help-27

[plush quarry]

(btw, what is this question for?)

[pure cedar]

idk bout the sum but ik that
$$ \int ln(u) dx = \frac{uln(u) -u}{u'} + c $$
when u is linear in terms of x

[woven radish]

Mr. Smith

[pure cedar]

idk if that sum is standard or not
but by the looks of it i'd try to relate it to an integral

[plush quarry]

The sum expands to 1-1/2+1/3-1/4+1/5-1/6+...

There is a power series that can help with that iirc

[pure cedar]

oh yeah

its the taylor polynomial for lnx ig

[devout snow]

@simple mesa Has your question been resolved?

[simple mesa]

i need the answer

[bronze void]

is this meant to be done without a calculator?

because just simplify the constants and bring them all together, then just solve the sum and integral no?

it simplifies nicely

[devout snow]

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[devout snow]

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[frigid tartan]

im having trouble figuring out how to sketch from a slope field

like it looks really overwhelming for me

[devout snow]

@frigid tartan Has your question been resolved?

[frigid tartan]

<@&286206848099549185>

[lunar harbor]

Just trace the segments in the slope field tbh

Something like this works

[frigid tartan]

hrrggg i guess

[coarse flume]

The graph below y=1 should be somewhat less vertical but you get the idea

[frigid tartan]

yea

i guess my second question is

for part b

is the line tangent to the graph that i drew on the slope field?

is dy/dx the graph i drew

[coarse flume]

I believe what you drew is f(x) itself

[frigid tartan]

is this what b is asking

do you think

idk the wording is really confusinh

[sacred sedge]

it is asking to solve the differential equation for y(x)

It is a seperable ODE

[frigid tartan]

like this

?

[sacred sedge]

yup

now use the inital condition to find C

or

first rearrange your equation

so you get y = ....

then getting C will be easier

[frigid tartan]

?

[sacred sedge]

yup

[frigid tartan]

wait are we still on part b or part c

[sacred sedge]

oh I read part c

as part b

hold on sorry

[frigid tartan]

nono its okay

[coarse flume]

👍

[frigid tartan]

i think ive had like 5 diff people come and go from here 😭

[sacred sedge]

ye sorry, this is correct

now solve for f(1.3)

[frigid tartan]

uhhh

[coarse flume]

Wait did you finish b yet

[frigid tartan]

i need to plug in 1.3

[coarse flume]

Into the line equation

[frigid tartan]

yea

i got 1.6

and for c

my answer is this one

thats the specific solution right

cuz i plugged in c

[sacred sedge]

I think it is good

[devout snow]

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[devout snow]

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[prime hemlock]

Prove tat if H is normal in G then the quotient group G/H is a group

[modest dune]

What field of math is this

[prime hemlock]

Group Theory

[modest dune]

Okay

Gl

[sand quarry]

Whars your question

Are you stuck? Need a nudge?

[prime hemlock]

Yep

Do I have to prove the quotient group operation or do I have to prove it's a group ?

[sand quarry]

What makes a group?

[sand quarry]

Both

[prime hemlock]

Ok
so how do I start 😅

[sand quarry]

Answer this

[prime hemlock]

Closure, Associativity Identity and Inverse

[sand quarry]

Yes

So you can define the binary operation to be
(aH)(bH) = (ab)H on G/H

Or something

[sand quarry]

Try to prove all four of these now

[sand quarry]

Is the operation well defined?

[prime hemlock]

Yes I'll try

[prime hemlock]

Alright?

[near jolt]

you have to show that this multiplication operation is well defined:

There are many ways to represent aH, what if a'H=aH but a\neq a'? you are saying that a'bH=a'HbH=aHbH=abH. It would be a big issue if a'bH is not equal to abH

[prime hemlock]

Ohh

[near jolt]

need to be careful in definition when elements have multiple representations.

[prime hemlock]

So apparently I just have to prove the operation is well defined b4 actually proving it's a group ?

[near jolt]

let X and Y be elements of G/H. We define XY to be equal to abH for ANY choice of a\in X, b\in Y.

Writing it like this, it should be evident exactly what you need to prove to absolutely ensure it is well-defined.

[prime hemlock]

Okae lemme try

[dim perch]

(as an extra nudge, recall that elements of G/H are cosets, i.e., sets, so showing abH = a'b'H you must show set equality)

[devout snow]

@prime hemlock Has your question been resolved?

[prime hemlock]

Phew finally
Correct ?

[prime hemlock]

,rotate

[devout snow]

@prime hemlock Has your question been resolved?

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[sullen karma]

can someone double check my answers

[devout snow]

@sullen karma Has your question been resolved?

[lean crater]

wait

yeah part a is incorrect

youre missing K to M and K to N

[sullen karma]

oh yeah just realised

alr ty

.close

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[forest stratus]

Four couples are to be seated around a circular table. All the couples are to sit together,except for Jeremy and Candace,in such a way that males and females alternate.How many seating arrangements are possible?

[lean crater]

!status

[devout snow]

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

[lean crater]

which part are you on^^

[forest stratus]

thinking part

[lean crater]

for a circular table (im assuming rotational symmetry doesnt add more)

you always fix someone as a frame of reference

[forest stratus]

yes!

[sharp adder]

-# oh ididnt think of that nice

[lean crater]

so, arbitrarily fix someone as a frame of reference as your first step!

[forest stratus]

i have fixed jeremy

[lean crater]

mhm, so since males and females must alternate seats, jeremy's position dictates the gender of every other seat

say jeremy is in seat one

what does that say about the odd and even seats

[forest stratus]

even seats must be females and odd males ?

[lean crater]

yep!

there are four couples so four seats
odd; 1,3,5,7 are male
even; 2,4,6,8 are female

now, Jeremy and Candace cant sit together

for some odd reason

the next step would be to seat candace

[forest stratus]

candace cannot sit in 8,2 so does that mean that there are 2 ways to seat her ?

[lean crater]

yep

seat 4 and 6

so, seat the remaining 3 couples

we have 3 distinct couples to place into these 3 distinct slots if candace is seated in seat 6

we have 3 distinct couples to place into these 3 distinct slots if candace is seated in seat 4

[forest stratus]

3! x 2 ?

[lean crater]

yep!

and thats your answer

[forest stratus]

thank you !

[lean crater]

anytime!

if youre done

type .close

[forest stratus]

.close

[devout snow]

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[devout snow]

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[lost laurel]

so here Im unsure of what exactly i need to do

I need to show integrate (-2)^x over (-\infty,0) and show it's equal to that of the inegral of ce^{-3x}?

opps

poission is discrete

[grand warren]

yep so you will use sums

[lost laurel]

so $\sum_{x=1}^{\infty} x(-2x)^{x}$?

[woven radish]

Wai

[lost laurel]

but that diverges

is that supposed to be (-2)^{X} or what

[sand quarry]

Thays nkt right

Like

Why is it multiplied with x

[grand warren]

uh no $\sum_{k=0}^\infty (-2)^k P(X=k)$

[lost laurel]

expectation

[woven radish]

tm (#1 number theory fan)

[lost laurel]

ooh

cool

thanks!

.close

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[dark sable]

well (f+g)(x)=f(x)+g(x) and (fg)(x)=f(x)g(x) and just direct substitution for 1 and 2. 3 is using the formal definition of a limit, 4. is basically just the same as what I said for 1 and 2

[covert mango]

yeah sry should clarify its only 3 I need help with

[dark sable]

search formal definition of a limit I'm not gonna explain the funny epsilon delta definition

[static cobalt]

@woven radish yo solve this

[dark sable]

!occupied

[devout snow]

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).

[static cobalt]

ih

oh

k

i shall leave

😔

[covert mango]

is it not like $\forall \epsilon > 0 \exists \delta > 0 : |x - l| < \delta(\epsilon) \implies |f(x) - f(l)| < \epsilon$
\\
means that $\lim_{x \to l} f(x) = f(l)$

[woven radish]

AnitaG

[covert mango]

or is this pointwise continuity

[dark sable]

I think someone better can explain this

lowk thought there was an /2 in the proof for limit s

[covert mango]

dont think this matters

\epsilon / 2 still satisfies requirement

arbitrary ig

[arctic yarrow]

this is the continuity of f in l

[covert mango]

whats limit thennnn

[arctic yarrow]

same but replace l by x_0 and f(l) by alpha in your case

$\forall \epsilon > 0 \exists \delta > 0 : |x - x_0| < \delta \implies |f(x) - \alpha| < \epsilon$

[woven radish]

Lin Xia

[covert mango]

how does that make a difference

as in \alpha doesnt have to be f(x_0)?

[arctic yarrow]

what you wrote is lim f(x) = f(x_0) that is the definition of f being continuous on x_0

[covert mango]

does limit exist imply continuity

[arctic yarrow]

not necessarily

[covert mango]

but not the converse?

[arctic yarrow]

if f is not continuous on x_0

for example take f(x) equal 0 for x < 0 and x > 0 and f equal 1 for x = 1

lim f(x) on 0 is 0 but f(0) = 1

[covert mango]

yeah makes sense

has limit at 0 but not continuous at 0

[arctic yarrow]

anyway it is not about continuity here

[covert mango]

wait no it is continuous as 0+ limit equals 0- limit

(=0)

[arctic yarrow]

we just assume that g as a limit on x_0 and this limit is \beta

[covert mango]

okay

[arctic yarrow]

my example you mean ?

[covert mango]

yes

[arctic yarrow]

no it is not because f(0) is different from the limit

continuity is really this

lim f(x) = f(x_0)

[covert mango]

alr got it

[arctic yarrow]

ok so for your question 3 you need to use the formal definition of limit for g

[covert mango]

so we have $|x - x_0| < \delta(\epsilon) \implies |g(x) - \beta| < \epsilon $

$\forall \epsilon>0 \exists \delta>0 : |x - x_0| < \delta(\epsilon) \implies |g(x) - \beta| < \epsilon $

[woven radish]

AnitaG

[arctic yarrow]

and you want |g(x)| > \beta/2. What can vlue of epsilon can you take to have this

[covert mango]

$|g(x)-\beta| \leq |g(x)| + |\beta|$

[woven radish]

AnitaG

[covert mango]

kinda lost, inequality sign goes the other way

[arctic yarrow]

yes you are right you need to use triangular inequality but not this form

[covert mango]

$|g(x)-\beta| \leq \left|g(x) - \frac{\beta}{2}\right| + \left|\frac{\beta}{2}\right|$

[arctic yarrow]

not this one either

[woven radish]

AnitaG

[arctic yarrow]

you need to find something in term of |g(x)| that is less than |g(x) - \beta|

do you know this formula : $$ ||a| - |b||\le |a -b|$$

[woven radish]

Lin Xia

[covert mango]

yeah idk how besides like trivially 0 is less than mod of g(x) - \beta

[covert mango]

I do now 😭

but no I didnt

[arctic yarrow]

you can show this from the standard triangular inequality

[covert mango]

makes sense conceptually

[arctic yarrow]

you can use this one now

[covert mango]

k so $||g(x)|-|\beta|| \leq |g(x)-\beta|<\epsilon$

[woven radish]

AnitaG

[arctic yarrow]

yes so now by definition of absolute value you can find a lower bound for |g(x)|

[covert mango]

0

so it reduces to |\beta|

on the LHS

[arctic yarrow]

well sure but i was thinking of this |x| < a iff -a < x < a

[covert mango]

how can I apply this

[arctic yarrow]

use it on $||g(x)| - \beta| < \epsilon$

[woven radish]

Lin Xia

[covert mango]

$-\epsilon < |g(x)|-|\beta| < \epsilon$

[woven radish]

AnitaG

[covert mango]

so just choose -\epsilon = 1.5|\beta|?

but that doesnt satisfy epsilon positive

wait

im dumb

epsilon is 0.5|beta|

[arctic yarrow]

exact

if you take epsilon = \beta/2 then |g(x)| > \beta/2

[covert mango]

r u missing mods bc lazy or are they just not needed

idt question mentions the sgn of beta

[arctic yarrow]

i though it was \beta > 0

ok so it is a bit different then

let's assume \beta > 0

[covert mango]

proof still works with negative beta u just use mod

brb

[arctic yarrow]

yes you re right

just do like this it is more concise

[covert mango]

.close

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[pure stone]

$(2xy^2 - 3y^3)dx = (3xy^2-y)dy$

[woven radish]

DeC∆rbonizeD

[sand quarry]

this screams uh

what was it called again

almost exact equation

find integrating factor?

[lean crater]

yep

[pure stone]

oh great it's in my book just never learnt it in class fsr

.close

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[raven forge]

Just like he tried two possible equations of pythagoras, why didn't he tried and solved the third equation which can be formed using pythagoras theorem?

[dark sable]

(t+1)^2+4^2=(t-1)^2?

try expanding it and see what value of t you get

[jagged harbor]

or more geometrically, the hypotenuse of a right triangle is the longest side

how could t-1 be the length of the hypotenuse if t+1 > t-1?

[devout snow]

@raven forge Has your question been resolved?

[raven forge]

Makes sense

Very much

[shy osprey]

so well then

[devout snow]

If you are done with this channel, please mark your problem as solved by typing .close

[raven forge]

Wait

I got another question, in this problem what will happen in the case of different parity?

[jagged harbor]

try it yourself, it probably doesn't imply the product is a multiple of 8

also you should limit a channel to one question, to make it easier for us to follow

[dark sable]

try if a,b, or c is/are odd/even

truth tables and logic 😱

[raven forge]

Alright

.close

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[stoic relic]

what should i do here

so far i've done this

not sure where to go from here

[faint gorge]

Notice that the leading coefficient is 0 and the parabola has roots, so for which intervals is the parabola non-negative?

[devout snow]

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[quaint gull]

static question idk if it counts

[lunar harbor]

discord.gg/physics

wrong server

[quaint gull]

mb thanks a lot

[devout snow]

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[devout snow]

@quaint gull Has your question been resolved?

[devout snow]

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[dark sable]

<@&268886789983436800>

[devout snow]

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[hasty cargo]

[dark sable]

anything for that better blue role

[livid geyser]

what is a god amongst mods

[dark sable]

prolly rice

[potent dirge]

Rice is god!!!

[devout snow]

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[celest lantern]

Suppose I have a finite dimensional vector space V with dim (V) = n, how would I show dim (W) <= n if W is a subspace of V?

[jade pecan]

form a basis of W and extend it

[celest lantern]

But I have no ideas what do you mean form one?

[jade pecan]

let {w1,..., w_i} be a basis of W

[celest lantern]

Do I have to construct independent vectors in W?

I asked my peers and other Artificial Intelligence tools, they said that I should pick independent vectors set.

[jade pecan]

u can always choose a basis of W

as i did

these vectors must be independent in V as well

[celest lantern]

Right, so if I have a basis, what can I move on from here?

Wait, dim({0}) = 0, right?

[jade pecan]

yes

[celest lantern]

Okay, I will pick the first base step whereas W only have a zero vector.

[jade pecan]

use the fact that this set (the basis of W) is independent in V as well

[celest lantern]

I still not understand this bit here...

Here is my workout so far:

Suppose $V$ is a finite dimensional vector space such that $\text{dim}(V)=n$.\
Also assume that $W$ is a dimensional vector subspace in $V$, such that $\text{dim}(W)=m$.\
We want to show that $m\leq n$.\
First, we check for zero vector, then $W={0}$ gives $\text{dim}(W)=0$, which the statement holds. Now, suppose $W\neq {0}$, then choose a vector $w_1\neq0$, then $W={w_1}$ and linearly independent.

[woven radish]

Mercury.

[jade pecan]

if W is not the trivial space then why do u conclude W={w1}

[celest lantern]

Suppose we only pick 1 vector in first which is differs from zero vector.?

[jade pecan]

we pick a basis of W, {w1,..,w_i}. this means dim(W) = i
since it is a basis in W, a1w1+...+a_iw_i = 0
but this same relation holds in V as well because W is a subspace of V and they share the same 0 vector. Hence this set is independent in V as well
use the first part of the corollary to conclude

[celest lantern]

I think I should use the first one.

Pick any vectors $w_1,w_2,... w_n$ such that they are linearly independent.

[woven radish]

Mercury.

[celest lantern]

Alright, thank you. I think I understood this.

.close

[devout snow]

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[devout snow]

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[fervent terrace]

This is what I did

I missed the probability?

[solid perch]

you used the number of heads as x, when I think you should be using the payout.

after all, the question wants the SD of the net gain.

[fervent terrace]

Okay that makes sense

Was I also supposed to do sum of (x-m)^2 *p(x)?

[solid perch]

yes, but take note: doing so gives you the variance, NOT the SD.

you'll need to square root this to get the SD.

[fervent terrace]

Ohh okay yes

[solid perch]

but otherwise, remembering the probability of each event, and using the net gain instead of the number of heads, should be enough for you to continue.

[fervent terrace]

,rotate

[solid perch]

,rcw

,rotate 180

[fervent terrace]

I got it wrong again

[solid perch]

incorrect.

you forgot that you have to pay $2 to even play the game to start.

[fervent terrace]

I put -2 on the end

[solid perch]

that does not correct for the squaring done inside the summation.

so your net gain must take into account this upfront payment from the start.

[fervent terrace]

Oh yes X=-2

I'm the case of loss

[solid perch]

yes.

[fervent terrace]

i keep getting it wrong

i keep getting variance as 5 or 7

[solid perch]

show your work.

[solid perch]

you correctly accounted for the loss, but the payouts in the first two events are still too big.

where's the 4 coming from?

in the event you win $4, you should only have gained $2 because you paid $2 up front.

likewise with the event where you win $2, you should have gained nothing.

[fervent terrace]

ohhhh okay i see

i should account for all of them the same way i did for loss

[solid perch]

yes!

[fervent terrace]

do i keep the negative sign when squaring the -2

[sand quarry]

squaring anything results in a positive number

except 0, i suppose

[bitter ledge]

squaring any real number gives a nonnegative number :)

[crystal dawn]

keep it anyway

if only to convince yourself and potentially your examiner that you remembered negative payouts exist

[devout snow]

@fervent terrace Has your question been resolved?

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[half lodge]

I’ve been struggling with finding the quadratic equation for this, I’m using an area model and have been using all the x-ints listed, but every time i put them into the area model, it never works out. which x-int should i use?

[half lodge]

the x-ints are the numbers near the middle in pencil

[spiral raptor]

Is this the x ints?

I would assume there would be a 1/2 too

[fierce heath]

what ints have you tried already

[devout snow]

@half lodge Has your question been resolved?

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[barren dune]

Where did I go wrong?

[barren dune]

The previous part isn’t needed

My final answers is correct except the signs are flipped smh

[hasty cargo]

Can you post the full question

This is only part b

[graceful cosmos]

,w integral of tan^(-1)(4x) between 0 and 0.25

[woven radish]

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Requested by kaynex

[graceful cosmos]

,w 1/16 pi - 1/8 ln(2)

[woven radish]

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Requested by kaynex

[mild sorrel]

I think the error is in the second last step.

[graceful cosmos]

Ah yes I see it now

You want upper - lower, you've got lower - upper

[barren dune]

Ohh

Thank you sm

[mild sorrel]

nvm

[lunar harbor]

.pin

[lunar harbor]

(In the future, please make the first message you send the actual question since that's what the bot pins. This saves us the effort of having to change the pin manually.)

[devout snow]

@barren dune Has your question been resolved?

[barren dune]

Thank you
And apologies

[devout snow]

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[devout snow]

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[neon wagon]

so ive got this

[neon wagon]

and f(x) = x^3 -4x

so

this is really tricky

can i just say that the function h is

between the origin and f

but i gotta prove this somehow

[alpine python]

you should show that the midpoint satisfies the equation h(x)

and you can use the information that P satisfies the equation f(x)

[neon wagon]

wdym satisfies

[alpine python]

that means you can do a substitution with h(x)

[faint gorge]

-# erfüllt die Gleichung

[neon wagon]

so

but what equation

[alpine python]

you can use some variables, like say that P is (a,b)

you know some information about (a,b) and you also can figure out the midpoint

[neon wagon]

hmm

can i just pick some random point

bcs i have a material with the sketch

of f(x)

[alpine python]

you can just pick a random point to help visualize it and stuff

but the argument needs to work for all points

[neon wagon]

hmmmm

let me write some stuff

[alpine python]

ok

[neon wagon]

wait can i say

f(x/2) = h(x)

[alpine python]

i think you want (1/2) f(x/2) = h(x)

[neon wagon]

oh my god

but i was close

[alpine python]

ye

[neon wagon]

but what did you mean by P(a,b

[neon wagon]

hold in also why 1/2 before f

[alpine python]

the midpoint will be half as high as f(x)

i mean, half as high as P

the (a,b) thing was my way of trying to think carefully about the problem

but what you said is fine

[neon wagon]

but wait shouldnt it be h(x/2)

because

or no

nope

[alpine python]

oh

(1/2) f(x) = h(x/2)

[neon wagon]

i basically did this rq

[neon wagon]

why the 1/2 before f tho sorry

[alpine python]

try this, pick some x and plot it on f(x)

and plot the midpoint, which will be (x/2, f(x)/2)

so (x/2, f(x)/2) needs to be on h

being on the graph always means h(the first thing) = the second thing
like if (a,b) is on the graph of h, that means h(a) = b

so (x/2, f(x)/2) being on h, means the same thing as h(x/2) = f(x)/2

[neon wagon]

what is even the midpoint

[alpine python]

the midpoint is the point halfway from the origin to P

[neon wagon]

oh so the y value is also halved?

[alpine python]

yeah

[neon wagon]

oh thats why u said 1/2?

[alpine python]

in the image you can see "divided by 2" on both of the coordinates

yeah

the y value of the point P is f(x)

P is at (x, f(x))

[neon wagon]

damnnn crazy

wait do i have to solve some stuff too or nah

[alpine python]

once you reach this point: (1/2) f(x) = h(x/2)
then you have to solve some stuff

to show that both sides of that equation are the same

[neon wagon]

ohhhh dang

ok ill try

[alpine python]

ok

[neon wagon]

OMG

wait

cant i just plugin x/2 for h

[alpine python]

yeah

[neon wagon]

i literally got 1/2x^3 -2x

thats proof

[alpine python]

yeah because that's half of f(x)

do you see what i mean?

that's equal to (1/2) f(x)

[neon wagon]

yeppö

but i hope i can do problems like theseon myown i just cant

[alpine python]

you might need to remember the midpoint formula

[devout snow]

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[whole wave]

Gang can someone explain this me i cant understand for some reason💔

[whole wave]

Why did they put dx/du there

[dull parrot]

what part do you not understand?

[whole wave]

Already the first part

[dull parrot]

they basically did something called a u substiution

[whole wave]

Bro its confusing me😔

[dull parrot]

basically let u = 2x^4 - 3

they differentiate with respect to d/dx to du/dx = 8x^3

once they did that

they rearranged dx to make it the subject

to get du/(8x^3) = dx

you can substiute that in into your dx

[flat bronze]

2999 - 3000 + 1

[dull parrot]

!occupied

[devout snow]

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).

[whole wave]

Ty gang i think ill just watch a yt video aboyt it🙏

[dull parrot]

yeah that would probably be better

https://www.youtube.com/watch?v=3eWxzBbsS9o&pp=ygUNdSBzdWJzdGl0dWlvbg%3D%3D

[whole wave]

🥰

[dull parrot]

my dumbass sending the image of the thumbnail instead of the link 💔

[dim perch]

how is the video so far?

[whole wave]

Im doing their example rn

I jope bro explains well

[dim perch]

come back if it still doesnt make any sense :)

[whole wave]

I feel like i can do the questions but i dont actually understand it😔

Like what actually is „dx” and how come you can jusy do that with it🙏

[dim perch]

you don't understand what dx means in an integral?

or what it represents?

[whole wave]

I though i did🥀

[dim perch]

well without delving into too much detail (unless you want it), dx just refers to the variable we are differentiating/integrating with respect to

[whole wave]

But how come u can just do du=4dx ir whayver

[dim perch]

well, when we're doing u-sub, we may rewrite a function of x in terms of u, provided that we may also rewrite du in terms of dx

as a simple example, consider $\int(x + 1)^{10} dx$

[woven radish]

blanketism

[dim perch]

we could multiply out x + 1 10 times, but that's a lot of algebra im sure we don't want to do

so what we do instead is we can do u-sub

i'll give you this one so i can explain it, but we may first set u = x + 1

what happens when i differentiate/take the derivative of u with respect to x? i.e., what is du/dx?

<@&268886789983436800>

[whole wave]

Oh wait ive done questions like this before we hust didnt use u

😭

[dim perch]

im sure you have

[dim perch]

.

[whole wave]

Wait im fonna write ut down

So its just a longer way to wtite it

[dim perch]

looks good!

id like to say that that's a shorter way of doing the problem since we dont have to multiply out x + 1 10 times lol

[whole wave]

when i got du=dx i was like what😭

[dim perch]

yeah, its more or less a trivial substitution

just wanted to stick with something simple for you to see it a little better

[whole wave]

Yeah ok🥰

[dim perch]

anyways

a more practical problem would be something like $\int\sin(x^2 + x)(2x + 1) dx$

[woven radish]

blanketism

[whole wave]

Then would u=x^2+x?

[dim perch]

correct!

what would du be?

[whole wave]

Ohh

Wait

I cant think unless i write it down🥀

[dim perch]

no worries, that looks good!

[whole wave]

It’s still weird that u can multply and divide by dx or du

[dim perch]

technically you cant :)

but its an abuse of notation

[whole wave]

Thats why it confuses me💔

[dim perch]

do you want to know what's actually occurring? or would that confuse you more, you think?

[whole wave]

You can try to explain to me🙏

[dim perch]

okay, sure

we claim the following:

$$\int f(g(x))g'(x) dx = \int f(u)du\big|_{u = g(x)}$$

[woven radish]

blanketism

[dim perch]

to see this, we use the chain rule. let $F$ be a function such that $F' = f$ i.e., the derivative of $F$ is $f$. then $(F \circ g)' = (f \circ g) \cdot g'$

[woven radish]

blanketism

[whole wave]

Wdym by the circle symbol

[dim perch]

composition

$(f \circ g)(x) = f(g(x))$

[woven radish]

blanketism

[whole wave]

Ok then

Never seen that🙏

[dim perch]

its more or less saying that if you see the derivative of some function, we may integrate with respect to that function

in the sin example i provided, we saw that the derivative of $x^2 + x$ was present, i.e., $2x + 1$ so we may rewrite the integrand in terms of $u = x^2 + x$

[woven radish]

blanketism

[whole wave]

When i write it it looks like fog💔

[dim perch]

lol its okay

[whole wave]

I think i kidn of understand

Well ty bro for explainibg🥰

Im gonna sleep now

[devout snow]

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[whole wave]

Mb gang i forgot

[devout snow]

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[ionic jasper]

yo guys, how do i Calculate the distance of point P from the line g

[devout snow]

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

[ionic jasper]

1

[trail eagle]

Intuitively, this distance should be the length of the vector going from P to a point on the line g perpendicularly.

Like in the 2D case.

Write down the parametric equation of a line going through P and a generic point in the line g Q(lambda).
This new line has a certain direction vector. Your goal is to pick lambda such this direction vector and the direction vector of g are orthogonal.

Then your distance will be PQ.

[ionic jasper]

okay 1 sec

i dont get it

[trail eagle]

Which part?

[ionic jasper]

Write down the parametric equation of a line going through P and a generic point in the line g Q(lambda).

idk what exactly to do

problem is also the translation, which i gotta do on every 2nd word

[devout snow]

@ionic jasper Has your question been resolved?

[faint gorge]

was geht

oh nvm azy already helping

[trail eagle]

I'll try a bit but you might explain it better than I do

[trail eagle]

Well say we pick a point on the line $g$. This point, let's call it $Q(\lambda)$ has the form $\begin{bmatrix}2-\lambda \ 1+3\lambda \ -3+\lambda\end{bmatrix}$.\
Then you can get vector from $P$ to $Q(\lambda)$.

[woven radish]

Azyrashacorki

[trail eagle]

You want this vector PQ(lambda) to be orthogonal to the direction vector of g.

[ionic jasper]

oh mb i didnt saw

the messages

i am not sure but is it meant like this ?

and then u do something with skalar product

like so

[trail eagle]

Yes essentially

So first I would write down what this vector PQ(lambda) looks like.

Then you can use the dot product of PQ(lambda) with the direction vector of g.

We want this to be 0.

So this will give you an equation involving lambda.

This lambda that you find then will give you the point on g which is closest to P.

[ionic jasper]

ye i understand now

thanks

.clsoe

.close

[devout snow]

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[cerulean ruin]

(l) ${f : \mathbb{R} \to \mathbb{R}: f$ is strictly increasing, i.e., $f(x) > f(y)$ whenever $x > y }$.

[woven radish]

toast

[cerulean ruin]

cardinality of this 🙂

any properties of rational?

instinct says c^aleph_0

wait maybe. not

[dim perch]

i also guessed this

[cerulean ruin]

these problems are killing me 💀

but i am not 100% sure what peroperties i should use

[dim perch]

yeah theres so many 💀

[cerulean ruin]

hmm

[dim perch]

i may think its just $\mathfrak c$ since we can have countably many points of discontinuities?

[woven radish]

blanketism

[cerulean ruin]

i think thats true

so can we make

uniquely identify a strictly incresing function

like before

(with continuous functions)

by a set of rationals

[dim perch]

i believe so

[cerulean ruin]

o rlly

[dim perch]

or rather, in the sense that we can associate the countably infinite jump discontinuities with a rational?

[cerulean ruin]

there gotta be a specific difference

ye

so maybe we look at limit from both sides

for each rational values

like limx->3-f(x) = 10 but limx->3+f(x) = 20

so we just consider both valuyes?

then its the same thing

idk

[dim perch]

hmm i think we wouldnt have to examine the limits, but just counting would be fine

i.e., for any jump discontinuity over some interval, just pick a rational in that interval (may also be a point)

[cerulean ruin]

hm

for the middle interval we just pick a rational?

and thats uniquely determinedd

bc

since the function is stirctly increasing

once a raitonal value is "picked" it will never be selected again?

[dim perch]

i don't think its uniquely determined

just some rational in the interval will do due to density

[frosty crescent]

wait is c^c = c or is that independent of zf(c)

[dim perch]

no, i think c^c is strictly larger

[frosty crescent]

yeah what i guessed

[cerulean ruin]

oops

mb

but

[dim perch]

c^c = (2^n)^c = 2^c = c?

[cerulean ruin]

the set of picked vaues

[dim perch]

i guess

[cerulean ruin]

will give us a unique increasing function?

actually

probably not necessarily

[dim perch]

not necessarily no

though i guess we could construct as such

[frosty crescent]

i'm quite sure that a function that is strictly increasing should be continuous a.e.

fact check me on that

[cerulean ruin]

so why are we selecting a single rational?

[cerulean ruin]

rlly?

idts

wait

maybe i am dumb

[frosty crescent]

and clearly we can make a strictly increasing function with countably many discontinuities

[cerulean ruin]

yes

[dim perch]

just so that we can associate the number of discontinuities is countable

[cerulean ruin]

oh ok

yes

that makes sense

[frosty crescent]

but iirc there are uncountable sets with zero lebesgue measure so idk

[cerulean ruin]

but now

we want to probably make an injection from increasing functions -> {f: Q ->R} right

[frosty crescent]

there's a quite canonical one

let (a1, a2, a3, ...) be a convergent sequence of increasing rational numbers to a limit r

then (f(a1), f(a2), f(a3), ...) should also be monotone

claim: the sequence is bounded above

proof: take any rational number q greater than r, sequence is bounded above by f(q)

so (f(a1), f(a2), f(a3), ...) has a limit by monotone convergence theorem

let this limit be f(r)

[cerulean ruin]

analkysis 😩 its been a year haha

[frosty crescent]

ofc check well definedness lol

[cerulean ruin]

hmm

[frosty crescent]

you can also claim that there is an injection from that set you described above to f : Q -> R i think

[cerulean ruin]

ye

i was thining of doing solemthing like that

[frosty crescent]

actually idk about that

[cerulean ruin]

if i have a way of uniquely identifying a increasing function by a set of rational numbers

[frosty crescent]

but the construction above should give some inspiration

[cerulean ruin]

then i can make an injection right

ic

[frosty crescent]

yeah

[cerulean ruin]

so why cant we just do the same thing as continuous functions, just the only difference we have to look at left and right limits

for some rational x

[frosty crescent]

yeah i think the construction i gave above admits the left limits but not the right

[cerulean ruin]

ohh

but i guess you would do something very similar to right limits

[frosty crescent]

yeah probably

[cerulean ruin]

find a lower bound?

ykk what this is my problem set

imma just bs that

🤣

then for the lower bound

i csn just use the set of linear functions

which is pretty straight forwards c

[frosty crescent]

actually show that the construction above probably uniquely identifies strictly increasing functions

[cerulean ruin]

this one?

[frosty crescent]

ye

that is, restricting any strictly increasing function f : R -> R to Q gives a function with the "left limit" f

idk if this is true tho

[cerulean ruin]

ah

my brain hurts

😔

ill have to read into this a bit

🤣

[frosty crescent]

actually it may be false

take a function defined by

$f(x)=\begin{cases}x&x<\sqrt2\x+1&x\ge\sqrt2\end{cases}$

[woven radish]

nAdat12

[frosty crescent]

since sqrt2 is irrational the restriction may not admit f as a left limit idk

[cerulean ruin]

hmmm

alr

😭

maye

mmaybe ngl

[frosty crescent]

actually they're kinda close enough in a sense

[cerulean ruin]

i tink so

[frosty crescent]

maybe mod out by "admit the same left limits"?

the hard part is that this may not have the same cardinality hmm

[cerulean ruin]

i think i am cooked haha

next i have to do some sort of induction regarding beth umbers

[frosty crescent]

rah

[cerulean ruin]

🗿

.sovled

[gloomy aurora]

its actually .close or .solved

[cerulean ruin]

let me just close this ig

i gave up on that problem since i dont have to provide a strict proof

so maybe providing intuition is enough

.close

[devout snow]

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[devout snow]

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[sleek gull]

Guys I want to know that how we k> sinx at last step was written k>1

[potent dirge]

Do you know what it means for a function to be increasing?

[sleek gull]

Yes it’s tangent should be >1

That’s mean the derivative of the function

Right?

[potent dirge]

No. But you’re very close

f’(x) > 0

The derivative just has to be non-negative

[sleek gull]

Isn’t it what I said

Tangent and derivative aren’t the slope of the f(x)

[potent dirge]

You said the derivative had to be grater than 1. Which, is technically true. A derivative greater than 1 would be increasing. But we say for a function to be increasing, its derivative has to be greater than 0

So, what you do, is you take you’re derivative, and create the inequality

If you start with

“kx + cos(x”

Then you’d get:

k - sin(x) > 0

Were you able to get to that step ok?

[sleek gull]

Yes

[sleek gull]

Then how they get k>1 from k-sinx >0

[potent dirge]

For that, we’re gonna do what’s called a sign analysis test

We’re gonna first figure out where it’s equal to 0, and then plug numbers on either side, to check if they’re positive, or negative

It’ll make more sense once we do it

[sleek gull]

Like this ?

[frank musk]

Umm ig he wanna know smthing else

[sleek gull]

Then we will put both the values and check which one holds true ?

Ig I am wrong

[frank musk]

Umm so k is greater than -1 and 1 at same time so....don't you get it ??

[potent dirge]

Ok, bro. That tones a little bit harsh

[sleek gull]

It means we take intersection

[frank musk]

Mb my english weak

[sleek gull]

That will give k>1 .

[frank musk]

Ye that what I mean smthing like that

[potent dirge]

Well, that idea is basically saying, what is the max of sin(x)

[frank musk]

Yeee

[sleek gull]

Okk

[potent dirge]

What is the biggest y value sin can give you?

[sleek gull]

1

[potent dirge]

So, in order for you to be greater than sin(x) always…..

What do you have to be bigger than to guarantee that?

K would have it be greater than 1 to be greater than sin(x) everywhere

[sleek gull]

Oo

Now it’s clear

Thanks

.close

[devout snow]

Channel closed

Closed by @sleek gull

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

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[sleek gull]

How the rate is 7.5 not 2.5km/h

[sleek gull]

We clearly see that ds/dt is 2.5km/h then why we add dl/dt

[summer summit]

the question is asking for the speed of the tip of the shadow, not just the shadow length

and the tip is at distance l + s

[sleek gull]

I guess we are adding it because as the man moves the shadow also moves

Isn’t it

[summer summit]

yes exactly

[sleek gull]

Ok

[summer summit]

because when the man moves, the shadow length grows and the tip moves

[sleek gull]

Yeah

I got it

Thanks

.close

[devout snow]

Channel closed

Closed by @sleek gull

Use .reopen if this was a mistake.

[summer summit]

of course!

[sleek gull]

Bruh can you tell me what is this ds/dt here

Like if the question wanted to ask ds/dt then what It would have asked

.reopen

[devout snow]

✅ Original question: #help-27 message

[sleek gull]

@summer summit

[versed glacier]

differenciation of s with respect to time

[shy osprey]

ds/dt idhar is the speed at which the shadow is moving

[sleek gull]

Og name btw

[shy osprey]

as in the rate of change of length of the shadow

[sleek gull]

So the rate of change of length of shadow and the rate of change of tip of the shadow aren’t the same

[versed glacier]

to summarize, It's the instantaneous velocity

[shy osprey]

The qsn would have been asking at what rate does the length of the shadow change

[shy osprey]

no

rate of change of tip of shadow is d(l+s) / dt

[sleek gull]

Ok

[devout snow]

@sleek gull Has your question been resolved?

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

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[amber bone]

is every even function with co-domain of real numbers an into function?
and is this opposite for odd functions or there is no correlation at all?
if not what are some examples ?

[gloomy aurora]

what's an into function?

did you mean onto?

[solid perch]

is "into" just another term for injective?

wait no, an into function is apparently a function whose codomain contains at least one element not mapped to by the function?

[amber bone]

no into as in if a functions codomain is not equal to range

[uncut crow]

[gloomy aurora]

I'd love if you could use the term non-surjective, because I doubt that the terms 'onto' and 'into' are used in higher contexts.

[amber bone]

oh ok i have no idea as i am learning these terms in school

[solid perch]

that's fine. let's see.

what's your definition of evenness of a function?

[gloomy aurora]

consider

[amber bone]

f(-x)=f(x)

[gloomy aurora]

$$f(x) =
\begin{cases}
\ln(x) & \text{if } x > 0 \
0 & \text{if } x = 0 \
\ln(-x) & \text{if } x < 0
\end{cases}$$

[solid perch]

cool. now what's a function you can come up with that is even?

keep it simple.

[woven radish]

Annie Maqionde

[amber bone]

mod?

[solid perch]

okay, let's use the absolute value function f(x) = |x|.

[amber bone]

yes

[solid perch]

is this non-surjective?

[gloomy aurora]

for your reference, this function is surjective, but its also even.

[amber bone]

yes

[solid perch]

there you have a counterexample for your first question.

[amber bone]

no

im sorry

[solid perch]

hm?

[amber bone]

but i asked if all

is every even function with co-domain of real numbers an into function?

[gloomy aurora]

wait wasnt his question all even functions are non-surjective

[amber bone]

here into

as

in

non surjective

[amber bone]

yes

[gloomy aurora]

that's false; i provided an example above

usually for all the functions you'll encounter at school it is true.

[solid perch]

oops, my bad.

[amber bone]

this example?

[solid perch]

but I can produce another CE either way (f(x) = x^2 sin(x^2)), so the conclusion still stands.

[gloomy aurora]

it's not true generally, but at your level of education it may be accepted to be true

[sand quarry]

Isnt that always dependent on how you define your codomain

[solid perch]

codomain in OP's question is R.

[sand quarry]

Ok

[solid perch]

so now onto your second question.

[gloomy aurora]

Yes.

[solid perch]

what's your definition of an odd function?

[amber bone]

f(-x)=-f(x)

[amber bone]

is this function continuous?

[solid perch]

fantastic. now take a look at f(x) = sin(x) and tell me if it is odd, and if it is surjective.

[gloomy aurora]

hint: consider trigonometric functions such as arctan, or tanh, or sin, cos, etc. they're odd. are they surjective?

[gloomy aurora]

no

[solid perch]

your question never needed the function to be continuous.

[gloomy aurora]

by functions, we don't imply continuity. As I've mentioned earlier, at your level of education, we may the first property holds.

[amber bone]

yes it is odd and but surjective on [-1,1] right?

[gloomy aurora]

your codomain as you mentioned is R.

[solid perch]

your codomain is R.

[gloomy aurora]

if not R, then you can always construct a codomain such that the function is surjective in it(usually, again)

[solid perch]

the range of sin is [-1, 1], yes. but again, your codomain is R.

[gloomy aurora]

So the first property holds usually, the second, not, if your codomain is R.

[amber bone]

so it will be non surjective?

[solid perch]

well, does sin(x) hit any value outside [-1, 1]?

[amber bone]

no

[solid perch]

there you have it.

[amber bone]

ohh

thank you sm it helped a lot

.close

[devout snow]

Channel closed

Closed by @amber bone

Use .reopen if this was a mistake.

[inner thistle]

F_yeet = ((m * a + kx) * J^2) / R

. open

how to open

i want a pic for the equation

how to open

[solid perch]

if you only want a picture for the equation, use #latex-testing.

[inner thistle]

ok ty

wont work

[solid perch]

then ask in #latex-help.

[devout snow]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[dim perch]

Prove that if $N$ is a normal subgroup of $G$ then $n_p(G/N) \leq n_p(G)$.

[woven radish]

blanketism

[dim perch]

i might just be tired, but im lowkey kinda tweakin

preliminary ideas are to take a mapping from $\text{Syl}_p(G) \to \text{Syl}_p(G/N)$ but other than that, idek

[woven radish]

blanketism

[dim perch]

or maybe the natural projection would do something here

only reason id like to work with this mapping is that showing surjectivity would be what i want

[cerulean ruin]

my goat 😔 hope someone can help u soon

[dim perch]

actually wait, i think i got it. if pi is the natural proj homomorphism, need to show |G/N : pi(P)| is coprime to p. |G/N : PN/N| = |G : PN|.

|G:P| = |G:PN||PN:P| so that |G:PN| divides |G:P| which is then coprime to p, hence pi(P) is in syl_p(G/N) hence map is well defined.

then surjectivity is sort of trivial right? for some $\bar Q \in Syl_p(G/N)$, then there exists $\bar g \in G/N$ such that

[woven radish]

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

[dim perch]

oops

Let $\pi$ denote the natural projection homomorphism. We need to show that $|G/N : \pi(P)|$ is coprime to $p$. Using the Second Isomorphism Theorem, we see that
\[|G/N : PN/N| = |G : PN|\]
Since $P \leq PN \leq G$, then $|G :PN|$ divides $|G:P|$ which is coprime to $p$, hence $\pi(P) \in \text{Syl}_p(G/N)$.

To show that $\phi$ is surjective, let $\bar Q \in \text{Syl}_p(G/N)$. By definition, there exists some $\bar g \in G/N$ such that
\[\bar Q = \bar g\bar P \bar g^{-1} = \pi(gPg^{-1})\]
By definition, $P \in \text{Syl}_p(G)$ implies taht $gPg^{-1} \in \text{Syl}_p(G)$, hence $\phi$ is surjective. The result follows.

?

[woven radish]

blanketism

[dim perch]

aight nvm i got confirmation, we chillin

ty

.solved

[devout snow]

Channel closed

Closed by @dim perch

Use .reopen if this was a mistake.

[devout snow]

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