help-36

[final saddle]

@left trail Has your question been resolved?

[left trail]

Ok so I choose the u of log (t-1)

Now the integral that I obtain is $\int_{\log(c-1)}^{\log(x-1)} \frac{e^{2e^u+2}e^u}{u}du$

[soft zealot]

BigBen

[scarlet sequoia]

wait so u = log(t-1)

[left trail]

Yes

It seemed to be the only way to have t in the denominator

Oh wait

I've been confusing myself

Let me work on the sub again then

The options seem to be u =t-1 and u= 1/t-1 but both don't seem to lead anywhere

[scarlet sequoia]

why not the first one

[left trail]

Well our bounds are off and the integrand is off

[scarlet sequoia]

don't worry about the bounds yet

let's just focus on integrand

what is "off" about the integrand?

[left trail]

Well to match c) we want e^u. We on the other hand have e^2u+2

[scarlet sequoia]

ok

well, first of all, e^(2u+2) = e^(2u)e^2

so, up to a constant, it's just e^(2u)/u, right

[left trail]

You took the constant out?

[scarlet sequoia]

yes, put it in front of the integral

it's a multiplicative constant

[left trail]

Ye I see that but I didn't go past that since c didn't have that

[scarlet sequoia]

ok

anyways

$e^2\int_{?}^{?}\frac{e^{2u}}{u}du$

[soft zealot]

Rafilouyear2026

[scarlet sequoia]

what do we need to do now?

to get e^t/t

[left trail]

Let me see what t=2u leads to

Also why question marks for the bounds? We know the bounda

[scarlet sequoia]

well Idk you didn't give them to me after substitution u = t-1, so I'm leaving it to you to fill them in

[left trail]

t=2u will get us the correct integrand

Oh ok. There just c-1 and x-1

So our integral right now is $e^2\int_{\log(2)}^{2x-2} \frac{ e^t}{t} dt$

[scarlet sequoia]

not accounting for constants in front, yes

[left trail]

Oh wait right

[soft zealot]

BigBen

[left trail]

Hmm. I just need to fix the upper bound now. But I cant make a sub since that would ruin our integrand. Does that mean that e^2 will play a role in correcting our upper bound?

[scarlet sequoia]

no

you know exactly what $\int_{\log 2}^{\log x}\frac{e^t}{t}dt$ is for ALL x

[soft zealot]

Rafilouyear2026

[scarlet sequoia]

or rather, to not confuse with this x here

you know exactly what $\int_{\log 2}^{\log X}\frac{e^t}{t}dt$ is for ALL $X$

[soft zealot]

Rafilouyear2026

[scarlet sequoia]

so...

we just need to find a correct X

such that the upper bound, 2x-2

becomes log(X)

[left trail]

Oh. e^2x-2

Wait

Oh just algebra. x= (2+logx)/2

But who says we are allowed to choose our x in such a fashion?

[scarlet sequoia]

huh

x and X are different

you had it earlier

log(X) = 2x-2

so X = e^(2x-2)

[scarlet sequoia]

I specifically called it X so you wouldn't confuse it with x

[left trail]

Why could I not just say this?

Aren't both of them choosing a specific x and X

[scarlet sequoia]

uh, no?

The small x you're working with has been arbitrarily fixed at the beginning

so you can't change it in any kind

[left trail]

Isn't the same with big X?

[scarlet sequoia]

No because 28c has already been proved

so you know, that for ANY x >= 2, that the integral [...] equals Li(x)

so you're allowed to plug in any value you want

you can change the name of the variable if you want to

so you know, that for ANY s >= 2, that the integral [...] equals Li(s)

now, you can apply it to a specific value, for example s = e^(2x-2), with the fixed x from question d

and would you look at that

[scarlet sequoia]

this is now e^2 Li(s), with s = ...

[left trail]

Ok I see c) is for any X and we just choose one. The one we choose has another x that is chosen but not given explicitly.

Thank you for the help

.solved

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[patent steeple]

How was a≤c+3 derived?

[gritty chasm]

$\mathrm{gcd}(b,c) \le c$ right?

[patent steeple]

Yes

[soft zealot]

Annie Maqionde

[patent steeple]

Ok I get it

[gritty chasm]

its given $a = \mathrm{gcd}(b,c) + 3$

[soft zealot]

Annie Maqionde

[patent steeple]

Yes

[gritty chasm]

from there you get it

[patent steeple]

Thank you

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[sour vale]

Can someone help me with this question. It is related to Arcs

[tropic crest]

Hint : ||try finding the radius with the pythagorean theorem||

[sour vale]

square root 8^2 + 5^2 = 9.43

r=9.43

[tropic crest]

Cool.

Now try finding $\angle BOQ$

[soft zealot]

multiplexer

[sour vale]

Thrugh sin rule right

32 degrees

-180

148 degrees

[tropic crest]

Hold up you meant BOA = 32°?

[sour vale]

BOC = 32 degrees

and hten i subtracted 180 to find BOQ right?

So BOQ = 148 degrees?

and then i can finmd the arc legnth of the whole PQ which is 29.63

and find the arc length of BQ now becuase i have the radius and degree

and i subtracted the arc length of BQ with PQ to find PB

[tropic crest]

ehh not really, i was thinking BOQ = BOA + AOQ

[sour vale]

hm im kidna confused

after i fidn the radius

I find the angle of BOQ right

then after i fidn the angle of BOQ i find the arc length og BQ

and subtract that with hte arc length of PQ

which will result in just the length oif PB

[tropic crest]

wait

you misread

you just need to find BQ

[sour vale]

ohh nvm thank you thank you

i understand it now

[tropic crest]

np

[sour vale]

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[coral zenith]

<@&268886789983436800>

[tropic crest]

<@&268886789983436800>

[tropic crest]

sniped /j

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[worldly spruce]

[strange pelican]

Im better

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[drowsy epoch]

skully

vryh

[soft zealot]

Renato

[drowsy epoch]

What have you tried

[gentle zephyr]

I proved right direction

need help with left direction

[spring haven]

The left direction is easy, just set B=Y

[gentle zephyr]

Proof: f(f^-1(B)) = B <=> f surjective

Proof1: f surjective => f(f^-1(B)) = B

(subseteq)
y in f(f^-1(B))
=> there exists some x in f^-1(B) such that y = f(x)
=> there exists some x in X such that y in B
(supseteq)
y in B
=> y in Y
Because f : X -> Y surjective
DEF: forall y in Y, there exists some x in X such that f(x) = y

=> there exists x in X such that y = f(x) in B
=> there exists some x in X such that f(x) in B
=> x in f^-1(B)
=>y = f(x)
=> y in f(f^-1(B))

[blissful meadow]

That direction seems good. How far along are you with the other direction considering what kheer mentioned?

[gentle zephyr]

first of all, can we modify this direction to make it less messy?

I did this at like 5am and now that I see it I didnt explained anything and just wrote what I had on my mind

please dont hold anything back, this proof sucks

[blissful meadow]

Hmmmm the argument will remain mostly unchanged. Your proof isn't bad. Maybe you can afford to jump a few steps?\

Like if $y\in f(f^{-1}(B))$, then there is $x \in f^{-1}(B)$ such that $y = f(x)$. From this you instantly have $y \in B$ since $x\in f^{-1}(B)$.\

[soft zealot]

Azyrashacorki

[blissful meadow]

At the end of the day it's mostly just writing your proof more in words rather than a series of implications.

[gentle zephyr]

yeah exactly this is better

[blissful meadow]

(In fact this direction holds regardless of whether f is surjective or not. If f^-1(B) is empty then f(f^-1(B)) is empty, so it's a subset of B.)

[gentle zephyr]

subseteq holds regardless of surjectivity of f but supseteq aswell you say?

[blissful meadow]

Oh no sorry I meant like the first inclusion $f(f^{-1}(B)) \subseteq B$.

[soft zealot]

Azyrashacorki

[gentle zephyr]

because in supseteq i did use that f is surjective

[blissful meadow]

Yeah that's what I mean

[gentle zephyr]

dont know how to start

[blissful meadow]

For the other direction you mean?

[gentle zephyr]

for the direction f surjective <= f(f^-1(B))

[blissful meadow]

Okok

Well as Kheer suggested, since f(f^-1(B)) = B holds for all subsets B of Y, try to see what happens if you take B = Y.

This gives you some equality between sets, so you may use it to show that f is surjective.

[gentle zephyr]

B = Y is just one case of when B subseteq Y

it does not prove the general case

[blissful meadow]

Your claim is $(\forall B \subseteq Y. f(f^{-1}(B)) = B.) \implies (f \text{ is surjective}.)$

[soft zealot]

Azyrashacorki

[blissful meadow]

So in your premise you're assuming that $f(f^{-1}(B)) = B$ for any $B \subseteq Y$.

[soft zealot]

Azyrashacorki

[blissful meadow]

This says something useful for the case B = Y, which you can use to show surjectivity.

So try and see what your premise says about the case B=Y and then try to find a preimage for any y in Y.

[gentle zephyr]

again, we are not proving the general case of forall B subseteq Y when we pick B = Y

you are saying without loss of generality that if we pick B = Y the logic still holds forall B subseteq Y?

[blissful meadow]

Your premise says that f(f^-1(B)) = B for any B you want. You know this to be true.

So in particular f(f^-1(B)) = B when B=Y.

There is no "without loss of generality" implicit in there.

[gentle zephyr]

we are losing generality though

what if B is a proper subset of Y?

[blissful meadow]

The premise is that f(f^-1(B)) = B for ANY subset of Y.

This necessarily implies that f(f^-1(Y)) = Y.

This would be true for any subset of Y, but using this you can prove that f is surjective.

[gentle zephyr]

yes but if we manage to prove that f(f^-1(Y)) subseteq Y implies that f is surjective

we might still encounter some B subset Y such that B ≠ Y and that f(f^-1(B)) = B but f is not surjective

[blissful meadow]

But you're not proving that f(f^-1(Y)) = Y. It's part of your premise.

You're assuming that f(f^-1(B)) = B for any subset B of Y, and you want to show that this means that f is surjective.

[gentle zephyr]

I dont think I follow

Say for example, we are given that the statement holds for primes p, with forall a,b in N

p | ab => (p | a or p | b)

we are not proving that this is true by showing that p = 2 works

AH, I see it now, we are not trying to prove this implication, we are trying to prove that assuming this implication holds then f is surjective

[blissful meadow]

f(f^-1(Y)) = Y is just a direct result of your premise essentially

Then from this you can show that f is surjective.

[gentle zephyr]

forall y in Y, there exists some x in X such that f(x) in Y?

[blissful meadow]

Well from $f(f^{-1}(Y)) = Y$, you can pick $y \in Y$. Then $y \in f(f^{-1}(Y))$. If you expand what it means for $y$ to be in $f(f^{-1}(Y))$, you should be able to find $x \in X$ such that $f(x) = y$.

[soft zealot]

Azyrashacorki

[gentle zephyr]

forall y in Y, there exists some x in X such that f(x) in Y. In particular there exists some x in X such that y = f(x), then y in f(f^-1(Y))

[gentle zephyr]

what are we doing here?

proving that f is surjective, that is, forall y in Y there exists some x in X such that y = f(x)

[gentle zephyr]

how easy from 1 to 10 is this exercise?

I have become quite acquainted with preimages and images of sets as of now, I think I can recite the definition from my memory almost instantly

[gentle zephyr]

this might look like a shocker but if you see the definitions, you see this is understandable

however the real challenge is using wording for the proof and omitting details that are not explictitly needed in the proof, such as making the proofs self containable as possible

[blissful meadow]

I'd say it's on the easier side in the sense that if you unravel the definitions of preimages and direct images, it follows pretty much directly; you don't need to come up with a trick or something.

[gentle zephyr]

thanks

i will take a break perhaps now

.solved

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[chilly raven]

hwk to simplify

[south dirge]

does a have any value or js that?

[vital crag]

,rotate

[chilly raven]

no

wait

new question

24

[chilly raven]

24

[polar spruce]

what have you tried

[chilly raven]

?

that a and y part is not good

[chilly raven]

.rccw

[chilly raven]

,rccw

[chilly raven]

ma progress

oye @south dirge

[south dirge]

What did you do after x=1??

[chilly raven]

that's one asnwer

[chilly raven]

i took out log x

[south dirge]

no i mean after x=1

[chilly raven]

log x =0

x =1

[south dirge]

i got tht bhai mera mtlb x=1 likhne ke baad kya kiya

[chilly raven]

log x =y log 2=a

[chilly raven]

phir ye aaya

glt hai shayad

[south dirge]

yeah

rhs mein 1 kahan se aagya

[chilly raven]

woh pura 0 hai start ka question

the log x (...)=0

[south dirge]

Okay

logx kat diya

uske baad 1 kahan se aagya

[chilly raven]

ha

[south dirge]

udhar to 0 hi rahega na

[chilly raven]

log x =0
10^0=1

[south dirge]

yeah but log(x*...) nahi hai log(x) * .... hai

[chilly raven]

suppose base 10 hai toh log hata ne keliye 10^0 hoga na
aur 2/(logx/2 )...

[south dirge]

Abhi log ko bhul jao

[chilly raven]

ok

[south dirge]

tell me smth a (b +c) = 0

if a isnt 0 then b +c =0 bolna theek hai kya.

[chilly raven]

a=0
b+c=0

[chilly raven]

ha

[south dirge]

Right

same cheez idhar

log(x) * (qwerty) = 0

[chilly raven]

oo

ok

[south dirge]

if log(x) isnt 0 then qwerty =0

[chilly raven]

lekin phir jo maine likha hai usme galat kya hua

[south dirge]

aapne qwerty =1 likh diya

[chilly raven]

oo

ok

ok

.close

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[chilly raven]

ty

[south dirge]

Your wlcm!
Have a nice day!!

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[kindred hare]

loga​b=loga/logb​

[final saddle]

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.

[kindred hare]

this is original problem. ArsH, right?

[blissful meadow]

They've closed the channel 2 hours ago and you haven't pinged them either way. If you don't need this channel for help, please close it with .close.

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@kindred hare Has your question been resolved?

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[sterile pebble]

ya so I dont know how to get theta value

isnt it sqrt(x^2+y^2) for magnitude?

and arctan y/x for direction

[fringe stone]

magnitude of the resultant would be sqrt((∑x)^2+(∑y)^2)

∑x would be adding the x component of the forces

so ∑x = Gcosa + F

∑y would be adding the y component of the forces

∑y = Gsina

and the angle theta = arctan(∑y/∑x)

[sterile pebble]

what

I do not understand these components

[fringe stone]

every force has an x and y component

like a diagonal line can be broken up into a horizontal and vertical component

in which if you use pythag on these 2 components you get that force

[sterile pebble]

so g and f give us the force of the pink line

but why is x Gcosa + F

<@&286206848099549185>

[grizzled atlas]

How i may help you?

[sterile pebble]

im wondering as how to find ∑x and ∑y

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[short vale]

can someone help me with linear and exponential handmodeling

[barren pebble]

Do you have a specific problem in mind?

[short vale]

YES

this is all it says for the assignment "linear hand-modeling: justify how to obtain initial value of dataset, slope calculation from 2 points, define all variables and numbers with units;
exponential hand-modeling: justify how to obtain initial value of dataset, relative change per year calculation from 2 points, define all variables and numbers with units;"

heres the data and i need to do it by hand

[short vale]

please asian guy help a wasian out

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@short vale Has your question been resolved?

[short vale]

help

[final saddle]

@short vale Has your question been resolved?

[compact laurel]

Joy isn’t happy rn?

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@short vale Has your question been resolved?

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[dense arch]

“Let P be a point inside a circle. Three distinct chords AB, CD, and EF pass through P. If AB=CD=EF, prove that P is the center of the circle.”

[timber leaf]

Let's draw it out

[dense arch]

i actually drew it out and i was trying to use power of a point to solve it

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@dense arch Has your question been resolved?

[charred agate]

I am not good at proof writing..

[timber leaf]

Then why do you state this one here? Is it relevant?

[charred agate]

I dont know, could be while discussing the approaches

[timber leaf]

If you don't know then don't say random stuffs here

[charred agate]

My bad

[timber leaf]

Chords of equal length are equidistant from the center, you can use this definition

[dense arch]

tysm!!

[charred agate]

!done

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[tall spruce]

can someone help me

[tall spruce]

what is 2+2=

[fringe stone]

4

[tall spruce]

TYSM

[gritty chasm]

is this spam

[tall spruce]

im so happy if someone says answer

[grave trout]

It probably is but its a gd reference

[gritty chasm]

not everyone can do that lmao

.close

[final saddle]

Channel closed

Closed by @gritty chasm

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[fringe stone]

shhh

[charred agate]

I think so, he asked me some very silly & standard stuff

[tall spruce]

say answer

[tall spruce]

alr forget it

GNG WHAT IS ANSWER I WANNA GET FREE WIFI

[gritty chasm]

<@&268886789983436800>

[tall spruce]

wym u mean?😭

[gritty chasm]

Based on past interactions, I have drawn a conclusion.

[tall spruce]

sorry

[gritty chasm]

hm mods not replying

[stray hill]

Come back tomorrow and dont troll help channels. You've done this in other channels too.

[crude rampart]

hahaha

this is a very interesting problem actually

once you see the trick, it’s actually quite beautiful, in my opinion

[stray hill]

.close

hmm

[gritty chasm]

its closed

[stray hill]

oh already done

nice, ty

[final saddle]

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[molten zinc]

guyss solve - Find the domain of the function:- |cos-1(x-4)| + 1/log(3-x)

[brazen breach]

is that an arccosine

[molten zinc]

its cos inverse

[crude rampart]

$cos^{-1}(x-4) + \frac{1}{\log{(3-x)}}$

[soft zealot]

9k

[crude rampart]

is this it?

just making sure i read the function correctly

[molten zinc]

yess

[brazen breach]

well for the 1/log(3-x), x<3 since the inside has to be positive
cos^-1 can only evaluate for when the inside is between -1 and 1 so
-1__<x-4<1 => 5>x>__3
hence there is no domain or whatever it's called

no x value satisfies the solution

no real x-value satisfies the equation

or gives a real solution idk how to word it

[molten zinc]

well bro i asked for domain not the range of the function

[crude rampart]

what is the domain of arccos?

[molten zinc]

its 1/(1+x^2)

i think

[brazen breach]

no x value satisfies the equation, it is literally x epsilon ()

{}

wait

[crude rampart]

no, i think you’re thinking of the derivative , also, that’s not the derivative of arccos assuming you are thinking of derivatives

can you tell me what a domain is?

[molten zinc]

{-1,1} i think

[brazen breach]

x!ϵ{R}
since x!ϵ{R}, then y!ϵ{R}
! indicating the opposite/negation of ϵ or just not a part of the set of real numbers or whatever i was

[crimson pike]

I feel like it'd just be the range/codomain of cos no?

[brazen breach]

that isn't even the derivative of arccos it's the derivative of arctan

[molten zinc]

yes its of tan

[brazen breach]

tell me, why would you need an equation of x to find the domain of a function of x already previously defined

am I saying this right I can't tell

[crude rampart]

yes, but you must restrict the domain of cosine.

[molten zinc]

umm i cant understand what you are saying

and all function have their own domain

i ve got a similar function of which i have already found the domain

[zinc verge]

did you got enough help

[molten zinc]

i think so

i am just trying on this one

[crude rampart]

@molten zinc we should start with the basics. what is a domain? can you define it?

[brazen breach]

what is 1/(1+x^2) telling you about x when you have a function |arccos(x-4)| + 1/log(3-x) to find the domain of?

[molten zinc]

f ( x ) = cos − 1 ( 2 − | x | 4 ) + ( log e ( 3 − x ) ) like i got the domain of this function

[molten zinc]

no i wass just trying

[ − 6 , 6 ] this was the answer i got for this one

[brazen breach]

is that |x| to the power of 4 of |x| times 4 because neither are the domain of f(x) you're describing

for x__>__3, the inside of the logarithm would be nonpositive hence invalid since the inside of logarithms no matter the base have to be positive

[molten zinc]

it was for the cos part

[brazen breach]

can you show a pic of the question?

[molten zinc]

this is the finsl answer [ − 6 , 3 ) − { 2 }

f ( x ) = cos − 1 (( 2 − | x |)/ 4 ) + ( log e ( 3 − x ) ) like i got the domain of this function

[brazen breach]

you forgot something

[molten zinc]

this is the full question that i have solved and got the answer and the first question is my doubt

[brazen breach]

|x| is the absolute value of x

[zinc verge]

seems like a jee question

[brazen breach]

there are 2 possible values for |x| for it to be equal to 2

-2 and 2

[zinc gazelle]

please help 😭

[molten zinc]

yeah mains 2024

[zinc gazelle]

😭

😭😭😭😭😭😭😭

[zinc verge]

i am also preparing for jee

let see

[molten zinc]

ohh nicee

but this question is not my doubt

[gritty chasm]

what is your doubt then?

[zinc verge]

then??

[molten zinc]

my doubt is :- Find the domain of the function:- |cos-1(x-4)| + 1/log(3-x)

this is similar to that one

[zinc verge]

ohh

[gritty chasm]

well what have you tried

[molten zinc]

a lot and also diffrentiation

[gritty chasm]

Find the domain of the function:- $|\cos^{-1}(x-4)| + \frac{1}{\log(3-x)}$

[gritty chasm]

how in the world does differentiattion help here?

[soft zealot]

Annie Maqionde

[gritty chasm]

and !show

!show

[final saddle]

Show your work, and if possible, explain where you are stuck.

[molten zinc]

yess

thats the question

[molten zinc]

umm maybe u wont be able to understand my approch because i am doing it in rough

[gritty chasm]

upload it anyways

[zinc verge]

indeed i can

[gritty chasm]

and explain it as well

[zinc verge]

theres no solution

[molten zinc]

maybe you are right

alight then i will close

[gritty chasm]

Please do so.

[molten zinc]

.close

[final saddle]

Channel closed

Closed by @molten zinc

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[final saddle]

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[rain patrol]

Hi everyone, I'd like to ask about the significance of Green's functions in the context of solving PDEs.

Here's what I understand, roughly speaking. An example is a Poisson problem $\nabla^2 \phi = f$ (I guess some people put a minus sign but I don't mind). To determine the solution for an arbitrary $f$, we use the Green's function to first solve it for $f = \delta (x - x_0)$ and then integrate the result times the real f over all the $x_0$.

However, I've also seen it used in the context of the heat equation to derive the heat kernel where we are setting not the inhomogeneous term but the initial condition to a Dirac delta. So I guess my question is, where does it go, and how does my Green's function give me the solution?

[soft zealot]

Cark Marney 🐢

[final saddle]

@rain patrol Has your question been resolved?

[final saddle]

@rain patrol Has your question been resolved?

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[ancient mural]

Is the question wrong cuz i dont think its solvable with 'different +ve int' (it is a memory based PYQ so.. high chances)

I Started with begger's method to get a starting point and then started making cases for 2 same, 3 same.. but that just didnt feel right

[gritty chasm]

why not?

[royal gust]

You normally solve these with stars and bars

[ancient mural]

whats that?? i applied beggers method.. thats what i was taught

[gritty chasm]

search it up, will you?

[ancient mural]

aight

[gritty chasm]

khan academy or libretext math

[royal gust]

Maybe a different name for the same thing, I don't know beggars method

[vital crag]

TIL they are the same thing
https://kingseducation.in/mod/book/view.php?id=290&chapterid=57

[royal gust]

Yeah they are the same lol

[vital crag]

https://brilliant.org/wiki/integer-equations-star-and-bars/

[ancient mural]

they r the same things nvm

yea ok but

[royal gust]

Okay so you don't need cases. Consider a different variable u1 = x1 - 2

[ancient mural]

beggers method says nothing about Different +ve integers

[gritty chasm]

or there was this coefficient-in-a-certain-thingy way thing, for this problem i think i'd be coeff of $x^{100}$ in (x^2+x^3...............)(x^1+.......)^3$\\
hm but i dont expect you to know this method, and correct me if im wrong.

[soft zealot]

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

[ancient mural]

i have seen this..

wait lemme

[gritty chasm]

this is one of the most elegant ways if you know it

[ancient mural]

i have it in my notes but yk,,, never really saw it

[gritty chasm]

its important 💀

[royal gust]

That's a useful method if you have a max and min, but with just a min, there's a much easier way

[ancient mural]

no but from what i remember

all these never mentioned 'distinct integers case

[ancient mural]

I saw this in my notes.. it is

The non negative integral solutions of x + y + z = 15 is equal to coefficient of a^15 in the expansion of

$(\sum_{i=0}^{15} a^i)^3$

[soft zealot]

Chetan???

[ancient mural]

but this too doesnt mention distinct +ve solution

[final saddle]

@ancient mural Has your question been resolved?

[final saddle]

@ancient mural Has your question been resolved?

[final saddle]

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[sturdy badge]

I don't understand why im wrong

[blissful meadow]

You put parentheses within the parentheses.

[sturdy badge]

oh.

.close

[final saddle]

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[rough nimbus]

isnt an operation involving a row already being operated on invalid?

[royal gust]

This is valid

Note one is happening after the other. They don't happen "at the same time"

[final saddle]

@rough nimbus Has your question been resolved?

[heavy plaza]

the point of doing row operations is that they dont effect certain properties of the matrix (i.e. the null space) so any number of these operations is perfectly valid as long as the first one is

[trail mango]

[rough nimbus]

youre talking about the using the new value of R2 when operating R3 right? but R3 in next step is oly possible using the value of R2 before it was operated on

[rough nimbus]

can you elaborate a bit on that please?

[versed crater]

That’s fine

You can think of it as R3 -> R3 - (R2 - R1)

[rough nimbus]

but it doesnt equate to the elements we are getting in the det? (check R3 in 2nd step of the attached pic, it does not form after R3 -> R3 -(R2 - R1))

[versed crater]

ah sorry

just do the 2nd one first

do R3 -> R3 - R2

and then do the R2 -> ... line

[rough nimbus]

Oh yess we can also think that way about it

that makes sense, thanks for the help

.close

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[rustic sable]

wait hold on

.close

[final saddle]

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[rustic sable]

i saw it-

sorry

[rugged merlin]

fastest help channel

[rustic sable]

[drowsy epoch]

I need this power

[timber leaf]

https://tenor.com/view/peppa-pig-meme-peppa-pig-sigma-face-george-pig-pig-gif-5140080418325330501

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[left trail]

For his proof of theorem 7.2 how can he say that a,b,c are equations involving two polynomials of the same degree. For b and c the Taylor polynomials are not of the same degree

[final saddle]

@left trail Has your question been resolved?

[left trail]

<@&286206848099549185>

[rare girder]

I agree, they are not necessarily polynomials of the same degree, but they are polynomials of degree at most n, n-1, or n+1 respectively

[left trail]

Ok so what he said is just wrong

[rare girder]

take f(x) = x for instance lol

[left trail]

I'm not seeing the connection?

[rare girder]

let me make sure I understand, is $T_n(f)=\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k$?

[soft zealot]

Flip

[left trail]

Yes

[rare girder]

ok great, so each T_n(f) is at most an n-degree polynomial

[left trail]

Why at most? It is a n degree polynomial?

[rare girder]

because the nth term may be 0 itself

the nth term of the summation in T_n(f) involves taking the nth derivative of f

but if f is a polynomial of degree less than n (like a monic polynomial), then that nth derivative is 0

[left trail]

But isn't f always a polynomial of degree n

[rare girder]

f could be exp(-)

f(x) = e^x

[left trail]

But an exponential is not a polynomial

[rare girder]

right, thus f is not always a polynomial

maybe this is a semantics thing

[left trail]

Ok. But the taylor polyomial will always be of degree n if the nth derivate of f is not 0

[rare girder]

yes

[left trail]

Ok so going back to the question we have Taylor polynomials of n,n+1 and n-1. Clearly they are not the same

[rare girder]

(a) seems fair I think, right?

you're adding two <= n-degree polynomials, whose sum is another <= n-degree polynomial

[left trail]

Yes that is but the other two I cant see. And the way he states it makes it seem as though we are applying it for a,b,c. I was worried that I was missing soemthing

[rare girder]

the left-hand side of (b) is the derivative of an <= n-degree polynomial, which is an <= n-1 - degree polynomial

[left trail]

Is it really ≤n since we define it as strictly n

[left trail]

Ok so then I am correct in stating his statement doesn't hold for b and c

[rare girder]

the inequality only covers the instance where the functions have n'th derivative 0 at x=0 or x=a for whatever agreed-upon a

but like, even in statement (a), f = -g and c1 = c2 could be the case

[rare girder]

you might see what they're trying to go for if you tried to prove it yourself

[rare girder]

still defending the vibe of their claim, the degree in question may not be the same for each of the three statements

like if you're saying the claim is false because n-1 is not the same as n, then you could be missing the idea

in (a), both sides are degree at most n
in (b), " " at most n-1
in (c), " " at most n+1

in either of the three statements, both sides of their respective equation involve polynomial expressions whose degree is bounded by a common number

[left trail]

So your saying it is the same after you apply the integration operator and the differniation operator

[rare girder]

I think I would casually say that this theorem is asserting that T_n is an operator that plays well with addition, differentiation, and integration

[left trail]

Yes I understand that but could his statement be explained then on the left side we have the deirvative of a taylor polynomial of degree ≤n. This of degree ≤n+1 which is the same degree for the one on the right and likewise for the integration

[rare girder]

just to be clear, I don't think the proof in the textbook is satisfactory, it only indicates what the idea is

it's easier to just use the definition of T_n in all three statements

[left trail]

Ok so for the first proof I was thinking setting up the polynomial as a sum and simply using the properties of the sums

[rare girder]

yes

[left trail]

The other two proofs I don't know how to formulate rigorously

[rare girder]

just take (b) as an example

I don't want to do two things at once lol

[left trail]

So let's say for b. We have a deirvative of a sum. It can be shown by induction that this is equal to the sum of the deirvative

So then we are applying the deirvative to each term in the sum and by the power rule this should lower each term by 1. But this to me seems more like an explanation than a rigorous proof.

What could I do about that?

[rare girder]

I would address the right-hand side

the k'th term in the sum T_(n-1)(f') instructs us to take k derivatives of f'

and crucially this is the same as taking k+1 derivatives of f

that at least motivates the next step in your proof starting from the left-hand side

[left trail]

Why are you calling it the k'th term is it not the n-1 term?

[left trail]

I'm not seeing it

[rare girder]

I'm saying when you write T_n(f) as a sum

like you have
\begin{align*}
\frac{\mathrm d}{\mathrm dx}T_n(f)&=\frac{\mathrm d}{\mathrm dx}\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k\
&=\sum_{k=0}^n\frac{\mathrm d}{\mathrm dx}\frac{f^{(k)}(a)}{k!}(x-a)^k\
&=\ldots
\end{align*}

[soft zealot]

Flip

[left trail]

Yes I see this. And then we want this to be a sum of n-1 terms where the constant is the same * k(x-a)^k-1.

[rare girder]

it should actually be n terms, but you can write it as n+1 terms all the same

you just realize that the first term is 0 and remove it from consideration

$\sum_{k=0}^n\frac{\mathrm d}{\mathrm dx}\frac{f^{(k)}(a)}{k!}(x-a)^k=\frac{\mathrm d}{\mathrm dx}\left(f(a)\right)+\sum_{k=1}^n\frac{\mathrm d}{\mathrm dx}\frac{f^{(k)}(a)}{k!}(x-a)^k=\sum_{k=1}^n\frac{\mathrm d}{\mathrm dx}\frac{f^{(k)}(a)}{k!}(x-a)^k$

[soft zealot]

Flip

[rare girder]

because d/dx(f(a)) = 0

then you fiddle with indices

after taking your derivatives as you mentioned

[left trail]

I see you say j=k-1 and we obtain everything we want in j and simple say k=j

[rare girder]

I like to write reindexing like that in my notes as $k\mapsto k-1$ so that $1\mapsto 0$ and $n\mapsto n-1$

[soft zealot]

Flip

[rare girder]

you are completely right though

[left trail]

Wait for integration shouldn't our sum start from 1?. Since there will be no constant

[rare girder]

no, I'm pretty sure we still must faithfully apply the definitions where they go

[left trail]

We are going to have $\sum_{k=0}^{n+1} \int \frac{ f^{(k)}(a)}{k!} (x-a)^k$

[rare girder]

that seems fine

maybe I'm just not seeing the problem because I haven't gone through the motions myself

[left trail]

So each term will gain an (x-a)

[rare girder]

ah, which side is this meant to be?

this looks like T_n(g)

[left trail]

Oh so my sum is wrong

[rare girder]

at least if it's meant to be read as the left side

[soft zealot]

BigBen

[rare girder]

I feel like I keep seeing it wrong sorry

$T_{n+1}(g)=T_{n+1}\left(\int_0^xf(t)\mathrm dt\right)=\sum_{k=0}^{n+1}\frac1{k!}\left[\frac{\mathrm d^k}{\mathrm dx^k}\int_0^xf(t)\mathrm dt\right]_{x=a}(x-a)^k$

[left trail]

Ok so I take the integral out and I have it for T_n+1

[left trail]

Why are you taking the derivative?

[rare girder]

because $g^{(k)}(a)=\frac{\mathrm d^k}{\mathrm dx^k}(g(x))|_{x=a}$ with $g(x)=\int_0^xf(t)\mathrm dt$

[soft zealot]

Flip

Flip

[rare girder]

that looks a little better to me

[left trail]

Doesn't the differentian cancel the integral

[rare girder]

kinda, the first one does this but not when k=0

[left trail]

Ok makes sense. So for all but k= 0 since it is no derivative.

[final saddle]

@left trail Has your question been resolved?

[final saddle]

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[harsh stratus]

yo I'm stuck

[vital crag]

nah your antiderivative is wrong

$\int \frac{dx}{x} = \ln |x| + C$ yes but not $\int \frac{dx}{f(x)} = \ln |f(x)| + C$

[harsh stratus]

oh

[soft zealot]

riemann

[vital crag]

use exponent rules

,tex .exp rules

[soft zealot]

riemann

[vital crag]

then power rule for integrals

[unborn cosmos]

¿Buenas tardes quisiera pregunta,como se desarrolla exactamente los vectores cuando hablamos de fuerza de newtons?

[harsh stratus]

podemos draw vectors like this

➡️

[compact laurel]

Oye, amigo, tienes que crear tu propio canal

[plain talon]

Dirígete a #❓how-to-get-help para obtener ayuda; busca los canales llamados «help» que no tengan ningún nombre junto a ellos.

[unborn cosmos]

a es que soy nuevo gracias👍

[vital crag]

,w int 0 to 1 x/sqrt(x^2 + 1)

[soft zealot]

Results provided by WolframAlpha

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

Requested by rie.mann

[compact laurel]

No te preocupes hermano!! Todo bien 👊

[vital crag]

correct

[harsh stratus]

thank you so much

.close

[final saddle]

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[loud sundial]

<@&268886789983436800>

[final saddle]

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[acoustic wharf]

hi i have a problem:
(67f)^3 = c mod n
with f^3 < n
f is the unknown
to solve this i thought of applying the inverse moduli of 67^3 both sides:
which would give:
f^3 = cinv(67^3) mod n

[acoustic wharf]

and since f^3 < n

f^3 would be equal to c*inv(67^3)

but the issue is that last is not a perfect cube

where is the mistake in my reasoning

[grizzled tusk]

because you needed c to be a cube mod n (a cubic residue) in the first place

(67f)^3 = c mod n means c is equal to a cube mod n

then if d is a cubic root of c mod n, ie d^3 = c mod n
f is d*inv(67)

you can't tell much more than that if you don't give a n and a c

[final saddle]

@acoustic wharf Has your question been resolved?

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[gentle zephyr]

3. Prove that for all $n \in \mathbb{N}$ it holds that $36 \mid 19^n - 18n^2 - 1$.

[soft zealot]

Renato

[gentle zephyr]

6 | 36 => 6 | 19^n - 18n^2 - 1

19^n - 18n^2 - 1 = 1^n - 0 - 1 = 0 (mod 6)

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[turbid bramble]

I’m not sure if I got the graphing correct I also don’t know how to find the b value

[turbid bramble]

[tame cipher]

spam vs skill

[tame cipher]

wait

i think

2pie • b = period

[sterile condor]

Can u make it blurrier

[tame cipher]

LOLOLOL

[turbid bramble]

Just for u baby

[tame cipher]

SON😭😭😭😭😭😭😭😭😭😭

[turbid bramble]

No seriously I need help

How do I find the b value

[gritty chasm]

What is the question? what are you graphing?

[tame cipher]

Bro you can’t find ANY value on this 😭😭😭😭😭😭😭

[gritty chasm]

!original

[final saddle]

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.

[gritty chasm]

!redir

[final saddle]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[turbid bramble]

I’m graphing a cos

Idk the name

[gritty chasm]

Post the original question, please.

With full context

If it's not in english, translate

[gritty chasm]

so what you are graphing is $y = 2\cos \theta + 1$

[soft zealot]

Annie Maqionde

[tame cipher]

yes

[turbid bramble]

Yes

[gritty chasm]

If I were you I'd take 5 values of $\theta$ and find the corresponding values of $y$, make a table, then plot.

[soft zealot]

Annie Maqionde

[gritty chasm]

5 cuz it says so in the question

[tame cipher]

wait lemme try to make a table

[gritty chasm]

We will from now on let OP answer

Since this is not your help channel

Please don't interrupt

[turbid bramble]

We’re in the same class sorry

That’s why they’re here

[gritty chasm]

It doesn't matter.

[gritty chasm]

Required person will copy from the OP later.

[turbid bramble]

Thanks for the help

.close

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[warped herald]

how do I start

[gritty chasm]

Well I may give some hint on (ii)

Try to show something is congruent

hint semicircle subtends right angle

[warped herald]

as in b?

[gritty chasm]

yes. sorry

[warped herald]

so here

I joined OQ

so the angle at the centre is twice the angle on the circumference

[gritty chasm]

which qs? a, b or c?

[warped herald]

for b

[gritty chasm]

no try to show QRP and SRP are congruent

[warped herald]

how tho

radius and common side matches

[gritty chasm]

look at the angles

[warped herald]

so QR is the bisector

so angle PQO and SQO equal

meaning POR and SOR too

that’s it right?

[gritty chasm]

observe RQP=RSP = 90 degrees

[warped herald]

yea

[gritty chasm]

and QRP = SRP

and theres a common side RP

[warped herald]

yes

[warped herald]

howd you get that

because of the bisector?

[south dirge]

Yeah

[gritty chasm]

yes

[warped herald]

ok

[final saddle]

@warped herald Has your question been resolved?

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[flint ibex]

can someone explain

[flint ibex]

where the minus comes from

what i did was

sechy=x/2

differentiate with respect to x

dy/dx(sechy multiplied tanhy) = 1/2

[rocky tusk]

you’re missing a - and also the derivative of x/2 is not x/2

[sly forum]

respect to y

[flint ibex]

mb typo

[sly forum]

it is function of x in terms of y

[flint ibex]

but where is the minus coming from

[rocky tusk]

derivative of sechy

dy/dx (-sechy tanhy)

[flint ibex]

if u derive sec u get secxtanx

[rocky tusk]

this isn’t sec

it’s sech

[flint ibex]

i dont see why it doesnt stay the same cos of the hyperbolic function

[sly forum]

dont you have to implict diff if you are doing that?

[flint ibex]

i know abt the one rule

[rocky tusk]

???

[flint ibex]

if u have like sinh^2 x u put a negative

infront

[rocky tusk]

refer to the exponential definition

[flint ibex]

but like most the time it randomly doesnt apply

[rocky tusk]

sec and sech aren’t the same function

[flint ibex]

yh

but most the derivatives of the normal function are the same

as its hyperbolic

[rocky tusk]

instead of just assuming the same rules from trig work for hyperbolic just use the exponential definition and figure it out yourself

[rocky tusk]

and?

[flint ibex]

i dont have that much time in an exam

this is a past paper dw

but still

[rocky tusk]

you do

it’s quite simple

[sly forum]

or yea just do it once

and you will remember it everytime a question comes up

[flint ibex]

yh fair point

[flint ibex]

this acc very useful

thank you

but do any of you abt the one rule

where if u have sinh^2x u put a negative in front

[sly forum]

still prove for yourself where that negative comes from

[flint ibex]

its like why it goes from cos^2+sin^2=1 becomes cosh^2-sinh^2=1

[sly forum]

because it is hyperbolic functions

not the

regular

[final saddle]

@flint ibex Has your question been resolved?

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[dry path]

What do you call these things in the calculator? Referring to "x10²⁷." There are some cases where that's also a negative (e.g. x10^-03). What actually are they and how do they work?

Do they have a specific term or name?

[formal trail]

it's scientific notation

it's literally multiplying by 10 raised to that power

[dry path]

Ohhh okay then, thank you! Just wanted to make sure

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[proper lion]

<@&268886789983436800>

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[proper lion]

also who removed my true role

[tropic crest]

True and false roles can be added and removed at will by a senior mod

[proper lion]

😡

who did

[formal trail]

true and false roles generally only last 1 day

[proper lion]

https://tenor.com/view/ahhh-gif-9549451645697661034

[final saddle]

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[potent nova]

Hi How do you do this no calc?

[severe canyon]

Integrals without calc? 🤔

[sturdy cypress]

yeah, if you do an integral that was calc

[timber leaf]

It's indefinite integral anyways

[chilly pollen]

They mean calculator but its indefinite anyways

[quiet python]

They might mean without a calculator

[timber leaf]

Elderly won't get this slang

[sturdy cypress]

calculators can;t do this as whipl said

[quiet python]

Wolfram Alpha is a calculator that can solve it, im assuming there will be various other ones that can as well

[worldly spruce]

$\sin^2(x) + \cos^2(x) = 1$

[worldly spruce]

Use this

@potent nova

[timber leaf]

Then it can be splitted into two smaller integrals, smart

[final saddle]

@potent nova Has your question been resolved?

[potent nova]

ah okay lm try it

[potent nova]

we can use ti nspire

to make integrals

[worldly spruce]

You might want to react to the bot

[potent nova]

I still dk jow to solve it

i tried the identity and it still didnt work

[final saddle]

@potent nova Has your question been resolved?

[potent nova]

??

[slate crater]

ngl i lurked theough the server and found ur integral and ive been trying stuff for a while but i cant get anywhere either 💀💀🥀

if only it were from 0 to π/2...

[timber leaf]

if sin^2 = 1- cos^2 then?

$(1-\cos(x)^2)\cdot \cos(x)^4$

This should help

[soft zealot]

Minλ

[timber leaf]

@potent nova @potent nova

[slate crater]

ik the solution (or one ig) from a calculator app i got and gaw damn

[timber leaf]

!nosols 😭

[final saddle]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[coral zenith]

its a lengthy integral anyways

[worldly spruce]

How in the world did you get that

[slate crater]

im sorry

[worldly spruce]

You are done in like 6 steps, I don't know what's the problem

[slate crater]

https://tenor.com/view/uma-musume-meisho-doto-umm-fingers-pointed-anxious-gif-3423823507605431942

[timber leaf]

Your sorry by umamusume have been forgiven because i like that game

[slate crater]

hell yeah 🔥🔥🔥

[timber leaf]

But anyways by expanding the cosines and split it out it wont be that bad

[slate crater]

i mean if u go from sin²cos⁴ to cos⁴-cos⁶ and manip a bit u end up with (sincos)²dx idk 🥀🤠

[worldly spruce]

How

[slate crater]

maybe im just making errors

[worldly spruce]

Show your work

[timber leaf]

She is not OP

[worldly spruce]

Oh

[timber leaf]

My skibidi blud

[slate crater]

tbf i might aswell by the second OP LMFAO

cuz im also having trouble 🥀💀

[timber leaf]

I dont think the question is that bad

[coral zenith]

if you are comfortable with reducing cos^6x then its not bad

[slate crater]

thats what i thought snd why i tried doing it in order to help w a hint

but erm

[worldly spruce]

Power reduction

A bit of manipulation

Done

[timber leaf]

Or cos^6 =cos^4cos^2

[worldly spruce]

Sure

[slate crater]

as long as u know the formula but ive never seen it 🐌

yeah w it is just straight forwars

[brazen breach]

meisho doto?!?!?!?

[brazen breach]

uhhh u can do the funny with
sin2x=2sinxcosx => sin2x / 2=sinxcosx and squaring, and cos^2x=(cos2x+1)/2

[brazen breach]

power reduction is a formula made up by IBP followers who believe derivatives can actually be treated as fractions(they can't trust) we shan't do that sin

[slate crater]

yeah p much

[slate crater]

😭😭istg ive never seen it in my life until now

[slate crater]

yes!!!! my dear baby shes the cutest

[final saddle]

@potent nova Has your question been resolved?

[worldly spruce]

What?

I meant the power reduction identity for cos^2(x)

(1+cos(2x))/2

"they can't trust"

It seems the OP has no intention of solving this anyway ...

[final saddle]

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[mint forum]

hi guys im really confused on how to proof by contradict its just a hard concept for me, and i was wondering if you could check if my answer is right and i didnt know whether to say " asuume that m^2 is not divisible by 2 or that m is not divisible by 2"

[vital crag]

the proof should start with the negation of the conclusion. The statement in question has the form
if P then Q

where P is the statement $m^2$ is divisible by 2. and Q is the statement $m$ is divisible by 2.

[soft zealot]

riemann

[mint forum]

Oohh okay

so you negate Q?

[vital crag]

The negation of Q is the negation of m is divisible by 2

[mint forum]

And is that always the case?

Where it says if P then Q

[vital crag]

for proof by contradiction yes

[mint forum]

Just negate Q alwyas

WOWWWWWW OMG

Wait let me find a new question just to test this

Sorry, what about these two?

these dont have the if P and Q so for example, the 2nd one i'll just say "assume that there are positive integers for p and q"

but for the first one i make sure that tanx - sinx < 0?

so like flip the sign orrr

[vital crag]

how about you focus on one problem at a time and write out your "if P then Q" here

[vital crag]

similar to what i did for you here

[mint forum]

As in one question at a time 🤣 Yeah sorry im getting ahead of myself

Okay so, for this question, "if tanx - sin x > 0 within 0 < x < 90, then tanx - sinx > 0 within a different range?"

[whole lion]

sorry to interrupt but i think i can solve question 16

[mint forum]

Icl this ones hard idek wht im saying

[mint forum]

Yes pleaseeee

[whole lion]

ok

[whole lion]

firstly it goes like this

note that, a perfect square when modulo 3 is either 0 or 1

4p^2 - q^2 = p^2-q^2 = 2(mod 3)

[mint forum]

where have u got modulu 3 sorry

[whole lion]

what do you mean

[mint forum]

like where have u gotten that from

[whole lion]

you consider 3 cases: x modulo 3 is 0 1 or 2

i mean my english(in math) sucks

forgive me

[mint forum]

ok so first are we provin that there are no positive integers for 4p^2 - q^2 =25

[mint forum]

lmao its fine 😭 im just in year 13

which is like senior year

so i dont wanna overcomplicate it

[whole lion]

i'm 14yrs old

[mint forum]

💀 oh damn

[whole lion]

it's kind of basic knowledge you know

[mint forum]

yeah i just got humbled

[whole lion]

nah

the system in my country is kinda complicated

[mint forum]

ok carry on thank uu

Yeahh maybe you're asian lol

Really smart

[whole lion]

you got it right 🙂

[mint forum]

🤣 Could tell

[whole lion]

for this to happen, there is only one case where p^2 is divisible by 3 and q^2 is not

p^2 is divisible by 3 and 3 is prime so p is divisible by 3, which means p = 3

wait

oops

[blissful meadow]

It's probably more straightforward at this level to factor 4p^2 - q^2 and consider factors of 25.

[whole lion]

lemme rethink for a bit

[mint forum]

Yeaaa i think it'll be simpler

[whole lion]

oh yeah i forgot 🙃

[mint forum]

Thank you tho logic guy

[whole lion]

sry for overcomplicating things

[mint forum]

Noo its ok thank u for trying anyways 🥰

So like i factor out 25 for 5 x 5

[blissful meadow]

Then try to factor 4p^2 - q^2. Should be an identity you've seen before.

[mint forum]

idk is it like difference of 2 squares

Omg yayyy

is it like that

[blissful meadow]

Yep