help-39

[golden fjord]

what were you solving for next?

[manic cape]

neo helpp

[golden fjord]

im not the brightest Merai D:

[pine temple]

The unmarked interior angle (x and 51 are the other interior angles)

[golden fjord]

Ohhh

Okay so

The angle there right?

to the left of 117? that is the unmarked interior angle?

[manic cape]

can someone help ?

[golden fjord]

Wait what is it that you need help with I am confused

[manic cape]

ok

[golden fjord]

OH MERAI

[golden fjord]

You are the one who needs help

[manic cape]

yes

[golden fjord]

I didn't understand that

Lmao

Okay hi Merai

Did you understand how you got z?

Or do you still think z = 12?

If I am confusing you please ask

or tell me rather

[manic cape]

alr

is the sum of the triangle 180 0r 360 ?

[golden fjord]

The inside is 180!

[manic cape]

the whole thing

[golden fjord]

When looking at geometric objects we really only look at the inside angles, and since there is only one triangle

The entire triangle would have 180 degrees total

There are angles on the outside

But they aren't part of the triangle

They are just between two lines, that happen to be connected to the triangle

Sorry if I have confused you, but yeah, a triangle sums up to 180 degrees, always

and a triangle always has 3 angles

[manic cape]

alr

so how would i solve for z x and y

[manic cape]

nw

[golden fjord]

okay right

im just gonna add that angle called w there

is that okay with you

like the angle was already there

but it just didn't have a name

now we can both talk about that as w

[manic cape]

which one

[golden fjord]

look at this image

[manic cape]

oh

[golden fjord]

im calling that angle right there w

[manic cape]

alr

[golden fjord]

and to be clear i didn't actually add it, im just giving it a name

it already existed

now you want to find x, y and z

to do this you need to recognize a few things

first that the angle of a triangle always sum up to 180 degrees

and to avoid any confusion with regards to the angles outside of the triangle

the only angles that are a part of the triangle are

51, x and w

do you follow so far??

[manic cape]

yes

[golden fjord]

nice

something else that would be good to recognize

is the angle of a flat surface

do you know what the angle of a flat surface is?

[manic cape]

no

[golden fjord]

okay that is integral to this problem, so without knowing that, this would be pretty difficult to solve

the angle of a flat surface is 180 degrees

let me draw what i mean

[manic cape]

ok

[golden fjord]

im just calling the angle h

it looks rather silly

[manic cape]

nw

[golden fjord]

but if you are given the information that we have a plane surface, a flat one

you can always determine that h = 180 degrees

[manic cape]

thats 180 digree

yes

[golden fjord]

yes that is 180 degrees

and you might wonder how this is useful

when we have a triangle and a bunch of unknown stuff outside it

but look again

now let me give you a clue

and remember we know that h is 180 degrees on a plane surface

[manic cape]

yes

[golden fjord]

and do you notice the line i added

[golden fjord]

in this picture

[manic cape]

yes

[golden fjord]

we can describe 180 in a lot of ways if you think about it

we could say that 180 is

123 + 57

maybe 90 + 90

maybe 70 + 110

1 + 179

[manic cape]

yes

they all equal to 180

[golden fjord]

yes

[manic cape]

right ?

[golden fjord]

yes they do!

now do you see anything in this image

that reminds you a little bit of this image

[manic cape]

the base of the triangle is 180

oh nvm

its the w

hmm

[golden fjord]

it is okay if you are confused but, i think looking where the w is

is good

that is where i wanted you to look

but be careful

when you say the base of the triangle is 180, it isnt that the flat surface isnt 180 degrees,

but it is kind of odd to say it has a base

since it has 3 sides it isnt entirely clear in all cases

[manic cape]

yes

[golden fjord]

but anyway

so yeah

they look kind of similar

dont they?

[manic cape]

yes

[golden fjord]

and for a problem like this, i think it is safe to assume the flat surface is actually flat

even if it isnt stated

it is just convention to not state it sometimes

i dont really understand why

[manic cape]

yes

[golden fjord]

anyway

do you see any way to proceed by looking at those two images?

i will keep feeding you clues

[manic cape]

no

[golden fjord]

okay no problem

remember what i said about

how we can represent 180

we can write it as 123 + 57

1 + 179

[manic cape]

yes

[golden fjord]

well isnt it kind of interesting

that we have two numbers here

one unknown

and one known number

[manic cape]

yes

what do we do now

[golden fjord]

i will tell you but it would be sweet if you were able to see it on your own

im gonna draw another clue real quick then i will just tell you unless you want me to keep it from you a bit longer, is that okay with you?

[manic cape]

I don't have time

[manic cape]

yup

[golden fjord]

im not much of an artist

do you remember how we said that the flat surface creates an angle of 180 degrees

[manic cape]

that nice

[golden fjord]

and to be clear it isnt specifically for that spot, it is for any spot that is flat, anywhere that it is flat

[manic cape]

yes

[golden fjord]

right

[manic cape]

right

[golden fjord]

so we have two numbers

w + 117

and we know they need to equal

180

[manic cape]

yes

[golden fjord]

how would you make them equal 180?

[manic cape]

117+ 63

[golden fjord]

yes

that's right

now we know that w = 63

[manic cape]

ok

[golden fjord]

and with two angles and only one missing angle inside of the triangle now

[manic cape]

great

[golden fjord]

we know that

51 + 63 + x = 180

because the triangle also adds up to 180 degrees

you can also find z

using the same line of thinking we used for w and 117

except now the unknown angle is on the outside

[manic cape]

51+ 63 + 65 =180

[golden fjord]

i dont think that's right because 3 + 1 + 5 is gonna get you to something something then 9

but 66 would work, instead of 65

[manic cape]

sorry 66

[golden fjord]

yes no problem

[manic cape]

ur smart

[golden fjord]

thanks you too

sorry if it feels like i was wasting your time

and then for the last unknown

you have y

and now you need to use this idea one last time

[manic cape]

nope thx for ur help

[golden fjord]

that a flat surface has an angle that adds up to 180 degrees

[manic cape]

alr

[golden fjord]

you might be confused, but you hopefully caught that we did find x

what is x?

just to confirm

[manic cape]

66

[golden fjord]

right

so instead of just x + y = 180

we are gonna get

66 + y = 180

and we can solve that

it is all about perspective in this image

the surface is flat

we just rotated it

[manic cape]

what do you mean by flat

[golden fjord]

it isnt curved

if that makes sense

it is definitely on an incline

right?

we even know the incline

it is on a 51 degree incline

but that side is still flat, isnt it?

i drew it a little big

just to get the point home that x + y = h

and we know h = 180

so we know x + y = 180

and we know x

66 + y = 180

[manic cape]

yes

[golden fjord]

do you see what i mean by flat?

i understand it might not be super obvious when you see one of these problems the first time around

[manic cape]

not really

[golden fjord]

no okay

would you agree it is straight?

maybe the word flat is confusing you unnecessarily

[manic cape]

yes

[golden fjord]

like these are what i mean by flat/straight

the colored parts

[manic cape]

yeah

[golden fjord]

yeah yeah ok

yeah so if it is straight/flat

it has an angle of 180 degrees

you can draw that angle

on the other side by the way

actually i should say the surface has an angle of 180 degrees

so if you look at both surfaces it would have 360

then you would just draw a circle around the line

i should apply for art college

i think i could make it big in the world of art

but we didnt draw a circle, because we were only interested in the surface there

since the unknown angle w and the known angle 117 was both on one side of the straight line

we didnt have to think about the other side/other surface

anyway if you have more questions ask them or otherwise you can do .close to free up this chat room 😄

[manic cape]

so what did y = ?

[golden fjord]

if you know x i trust you can figure out what y is

[manic cape]

y = 114

[golden fjord]

yes!

that's right

and you found z already as well?

[manic cape]

i'm not sure if i would be able to solve the same way on the test

[manic cape]

51 i think

[golden fjord]

hmm that isnt correct for z

[manic cape]

what was it

[golden fjord]

do you remember how we found w?

and do you remember how you found y?

[manic cape]

51+66

[golden fjord]

Could you show me your work?

[manic cape]

66+ y = 180

y=114

[golden fjord]

okay so not 51 + 66 for y

[manic cape]

yes

[golden fjord]

this is correct

[golden fjord]

and this is correct

and it makes sense how you got 114

[manic cape]

yes

alr thx for your help

[golden fjord]

!done

[pearl pond]

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

@manic cape Has your question been resolved?

[pearl pond]

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.

[slim matrix]

.close

[pearl pond]

Channel closed

Closed by @slim matrix

Use .reopen if this was a mistake.

[slim matrix]

.reopen

[pearl pond]

✅

[slim matrix]

.close

[pearl pond]

Channel closed

Closed by @slim matrix

Use .reopen if this was a mistake.

[oak ivy]

Bruh

open close open close

[slim matrix]

I got the answer sorry 🥲i thought i didn't get it so i reopened

[oak ivy]

ok

[slim matrix]

But yea i fot it now

[pearl pond]

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.

[sharp smelt]

Trying to show this converges

[sharp smelt]

now I know that $\frac{1}{n^n}$ converges

[jolly parrot]

What a wonderful world !

[sharp smelt]

so $\int \frac{1}{n^n}$ converges

[jolly parrot]

What a wonderful world !

[sharp smelt]

Now let ln(n)= t

so $\int \frac{e^t}{ t^t}$ is what I want to check for conevrgence

[jolly parrot]

What a wonderful world !

[sharp smelt]

so this is equivalent to showing $e^t/t^t$converges, which it does via the root test

and we're done?

[tropic saddle]

didnt we do this series literally 3 days ago?

[sharp smelt]

no, I wasn't abel to finish it

[tropic saddle]

but why are you now doing a different approach

[jolly parrot]

What a wonderful world !

[sharp smelt]

Just thought of this, so might as well try it?

[tropic saddle]

you should also try the old approach

[sharp smelt]

will do it now

Does this work though

I don't get this

[autumn fossil]

The point is that loglogn will eventually be greater than 2

so you can compare it with 1 / n^2

$\frac{1}{n^{\ln\left(\ln\left(n\right)\right)}}\le\frac{1}{n^{2}}$

[jolly parrot]

MathIsAlwaysRight

[autumn fossil]

for sufficiently large n, you should have this

[sharp smelt]

hmm

yea

but I'm looking at log^log

[autumn fossil]

$\left(\ln\left(n\right)\right)^{\ln\left(n\right)}=n^{\ln\left(\ln\left(n\right)\right)}$

[jolly parrot]

MathIsAlwaysRight

[autumn fossil]

ann already told you this yesterday, try proving it

[sharp smelt]

I see

Yea, makes sense

thanks

[wet osprey]

What in the world…

[autumn fossil]

Did i typo somewhere

[wet osprey]

Idk I’ve never seen this before

[autumn fossil]

Its jusr log rules

[sharp smelt]

Now trying to determine if this sequence is monotone and bounded.
\
To do so , we look at $a_{n+1}-a_n = \frac{ 2n-1}{3n+7}- \frac{2n-3}{3n+4} = \frac{17}{9n^2+33n+28}$, whose roots are both negative, for $n>0$, the function is strictly positive. and decreasing. Thus it converges

[jolly parrot]

What a wonderful world !

[sharp smelt]

It converges to 2/3

.close

[pearl pond]

Channel closed

Closed by @sharp smelt

Use .reopen if this was a mistake.

[pearl pond]

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.

[sharp smelt]

We wish to find the radius and interval of convergence, once more by the ratio test, we find that we have $\frac{x}{n+1}$, which goes to zero for all n, thus the intevral of convergence is $\R$

[jolly parrot]

What a wonderful world !

[rough forge]

ok

its just the power series of e^x anyway

[sharp smelt]

yea

[pearl pond]

@sharp smelt Has your question been resolved?

[inland ivy]

you mean for all x

[pearl pond]

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.

[sharp smelt]

yea, oops

[sharp smelt]

Anyway, I'd now like to find the radius of convergence for this

To do so we, use the comparison test, followed by the ratio test

$\frac{ (x-2)^n}{n^2+1}< \frac{(x-2)^n}{n^2}$
\
Using the root test for the latter, we find $\abs{x-2}<1$, so $1<x<3$

[jolly parrot]

What a wonderful world !

[sharp smelt]

Is this fine so far

[rough forge]

looks ok

but why would you use comparison test

[sharp smelt]

fair

yeah, (n^2+1)^{1/n} is 1 anyway

[rough forge]

not my point actually, just havent seen it before

[slow grove]

ratio test

is the standard

[rough forge]

my concern is what if you lose information by that

[sharp smelt]

yea, but root test works too\

[sharp smelt]

fair

I'd like to do this next( Radius of convergence , again)

Ratio test

[rough forge]

sounds good

[sharp smelt]

$\frac{(n+1)! (2x-1)^{n+1}}{ n! (2x-1)^n}$

[jolly parrot]

What a wonderful world !

[sharp smelt]

so $n \cdot (2x-1)$

[jolly parrot]

What a wonderful world !

[sharp smelt]

Which diverges everywhere but x=1/2

Does this work

.close

[pearl pond]

Channel closed

Closed by @sharp smelt

Use .reopen if this was a mistake.

[pearl pond]

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.

[cobalt acorn]

okay so this is like a really easy question but im not sure why i keep getting wrong answers, I've redone this 9 times and im not sure at all like whats wrong or if my starting equation is even right

[sweet oxide]

you ignored the parantheses

when multiplying

This rule has a fancy name, which name I forgot wait

[cobalt acorn]

oh waiy

oh my gosh

im sorry for wasting so much time

im like

half asleep rn

i dont know how i glossed over that

[sweet oxide]

distributive law

[cobalt acorn]

is the starting equatoin right?

like in relation to the problem im solving

[sweet oxide]

yea

[cobalt acorn]

okkk

omg i understand now

thank you so much like actually

i don't know how i did that wrong so many times but we ball

.close

[pearl pond]

Channel closed

Closed by @cobalt acorn

Use .reopen if this was a mistake.

[pearl pond]

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.

[last meteor]

We select three points at random on the circumference of a circle. What is the probability that ∆ABC
contains the centre O in the interior?

pls help

[brave hawk]

!stage

what

!stages

[errant cedar]

!status

[pearl pond]

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

[brave hawk]

status

right

thanks

[brave hawk]

in such problems it always helps to fix one point

so you pick point A and rotate the circle so that it's on top on on the right or whatever

!occupied

[pearl pond]

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).

[last meteor]

right but then there's a third point id need to account for

[brave hawk]

so you now have a problem of only 2 random points

that was the easy bit

the next task is to find a condition for the center to be in the traingle

lemme think

[last meteor]

yup this is the problem

two random points, both dependent on each other

[brave hawk]

hm

you can also assume that B is on the right half on the circle

now draw the lines AO and BO and continue them to the other side of the circle

[last meteor]

and what about the last point

[brave hawk]

what happens if C is inside the blue region, what happens if it isn't

[pearl pond]

@last meteor Has your question been resolved?

Channel closed

Closed by @last meteor

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.

[mortal spade]

Q23

[pearl pond]

@mortal spade Has your question been resolved?

[compact ridge]

you can clearly rule out some values of n above the minimum possible

[sonic hound]

IOQM 2024.

[mortal spade]

I am not quite sure how to do that

[compact ridge]

okay, let's see what happens if n = 14

what would 13^4 mod 14 be equal to?

[mortal spade]

It would be 1

Which is equal to 1⁴

[compact ridge]

yeah, because 13^4 mod 14 = (-1)^4 mod 14 = 1^4 mod 14

this is the key bit

now what would happen mod 15?

[mortal spade]

14⁴ will be = 1mod 15

[compact ridge]

great, and how about mod 16?

[mortal spade]

Fo mod 16 it is =8

Wait no

It is divisible

[compact ridge]

well I mean if you want to generalise this like before

14^4 is the same as which other number mod 16?

[mortal spade]

2

[compact ridge]

2^4 but yep

[mortal spade]

So 14⁴ will be = 3⁴ ???

For 17

[compact ridge]

yep, if you try mod 17, correct

[mortal spade]

So this pattern continues until 28 ??

[compact ridge]

yep, so you've identified the key number to check, 28

it turns out that n = 28 also doesn't work

with 13^4 mod 28, well we can split it up into 13^4 mod 4 and 13^4 mod 7

likewise consider 15^4 mod 4 and 15^4 mod 7

[mortal spade]

For 29 it wont work

[compact ridge]

and well I know you won't have Wolfram Alpha in the exam, but n = 29 works similarly to n = 28: it also doesn't work

I can't explain why exactly this is but all composite n don't work

yeah it must be similar to the above reasoning I gave

[compact ridge]

turns out n = 31 works

[mortal spade]

So is 31 the answer ???

[compact ridge]

yep

[mortal spade]

Oh

Thanks for the help

I will try to understand it in abit more detail

The composite part

Tysm

[compact ridge]

yeah Google AI is wrong a lot of the time, but it's suggesting $x^4 - y^4 = (x - y)(x + y)(x^2 + y^2)$ mod n, and then showing that their product is never zero

[jolly parrot]

south

[compact ridge]

when you have n being composite, there are more ways to achieve this

as in, 2 * 3 * 5 = 0 mod 30, even though none of 2, 3, 5 are divisible by 30

when you have n being prime, that forces at least one of the brackets to be divisible by n

[mortal spade]

I am unable to understand it right now

I maybe need to learn some theory before

.close

[pearl pond]

Channel closed

Closed by @mortal spade

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.

[last meteor]

help

[broken fossil]

strange ahh limit

[last meteor]

ikr

i cud simplify it to 2x * sin²x

but that dont help shit

[broken fossil]

maybe just taylor series expand

[last meteor]

oh ywah

that could help

lemme try

but... even then the x terms would be in the numerator :/

[compact ridge]

how? that looks wrong

[last meteor]

oh right

nah i fkd up wait

yeah i cant really make any headway any help is appreciated

[broken fossil]

what do you mean by this

[last meteor]

strike that, i made a mistake

[compact ridge]

it is true that you can let $u = \frac{1}{\sqrt x}$ however and then you need to find $\lim_{u \to 0^+} \left(\frac{1}{u^2} - \frac{1}{u^2} \cos u \right)$

[jolly parrot]

south

[last meteor]

even that is tedious, no?

[compact ridge]

and this should look much more familiar

[last meteor]

lemme try

[compact ridge]

no, it's a standard limit now

[last meteor]

oh youre right

ahhh

my goat south

thank you

.close

[pearl pond]

Channel closed

Closed by @last meteor

Use .reopen if this was a mistake.

[last meteor]

is the limit 1/2 @compact ridge

[compact ridge]

indeed, yes

[last meteor]

W

[pearl pond]

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.

[gentle lichen]

the gravity down an incline is mgsinθ

So the acceleration of the stone will be gsinθ

Use that for the first question

[pearl pond]

@stoic imp Has your question been resolved?

[stoic imp]

WTF?

[stoic imp]

we dont know the mass of the stone broski

????

.close

[pearl pond]

Channel closed

Closed by @stoic imp

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.

[vital slate]

I'd like your help on understanding the following linear algebra problem on a conceptual level. Given a polynomial of degree 3, I'd like to find a least-squares approximation in degree 2. For the example, let's say my polynomial is (p(t) = t^3). I need to find some (q(t) = a_1 + a_2 t + a_3 t^3) that best approximates (p(t))

Let the degree 3 polynomial be in an inner product space V, spanned by the basis ({1, t, t^2, t^3}). I need to find a polynomial in the subspace U, spanned by the basis (1, t, t^2).

I already know two approaches to solve this. The first approach uses the definition of best approximation, which states:

A vector (u \in U) is a best approximation of (v) if and only if (v - u \perp U). This is analogous to saying that the inner product of (v-u) and the subspace (U) must be zero, i.e. the error exists only in the orthogonal direction.

I use this definition by defining a system of equations with the inner product and the vectors of the "destination" basis U. I set up the equations:

\begin{align*}
\langle p-q, 1 \rangle &= 0 \
\langle p-q, t \rangle &= 0 \
\langle p-q, t^2 \rangle &= 0
\end{align*}

This sets us up for a system of equations, of which we can then solve for the coefficients (a_1, a_2, a_3). It works perfectly fine.

[vital slate]

The second approach that I know uses projections more explicitly. It involves taking the basis of (U) which is ({1, t, t^2}), and then orthonormalizing it using Gram-Schmidt orthonormalization. This allows us to derive an orthonormal basis for the inner product space of degree-2 polynomials. That orthonormal basis is ({1, 2\sqrt{3}t - \sqrt{3}, 6\sqrt{5}t^2 - 6\sqrt{5}t + \sqrt{5}}). Then, I can simply project the polynomial (p(t)) upon said basis. This also works, I have verified it and obtained the same answer as the first approach.

However, I know there's a third approach, but I'm not quite sure if I know how to use it. I know that there is a way to obtain the Singular Value Decomposition of a matrix, and then calculate it's pseudoinverse. Presumably, that would first involve setting up a 4 by 3 matrix that represents the mapping from degree 3 to degree 2. Using the pseudoinverse, I would be able to find the least squared approximation, but my knowledge gets hazy here.

So my question is, given my goal of finding a degree 2 polynomial approximation of a degree 3 polynomial, how do I do so using SVD and the pseudoinverse?

[jolly parrot]

RiverineDolphin

RiverineDolphin

[vital slate]

Alright, I finished editing the latex.

Looking forwards to hearing your advice.

To provide a little bit of context, this is the formula I'm familiar with in class.

Given an arbitrary m by n matrix A in field F, I'm comfortable computing it's singular value decomposition, and it's pseudoinverse, etc.

And I can setup the system x = A^{\dagger}b without any trouble.

But I'm stumped when I'm trying to apply this knowledge to a more "real world" example of approximating polynomials.

Like, if I have a degree 3 polynomial p(t) = t^3, and I want to approximate it using a degree two polynomial, how do I do it using the pseudoinverse?

I must set up some sort of matrix A, but I don't know how.

I know how to find approximations using the 1st approach and the 2nd approach, as I've written earlier. But I don't know how to do it using SVD

<@&286206848099549185> Please help, dear math gods. I even typed out my question nicely in LaTeX for you! 😭

[pearl pond]

@vital slate Has your question been resolved?

[hazy pilot]

#numerical-analysis

[pearl pond]

@vital slate Has your question been resolved?

[pearl pond]

@vital slate Has your question been resolved?

[karmic night]

I'm not any kind of expert here, but I think you might have to sample your functions.

[vital slate]

That was what I figured out as well.

Thank you 🙂

.close

[pearl pond]

Channel closed

Closed by @vital slate

Use .reopen if this was a mistake.

[pearl pond]

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.

[deft hinge]

can someone help idk what its trying to ask

[pearl pond]

@deft hinge Has your question been resolved?

[pearl pond]

@deft hinge Has your question been resolved?

[deft hinge]

uhh

<@&286206848099549185>

[tiny lintel]

Can you break the red curve in 2 to compute part A ?

Upper red and lower red are 2 different functions, can you find their expressions

[deft hinge]

ohh

[deft hinge]

wait how do i do part a

is it like

the same for part a and b

[tiny lintel]

What are the 2 red functions first ?

[deft hinge]

x=(y-3)^2 - 2 and y=x^2+1

[tiny lintel]

No

I want y = f(x)

[deft hinge]

ohh

uhh would it be y=sqrtx+2 + 3

[tiny lintel]

That’s one part of It Let’s call it f

What’s the second part

[deft hinge]

would it be just x^2 + 1

for second

[tiny lintel]

No

[deft hinge]

or would i make it into x

[tiny lintel]

If x = (y-3)^2 + 3 . What is y ?

[tiny lintel]

You’ll get 2 expressions for y just for this one

[deft hinge]

i got sqrtx-3 and 3

y=3+ sqrtx-3 and y=3-sqrtx-3

uhh

[tiny lintel]

Tell me if there's something you don't understand

[deft hinge]

uhh

the 1 thats like

the integral h(x) - f(-2)

[tiny lintel]

You should also find that A = B. Because your graph is symmetric along the line y = x + 3

[tiny lintel]

no it's the integral of h(x) - f(x). you need to find the intersection point

[deft hinge]

ohh

wait for like the 1 thats around b would it be lower limit - upper

[tiny lintel]

there's not limit here ... It's just that you have 3 functions (and not 2)

I showed you that the red curve was actually 2 functions = yellow + pink

[deft hinge]

ohh

wait would i have to like add the expressions together to show the limits of integration like integral of a + integral of b

[tiny lintel]

No, I gave you the three integrals that you need to compute

They're written in black

you just have to find the limits for each

and that's where the functions intersect

FOr the 1st integral for instance

find the points where f(x) = g(x)

and that's your integration bounds

for the second one, it's when f(x) = h(x)

[deft hinge]

oops back

[deft hinge]

would like the upper bound for first integral be -1 and like it would be - 1-sqrt5/2

[pearl pond]

@deft hinge Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

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.

[sharp smelt]

I'm trying to solve this using the method of Lagrange multipliers

[sharp smelt]

we have $(4,6) = \lambda(2x,2y)$

[jolly parrot]

What a wonderful world !

[sharp smelt]

so $4 = 2 \lambda(x) ; 6= 2\lambda y$
\
We thus have, on squaring and adding,
\
$52 = 4 \lambda^2(13)$

[jolly parrot]

What a wonderful world !

[bitter herald]

So, what do you need help with?

I'd also try to eliminate the lambda by the way

[rough stream]

You look like you're doing quite well on your own, you don't even need us

You don't even need me

[bitter herald]

Being independent is important

[rough stream]

The world is scary alone

[sharp smelt]

so $\lambda =\pm 1$, right

[jolly parrot]

What a wonderful world !

[bitter herald]

indeed

[sharp smelt]

I like solving stuff on mathcord, makes me feel safe

😭

[bitter herald]

Hence, x = \pm 2 as well

[rough stream]

See? Alone is scary

[sharp smelt]

hmm

and y = \pm 3

[bitter herald]

Seems like you are done

[sharp smelt]

I now evaluvate the function at (2,3)

[rough stream]

Where they all have the same sign of course

[sharp smelt]

and (-2,-3)

it's evident (2,3) should be the maximum(26)

and the minimum would be -26

[sharp smelt]

yea, thanks

I'd like to do one more problem, with 2 constraint this time

here we have $(y,z+x, y) = \lambda(y,x,0); (y,z+y,y) = \mu (0,2y,2z)$

[jolly parrot]

What a wonderful world !

[sharp smelt]

so we have y=0

as 0=2 \mu z

we have z=0

I'm confused

what am I doing wrong

.close

[pearl pond]

Channel closed

Closed by @sharp smelt

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.

[modest tartan]

what to do now?

[modest tartan]

its solve for x

is the ans 10?

[toxic lichen]

$x^{\log(x)} = (10^{\log(x)})^{\log(x)} = 10^{\log^2(x)}$

[jolly parrot]

Ann

[modest tartan]

take log on both sides and get log^2(x)= 1

[toxic lichen]

indeed

[modest tartan]

then logx = 1

x= 10

got it

thx

[toxic lichen]

log(x) = ±1

not just 1

[modest tartan]

-1 rejected?

base greater that 1

[toxic lichen]

log(x) = -1 not x = -1

[modest tartan]

oh

the x=1/10

man i forgot since we just started taking root x^2 as modx

thx

.close

[pearl pond]

Channel closed

Closed by @modest tartan

Use .reopen if this was a mistake.

[pearl pond]

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.

[pearl pond]

@stoic imp Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

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.

[void grail]

"Solve these equations for the u,v, unknown polynoms in the given polynom rings."

Please help with this, because I find it difficult to understand polynoms. What is the motivation, are there good ways t ounderstand, relate polynoms? Maybe through applications or significance in maths, or other fields?

[pearl pond]

@void grail Has your question been resolved?

[late spindle]

@void grail you again! the motivation for polynomial rings is simple: to generalise normal polynomials we learn in basic algebra and learn something(hopefully) by zooming out. it consists of a dummy variable and a coefficient ring.

oh, and a evaluation function that evaluates given polynomials at a certain point

[void grail]

Where can you learn about this, from textbooks of algebra? I found there is a section in Serge Lang Basic Mathematics, I'll start there

Thank you @late spindle

[late spindle]

i read from a book, dont remember title unfortunately, let me look up sth

[void grail]

Oh and they have to do with modules, which seem to be very interesting

I think they do

A polynomial ring is a module?

[late spindle]

umm im not strong on this area, but heres a link https://www.reddit.com/r/math/comments/z1pd66/good_intro_to_abstract_algebra_books/
and a linkhttps://en.wikipedia.org/wiki/Module_(mathematics)#Examples

[void grail]

Nice, thank you

[plucky python]

well Z[X] is a Z-module, but something like R[X] is an R-module, otherwise known as an R-vector space

[void grail]

What do you mean by zooming out? Generalising, abstracting, and seeing how the pattern applies to different perspectives / fields, in different contexts? Like seeing that R[X] is a R-vector space and there is a relationship of vector spaces and polynomial rings? Polynoms and vectors are the same /similar? This is what you mean?

[late spindle]

in a word,

YES

[void grail]

[pearl pond]

@void grail Has your question been resolved?

[void grail]

I thought I share this, which seems interesting https://ncatlab.org/nlab/show/polynomial

there is a "deeper objective meaning of definable operations which of course is the actual point of it all."

What is this objective meaning of definable operations?

[pearl pond]

@void grail Has your question been resolved?

[scenic prism]

Looks like it’s related to computing the GCD

By the Bezout’s theorem, the right hand side must be the multiple of the GCD of the two known polynomials on the left hand side

You can calculate (a, b) and u, v satisfying au + bv = (a, b) using the extended Euclidean algorithm

Then multiply them by RHS / (a, b) to get the answer

[void grail]

Thanks so much @scenic prism

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

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.

[quasi turtle]

The third direction kinda dominates i believe here....

so if i was just to minimize that couldnt that get me close to the solution?

for example i could go -1 - \alpha 1024 = -5

then \alpha = 5/1024 = .0039. this is close to the secant methods solution?

[pearl pond]

@quasi turtle Has your question been resolved?

Channel closed

Closed by @quasi turtle

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.

[agile parcel]

hey how do i properly find the bounds for u?

[agile parcel]

.close

[pearl pond]

Channel closed

Closed by @agile parcel

Use .reopen if this was a mistake.

[pearl pond]

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.

[proven arch]

Hi, could someone explain why we use 1 for the upper bound? I don't understand why we cant just use -5<=x<3 and use 3 for the upper bound and -5 for the lower bound to solve it

[proven arch]

uh one sec

[toxic lichen]

3 is the y-value, not the x-value

you need to involve 1 as an integration bound in order for the output of FTC to tie back to something you know (which is f(1))

$\int_{-5}^3 f'(x) \dd{x} = f(3) - f(-5)$ and you simply cannot do anything with that!

[jolly parrot]

Ann

[proven arch]

so like

[proven arch]

if we just used geometry to solve this, we wouldnt get what f(-5) equals?

[lilac jackal]

u dont know f(3) so u will get stuck

[proven arch]

if i use geoemtry just to get what f(x) equals over -5<x<3 then it wouldnt be possible to get what it equals at the point f(-5)?

[verbal whale]

But how would you use geometry?

[proven arch]

kinda like this but also including 4 from 1 < x < 3

[lilac jackal]

sure u can use geometry to integrate but u forget that the other part of FTC is specified by known values of f

see where they use f(1)=3

[proven arch]

so if I did f(x)= (-9-3/2+1+4), I integrated it for f(x) but that gives the total area, but not at f(-5).

in order to find f(-5) using FTC we have to have atleast one specific value?

or the area under it doesnt matter because its for y value

or like

[lilac jackal]

yes. in general if u have 3 things in an equation u need to know or be able to find 2 of them to find the last

f(1) is known and integral is easy to find, so f(-5) can be found

[proven arch]

ed

wait

i think i understand it now

ty

.close

[pearl pond]

Channel closed

Closed by @proven arch

Use .reopen if this was a mistake.

[lilac jackal]

np

[proven arch]

sry rq would it be possible to find f(-5) if they didnt provide f(1)?

[lilac jackal]

no

[pearl pond]

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.

[silent rock]

hiker travels from point A to point B and back, reaching B at 16;15
from A to B - 9km distance travelled
from B to a - 12,5 km distance travelled, 1km quicker speed, 15minutes (1/4 hours) longer time
whats the time when hiker is at point A before he starts the hike

ive managed to get to the point that x = both 4 and 9, which i dont see how its possible
(9/x = 12.5/x+1 + 1/4 is my equation)

[pearl pond]

@silent rock Has your question been resolved?

[silent rock]

<@&286206848099549185>

[covert wolf]

Show ur working out

[silent rock]

no joke i just figured it out

[covert wolf]

And i dont think the q makes sense how can A to B be 9km but b to a be 12.5km

[silent rock]

takes a longer path

for whatever reason

[covert wolf]

Oh right okay

Aight so u don't need hlel?

Hell

Help

[silent rock]

yeah not anymore

thanks for coming in though

.close

[pearl pond]

Channel closed

Closed by @silent rock

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.

[covert wolf]

Can someone help me with this

My first thought was to do (cos(theta) +isin theta) ⁴ binomial expansion

[inland ivy]

That sounds like a good idea

[covert wolf]

Giving me cos4thsta and sin4theta expressions

But then it got alot more complicated tbh and thought there probably is a better way to go about it

And I've got no clue on part b

I mean i see the pattern but can't figure out how to do it at all

[inland ivy]

it shouldn't be too complicated

[covert wolf]

Mabye not but alot of simplification and matching trig identities

[inland ivy]

you just divide the expressions you get for sin4theta and cos4theta

to get tan4theta

[covert wolf]

Yh ik but the RHS When simplifying

Is long as hell

[inland ivy]

It should only be 5 terms

[covert wolf]

Okay mabye its doable forsure ik i csn do it, is there an alternative method+

And mainly part B is what i got no clue on

[inland ivy]

The basic concept will be the same

[covert wolf]

Yh i was thinking if i can write some sort of complex number in terms of tantheta couldn't think of anything

[inland ivy]

You might be able to simplify the calculations a little bit by writing $$\cos 4\theta+i\sin 4\theta=(\cos\theta+i\sin\theta)^4=\cos^4\theta(1+i\tan\theta)^4$$

[covert wolf]

So i did the first method as a last resort

[inland ivy]

so that you can only work in tan's

[jolly parrot]

kheer257

[covert wolf]

Hmm seems reasonable tbf

Aight what abt part B+

[inland ivy]

that's a good question

maybe you can rearrange the equation to get something of the form $$c=\frac{4x-4x^3}{1-6x^2+x^4}$$

[jolly parrot]

kheer257

[covert wolf]

Yh these qs are fked lmao

Harder than usual exam qs

[inland ivy]

Are these A levels?

[covert wolf]

Yh

Alevel furthermaths but selectively chosen to be hard

[inland ivy]

This is a pretty fun question

[covert wolf]

Oooh i see

Smart trick lmao

Think ur way is the way to go

Bet i got another q

Was confused abt this one

I can't really apply a graph transformation can I?

[inland ivy]

a general method of solving these is to set $y=x^2-1$ (or whatever other function) and solve for $x$ in terms of $y$, then substitute that into the equation to get an equivalent equation in $y$

[jolly parrot]

kheer257

[inland ivy]

it doesn't quite work here because the roots can be both positive and negative

[covert wolf]

I tried that it didn't give me a quartic

[inland ivy]

but you can try playing around with this

[inland ivy]

I'm assuming you have to bring all the square root terms to one side and square it

[covert wolf]

Hmmm would that maintain the same solutions?

Nvm ofc it does.

I was thinkign against it bc i thought it would change tje solution

[inland ivy]

actually only after squaring will it have the same solutions

(I think)

because the two equations with the positive and negative square root each will have 2 roots but when you square they become one

[covert wolf]

Ik the method is there a understanding behind it. For general qs like new eq with roots 2(alpha,beta gamma...)

For that i simply think of graph transformations and it makes sense

F(x/2) stretches graph factor 2

So root will be double

That's how i think of it. Why does your general method work and what's the idea?

[inland ivy]

it's a similar concept here

when you set x = f(y) you go from solving P(x)=0 to P(f(y))=0, so the roots for the new equation in y become f^-1(roots of P)

[covert wolf]

The reason i struggled with that q is because ive never encountered a graph transformation with square roots

[covert wolf]

I kinda understand it thanks. Makes alot of sense tbh

Leme see if i hsd another q 1sec

I did a 3hr hard paper

Tbats y

So essentially im pretty sure i got it right

But i want some intuition

The idea is i do A x general point x, mx+c

And it forms a new point X, mX+c

[inland ivy]

essentially you just want to show that A has no real eigenvalues

[covert wolf]

Egen value is not in my specification

They use discriminate

[inland ivy]

the concept is the same as what you're saying

[inland ivy]

one of the ways to find eigenvalues is through a discriminant, yes

[covert wolf]

But uhhh what they do is multiply thru get a expression interms of x m c eliminating X

Then they use discriminant on a equation interms of M and X and show m has no real solutions

They've kind of ignored C not sure why

Can u take a look at 23a

[pearl pond]

@covert wolf Has your question been resolved?

[errant cedar]

i think you interpret the derivative geometrically and that’s it

[covert wolf]

Wdym

Takw the derivative

Sub x=4

Then tan-1(dy/dx atx=4)?

[errant cedar]

seems like a plausible way to get the angle

[covert wolf]

Bet thanks

[errant cedar]

they might want the blue angle tough

[covert wolf]

Ahhh okay so 90-thsta

[pearl pond]

@covert wolf Has your question been resolved?

[pearl pond]

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.

[woeful harness]

can anyone tell where i went wrong using gram schmidt ?

[woeful harness]

v1 and v2 are correct but v3 is wrong

[pearl pond]

@woeful harness Has your question been resolved?

[pearl pond]

@woeful harness Has your question been resolved?

[pseudo plank]

looks like you just added/subtracted wrong in that last step

also dont forget to normalize your vectors for the final answer

[pearl pond]

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.

[azure ether]

Help

[azure ether]

<@&286206848099549185>

[plush bramble]

!15m

[pearl pond]

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

[azure ether]

Can u help me with that

[deft hill]

draw a graph

[karmic pier]

find the derivative

[plush bramble]

#help-39 message

[clever zenith]

In what the graph would help, we rate of change (rising or falling)

[azure ether]

How

[deft hill]

It js helps me visualize these kinds of problems sry

[azure ether]

I got like an equation without t

lol I cooked

[karmic pier]

https://tenor.com/view/dr-house-gif-13390470055447068071

brother

[azure ether]

https://tenor.com/view/baffling-gif-24471655

[karmic pier]

expand the expression first

(f(x) + g(x))’ = f’(x) + g’(x)

use that

[azure ether]

I did that?

Derivative of t is just 1

Right?

[karmic pier]

oh you did im an idiot

sorry its late

[azure ether]

It’s ok

[karmic pier]

aight let me look this over

[azure ether]

Yeha

[gray lantern]

chain rule 🙏 🙏

[karmic pier]

oh

[clever zenith]

also use (f(g(x)))'=g'(x)*f'(g(x))

[karmic pier]

yeah

[azure ether]

Find my mistakes

[karmic pier]

chain rule

you didn’t do that

[azure ether]

But 0.503 is not a function 💀

It’s a constant

[karmic pier]

yes it is

well

it is also a constant

[azure ether]

Ok let me try

[clever zenith]

your mistake is in the 4th line

[karmic pier]

a constant function, if you will

[azure ether]

I need to use the chain rule instead of product rule?

[karmic pier]

cos(something) requires chain rule

if that something isn’t x

[azure ether]

Okay

[gray lantern]

sine is a function

in itself

[karmic pier]

cos(x) = f(x)
cos(whatever) = f(g(x))

use chain rule

[azure ether]

Got it

Like that?

[clever zenith]

also there is a function (not a constant function) it's not 0.503, it's 0.503(t-6.75)

[azure ether]

Agreed

[clever zenith]

seems good

no my mistake, why is the 5 not connected to the sin? (by multiplication)

[azure ether]

Yeah it’s multiply

-sin

[clever zenith]

why not use brackets?

[azure ether]

Cuz I forgot

[azure ether]

Like this?

[clever zenith]

yes, but why not compute 5*0.503?

[azure ether]

Oh.

[azure ether]

Nvm

[clever zenith]

just calculate with the formula D'(noon)

[azure ether]

Oh Yeha can u help me with another question?

[azure ether]

Ik just plug in 12

[clever zenith]

I can try but no promises

[azure ether]

I just found out that my teacher didn’t provide example 5 here

I will skip c

[clever zenith]

do you need help with all three questions, or did you solve some of them? (also I don't know what is isothermal compressibility)

[azure ether]

Only b

[clever zenith]

If I'm not mistaken this is the graph:

So by this I would say it is decreasing more rapidly at the start but I'm not sure as I don't know a lot about physics, so take my word with a grain of salt.

[azure ether]

When p is smaller dv/dp is larger

And when p is larger dv/dp is smaller

I got it

Thx mate

.close

[pearl pond]

Channel closed

Closed by @azure ether

Use .reopen if this was a mistake.

[pearl pond]

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.

[hard crystal]

How to approach this problem?

[toxic lichen]

,rccw

[toxic lichen]

what is this symbol next to the k?

is it pi?

[hazy pilot]

looks like korean j

but probably pi

[toxic lichen]

it looked like korean j to me too that's why i asked

for the record i dont see a good way to do this except maybe to write cos(kpi/n) as 1/2 (e^(ikpi/n) + e^(-ikpi/n)) and somehow rawdog the double summation

[pearl pond]

@hard crystal Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[pearl pond]

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.

[fathom valley]

The answer is wrong, but I have no idea where I went wrong

[fathom valley]

Controlled the answer from the back of the book, it should be x=7

[toxic lichen]

4(x-2)^2 ≠ 4x^2 - 4

i have a feeling you made 2 different mistakes there

[fathom valley]

could you help me with them?

i have been at it for 30 mins and still cant figure it out 💀

[toxic lichen]

30 minutes is a lot.

ok, so just so i know how best to explain it, what level of education are you at?

[fathom valley]

9th grade

[toxic lichen]

ok

let's forget about all the other shit in the equation and just focus on one bit

(x-2)^2

can you tell me how to expand this properly?

[fathom valley]

alr

well its a equation

so

[toxic lichen]

nope

(x-2)^2 is an expression not an equation

[fathom valley]

got it

english is not my first language, sorry

[toxic lichen]

what is your first language?

[fathom valley]

estonian

[toxic lichen]

ok, nevermind

was asking to see if i speak it

i don't

ok so

(x-2)^2 = ?

[fathom valley]

understandable

x^2-2x+4