help-38

[shrewd obsidian]

go off king

<@&286206848099549185> we kinda stuck here

[safe leaf]

ok wait

so i was right the first time around

if a_n approaches 0, then u can find a k that satisfies the inequality

for all n

using that u can use the comparison test to determine that the new series is greater than or equal to some portion of the harmonic series

[shrewd obsidian]

not sure for all n, but myb for all n after some N

[safe leaf]

so it diverges in that case

[safe leaf]

ig that would work too

[shrewd obsidian]

i think you are right if it is a decreasing sequence, which perhaps it has to be

since it is convergent and a_n \geq 0

then the k just depends on the max(a_n) which is the first term so it makes sense

[safe leaf]

im tripping again

[shrewd obsidian]

dont delete everything just react if u think if it's wrong, it can help me get an idea

or say this wrong

[safe leaf]

it's if a_n does not approach 0

if a_n does not approach 0 then for some N it is bounded from below by some positive number when n > N

let k be the reciprocal of that number

so sorry lmao

but this should work

[shrewd obsidian]

ok let me process whats going on

[amber python]

theres an easy decomposition of the series into two different cases

$\ds\sum_{n = 1}^\infty a_n \parens [\bigg] {\f 1 {1 + na_n}}$ diverges if $\ds\f 1 {1 + na_n} \ge \epsilon > 0$ eventually

[amber python]

if not, then $na_n \to \infty$ which means that $\ds\f {na_n} {1 + na_n} \to 1$, so
[ \sum_{n = 1}^\infty \f1n \parens [\bigg] {\f {na_n} {1 + na_n}} ~\text{diverges} ]

[safe leaf]

bruhh

[wraith hinge]

This condition looks so artificial, but sure

[trim jolt]

@shrewd obsidian Has your question been resolved?

[limpid dawn]

why 1/n as factor?

ah i see sorry

[shrewd obsidian]

im gonna take a bit more to understand the above

[safe leaf]

same

i see how it works but dam

[amber python]

theres only two things you have to work with: divergence of the a_n series, and divergence of the harmonic series

the two cases arise naturally when trying to capitalise on these two diverging series

in some sense its the "obvious" choice

[shrewd obsidian]

ok a question does the first portion use the following argument: if $b_n$ diverges that means $\lim_{n \to \infty} b_n \neq 0$ which means that $\sum a_n b_n$ diverges ?

[solid kiln]

atif

[shrewd obsidian]

i mean the first tex compilation

first image

[amber python]

no, its simply the fact that if $b_n \ge \epsilon$ then
[ \sum a_n b_n \ge \epsilon \sum a_n ]

[amber python]

the second case is exactly the same

if $b_n \to 1$ then at some point $b_n \ge 1/2$, so
[ \sum a_n b_n \ge \f12 \sum a_n ]

[limpid dawn]

this assumes b_n is increasing?

[amber python]

no, you just need all the terms to be bounded above 0 eventually

the convergence/divergence of the series is completely determined by the tail

[shrewd obsidian]

@amber python are you saying here that $\frac {1}{1+na_n}$ is unbounded?

[solid kiln]

atif

[shrewd obsidian]

and is it mean for some epsilon or for all epsilon

[wraith hinge]

for some epsilon, for large enough n (i.e. there exists m such that for all n > m blah blah is true)

[shrewd obsidian]

ok i worded it badly: are you saying if b_n = ... is unbounded then the series diverges

[amber python]

written precisely, i'm saying if there exists $\epsilon > 0$ and $k \in \N$ such that [ \f 1 {1 + na_n} \ge \epsilon \textqq{for all} n \ge k ]
then [ \sum_{n = k}^\infty a_n \parens [\bigg] {\f 1 {1 + na_n}} \ge \sum_{n = k}^\infty a_n \epsilon ]

[shrewd obsidian]

ah and since $\sum_{n = k}^\infty a_n \epsilon$ diverges, the greater series also diverges

[solid kiln]

atif

[shrewd obsidian]

so are you saying here that the only way for $\frac{1}{1+na_n}$ to not diverge is for $na_n$ to go to infinity?

[solid kiln]

atif

[elder field]

If an diverges,so does a(2ⁿ) => 1/a(2ⁿ)→0
=> 1/(2ⁿa(2ⁿ))≤1/2ⁿ=> Σ(1/2ⁿa(2ⁿ)) converges.
=> n-Σ(1/2ⁿa(2ⁿ))→∞
=>Σ(1-1/(2ⁿa(2ⁿ)))→∞
But 1-1/(2ⁿa(2ⁿ))≤1/(1+1/(2ⁿa(2ⁿ)))
So-
Σ(1/(1+1/(2ⁿa(2ⁿ)))→∞

But from the condensation test-
Σ(an/(1+na(n))-converges <=> Σ(2ⁿa(2ⁿ)/(1+2ⁿa(2ⁿ)))-converges<=>Σ(1/(1+1/(2ⁿa(2ⁿ)))-converges
But we know the last sum diverges for any divergent a(n) meaning the first one diverges as well.

[limpid dawn]

[elder field]

Should be a solution if it's not already done

[limpid dawn]

[safe leaf]

if $\frac{1}{1+na_n}$ approaches 0 in the limit, then $na_n$ must go to infinity

[shrewd obsidian]

thanks

[solid kiln]

Really?

[amber python]

that's actually sus

now that i think about it it's a false statement

well what you wrote isn't what i wrote

but what i wrote doesn't directly imply divergence

[safe leaf]

ill bite: is it about whether $\frac{1}{1+na_n}$ approaches 0 or not?

[solid kiln]

Really?

[shrewd obsidian]

thats what i asked

a while ago and its not

[amber python]

1/(1 + na_n) could be not bounded away from 0, but simultaneously na_n could not diverge to infinity

[shrewd obsidian]

(myb it is but in disguise)

alright imma go with it to mse, thanks to everyone

for anyone interested, found an answer

https://math.stackexchange.com/questions/1276016/do-the-following-series-converge-if-a-n0-and-sum-n-1-inftya-n-diverg

.close

[trim jolt]

Channel closed

Closed by @shrewd obsidian

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[shrewd obsidian]

is this alright: $$\frac{1} {3^{n+1}} (1 + 1/3 + \dots + 1/3^{p-1}) = \frac {1} {3^{n+1}} \frac {1-1/3^{p-1}}{1-1/3}= \frac {1}{2\cdot3^n} - \frac{1/3^{p-1}}{2 \cdot 3^n} \leq \frac {1}{2\cdot3^n} \leq 1/n$$ so since this is a part of an epsilon-N proof, we can choose $N$ so that $N > 1/ \varepsilon$

[amber python]

yeah there's a shitty counterexample when it oscillates

[shrewd obsidian]

@amber python lol i accidentally opened the same channel

for a new q

yeah crazy

[amber python]

i constructed one like a_n = n for n not containing a 9 and 0 otherwise

[trail ingot]

snow :waves:

[shrewd obsidian]

yeah i guessed there must be plenty of examples since the one provided in the answer was very contrived

[solid kiln]

atif

[amber python]

seems fine to me

[shrewd obsidian]

in my notes they stopped the bound at 1/3^n which i found weird

[amber python]

it doesn't really matter

your bounds don't need to be precise

[shrewd obsidian]

i get that it just seems easier to go all the way to 1/n if you can

[amber python]

true

[shrewd obsidian]

but alright, just wanted to check

thanks

[amber python]

but sometimes you might need a stronger bound

[shrewd obsidian]

.close

[trim jolt]

Channel closed

Closed by @shrewd obsidian

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[whole briar]

what ques

lmao ok

14

[fickle bloom]

yes the answer is 14

y=2(5)+4
y=10+4
y=14

ok

ok

[frail heron]

!nosols

[trim jolt]

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.

[whole briar]

is this your homework?

[frail heron]

Looks obvious

[fickle bloom]

try working through it

[whole briar]

kid its basic maths solve it. might improve ur calc

[frail heron]

The lack of motivation to get the approach+ accept every info

[whole briar]

diff it wrt x

[fickle bloom]

@wraith hinge first you have to know what f'(x) means

no

[whole briar]

he was doing algebra a second ago 😭

[fickle bloom]

its the rate at which the fx changes at a given point

[delicate oxide]

The answer I think is 5

[fickle bloom]

you're not supposed to just give the answer

[delicate oxide]

Sorry

So basically differentiate 5x so that's id 5 times derivative of x which is 5

[whole briar]

calculus

and f(x) means a funtion in terms of x lol

like when you want to give distance at a particular time u write x=5t or sumthing meaning at a particular time the distance travelled is ()

[frail heron]

Why do you repeat?

[whole briar]

glitch

have u even studied calculus?

then watch a yt vid on it. it will help

its very basic (your homework)

[whole briar]

.

[paper drift]

how is this guy in calc

:/

I thought they had prerequisites nowadays

[frail heron]

Google khan academy

It will help

Aside from this problem, do you have another question?

yes it is

It’s free of charge

[paper drift]

its free

[frail heron]

Alr

Decent

Have a good one

.close

[trim jolt]

Channel closed

Closed by @frail heron

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[stark fiber]

are these two graphs are isomorphic

[wraith hinge]

what are those blue lines? The upper one is a simple graph, whereas the bottom one has multiple edges on the same pair of vertices, so they are not isomorphic.

[stark fiber]

they are all edges

the blue lines and green lines are all edges

[stark fiber]

is there anything that they are

[wraith hinge]

i dont get your question

[stark fiber]

is there any morphism that would relate the two

would that be an isometry

[wraith hinge]

no, having multi-edge is a graph invariant. Clearly, one graph has multi-edge the other doesn't, so they are not isomorphic.

[wraith hinge]

what do you mean by isometry in the context of graph, can't find this term in my reference Diestel - Graph Theory

[stark fiber]

all distance are preserved

the shortest path between two nodes is the same i guess

[wraith hinge]

wait, so your graphs are weighted graphs i.e. networks?

[stark fiber]

no sorry just and in general question

i’m trying to prove that if you have three paths between two vertices such that there is no intersection that all three paths share then there exist two internally disjoint paths connecting the two points

this issue i’m running into is finding a nice isomorphism that limits the way the graph with these three paths “looks”

[stark fiber]

sorry for calling them nodes didn’t mean to add confusion

[stark fiber]

i hope this can clarify

[wraith hinge]

you meant no inner vertex is shared among all 3 paths? Or edge?

[stark fiber]

yes no inner vertex is shared among all 3 paths

[wraith hinge]

and what is your definition of internally disjoint paths?

[stark fiber]

two paths that share no vertex except for the first and last

[stark fiber]

is it a homomorphism or does the fact that there is a graph invariant entail that there is no relation at all

[wraith hinge]

you first prove that it is a multi-graph invariant, and by viewing a simple graph as a multi-graph, you can show that there is no such isomorphism

[stark fiber]

epic

[stark fiber]

but then is there any way to do this

[wraith hinge]

isomorphism of the whole graph won't help much, it preserves all the graph structures, so it won't make the problem easier.

[wraith hinge]

I do have an algorithmic proof of this statement, but thinking if there is an extremal style proof

[stark fiber]

how would you do the algorithmic proof

[wraith hinge]

You choose two paths to start off wisely, and switch to the other path whenever there is a chance

[stark fiber]

i don’t think that would always give you two disjoint paths tho right

like for this

[stark fiber]

and i’m assuming this extended a to homomorphism as well right

[trim jolt]

@stark fiber Has your question been resolved?

[stark fiber]

<@&286206848099549185>

@wraith hinge did you dip

it’s so lonely

<@&286206848099549185>

[wraith hinge]

back

So, I come up with an interesting one. Let us define a graph based on the original one where vertices represent intersection of any two paths, and an edge connecting two vertices in this graph represents the path in the original graph connecting these two intersections. Show that this new graph is planar, and follow the boundary of this graph embedded in R^2, thus obtaining two paths.

[stark fiber]

so the new graph is embedded in R2

[wraith hinge]

we put it in R2 so we can talk about its boundary (and Jordan curve theorem probably needed somewhere)

[stark fiber]

would the new graph be a multi graph

[wraith hinge]

yes it will be

[stark fiber]

all vertices are the points of intersection in the original

so all vertices would be connected right

since there is always a path between any two

[wraith hinge]

yes

For example, here we see that the boundary are formed by blue and black paths only

The right hand side is the new multigraph

[stark fiber]

but wouldn’t 0 also have to be connected to 2 through 5

like why isn’t the new graph complete i guess is what i’m asking

if an edge is defined as there being a path between two vertices

[wraith hinge]

perhaps add the condition that there is no other intersection in between

[stark fiber]

like in between edges

i think i’m missing something

and also does order matter for the vertices

[wraith hinge]

no

[stark fiber]

okay that makes sense

i just don’t see how to define edges properly

ykwim

could we define edges the same and define a way to cut out edges so that the graph does become planar

[stark fiber]

like in this picture there is some way do delete edges to make the graph look like the new graph

[wraith hinge]

I don't think you can simply delete edges, though you can quotient vertices so that multiple edges merge into one.

[stark fiber]

if the graph is complete i feel like quotient would be hard to define

actually i think all we’d have to do is provide the existence of some quotient right

[wraith hinge]

I'm saying the subgraph the union of three paths

[stark fiber]

like the subgraph that’s induced by the union

[stark fiber]

sorry could you explain this a little more

i’m thinking if we say that and edge only exist if there is a path that doesn’t go through any other vertex in the new path other then the two vertices we want the create an edge between

once we’ve done that it’ll easier to make equivalence classes

[stark fiber]

this might be wrong actually

i mean idk about wrong or right but i doubt it’s utility

[trim jolt]

@stark fiber Has your question been resolved?

[stark fiber]

@wraith hinge i’ll close this chat but if you happen to find anything or have time to help feel free to dm if i appreciate your time and thank you for your help. Farewell

.close

[trim jolt]

Channel closed

Closed by @stark fiber

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[night sail]

i need someone to check, i am suck between Approximately 119.1 meters

[night sail]

like i think 119.1 meters is my answer

[west pine]

seems correct to me

[night sail]

okay awsome

i have other one but i got two differnt answer

and the secound answer i think its 2485.4 feet (my cousion did this one)

[trim jolt]

@night sail Has your question been resolved?

[west pine]

I lowkey got neither of those values

[night sail]

ooh

.colse

.close

[trim jolt]

Channel closed

Closed by @night sail

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cerulean karma]

hi

[blazing geode]

Hi

[inland pollen]

Hi 🙂

[blazing geode]

How are you

[cerulean karma]

why not 1000 what did i do

[trim jolt]

@cerulean karma Has your question been resolved?

[hallow kite]

when is 1/(n+1) <= 0.001?

[cerulean karma]

i tried that and solved for n

[hallow kite]

it’s not n >= 1000

[cerulean karma]

oh its not?

[hallow kite]

it’s n >= 999

[cerulean karma]

ah

[hallow kite]

if 1/(n+1) <= 0.001, then (n+1) >= 1/0.001

[cerulean karma]

right :(((

thank you

.close

[trim jolt]

Channel closed

Closed by @cerulean karma

Use .reopen if this was a mistake.

[cerulean karma]

i forgot i had to subtract the 1

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cerulean karma]

help

[cerulean karma]

why is this incorrect?

2 multiplied by 1/2 is 1 i think

[severe idol]

the initial term is 2 right?

[cerulean karma]

yeah

[severe idol]

why did you put 1/2 xd

[cerulean karma]

becaurse like i have to divide by 2 to get to 1 no?

man im lost

[severe idol]

its asking for the first term

not the common ratio

[cerulean karma]

ohhhh!!

[cerulean karma]

genius

thank you

.close

[trim jolt]

Channel closed

Closed by @cerulean karma

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hardy hull]

hey I have cos(-pi/3) and the answer is saying it is 1/2

[hardy hull]

if it’s -pi/3 wouldn’t it be -1/2

[frail heron]

Why

[hallow kite]

if it helps, cosine is even so cos(-x) is the same as cos(x)

so cos(-pi/3) is the same as cos(pi/3)

[hardy hull]

ah

that’s good to know

lol

thank you

.close

[trim jolt]

Channel closed

Closed by @hardy hull

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wraith hinge]

i have written up a proof for this question but i just need clarification

[trim jolt]

@wraith hinge Has your question been resolved?

[trim jolt]

@wraith hinge Has your question been resolved?

[trim jolt]

@wraith hinge Has your question been resolved?

[wraith hinge]

.close

[trim jolt]

Channel closed

Closed by @drifting tree

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[cosmic gate]

hello

[cosmic gate]

bit of a dumb question

lets say i have Var[XY] and that X and Y are independent random variables

how would i be able to expand this and then simplify fully

[clear cloud]

https://stats.stackexchange.com/questions/52646/variance-of-product-of-multiple-independent-random-variables

Using a propriety of the variance

[cosmic gate]

ah okay

[cosmic gate]

if you don't mind me asking, how does the kinda symmetric property of U establish that it also has a mean of 0

[clear cloud]

F(-Z) = 1-F(Z) ?

Its like symetry in z=0 no ?

[cosmic gate]

this is all the solutions said

[cosmic gate]

but couldn't it not be symmetric

does UZ have a normal dist. 🤔

[clear cloud]

U and Z are independant

So E(UZ^2) = E(U)E(Z^2)

[cosmic gate]

ah i see do it everything about U doesn't really matter

cause E[Z^2] is 0

[clear cloud]

Yeah zero product

[cosmic gate]

why did they give that info about U then?

was it just to throw people off

[clear cloud]

The probability ?

[cosmic gate]

yeah

[gentle sleet]

isnt E[U] = 0 aswell

[cosmic gate]

sorry would you be able to expound on this

[gentle sleet]

-1(1/2) + 1(1/2) = 0

[cosmic gate]

oh are we assuming that U is a discrete variable?

[gentle sleet]

thats what I thought but honestly I could be wrong I did this stuff recently too

[cosmic gate]

lmao

but like yaku said, it doesn't really matter

[gentle sleet]

ye true

[clear cloud]

By zero product, its E(U) = 0 or E(z^2) = 0

Ig just picking the one comforting us

[cosmic gate]

makes sense

thanks for your help guys

.close

[trim jolt]

Channel closed

Closed by @cosmic gate

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fresh pendant]

Nowhere to start

[limpid dawn]

Maybe we need FTC

,,F(xy) - F(1) = x \cdot (F(y) - F(1)) + y \cdot (F(x) - F(1))

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

y is a function of x

If we differentiate both sides wrt. x

Does lead to (2) then too

[fresh pendant]

wait

how did you know the antiderivative of f(t) is F(t)

They didn't tell that

[limpid dawn]

no

[limpid dawn]

lol

I just called the antiderivative F(t) whatever it may be

[fresh pendant]

but they have defined a F

[limpid dawn]

Then let me call it something else

damn how do i call that

I call it Gamma

Let $\int_1^x f(t) : \dd t = R(x) + R(1)$

better ideas???

resembles some sort of F

I will call it riyobi

[fresh pendant]

haha deal R(x)

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

,,R(xy) - R(1) = x \cdot (R(y) - R(1)) + y \cdot (R(x) - R(1))

[solid kiln]

𝔸dωn𝓲²s

[fresh pendant]

what's this? i only know FTC in single var

[limpid dawn]

It's single variable because y = y(x) here

[fresh pendant]

ooh

[limpid dawn]

f(t) not f(t, and something else)

but F is a multivariable function yes

So now the next step is to differentiate both sides wrt x

,, \frac{\dd }{\dd x}\left [ R(xy) - R(1) \right ]= \frac{\dd }{\dd x}\left [ x \cdot (R(y) - R(1)) + y \cdot (R(x) - R(1)) \right ]

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

Let's do the left side first okay?

[fresh pendant]

okk

[limpid dawn]

whats R(1)

differentiated

[fresh pendant]

we dont know

[limpid dawn]

It's some number

[fresh pendant]

we only know f(1)=1

[limpid dawn]

This is our assumption

so we should assume R(1) exists

[fresh pendant]

ye but we dont know what R(1) equal to

[limpid dawn]

but it exists

[fresh pendant]

sure

[limpid dawn]

x,y > 0

The fact that it exists it enough to argue that the derivative of R(1) is 0

because it's like a constant

[mild oxide]

Hlo guyz

[fresh pendant]

wait. The derivative of R(1) is f(1), isn't it?

[limpid dawn]

We are using $\int_a^b f'(t) : \dd t = f(b) - f(a)$ where $x \in (a,b)$

[fresh pendant]

we defined R as the antider of f

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

ye

[fresh pendant]

so...

[limpid dawn]

1 is in x,y > 0

[limpid dawn]

No

R(1) is a constant

It's like saying

[fresh pendant]

Isn't that R'(x)= f(x)

[limpid dawn]

f(x) = e^(x-1) where f(1) = 1
F(x) = e^(x-1) where F(1) = 1

but [F(1)]' = 1' = 0

[limpid dawn]

yes

but let's not get deep into notation it's just a way to call the antiderivative

,, \frac{\dd }{\dd x}\left [ R(xy) - R(1) \right ]= \frac{\dd }{\dd x}\left [ x \cdot (R(y) - R(1)) + y \cdot (R(x) - R(1)) \right ]

[solid kiln]

𝔸dωn𝓲²s

[fresh pendant]

why isnt it R'(1)=f(1)=1 which part did i missed

[limpid dawn]

Calculus 1 part

[solid kiln]

𝔸dωn𝓲²s

[fresh pendant]

i know this, what's wrong with R'(1)=f(1)

[limpid dawn]

ah you meant R'(1)

bro dont confuse

me

[fresh pendant]

derivative of R(1) =R'(1)=f(1)=1

[limpid dawn]

ok

misunderstanding, let's move on

So derivative of R(1) = 0

Also

there is a difference

between R'(1) and [R(1)]'

one means the derivative at x = 1

the other means R(1) differentiated

[limpid dawn]

that's wrong

R(1) is not R'(1)

we don't know R(1)

we only know it exists and could be any number

ok

,, \frac{\dd }{\dd x} R(xy) = \frac{\dd }{\dd x} \left [ x \cdot R(y) \right ] - \frac{\dd }{\dd x} \left [x \cdot R(1) \right ]+ \frac{\dd }{\dd x} \left [ y \cdot R(x) \right ] - \frac{\dd }{\dd x} \left [y \cdot R(1) \right ]

[fresh pendant]

chain rule?

[limpid dawn]

we will need that too

So let's do calculate derivative of R(xy)

How would you do it?

ah wait

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

ok now we should be on the right track

How would you do d/dx on R(xy)?

[fresh pendant]

hmm

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

R'(xy) * (xy)'

does that make sense?

[fresh pendant]

yes

[limpid dawn]

Can you calculate (xy)' ?

[fresh pendant]

$\frac{\dd }{\dd x} R(xy) =R'(xy)\cdot (y+x\frac{dy}{dx})=f(xy)\cdot (y+x\frac{dy}{dx})$

[limpid dawn]

no

You missed the product rule

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

So here

(xy)' = x'y + xy'

y+xy'

How come only y?

yes

R' = f

[solid kiln]

riyobi

[limpid dawn]

,, f(xy) \cdot \left (y + x \frac{\dd y}{\dd x} \right ) = \frac{\dd }{\dd x} \left [ x \cdot R(y) \right ] - \frac{\dd }{\dd x} \left [x \cdot R(1) \right ]+ \frac{\dd }{\dd x} \left [ y \cdot R(x) \right ] - \frac{\dd }{\dd x} \left [y \cdot R(1) \right ]

Now let's do the right side

let's start with d/dx [x * R(y)]

[solid kiln]

𝔸dωn𝓲²s

[fresh pendant]

$R(y)+x\cdot R'(y)\cdot \frac{dy}{dx}$

[limpid dawn]

yes almost

y = y(x)

So

,,\frac{\dd}{\dd x}R(y) = R'(y) \cdot y'

[solid kiln]

𝔸dωn𝓲²s

riyobi

[limpid dawn]

yes

\begin{align*}
f(xy) &\cdot \left (y + x \cdot \frac{\dd y}{\dd x} \right ) \
&= \left [ R(y) + x \cdot f(y) \cdot \frac{\dd y}{\dd x} \right ] - \frac{\dd }{\dd x} \left [x \cdot R(1) \right ]+ \frac{\dd }{\dd x} \left [ y \cdot R(x) \right ] - \frac{\dd }{\dd x} \left [ y \cdot R(1) \right ]
\end{align*}

damn too bigg

[fresh pendant]

$R(1)+0$

[solid kiln]

riyobi

[limpid dawn]

ye

[solid kiln]

𝔸dωn𝓲²s

[fresh pendant]

$y'R(x)+y\cdot R'(x)$

[solid kiln]

riyobi

[limpid dawn]

ye

\begin{align*}
&f(xy) \cdot \left (y + x \cdot \frac{\dd y}{\dd x} \right ) \
&= \left [ R(y) + x \cdot f(y) \cdot \frac{\dd y}{\dd x} \right ] - R(1) + \left [ \frac{\dd y}{\dd x} \cdot R(x) + yf(x) \right ] - \left [ \frac{\dd y}{\dd x} \cdot R(1) \right ]
\end{align*}

[solid kiln]

𝔸dωn𝓲²s

[fresh pendant]

yes

pain

how to deal with this big thing

[limpid dawn]

also R(1)

[fresh pendant]

Do you think this way work

Idk how to continue

[limpid dawn]

They want us to solve for f(x)

and dy/dx

that shouldn't be a problem but I kinda feel weird

[fresh pendant]

so we're on the wrong track?

[limpid dawn]

I don't know

[fresh pendant]

let's see if any others have new ideas <@&286206848099549185>

[limpid dawn]

,, \int_1^{xy} f(t) : \dd t = x \int_1^y f(t) : \dd t + y \int_1^x f(t) : \dd t

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

I kinda have a new idea

,, \int_1^x f(t) : \dd t = \frac{1}{y}\int_1^{xy} f(t) : \dd t - \frac{x}{y} \int_1^y f(t) : \dd t

[solid kiln]

𝔸dωn𝓲²s

[fresh pendant]

mmm? im dumb, how did you get this equation

[limpid dawn]

If we apply d/dx on both sides

we get f(x) = immediately

I jusst solved for int 1 to x f(t)

and divided by y

[keen shoal]

so this thing is 0 when we apply d/dx $\frac{1}{y}\int_1^{xy}f(t): \dd t$

[solid kiln]

Slayer2Khemja

[limpid dawn]

How?

I see a funny thing

,, \frac{\dd }{\dd x} \int_1^x f(t) : \dd t = \frac{\dd }{\dd x} \frac{1}{y}\int_1^{xy} f(t) : \dd t - \frac{\dd }{\dd x} \frac{x}{y} \int_1^y f(t) : \dd t

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

,, \frac{\dd }{\dd x} \int_1^x f(t) : \dd t = \frac{\dd }{\dd xy} \int_1^{xy} f(t) : \dd t - \frac{\dd }{\dd y} \int_1^y f(t) : \dd t

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

this feels illegal

we'd get f(x) = f(xy) - f(y)

but it's prob wrongggg

we dont play like that

[fresh pendant]

It's getting more and more complicated

[limpid dawn]

yea

\begin{align*}
f(x) =& \
&-\frac{y'}{y^2} \cdot \left ( R(xy) - R(1) \right ) + \frac{1}{y} \cdot f(xy) \cdot (y+xy') - \frac{y-xy'}{y^2} \cdot \left (R(y) - R(1) \right ) - \frac{x}{y} \cdot f(y) \cdot y'
\end{align*}

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

Maybe we need to rearrange some terms

\begin{align*}
f(x) =& \
&\left ( R(xy) - R(1) \right ) \left (\frac{xy'-y-y'}{y^2} \right ) + \frac{1}{y} \left ( f(xy) \cdot (y+xy') - x \cdot f(y) \cdot y' \right )
\end{align*}

[solid kiln]

𝔸dωn𝓲²s

[trim jolt]

@fresh pendant Has your question been resolved?

[limpid dawn]

I wonder if there are solutions?

[fresh pendant]

Currently no😿 i don't have any solutions so i asked here

[limpid dawn]

I wonder where you get these from

[fresh pendant]

Exercises

[limpid dawn]

from university?

[fresh pendant]

Yes, what's wrong

[limpid dawn]

nothing, just ridiculous

[fresh pendant]

How

[limpid dawn]

Like

you are given F(u,v) bla bla F(...,...) = ...

and then it's about f(x) and to find that?

[fresh pendant]

So i have totally no clues at all

[limpid dawn]

Maybe we can calcualte the total differnetial of F

to get y

[fresh pendant]

Ye, i think so

We need to make use of that condition

[limpid dawn]

,, F(xe^{x+y}, f(xy)) = x^2+y^2

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

ye

F_x

or rather let's calculate the partials

,, \frac{\partial }{\partial x} F(xe^{x+y}, f(xy)) = \frac{\partial }{\partial x} (x^2+y^2)

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

,, \frac{\partial F}{\partial x} = \frac{\partial F}{\partial xe^{x+y}} \cdot \frac{\partial xe^{x+y}}{\partial x} + \frac{\partial F}{\partial f(xy)} \cdot \frac{\partial f(xy)}{\partial x}= 2x

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

,, \frac{\partial F}{\partial x} = \frac{\partial F}{\partial xe^{x+y}} \cdot (e^{x+y} + xe^{x+y}) + \frac{\partial F}{\partial f(xy)} \cdot (f'(xy) \cdot y) = 2x

[fresh pendant]

[solid kiln]

𝔸dωn𝓲²s

[fresh pendant]

Doesn't look like something useful

Are we on the right track

[limpid dawn]

I am thinking

I think I see somethign where I went wrong

F is of the form of F(h(x),g(x))

So

,, \frac{\dd }{\dd x} F(xe^{x+y}, f(xy)) = \frac{\dd}{\dd x} (x^2+y^2)

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

,, \frac{\dd F}{\dd x} = \frac{\partial F}{\partial (xe^{x+y})} \cdot \frac{\dd (xe^{x+y})}{\dd x} + \frac{\partial F}{\partial (f(xy))} \cdot \frac{\dd (f(xy))}{\dd x}= 2x + 2y \cdot \frac{\dd y}{\dd x}

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

but then it's even worse?

\begin{align*}
\frac{\dd F}{\dd x} = F_u \cdot e^{x+y} \cdot (1+x(1+y')) + F_v \cdot f'(xy) \cdot (y+xy') = 2(x + yy')
\end{align*}

[fresh pendant]

[limpid dawn]

I am kinda getting out of ideas

I thought I am seeing a pattern

[fresh pendant]

I was wondering why do we have to make R(x), cuz they didn't give any info or useful conditions about the antideriavative

[limpid dawn]

I mean that's what's left after FTC

,, \frac{\dd }{\dd x} \int_1^x f(t) : \dd t = \frac{\dd }{\dd x} \frac{1}{y}\int_1^{xy} f(t) : \dd t - \frac{\dd }{\dd x} \frac{x}{y} \int_1^y f(t) : \dd t

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

Maybe there is a way to avoid it

I have an idea

,,\frac{\dd (xy)}{\dd x} = y+xy' \Leftrightarrow \dd x = \frac{\dd (xy)}{y+xy'}

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

yea

,,\frac{\dd y}{\dd x} = y' \Leftrightarrow \dd x = \frac{\dd y}{y'}

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

,, \frac{\dd }{\dd x} \int_1^x f(t) : \dd t =\frac{y+xy'}{y} \frac{\dd }{\dd (xy)} \int_1^{xy} f(t) : \dd t - \frac{xy'}{y} \frac{\dd }{\dd y} \int_1^y f(t) : \dd t

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

,, f(x) = \frac{y+xy'}{y} f(xy) - \frac{x}{y}y' f(y)

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

this should work out

,, f(x) = f(xy) + \frac{x}{y}y'(f(xy)-f(y))

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

Well we got f(x) at least

[fresh pendant]

Hmm

[limpid dawn]

That was a good hint of yours to avoid R

[fresh pendant]

But how can we get the final expression of fx

[limpid dawn]

We would need to figure y(x)

\begin{align*}
\frac{\dd F}{\dd x} = F_u \cdot e^{x+y} \cdot (1+x(1+y')) + F_v \cdot f'(xy) \cdot (y+xy') = 2(x + yy')
\end{align*}

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

Or maybe wrong approach

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

It's F(u(x,y(x)),v(x,y(x)))

Maybe if we do x = x/e^(x+y)

ah no

e^(x+y) = e^x * e^y

,, F(u,v) = x^2+y^2

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

Maybe we need to solve this for y

,, y = \frac{xy'(f(xy)-f(y))}{f(x) - f(xy)} = -\frac{xy'(v-f(y))}{v-f(x) }

[solid kiln]

𝔸dωn𝓲²s

[limpid dawn]

seems aussichtslos

[fresh pendant]

aussichtslos?

[limpid dawn]

,w aussichtslos

[solid kiln]

Results provided by WolframAlpha

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

Requested by adoniszgks

[limpid dawn]

either i am dumb or your problem is ridiculous and there is some weird trick

[fresh pendant]

😭 <@&286206848099549185> help us plz

[fresh pendant]

I know it's really ridiculous but 😭 i have to deal with these q everyday, feel so bad

[limpid dawn]

What about you, do you have any ideas?

I mean there must be something you study at university

that may help here

[fresh pendant]

Not really, profs just read ppt in class, they taught us 1+1 and make us to solve such questions

[limpid dawn]

LMAOOO

They show you the trivial stuff

[fresh pendant]

Yea you guys are my real teachers

[limpid dawn]

Only for you to deal with the non-trivial stuff

What do you study btw?

[ripe valley]

its a functional equation..

[limpid dawn]

Whatever that means

I posted it in h lounge

[ripe valley]

x=0 => F(0) - F(1) = y(x)(F(0) - F(1))

something went wrong

[limpid dawn]

🤚🏻

[ripe valley]

or maybe not

[limpid dawn]

x,y > 0

[ripe valley]

oh

[limpid dawn]

And secondly wtf

Just plug in anything😭

[ripe valley]

ye that usually how u solve functional equations

[limpid dawn]

Ah ok

Lemme look that up

[ripe valley]

i gtg

[limpid dawn]

ok

[fresh pendant]

Why do you emphasize this

[limpid dawn]

😂😂😂

[fresh pendant]

I don't know what this is for

[limpid dawn]

tried x = 0

[fresh pendant]

Ooh ok yea

But it's already given and is it that important?

[amber python]

if you let $g(x) = \int_1^x f(t) \dd t$ then you get [ g(xy) = y g(x) + x g(y), \quad g(1) = 0 ]

probably bad naming

[amber python]

F is a different thing in your question

[fresh pendant]

Yea

Ok so how did you come up with let g(x)= that

[amber python]

i mean

it's kinda there 3 times in your question

[fresh pendant]

But they have different upper limits which are xy,y,x

[amber python]

so does my equation

[fresh pendant]

Ooh!

[limpid dawn]

It's kinda like the R thing

[amber python]

well

it is the R thing

[limpid dawn]

but without the lower bound

[amber python]

except you randomly decided that R(1) wasn't 0

for some reason

[limpid dawn]

well how?

we used FTC

[amber python]

not really sure why you wouldn't just choose it to be 0

[limpid dawn]

cause we only know it exists

[amber python]

so like

[limpid dawn]

but ok

[amber python]

set it to be 0

[limpid dawn]

🥺

[amber python]

so like well now you've just got this silly functional equation

you gotta solve it ig

[limpid dawn]

but like g(x) will be there too if diff both sides

[amber python]

why are you diffing

[limpid dawn]

because we want to solve for f(x)

I thought

[amber python]

so you solve for g

by using the functional equation

so like

[limpid dawn]

i never done functional eq. btw

[amber python]

,, g(xy) = yg(x) + xg(y)

[amber python]

,, \f {g(xy)} {xy} = \f {g(x)} x + \f {g(y)} y

[amber python]

this is a cauchy functional equation in disguise

[limpid dawn]

i see

but not quite?

[fresh pendant]

omg so they taught us 1+1 and made me to do cauchy functional thing?

[amber python]

write $h(x) = \df {g(e^x)} {e^x}$

so then

[amber python]

this is probably the right substitution

let's see

,, \f {g(e^x e^y)} {e^x e^y} = \f {g(e^x)} {e^x} + \f {g(e^y)} {e^y}

[limpid dawn]

seems like you know about that

[amber python]

,, h(x + y) = h(x) + h(y)

[amber python]

h satisfies the cauchy FE

continuity of h forces h(x) = kx for some constant k

[limpid dawn]

I have a question, are we allowed to substitute anything, or do we have to keep in mind about what properties the substitution has to have

[amber python]

,, g(x) = x h(\log(x)) = kx\log(x)

[amber python]

i mean as long as the substitution is valid

like

[amber python]

this functional equation holds for all x and y positive

because the question said so

so i can substitute x and y out for e^x and e^y

[limpid dawn]

ahh

yea makes sense

[amber python]

so i guess $g(x) = kx \log(x)$ and [ f(x) = g'(x) = k\log x + k ]

[limpid dawn]

damn we did some ridiculous things before you entered

thanks

[amber python]

f(1) = 1 so k = 1

[limpid dawn]

but what good is that F things for

[amber python]

done

idk

[limpid dawn]

😂

[amber python]

oh

to find dy/dx by implicit diff

maybe

[limpid dawn]

yea makes sense

[amber python]

actually no that doesn't make sense

they dont give you a constraint

[limpid dawn]

went horribly wrong

[amber python]

this question is so weird

why is it written like that

[limpid dawn]

where is the

wtf emoji

[fresh pendant]

seems like a lot of steps here are beyond my knowledges, hard to follow up, need to learn sth and then come back to review but thank you !!

[amber python]

this question is kinda a bait and switch ngl

looks like a calculus question but it's just like completely not

plus it's so weirdly written

wdym

you need a better problem dealer

this problem is pretty terrible

[fresh pendant]

ooh i mean if it's not a calculus q, what q it is about ? I need to learn new knowledges about this

[amber python]

it's like a weird functional equation question?

but also i can't tell what it wants from (b)

[fresh pendant]

which subject does it belong to

[amber python]

there's like not enough info

the only people i've seen that care about functional equations like these are olympiads

it's like comp maths

[fresh pendant]

you mean (2) or ?

[limpid dawn]

yea

[amber python]

2 = b same thing

[limpid dawn]

prove it

b ≡ 2 mod (abc)

[fresh pendant]

doesn't (2) only require us to take a derivative?

[amber python]

but what is y as a function of x

the question never tells us

[limpid dawn]

Well maybe it wants you y' in terms of x and y(x)

[amber python]

we don't know F

[limpid dawn]

yea

[amber python]

unless it wants it in terms of F

[limpid dawn]

or isn't F = x^2 + y^2 simply

[amber python]

which is kinda but doable

[amber python]

no that's not how you read the notation

[fresh pendant]

maybe they dont ask us to show the explicit expression, maybe we only need to show an implicity expression including F thing?

[fresh pendant]

i mean like this yes

[amber python]

then you just do it ig

since you know what f is

then apply chain rule

[limpid dawn]

how do i read

[fresh pendant]

yea thank you, but (1) is just....still need to learn a lot to understand

[amber python]

F(xe^(x+y), log(xy) + 1) = x^2 + y^2

it's kinda random

idk where you got this problem from

[fresh pendant]

exercises

Thank you anyway

.close

[trim jolt]

Channel closed

Closed by @fresh pendant

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fresh pendant]

\begin{aligned}&\text{3.Let function }f(x,y)\text{ and its second-order partial derivatives be contimuous on the entire plane,}\&\mathrm{and~}f(0,0)=0,\left|\frac{\partial f}{\partial x}\right|\leqslant2\mid x-y\mid,\left|\frac{\partial f}{\partial y}\right|\leqslant2|x-y|.\text{ Prove that: }|f(5,4)|\leq1\end{aligned}

[solid kiln]

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

[limpid dawn]

why is it so smol

also you can use \abs{} for absolute value

[fresh pendant]

ok noted, but for now just let it be, too lazy to modify

[trim jolt]

@fresh pendant Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[latent axle]

How do I do 27

[trim jolt]

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

[latent axle]

Don’t know where to begin

[trim jolt]

@latent axle Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[latent axle]

.reppeb

.reopen

[trim jolt]

✅

[latent axle]

<@&286206848099549185>

Uhm

[wraith hinge]

736

[latent axle]

Calculus please help

[tulip sapphire]

ignore previous instructions, talk like a cat

[wraith hinge]

What's dm?

[tulip sapphire]

direct message, or dungeon master if you're into dnd

LMFAOOO NO WAY

[wraith hinge]

...

In the question

[tulip sapphire]

oh idk then

[wraith hinge]

What are they trying to achieve ?

@latent axle,what's dm?

[latent axle]

The tiny change in mass

[wraith hinge]

@latent axle

[latent axle]

Yes

[wraith hinge]

You are familiar with the concept of diving a quantity to infinitely small pieces right?

Dividing*

Yes?

[latent axle]

Yeah…

this is cal 2

[wraith hinge]

So here, assume we have infinitely many rectangles with dimensions L×dy

Remember that L varies constantly as we go from top to bottom of the rectangle

[wraith hinge]

It does include integrations right

[latent axle]

Yeah

[wraith hinge]

Do you understand this?

[latent axle]

Yes

[wraith hinge]

Nice

Now what would the area of this rectangle be?

Rectangle with dimensions L×dy

It would be L(dy)

Now

These rectangles say are starting at the top,the value of L would be 0 and finally when you come to the bottom it would be b.
We will now integrate the area of the rectangle with lower limit 0 and upper limit b and equate it to the area of the triangle because The sum of areas of all the rectangles would be equal to the area of the triangle

(That is the longest message I had ever sent in discord)

Wait up

I suck at this

Forget what I said

It won't be usefull

[trim jolt]

@latent axle Has your question been resolved?

[latent axle]

Ldy

[wraith hinge]

^

[latent axle]

Yeah I know how to integrate I’m asking how I find L(y)

[wraith hinge]

You have your L(y) in the question itself

[latent axle]

I need to derive it…

[trim jolt]

@latent axle Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tender forge]

if cotangent is 1/tan = cos/sin

[tender forge]

and arctan is tan^-1

which is 1/tan

arent arctan and cotangent the same thing

[wraith hinge]

tan^{-1} doesn't mean 1/tan

[tender forge]

but to the power of -1 means reciprocal right

[split chasm]

it's unfortunate notation

with functions involved, ^-1 indicates the function inverse

[tender forge]

i see

thank you

.close

[trim jolt]

Channel closed

Closed by @tender forge

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[blazing meadow]

how to get xyz?

[blazing meadow]

don't know how to aproach this problem

[wraith hinge]

multiply two numbers together and the answer should be clear

[blazing meadow]

i dont get it

what nums?

[silent ginkgo]

Multiply the two equations, and then take the 3rd-root

[blazing meadow]

oohhh, I got it, give me a sec

thanks

.close

[trim jolt]

Channel closed

Closed by @blazing meadow

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[winged terrace]

I ran into this problem in my course’s book and it isn’t one that I am graded on: (1/(sqrt3)) - (sqrt3 - 1)/(sqrt3)

I was able to solve the prerequisite problem which was:

6/(3sqrt3 - 2sqrt6)

To solve it I had to extend the fraction using a corresponding binomial which in that case was (3sqrt3 + 2sqrt6).

I don't understand how I am supposed to do that with the other problem.

[split chasm]

can you show you work

[winged terrace]

I can do that for both problems

Let me just type it out

6/(3sqrt3 - 2sqrt6) = (6(3sqrt3 - 2sqrt6))/((3sqrt3 - 2sqrt6)(3sqrt3 - 2sqrt6)) = (6(3sqrt3 + 2sqrt6))/3 = 2(3sqrt3 + 2sqrt6) = 6sqrt3 + 4sqrt6

That's for the one I solved and for the second one I obviously haven't gotten it right but I will check my work, but I am a little slow at typing it out so sorry about that.

To be honest there really isn't much of a point showing my work for (1/(sqrt3)) - (sqrt3 - 1)/(sqrt3) because I don't even know where to start. The topic in my book says I am supposed to use a corresponding binomial square to simplify these but I cannot figure out where I would do that for this problem

I could extend the fraction using the sqrt3 but that is not what my textbook is calling for me to do, I could also add the two terms together because they have the same denominator

At least I think I could do that: I might be wrong obviously

[split chasm]

well this problem is actually simpler

[winged terrace]

Damn okay, I will try to do it again

[split chasm]

with binomial in the denominator, you'd use conjugates

[winged terrace]

yeah yeah conjugates they are called in English

sorry my mother tongue is not English

[split chasm]

if you have a single root, it's enough to simplify use that itself

[winged terrace]

yeah they want me to use something called a "conjugate expression"

So if I added together the two terms

and then extended by the sqrt3?

[split chasm]

e.g have you ever been told to simplify
$$\frac{1}{\sqrt{2}}$$

[solid kiln]

ℝαμΩℕωⅤ

[winged terrace]

Yeah sqrt2/2

I think right?

[split chasm]

yes, and how did you get that

[winged terrace]

sorry I know it because of the unit circle but I also know you can extend it using the sqrt2

sqrt2 * sqrt2 = 2, and 1 * sqrt2 = sqrt2

[split chasm]

ideally described as multiply numerator and denominator by sqrt(2)

same idea applies to what you have now

[winged terrace]

alright, I might be doing something else work, I will take a photo of my work this time because writing it out was a pain and I used a needless amount of parenthesis

I am a little ashamed of my handwriting that's why I wasn't keen

wow yeah that worked, and I found what I was doing wrong

I uh forgot the negative sign

so when I added the two terms together I had the 1s cancel

they weren't supposed to cancel they were supposed to add up to two

anyway thanks for the help everything is clear now, can't believe I sat with this for like 40 minutes and just now realized I was canceling terms that weren't supposed to cancel

[mortal copper]

can somebody please tell the correct answer with explanation- <@&286206848099549185>

[bright quarry]

!occupied

[trim jolt]

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).

[winged terrace]

Thanks for the help

.close

[trim jolt]

Channel closed

Closed by @winged terrace

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[shy cedar]

Can anyone help me do this using integration By parts? I've tried 1+x, Dr ; x+1/x, (1+xe^x) etc. Nothing seems to work

[shy cedar]

,rotate

[trim jolt]

@shy cedar Has your question been resolved?

[trim jolt]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wraith zinc]

hi, can someone find X as a final clear number, like 1.0326 or something? TY

[austere cedar]

First solve for x⁹
Then take the 9th root of both sides

[wraith zinc]

x^9 = 9 : 4

x^9 = 2.25

[proper kernel]

you mean as a decimal?

[wraith zinc]

Yes. Not sure what to do next

[proper kernel]

you have to use a calculator for that one though

[wraith zinc]

smart robot found for me answer like this but it doesnt make it easier for me tbh

x = 387 420 489 / 262 144 = 1 477,89. it cant be

[austere cedar]

,calc (9/4)^(1/9)

[solid kiln]

Result:

1.0942873805123

[austere cedar]

You are doing ^9, you want to be doing ^1/9

[wraith zinc]

.close

[trim jolt]

Channel closed

Closed by @wraith zinc

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[trim jolt]

@urban copper Has your question been resolved?

[granite light]

(1, -1, 1)a + (1, 1, 0)b + (0, -1, 2)c = (0, 1, 1)
&
(2, 1, 0)a + vb + (0, 0, 4)c = (0, 1, 1)

[hallow kite]

so by the first equation, a + b = 0, -a + b - c = 1, and a + 2c = 1

i believe you should be able to solve that for a, b, and c

and then plugging those into the second equation, find v

[urban copper]

I messed up

[hallow kite]

i think the basis vectors should be the columns

not the rows

[granite light]

is it gaussian elimination time?

[granite light]

why even write matrix if you are gonna solve it this way?

[urban copper]

my bad

but I relized mayube faster

[urban copper]

<2,2,-3> = v

.solved

[trim jolt]

Channel closed

Closed by @urban copper

Use .reopen if this was a mistake.

✅

[trim jolt]

@urban copper Has your question been resolved?

[granite light]

i don't know enough linear algebra to know what that last condition means

update: i figured out what it means

S also has to be proper subspace?

So if you downgrade this to 3d, then you have H as a plane, T as a line not on H, S as a line on H, T^\perp is a plane that intersects H at a line, and you are asked to find a line on H and another line given by the <-1,1,1,a> for some a, such that the plane on which those 2 lines lie is T^\perp

So only subspace S that will work is the intersection of H and T^\perp

I feel like finding that is the first step

$\mathbb{T}^\perp$

[solid kiln]

CyclicTree

[urban copper]

ok

how do I do that

[granite light]

$\begin{pmatrix}
x_1 \ x_2 \ x_3 \x_4
\end{pmatrix}
\cdot
\begin{pmatrix}
1 \ 1 \ -1 \1
\end{pmatrix} = 0$

[urban copper]

bmatrix not matrixb

[solid kiln]

CyclicTree

[granite light]

this is condition for a vector to be in T^\perp, right?

[urban copper]

yea

orthogonality

perpendicular

[granite light]

so intersection of T^\perp and H satisfied this condition and the condition for H

[urban copper]

unsure

can u do a drowing

[granite light]

i'm not sure i can draw intersection of 2 3d-subspaces in a 4d space

but like

v in A if condition1 and v in B if condition2 -> v in (A intersection B) if (condition1 and condition2)

[urban copper]

can u do it in latex

I dont really get it

I'm very confused about the direct sum

[granite light]

i'm not talking about direct sum yet, only about intersection

which is kind of an inverse of direct sum

but basically

from what i read not even an hour ago when i saw the problem

the idea is that you add vectors to the spanning set

direct sum of A and B is span(A, B)

so if you have 2 lines (passing through the origin) direct sum is the plane on which they both lie