help-13

[cedar kiln]

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[rare nova]

I am not really working on something, I just don't understand this equation.

[rare nova]

the square root of (the coefficient ratio of specific heats × the pressure of the gas / the density of the medium).

I don't know what that means

I know it's for calculating the speed of sound

But I don't understand coefficients or square roots

From what I found, the coefficient ratio of specific heats is Specific Heat Capacity

Which i also don't understand

This is apparently the formula for heat capacity ratio

I am all over confused

Is this the right place to ask this question? Because it is some complex mathematics, so I came here first.

[dire geode]

Where's the expression that uses square root

[rare nova]

Let me find that

I can’t find a visible expression of it

You know what, I think I’m just trying to understand something when I don’t even understand what I need to understand it. I’ll be closing this channel and doing something else about my problem.

.close

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[merry nacelle]

Would you say this is the right solution x,y,z = 0

For the equation on the picture in the gradient

is the way I solved it correct or am I just doing magic is what I’m asking I know the answer is 0,0,0

[cedar kiln]

@merry nacelle Has your question been resolved?

[cedar kiln]

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[upper quartz]

I have no idea what to d9

.close

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[twilit cliff]

i am having problems with this question because i dont think we have ever done anything with anything multiplied by theta and the notes had nothing to do with this

[twilit cliff]

sorry for cutting you off kami

[clear umbra]

double angle identities

[twilit cliff]

I have tried turning cos(2theta - 30degrees) into cos2thetacos30 + sin2thetasin30 but that didnt work

we arent onto those yet

hold on lemme look just incase i forgot to put those in my little formula sheet thing

[clear umbra]

or you could try complementary angles?

sin theta = cos (90 - theta)

[twilit cliff]

havent seen those yet

maybe i could

but i just end up turning it into like a slightly more complicated version of what i have now? or am i entirely missing something

[clear umbra]

i mean the standard way to solve for this is to use the sum and difference identities then double angle

[twilit cliff]

well the problem is we havent learned double angle identities yet

although i can try them

i am not sure

i can also just straight up tank the 2 missed questions out of 20 but thats not math of me

like literally theres this compared to the next question being literally tan(pi/theta)

[clear umbra]

wait i dont think the double angles help much

so the only thing i left i could think of is just using the complementary angles

[twilit cliff]

yea i might

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[pine crescent]

I need help

am i doing it right?

[upper abyss]

Forgot to take the integral of kx

Expanding (x + 2)² is a strange choice, and it only makes more work

But yes, the idea is correct. You want to find the k that makes this integral 1

[pine crescent]

wait wdym iI added it

[upper abyss]

Whoops no you're right

[clear umbra]

4th line is unnecessary tbh

[pine crescent]

so just solve now?

[upper abyss]

And grab k!

[pine crescent]

aight sounds good

ty

i kinda need help isolating 😭

[upper abyss]

Factor out k

[pine crescent]

ahh

i see

like this?

[civic eagle]

what software is that

[pine crescent]

its google docs

is this correct?

[cedar kiln]

@pine crescent Has your question been resolved?

[digital cliff]

.close

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[turbid mesa]

Try asking in #book-recommendations

[cedar kiln]

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[fervent charm]

Question If we use a p value cut off of 10% do we reject? Fail to reject? Or are we unable to tell using our confidence interval?

[fervent charm]

Help?

[cedar kiln]

@fervent charm Has your question been resolved?

[cedar kiln]

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[somber zenith]

2 col proofs

I need someone to teach me this badly

I know everything on my geo test expect 2 colum proofs

(geometry)

What mainly confuses me with this certain problem is
I dont know what comes after in the 1st row so
Therefore, I cant solve the 2nd colum which results to me failing this problem

[cedar kiln]

@somber zenith Has your question been resolved?

[fervent charm]

No

[cedar kiln]

@somber zenith Has your question been resolved?

[cedar kiln]

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[old fjord]

how do i work this out?

i tried using bodmas

got the wrong answer

(index laws)

[crimson sedge]

remember [
a^m\cdot a^n = a^{m+n}
]

[old fjord]

OMG!!!!!!!.!!!!!

ily

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[obtuse sierra]

general question, need help with studying

[obtuse sierra]

have midterms tomorrow, doing computer engineering

i have to study these, by tomorrow

how would i go about it?

[proven current]

thats kinda hard to answer without knowing how comfortable you are with the material

[obtuse sierra]

.close

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[snow leaf]

find the critical points of
f(x) = x^4 - 6x^2 + 8x

[snow leaf]

I have found the first derivative

and used the first derivative test

4x^3 - 12x + 8 = 0

x^3 - 3x + 2 = 0

x^3 - 3x = -2

x(x^2 - 3) = -2

but then

i get to x^2 - 3 = 02

-2

and its

x^2 = 1

but then the x's in the answer are only 1 and -2

[undone epoch]

Try some values.

[snow leaf]

?

[undone epoch]

Substitute small numbers to see if the equation works.

[snow leaf]

oh shoot

so i gotta

subsistute them after

[crimson sedge]

why dont you use x ≈−1.39?

[snow leaf]

idk

[summer lintel]

yo

[snow leaf]

how come i have to subsitute them afterwards though

[undone epoch]

Not after, you just try out a value to see if it works.

[snow leaf]

oh

does that mean i cant solve the way i did

[undone epoch]

It wouldn't be that easy to solve this equation.

[snow leaf]

oh ok

tysm

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[cedar kiln]

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[undone epoch]

After finding a value that works, use long division.

[snow leaf]

shoot

.reopen

[cedar kiln]

✅

[snow leaf]

how do i do this one

i used

first derivative test

and then reached

2x = 0

and then i got (0,0) as the relative minimum

i dont understand how they got

(2,4) as the relative max

do you just plug in the values from [-1,2]

into the original function

[undone epoch]

If the function didn't got restricted by interval, then the function would be as large as it could be towards -inf and inf. And those wouldn't be defined.

But when a function is restricted by interval, then the end values are included.

[snow leaf]

but then

i donget ti

dont get it

since

when i was finding the critical points i only got 0

[undone epoch]

You do indeed is correct, and this does apply to this question too. But this question also want's the relative and absolute max, and thus you need to see if the function is restricted by interval, and it does.

[snow leaf]

what do i do if it is restricted

[undone epoch]

You just see if the end points are usable.

[snow leaf]

so i make a table

and then i just see

which points are

highest / lowest

[undone epoch]

That works, and so does drawing a graph to visualize.

[snow leaf]

how come (2,4) is a relative maximum though?

[undone epoch]

A relative max is when the function turns from increasing to decreasing. And the end point is a absolute max, that's the same as a relative max in regular function, and this then is the same points.

[snow leaf]

oh

tysm

[undone epoch]

I feel the explanation this time is pretty confusing

[snow leaf]

waittt

if a relative max is when a function goes from incerasing to decreasing

in this one

doesnt the function

not decrease

after (2,4)

[undone epoch]

https://www.youtube.com/watch?v=GaltHJF6WyI

I'm not sure if this helps. If other has a better explanation, sure do explain.

[snow leaf]

bro i am so confused

now i went tothe next questions

and theres 2 absolute maximums

but then i searched

and it said there can only be 1 absolute maximum

nvm

all my classmates said my teacher made mistakes

thx

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[cedar kiln]

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[cedar kiln]

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[worn geode]

What does it mean for the system to be 'consistent' here?

[tropic oxide]

we say a system of linear equations is consistent if it has at least one solution.

[worn geode]

Ok and how would I figure out if it is consistent or not in the state is presented? Obviously if it has something like 0 = 8 we know it is impossible (inconsistent?), but here or in other similar situations is it that obvious that it is possible?

[tropic oxide]

well here they'vee straight up solved it.

[worn geode]

You are right, idk wtf I was thinking my brain just disconnected I guess

Thank you

.close

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[fallen hemlock]

i plugged into things and it told me x was 0 but i dont know all the steps

[peak minnow]

id start by taking LCM or LCD to subtract the fractions together

[fallen hemlock]

i multiplied both fractions by the others denominator

ill senda picture of my working

[peak minnow]

,rccw

[peak minnow]

shift denominator to right side

to get rid of sqrt , take square on both side and solve

but it might be tough cuz so many terms

[fallen hemlock]

yea thats the step idk how to do

[peak minnow]

actually nvm theres only 2 terms and use the (a-b)^2 formula

[fallen hemlock]

put brackets around the enitre thing and square it

oh

[fallen hemlock]

is that difference of 2 squares

(a-b)(a-b)

also how did you spot that thers so many terms

wait i think i can factorise some of it

[peak minnow]

the questions gonna be lengthy

[fallen hemlock]

exam marke rbe like yeah im not reading all that shit

[peak minnow]

xd, they would probably read the answer only

also i think u get atleast 4 solutions for this question

[fallen hemlock]

apparently the answer is 0 so i guess 3 of those just don work because of fraction undefined

[peak minnow]

maybe

[fallen hemlock]

i checked the answer on symbolab and it goes up to the super long thing with the denom shited to the right and then it just says its "its 0 bro idk what to tell you"

ig i just pray this isnt in my exam tomorrow lol thats gonna eat up so much time

thanks for ya help

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[cedar kiln]

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[cedar kiln]

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[woeful dagger]

I have question about complex numbers.

[tropic oxide]

!da2a

[cedar kiln]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[tropic oxide]

you should ask your question directly @woeful dagger

[woeful dagger]

So i have exercise
Re( Zcon * Z1 ) = 1 and
Im (Z/Z1) = - 3 /5

and i have a problem later on

[tropic oxide]

picture?

!original

[cedar kiln]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[woeful dagger]

okay 1 sec

my handrwirtnig sucks tho

so here it is

this is what i get for first one

And now im stuck in the second one when i wnat to exchange x or y

The i confuses me here alot

Like if i want tto do Y=1-2x

i have 2yi
do i do 2(1-2y)*i then its 2(i-2xi) -xi / 3
2i-4xi-xi/3
2i-5xi/3 and then i dont know what to do

[tropic oxide]

ok fuck some of these are rotated

[woeful dagger]

yeah wait a sec

they arent in order tho but yeah

[tropic oxide]

also the separation between problem statement and work is just gone

[woeful dagger]

what do you mean

[tropic oxide]

i don't know where the problem data ends and your work begins

[woeful dagger]

So basically

Here i got the Re part needed

here i got the Im part

and now i need to find x and y bu ti dont know how

i got this, but i dont know how to do it after that

I dont know how to deal with i's here

<@&286206848099549185>

ping me if answer

[tropic oxide]

fuck sorry i am busy in class

[woeful dagger]

no worries all good

[cedar kiln]

@woeful dagger Has your question been resolved?

[cedar kiln]

@woeful dagger Has your question been resolved?

[cedar kiln]

@woeful dagger Has your question been resolved?

[woeful dagger]

<@&286206848099549185>

[cedar kiln]

@woeful dagger Has your question been resolved?

[cedar kiln]

@woeful dagger Has your question been resolved?

[woeful dagger]

.close

[cedar kiln]

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[cedar kiln]

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[glacial moth]

Is this accurate?

[glacial moth]

Won't the sum end up being negative which makes no sense?

[upper abyss]

That works for -1 < a < 1

[glacial moth]

Okay

[upper abyss]

We call it a geometric series

[glacial moth]

That wasn't mentioned. Good to know

Thanks @upper abyss

[upper abyss]

Np. Feel free to ask if you have anything else

[glacial moth]

is there a way to find the nth factorial without having to calculate the n-1th factorial @upper abyss

[vast pike]

stirling formul to find its grown

if you want the exact value the answer is no

[glacial moth]

what even is the "grown"

[upper abyss]

As mentioned, Stirling's Approximation is bae, and can be an easy way to get approximations of larger factorials

But no, multiplying them out is the only way to calculate factorials exactly

Do you have a series with a factorial in it?

[glacial moth]

No

that sounds miserable though

like series from i=0 to n i!

How do we get this?

I know this

But idk how they adapted it for n^2

is it as simple as changing n to n^whateverPower ?

@upper abyss

[upper abyss]

That formula works for any n, including n², cos(n), etc.

[glacial moth]

gotcha

was about to say I can sub n for n^2 and pretend they are different names

[cedar kiln]

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[cedar kiln]

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[royal loom]

I'm having trouble understanding this proof, specifically at the step where they 'use the 1D mean value THM in both coordinates....'

I’m not seeing where the MVT is applied here. Could anyone help explain?

[cedar kiln]

@royal loom Has your question been resolved?

[royal loom]

@hollow osprey ?

<@&286206848099549185>

[hollow osprey]

Damn he didn’t write it out

It’s basically defining g(x) = f(x, a_2) then using the MVT on g to get some bounds on it

I think you can even do equality instead on that third line

@royal loom

[royal loom]

I see what you said. I’ll have to think about it

[cerulean sail]

If you e.g. defined $g(x) = f(x, a_2 + h_2)$ and applied MVT on $[a_1, a_1 + h_1]$ then there’s some $c_1$ between those points st $\frac{ g(a_1 + h_1) - g(a_1)}{h_1} = g(c_1)$
(had this mostly written out lol)

[wraith dagger]

@cerulean sail

[cedar kiln]

@royal loom Has your question been resolved?

[royal loom]

do you mean g'(c1)

I think so

but either way, I still don't see how that gives us the next step

@hollow osprey Could you maybe elaborate some more ? I don't understand still

[cerulean sail]

Yeah mb, and rather the c1 should be replaced with a1 + c1 too

[royal loom]

$g(x) = f(x, a_2 + h_2)$ and applied MVT on $[a_1, a_1 + h_1]$ then there’s some $c_1$ between those points st $\frac{ g(a_1 + h_1) - g(a_1)}{h_1} = g'(a_1+c_1)$

[wraith dagger]

Austin

[royal loom]

like so?

[cerulean sail]

yea but basically the c1 is positive and less than h1, so that you get a1 + c1 between those points (rather than c1 itself, me )

Of course then as $h_1$ goes to zero, your $c_1$ goes to zero, as you had $0 < c_1 < h_1$

[wraith dagger]

@cerulean sail

[cerulean sail]

@royal loom

[royal loom]

oh sorry I'm back

sorry working on multiple things here

Okay so I understand that
$$f(a+h)-f(a)=$$
$$=f(a_1+h_1,a_2+h_2)-f(a_1,a_2+h_2)+f(a_1,a_2+h_2)-f(a_1,a_2)$$

[wraith dagger]

Austin

[royal loom]

then we want to apply the MVT to this

so like just going one step at a time

let's do the one that looks likely easier

applying it to

[wraith dagger]

Austin

Austin

[royal loom]

Sure

understood

but then what?

this still doesn't look anything like what the proof just skips to next

atleast not to me

[cerulean sail]

If it's for this one here, the $f(a_1, a_2 + h_2) - f(a_1, a_2)$ it's kinda similar in that notice how the $a_1$ is fixed between both terms, and yada yada apply MVT on $[a_2, a_2 + h_2]$ on similarly defined $g(x) = f(a_1, x)$

[wraith dagger]

@cerulean sail

[cerulean sail]

Gonna relabel your $c$ as $c^$ for a moment, multiply out by the $h_1$ and you get the equality $g(a_1 + h_1) - g(a_1) = h_1 g'(c^)$

[royal loom]

chartbit

I appreciate your help a lot

[wraith dagger]

@cerulean sail

[royal loom]

but I'm starting to feel like this is above my head

I'm not sure I'll understand

so I don't want to waste your time

because I'm a bit lost already

I can ask my professor on Monday about this I think might be best to do in person

if that's fine

[cerulean sail]

Sure if that would work better it's really not too bad overall but whatever works best for you

[royal loom]

I'd be willing to try one more time if you want?

I'm just getting frustrated with this proof

and I don't want to take up your time if I'm not going to get it

[cerulean sail]

sure sure

[royal loom]

Okay

So

we want to show in 2 dimensions that if all the partials of f exist and are continuous in a neighboorhood of a, then f is differentiable at a

[wraith dagger]

Austin

Austin

Austin

[royal loom]

and then we go to apply the MVT to the two separate pieces

Which piece do you want to do first?

[cerulean sail]

We can do whichever, they're basically the same idea for both

[wraith dagger]

Austin

[royal loom]

we could simplify by subbing b=a2+h2 and ~~d=a1 ~~if that might help. Or what now?

[cerulean sail]

Yep, for ease we can write $b = a_2 + h_2$ (noting that it's just a constant) so then we have
[
f(a_1 + h_1, b) - f(a_1, b)
]

[wraith dagger]

@cerulean sail

[cerulean sail]

Of course spying that the b is the same between them, but the first coordinate changes

[royal loom]

Right

if we make g(x)=f(x,b)
then g(a1+h)=f(a1+h1,b)

so we can do that

make g(x)=f(x,b)

then we have

g(a1+h1)-g(a1)

by the MVT there exists a C in (a1, a1+h1) such that g'(C)*(h1)=g(a1+h1)-g(a1)=f(a1+h1,b)-f(a1,b)

so we can make that replacement. But g'(C) what does this represent?
f'(C,b)

so we can replace in the sum

with

f'(C, a_2+h_2) * h1

for that term

and then similarly for the other make a

f'(a1, D) * h2

replacement?

just guessing, I can repeat the process if it worked for the other one

[cerulean sail]

last bit is - instead of +

[cerulean sail]

yep it's pretty much the same thing, also a tiny point too

[cerulean sail]

We have our $C\in (a_1, a_1 + h_1)$, but basically in theirs they chose to note that because it's in that interval you could instead write $C = a_1 + c_1$ for some $c_1\in(0, h_1)$

[wraith dagger]

@cerulean sail

[royal loom]

sure

[cerulean sail]

Either way, for the most part, that difference doesn't matter much tbh

[royal loom]

I think let's keep the big C

they skipped so many steps and subsitutions omg

no wonder I could not follow that

Okay so then

f(a+h)-f(a)=................ big thing

and big thing

we've shown

is

$$f(a+h)-f(a)=f'(C,a_2+h_2)\cdot h1 + f'(a1, D)\cdot h2$$

[wraith dagger]

Austin

[royal loom]

this is true right?

Okay

so we are missing

the gradient term

and divided by the quotient of h

What happens next?

[cerulean sail]

From that point, they basically put that into the thing here

[royal loom]

$$\frac{f'(C,a_2+h_2)\cdot h_1+f'(a_1, D)\cdot h_2 -\langle c, h \rangle}{||h||}$$

[wraith dagger]

Austin

[cerulean sail]

But then the $\ip{\vec{c}}{\vec{h}} = \partial_{1}f(a_1, a_2) h_1 +\partial_{2}f(a_1, a_2) h_2$

[wraith dagger]

@cerulean sail

[cerulean sail]

Oh also remember that as f is a function of two variables here

[royal loom]

$$\frac{f'(C,a_2+h_2)\cdot h_1 +f'(a_1, D)\cdot h_2 -\partial_1f(a_1, a_2)\cdot h_1+\partial_2 f(a_1,a_2)\cdot h_2}{||h||}$$

[wraith dagger]

Austin

[cerulean sail]

First term is $\partial_1 f(C, a_2 + h_2)$

[wraith dagger]

@cerulean sail

[royal loom]

$$\frac{\partial_1f(C,a_2+h_2)\cdot h_1 +\partial_2f(a_1, D)\cdot h_2 -\partial_1f(a_1, a_2)\cdot h_1+\partial_2 f(a_1,a_2)\cdot h_2}{|h|}$$

[cerulean sail]

Yep (norm of h for the denom)

[wraith dagger]

Austin

[royal loom]

yet

hmph

looks nice

but

[cerulean sail]

like terms

[royal loom]

$$\frac{h_1(\partial_1f(C,a_2+h_2)-\partial_1f(a_1,a_2))+h_2(\partial_{2}f(a_1,D)+\partial_2f(a_1,a_2)}{||h||}$$

[wraith dagger]

Austin

[cerulean sail]

yep, then what they did was just split the fractions into separate bits

[royal loom]

sure

[cerulean sail]

What they then did was say that because you know the partial derivatives all exist and are continuous, in particular the $\partial_1 f$ and $\partial_2 f$ are continuous

[wraith dagger]

@cerulean sail

[royal loom]

something to do with a limit here

yea

not seeing what the limit is though

lim of partial -> a point should equal partial at the point so we can make the top become 0

I think

[cerulean sail]

Yep the top can become 0 as a result of that

[royal loom]

but what is the actual limit we take to do that?

formally

[cerulean sail]

They then send $\norm{h} \to 0$

[wraith dagger]

@cerulean sail

[royal loom]

oh as norm h-> h1/h2 -> 0

[cerulean sail]

Which implies that you have to have $h_1 \to 0$ and $h_2 \to 0$ both

[wraith dagger]

@cerulean sail

[royal loom]

indeed

[cerulean sail]

Yep

[royal loom]

but

we have C and D still

as norm h to 0 we have partial1 (C,a2)- partial1 (a1,a2) and something similar for the other one

how do we know C -> a1?

because

C was in (a1, h1+a1)

as norm h -> 0 h1 -> 0 C-> a1

and likewise

OKAY

U r the best chartbit

now level with me here

the proof presented in my lecture notes

complete ass?

[cerulean sail]

Think they have a "you must struggle by yourself" philosophy (which can't lie I find annoying a lot of the times)

[royal loom]

struggle I did

but thanks to u I got it figured out

gonna write this all out formally now hopefully

ty very mcuh

[cerulean sail]

my pleasure, happy it's all good!

[royal loom]

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[dawn quest]

I'm having trouble with P1:a. I simply don't get how to solve it, do I solve it like a quadratic and apply log rules? Please help.

[carmine dock]

See e^2x as (e^x)^2

[cedar kiln]

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[lapis laurel]

what is the smallest triangle the tangent of 9-x^2 can draw with the coordinate axes

[remote zinc]

Smallest area?

[lapis laurel]

yes

and i was thinking can i create a function from this that has an input x then creates the tangent line at that given point on the function an outputs its zero point

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[random kelp]

Since the 3rd row and 3rd column have all zeros, does that mean my variable x3 would be a free variable?

[crimson sedge]

x3 is a free variable yes

[random kelp]

explain

[crimson sedge]

but only because it isnt leading

and wdym 3rd row and 3rd column have all zeros

[cerulean linden]

Both x3 and x4 are free variables actually

[cerulean linden]

^

[random kelp]

yeah true

[crimson sedge]

is this an augmented or coefficient matrix

[random kelp]

but im confused regarding the column filled with zeros

[random kelp]

augmented

[cerulean linden]

Ohhh, then ignore what I said

The column to the right represents what the linear equations are equal to. You’re dealing with a homogenous system

[crimson sedge]

^

[crimson sedge]

do you see how you identify free variables

x1 x2 both have leading entries

[cerulean linden]

Exactly. A variable is a free variable if it doesn’t have what we call a “pivot” when put in rref

Which is just a leading 1

[random kelp]

oh shit im just being dumb

[cerulean linden]

Nah, linear algebra can be a bit hard intuitively at first. Don’t beat yourself up :)

[random kelp]

hold up

i get

x1=-3x3

x2=-5x3

x3=x3

its telling me that the co-efficients for v1 v2 & v3 are exactly the opposite

[cerulean linden]

If you want to solve for whether v1, v2, and v3 are linearly independent, and if not, give a nontrivial solution to it, then I suggest putting the vectors in a matrix where the entries are put in separate columns

By that I mean a matrix with the following entries:
((-18, -7, 12),
(8, 10, -13),
(-14, 29, -29))
If the matrix row reduces to the identity matrix, then the vectors must be linearly independent. Otherwise, keep track of the elementary row operations you use to get a row of all zeros. The linear combination that leads to that all zero row is your answer

[random kelp]

i actually had the right solution the entire time 😭

but thanks

I got a question though

[cerulean linden]

Fs :) what’s up?

[random kelp]

for a matrix to span all of all R^n, does its RREF form have to be an identity matrix?

[cerulean linden]

Well not necessarily. A matrix can span R^n but have m rows such that m>n

And that certainly can’t row reduce to the identity matrix

[random kelp]

but this doesnt have a pivot in every column

so its not a basis

[cerulean linden]

Correct. If you have a set of vectors who’s cardinality is equal to the dimension of the vector space, then those vectors can only span the vector space if the matrix form of those vectors row reduces to the identity matrix

That way the matrix has to be n by n (given that the vector space is n-dimensional)

[cerulean linden]

A better justification for why W2 isn’t a basis is that it has a 0 vector in it. Any set of vectors with a 0 vector must be linearly dependent, because a non-trivial solution is just multiplying the 0 vector by any non-zero scalar

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

Could someone check my answers for some simple year 9 math, thanks.

[cedar kiln]

Please don't occupy multiple help channels.

[royal loom]

.close

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[potent vector]

Does this constitutes to a quadratic equation?

[patent hearth]

nope

do you know why?

[potent vector]

my intution would also say no, because there's still ln(a) in the second term

but I'm not sure how to solve it

which technique

[dusk finch]

It seems like rather uneasy thing to do

lemme check if its even possible

Wolfram didnt find any closed form solution

[potent vector]

what do u mean closed form solution

[dusk finch]

closed form means basically that it can be writen using all the symbols you know

If it doesnt have closed form, then you can just aproximate the solution

[potent vector]

So that means it's irrational? and the only way is to appoximate it

[dusk finch]

Yeah, its definitely irrational

but furthermore, it cannot be also written using logarithms, roots and all these things

its around 3.369

[potent vector]

Seems like it could be a mistake in the calculation then

this equation actually stems from this question

https://cdn.discordapp.com/attachments/1129615478579527741/1166790303223644282/20231025_135611.jpg?ex=654bc526&is=65395026&hm=8b27b6babf827c832bc57747853d73b276e4ad19d3d06a7d8643ca105b0548cd&

where u would rotate the area between f(x) and a around the x-axis

[dusk finch]

rotate the area between f(x) and a around the x-axis or around the a-axis?

[potent vector]

x-axis

[dusk finch]

Alright

So how did you set up your integral?

[potent vector]

I just figured I would subtract f(x) with a. so I would get g(x) = e^x - a

[dusk finch]

a-e^x might be better option

e^x - a would be negative

but it actually doesnt matter as much, since you will square it in the process

[potent vector]

oh okay

[dusk finch]

So what would you do then?

[potent vector]

so I would use the formula of integration around the x-as and have V= pi * integral (a-e^x)^2 dx

[dusk finch]

yep

thats right

[potent vector]

I tried subtitute, but that didnt really work, so I just workout the brackets and split them in multiple integrals

[dusk finch]

yeah, thats probably easier

[dusk finch]

Just one more thing, what are the bounds of that integral?

[potent vector]

yeah the bound go from 0 until the point where e^x and a cross and that point is ln(a)?

[dusk finch]

yep

so $V=\pi\int_{0}^{\ln\left(a\right)}\left(a-e^{x}\right)^{2}dx$

[potent vector]

right so

[wraith dagger]

EmilyIsAlwaysRight

[dusk finch]

how did you solve that integral?

or what solution did you get

hmm it actually seems like you got it right, does your textbook provide solution?

Or can you send photo of the original question? To make sure you didnt misinterpret something

[potent vector]

this was on an exam, so unfortunately I only have what I remember

How would one solve this if it doesnt have closed form?

[dusk finch]

If it doesnt have a closed form, then you cant solve it. You can only aproximate the solution

[potent vector]

what techniques would you use? taylor series?

[dusk finch]

for the aproximation?

probably newtons method, i dont know much aproximation techniques

taylor series would give you polynomial of a large degree which would be probably unsolvable as well

[potent vector]

oh okay

I must've misinterpret the question to be getting an unsolvable equation

[dusk finch]

Yeah probably

[potent vector]

appreciate the help

[dusk finch]

yw, if you dont have any further questions you can .close this

[potent vector]

thanks

.close

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[mighty hawk]

Hi, everyone. I'm currently learning fuzzy logic (fuzzy sets, specifically), and I'm encountering this question:

[mighty hawk]

I feel like my proof down there is not really much of a proof. I do know that (a) is true, but I don't know what to write. I would really appreciate some help, or at least a clue on what I should write in the proof

[cedar kiln]

@mighty hawk Has your question been resolved?

[mighty hawk]

<@&286206848099549185>

[cedar kiln]

@mighty hawk Has your question been resolved?

[mighty hawk]

<@&286206848099549185> anyone?

[mighty hawk]

<@&286206848099549185>

[queen ice]

Maybe write them using set builder notation, then use a union symbol instead of the +

I don’t know what it means to multiply two intervals

But whatever it means, you can probably solve it in a similar fashion

[mighty hawk]

Multiplying the two intervals lead to this:

[mighty hawk]

I'll try doing that, thanks

[ripe siren]

all you need to show is that if x\in A.B then x \in E.F
A.B:={a.b|a\in A, b\in B}
then x\in A.B, means x=a_1b_1 for some a_1\inA, b_1\in B
and a_1\in E, b_1\in F
thus a_1b_1\in E.F
hence x \in E.F

similarly do the rest

[cedar kiln]

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[crisp rover]

Hi everyone. I'm working with limits at Infinity. I'm stuck with the 5^x. I don't know if I need to apply Euler or distribute the limit to lim (5^x) - lim (1/x)

[opal basin]

Distribute the limit, both these limits exist so you are allowed to do it

[tropic oxide]

wdym "apply euler"

[opal basin]

probably write as e^ln(..)

[crisp rover]

yeah

but 5^x is an indeterminacy right?

[tropic oxide]

is it tho

[opal basin]

x goes to -infinity, notice the negative sign

[tropic oxide]

x goes to minus infinity

[crisp rover]

Ohh ok. It leaves to 0 - 0= 0 right?

[opal basin]

Yes

[crisp rover]

thanks!

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[long walrus]

∂f/∂x * ∂g/∂z * ∂j/∂s + ∂f/∂y * ∂h/∂z * ∂j/∂s + ∂f/∂y * ∂h/∂s + ∂f/∂z * ∂j/∂s + ∂f/∂s

is this correct?

[cedar kiln]

@long walrus Has your question been resolved?

[sullen saffron]

I get to start with:
w(s, t) = f(g(j(s), t), h(j(s), s, t), j(s), s)

[long walrus]

?

[long walrus]

this is what I got from that

[sullen saffron]

I start with z = j(s) and then put that into y(s, t) = h( j(s) , s , t) and x(s, t) = g( j(s), t). Hence for f(x, y, z, s) we obtain
w(s, t) = f(x, y, z, s) = f( g(j(s), t), h(j(s), s, t), j(s), s).

I tried to put spaces to avoid creating a massive jumble of letters.

[long walrus]

yeah I think I got that

[sullen saffron]

$\pdv{w}{s} = \pdv{f}{g} \left(\pdv{g}{s}\right) + \pdv{f}{h} \left( \pdv{h}{s} \right) + \pdv{f}{j} \dv{j}{s} + \pdv{f}{s}$.

[wraith dagger]

stabulo

[long walrus]

should you have ∂f/∂x * ∂g/∂z * ∂j/∂s in the first term

since g has z and then z has s

[sullen saffron]

w = f( g(j(s), t), h(j(s), s, t), j(s), s)

We want to write x in terms of f, g, h, or j in some way due to the problem wanting that.

x = g so you could write both but they want it in terms of the above.

[long walrus]

yeah I can change the notation but I don’t get why it’s just ∂f/∂g * ∂g/∂s

shouldn’t there be another thing in the product

[sullen saffron]

I see now.

The terms I put in big brackets expand themselves.

[long walrus]

so expanding them would give you ∂f/∂x * ∂g/∂z * ∂j/∂s + ∂f/∂y * ∂h/∂z * ∂j/∂s + ∂f/∂y * ∂h/∂s + ∂f/∂z * ∂j/∂s + ∂f/∂s right

[sullen saffron]

$\pdv{g}{s} = \pdv{g}{j} \dv{j}{s}$

[wraith dagger]

stabulo

[long walrus]

ignoring the letters

[sullen saffron]

I'll write it out in full and you can translate it I suppose to compare.

[long walrus]

oh yes please

[sullen saffron]

$\pdv{w}{s} = \pdv{f}{g} \pdv{g}{j} \pdv{j}{s} + \pdv{f}{h} \left( \pdv{h}{j} \dv{j}{s} + \pdv{h}{s}\right) + \pdv{f}{j} \dv{j}{s} + \pdv{f}{s}$.

Let me just do a bit of error checking though, etc.

[long walrus]

yeah I think that’s what I got

Ty!

[wraith dagger]

stabulo

[sullen saffron]

I think that's right.

[cedar kiln]

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[fervent arch]

hii

[fervent arch]

guys i have a question

if you have this function

Can you say that the derivative of this function is equal to the derivative of (x(x-2)(x-1))/(1-x) , -x(x-2) = -x^2 + 2x = -2x + 2 right?

[brisk dirge]

Yeah you can

[fervent arch]

but if you do that , now 1 belongs to the domain of the derivative isnt that an issue?

[brisk dirge]

It does not because that function would already be defined for x≠1

[fervent arch]

so imagine you have this function

And you want to study differenciability

( Other points besides one dont matter because for sure this function is differenciable there ) , what about in the point x = 1?

[brisk dirge]

the =1 case is already included in the other part of function

[fervent arch]

yeah you right

[brisk dirge]

so if f'(1) is equal on both, the function is differentiable

[fervent arch]

so one more question , if you do the derivative by definition would you get the same thing as -2x + 2?

[brisk dirge]

Yes

The denominator would still give you the factors to reach at -2x+2

You can try it later

[fervent arch]

ok ty

[cedar kiln]

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[split pier]

P and Q are subgroups of a group G and |P|, |Q| are relatively prime, prove that P∩Q = {e}

[opaque harbor]

consider the order of an element in both

[cedar kiln]

@split pier Has your question been resolved?

[split pier]

😭😭 thank you, oh damn it was this simple

.close

[cedar kiln]

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[crimson sedge]

Hi I need help !

[crimson sedge]

Limits

I have a question *

[lyric narwhal]

[crimson sedge]

How do you think when solving limits

[sleek condor]

wdym solving limits?

[crimson sedge]

Finding limits

solving limit problems

Idk how do you guys call that in America

[sleek condor]

im french lmao

i meant what kind of problem?

well it depends

if you are talking about an epsilon delta proof

or just plugging in value and factorizing

[crimson sedge]

@sleek condor factorizing it

idk how should I think of solving the problems

I can factorize it in many diffrenent ways

[sleek condor]

give an example

[crimson sedge]

Lim x -> -5 = x^2 +3x-10 / x^2 + 6x + 5

I can write that like x^2 +3x-10 / x^2 + 3x + 3x + 5

[crimson sedge]

and then x(x +3) -10 / x(x+3) + 3x + 5

and then -10 / 3x + 5

-10 / -10

= 1

But thats not correct

idk why

[sleek condor]

this doesnt cancel out

[crimson sedge]

why not

[sleek condor]

a+b/a+d =/= b/d

thats just not how fractions work

[crimson sedge]

oh

my algebra is weak

[cedar kiln]

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[verbal terrace]

Need help finding domains of the above problems

Im so confused

[cedar kiln]

@verbal terrace Has your question been resolved?

[verbal terrace]

<@&286206848099549185>

[cedar kiln]

@verbal terrace Has your question been resolved?

[cedar kiln]

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[peak salmon]

number 4

its in romanian but you can probably decipher it

but what I dont get is

the woman says (as u can see on the second from last line) that because x is a part of R

then then the bigger parantheses (x+3-1)

cannot be equal to 0

what is the logic to this

I personally dont understand how thats possible

[peak salmon]

determine the real number a for which a o x = a (the 'o' operation is described at the top of the image) for any real number x

<@&286206848099549185>

[limber zephyr]

can you clarify the question are you trying to find an a when x is real?

well first step would be to plug in the a and x

[peak salmon]

im trying to find an a so that a o x = a for any x, real number

im not askign about the steps since ik them

[limber zephyr]

oh ok well still plug in a and x

so then...

[peak salmon]

im asking about the question above

[peak salmon]

.

from here

[limber zephyr]

I dont really get it either

I believe it saying if the x value doesnt make that zero then we need to find something to make the other one zero so the whole thing works out

So basically you want the thing to equal zero no matter what x is, so if x+3-1 is not zero, then what would a have to be to make it zero

[peak salmon]

yeh a friend i messaged just got back to me

its pretty much what you just said

[limber zephyr]

ok

[peak salmon]

since x is a variable therell be cases where it isnt 0

and since a is constant

[limber zephyr]

right

[peak salmon]

then its only logical that its that

thanks

[limber zephyr]

np

[peak salmon]

.close

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[cedar kiln]

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[scenic vale]

I know that x_n = ((x_n)(x_n - 1)+1) and x0 = 2, x1 = 3, but is there any way to simplify more?

I was thinking something with factorials because the since xn is defined by 1/(x_(n-1))(x_(n-2))...(x_2)(x_1)(x_0) + 1

But I got stuck there

[opaque harbor]

I would try find an induction

[scenic vale]

How does that help though

I dont even know what I would put for my inductive hypothesis

[opaque harbor]

you're not being asked to find every solution, and there's obviously a solution when n = 1

[crimson sedge]

Yeah I was gonna say just take 1

[scenic vale]

What do you think I should do for the hypothesis then?

I know base case is easy

but then what

[opaque harbor]

play around with the n+1 case until you get something helpful

[scenic vale]

well yeah thats what im saying I got stuck there

[opaque harbor]

I've not managed to find something myself yet

[scenic vale]

alright

[crimson sedge]

I’ve not tried to get something and wish you good luck as I m falling asleep lmao

[scenic vale]

alright thanks

[vast pike]

cute do you want help ^

?

[scenic vale]

yes please

[vast pike]

mh...

[scenic vale]

alright

[vast pike]

for n=1 you can use a=2

[scenic vale]

what do you mean by a

[vast pike]

a_1=a

[opaque harbor]

I was wrong base case

oh

yeah

[scenic vale]

yeah I know

[opaque harbor]

x1 = 2

[scenic vale]

I know

x1 = 2, x2 = 3, x3 = 7, x4 = 43, x5 = 1806

or sorry x5 = 1807

the pattern is x_n is defined by (x_(n-1))(x_(n-1) - 1) + 1

[opaque harbor]

oh fml

I'm going to bed

[scenic vale]

alright thanks for trying

[opaque harbor]

I missed something

[opaque harbor]

aka this question

I thought you needed help finding xn at all

[scenic vale]

nah

for anyone else I think there is something with factorials because xn also equals (x_n-1)...(x_2)(x_1)

but of course the xns dont decrease by 1 so I dont know how you would define it with a factorial so maybe im wrong but it looks a lot like a factorial

<@&286206848099549185>

[vast pike]

very difficutl

good nighe @opaque harbor

[scenic vale]

Yeah I took a break but i feel like there are too many patterns for there not to be a way to simplify

[vast pike]

however its very simmetric

[scenic vale]

Yeah

I guess you could rewrite as ((x1 + x2 + ... + xn) + ((x1)(x2)...(xn))^2)/((x1)(x2)...(xn)) but I dont see anything to do from there

[cedar kiln]

@scenic vale Has your question been resolved?

[vast pike]

I think $(x_n)$ follows $x_n=x_1\cdot\ldots\cdot x_{n-1}+1$

[wraith dagger]

everg

[vast pike]

$x_1=2$

[wraith dagger]

everg

[scenic vale]

Yeah it does I wrote that already

I was thinking though that maybe there is a way to have everything in terms of x1

Idont know maybe theres not

thanks for trying im probably just gonna give up

.close

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[vast pike]

you said x_n=1/...

[cedar kiln]

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[scenic vale]

sorry replied to wrong message it was here

[scenic vale]

.close

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[scenic vale]

its the same thing just simplified more

because x_n-2 is equal to (x_(n-1 )- 1)

[vast pike]

yeah

in terms of $x_1$ you can write $x_n=(((x_1+1)x_1+1)x_1+1)x_1+1$

[wraith dagger]

everg

[vast pike]

where you put as many x_1 as it is n-1

[scenic vale]

No its more exponential

[vast pike]

in this case n=5

try it

[scenic vale]

Doesnt work

it should be 1807

it expands exponentially I think thats what you missed

$x_2 = x_1 +1$

[cedar kiln]

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[wraith dagger]

CuteAnimalLover52

[scenic vale]

and $x_3 = (x^2)_1 + x_1 +1$

[scenic vale]

.close

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[scenic vale]

and $x_3 = (x_1^2) + x_1 +1$

[wraith dagger]

CuteAnimalLover52

[scenic vale]

$x_4 = x_1^4 + 2x_1^3 + 2x_1^2 + x_1 +1$

[wraith dagger]

CuteAnimalLover52

[scenic vale]

so its exponential not linearly increasing x_1 +1 terms

[vast pike]

okk

[cedar kiln]

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[gleaming glade]

<@&286206848099549185>

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[gleaming glade]

<@&286206848099549185>

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[gleaming glade]

<@&286206848099549185>

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[neat dune]

.

[gleaming glade]

pls help woth both of them

[neat dune]

10 more mins u spammed the helpers role too much for me :p

[plucky owl]

Can you stop abusing the ping? You pinged early and spammed it 3 times in a row

[gleaming glade]

just help

[cedar kiln]

Please wait patiently, and do not interrupt other channels with your question. Helpers in this server are volunteers, and the server cannot guarantee that someone will be able to help you. By being impatient or begging, you will only turn potential helpers away.

In the meantime, please make sure your channel contains the original question, clearly describes what you have already tried, and states exactly what you are having trouble with. This increases your chances of getting a good response.

[gleaming glade]

.close

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[undone star]

Find the locus of z = (w-i)/(w-2) where w is purely real

[undone star]

Then find z if w moves around the unit circle

[undone star]

I made w the subject and got the line y=1/2 - x/2

I’m not very sure how to interpret when w moves around unit circle

.close

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[swift bridge]

I just don't understand this in general

<@&286206848099549185>

Help??

[cedar kiln]

@swift bridge Has your question been resolved?

[swift bridge]

is anyone gonna help me?

[scenic vale]

Take the second derivative

[swift bridge]

can you explaine?

[scenic vale]

Do you know what derivatives are or na

[swift bridge]

yeah

[scenic vale]

Like how dy/dx is the slope of tangent line and d^2y/dx^2 is change in slope in the tangent line?

If you dont understand that I think it would be best to just watch a video explaining why

If you do understand that then I can try and explain how to apply it

[swift bridge]

I dont understand what a tangent line is

but try and explaine

if you can

[scenic vale]

Its best if you just look up what a derivative is, I think the most intuitive for this example definition is graphically, which is hard to show over text

Khan Academy always has good videos

But if you need to do it now, basically the if the second derivative is negative the graph is concave down, and if it is positive it is concave up

[swift bridge]

i know what a derivative is though

yes i need it done now haha

[scenic vale]

Well if you dont know what a tangent line is, you dont know what a derivative is fully

if the second derivative is negative the graph is concave down, and if it is positive it is concave up, and inflection points are when it is 0

do you know how to take second derivatives or na

[swift bridge]

I understand the tanget line is but i dont know its significance

if that makes sence

like how to find it

[scenic vale]

Think of a parabola for example

Does it have a constant slope or no

[swift bridge]

idk what parabola is but no?

😦

[scenic vale]

uhhh thats not good

if youre in calculus you really should know it

[limber zephyr]

it's the graph of a quadratic if that helps

[scenic vale]

Just take the second derivative and if it is negative the graph is concave down, positive concave up, 0 means inflection point if you need to get it done quick

[swift bridge]

i just dont know how to find them

[scenic vale]

but you really should look at all this stuff later

[swift bridge]

link me?

[scenic vale]

https://www.khanacademy.org/math/differential-calculus/dc-diff-intro/dc-diff-calc-intro/v/derivative-as-a-concept

Everything in that playlist will be helpful

[swift bridge]

okay

[scenic vale]

also you should look up and learn what the equation of common equations look like

like polynomials, exponentials, trig functions, etc

[swift bridge]

link me that too or already on khan?

[scenic vale]

memorize this

also look up the graphs for e^x and logx and memorize those too

and you should be fine for a while

[swift bridge]

so for my current question mind helping me with this so i can get it done?

[scenic vale]

Yeah ive said multiple times

Just take the second derivative and if it is negative the graph is concave down, positive concave up, 0 means inflection point if you need to get it done quick

[swift bridge]

okay

[swift bridge]

wdym by second derivative if you can simple explain?

please

[scenic vale]

so you know how to take derivatives right

so you take the derivative of the equation and you should get another equation

then take the derivative of that

and its second derivative