help-41

[floral totem]

Az+ Bz- + C =0
What are the conditions A,B,C need to meet so that the equation above represents a line in the complex plane

[ripe cipher]

do you perhaps mean $Az + B\overline{z} + C = 0$

[grizzled pagoda]

Ann

[floral totem]

Yes

[ripe cipher]

ok, and are A, B and C supposed to be real?

[floral totem]

Sorry i am not aware of how we symbolise conjunctive in text

[patent raptor]

z*

[ripe cipher]

conjugate*

[floral totem]

That is not stated no

[ripe cipher]

yeah your best bet is z* or maybe something like conj(z)

[floral totem]

Alr

So do you guys have any idea about this problem

[ripe cipher]

it's poorly stated and we're not even allowed to assume A, B, C are meant to be real numbers.

are we at least allowed to assume they're complex somehow

[floral totem]

Idk tbh

I told you how its stated

Well if its real then we just need (A-B) =/ 0 OR (A-B)i =/ 0

But idk what to do if A,B,C is complex

[floral totem]

You have any idea for this case ?

[patent raptor]

Did you try to isolate y?

[floral totem]

How

Splitting A,B,C to real/Imaginary parts?

Because since we got a complex equation it splits into 2 Real ones

And idk what to do with the 2 equations

[amber wasp]

@floral totem Has your question been resolved?

[patent raptor]

to me it seems only way is to multiply the equation by some nonzero complex number to make A,B and C somehow real again

[amber wasp]

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[wanton harbor]

Is my answer valid?

[wanton harbor]

✅

.close

[amber wasp]

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[plain patrol]

yo

[plain patrol]

example of sigma algebra on {1,2} would be:
{{1,2},{1},{2},{empty}}?

{1,2} would be another sigma algebra right ?

or rather

{{1,2},{empty}}

[fluid wren]

Yes

[plain patrol]

do all sigma algebras have to have even number of sets since they always contain complement ?

And complement is always different ?

[fluid wren]

On finite sets, yeah

Unless it’s on the empty set

[plain patrol]

ohh

u one smart mf

[plain patrol]

if in the example i am asked for definition of sigma algebra

would this be a valid answer ?

A sigma algebra on a set X is:

A collection of sets which contain X

contain the complemenet of any set in said collection

contains the countable union of all sets in said collection

[fluid wren]

Yeah, but to avoid ambiguity here I would add “as an element”

Someone could read that as ”is a superset of X”

[plain patrol]

what is a superset ?

[fluid wren]

A is a superset of B iff B is a subset of A

[plain patrol]

aight thanks

.close

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[amber wasp]

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[fast jackal]

I need help regarding some questions plz

[amber wasp]

Please don't occupy multiple help channels.

[fast jackal]

Anyone here?

@fast jackal

Here

[frozen tinsel]

Okay so how thing works here is

[cerulean forge]

!da2a

[amber wasp]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[frozen tinsel]

You post the question

And wait

Wait for someone to help, now post the question

[autumn lark]

#help-35

[fast jackal]

Q3 Or part And Q4 first part plz

[autumn lark]

please only occupy 1 channel at a time @fast jackal. type .close to close your other channel

[fast jackal]

Ok

[frozen tinsel]

Are you doing the test rn?

[fast jackal]

No

[frozen tinsel]

okay sure

which one you want to go first

[fast jackal]

Q3 both parts

As same for Q4

??

[ripe cipher]

ok so 3a

[fast jackal]

Yea

[ripe cipher]

do you know the relationship between factors and roots of a polynomial

[fast jackal]

no

[ripe cipher]

if (x+3) is a factor of your polynomial then one certain root of it is?

[fast jackal]

?

[ripe cipher]

mm okay so you dont know/remember.

let's try another way

[fast jackal]

Ok

[ripe cipher]

the equation (x-420)(x-69)=0, its roots are?

(you don't need to do any calculations)

(if you are trying to complete the square or apply QF that is wrong don't do it)

[fast jackal]

Quadratic equation?

[ripe cipher]

this is a quadratic equation

[fast jackal]

Yea

[ripe cipher]

but, i will repeat, if you're trying to jump into the b^2-4ac shit right now im going to cut that off and say you're doing it wrong

[fast jackal]

Ok

[ripe cipher]

you should be able to answer immediately

based on your knowledge of how to solve equations in factorized form

[fast jackal]

Roots are 420 and 69??

[ripe cipher]

exactly.

[fast jackal]

Ok

[ripe cipher]

a number $r$ as root corresponds to $(x-r)$ as factor.

[grizzled pagoda]

Ann

[ripe cipher]

so factor (x+3) gives root as?

[fast jackal]

As x-3?

Ahhh then?

[ripe cipher]

-3 only.

or x = -3

[fast jackal]

Ok

[ripe cipher]

but equals sign is undroppable

[fast jackal]

Right

[ripe cipher]

anyway, put x=-3 into P(x).

answer should come as 0.

[fast jackal]

Ok

[ripe cipher]

that'll be your part a.

[fast jackal]

Of 3?

[ripe cipher]

yes we're doing question 3, top left box.

[fast jackal]

Ok

Now right side?

[ripe cipher]

ok so you decided to simply skip entire parts b, c, d yes? am i understanding correctly?

or you can do them yourself now?

[fast jackal]

Yes

[fast jackal]

Correct

[ripe cipher]

ok

then yes i guess it's time for Q3 top-right box

[fast jackal]

Yes

[ripe cipher]

a × b means cross product of a and b, do you know how to find cross products generally?

sometimes also called vector product

[fast jackal]

Ok

cross product
a×b??

We solve it in matrix form?

[ripe cipher]

there's a matrix-looking thing that can help you calculate it, yes

[fast jackal]

ok

Next?

[ripe cipher]

so you calculated cross product?

[fast jackal]

w8

a×b=⟨2⋅6−3⋅5,3⋅4−1⋅6,1⋅5−2⋅4⟩=⟨−3,6,−3⟩?

[ripe cipher]

,w (3,-2,4) × (1,2,-1)

[grizzled pagoda]

Results provided by WolframAlpha

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

Requested by ann_dec

[fast jackal]

Oh ok

[ripe cipher]

..how did any of this come

where is 6 from, where is 5 from

and so on

what the hell did you do

did you try to do (1,2,3) × (4,5,6) ?

[fast jackal]

Its like i did multiply in matrix form 1 by 1

[ripe cipher]

show paper with work

[fast jackal]

ok

oh yea you right

[fast jackal]

I did this my bad

[ripe cipher]

yeah dont

[ripe cipher]

...

!nogpt

[amber wasp]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[fast jackal]

ok

[ripe cipher]

if you are going to beg AI to give answers to you then why even come here, why bother?

[fast jackal]

No i was trying to solve there

Which you told me to do

[ripe cipher]

well you were supposed to do it yourself and not run to the "give you lies and smile" machine

[runic briar]

Trust deepseek

[ripe cipher]

i think i need to go anyway so someone else will have to take over

[ripe cipher]

!noai

[amber wasp]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[fast jackal]

Ok as you wish

[amber wasp]

@fast jackal Has your question been resolved?

[amber wasp]

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[cold beacon]

Can someone help

[cold beacon]

How do I do these problems

<@&286206848099549185>

The first problems we did with the class but I'm still confused

<@&286206848099549185>

[nimble shuttle]

if we take this one

do you know what 9 means?

where do you start to get confused

[cold beacon]

So 8 ・ 9 is 72 plus 1 is 73

[nimble shuttle]

correct so far

[cold beacon]

And 2 ・ 16 Is 32 plus 1 is 33

But I don't get what I do next

[nimble shuttle]

so now you got

do you know what you have to do next

[cold beacon]

No

[nimble shuttle]

do you know what you call all the stuff in fractions?

[cold beacon]

No

I don't think so

[nimble shuttle]

the bottom part is the denominator

when adding/subtracting fractions you have to have the denominator

[cold beacon]

Oh yeahhh

[nimble shuttle]

how do you express 73/8 without changing the value of it?

so you can subtract 33/16

[cold beacon]

19

I mean 17

33-16 is 17

I mean the least common denominator is 16 right?

[nimble shuttle]

Yes.

[cold beacon]

So now what do I do with 16

[nimble shuttle]

^^

8 and 16 are not the same

[cold beacon]

Wdym

[nimble shuttle]

This is what you now (based off of what you calculated)

[little hornet]

before you add or subtract, the denominators need to be the same

[nimble shuttle]

In order to calculate +- fractions the denominators need to be the same

[little hornet]

how would you change the first fraction so both have the same denominator?

[cold beacon]

8 ・ 2

Is 16

[nimble shuttle]

Yes

Good

Now what?

[cold beacon]

Idk

[nimble shuttle]

do you still have the same fraction?

is 73/8 and 73/16 really the same?

what did you forget?

[cold beacon]

Wait what

Do I times it by 2

The numerator

[nimble shuttle]

Give me a moment to see if I can visualize it for you

Lilian if you got a good explanation please help out

[little hornet]

in fractions, what you do to one side, you must do to the other

[nimble shuttle]

see how 1/2 and 2/4 has the same value?

[pastel basin]

@nimble shuttle
1/2 + 1/4 =
(1x2)/(2x2) + 1/4 = 2/4 + 1/4 = 3/4

whose question is this

whoops

I tagged wrong

[nimble shuttle]

xfil

[pastel basin]

we even got the same example

[nimble shuttle]

@cold beacon how's it going?

I think it's clear that these problems are a tidy bit too hard for you currently.

I'd advise you to try smaller numbers.

Maybe 1/2+2/4=?

[cold beacon]

It's going good

So 73 x 2/8 x 2 = 146/16

[little hornet]

yup

[nimble shuttle]

https://tenor.com/view/coach-josh-wood-coach-josh-gold-star-gold-star-gif-593378152944786379

[cold beacon]

Then I subtract the fractions 146/16-33/16 = 113/16

Then divide 113/16

[little hornet]

yeah

[cold beacon]

So is it

7 left over with 1/16?

Or 7 1/16?

[little hornet]

both are correct, but 7 1/16 is probably recommended

[cold beacon]

Yayyyyyyyyyyyyy

I got it

Tyyyy

[little hornet]

🎉 🎉 🎉

[nimble shuttle]

try the next one

[cold beacon]

Ok

4・3 + 2 = 14/4 12・4 + 2 is 50

First I start with 14/4

Find common denominator which is 12

Add fractions

42/12+50/12 = 92/12

92 divides by 12 is 7 8/12

Do i simplify

?

[nimble shuttle]

sure

[cold beacon]

Their common thing is 4

I think

Is 7 2/3?

[nimble shuttle]

ye

or 23/3

[cold beacon]

Yayyyyyy

Ok ty I know both subracting and adding

[nimble shuttle]

what about division/multi?

[cold beacon]

Nah we aren't on that yet

But it's late for me

[nimble shuttle]

alr

[cold beacon]

Byeee

[little hornet]

good night!! (i think)

[nimble shuttle]

bye

[cold beacon]

Yes

Good night

[nimble shuttle]

pretty crazy to think that someone that young turned to discord instead of asking his/her parents

[little hornet]

hmm, maybe his parents weren't much help?

[nimble shuttle]

maybe but it'd be sad for parents to fail that level of math

[little hornet]

yeah, he probably had a reason though

[amber wasp]

@cold beacon Has your question been resolved?

[amber wasp]

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[jaunty bobcat]

.reopen

[peak kestrel]

.close

[ionic loom]

I didn't understand how they the third step is written?

[ionic loom]

can you explain that

.close

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[fresh hatch]

$$Y(s) \left[ (s-1)^2 \right] = \frac{ 6(s-2) }{ (s-2)^2 + 1} + 3$$
I was struggling a bit to find the inverse Laplace of Y(s).
I first divided the equation as follows:
$$Y(s) = \frac{ 6(s-2) }{ (s-2)^2 + 1} \cdot \frac{ 1 }{ (s-1)^2 } + \frac{ 3 }{ (s-1)^2 }$$
Then I tried to use convolution for the first part.
But I found that it is really hard to find the convolution of this part.
So I tried to think of any property that may be used, but couldn't find one,
So I came to do a partial fraction for the first part, which required 4 constants to be found.
Which came in the end to be.
$$\frac{6(s-2)}{(s-2)^2 + 1} \cdot \frac{1}{(s-1)^2} = \frac{0s+3}{(s-2)^2+1}-\frac{3}{(s-1)^2}+\frac{0}{s-1}$$
As appears from the answer a lot of zeros has appeared, which i think suggests that there would have been an easier solution to that I did here.
I reached to the answer using this method and it was correct.
but I just wanted to check if there were any easier solutions.

[grizzled pagoda]

Sherif Player

[amber wasp]

@fresh hatch Has your question been resolved?

[fresh hatch]

hmmm

.close

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[raven mortar]

27th
Here is my work I'm stuck😭

[sonic jasper]

Sin 50 and sin 70 can be changed to cos

What will it be ??

[raven mortar]

I did it didn't I?

[raven mortar]

,rccw

[sonic jasper]

Domr close sin 40

Keep it 2sin20cos20

[raven mortar]

Now?

[sonic jasper]

I dont getvthe second step

[raven mortar]

Btw Final answer is 80

[raven mortar]

In attachment?

[sonic jasper]

I couldnt solve it sorry

[raven mortar]

No worries man

<@&286206848099549185>

[rapid hornet]

spam it

@sudden flare

MB

I MEANT TO SAY HELPERS

OML

FRIKING HELL

<@&286206848099549185>

[raven mortar]

Calm down man

[torpid hull]

Use the following identity
[\rb{
\sin(\ta)\sin(60-\ta)\sin(60+\ta)=\f14\sin(3\ta)
}]

Hm

[grizzled pagoda]

Cooly

[raven mortar]

Where is it applicable
I had thought to apply it

But found ni where to apply it

[raven mortar]

Oh i see

[torpid hull]

Helpers here are all volunteers. Please refrain from assuming entitlement to their assistance.

[rapid hornet]

cooly

It was an accident

[red spade]

if you found how you can apply it, can you share the calculations? im solving in tandem and cant quite get it

[true jackal]

We don't believe it. You did the same in another channel...

[round reef]

quick detour: don't ping helpers for others.

[rapid hornet]

I apologize

[red spade]

@raven mortar did you get it?

[torpid hull]

Rewrite the original equation as below:
[\rb{\sin(10)\sin(20)\sin(40)\sin(10)\sin(50)\sin(70)
}]

[grizzled pagoda]

Cooly

[red spade]

done

[torpid hull]

You can simplify sin(10)sin(50)sin(70) using the identity I mentioned earlier.

[red spade]

that simplifies to sin(30) * 1/4

ok

right

ill try after that

[raven mortar]

But

[raven mortar]

Where is sin60 here

[red spade]

you dont need a sin60

[raven mortar]

But identity is based on sin60

[red spade]

you need to sin functions which are equally deviated by theta, here those are sin 50 and sin 70

[red spade]

it is based on 60- theta and 60 + theta

theta here is 10

[raven mortar]

Then divide and multiply be sin60?

[red spade]

in this identity

you dont need a sin 60

[raven mortar]

Oh i get it

[red spade]

take theta = 10 for the dientity and you will find the pattern

[raven mortar]

Oh god I just forget it a hell

[raven mortar]

Oh now then?

[red spade]

well

i wouldnt know

im trying myself

try to call cooly

[raven mortar]

Angles are in AP but in multiplication

[red spade]

the identity might just be a suggestion by cooly

[torpid hull]

Then you can use this identity below to simplify sin(20)sin(40)

2sinAsinB = cos(A-B)-cos(A+B).

[raven mortar]

Wait I get smth

[red spade]

im trying to simplify and prolly getting somewhere

[torpid hull]

And after some simplification, the last identity you’ll need is:

2sinAcosB = sin(A+B)+sin(A-B)

Ping me if you have a question.

[raven mortar]

Got SMTh?

[red spade]

yep

just a sec

im tired so my handwriting is a bit messy but i can explain my logic and thought process if you want

line 4 and 5 are especially atrocious fk me

[raven mortar]

Still ur writing is better than mine😭

[red spade]

you need an explanation?

of what properties i used?

[raven mortar]

No but ig u did it in a different way then cooly said

[red spade]

the first line expression is what cooly showed us

i solved after getting 1/4sin10sin20sin30sin40

this step i showed above only uses like the sinAsinB identity

then the uh

cosAcosB identity

[red spade]

this one

[red spade]

and this

there is probably a more efficient way tho

using some other identities

[raven mortar]

Oh got it ty

.close

[amber wasp]

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[raven mortar]

9th idk where to begin

[royal spoke]

Not sure if this counts, but have you tried the completing the squares method? Maybe that would help.

[raven mortar]

No

Lemme try

I don't think so

[royal spoke]

Hint: the expression can be written as a sum of three squares.

[raven mortar]

I mean r u sure if that works?

[royal spoke]

Yes. I think they are (x-2y)^2 + (y-z)^2 + (y-1)^2 = 0

[raven mortar]

Wow

[crystal mantle]

Try each case of x + y + z = a where you write any of the x,y,z in terms of (y,z),(x,z),(x,y) whatever u want then see if the equation would hold depending on the value of a

[royal spoke]

And since the sum of squares is zero only when each square term is zero, you get your answer.

[raven mortar]

Answer is 4

[royal spoke]

yh, I think the answer is 4.

[raven mortar]

Yea guessed

By guys

Thank You so much Annie
Anne Frank

.close

.close

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[coral totem]

if I have:

x > a - 0.7
y > b - 0.7

What is x - y?

[indigo cloud]

unknown

you cant say anything

!xy

[amber wasp]

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

[plain sun]

x-y has no bound

[coral totem]

what if it had a greater than or equal sign

[torpid hull]

Please show the original question

[true jackal]

Wdym by "it"?

[indigo cloud]

= in place of those > would still not allow anything

(that would be weaker conditions and we already cant say anything with stronger conditions)

[amber wasp]

@coral totem Has your question been resolved?

[amber wasp]

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[fast vault]

Hi

[fast vault]

Why do we set -3 on the worked example

For the domain

[amber wasp]

@fast vault Has your question been resolved?

[signal furnace]

You just need some region to the left of 0 and some region to the right of 0

The choices of -3 and 3 are fairly irrelevant

[fast vault]

And is it like it might not be zero?

Like it doesn’t have to be zero at the middle

[fast vault]

I see, thanks

[signal furnace]

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[woven plinth]

Newbie question here
The array consists of sum of all the integers upto that index(the actual array is full of 1s)
And i was trying to get the sum of the square
It should be 4 but i am getting 0
(Sorry if it's a stupid question i suck)

[woven plinth]

Where did i mess up?

[indigo cloud]

its hard to understand whats happening

you marked a square but in your calculations the numbers 9,10,14 didnt even show up?

also why are you randomly subtracting some of the values?

[woven plinth]

Sorry i am bad
I will try to explain
I have a 3X5 matrix which have all 1s
I am trying to get the sum of a subarray within that matrix(the squared area)
Because the matrix is all 1s, the square must be a 2x2 matrix of 1s whose sum should be 4 but i am getting 0

I got the formula S(a)-S(b)-S(c)+S(d) from somewhere where S(x) is the sum of all the values up until x(x is the index of that value)

a is the index at the bottom right of the box b is the index just left of the box(same row as a), c is just above the box(same col as a) and d is the index at the top left of the sqyare

Let me send the source of the algorithms

[indigo cloud]

if you do 15-13 you are essentially counting the two 1s in cells 14 and 15

if you next do 10-8 you count the two 1s in cells 9 and 10

is that what you actually wanted with your formula?

[woven plinth]

Hmm

[indigo cloud]

well something is certainly wrong with that formula

imagine moving both A and B down one row

then S(A)-S(B) doesnt change and so the value of the formula also doesnt change

[woven plinth]

I see

What formula do you think i should use?

I want to get the sum of all the values in the subarray

[indigo cloud]

oh wait

my interpretation of S(X) was wrong

and so was yours

the cell b in your original picture should be a 9

[indigo cloud]

read the last sentence again

[woven plinth]

Hm?
Chatgpt said the same i dfnt believe it

[woven plinth]

You mean the rectangular part?

[indigo cloud]

S(X) does not count cells that are "to the right" of X

which your definition does

[woven plinth]

Oh

I get it i think

So b should be 9 because the row shouldn't be larger than 3(b)

So the value at b should be like--
Row 1--- 1 2 3
Row 2--- 4 5 6
Row 3--- 7 8 9
?

Wait did i do something wrong?

[indigo cloud]

your fifteen cells should be:
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15

[woven plinth]

Ohh i get it

Thank you bro

The formula now works correctly

[woven plinth]

Sorry for asking stupid questions🙏🙏
I should have read the line carefully

[indigo cloud]

dw

happens

[woven plinth]

.close

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[grizzled crystal]

I give Wolframalpha an equation but it treats letters like c, G, M and so on not as variables but as some weird constants. Can I change this?

[ripe cipher]

it should say something like "assuming c means the speed of light [Use as a variable instead]"\

does it?

[grizzled crystal]

the option to use as variable is not there 🙁

[shadow stump]

select the more option?

[grizzled crystal]

then only other constants appear

[tiny juniper]

what’s the equation you’re trying to input?

also if they’re just variables you can replace em with 𝑥 or 𝑎 blah blah blah

[hoary rover]

Honestly? Just ask chatGPT to do it.

[tiny juniper]

!nogpt

[amber wasp]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[grizzled crystal]

yeah no gpt

[hoary rover]

It seems like you just want to use it as a calculator though, not to explain anything

[grizzled crystal]

gpt tries to sell me this crap hahaha

[grizzled crystal]

that did work, but it would be nice if there would be a nicer way

[tiny juniper]

unfortunately wolfram is just buggy sometimes. You could input, “C” and it would pull up “Crewtons scallywag dingleberry constant”

[tiny juniper]

[fiery fossil]

hello infinium long time no see

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[tough mica]

I need some help with estimating the error of an taylor polynomial of second degree using lagrange remainder, this in multivariable

[tough mica]

this is the exercise

this is my progress

This was done in typst btw :D

if someone comes to help feel free to ping

[amber wasp]

@tough mica Has your question been resolved?

[amber wasp]

@tough mica Has your question been resolved?

[cedar rose]

okay do you know like

what the lagrange remainder usually looks like?

[tough mica]

in single variable yes, its an upper bound for the error

like third order taylor polynomial or something

[cedar rose]

yeah exactly

[tough mica]

at a point, that point usually is like, in between something

[cedar rose]

now the way this is written out is a little bit ugly

hold on actually this will be really ugly

what you want to get is something like this

you wanna write taylor's theorem in something more managable

so let's write it using the gradient

we'll write it like this

$f(x) = f(0) + Df(x_0)(x-x_0) + \frac{1}{2}D^2f(x_0)(x-x_0,x-x_0) + R_2(x)$

$Df(x)$ here is the jacobian of your function if you know what that is

[grizzled pagoda]

Irony

Irony

[cedar rose]

D^2 is called the hessian but that's less important

now

what your remainder will look like is this

you have a bunch of this order derivatives multiplied with (x-x_0) terms right

[tough mica]

I am very familiar with jacobian

[tough mica]

this is for single variable or what?

you are showing the taylors theorem for single variable case?

[cedar rose]

so it'll be like $R_2(x) = \frac{1}{3!} \left(\frac{\partial^3 f}{\partial x^3} + 3\frac{\partial^3 f}{\partial x^2 \partial y} + 3 \frac{\partial^3 f}{\partial x \partial y^2} + \frac{\partial^3 f}{\partial y^3} \right)$

[cedar rose]

this is for multivariable

[grizzled pagoda]

Irony

[cedar rose]

you can write the multivariable taylor's theorem using the jacobian and higher order differentials

[tough mica]

looks single variable to me

[cedar rose]

but it isn't!

x_0 is a vector

so is 0

[tough mica]

care to elaborate

[cedar rose]

yeah so let's go term by term

you start out like so

f(x,y) = f(0,0) + ... right

[tough mica]

can we

first go through the single variable taylor theorem

[cedar rose]

yes okay ofc

[tough mica]

and the lagrange reminder for the single variable case

and then generalize this to more variables

[cedar rose]

yeah ofc

[tough mica]

because I need a refresher

if you know what I mean

this lagrange reminder is kinda confusing even in single variable

[cedar rose]

so the single variable taylor's theorem says the following

if you have a function that's of class C^{k+1} (do you know what this means?)

[tough mica]

yes, a function of the class C^k is a function that is differentiable and his derivative is continous on a neighborhood

i mean is k times differentiable

and the kth derivative is continous

[cedar rose]

yes perfect
So now taylor's theorem says you can write your function as a k-th degree polynomial plus an extra error term

and the polynomial looks like this

$f(x) = f(x_0) + \frac{df}{dx}(x_0)(x-x_0) + \frac{1}{2}\frac{d^2f}{dx^2}(x_0)(x-x_0)^2 + \frac{1}{3!}\frac{d^3f}{dx^3}(x_0)(x-x_0) + \dots + \frac{1}{k!}\frac{d^kf}{dx^k}(x_0)(x-x_0)^k + R_k(x)$

where $R_k(x)$ is the Lagrange remainder

[grizzled pagoda]

Irony

[cedar rose]

r u with me so far?

[tough mica]

Rk involves the k+1 deriv

of f

[cedar rose]

yes and it looks like this

$R_k(x) = \frac{1}{(k+1)!}\frac{d^{k+1}f}{dx^{k+1}}(c_x)(x-x_0)$ where $c_x$ is some values in between $x$ and $x_0$

[grizzled pagoda]

Irony

Irony

[tough mica]

ok, now going back to multivar, we can generalize this to multivar with hessian and jacobi

[cedar rose]

so you evaluate the k+1th derivative at some mean value (this is a result of the mean value theorem because otherwise you can write the remainder as an integral)

[tough mica]

?

[tough mica]

also, this reminder is theoretical, because we can never found it unless we use a calculator, but we can find an upper bound to this remainder

correct?

[cedar rose]

yes

the bound isn't hard to find tho since f is of class C^{k+1} it's continous so it's k+1th derivative has a minima and maxima on the closed interval between x_0 and x

so you can just take the maximum of d^{k+1}f/dx^{k+1}(x) and that's your bound!

[tough mica]

what would be the form of the upper bound for the reminder

[cedar rose]

it looks like $|R_k(x)|\leq \frac{1}{(k+1)!}(x-x_0) \max_{y\in [x,x_0]} \frac{d^{k+1}f}{dx^{k+1}}(y)$

[grizzled pagoda]

Irony

[cedar rose]

here I'm assuming that x_0 > x but it's the same if x_0 < x

now in the multivariable case the expansion is a lot more complicated but let's write it out

I think it's best if we forget the jacobian/hessian stuff and write all the coordinates out

so the 2nd order taylor polynomial you already wrote out

now imagine you're writing out the third order term in that expansion

[cedar rose]

it would look like this right?

the reason you need all the extra derivatives is that now you have multiple directions your function can move around in

and you have to account for every one of those when approximating it

[tough mica]

multiple directions in single variable?

[cedar rose]

oh no sorry I was talking about the multivariable taylor theorem now

[tough mica]

i think so, but isnt the absolute value should be for the reminder itself and not on the Rk

[tough mica]

i see

what about the single variable case

why do we need k + 1 derivs

[cedar rose]

it should be on both but yeah you're right I forgor to put it on the RHS
JUst imagine everything is in absolute values lol

[tough mica]

ok

[cedar rose]

it's a result of the mean value theorem

I can show you the proof if you want

[tough mica]

please

are you a first year student? it seems to me nobody knows Taylor's theorem in other than the single variable case

[cedar rose]

no lol I'm finishing my undergrad rn

[tough mica]

intuition should be similar for the multivariable case

[cedar rose]

yeah it is

[tough mica]

but now we got hessian and jacobi

gradients and multiple directions

[cedar rose]

yeah so this is a bit harder to picture

because derivatives of multivariable functions are very complicated

this is why things like the jacobian exist

think of it like this
The multivariable taylor theorem says in first order you take the gradient of your function and approximate it like that right?

in second order you take the differential of the gradient
in third order you take the differential of this and so on
So it's the same thing as the single variable case except you're taking gradients/differentials instead of regular derivatives

so just think everything gets replaces by vectors

[tough mica]

sure

but, wdym my differential

for second order taylor poli we need gradient and jacobi

or just, jacobi

[cedar rose]

hm okay this is the language we used in my class but it may be different elsewhere

[tough mica]

well im not from america, im sorry for my bad English

[cedar rose]

essentially think of the differential as a matrix that contains all the derivatives of your function to some order

[cedar rose]

dw it's fine

[tough mica]

so like, a generalisation of, jacobian and hessian

[cedar rose]

yes exactly

[tough mica]

what would be the third order differential of f(x,y) for example

matrix of 3x3?

where f(x,y) is a vector field

[cedar rose]

it's better not to think about what the 3rd order one would look like

because it would be a 3x3x3 matrix

and we don't like to think about that

it's better to think about what the elements of the matrix are

anyways let's write out the taylor expansion

[tough mica]

[cedar rose]

$f(x,y) = f(x_0,y_0) + \frac{\partial f}{\partial x}(x-x_0) + \frac{\partial f}{\partial y}(y-y_0) + \frac{1}{2}\left(\frac{\partial^2 f}{\partial x^2}(x-x_0)^2 + 2 \frac{\partial^2 f}{\partial x \partial y}(x-x_0)(y-y_0) + \frac{\partial^2 f}{\partial y^2}(y-y_0)^2\right) + R_2(x,y)$

now we know what the remainder looks like in the single variable case

and we know what the terms of the taylor expansion in the multivariable case look like

so we can conclude the following

[tough mica]

what is this? a multidimensional matrix?

thats so messed up

[cedar rose]

yeah essentially it's a bit complicated and annoying to work with so don't worry about it lol

so now the error term in the multivariable expansion looks like this

[tough mica]

we need the third order differential for the R2

[grizzled pagoda]

Irony

[cedar rose]

Yesssss
$R_2(x,y) = \frac{1}{3!} \left(\frac{\partial^3 f}{\partial x^3}(c)(x-x_0)^3 + 3\frac{\partial^3 f}{\partial x^2 \partial y}(c)(x-x_0)^2(y-y_0) + 3 \frac{\partial^3 f}{\partial x \partial y^2}(c)(x-x_0)(y-y_0)^2 + \frac{\partial^3 f}{\partial y^3}(c)(y-y_0)^3 \right)$

[tough mica]

where c is?

c in [x,x0]?

[tough mica]

there is a typo i think

[cedar rose]

c is in some circle that surrouns (x,y) and (x_0,y_0)

[cedar rose]

errr yes I see it lol

[tough mica]

you wrote x0^2

[grizzled pagoda]

Irony

[cedar rose]

yeah fixed

[tough mica]

i am so cooked bro

[cedar rose]

now to bound this you need to like calculate all four derivatives

[tough mica]

things are not clicking

[cedar rose]

yeah sorry

so this is the remainder because the remainder looks like the k+1-st term in your expansion evaluated at some point (which we don't a priori know what that point is)

so we just evaluate all the derivatives at some point c

[tough mica]

i dont think I follow, so how do I continue with my exercise

[cedar rose]

tho okay wait a second

what does your exercise say exactly

[tough mica]

did you checked my progress?

[cedar rose]

does it just say "describe the lagrange remainder"?

[tough mica]

like we might need

[tough mica]

it says write the form of the lagrange reminder

[cedar rose]

yeah I read it and it looks good!

[cedar rose]

yeah okay so then you don't need to bound it!

[tough mica]

yeah we just need the theoretical one from taylors theorem, but in lagrange form?

dont think I understand

the lagrange one is the one that does not require integral

[cedar rose]

yeah the lagrange one is the one that I wrote up there

with all the 3rd order derivatives

[tough mica]

.

[cedar rose]

this is the lagrange remainder

you just have to sit down and calculate out all these derivatives

which uhhh sounds like a pain in the ass to say the least lmfao

[tough mica]

jajaj

[cedar rose]

also this is wrong

f_xy and f_yx have to be identical

you got the correct answer in the end cuz you were lucky but check your work

[tough mica]

no its not

[cedar rose]

oh shit

wait I am a adumbass

yeah everything is correct

man

everything is written out in so much detail I didn't realize where the result actually was lol

[tough mica]

can I add you ? what are your pronouns and how can I call u?

i dont plan to dm you with doubts I just find you cool

[tough mica]

dont worry abt it, I spent too much time writing this solutions but I think I get the idea of what I have to do now, i appreciate the help irony

[tough mica]

no you are not, though its kinda redundant to calculate partials yx and xy separetely since they are always symmetric I think

[cedar rose]

yeah you can add me haha and you can call me irony

[cedar rose]

yess they are!

so okay to finish the exercise

[tough mica]

by clairauts theorem I think

[cedar rose]

calculate the third derivatives and just say it looks like "R_2(x) = 1/3! (all the derivatives worked out)"

it'll be a bit messy just because you'll have to evaluate the derivative at this mystery point c

[tough mica]

I appreciate it, thank you

.solved

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[autumn pumice]

Can somebody help me with question a? I dont understand how to find the missing dots

[autumn pumice]

.solved

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[limber lynx]

Here… i got this question wrong on a test, and just why?

[limber lynx]

I did (5(5x-6))/10 - (2x)/10 = (16)/10.

I did like this

[pallid canopy]

show the whole thing what you did

[pallid canopy]

this is fine

[amber wasp]

@limber lynx Has your question been resolved?

[limber lynx]

But they told me to multiply EVERYTHING with 10?

Da fuck

[pallid canopy]

matter of grader's choice ¯_(ツ)_/¯

[amber wasp]

@limber lynx Has your question been resolved?

[limber lynx]

But how does it matter?

10 is bigger than 5

Or 2

[pallid canopy]

i can't read your grader's mind

ask them

[limber lynx]

@pallid canopy no, I’m asking what’s the difference by making it times 5 and 2 instead of doing everything 10?

[drowsy lava]

both are correct, making it LCM is generally simpler than just multiplying them tgt

marker's choice

just do what they accept lmao

[limber lynx]

so i'd have to do

10 * 5x-6 - 10 * X = 10 * 8?

and they all would have 10 as numerator

[rich frigate]

Did you get the final value of x right? If you get it right then you are good.

[amber wasp]

@limber lynx Has your question been resolved?

[amber wasp]

@limber lynx Has your question been resolved?

[modest linden]

What's the question ❓

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[stray knoll]

Are my steps ok?

I need to prove the green brackets

[amber wasp]

@stray knoll Has your question been resolved?

[astral relic]

Probably it comes from the fact that i cant read the language, but i feel like im missing some steps

Just to go over, you showed that $\frac{(k+2)^{k+1}}{2} > \frac{(k+1)^{k+1}}{2}$

[grizzled pagoda]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[astral relic]

From there, you have to show that
$\frac{(k+1)^{k+1}}{2} > (k+1)!$

[grizzled pagoda]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[astral relic]

Thats basically what you did?

Im ignoring the whole induction logic, the first element thing was alright

[astral relic]

tbh, from this one you are quite literally one step away from solving it

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[tough mica]

i need some help with b)

find all the solutions to 56X = 28 (mod 35)

21X = 28 (mod 35)

gcd(21,35) = gcd(21, 14) = gcd(7,14) = gcd(7,0) = 7

[solar gust]

You can divide by 7

[tough mica]

i dont think I follow

[solar gust]

21x = 35k + 28

[tough mica]

3x = 5k + 4

[solar gust]

So 3x = 4 mod 5

[tough mica]

3X = 4 (mod 5)

[solar gust]

Now what do you do

[tough mica]

gcd(3,5) = gcd(3,2) = gcd(1,2) = gcd(1,0) = 1

so we can use bezouts

[solar gust]

What does it says

[tough mica]

no,srry

bezouts lemma idk

[solar gust]

au + bv = 1 ?

[tough mica]

ye

so like

3X + 5Y = 1
3X + 5Y = 1 (mod 5)
3X = 1 (mod 5)

wait, I am tripping hard, feel free to ignore

[solar gust]

You proved the invertible thing i showed you yesterday kek

[tough mica]

sorry I am tripping hard

[solar gust]

How so

You just need to find a x such that it works and then you can generalise it

[tough mica]

X = 3^-1

[solar gust]

3x = 1 mod 5 is seeking for the inverse of 3 mod 5

It exists since 3,5 are coprime

[tough mica]

dude we have 3X = 4 (mod 5)

[solar gust]

And then you need to multiply by this inverse

[solar gust]

Never said you was done after this

[tough mica]

wait a second dude

we have

3X = 4 (mod 5)

by bezouts because gcd(3,5) = 1, then

3a + 5b = 1 exists and is solvable

so

3a = 1 (mod 5)

so a = 3^-1 is the inverse

and then, X = 4a

so X = 4 * 3^-1

[solar gust]

4a mod 5

[tough mica]

and then, X = 4a (mod 5)

so X = 4*3^-1 (mod 5)

thats it?

[solar gust]

Sure

Whats 3^(-1)

[tough mica]

like, we have
3X = 4 (mod 5)

[tough mica]

thats a in 3a = 1 (mod 5)

[solar gust]

It has a value

[tough mica]

multiplicative inverse of 3 (mod 5)

[solar gust]

Which is

[tough mica]

3^-1 (mod 5)?

[solar gust]

It has a value

3^(-1) is not a value you can use to calculate, its mostly a notation

[tough mica]

i am so lost

[solar gust]

And yes it is the inverse of 3 in Z/5Z

[solar gust]

You just have to solve 3a = 1 mod 5 as you said

[tough mica]

is just a = 3^-1 (mod 5)

a is the inverse of 3 (mod 5)

[solar gust]

3a = 1 mod 5

a is a value that is in (Z/5Z)* = {1, 2, 3, 4} (group of invertible mod 5)

[tough mica]

dude but a = 3^-1, wdym?

[solar gust]

its not a number

[tough mica]

what does the notation mean

[solar gust]

the set of numbers that has an inverse mod 5

[tough mica]

yes is not an integer

[true jackal]

Huh??

[tough mica]

well we only care the inverse exist

no?

[true jackal]

You're working with integers, how can it not be an integer?

[tough mica]

so we can multiply by 3^-1

[solar gust]

3^-1 isn't 1/3

its the inverse of 3 mod 5

so the number multiplied by 3 that gives 1 when you do euclidian division by 5

[tough mica]

yes

[solar gust]

i do agree that a = 3^-1 mod 5 but you need to find what is 3^-1

imagine

AX = B

with matrices

given matrices

do you say X = A^-1 B ? or do you calculate explicitely what A^-1 is

[tough mica]

well both?

[solar gust]

and here you are telling me you only want to do the first one

[tough mica]

ok so, what do we do, we apply the extended euclidean

[solar gust]

actually just find one a where it works

[tough mica]

ok

[solar gust]

and multiply the other side by this a

[tough mica]

for 3a = 1 (mod 3)?

[solar gust]

3a = 1 (mod 5) yes

[tough mica]

a = 2

6 = 1 (mod 5)

[solar gust]

perfect

so now 3x = 4 mod 5

you multiply by the inverse both sides

[tough mica]

what?

[solar gust]

we found the inverse of 3 mod 5

so now we solve the equation

[tough mica]

x = 8(mod 5)?

[solar gust]

which is

[tough mica]

x = 3 (mod 5)

[solar gust]

perfect

is there a moment you find hard or not

[tough mica]

x = 5k + 3

[tough mica]

everything!

i had to use my brain like 99% of the time

[solar gust]

go on the next one, im watching

think about the method

• try to make gcd = 1

• find inverse

• say the result

[tough mica]

56X = 2 (mod 884)
56X = 884k + 2
28X = 442k + 1
28X = 1 (mod 442)
442 - 28x15 = 22
gcd(28,442) = gcd(28, 22) = gcd(6,22) = gcd(6,4) = gcd(2,4) = gcd(2,0) = 2

[solar gust]

what can you conclude

[tough mica]

X = 28^-1 (mod 442)

[tough mica]

X is the inverse 28 (mod 442)

[solar gust]

think about this

especially remember about conditions of existance of the inverse

[tough mica]

X is the multiplicative inverse of 28(mod 442)

we have 28x = 1 (mod 442)

so like, if the multiplicative inverse exists for 28 (mod 442), then there exists some a in Z such that gcd(a,28) = 1

[solar gust]

ax = 1 (mod n), x exist only if gcd(a,n) = 1

[tough mica]

this

[solar gust]

what did you say last line here

[tough mica]

but where is 28 here and 442

[tough mica]

gcd(28,442) = 2

so the inverse doesnt exist

[solar gust]

great, what can you say about solutions then?

[tough mica]

no solutions

[solar gust]

perfect

next one

same methods

different numbers

[tough mica]

78X = 30 (mod 12126)

78X = 12126k + 30

we divide by 6

13X = 2021k + 5

13X = 5 (mod 2021)

2021 - 13*155 = 6
gcd(13,2021) = gcd(13,6) = gcd(1,6) = gcd(1,0) = 1

so, since gcd(13,2021), the multiplicative inverse of 13 (mod 2021) exists

having said that,

a(13X) + b(2021) = 1
a(13X) = 1 (mod 2021)
so a is the multiplicative inverse of 13X

[solar gust]

||here you need the extended euclidian algorithm to find a||

[tough mica]

ok

2021 = 13*155 + 6
13 = 6*2 + 1

so

6 = 2021 - 13*155
1 = 13 - 6* 2
1 = 13 -(2021 -13* 155) * 2

1 = 13(1-2*155) + 2021(-2)
1 = 13(1-310) + 2021(-2)
1 = 13 (-309) + 2021 (-2)

[solar gust]

no

becareful to signs

[tough mica]

yeah my bad

[tough mica]

1 = 13 -(2021 -13* 155) * 2

1 = 13 - 2*2021 + 13* 155 * 2
1 = 13(1+155*2) + 2021(-2)
1 = 13(311) + 2021(-2)

now what?

the inverse is 311?

[solar gust]

you want the 2021 to disapear

[solar gust]

yes and if you mod 2021 the equation you find this

[tough mica]

we have
13X = 5 (mod 2021)

and because gcd(13,2021), the multiplicative inverse of 13 (mod 2021) exists, and is 311

so we just multiply both sides by the inverse

and we get

X = 1555 (mod 2021)

[solar gust]

perfect

[tough mica]

or what?

[solar gust]

does it seems still hard now ?

[tough mica]

so all X is of the form
X = 2021k + 1555

with k in Z

[tough mica]

well, it certainly is, idk

correct?

nothing like the previous maths I have done that was plug and chug and having a race with your calculator

[solar gust]

you do this with calc?

[tough mica]

wdym?

[solar gust]

the division and all

[tough mica]

the euclidean div no

i don't think there exists an calculator that can do that

[solar gust]

well

you rather want to keep the solution that are distincts but ig its not necessarily asked here

[tough mica]

we can encapsulate it in a set

[solar gust]

is there a shady part ?

[tough mica]

math is hard but sexy

[tough mica]

well

i just need more practice i think, but the theory i think I grasp it a bit more

[solar gust]

its basically hidden group theory

[tough mica]

though everything regarding the inverse is shady

asfk

seems like black magic to me

well not really black magic but the idea we can use euclidean extended to find inverse is mind boggling

[tough mica]

care to elaborate?

[solar gust]

the multiplicative group of the invertible in Z/nZ is denoted (Z/nZ)* and is made of the numbers between 1 and n that are prime with n

(thats where the gcd = 1 comes from in some way)

basically if its not prime with n, then its not in the set of invertible, so theres no inverse to it

but thats for the culture, you don't really need this explicitely here

[amber wasp]

@tough mica Has your question been resolved?

[solar gust]

also it is the method to find it, i just don't want you to do it for such small number as 3x = 1 mod 5 for the one we did

it quite jumps to the eyes that it is 2 mod 5

[tough mica]

yes

[tough mica]

interesting

[tough mica]

I appreciate the help @solar gust

.solved

[amber wasp]

Channel closed

Closed by @tough mica

Use .reopen if this was a mistake.

[amber wasp]

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[keen pawn]

Maybe it's just too late, but here why does the minamlty of s impliy r=0

[keen pawn]

as then bx^r would be in H right

<@&268886789983436800>

[cunning birch]

s is the smallest positive integer such that h_n = (something in A_(n-1)) * x_n^s

[keen pawn]

yup

[cunning birch]

But you just showed 0 <= r < s is such that h_n = (something in A_(n-1)) * x_n^r

[keen pawn]

so basically this

[cunning birch]

not really

the "something in A_(n-1)" doesn't have to be specifically b

[keen pawn]

oh right, makes sense

[cunning birch]

we just denoted this "something in A_(n-1)" as b

[keen pawn]

that was irking me earlier on, glad to know my suspicsions were right

thanks

TYSM again, I'll have to sleep now

,ti

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The current time for math_rocks is 01:31 AM (IST) on Tue, 03/02/2026.

[keen pawn]

.close

[amber wasp]

Channel closed

Closed by @keen pawn

Use .reopen if this was a mistake.

[zealous nymph]

YO WTH

i just realized is 1:31

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[tawdry moon]

I am stuck here. So basically it says let's say f(x)=x sin x. Determine function f graph on the point of x = pi/4 set tanget's slope

[tawdry moon]

,ti

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[tawdry moon]

,ti--set

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