848099549185> :(

[calm coral]

@austere moth Has your question been resolved?

[austere moth]

.close

.close

[calm coral]

Channel closed

Closed by @austere moth

Use .reopen if this was a mistake.

[calm coral]

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[silent sinew]

hello, i am trying to prove {0,1} is an albien group under ⨁.

[modern peak]

you mean

abelian?

[ancient helm]

do you mean $\mathbb{Z}_2$?

[potent lotus]

chebyshev's infinite pee norm

[silent sinew]

yesh

[ancient helm]

because that's defintely not closed

[silent sinew]

maybe?

[modern peak]

fyi {0, 1} isn't really a group

[silent sinew]

uh oh

[modern peak]

you need to define the operation ;-;

it could be addition, multiplication, anything...

[silent sinew]

i just got into discrete srry

[modern peak]

we can't really figure out what u mean

[silent sinew]

this is what the connective means

[modern peak]

same i started grp theory like a week and a half ago

[modern peak]

...

[ancient helm]

it's not the same

[silent sinew]

its logic

[ancient helm]

unless

1 = true

and 0 = false

[silent sinew]

yes its logic

[ancient helm]

ok

[silent sinew]

groups in logic

[ancient helm]

it's just xor

[silent sinew]

so for set A ={1,0} is {A, ⨁} a abelian group

[silent sinew]

yes i think so

[ancient helm]

it is

[silent sinew]

ok

so for abelian group

[ancient helm]

what have you done so far

[silent sinew]

give me a sec breh

we need it to be closed

which it is because its always either gonna be 1 or 0

associative

which idk how to prove but maybe if i do truth tables for theta ⨁ phi and then phi ⨁ theta and they equal thats enough proofg

identity i really dont know how to do

inverse i really really dont how to do

[ancient helm]

i think i would just write out the multiplication table if i were you

[silent sinew]

multiplicatiom table?

or truth table

[ancient helm]

yeah truth table

[silent sinew]

yeh done that

[ancient helm]

show

[calm coral]

@silent sinew Has your question been resolved?

[silent sinew]

srry am back

one sec for phi ⨁ theta trusth table

phi ⨁ theta is same truth table

so it is commutative

composition table is wrong

is diagonal

but anyway

what is the identity element

@ancient helm come back 😢

[ancient helm]

,rotate

[ancient helm]

please fix the tables on the right but yes

[ancient helm]

you tell me

there's only two options, true or false

[silent sinew]

yeh

identity means that for any element in the group

if it undergoes the operation then it stays the same

oh it’s 0

just by looking at the truth tables

[ancient helm]

if it undergoes the operation then it stays the same

if it undergoes the operation with the identity then it stays the same

[silent sinew]

yeh srry

row 2 and 4 it goes unchanged

so 0 is the identity

[ancient helm]

yes identity is 0

[silent sinew]

okay nice

now inverse

for inverse when it goes through the operation it goes to the identity

and we know the identity is 0

first and last row

and it is the same

so it’s inverse is itself?