help-23

[vestal bear]

Anything more to ask?

[solemn kestrel]

yeah this seems very confusing

[vestal bear]

You have idea about pre image?

And rotation

[solemn kestrel]

yes

OH WAIT

im gonna tyr and do this

myself

is this the correct answer @vestal bear

.close

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[lean otter]

Odd question, but just wanted to verify. My teacher said something along the lines of "the sum of two divergent series can be convergent". Are there any examples of this?

[lean otter]

It's probably something very simple but I just don't see it

[solid vector]

yes

it is simple

[marsh walrus]

several people are typing

[tall bough]

Sum(1/n + 1/n) do they mean this with sum of two divergent series ?

[solid vector]

1 + 1/2 + 1/4 + 1/8 + ...

[marsh walrus]

you can construct a trivial example

[solar hazel]

sum (-1)^n and sum (-1)^(n+1)

[marsh walrus]

you should be able to construct less trivial examples using notions of absolute convergence

[solid vector]

1 + 1/2 + 1/3 + 1/4 + 1/5 +...

[lean otter]

Yeah i have no idea if I should be worrying about the whole gang pulling up everytime I open a channel

[marsh walrus]

it is a blessing

[lean otter]

$-\frac{1}{n}$ and $\frac{1}{n}$ is an intuitive example

[flat frigate]

Ø

[marsh walrus]

my first thought

[lean otter]

Oh, right this makes sense

[marsh walrus]

but in general i mean compare $\sum \frac 1n$ and $\sum \frac{(-1)^n}{n}$

[flat frigate]

jan Niku

[marsh walrus]

this is probably only a subset of the situations where this can occur

but adding in negative terms is just where my brain goes

i guess nah itd have to be that right

i feel like

there are analogous things with sequences

but idk you didnt ask so i wont babble

[lean otter]

It's okay we can take this to discussion later

Thanks lads

[wild basin]

ok

[lean otter]

And gals ofc

[wild basin]

hi lexqa

[lean otter]

.close

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[solar hazel]

sequences open up a lot more examples

[lean otter]

hi

[wild basin]

:).

[lean otter]

what happened to ur nails

[toxic stratus]

series are just sequences

but you construct them from partial sums

[solid vector]

.close

[lean otter]

Yeah that makes sense

[marsh walrus]

i guess im thinking of every sequence has a convergent subsequence

it makes it clear what like

i mean the image in your head of just pulling off terms

can be thought of as the sum

make the sequence in question the sops

[marsh walrus]

wait this is not true is it

[clear blade]

true in R^n

[solar hazel]

it is not :c

[clear blade]

bolzano weierstrass

[junior smelt]

Bounded? [if limits must be finite]

[clear blade]

ah yep

ofc

[junior smelt]

Or in a compact space

[clear blade]

that holds true in any sequentially compact topological space

[safe radish]

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[marsh walrus]

yea obvious missing piece analysis falls out of my head

[marsh walrus]

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[junior smelt]

Totally, bounded

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[ember sphinx]

https://gyazo.com/4e1a17ef36b6c09a27304ebe7aa214b5

[ember sphinx]

is there a cheat way to find the maclauren series of this

coz no way i just take like 4 derivitves for 2 marks

[lean otter]

do u at least have the popular ones memorised

[ember sphinx]

i know

yk

sin x

cos x

e^x

thats about it

[lean otter]

okay do u know what 1/(1+x) is

the taylor of that

[ember sphinx]

erm

no not really

is it like

(-1)^n x^n

[clear blade]

yes

[ember sphinx]

ok

so how does it help us

[clear blade]

make a variable substitution

u = x^2

[ember sphinx]

so its 1/1 + u

[tall bough]

yes

[ember sphinx]

so it'll still be (-1)^n u^n

(-1)^n x^2n

[clear blade]

yes

[ember sphinx]

omg

its right

https://gyazo.com/1b299d432c77bef22b51f6c3ff03cb76

what about this

one

all i know about this is that

thats ln(1+x^2) when integrated

[clear blade]

ln(1+x) also has a well-known maclaurin series

[ember sphinx]

and we literally just had that

[shut inlet]

i made a little mistake

[toxic stratus]

just multiply this by 2x

[ember sphinx]

gets us the 2nd image i posted

which is my 2nd question

[safe radish]

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[cosmic dune]

https://media.discordapp.net/attachments/1059467107814866944/1059467107978465280/20230102_191320.jpg

[cosmic dune]

I just want for spherical

[safe radish]

@cosmic dune Has your question been resolved?

[clear blade]

,rccw

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@cosmic dune Has your question been resolved?

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@cosmic dune Has your question been resolved?

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[languid vigil]

what do i start with?

[desert inlet]

you can start by sending the 3x+4 to the right side

[gleaming wigeon]

step 2
||then converting it into a quadratic equation||

[desert inlet]

i think she's bussy

[languid vigil]

sorry I'm back

[desert inlet]

and what did you find

[languid vigil]

by sending it to the right do i then multiply the quadratic by the reciprocal of 3x+4?

[desert inlet]

yes

[languid vigil]

it goes back to the question

[desert inlet]

wait i'm kinda confused rn what did you do exactly?

if you bring 3x+4 to the right it should then be (3x+4)(3x-4)

[languid vigil]

ohh yeah yeah

so then it's ax^2+bx+c=9x^2-16

[desert inlet]

yeah

but what happened to the x in the right side?

[languid vigil]

well it was -12+12 so they cancelled...

[desert inlet]

so it becomes?

[languid vigil]

0

oh i just solved it lol

[desert inlet]

yes

[languid vigil]

noice

[desert inlet]

you know that b=0

[languid vigil]

yeah heh thanks again dude 😄

.close

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[unique valley]

In the last line I had factored out lambda from the entire bracket

Is that valid to do?

[nova creek]

Yes

[unique valley]

Did they make a mistake with the ms

Surely it should be S•T that they should be showing

[nova creek]

Wdym

Oh yeah, it seems they're showing T○S is linear rather than S○T

[clear blade]

lol the inputs come from V in the first part and from U in the second

[hot remnant]

so T is a transformation taking U to V?

[safe radish]

@unique valley Has your question been resolved?

[unique valley]

I’m confused? Are you answering my original question

[unique valley]

Yes

[clear blade]

not answering anything

pointing out a discrepancy in the solution

[hot remnant]

what is ur question? the Mark scheme seems to be wrong

[safe radish]

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[timid plank]

Hello , need help in composite functions in math. I got stuck on this question and I was wondering if anyone knows how to do this.

F(x) = x³ + 1 g(x)=2x-5

A) find ff(x)

[timid plank]

My working

I seem to have gotten stuck

And not sure what to do

[hasty folio]

f(x) = x^3 + 1

f(f(x)) = (x^3 + 1)^3 + 1

[lean otter]

,rotate

[lean otter]

uh is f meant to be capitalised to F(x)?

or is that a typo?

because F(x) is different from f(x)

[timid plank]

Oh just typo

[lean otter]

oh okay phew good

also yeah what supah did you need to self-compose basically

[timid plank]

What do you mean

[hasty folio]

(x^3 + 1)^3 + 1 you wrote this in ur paper

but u expanded the parenthesis wrong

[lean otter]

u r right it is $(x^3 +1)^3 + 1$ but how did u conclude that it is $(x^3 +2)$ from that

[flat frigate]

♡LexQa♡

[lean otter]

like

[timid plank]

Hmm I didn't expand it

[lean otter]

where did the +2 come from

rather

[hasty folio]

^

[timid plank]

I added the 2 no 1s to make 2

[lean otter]

oh so like

[timid plank]

Basically I didn't know what to do , do I add a power ³ outside the bracket ?

[lean otter]

you added $(x^3 + {\color{red} 1})^3 +{\color{red} 1}$ the two numbers highlighted in red?

[flat frigate]

♡LexQa♡

[timid plank]

Yes

Yes I did

[lean otter]

mmm

thats an illegal operation

[timid plank]

But I'm not sure if I'm right

So what do you think I should do

[nova creek]

$(a + b)^n + c \neq (a + b + c)^n$

[flat frigate]

Gamer Dio

[lean otter]

your answer is fine as is here

but

you can expand the binomial if you prefer

[timid plank]

Also when x³ is multiplied with ³ do you multiply only the powers?

[lean otter]

make note of the fact that
[
(a+b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3
]

[flat frigate]

♡LexQa♡

[lean otter]

what do you mean? can u show me what you mean?

[timid plank]

Okay like wait I'll show you

[lean otter]

suresies

[hasty folio]

I honestly don't think the question requires binomial expansion

But its a fun exercise i guess 😄

[lean otter]

the answer is fine as is there is no need for expanding it yeah

but might as well ig

[timid plank]

Like here do you multiply only the powers when you want to expand them

[nova creek]

$(a + b)^3 \neq a^3 + b^3$ and $1^3 \neq 3$

[flat frigate]

Gamer Dio

[lean otter]

alright so basically what gamer dio is said

exponents are NOT distributive over terms that are added

but they are over terms being multiplied

so like

[timid plank]

Is it like how you do in indices

Cuz ik how to do indices

[lean otter]

[(a\cdot b \cdot c)^n = a^n \cdot b^n \cdot c^n]
[(a+b+c)^n {\color{red} \neq } a^n + b^n + c^n ]

[flat frigate]

♡LexQa♡

[timid plank]

I'm sorry but I don't know these symbols

[lean otter]

oh a, b ,c are just any number you want

like

what are u used to?

variables-wise

[timid plank]

Yeah but x³ has a power

Then what?

[lean otter]

same story

[timid plank]

Like do you add them or multiply them

The powers I'm saying

[lean otter]

are you talking about like how to compute $(x^3 +1)^3$?

[flat frigate]

♡LexQa♡

[timid plank]

Yes

[lean otter]

like how spread the exponent in?

alright

[timid plank]

Like to open the brackets yes

[lean otter]

so you need to know the binomial expansion of the third power, which is this

you cannot say

$(x^3 +1)^3 \enspace {\color{red} \neq} \enspace x^{3 \cdot 3} + 1^3$

[timid plank]

So

[flat frigate]

♡LexQa♡

[timid plank]

Okay so you do multiply them

Okay

[lean otter]

oh that was ur question

okay so like

[timid plank]

That makes sense

[lean otter]

$(a^{n})^m = a^{m \cdot n}$

[flat frigate]

♡LexQa♡

[lean otter]

but this is wrong regardless

like the expansion

[timid plank]

What

Why

[timid plank]

Indices , yes

This is what I'm talking about

Since they have powers , do you multiply them

[lean otter]

the correct expansion would be $x^9 + 3x^6 + 3x^3 + 1$

[flat frigate]

♡LexQa♡

[lean otter]

yeah sure, but

just take those following rules/properties as facts:
\begin{align*}
& (a^{m})^n = a^{m \cdot n} \
& (a+b)^n \enspace {\color{red} \neq} \enspace a^n + b^n \
& (a+b)^2 = a^2 + 2ab + b^2 \
& (a+b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3 \
\end{align*}

there are derivations and explanations as to why all of that is the way it is online

[flat frigate]

♡LexQa♡

[timid plank]

I see alright

Thanks

[lean otter]

ofc do u have any other questions?

[sand minnow]

you should probably write it out as $$(x^3+1)^3=(x^3+1)(x^3+1)(x^3+1)$$and do the standard bracket multiplication from there to convince yourself why these are the case

[flat frigate]

Duh Hello

[safe radish]

@timid plank Has your question been resolved?

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[lean otter]

Ok

[past hearth]

But the answer is x^2y

So I’m a bit confused

Sorry it should be a dx on the last one

[final halo]

y is a function of x so you can't just treat it as a constant when doing the integrations

Instead you should think about "undoing the product rule" inside the original integral

[past hearth]

how so?

[final halo]

It looks like the product rule applied to something

Figure out what that something is

[past hearth]

so its integration by parts?

[final halo]

No

[past hearth]

oh

well how do u figure out what that something is

lol

[final halo]

Write out what the product rule says in general for two arbitrary functions, and compare with what you have

[past hearth]

so the product rule for f(y)f(x)

[final halo]

No you want it for two functions of x

[past hearth]

oh ive got it

f(y)f(x) differentiates to f'(y)f(x) + f'(x)f(y)

so f(y) is just y

and f(x) is x^2

[final halo]

You should be doing f(x)g(x)

[past hearth]

but then that doesnt give any dy/dx terms in the answer

[final halo]

It will tell you that f(x)=x² and g(x)=y

The way you've written it you have two of the same function bc you wrote f twice

[past hearth]

true

[final halo]

And you shouldn't restrict yourself to a function of y

[safe radish]

@past hearth Has your question been resolved?

[safe radish]

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[sharp warren]

hey

[sharp warren]

We propose to approximate the length of a curve portiòn using an algorithm.
f being a function defined on an interval a ; b, the method consists in approximating its representative curve in an orthonormal frame of reference by segments whose extremities are points of the curve and to sum their lengths. The selected points of the curve have their abscissas equidistant on the interval [a; b].
Thus the greater the number of divisions of the interval a; b], the more precise the approximation will be.

1. Calculate the length of a segment
   In an orthonormal frame of reference, A(xA : YA) and B(X ; Xg) are two points.
   a) Express the distance AB in terms of the coordinates of A and B.
   Complete this program in Python language
   the function Distance which has for parameters the coordinates of points A and B and which returns the distance AB.
   We subdivide the interval [a : b] in N intervals of the same amplitude.
   (a) Express the amplitude of each interval as a function of a, b and N.
   (b) If u is the abscissa of one of the selected points on the curve, what is the abscissa of the next point?
2. Sum and test
   We propose to approximate the length of a portion of a parabola representing the square function.
   a) Complete the following Length function written in Python language which returns the sum of the lengths of the segments thus formed.
   b) Enter these functions one after the other and test them by approximating the length of the parabola representing the square function on the interval [-1 ; 1]

what i do:
1A
Root( (xb-xa)2 + (yb-ya)2 )

1B
Instead of the star: sqrt
And in the box : sqrt((xB - xA)2 + (yB - yA)2)

2A
(b-a)/N

2B
u + (b-a)/N

3A
Again, several ways to do it
I would put v = u+(b-a)/N
And Distance(u, u2, v, v2)

[safe radish]

@sharp warren Has your question been resolved?

[low pasture]

@sharp warren Just a little tidbit,

from math import *

This means that everything jn the math library will be imported. So instead if typing math.sqrt you can type sqrt. Now, replacing the star with sqrt shouldn't truly change anything provided you only use the sqrt function from the library. Just wanted to point out that the question did not ask to replace star.

[safe radish]

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[lean otter]

for which integer n>1 is 2023/n^2 an integer?

[lean otter]

I don't get it

why can't it be 1?

because 2023/1^2 = 2023 which is an integer

[sonic sierra]

n>1

[lean otter]

oh right

[sonic sierra]

A condition is given

[lean otter]

so how would one solve this

[sonic sierra]

Maybe prime factorisation of 2023

?

[compact ferry]

just factor 2023

[lean otter]

i c

thanks

.close

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[echo jolt]

does tr(A^2) = tr(A)^2, where tr(A) is the trace of a matrix?

[cosmic grove]

no

[echo jolt]

I have to prove that A^2 = tr(A) * A for every rank 1 matrix

[quasi bison]

the key word is rank 1.

[echo jolt]

it looks a bit like Hamilton Cayley

but the matrix has n rows and n columns

[quasi bison]

a matrix of rank 1 has only a single nonzero eigenvalue, and it occurs with multiplicity n-1

so you could apply cayley-hamilton

unless your class uses that name only for the 2 by 2 version...

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[bright turtle]

is this right

[bright turtle]

??????

[fossil owl]

Yup! Looks good 🙂

[bright turtle]

ok thx

.close

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[lilac jacinth]

I need help with this question

[novel sandal]

what answer are you currently getting

[worthy hemlock]

What did you try?

[lilac jacinth]

i tried doing this: -9 x -1/625

and I dont know how to solve that

[worthy hemlock]

Just multiply fractions

[lilac jacinth]

i got it

9/625

[worthy hemlock]

That's it

[lilac jacinth]

alr thanks

.close

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[pine summit]

if v is 8.0 l.e. what is v + u?

[wheat cave]

hey, I've seen this before

[pine summit]

nice

[wheat cave]

first find the ratio of v to u

or straight up the lenght of u

[pine summit]

is is 5

[wheat cave]

did you just guess that?

start by getting the size of a square in the backround

aka, one of those blue lines

[pine summit]

from the -> of v to u it is 5 squares

[wheat cave]

yes, but square != lenght unit

like, your v is $

[pine summit]

idk what ur saying right now

can u simplifie

[wheat cave]

you have to find the lenght of v+u in l.e, so it doesn't help that you count the squares just like that

[pine summit]

oh ok

the v is 8l.e and the length of it is 4x and 1y so im not sure what to do

[wheat cave]

whats the formula for the lenght of vertices in a coordinate system?

[pine summit]

x,y

[wheat cave]

thats not even a formula

[pine summit]

im bad at english

let me translate fast

[wheat cave]

whats |v|

[pine summit]

u mean the v in the question?

[wheat cave]

any vector tbh

[pine summit]

it is a vector

[wheat cave]

when you know x and y

no, the lenght

'betrag'

[pine summit]

the lenght of v?

[wheat cave]

yes

whats the formula for that

[pine summit]

f(x)

?

[wheat cave]

thats not a formula

[pine summit]

let me translate

[wheat cave]

its $|v|=\sqrt{x^2+y^2}$

[flat frigate]

Jigglyproff

[wheat cave]

with v=(x,y)

[pine summit]

v = 4x + y

[wheat cave]

so whats x

[pine summit]

that idk

because the question is with l.e

[wheat cave]

so tell me what x is

you know that |v|=8 l.e

[pine summit]

so 4x + y = 8l.e

can i do 8/4?

so then x = 2 l.e

[wheat cave]

that made zero sense

[pine summit]

why

[wheat cave]

where the y go

[pine summit]

it is only 1

[wheat cave]

also, its still squared

https://cdn.discordapp.com/attachments/903481106819612714/1060688307106893966/266573406919589888.png

[pine summit]

ok so then

[wheat cave]

so $8=\sqrt{x^2+y^2}$

[flat frigate]

Jigglyproff

[pine summit]

yes

after that then?

[wheat cave]

find x?

[pine summit]

how

[wheat cave]

you have an equation with one variable?

https://cdn.discordapp.com/attachments/903481106819612714/1060686451043479682/Screenshot_2023-01-05_232248.png

what can you tell me about x and y

what are x and y ?

[pine summit]

on v it is 4x + y and u is 2x +4y

[wheat cave]

i dont care

I want to know what you can tell me about x and y

[pine summit]

one x and y is one square

[empty gyro]

I think a better way to ask this is, how many units to the right and how many units up v from its base?

[wheat cave]

so... what are x and y in relation to each other?

[pine summit]

they make an vector

[wheat cave]

bre das wird unterkurs

[pine summit]

english pls

[wheat cave]

if x and y make a square, what are the lenghts related to each other

okay

x=y

[pine summit]

yes

[wheat cave]

y must be the same lenght of x

so

[pine summit]

yes

[wheat cave]

https://cdn.discordapp.com/attachments/903481106819612714/1060688984184983612/266573406919589888.png

solve this please

[pine summit]

ill try

16x + y

[wheat cave]

WHY Y

jesse

[pine summit]

because it is one y?

[wheat cave]

we know that y=x

???????

[pine summit]

so then its 5x

0 y

[wheat cave]

how is it 5 x

is squared

[pine summit]

then its 5 squares

[wheat cave]

???

[pine summit]

bruh

if x = y

then 1 y = 1 x

then 4 x + 1 y = 5x

[wheat cave]

$8=\sqrt{(4x)^2+(1y)^2}$

[flat frigate]

Jigglyproff

[wheat cave]

thats your formula

[pine summit]

yes

[wheat cave]

how are you getting 5

[pine summit]

but 1 y * 1y is still 1 y

[wheat cave]

ok

and 4x*4x is 4 x???

[pine summit]

no that was why i was saying 16x + 1 y

then root

[wheat cave]

thats what you said

[pine summit]

look

[pine summit]

this

[wheat cave]

thats y

and not x

just make x

[pine summit]

wdym

its is 16x + y

[wheat cave]

you have 16x+y, you know y=x, then you say it 5x?

[pine summit]

then if u do the root of it all its gonna be 4x + y

[wheat cave]

its not

[pine summit]

how

[wheat cave]

thats not how roots work

[pine summit]

what?

[wheat cave]

$\sqrt{a+b}\neq \sqrt{a}+\sqrt{b}$

[flat frigate]

Jigglyproff

[pine summit]

oooh

so i need to add 16x + y then do the root

so its gonna be 4xy

[wheat cave]

its $16x^2+x^2$ btw

[flat frigate]

Jigglyproff

[wheat cave]

no its not

[pine summit]

where is the y

[wheat cave]

y=x

[pine summit]

so u change y to x

[wheat cave]

just replace, thats what I was saying 20 messages ago

[pine summit]

oh

u could just say that

[wheat cave]

its so logical I thought you know

[pine summit]

im 12

ok so

16x^2 + x^2 is 17x^2

then root

its gonna be 17x

yes?

[wheat cave]

no its not

[pine summit]

why

[wheat cave]

<@&286206848099549185> I can't do this no more

[pine summit]

u are noob

[wheat cave]

you're trolling

[rancid dagger]

bro 💀💀

[pine summit]

no im not

[tall bough]

He's 12

[pine summit]

im actually trying

[tall bough]

<@&268886789983436800>

[stray socket]

Wdym he's 12

[pine summit]

i cant be 12?

[stray socket]

What

[tall bough]

U can't here

[stray socket]

What is l e

[pine summit]

why?

[glad silo]

@wheat cave what's that?

[wheat cave]

12 is below minimal age for discord, the mods have to kick you

sorry

[pine summit]

im turning 13 this year

[empty gyro]

Funny typo.

[wheat cave]

<@&268886789983436800> someone breaking discord TOS, please remove, they claim to be 12

[lament creek]

Come back then

[pine summit]

help me with math pls

[stray socket]

Ope

[wheat cave]

I swear they are just trolling me tbh

[empty gyro]

Would love to but discord TOS requires you be 13

[final halo]

Green names

[warped roost]

.close

[safe radish]

Channel closed

Closed by @warped roost

Use .reopen if this was a mistake.

[empty gyro]

If trolling, why would they say they are at a bannable age? That ends their trolling

[warped roost]

please move on

[wheat cave]

they didn't know about the age limit I guess

ok sorry

[safe radish]

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[lean otter]

how do u divide integers

[desert inlet]

that's a really obscure question

[grand inlet]

On paper? Manually?

[tall bough]

Give us an example

[lean otter]

like

-7 divided by -20

or -2 divided by +9

[tall bough]

How old are u?

[lean otter]

i just need to know

i forgot how

[tall bough]

Watch a youtube video

[lean otter]

my teacher provided one but he was a dumbass

[pallid nymph]

then find another one?

[lean otter]

js tell me bro

[stray socket]

Fractions go crazy

Surely you can use a calculator

[lean otter]

we arent allowed calculators

[stray socket]

Yet you want us to tell you ... when a calculator can

Why not just whip out your calculator rn then

[pallid nymph]

31.75
4)127.00
12 (12 ÷ 4 = 3)
07 (0 remainder, bring down next figure)
4 (7 ÷ 4 = 1 r 3)
3.0 (bring down 0 and the decimal point)
2.8 (7 × 4 = 28, 30 ÷ 4 = 7 r 2)
20 (an additional zero is brought down)
20 (5 × 4 = 20)
0

127 divide by 4

[stray socket]

LOL

[tall bough]

Do u know what's 2/2?

[lean otter]

no-

[tall bough]

Nice try he doesn't even know 2/2 xd

[pallid nymph]

heres my question

[stray socket]

If you're on discord, go use an online calculator if you just want the answer

[pallid nymph]

how do you call someone a "dumbass" and cant divide

[lean otter]

i can

i mean

[stray socket]

Did you know google can tell you the answer

[lean otter]

i just dont remmeber

grade 3

[pallid nymph]

how does a grade 3

swear...

[lean otter]

im not in grade 3

[tall bough]

What grade are u in

[lean otter]

i learnt how to divide

in grade 3

and forgot

[tall bough]

How can u forgot division, well whatever, watch a YouTube video

[willow urchin]

to be fair we dont really do a lot of manual division in high school

[lean otter]

how tf does 2 divided by 2 equal 1

[glass carbon]

you learnt and you didn't use this knowledge in further maths?

[willow urchin]

ok nvm he is just an idiot

[lean otter]

js didnt care

[desert inlet]

he's trolling

[grand inlet]

Seems so

[tall bough]

He is

[glass carbon]

https://tenor.com/view/fake-account-troll-detected-trol-gif-24822648

[tall bough]

His first message was whats 2+2

.close

[safe radish]

Channel closed

Closed by @tall bough

Use .reopen if this was a mistake.

[lean otter]

,reopen

, reopen

. reoped

[desert inlet]

.reopen

[lean otter]

OK BUT LIKE

I JUST KEARNT HOW TO DIVIDE

SO LIKE

2 groups of 2

no

that mulitoliucation

2

[pallid nymph]

2 groups of what is 2

[lean otter]

divided

into 2 groups

is 1

bc theres one

in each group

so 14

divided into 3

2

[safe radish]

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[paper arch]

I need to display graphics for a line that goes through these dots: [{-5, -4.5}, {-4, -3.8}, {-3, -3}, {2, -1.8}, {-1, -0.7}, {0,
0.1}, {1, 1.11}, {2, 1.98}, {3, 3.12}, {4, 3.8}, {5, 4.7}]. I don´t know how to show it in wolfram.

[frank glen]

,w regression line of {(-5,-4.5), (-4,-3.8), (-3,-3), (-2,-1.8), (-1, -0.7), (0,0.1), (1,1.11), (2, 1.98), (3, 3.12), (4,3.8), (5, 4.7)}

This is a regression line question

[flat frigate]

Results provided by WolframAlpha

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

Requested by VulcanOne#5806

[frank glen]

@paper arch

[paper arch]

ty

[frank glen]

If you wanna do it manually, you do this

$\sum y = a\sum x + nb \
\sum xy = a\sum x^2 +b\sum x$

[flat frigate]

VulcanOne

[paper arch]

need to do it in matematica

[frank glen]

Well should be the same thing as I did with wolfie

[paper arch]

Cant find the regression comand

[compact ferry]

you can also make an 11th order polynomial that fits the points perfectly

[frank glen]

[paper arch]

Is this possible to do in mathamatica?

if so how?

[frank glen]

Yeah that's gonna be interpolation

Either using least squares or using lagrange or using newton's forward/backward differernce

Using least squares will need you to know how many points you want and the degree of the polynomial will be n-1

[paper arch]

could u try it in wolfram with my example?

[frank glen]

,w interpolate {(-5,-4.5), (-4,-3.8), (-3,-3), (-2,-1.8), (-1, -0.7), (0,0.1), (1,1.11), (2, 1.98), (3, 3.12), (4,3.8), (5, 4.7)}

[paper arch]

hmm..

[flat frigate]

Results provided by WolframAlpha

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

Requested by VulcanOne#5806

[frank glen]

@paper arch

[paper arch]

so ,w interpolate {(-5,-4.5), (-4,-3.8), (-3,-3), (-2,-1.8), (-1, -0.7), (0,0.1), (1,1.11), (2, 1.98), (3, 3.12), (4,3.8), (5, 4.7)}

[frank glen]

That's for the TeXit bot

It can do wolfram stuff

[paper arch]

ok thanks 🙂

[frank glen]

You can test it in #bots

And react with more for a bigger output

[paper arch]

ok thax 🙂

[frank glen]

[paper arch]

.close

[safe radish]

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[safe radish]

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[lone arch]

If $f(x) = log_a(x)$, then $f'(x) = \frac{f'(1)}{x}$ \\ How do you arrive at this?

[lone arch]

$f'(x) = \lim_{x \to a} \frac{log_a(x) - 1}{x - a} = \lim_{h \to 0} \frac{log_a(x + h) - log_a(x)}{h}$

$\lim_{x \to 1} \frac{\log_a(x) - 1}{x - 1} = f'(1)$, so $\lim_{x \to a} \frac{\log_a(x) - 1}{x - a} = \lim_{x \to a} \frac{\log_a(x) - 1}{x - 1} \cdot \frac{x - 1}{x - a}$

But x is going to a

[glass carbon]

you can find limit with h

pretty easy

so I've got

yes

it was right

so 1/x * log_a(e) which is 1/(xln(a))

[lone arch]

$\lim_{h \to 0} \frac{log_a(x + h) - log_a(x)}{h}$, substitute $u = \frac{1}{h}$, then
we get $\lim_{u \to \infty} \frac{log_a(x + \frac{1}{u}) - log_a(x)}{\frac{1}{u}}$

[glass carbon]

notice that:

[glass carbon]

$$\log_a (x+h) - \log_a x = \log_a \Big(1+\frac{h}{x}\Big)$$

[lone arch]

How would you proceed?

[glass carbon]

then

[lone arch]

Oh, logarithm identities

[glass carbon]

$$\frac{\log_a \Big(1+\frac{h}{x}\Big)}{h}=\frac{1}{h}\log_a \Big(1+\frac{h}{x}\Big) = \log_a \Big(1+\frac{h}{x}\Big)^{\frac{1}{h}}$$

[flat frigate]

Modus

Modus

[glass carbon]

now def. of e

and you're done

[lone arch]

$\lim_{x \to \infty} \ln(1 + \frac{1}{x})^x = e$

[glass carbon]

rather (1+1/x)^x only, with ln it gives ln(e) = 1

if you don't see next step here then

[flat frigate]

Modus

[safe radish]

@lone arch Has your question been resolved?

[lone arch]

Thanks

[safe radish]

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[viscid tartan]

im stuck on a hard limit

[viscid tartan]

it reduces to a product of notable limits

but cant figure it out

[glass carbon]

maybe l'Hospital, but it's painful a bit

[viscid tartan]

yeah i tried makes it harder but markscheme , says its a product of notable limits like sinx/x =1 , e^x-1/x=1 , 1-cosx/x^2 1/2

as x->0

[safe radish]

@viscid tartan Has your question been resolved?

[clear blade]

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

[flat frigate]

tushar

[safe radish]

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[gritty depot]

I have some scalar multiple of the identity matrix, call it matrix A
now I want to write A = scalar * Id
What is a nice way to write it in terms of the ij coordinate? How is that usually done, if my attempt is uncommon (even if its correct, which im not sure about)?
A = ( a_ij * δ_ij)_ij = ( (scalar * 1 )_ij * δ_ij )_ij = scalar * ( 1 * δ_ij)_ij = scalar * (δ_ij)_ij = scalar * Id

[gritty depot]

.close

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[glass carbon]

y isn't y^0

it's y^1

y^0 = 1 (and generally a^0 = 1)

also you can do this like that

[flat frigate]

Modus

[glass carbon]

now it's correct

[safe radish]

@undone swift Has your question been resolved?

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[woven terrace]

Can I simplify these any further

[woven terrace]

Like is partial(rhoV_z) the same thing as rho*partial(v_z) since rho is a constant?

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[lean otter]

By experiment, you find that a force field $\bold{F}$ performs only half as much work in moving an object along path $C_1$ from $A$ to $B$ as it does in moving the object path from $A$ to $B$. What can you conclude about $\bold{F}$?

[frank glen]

[lean otter]

what

what is making it break

[frank glen]

By experiment, you find that a force field $\bold{F}$ performs only half as much work in moving an object along path $C_1$ from $A$ to $B$ as it does in moving the object path from $A$ to $B$. What can you conclude about $\bold{F}$?

[lean otter]

well whatever

[flat frigate]

VulcanOne

[frank glen]

fi was not defined in the English unicode

[flat frigate]

♡Lex♡

[lean otter]

oh i see now lol i must've had turkish enabled

well anyways

i guess because $\int_{C_1} \bold{F} \cdot \dd{r} = \f12 \int_{C_2}\bold{F}\cdot \dd{r}$

[flat frigate]

♡Lex♡

[lean otter]

we just say that the field is neither conservative or path independent ig?

[frank glen]

It is definitely path dependent

[lean otter]

yeah makes sense ty

.close

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[frank glen]

[cosmic dune]

I just want for q.69 ,find the volume of the solid

[frank glen]

Well you can use cylindrical coordinates

[lean otter]

pain

[crimson field]

These questions seem like slightly different variants of the same one. What's causing problems?

Yep.

[lean otter]

yeah okay makes sense

[frank glen]

Ye and you will just integrate sqrt(1-r^2)

[crimson field]

We are bounded by the rays theta = 0 and theta = pi/2. Remember the polar angle is measured from the positive x direction in the direction of the y-axis as positive.

[lean otter]

uh

but like

isnt it

[crimson field]

$V = \iint_{R} \sqrt{1 - r^2} \dd{S}$.

[flat frigate]

stabulo

[crimson field]

Yes. The triple integral will reduce to the one I posted above if you evaluated it using polar coordinates.

[lean otter]

oh i see yeah fair

i guess i will delete this just to let OP figure it out

[crimson field]

Ok. I'm not sure what's causing problems for them as they literally basically did the exact same problem before. 🙂

[frank glen]

I guess they're just making sure

[crimson field]

I think they are answer fishers instead of the process.

[lean otter]

they're not really providing anything on their end though

[lean otter]

yeah

[crimson field]

Nope. I won't help with further problems I don't think. 🙂

[lean otter]

yeah do as you desire

[safe radish]

@cosmic dune Has your question been resolved?

[cosmic dune]

Im new to this topic,so even if i am able to docit on my own,i would like to cross check with people who are well versed with this topic

[cosmic dune]

Polar coordinates wont be necessary for thos problem

[cosmic dune]

The problem ,i would be like spending an whole hour trying to figure out on my own only to end being stuck ,and takes more time to solve it here , so i would want correct answers first ,so when i do the problems on my own ,each step by step i would check here ,so that would make time less consuming

[cosmic dune]

Alright ,next time when i post the questions ,i will also provide eoth what i am thinking to do

So taking cylindrical coordinates

Z will be 0 to root (1- r^2)

R will be 0 to r sin theta

[cosmic dune]

And theta will be this ,thnx

Will integrate and check

[safe radish]

@cosmic dune Has your question been resolved?

[safe radish]

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[round tinsel]

Are these answers for limits correct? I do not have an answer key to check. Any help would be appreciated. Thanks in advance 🙏

[unreal fiber]

Yeah all correct

[deep depot]

I Don't see any mistakes there

[round tinsel]

Thankyou

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[ember sphinx]

https://gyazo.com/d73c33e1988214362afb0413fa4de436

[ember sphinx]

i need ze help

[light shoal]

maybe start by computing x_n - 1/3 and finding a bound for its absolute value?

[ember sphinx]

u mean someting like

|x_n - l| < epsilon

[plucky elk]

https://www.milefoot.com/math/calculus/limits/DeltaEpsilonProofs03.htm

[ember sphinx]

thanks

its delta

in our case is big N the same as delta

[woeful plume]

Probably used in ∀n > N

[ember sphinx]

yea we do use that

but i dont know how to give proof

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[tender mountain]

How do I do this? Please could you do it with me?

[plucky elk]

Just do the fraction additions one at a time

And share your work here so people can check it

[lean otter]

what an absolute pain

i wonder if you can make an algorithm to make this faster

[zinc token]

wolframalpha time

[plucky elk]

Simplify 2+1/2 first

[quasi bison]

just do it bottom-up.

[plucky elk]

Then plug that in

Oh Ann's interested.

[lean otter]

calculate the answers and see which is closest to 1.41

[lean otter]

thats exactly why it is pain haha, but i guess there is no escaping it

[quasi bison]

there's also a way to save yourself some trouble writing out the long-as-shit fraction,

but it requires some abstraction

namely, if you let $f(x) = 2 + \frac{1}{x}$ and define a sequence by a recurrence relation as $x_{n+1} = f(x_n)$ and $x_0 = 2$, then the expression we're looking for is $x_6 - 1$.

[flat frigate]

Ann

[lean otter]

oh thats smart! thats what i was looking for

[quasi bison]

the idea is $x_0 = 2, x_1 = 2 + \frac{1}{2}, x_2 = 2 + \frac{1}{2 + \frac{1}{2}}, x_3 = 2 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2}}}, ...$

[flat frigate]

Ann

[quasi bison]

i'd need OP to be on board with this though.

[clear blade]

this is how you'd solve it if it were a continued fraction and look for the limit

[sharp tiger]

[quasi bison]

@tender mountain does my suggestion make any sense to you?

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@tender mountain Has your question been resolved?

[quasi bison]

interesting. many times now, when i've asked kangaroo rat a question, they just conveniently appear to vanish into thin air.

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[pliant merlin]

is S-1 a function?

[sly field]

can u translate it

[junior smelt]

Oh I think I'm all read up now

Your previous argument should work I would think: S^{-1} won't be a function because e.g. the [strictly] negative numbers don't get mapped anywhere

For a strictly negative x, x S^{-1} y is the statement x = y^2, which has no real solution y

• as you said, for a strictly positive x, there are two possible choices of y values for it

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@pliant merlin Has your question been resolved?

[pliant merlin]

ok, thinks for confirming ☺️

*thanks

[junior smelt]

Didn't even notice that

Anyways bedtime for me 🛌

[pliant merlin]

goodnight

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[marsh cloud]

hi i need help

[marsh cloud]

what is the common of πx^2 + 10πx+ 25π

?

[junior smelt]

Common?

Note in all of them you have a pi so factor that out

Then it’s something nicer to deal with?

[marsh cloud]

is common of them is 5pi ?

[stray socket]

No

Last time I checked, 5pi is not a factor of pi

[marsh cloud]

common of these three is what?

pie?

is common?

[junior smelt]

Well, pi the constant, not the actual food pie, but you get my point

[marsh cloud]

ik

so eqaution will

pi(x^2+10x+25)

right?

[junior smelt]

Then you can factor x^2+10x+25

[marsh cloud]

ok

so

i am in 8th grade according my grade

it will become \

$root by 5x^2+10x+25x$

[flat frigate]

Rock Man

[marsh cloud]

idk

bye

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[fiery elbow]

can someone help me get started?

i think i need to use intergrals

making it like, intergral with the limits of 1 and -2 for the first wave?

But it gets confusing when i need to somehow incorporate the straight line so it would give the only the area between those two things instead of the full wave

also it looks symmetrical meaning i can find the midpoint easily (1,0)

[indigo brook]

hello guys

[fiery elbow]

hi

wrong chat?

[dull sequoia]

Ok we need integral you’ve got that much

We can also simplify it to just twice 1 section

Since it’s symmetrical we just need to solve for 1 of the sections

Now we know integrals is adding a bunch of very thin boxes

The height of the boxes defined by the function from that line to the x axis

If we instead define it as from f(x) to g(x) where g(x) = 0 (the x axis is g(x) )

Then the height of the boxes in an integral would be f(x) - g(x)

This finds the difference in y value and thus the height of our boxes

So simply put, let f(x) be the cubic, g(x) be the linear function

Integrate from intersection to intersection, of f(x)-g(x) with respect to x

Then double it

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@fiery elbow Has your question been resolved?

[fiery elbow]

just saw this^

hold on

Thanks ill get on it

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[wind osprey]

Hey guys could someone try and help me understand why when we change the variable to y = 1-x, the integral becomes the opposite ? ( -1* the new integral ) not the best wording sry

Oh shit i just understood why

Lol sorry ❤️

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[unreal fiber]

I think factoring x²+y² helps

(hint you use complex numbers for that)

well basically

you know the difference of squares right?

yea

listening

[obtuse jackal]

They give you the hint so that the exercise is simple. Otherwise there are many proofs that lead to the hint, from trigonometry or complex numbers for example

[steep pewter]

That is tho

[zinc token]

they are infinite

[obtuse jackal]

That's infinitely many (n, m) pairs. Just show that creates infinitely many (x, y, z) triplets

[zinc token]

those are pythagorean triples

[unreal fiber]

https://youtu.be/n6vL2KiWrD4 I think this video comes to a similar conclusion

[steep pewter]

Put a m get a n

Infinite times

[obtuse jackal]

For instance for every complex integer z = a+ib, you have z^2 = a^2 - b^2 + 2iab
Taking the modulus yields a Pythagorean triple:
|z^2| = |z|^2 = (a^2 - b^2)^2 + (2ab)^2
In case you wanted a short proof of it, though that uses complex numbers

Why do you want 2mn = 1 ?

[steep pewter]

Umm is 2.1 a integer

[obtuse jackal]

Idk where that first equality comes from

[steep pewter]

[steep pewter]

Proof by contradiction i think

Or the other way

Proof by no contradictions

[obtuse jackal]

It works

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[steep pewter]

No

Are fractions integers?

[obtuse jackal]

Btw these are the only solutions to the equation

[safe radish]

✅

[steep pewter]

[obtuse jackal]

No

[steep pewter]

Hmm ok thanks

[zinc token]

for b) have you seen density of Q in R

between any 2 real numbers there is a rational number

I mean there's a hilariously easy way to do b) btw

[steep pewter]

Try manually show it is reapeating, square the mean of the two numbers add one then root it, same again

[zinc token]

w h a t

sure

suppose we have a, b as our rational numbers

and suppose WLOG a < b

can you turn them irrational?

ok ok hmm

let's try a different way

let m be

a + |a-b|

so basically m is exactly in the middle

great great

so add pi or something

or sqrt(2)

[obtuse jackal]

I think it might work though

[steep pewter]

Bro wtf

Add pi?

Y?

[zinc token]

you can add whatever irrational number you want to a rational number