help-17

[lyric relic]

if you have some problem like this don t hesitate to write to me ...when i 'll have some times i 'll check them

[cedar burrow]

Thanks. May I ask how long you’ve been in Uni? I’m kinda curious, cause you know a lot for an undergraduate

[lyric relic]

i reality now i get bachelor last year

I’ve been at university for 4 years

[cedar burrow]

Interesting okay

[lyric relic]

but before the university i didn t do maths in the free time

[cedar burrow]

Do you stick out among your peers or are most at your level?

[cedar burrow]

That’s wild lol

[lyric relic]

they are almost like me ...but we are 20 or less... in the previous years we were 100 or more

[cedar burrow]

Okay yeah, I would imagine most that are not as smart don’t continue for as long

[lyric relic]

yes i didnt have that hobby before the university ... in fact when i have time i try to do some contest math

[cedar burrow]

How long have you been doing that for?

[lyric relic]

i would not talk about smartness ... i think its all about the passion that you feel when you do maths ...somebody doesn't feel that passion as i feel it

[cedar burrow]

I do think those two strongly correlate though. Someone who has a hard time with math probably doesn’t feel as passionate about it

Where are you from btw?

[lyric relic]

i think i see some contest math 4 hours a month

because i have no time for that

[cedar burrow]

Understandable

[lyric relic]

italy

[cedar burrow]

Ah nice

I’m from Germany, have you ever been there?

[lyric relic]

yes but i promised to me that i ll be a math olympiad tutor for high school ..maybe after the phd

[cedar burrow]

Your students would be very lucky

[lyric relic]

no

i didn't

are you doing math in uni ?

[cedar burrow]

No I’m still in high school

I’m trying to learn the Uni stuff on the side though cause it’s a lot more interesting

[lyric relic]

what will you study in uni?

[magic wasp]

I don't want to sound annoying but this is still @old niche 's help channel

[cedar burrow]

Pssssst

[cedar burrow]

I’m not sure yet. I think I’ll decide once I’ve seen some more stuff

[lyric relic]

ok ...if you study math i'll help you to learn some complex argument if you want (like ring theory, differential geometry, algebraic geometry ec ec)

[cedar burrow]

Thank you

Do you like set theory? It’s the topic I’m most interested in at the moment

[lyric relic]

thanks but it is also an opportunity for me to explain that complex thing to you and "to me"

[cedar burrow]

Ahhh got it yeah

[lyric relic]

so i think this could be a good project for both

[cedar burrow]

Possibly, though I will admit that geometry is the one thing about math that I don’t like 😅

[lyric relic]

no sorry xD .... i think its very fundamental math (like logic) ... I want to study some complex construction

[cedar burrow]

I think our interests are quite different then

[lyric relic]

you ll see what type of geometry it is ..is not about euclidian or cartesian geometry

when i start to studing math (4 years ago) i try to interst me to everything ...so at the start i studied set theory by my own ...and my interst changed with time

and more skilled you are more you can jump through different fields

[cedar burrow]

Yeah, I probably like set theory cause it’s so new to me. Different stages different interests

What do you want to do with your phd in the future?

[lyric relic]

i want to remain in the academia

[cedar burrow]

So teach the things you learned to others?

[lyric relic]

do research

[cedar burrow]

Oh okay

Within mathematics? So discovering new proofs and such?

[lyric relic]

yes absolutely

[cedar burrow]

That sounds very cool

[lyric relic]

this is very exiting for me

[cedar burrow]

Yeah it’s a lot of fun

But you need to be very good at it if you want to make a living with it

But I think that shouldn’t be any problem for you

[lyric relic]

because years ago it seems imposible that a guy could prove thing...now it is possible every time more

[lyric relic]

i hope so ... i will do my best

[cedar burrow]

Yeah it’s interesting to see how we learn to get better at proving things

The connections we form in our brain every time we see or create a proof

That allow us to see patterns in problems that let us prove them so much faster

[lyric relic]

now its late for me ... i enjoy chat with you ..goodnight

[cedar burrow]

Good night!

[lyric relic]

🙂

[cedar burrow]

@old niche are there still any open questions?

[vocal sleet]

@old niche Has your question been resolved?

[vocal sleet]

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[marble wharf]

ok

[small quarry]

?

Any math question?

[marble wharf]

yes

a is how much cuz a+b=18

[livid tapir]

need more info

[marble wharf]

ok

[outer warren]

best you could do here is isolate a
which will give you an expression for it in terms of b

[vocal sleet]

@marble wharf Has your question been resolved?

[vocal sleet]

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[wooden sphinx]

does dividing by 2 continous functions

[vocal sleet]

Please don't occupy multiple help channels.

[wooden sphinx]

get u a continous function as well?

like

sinx/cosy

no right

[calm light]

not necessarily

key example, 1/x

[wooden sphinx]

yea

[vocal sleet]

@wooden sphinx Has your question been resolved?

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[hollow prairie]

Can someone help with this question?

[hollow prairie]

<@&286206848099549185>

[vocal sleet]

@hollow prairie Has your question been resolved?

[hollow prairie]

<@&286206848099549185>

[mellow rampart]

Whats your base case

[vocal sleet]

@hollow prairie Has your question been resolved?

[hollow prairie]

I’ll send a pic of what I did

[hollow prairie]

Did I do it right?

[stiff arch]

Your assumption is misplaced

Threw me off

[hollow prairie]

I’m confused. Wdym?

[stiff arch]

N=28 to n=32 are the base cases

Induction step comes after the base case

Which your induction step is the paragraph you wrote

[hollow prairie]

This is how the prof did a similar problem

[stiff arch]

How peculiar

Ive never seen anyone put the base case inside the induction step

[hollow prairie]

I decided to follow her since she would be the expert, and most examples were done in the lecture

[stiff arch]

I mean, it doesnt change the argument, so it makes no difference

What exactly do you need help with here?

[hollow prairie]

Oh I was stuck at the beginning, but now I wanted to verify if I did it correctly

[stiff arch]

It looks good

[hollow prairie]

yea thanks for the clarification, I will ask her about it. Appreciate it

.close

[vocal sleet]

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[fallen sandal]

Idk if I made a mistake

[patent nymph]

,calc 5sqrt(14) 4sqrt(21)

[twin meteor]

Result:

342.92856398964

[patent nymph]

,calc 5sqrt(14) 4sqrt(21)/sqrt(3)

[twin meteor]

Result:

197.98989873223

[fallen sandal]

Wat

[patent nymph]

The 2 disappeared somewhere in the third line

[patent nymph]

You made a mistake

[fallen sandal]

Oh

[patent nymph]

That’s the only mistake you made

[fallen sandal]

Is

140z^2sqrt(6z) correct

@patent nymph

@patent nymph

[patent nymph]

,calc 5sqrt(14) 4sqrt(21)/sqrt(6)

[twin meteor]

Result:

140

[patent nymph]

Yes

[fallen sandal]

OK

.close

[vocal sleet]

Channel closed

Closed by @fallen sandal

Use .reopen if this was a mistake.

[fallen sandal]

.reopen

[vocal sleet]

✅

[fallen sandal]

Is this right

[patent nymph]

You can still simplify sqrt(28) even more

[fallen sandal]

Oh

Oh right

[patent nymph]

But it’s right

[fallen sandal]

Wat abt

4a^3sqrt(7)

Uhh

I might've messed up

Idk

[soft walrus]

you forgot something :)

| | are extremely important!

[fallen sandal]

Wat

[patent nymph]

the assumption is that a is positive, if it’s the same as the other questions they’ve been posting

[patent nymph]

This should be the answer

[soft walrus]

ahhh ok sorry didn't have that context

then yes all squared away captain

[fallen sandal]

Yay I got it right

This feels off

[patent nymph]

You’re right

42 doesn’t have any square factors

[fallen sandal]

Omg

YAY

Omg

[fallen sandal]

These things looks so cool

.close

[vocal sleet]

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[regal edge]

part b)

im not sure how i would approach this

for my equation of dy/dx i have (y-x^2)/(y^2-x)

in desmos the graph will look like this

[north hamlet]

yes

which part of 'b' did you not understand?

[regal edge]

how to find the points

[north hamlet]

yes

so first

how to find the point where tangent lines are horizontal

do you know implicit differentiation

[patent nymph]

Incidentally this is the name of the shape https://en.wikipedia.org/wiki/Folium_of_Descartes

[regal edge]

i did implicit differentiation to get the equation i wrote out

dy/dx = (y-x^2)/(y^2-x)

[north hamlet]

yes

now if the tangent lines are horizontal it means the slope is zero at that point

the slope is just dy/dx right

[regal edge]

correct

[north hamlet]

so

y= x^2

then dy/dx is zero

[regal edge]

how would you justify getting to y=x^2

[north hamlet]

because for the tangent line to be horizontal, dy/dx is zero

[regal edge]

since the slope can't be undefined we can ignore the denominator?

[north hamlet]

we can ignore it as long as x is not equal to y^2

[north hamlet]

since the denominator doesnt become zero for all other cases except that

[north hamlet]

the slope can be undefined, better to say infinity which gives a vertical tangent line but that is not what we want right

[regal edge]

correct

[north hamlet]

but what we need is the slope to be zero

for a horizontal one

solving for dy/dx= 0 we get the relation y=x^2

[regal edge]

hmm ok

[north hamlet]

see the graph intersects y=x^2 at a point

if you can imagine a tangent there, it will be horizontal

[regal edge]

yes

but that would be at the origin no?

[north hamlet]

see in the origin

if we zoom in

theres a vertical and horizontal line

[regal edge]

in terms of y=x^2 would it be correct to say that the horizontal tanget when x=2 y=4

[regal edge]

so the horizontal tangent would be at (2,4)

[north hamlet]

how did you get x=2 y = 4

[regal edge]

just pulling numbers in a hypothetical

sorry i didnt clarify

[north hamlet]

we need to solve x^3+y^3-3xy=0 and y=x^2

so we get a point right

that is the point of intersection of the blue graph and red graph

y=x^2 has many points and x^3+y^3-3xy has many but we need a point which satisfies both the equations

which means it lies on both of them

[regal edge]

yes

[north hamlet]

so we solve them

we cannot guess that it may be (2,4)

[regal edge]

OH

OH

MY

GOD

[north hamlet]

it lies on y=x^2 but not on other right

did you get it

[regal edge]

I SUB IN X^2 FOR Y IN THE ORIGINAL EQUATION

[north hamlet]

yeah

[regal edge]

YOU'RE A PSYCHO

[north hamlet]

will you be able to solve it?

[regal edge]

i will try

i guess ill keep this open

and @ when i need help

[north hamlet]

i dont think they asked you for the point right?

okay

also for the second part

at (0,0)

theres two lines right

[regal edge]

yes

vertical and horizontal

[north hamlet]

so we get two slopes

one 0 and one infinity

but we cannot define ONE and only one slope there

so since dy/dx gives one slope at a point

we cannot say dy/dx is zero or infinity

because theres two slopes the idea of dy/dx is wrong

so it does not exist

if you want a mathematical proof you can put x = 0 and y = 0 in the expression for dy/dx and you will get zero divided by zero

which is indeterminate or not defined

[regal edge]

@north hamlet

i have 0=-4x^3 + x^6

after subbing in y=x^2

but now i dont know what to do

[north hamlet]

yes

now

we can factor out the x^3 right

[regal edge]

yes

[north hamlet]

so we get

[regal edge]

x^3(-4+x^3)

[north hamlet]

wait you did a mistake

the original equation is x^3 +y^3-3xy=0

we put y=x^2

x^3 + (x^2)^3 -3x(x^2)=0

[regal edge]

yes

[north hamlet]

x^3 + x^6 -3x^3 =0

understood till now?

[regal edge]

yes

[north hamlet]

we can simplify further

x^6-2x^3=0

[regal edge]

ouhh

[north hamlet]

x^3 - 3x^3 is -2x^3

[regal edge]

mb i wrote -4

[north hamlet]

yeah

so now

x^3 ( x^3- 2) = 0

we get x=0 and x= cube root of 2

now

it is partially right since at (0,0) there is a horizontal line so at x=0 we have a horizontal tangent line

and also the other point is x = cube root 2

the green line is tangent at x = cube root 2

and since we know y = x^2 we can find y as we know x

[regal edge]

yes omg

im stuck again

when i sub x=3rdroot(2)

[north hamlet]

yes

[regal edge]

i get to 0= 2 + y^3 - 3(3rdroot(2))y

im not sure how to isolate for y now

[north hamlet]

why do we need that

we have a simpler equaiton

y = x^2

remember that

[regal edge]

OHHH

SHITTT

[north hamlet]

we can use that instead

a better path is easier

[regal edge]

so y = (3rdroot(2))^2

[north hamlet]

yes

the equations of tangent is easy

if you need it

[regal edge]

would i have to smplify that

[north hamlet]

if you want to then we gotta use a calculator

[regal edge]

ok ill just leave it

for simplicity sake

[north hamlet]

1.25992104989 , 1.58740105197

dont freak out

[regal edge]

ewwww

[north hamlet]

(1.26 , 1.59)

approximately

[regal edge]

im pretty sure i dont need the equation of the tangent right

[north hamlet]

yeah

okay

[regal edge]

but im curious

do you have it?

[north hamlet]

sure

its easy

so we need the equation of a horizontal line

that means the y coordinate is same right? at all points on the line

visualise it

[regal edge]

yes

[north hamlet]

yes so

we got the point

the general equation of a horizontal line is y = k

as k varies the line shifts up and down but remains horizontal

here k is the y coordinate of any point on the line

here we got the y as (cube root 2)^2

which is about 1.59

so the equation will be y = 1.59

[regal edge]

WOW

lol

[north hamlet]

easier to understand

[regal edge]

thank you so much

[north hamlet]

yeah no problem

[regal edge]

.close

[vocal sleet]

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[random sundial]

I think my brain just blew a fuse..

[random sundial]

Does this look reasonable?

Theta(log phi n)

<@&286206848099549185>

[onyx fog]

what?

[random sundial]

This is my question. I think that the answer may be Θ(logϕ
n) ,but I'm not sure.

[onyx fog]

lemme read

[random sundial]

Ok, Thank you

[patent nymph]

I think so

[random sundial]

I've been using Chegg and Numerade to get some help with Algorithms.. Are there any other sites that you'd recommend for help.

[vocal sleet]

@random sundial Has your question been resolved?

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[vocal sleet]

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[errant peak]

Can someone help me with this

@spice sune

<@&286206848099549185>

[proven garden]

!15mins

[vocal sleet]

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.

[dull geode]

can you give a more clear picture?

[vast shale]

how comfortable are you with elementary matrices

also can you take a proper pic of the elementary row operations

[errant peak]

Hii

[errant peak]

I think you did it?

[errant peak]

Pretty comfortable

[vast shale]

is that all? nothing else is missing?

[errant peak]

Nopee

[errant peak]

This is what i got?

[vast shale]

,rccw

[errant peak]

The next part of the question

I might have made a mistake

[vast shale]

yh but those are the operations, they want the elementary matrices that we will multiply to perfom those operations

for example if we wanted to R3 -> 3R1 + R2 subtract 3 times the first row you would have something like
$$\begin{pmatrix}
1 & 0 & 0\
-3 & 1 & 0\
0 & 0 & 1\
\end{pmatrix}$$

[vast shale]

have you seen these before?

[errant peak]

Owh

[errant peak]

I don't think so 😬

[vast shale]

best to look up a tutorial on them first

[errant peak]

Wait what does an Elementary matrix mean

[vast shale]

see the way you did R2-> R2-R1 up here, you can get multiply by an elementary matrix to get that not a definition ik

@errant peak https://www.youtube.com/watch?v=QVKj3LADCnA&

this is how i got introduced

[errant peak]

Oh so instead of the row operation we multiply by a matrix?

[vast shale]

yh

a slight variation of the identity matrix

[vast shale]

you can probably find a shorter vid idk but best to be comfortable with them first

[errant peak]

Ok sure lemme go through this then

[vocal sleet]

@errant peak Has your question been resolved?

[vocal sleet]

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[vast shale]

how many integer values of x exist such that $\frac{(x^3 - 4x^2 + 3x + 16)}{(x-16) }$is an integer

[twin meteor]

AgentSpiely

[vast shale]

i dont know if it is the most optimal method, but you can do polynomial division

Yeah I did exactly that

then write it in the form Q(x) + R(x)/(x-16)

Q(x) being your quotient

R(x) being your remainder

Q(x) obviously is always an integer for x integers

so you have to verify when R(x)/(x-16) is an integer

Indeed!

So I think my answer is partially right

i think for the R(x)/(x-16) you just have to check all the numbers taht divide R(x)?

,w factors of 3136

[twin meteor]

Results provided by WolframAlpha

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

Requested by lexqa

[vast shale]

oh i guess thats it lol

i think you can work out the rest from this on your own

So there are ||21|| values of x right

[vast shale]

i was thinking ||256|| lol

How you got there?

Or wait maybe its ||42 because you have negative values too||

Yes its ||42 || I think

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

yep u r right

.close

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[gloomy fulcrum]

I'm not even sure of where to begin, please help!

[vocal sleet]

@gloomy fulcrum Has your question been resolved?

[crimson jetty]

#real-complex-analysis probably better

[vocal sleet]

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[vocal sleet]

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[vocal sleet]

Please don't occupy multiple help channels.

[forest maple]

surely someone can help

[vocal sleet]

@forest maple Has your question been resolved?

[cedar burrow]

No square number ends on 2.

[vocal sleet]

@forest maple Has your question been resolved?

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[low igloo]

I'm converting series to sigma notation
I have the N already but I forgot how to get the constant

[low igloo]

the d = 4

.close

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[graceful nimbus]

I need help with this question

[graceful nimbus]

[vocal sleet]

@graceful nimbus Has your question been resolved?

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@graceful nimbus Has your question been resolved?

[graceful nimbus]

<@&286206848099549185>

[grizzled halo]

!status

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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.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

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[lost basalt]

May I plz have some help

[lost basalt]

I need to put the equations above into a a matrix

the matrix should follow Az=b

I understand what z should be but not A

so if i could have help constructing A that would be appreciated

.close

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[silent turret]

hmm

[silent turret]

Could I get a hint on how to do this?

It is probably really obvious, but I can't quite understand. $$\frac{e^{iz}}{e^{z} + e^{-z}}$$

[twin meteor]

Konglekongen

[silent turret]

thats the integral im supposed to calculate, from real x -infinity to infinity

then the real part of the e^iz will be our answer.

[dull bear]

How did you get this, may I ask?

[cyan crest]

I think he used the e^i definition of cos(x)

and the definition of cosh(x)

[silent turret]

e^ix = cosx + i sin(x)

maybe I made a mistake in deriving that expression

[dull bear]

I would think so

[cyan crest]

uh, if you use that identity then you need to somehow get rid of that isin(x)

$$\cos x = \dfrac {e^{i x} + e^{-i x} } 2$$

[twin meteor]

_Kookie

[dull bear]

Anyways, assuming the way they'd want you to do it is through CRT

(for which it may be easier to remember that $\cosh(iz) = \cos(z)$)

[twin meteor]

@dull bear

[silent turret]

hmm

What will the residues be on that domain?

[dull bear]

Well you have that you're integrating $\frac{\cos(z)}{\cosh(z)}$ basically, if that reminds you of anything...

[twin meteor]

@dull bear

[dull bear]

If they've given you anything for fractions of that form

[silent turret]

hmmm

argument principle?

[dull bear]

Might be called that it’s something like if f(z)/g(z) is such that g(z0)=0 and g’(z0) isn’t zero, then the residue of that at z0 is(…)

[silent turret]

its f(z0)/h'(z0)

right?

hmm

[dull bear]

Yep yep

[silent turret]

So, then we can get the integral of the rectangle, using residue theorem, computing the residues where cosh(z) is zero, using the expression above.

Then, hmm.

Usually they show that the sides of the rectangle go to zero, or something like that.

And that the top and the bottom are equal.

I don't understand how to do that here.

[dull bear]

hmmm

[silent turret]

cosh(z) will be really big when the real part goes to infinity, right?

so maybe the sides do go to zero

And maybe e^z, the integral of the top of the rectangle will be -1 * the integral of the bottom, because e^i pi is -1

actually, that doesn't make any sense, then the entire integral would be zero

[dull bear]

let me try it out, I'm thinking it's probably that the other three sides other than the real axis one will go to zero (that's usually how it goes!)

[silent turret]

ok

side note: you really love to post cat emojis

[dull bear]

I do for sure

At least for the side between $R$ and $R + i \pi$, you could parameterise out that line segment, and eventually you can bound the absolute value of that integral above by like $e^{-R} \int_{0}^{\pi} e^{t} + e^{-t} \dd t$, which goes to zero as $R\to\infty$

[twin meteor]

@dull bear

[dull bear]

I think at least

[dull bear]

In any case it's along the lines of this and that cos(z) is bounded on those sides

[silent turret]

ok, thanks a lot for the help. i think I can solve it now

[dull bear]

No worries hope it goes well!

Also for the top side "exponentials beat polynomials" may turn out to be useful, just in case, to show it goes to zero too

[dull bear]

Oh wait also I think I see why they suggested this one actually, my bad
Think their idea was still CRT but then you'd compare the real and imaginary parts when you apply it on what they gave, noting that e^{iz} = cos(z) + i * sin(z)
Showing the "other sides" go to zero should be similar though at the very least
[edit: the top doesn't actually go to zero]

In fact I think showing those sides go to zero probably is much easier in that case!

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[orchid pier]

how do i get the particular and homogenous solutions out of this?

[cyan talon]

hey @orchid pier

u(t) is the step function right ?

[orchid pier]

yeah

i dont get it

t>0 is 1
t<0 is 0

thats all i know

[cyan talon]

yeah that's it

you don't need to know much more lol

[orchid pier]

oh

but it doesnt say what t is

[cyan talon]

it's a function

the value depends on t

[orchid pier]

hmm

[cyan talon]

here's what it looks like when you plot it

anyway, have you found the homogeneous solution then ?

or, what methods do you know for solving DEs ?

[orchid pier]

no

its like dy/dx + y =0 right?

and then it becomes Ke^-at

[cyan talon]

well just Ke^-t

[orchid pier]

but i dont know how to find it

looking at the DE

[cyan talon]

what have you done in your class then ?

that's literally the first DE you try to solve usually

@orchid pier

[orchid pier]

i skipped that lecture 😩

wait im going to try to solve it myself

@cyan talon so dv/dt =-v

[cyan talon]

yes

@orchid pier

[orchid pier]

okayokay

ln|v| = -vt

v =ke^-vt? @cyan talon

[cyan talon]

how did you get that

you should have -t, not -vt

[orchid pier]

oh but the intergral of 5 is 5x right

i.e

[cyan talon]

yeah ?

[orchid pier]

so why is the v gone

[cyan talon]

how did you integrate ?

what steps have you done ?

@orchid pier

[orchid pier]

dv/dt becomes ln|x| + C

-v becomes -vt

@cyan talon a being v

[cyan talon]

yeah that's not at all what they're doing

a is -1

not v

so dv/dt = -v ok

then they divide by v

(dv/dt) / v = -1

[orchid pier]

oooh

they skipped steps

wtg

wth

[cyan talon]

they skipped one step lol

[orchid pier]

omg

[cyan talon]

it's not completely insane

[cyan talon]

but I get it

[orchid pier]

but what are they doing with B=-1

[cyan talon]

all the solutions of the equation look like A + Be^-t (with A and B being some numbers)

but if you give a value of the function at a certain point (here v(0) = 0)

you can find the corresponding A and B

and get just one solution

[orchid pier]

how is the particular solution 1? wth

[cyan talon]

idk how they got A=1 here tbf

is it the sol to the same problem you posted ? @orchid pier

[orchid pier]

yea

[cyan talon]

ah yeah it's the particular solution

v(t) = 1 solves the equation with u(t) on the right hand side

you get dv/dt + v = 0 + 1 = 1 = u(t)

[orchid pier]

but u(t) can be 0 too

[cyan talon]

they're talking about the t>0 case specifically here

they're covering the t<=0 case just after I guess

[orchid pier]

no 💀

its just this

[cyan talon]

wtf is that book

[orchid pier]

fundementals of electrical circuits hahaha

[cyan talon]

well maybe they just don't give a crap about t <= 0 in electricity

idk

[orchid pier]

one last question tho @cyan talon

how is dv/dt 1/v

or dv/dt /v

[cyan talon]

?

wdym how is ?

there's no equation in what you wrote

@orchid pier

dv/dt = -v

divide both sides by v

(dv/dt)/v = -1

[orchid pier]

(dv/dt)/v yeah how does this equal 1/v

[cyan talon]

it doesn't

it equals -1

[orchid pier]

no when you intergrate it

integration becomes ln|v|(left side)

so left side will be 1/v

[cyan talon]

nope

you forgot one thing

chain rule

ln(v), v is a function of t inside

so $$\dv{\ln(v(t))}{t} = \mathbf{v'(t)}\cdot \ln'(v(t))$$

[twin meteor]

aPlatypus

[cyan talon]

which gives you dv/dt * 1/v

or (dv/dt)/v

[orchid pier]

ah okay

ty

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[humble patrol]

Complete each statement with a variable expression

You buy (m+5) bagels at 35 cents each. The cost is ________ dolars

[humble patrol]

The work I have done so far is that I thought of multiplying (m+5) by 35

I'm not sure if it's right but even if it is I wouldn't understand why can someone help

nevermind I figured it out

.close

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[proven garden]

how can I convert those permutations to 2 row notations? σ = (763)(245), τ = (16)(3427).

[vocal sleet]

Please don't occupy multiple help channels.

[tidal dock]

what's a 2 row notation?

[terse forum]

!1c

[vocal sleet]

Please stick to your channel.

[tidal dock]

is this group theory?

[tidal dock]

close previous channel

[proven garden]

Sorry I thought it was closed

[proven garden]

like this

[tidal dock]

oh okay

[proven garden]

<@&286206848099549185>

[vocal sleet]

@proven garden Has your question been resolved?

[timid briar]

so, 2 row notation tells us for each number where it goes after the permutation

what does the cycle (763) represent? where does each number go?

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@proven garden Has your question been resolved?

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[dire island]

How would I solve this

[queen scaffold]

Have you learned similarity?

In similar triangles, the sides will have a constant ratio.

this is a good image that explains it without overcomplicating it

the ~ sign represents similarity

and the equal sign with ~ represents congruency (exactly the same)

[dire island]

Wait so how would I find the value of x?

because I dont understand how to put the x+5 and the 30 into a ratio

oh wait

just cross multiply

so would it be 75X+375=-750+300x?

[queen scaffold]

U cna also compare the same sides

In the image i sent 4/5 = 8/10 right

U can apply the same logic in ur problem

[dire island]

so did I get it right then?

[queen scaffold]

The left side is correct but the right side should be 120x-300

[dire island]

oh yeah I see where I messed up on that. Thank you lol

[queen scaffold]

Np

[dire island]

do I do !done

!done

[vocal sleet]

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[dire island]

.close

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[sacred lantern]

Please help with my geometryproofs.

[sacred lantern]

pretty much did most of it just havr that little bit left

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[sacred lantern]

<@&286206848099549185>

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[vital wave]

how do i draw the graph?

[vital wave]

how to find the y intercept

[gaunt sparrow]

The y intercept is whenever x=0

That shouldn't be too hard

[vital wave]

yes it is

[gaunt sparrow]

I mean

Just let x=0

What is the value of the function?

[vital wave]

what are you tlking abt

[gaunt sparrow]

You said you wanted to find the y intercept

[vital wave]

ok

[gaunt sparrow]

I told you that you should plug in x=0

[vital wave]

i mean how do i make the gfraph though

[gaunt sparrow]

Because by definition, the y intercept happens when x=0

[vital wave]

it says i did the graph worng

yesyes u r right

First identify the basic characteristics of a parabola. Begin by finding the vertex of the parabola. When the equation of the parabola is in the form ​f(x= )​a(x​-h) ^2k, the vertex is the point​ (h,k). Then determine whether the parabola opens up or down using the sign of the leading​ coefficient, a.​ Next, determine the axis of​ symmetry, x- and​ y- intercept(s).​ Finally, draw the graph of parabola using all the information.

it says this for what i did wrong

[gaunt sparrow]

I mean there are TONS of parabolas that go through the (0,9), so you can't just make it go there and pray it'll be the good one.

Right so is the parabola upwards?

[vital wave]

ya

[gaunt sparrow]

Good

Can you put it in vertex form?

That would involve completing the square.

[vital wave]

thats what im having trouble doing

idk how to do that

what is h and k?

[gaunt sparrow]

They are the coordinates of the vertex.

[vital wave]

it would be like f(x) = 1/3(x-h)^2 + k

[gaunt sparrow]

Yes indeed

[vital wave]

yea i know but i mean how do i find that from what they provide

i know i got -18 is correct

but i dont know how wide to make the graph

here look:

[gaunt sparrow]

Once your vertex is placed, you can "open" the parabola up until it fits with the y-intercept you found

ax^2 + bx + c = a(x+h)^2 + k

The formula is usually h = b/(2a) and k = c-b^2/(4a).

[vital wave]

oh

[gaunt sparrow]

You can get those values by completing the square, but that's up to you

[vital wave]

so h is -9

[gaunt sparrow]

Yes.

[vital wave]

uh k is getting weird r u sure that is the formula? c - b^2 / 4a?

[gaunt sparrow]

k = -18

As you had before

Now if you place the vertex at (-9, -18) and open the parabola so that it intersects the y axis at 9

[vital wave]

ohh

[gaunt sparrow]

You will have it open enough

[vital wave]

ooooo i see now

their graph is intercepting at 0,9

[gaunt sparrow]

Yes

[vital wave]

damn thank u

couldva said

lol

thank u 4 halp

.close

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[twin meteor]

Determine the speed at which water would flow through a point in the pipe that has a cross sectional area of 0.25 m^2 and a volume flow rate of 5.0 m^3/s

[shut salmon]

<@&286206848099549185>

<@&268886789983436800>

[lost yarrow]

?

[shut salmon]

I need help

[lost yarrow]

cool, read #❓how-to-get-help and ask in an unoccupied channel

[hasty lark]

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[twin meteor]

Hello Xenophan,

ive been working abit on it

i believe Q = A * v

where q is the volume flow rate

[hasty lark]

Yep

[twin meteor]

and A cross sectional area

and v = velocity

i divide the volume flow rate by the area

to solve for velocity

[hasty lark]

Yes

[twin meteor]

so my equation would look like V = (5.0 m^3/s) / (0.25m^2)

so i believe with those units im left with V= 20 m/s

[hasty lark]

I think you're entirely correct

[twin meteor]

so its just normal division with those units?

when i have m^3/s and i divide it by m^2 im just basically removing m^2

[hasty lark]

You're cancelling, yes

Just like a/a=1

m^3/m^2= m

[twin meteor]

and since i cant cancel seconds

with anything it remains as m/s

[hasty lark]

Yes

And the units of velocity are meters per second

So it works out

[twin meteor]

thank you xenophon sometimes i doubt my logic! thank you for making me less uneasy about this.

[hasty lark]

No worries. Good luck with fluids or whatever you are studying

[twin meteor]

yeah just physics in the fluid dynamic chapters haha @hasty lark

[hasty lark]

Remember to close the channel when you're done

[twin meteor]

@hasty lark gotcha

.close

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[slim oracle]

what is the modulus and argument or z when |z| takes the least value and arg(z+3) = pi/3

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[wooden sphinx]

Is the idea of second order mixed partial derivative Fxy just evaluating the change of the slope in the x direction as y changes

[wooden sphinx]

So basically if I wanted to find my partial derivative when x=3 at some fixed y value

And I find it

And then if I were to basically

Find the slope at x=3 for several fixed y values

And see how that slope changes

Is that the meaning of my mixed partial derivative

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[mild flower]

every time i see your pfp i think it's a pair of pants

also yes

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[craggy hamlet]

what is the trig identity of $\frac{cosx-1}{x^2}$

[twin meteor]

putridplanet

[craggy hamlet]

i need a trig identity

[vocal sleet]

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

[dull bear]

what’s the aim? You’re presumably finding the limit of that as x goes to zero?

[craggy hamlet]

yes

[dull bear]

what are you allowed to use for that? Assuming the mention of trig identities implies you want to use one?

[dull bear]

interesting they did it that way, I had another idea…

But basically one of the cos double angle but replacing x with x/2

[craggy hamlet]

the answer sheet says -1/2

[dull bear]

Hmm, unless I’m missing something

[stark plume]

Looks like it should be 1/2 to me

[dull bear]

,w lim (tan(x) - sin(x))/x^3, x to 0

[twin meteor]

Results provided by WolframAlpha

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Requested by chartbit

[craggy hamlet]

wait can you just keep using l'hopitals rule until it isnt undefined when you plug in zero

[dull bear]

The workings here seem all valid

[craggy hamlet]

my one is sinx-tanx

[dull bear]

Well if they gave you this one instead, then it’s -1/2 yeah

[dull bear]

And in theory you should be able to I think, though I haven’t tried it to make sure the 0/0 conditions hold at each step

[dull bear]

As for the comment about another way here, you can multiply top and bottom by (cos(x) + 1) as well and use identities and rearrange

[craggy hamlet]

what is the derivative of -sec^2x

[vocal sleet]

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[stable sigil]

Can someone help me convert this two-parameter integral into a one-parameter integral?

[stable sigil]

,rotate ccw

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<@&286206848099549185>

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[wind imp]

does anyone know how to do these questions

[wind imp]

first i split the powers e.g. 2^n+2 = 2^n * 2^2

[loud pivot]

What's the objective?

[wind imp]

oh right sorry

to simplify

and express answers in positive integers

[vast shale]

I don't think this simplifies nicely

unless you mean as n is being tended to some limit?

[worthy citrus]

The first does you just have to factor

[wind imp]

this is what they got as answers (ignore 130)

[worthy citrus]

This is good

[wind imp]

.close

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[fast quarry]

im stuck on part c, pls could i get assistance?

[fast quarry]

I have some definitions that will help

(i.e. such that x ∈ S◦).```

member of X (i.e. a set S ⊂ R such that X ⊂ S◦.)```

The interior of a set S is the set of all interior points of S. It is usually denoted S◦, but it can also be denoted Int S.

δ > 0 such that
(x − δ, x + δ) ⊂ S.```

not sure where to start though

[high maple]

How about

So you want to show that the set is infinite.

But infinity is hard to handle

Then, how about proof by contradiction?

[fast quarry]

I'm not sure we are meant to use that method for these questions

but maybe you're right

[high maple]

Did your prof disallow proof by contradiction?

Most topological problems heavily use contradiction method

So idk if that would be problematic

[fast quarry]

i think we are meant to be working from definitions to prove it

[livid tapir]

you cna do that with contradiction ofc, but you can do a direct proof with this one

[high maple]

Oh hmm. True

[livid tapir]

like if U is a neighborhood of l then (l-epsilon, l+epsilon) \subseteq U for some epsilon right? And then l is the infimum so something something l+epsilon....

[high maple]

Well technically

[livid tapir]

Right R too

I mean

[high maple]

It is that U includes epsilon-nbhd of l.

[livid tapir]

Has a subset

Yeah

[fast quarry]

yes this is the correct path i think

[high maple]

Yea I guess the key here is, you have two places with "is infinite" here.

Gotta match that puzzle

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[misty surge]

How do i go about answering this

[scarlet lichen]

which one?

theres 10 questions xD

[misty surge]

like in genreal

[scarlet lichen]

find the lim as x --> a+ and x-->a-

like for question one, you can try and plugin -3.99 ( for the negative) and -4.01 ( positive)

[misty surge]

you mean negative 4.01?

[scarlet lichen]

yeah mb

-4.01 but its approaching 4 from the positive

[round plover]

-4.01 would be the left hand limit

[scarlet lichen]

yeah since the values are moving towards -4

which is less than -4.01

[misty surge]

and is that the only method?

[scarlet lichen]

only? no. efficient and a safe guess? sure

[round plover]

the modulus function is defined piecewise, so you'll need to split cases anyhow

[misty surge]

what does that mean?

[scarlet lichen]

positive when x>=-4 and negative when x<-4

[round plover]

as in |x| has two separate parts in its definition depending on the sign of x, so you'll need to do the two cases separately

[misty surge]

and so if -3.99 and -4.01 give me the same limit it means its two sided?

[scarlet lichen]

yep

think of it as getting closer and closer to the asymptope in the graph

since x =/= -4

hope that helps

try and graph thi

so much more easier to visualise whats happenings

heres the graph of the first equation

so you plugin values like -3.9999 and 4.0001

the value approaches +inf

[misty surge]

alright thanks

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[knotty kestrel]

um in the last sentence isn't it supposed to be concave down?

i thought if it was > 0 it was up

and vise versa

am i mistaken

idk

.close

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[sand hedge]

Can I do this

[hard atlas]

can you do what

ignore the indices?

[sand hedge]

Yes

[hard atlas]

yes

[sand hedge]

Thank you

Doing the proof with them was really annoying

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[hard atlas]

I can imagine

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[misty surge]

how do i do 4?

[knotty kestrel]

what happens to f(x) when x approaches infinity

since the x's have the same power....

[misty surge]

plus or minus infinity though?

how do i do both?

[knotty kestrel]

plus or minus infinity is infinity and -infinity

[misty surge]

so i dont understand do i repeat it twice fo + and -

[knotty kestrel]

well just do one at a time

[knotty kestrel]

this is a hint

[silent burrow]

there is no double-sided limits here, if thats your question, only 2 one-sided

because you cant go above infinity

[misty surge]

but how do i show that in math notation

[knotty kestrel]

just.... solve for positive infinity and negative infinity

what is x^2/x^2

idk if im explaining this right prob not

[misty surge]

but i also have it solve it from the left and rigth

and i just dont understand how that works with infinity

[silent burrow]

it does not

[misty surge]

1

[knotty kestrel]

yes

so would it be fair to say that f(x) = (x-4)^3/(x+4)^3 = 1

you can plug any value for x and it will approach 1

[misty surge]

how so x+4 does not equal x-4

[knotty kestrel]

um idk how to explain this but really only the x^3 matters

when the function approaches infinity (an infinitely large number) a number as small as 4^3 doesnt really contribute anything

so its really just x^3/x^3

[misty surge]

ah okay i understand the 4 is negligible relative to infinity so the two sided limit exists

[silent burrow]

well, you can expand (maybe do it with squares not cubes) and you can divide every term by x^3, so you end up with two 1s and a bunch of terms which approaches 0