help-42

[lilac nimbus]

ok so you have a circle of radius 1

an you draw a line from the center to the edge of the circle

[regal patio]

does it matter if u explain it w.o a triangle

?

[lilac nimbus]

We are going there

there is a triangle on the photo hehe

[regal patio]

i dont know why u didnt just get a triangle but i mean u know ur stuff and i dont so

go on

[lilac nimbus]

If you see

A triangle is made out with the horizontal line

the vertical line

and the line that we draw

So

The cosine will be the length of the horizontal line

And so the sin the lenght of the vertical one

[calm coral]

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[mossy flare]

How did they go from the first line to second, and then the third

[calm coral]

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[simple musk]

let $P(x) = 4 - 7(x - 1) + (x-1)^2$ be the taylor polynomial of degree $2$ centered at $x_0 = 1$ of a function $f$.$\\$ Let $g(x) = 2x -\int_{1}^{x} t^2 f(t) dt\\$ Find the taylor polynomial of degree $2$ of $g(x)$ centered at $x_0 = 1$

[potent lotus]

calc_and_real_anal

[calm coral]

@simple musk Has your question been resolved?

[remote mural]

Plug P(x) as f in the integration

And write the final form in terms of the powers of (x-1)

[calm coral]

@simple musk Has your question been resolved?

[simple musk]

can you elaborate

[lean hemlock]

what’s the derivative of g(x)?

[simple musk]

2 - x^2f(x)

haha

[lean hemlock]

u got it?

[simple musk]

wdym

[lean hemlock]

oh nvm, i’m not sure either; sorry

but i think it has some pattern

like its derivatives

[remote mural]

Plug P(x) in place of f(x)

[simple musk]

wdym

[remote mural]

We have given the function f in terms of Tylor polynomial

We can basically use the Tylor polynomial as the function itself

[simple musk]

,, P_2(x) = 4 - 7(x-1) + (x-1)^2 = f(1) + f'(1)(x - 1) + \frac{f''(1)}{2!} (x-1)^2

[potent lotus]

calc_and_real_anal

[remote mural]

Yeah

So you want to find g(1) , g'(1) ..?

[simple musk]

exactly, how

g(1) = 2 - 0

because $\int_{1}^{1} . . .$

[potent lotus]

calc_and_real_anal

[remote mural]

Plug 1 and you'll get g'(1)

[simple musk]

,, g'(1) = 2 - f(1)

[potent lotus]

calc_and_real_anal

[remote mural]

Yes

And f(1) is given

[simple musk]

f(1) = 4

[remote mural]

Coefficient of (x-1)^0 in Taylor poly

Yes

[simple musk]

g'(1) = -2 .

[remote mural]

Yes

[simple musk]

so confusing tbh

[remote mural]

Why?

Same way you can get g''(1)

[simple musk]

,, g'(x) = 2 - x^2f(x) \ g''(x) = -2xf(x) + -x^2f'(x)

[remote mural]

Yes

No

You'll have to use chain rule

[simple musk]

true

[remote mural]

Yea

Product rule

[potent lotus]

calc_and_real_anal

[remote mural]

Yea

[simple musk]

,, g''(1) = -2f(1) -f'(1)

[potent lotus]

calc_and_real_anal

[simple musk]

f(1) = 4

[remote mural]

Yea

[simple musk]

f'(1) = -7

,calc -8+7

[remote mural]

What

[simple musk]

?

[remote mural]

f'(1) is -7

[simple musk]

my bad

[potent lotus]

Result:

-1

[simple musk]

now what

[remote mural]

That's the g'(1)

[simple musk]

g'(1) = -2

[remote mural]

Yea

Now we need g"(1)

[simple musk]

,, \begin{cases} g'(1) &= -2 \ g''(1) &= -1 \ g(1) &= 2 \end{cases}

[potent lotus]

calc_and_real_anal

[remote mural]

Yeah

[simple musk]

,,Q_2(x) = g(1) + g'(1)(x-1) + \frac{g''(1)}{2!}(x-1)^2 \ Q_2(x) = 2 + -2(x-1) - \frac{1}{2!}(x-1)^2

[remote mural]

Yea

[potent lotus]

calc_and_real_anal

[remote mural]

Yeah That's the polynomial

[simple musk]

how do I check my solution

[remote mural]

You can put P as f in your integration

And do the integration

[simple musk]

how to check with wolfram

[remote mural]

I'm not sure this will work, but it should

[simple musk]

how

[remote mural]

Dunno how to call it here but

Can go to the site and do it

[simple musk]

",w "

[remote mural]

Basically f(t)=P(t)

[simple musk]

how do I check my answer with wolfram

[remote mural]

Ask it to integrate the function in your question

Write polynomial P in terms of t at the place of f(t)

[simple musk]

can you ask it in wolfram website and send screenshot

[remote mural]

Solution is in fact correct

Sorry for the delay

[simple musk]

lol

nono its alright

how did you asked in wolfram though?

[remote mural]

Asked it to integrate

[remote mural]

This function g

I mean the intention part of it

Got it?

[simple musk]

mmm kinda

[remote mural]

Check it here

If you want

,w $g(x) = 2x -\int_{1}^{x} t^2 ( 4 - 7(t - 1) + (t-1)^2 )dt$

[simple musk]

it was good but then its not

[potent lotus]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by kulusevskii

[remote mural]

Yeah

[simple musk]

?

[remote mural]

It somehow dosen't do two things together

[simple musk]

lmaoo haha

[remote mural]

Doing the expansion and integration

[simple musk]

icic

[remote mural]

Now you got it?

Just say ",w Taylor expansion g(x) "

Write the g(x) there

[simple musk]

okay thank

.close

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[calm coral]

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[alpine stone]

The simplest way is to just state that x is a prime number

I don't believe there is some conventional notation

[swift vortex]

You can just complicate it and say x has only 2 factors: 1 and itself

Yes

[calm coral]

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[wheat briar]

Can someone please check my solution I'm in doubt

[calm coral]

@wheat briar Has your question been resolved?

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@wheat briar Has your question been resolved?

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[wheat briar]

.reopen

[calm coral]

✅

[wheat briar]

<@&286206848099549185>

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[balmy flame]

alright I have a big issue. I have a ti 84 plus ce. One day, I was taking my algebra 2 test right, my graph was acting funky so i reset my calculator. i somehow deleted absolutely everything on this bad boy. Every app i've ever downloaded and every app that was already on it. I have no programs, no nothing. Now when I try to connect to ti connect ce, it doesnt reconginze my calculator. Ive gone to device manager and tried to update my drivers but it doesn't show up. This cord is new and highly doubt it has anything to do with it because its been fine until this problem occured. I dont know what to do and Id like the basic apps back on my calculator as long with the ones i've downloaded. PLEASE HELP ME! I know this isn't necessarily math, but this is the only server I know that could have even the slightest idea of what to do.

[strange lichen]

u need to download ur programs again

by resetting, u wiped it all

[balmy flame]

How do i do that when ti connect ce cant recongnize my calculator

[strange lichen]

maybe reset again, or try reinstalling connect

or do both

[balmy flame]

ive reset it but i can try to reinstall connect ce

it didnt work

[calm coral]

@balmy flame Has your question been resolved?

[calm coral]

@balmy flame Has your question been resolved?

[balmy flame]

.close

[calm coral]

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[unique knoll]

.reopen

[unique knoll]

how does the first circel turn into the 2nd circel

[tiny monolith]

1 1/2 = 3/2

the exponent is changed from negative to positive and moved to the bottom

[unique knoll]

but why does the -1.5 turn into a 3

and the x randomly gets a 2 added?

in the second circel

[tiny monolith]

-1 1/2 = -3/2

[unique knoll]

where do is see a 11/2?

[tiny monolith]

the left most thing u circled

[unique knoll]

ooooo wait i thought you meant 11/2 but i understand now that its 1.5

and 1.5 is also 2/3

thxxx

.close

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[tiny monolith]

3/2

np

[unique knoll]

yes that

[calm coral]

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[blazing coyote]

.close

[calm coral]

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[calm coral]

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[remote mural]

Is this correct?? And how do I continue it

[remote mural]

wait

ok the first one is good

lets start with the first one

okay

$7^{x+2y} = 1$

[potent lotus]

.

[remote mural]

what do you think the value of x and y are

[tidal grotto]

,tex .exp rules

[potent lotus]

🫎 Moosey 🫎

[remote mural]

0?

[remote mural]

nice

so thats the final answer?

[tidal grotto]

he applied power of power rule and then product rule

[remote mural]

yes for the first one

now the second

okay

[tidal grotto]

not necessarily

you only know

x+2y=0

so you have one equation

you can use the other equation to get the other equation involving x and y

[remote mural]

oh are the equation related

yeah

[tidal grotto]

they both have to have the same x and y

[remote mural]

then how do i solve it

you already have x+2y = 0

now lets solve the second one

what can you write 1/25 as

okay

1/4

no

no

no

wait

hahaha

0.04

no

im speaking

like

to get the same base

5

i dont

know

25^-x

no

😭

where do you have x in 1/25

????

in the formula

i think...

i give up

.close

[calm coral]

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[topaz kiln]

Hey could someone teach me how to do question 7? I’m very confused

[topaz kiln]

Nothing

@unique jackal

I have no idea what to do

[warm warren]

Did you read the question

[topaz kiln]

Yes😭

[tacit lark]

To be fair, this question is garbage

And confusing

And the entire sheet is garbage (boring)

Use it as toilet paper for example

[topaz kiln]

I understand that much

[tacit lark]

Yes

Now put the angle on the diagram

[topaz kiln]

Please help bruh

[tacit lark]

Remember

3x+5y = 0 means

y = -3x/5

So the first line is not supposed to look like that

[topaz kiln]

Ohh

Ok lemme fix

[tacit lark]

But we dont care

The idea is

[topaz kiln]

Oh

[tacit lark]

This angle theta

Is related to

The slope

Of the line

This is the result you need

[topaz kiln]

Hmmm

[tacit lark]

The tan of the angle is equal to the slope

You probably have seen this

In the course

[topaz kiln]

I didn’t know that

I’m hearing it from u first time

[tacit lark]

What do they teach you at school everyday then

Tiktok

Tbh i think you already have seen it

And maybe you forgot

[topaz kiln]

Uhhh

[tacit lark]

Otherwise such an exercise would not pop out

Out of nothing ?

The teacher came one day and said hi do this sheet

?

Or there was an explanation before

I usually explain things to my students

Then give them a sheet

Try to look back into the lesson

[topaz kiln]

I was not there maybe that day

I was absent a few days

[tacit lark]

That explains it

[topaz kiln]

Pls help

[topaz kiln]

Yes I understand why

[tacit lark]

Unfortunately I dont feel like doing this exercise its too boring for me 😦

[topaz kiln]

please

[tacit lark]

My life is boring as it is

But im sure someone else will

Plenty of competent helpers here

[topaz kiln]

@tacit lark

Am I on the right track

[calm coral]

@topaz kiln Has your question been resolved?

[calm coral]

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[idle fractal]

Yes i am asking why?

[idle fractal]

Why l'hopital fails

0/0 form

1/2√(1+x) +1/2√(1-x) ==0

[alpine stone]

Why do you think l'hopital fails?

[alpine stone]

Are you sure you differentiated sqrt(1-x) correctly?

[idle fractal]

Ohh

Got the point

It is one

Thanks sir

.close

[calm coral]

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[idle fractal]

Hints please @alpine stone

[alpine stone]

I would try the conjugate

[calm coral]

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[uncut island]

did you solve it @idle fractal

[calm coral]

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[calm coral]

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[terse swan]

Can someone check if these are right? Pls ping me

[calm coral]

@terse swan Has your question been resolved?

[hollow lion]

,w integrate x^2 cos(4x)

[potent lotus]

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Upgrade to WolframAlpha Pro!

Requested by shsgd

[hollow lion]

Looks good 👍

[calm coral]

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[calm coral]

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[icy oriole]

how to multiply 48 by √2/2?

[icy oriole]

im hella stuck

on this

i thought it'd be 24√2 but its stupid isnt it

[potent igloo]

That's exactly what it is

[icy oriole]

bro u fr

[potent igloo]

Yes

[icy oriole]

fuck me

[potent igloo]

No

[icy oriole]

not in that meaning

[cobalt basalt]

dam

[icy oriole]

alr alr one more how do i solve an area of an equilateral triangle with height 12?

[cobalt basalt]

a

height was somethin like underroot 3/4 a

[icy oriole]

eh?

[cobalt basalt]

see

[potent igloo]

Cut the base into two a/2 sides

[tall moon]

using pythag, 12^2+(a/2)^2=a^2

[icy oriole]

yeah

[potent igloo]

This, then solve for a

[icy oriole]

how am i even supposed to solve (a/2)^2

wut

the heck

is it just a=4

144+16=16????????????????????????????????????????????????? that doesnt make sense to me im stupid

[potent lotus]

faiyrose

[icy oriole]

e

144 + a^2/4 = a^2/4

[tall moon]

no???

[potent lotus]

faiyrose

[icy oriole]

oooooooooooooooh

i almost get it

yeah okay i get it

so

wait no

fuck

[buoyant crystal]

hi i dont really know much about latex but i hope you can get this
a^2 = (a/2)^2 + 12^2
a^2 = (a^2/4) + 144
a^2 - (a^2/4)= (a^2/4) - (a^2/4) + 144
4a^2/4 - (a^2/4)= (a^2/4) - (a^2/4) + 144
3a^2/4 = 144
( 3a^2/4 )/( 3/4 ) = 144/( 3/4 )
a^2 = (find the answer lmao)

[icy oriole]

okay im clueless at this point

[cobalt basalt]

wat da hec

[icy oriole]

i slept like 3-4 hours for past 2 weeks

[cobalt basalt]

get some sleep

[icy oriole]

wghat have i done XDDDDDDDDD

[buoyant crystal]

Hold on im gonna try and get tthe notation

[tall moon]

$$a^2=\frac{a^2}{4}+144$$
$$\frac{3a^2}{4}=144$$
$$a^2=(\frac{4}{3})144$$

[icy oriole]

erm

[potent lotus]

Skill_Issue

[icy oriole]

gotcha

why 3a and not 2a tho

[potent lotus]

faiyrose

[icy oriole]

im

clueless

im fucking stupid ok idk whats happening XDDDD

literally

oh wait

no

idk

[potent lotus]

faiyrose

[icy oriole]

yes

[potent lotus]

faiyrose

[buoyant crystal]

honestly just do step by step

[icy oriole]

divide by itself?????????

yeah fuck it i guess

annihilation

https://tenor.com/view/natalie-portman-firing-annihilation-annihilation-gifs-gun-fire-gif-9866986

[potent lotus]

faiyrose

[icy oriole]

where the fuck does 3 come from then

what

okay no

thank you, i think im done with that for today

[buoyant crystal]

$$a^2=\frac{a^2}{4}+144$$
$$a^2-\frac{a^2}{4}=\frac{a^2}{4}-\frac{a^2}{4}+144$$
$$\frac{4a^2}{4}-\frac{a^2}{4}=144$$
$$\frac{3a^2}{4}-=144$$
$$\frac{\frac{3a^2}{4}}{\frac{3}{4}}=144$$

[potent lotus]

bconfirm_ed

[buoyant crystal]

Mb

[icy oriole]

not a single clue

[calm coral]

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

[icy oriole]

1x3

[buoyant crystal]

$$a^2=\frac{a^2}{4}+144$$
$$a^2-\frac{a^2}{4}=\frac{a^2}{4}-\frac{a^2}{4}+144$$
$$\frac{4a^2}{4}-\frac{a^2}{4}=144$$
$$\frac{3a^2}{4}=144$$
$$\frac{\frac{3a^2}{4}}{\frac{3}{4}}=\frac{144}{\frac{3}{4}}$$

[potent lotus]

faiyrose

bconfirm_ed

[buoyant crystal]

^

[icy oriole]

oh mk

that makes sense

1/1/3

wha

3/1/1*

3

.

[potent lotus]

faiyrose

faiyrose

faiyrose

faiyrose

[icy oriole]

lcm's an abreviation?

abbreviation

oh fucking hell

1

????????????????????????????????????????????????????????????????????????????????????????????????????

a?

4x1 1x1

1

wha

wdftm no

oh

4

[potent lotus]

faiyrose

faiyrose

faiyrose

[icy oriole]

OOOOOOOOH

[potent lotus]

faiyrose

[icy oriole]

hold on

ill trty

nah

144x4 = 576

?

[potent lotus]

faiyrose

[icy oriole]

i will end up will some bulshit with infinite numbers after .

wait

192

[potent lotus]

faiyrose

[icy oriole]

yes

im doing it rn

3/2

no?

no

bnvm

3/8

[potent lotus]

faiyrose

[icy oriole]

no

wait its backwards

192:8 is 24, 24:8 is 3

8 doubles

[potent lotus]

faiyrose

[icy oriole]

so it stays

alright i got it

thank you

now i gotta get the f out of my place and run to school that is 17 miles away

/close

!close

.close

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[delicate spire]

when given a general quadratic curve and a point in which a tangent of the curve will pass through, how do you find the equation for the tangent?

[marsh valley]

Well you first need the slope of the tangent line, which will probably require calculus. Is that something you're familiar with?

[delicate spire]

i wouldnt have that question if i didnt know a derivative

actually i know now

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[marsh valley]

Alright alright, just making sure. Some people can be learning things in a different way and there are ways to go about this geometrically.

[delicate spire]

one of the coords are given, one is (x,f(x))

you've got the slope

silly me

[marsh valley]

Exactly :p

[calm coral]

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[pallid halo]

any thoughts so far?

[idle fractal]

No sir

Given answer is C

Kenal means number of elements in a set

[marsh valley]

No. The kernel is the space of all vectors that get mapped to (0,0,0) by F.

[idle fractal]

(0,0,0)=(0,0)?

So it is C

[marsh valley]

Yeah meant that

[idle fractal]

Tq sir

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[copper notch]

is real power of unitary matrix unitary?

[calm coral]

@copper notch Has your question been resolved?

[copper notch]

<@&286206848099549185>

[teal drift]

$I^n = I$ , yeah

[potent lotus]

Alberto Z.

[copper notch]

alrihgt

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[blazing coyote]

so so start let's define the function

$f(0)=f(0+p) \forall p \in R$

[potent lotus]

ƒ(Why am. I here)=lin-alg

[blazing coyote]

where f(x)=0

basicallly the zero element

let f(x) and g(x) be periodic funnctions

then f(x)+g(x) is perodic too

as is $\lambda f(x)$

[potent lotus]

ƒ(Why am. I here)=lin-alg

[glass heart]

why

[glass heart]

why

[blazing coyote]

$\lambda$ just affects the amplitude of the function

[potent lotus]

ƒ(Why am. I here)=lin-alg

[glass heart]

too imprecise

[blazing coyote]

how else do I express it?

[glass heart]

by using the definition of what periodic means

[blazing coyote]

$f(x)=f(x+p)$

[potent lotus]

ƒ(Why am. I here)=lin-alg

[blazing coyote]

where $p$ is the period

[potent lotus]

ƒ(Why am. I here)=lin-alg

[blazing coyote]

then we have

$\lambda f(x) =\lambda f(x+p)$

[potent lotus]

ƒ(Why am. I here)=lin-alg

[glass heart]

ok

better would be (lambda f)(x)=lambda f(x)=lambda f(x+p)=(lambda f)(x+p)

but anyway

[blazing coyote]

now to prove that the sum of two periodic functions is periodic

[glass heart]

now this

[blazing coyote]

hmm

not too sure

ok

so let's consider f(x)+g(x)

hmm

in the simplest case they have the same period

so f(x+p)+g(x+p)= f(x)+g(x)

[blazing coyote]

I'm not sure how to prove the other case

when they have different periods

<@&286206848099549185>

I guess a counter example could be sin(x)+ frac(x)

,w is sin(x) + x-floor(x) periodic

[potent lotus]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by physicsrocks

[marsh agate]

You do want to look for periods of irrational quotient yes

But finding a function for which it's easy to prove it's aperiodic is hard

[blazing coyote]

as this is the OG question, I guess a counter example is enough

[marsh agate]

Yeah but proving smt is a counter example is hard

[blazing coyote]

hmm

so what do I do now?

anyone?

<@&286206848099549185>

.close

[calm coral]

Channel closed

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[warm warren]

Bruh

Have a little patience I’m trying to sort out what you’ve done

[blazing coyote]

.reopen

oops

[calm coral]

✅

[blazing coyote]

sorry

[warm warren]

Hmm, suppose f has a period p₁ and g has a period p₂, and for now let p₁ and p₂ be ℤ⁺

What can you say about the periodicity of f + g?

[blazing coyote]

it doesn't necessarily have to be periodic

example :- Sin(x)+frac(x)

[warm warren]

Uh

You just ignored the last part

[blazing coyote]

ah

[warm warren]

For now let p₁ and p₂ be ℤ⁺

[blazing coyote]

the period will be the LCM of them

[warm warren]

Prove it

[warm warren]

Not necessarily the lcm, any multiple of the lcm works as well

[blazing coyote]

not sure of how I'd do that

I mean

ok

makes sense

[warm warren]

Use the definitions of periodicity you’re given

[blazing coyote]

so $f(x+mp_1)+g(x+np_2)$

[potent lotus]

ƒ(Why am. I here)=lin-alg

[blazing coyote]

where $mp_1=np_2$

[potent lotus]

ƒ(Why am. I here)=lin-alg

[warm warren]

Yeah that’s the idea but write it better

Show that mp_1 is in fact a valid periodicity for f

And similarly for g

From what is given

[blazing coyote]

what

okay

[blazing coyote]

wait, why wouldn't it be?

[warm warren]

It is, show it

[warm warren]

You have a habit of just writing expressions, they mean nothing in this context

[blazing coyote]

$f(x+mp)=f(x) \forall m \in Z$ by definition

[potent lotus]

ƒ(Why am. I here)=lin-alg

[blazing coyote]

I'm new to proofs, sorry

[warm warren]

I want closure of my operation this means I want to show it’s equal to something

[warm warren]

Yeah it’s fine I’m just saying it because you do it and it might not be helping you organise your ideas for the proof

The whole shtick of a proof is chaining implication signs

To prove A => B you go well A => C => D => … => B

Aha, then A must imply B

[blazing coyote]

$f(x+mp_1) \implies f(x)=f(x+mp_1)$

[potent lotus]

ƒ(Why am. I here)=lin-alg

[blazing coyote]

I don't know what else to write

[warm warren]

See no the left side is just a function

It should be $f(x) = f(x + p_1) \implies f(x) = f(x + mp_1) \forall m \in \mathbb Z^+$

[potent lotus]

Frosst

[blazing coyote]

oh

okay

[warm warren]

Can you prove this?

[blazing coyote]

isn't this true by definition

[warm warren]

No

[blazing coyote]

I mean

ok

$f(x)=f(x+p_1) \implies f(x+p_1)=f(x+2p_2).......$ this is repeated m times

[potent lotus]

ƒ(Why am. I here)=lin-alg

[warm warren]

Yeah

[warm warren]

Formally I’d use induction to represent this idea

The idea is correct though

Now, same for g

Write it out

[warm warren]

Like this

[blazing coyote]

got it

now what?

[warm warren]

Now show f+g is periodic

[blazing coyote]

the same logic follows

but this conjecture is wrong

[warm warren]

It is true in this case

[blazing coyote]

yeah, I know

[warm warren]

You need to actually do this part

[blazing coyote]

okie

[warm warren]

Because it uses a thing that doesn’t extend to non integer p’s

And that’s how we’ll end up showing it’s not true

[blazing coyote]

$f(x+p_1)+g(x+p_2) = f(x + 2p_1)+g(x+2p_2)= f(x+(p_1)(p_2))+g(x+P_1P_2)$$ repeated $p_1 \cross p_2 time$

[warm warren]

What

Left side is an expression

Right side is an expression

Expressions don’t imply one another

Statements do

Statements can be either true or false

Well, technically they’re called propositions but that’s logic talk not important

[potent lotus]

ƒ(Why am. I here)=lin-alg
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[warm warren]

We need to start with f(x) + g(x)

[warm warren]

It’s similar to this

[blazing coyote]

to both we add the LCM of their periods

[warm warren]

Who cares about the lcm

As long as it’s a multiple of the lcm

[blazing coyote]

ok , a multiple

[warm warren]

That’s good enough

That’s why we can just go p₁p₂

This is in general not the lcm

But if it’s a multiple of it

And that’s good enough for us

[warm warren]

Because of this

[blazing coyote]

$f(x)+g(x)=f(x+p_1p_2)+g(x+p_1p_2)$

[warm warren]

Aha and then?

No no just p₁p₂ is fine you don’t need the m

[potent lotus]

ƒ(Why am. I here)=lin-alg

[blazing coyote]

if $p_1 $ and $P_2$ aren't integers , we can't use induction like we did earlier

[potent lotus]

ƒ(Why am. I here)=lin-alg

[warm warren]

Now just so the next part is easier, this is \emph{also} $=f(x + mp_1p_2) + g(x + mp_1p_2), \forall m\in\mathbb Z$

[potent lotus]

Frosst

[warm warren]

For this part we don’t need this but we will for the next part

[warm warren]

So we’ve shown this now

We relied on the fact that p₁ and p₂ are integers

Then we used m = p₂ and f(x + p₂p₁) = f(x + mp₁) = f(x)

Now suppose p₁ and p₂ are rationals

Let’s say $p_1 = q/r$ and $p_2 = s/t$

[potent lotus]

Frosst

[warm warren]

Can we still find a “common” periodicity for f and g?

[warm warren]

(Now we need this)

[blazing coyote]

no, we can't

[warm warren]

Yes we can

Try do the same steps

[warm warren]

We end up needing this to show it is possible to find a common periodicity

[blazing coyote]

$f(x+mp_1P_2)=f(x)$ again

[potent lotus]

ƒ(Why am. I here)=lin-alg

[warm warren]

Woah

How do you know that’s true

You need to show me this f(x + mp₁p₂) is of the form f(x + kp₁) where k is an integer

Before you can claim it is equal to f(x)

[blazing coyote]

let m be chosen such that it's a multiple of t

[warm warren]

How do you know such an m exists

[blazing coyote]

r is an integer here

so as must m be

[warm warren]

Remember, m here needs to be an integer, we only know this statement to hold for integer m’s

[warm warren]

So which integer do I pick

[blazing coyote]

t

[warm warren]

Uh, f has periodicity p₁

Oh

Yes

That’s right

If m = t, then mp₂p₁ = sp₁

And s is an integer

Aha then f(x + mp₂p₁) = f(x + sp₁) = f(x)

[blazing coyote]

yup

[warm warren]

(You should be the one writing this out lol)

Now do the same for g

[blazing coyote]

g(x+mp_1p_2)=g(x+qp_1)=g(x)

[warm warren]

Woah

But g has a periodicity of p₂

[blazing coyote]

I meant $p_2$

[potent lotus]

ƒ(Why am. I here)=lin-alg

[warm warren]

Who knows if qp₁ is a multiple of p₂

[warm warren]

Write this step out clearly

What are you picking for m

[blazing coyote]

m=r

thanks for spending so much time with me on this one problem

[warm warren]

Np it’s a interesting one

Ok so we’ve run into a slight problem now

[warm warren]

Over here you didn’t quite finish the proof

The last step here would’ve been $=(f+g)(x + mp_1p_2)$

[potent lotus]

Frosst

[warm warren]

Hence f + g has a periodicity of mp₁p₂

In full:

$(f+g)(x) = f(x)+g(x)=f(x+p_1p_2)+g(x+p_1p_2) = (f+g)(x + p_1p_2)$

[potent lotus]

Frosst

[warm warren]

This proves that f + g as a function in periodic

[blazing coyote]

I see, thanks

[warm warren]

Now look at what you’ve got

$(f+g)(x) = f(x)+g(x)=f(x+tp_1p_2)+g(x+rp_1p_2) = …?$

[potent lotus]

Frosst

[warm warren]

Do you see a problem

[blazing coyote]

yes

[warm warren]

What’s the problem

[blazing coyote]

t and r have to be integers

[warm warren]

That is indeed true

They were the denominators of rational numbers

That’s no problem it’s true by assumption

[blazing coyote]

I see, thanks

[warm warren]

What is done on the last step

[blazing coyote]

you have re-written it as a composition

[warm warren]

I can only do that because the arguments are the same

By definition of adding functions

[warm warren]

Now what’s the issue here

[blazing coyote]

the arguments are different

[warm warren]

Can we fix that

(This is a yes for rationals and a no for irrationals, this is the entire crux of the proof)

[blazing coyote]

yeah, for rationals

[warm warren]

How

How can I choose the m’s differently

[blazing coyote]

by multiplying by a common factor

[warm warren]

Which one

[blazing coyote]

$p_1p_2$?

[potent lotus]

ƒ(Why am. I here)=lin-alg

[warm warren]

Nah

[blazing coyote]

then what?

[warm warren]

Look, we picked m = t for f, and m = r for g

[blazing coyote]

yes

[warm warren]

We need to pick the same thing

[blazing coyote]

oh

tr

[warm warren]

Yep!!

That’s it

[warm warren]

Fix this

And prove it’s periodic

[blazing coyote]

hmm

not sure

sorry, kind of down today, not able to think much

[warm warren]

$(f+g)(x) = f(x) + g(x) = f(x + rsp_1) + g(x + tqp_2)$ since $r, t, s, q$ are integers, they multiply to still be integers

[blazing coyote]

yup

[potent lotus]

Frosst

[warm warren]

$= f(x + rt\frac{s}{t}p_1) + g(x + rt\frac{q}{t}p_2 = f(x + rtp_2p_1) + g(x + rtp_1p_2) = …?$

[potent lotus]

Frosst

[blazing coyote]

can I do this later

[warm warren]

You should be able to finish it at this point

[blazing coyote]

not really feeling it right now

ok

I'll try

[warm warren]

All g

[blazing coyote]

$(f+g)(x+rtp_2p_1)$

[potent lotus]

ƒ(Why am. I here)=lin-alg

[warm warren]

And this hinges on being able to find an integer m (in this case tr) such that mp₁ and mp₂ were integers

Think about why this wouldn’t be possible if p₁ and p₂ were allowed to be irrational now

[blazing coyote]

because tr wouldn't necessarily be an integer then

[warm warren]

Well p₁ and p₂ wouldn’t be expressable as q/r and s/t anyway

[blazing coyote]

yea

[warm warren]

The point is that if p₁ is irrational, then mp₁ is irrational for all integer m ≠ 0

[blazing coyote]

yeah

makes sense

[warm warren]

So this m doesn’t exist

[blazing coyote]

sorry for being so dense today

[warm warren]

So there exist no such periodicity for f + g

There’s a bunch of nuance in this proof

It sorta gets easier since once you’ve seen this methodology it gets used elsewhere as well

You just go oh this is the same method as that other proof I did earlier, I don’t have to go through it

[blazing coyote]

got it, thanks!

can I close this now?

[warm warren]

Sure

[blazing coyote]

thanks so much !

.close

[calm coral]

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[onyx quiver]

what do i do?

[onyx quiver]

why would i do cross product between the

[remote mural]

i think its because cross product of any two vectors is always perpendicular to both the vectors

and its asking for a equation that is perpendicular to that plane

im not sure if that's correct tho

i can be wrong

[remote mural]

that would give you the directional ratio of the normal to the plane that you want to find

[onyx quiver]

ok ty

.close

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[hidden plume]

I genuinely don’t know where to start working on these problems

[glad rain]

differentiate and put the value

dy/dt = dy/dx * dx/dt

that's what you need

[strange lichen]

^ I’m Stephen and I endorse this message

[hidden plume]

so would it just be dy/dx = (dy/dx) * 4

[glad rain]

read what you wrote again (left side)

[hidden plume]

dy/dt = (dy/dx) * 4

[glad rain]

yeah

[hidden plume]

I’m still confused

Wait would I put the x = -1 in for the dx

[glad rain]

dy/dx is function in x

you put x there

[hidden plume]

Im sorry Im just completely lost right now

[glad rain]

find dy/dx first and write what u found

[hidden plume]

so wouldn't I just divide by 4 since it is multiplied by the dy/dx and then on the left side it would be dy/dt/4 ?

Im sorry I'm just completely lost right now

[glad rain]

read question what is being asked

and what you can find

[hidden plume]

I am more confused

[glad rain]

you have to find dy/dt

for question

but you can find dy/dx only because y is a function of x

find dy/dx, multiply it by 4 to make it dy/dt

and put values of x as per the question

you can read it if you need more help

or ping helpers

[hidden plume]

I read it and somehow got this, am I even on the right track now?

I subbed in -1 for x and got 28, tried it and it is incorrect

[glad rain]

wait let me check

what you wrote there?

why in the world did you write that?

where is dy/dx and why you placed y instead of dy/dx in your last equation

[hidden plume]

That is kinda how the book showed how to work it

I might just need a break cause I am at the point where I am confusing myself way too much

[glad rain]

dy/dx = 8x
dy/dt = 8x * 4 = 32x

now solve it

[hidden plume]

ah ok, so dy/dt when x=-1 is -32 ; when x= 0 it is 0 ; when x=-1 it is -32

[glad rain]

yeah

[hidden plume]

I kinda get it now. just a little bit

[calm coral]

@hidden plume Has your question been resolved?

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[kindred garden]

i dont understand why the x² inside the square root turns into a -x

[swift laurel]

,, \sqrt{x^2} = \abs{x}

[potent lotus]

cloud

[kindred garden]

indeed

does it have something to do with the negative infinity limit?

[swift laurel]

and we have that [ \abs{x} = \begin{cases} x & x \ge 0 \ -x & x < 0 \end{cases} ] and for the limit going to $-\infty$ we care about very large negative values of $x$

[potent lotus]

cloud

[kindred garden]

okay i understand that

but why not the x upstairs too?

[swift laurel]

x is just x, which happens to be negative. we had to add a negative sign on the denominator due to the absolute value

[kindred garden]

so you just leave it be

but with the sqrt you would normally have to show both -x and x right, as you say?

but because of the minus infinity we already get rid of the +x

[swift laurel]

if we didn't know which sign x should be we would have to leave it as |x|, but since we do know x is negative (due to the limit), we can say |x| = -x

[kindred garden]

okay

thank you

very much

.close

[calm coral]

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[native dragon]

how should i graph
|x| + |y| = 1
|x+y| = 1

[gray temple]

Consider the different possibilities for the sign of the expressions within the absolute value

[native dragon]

right right

but how do i consider the second one

like am i supposed to do it like

-(x+y) when x+y < 0

?

[tidal grotto]

well you have |x+y|=1

[tidal grotto]

so yes

[native dragon]

wait up lemme check once

[tidal grotto]

well

think about it more like this

|-a|=|a|=1

you don't have to deal with inequalities here

@native dragon

[native dragon]

how is

|-a| coming

[tidal grotto]

|-2|=|2|=2

i'm just using it as an example

a in this case is x+y

[native dragon]

oohh

so like i just substitute some value

[tidal grotto]

so if x+y=1 that would work

since |x+y|=|1|

but what if x+y=-1

[native dragon]

wait i got it

i think

i just put like

x+y as t

and then |t| = 1

and from there

im getting

x + y = 1
and x + y = -1

[tidal grotto]

yes

[native dragon]

so 2 lines

[tidal grotto]

mhm

[native dragon]

right

thank you

this works right

.close

[calm coral]

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[dusk osprey]

claim

[dusk osprey]

i dont understand how to solve for s… but i have a general understanding for x

[pearl sail]

wouldnt you just add up all the angles inside the trangle and set it = to 180 and solve from there

nvm im slow

[dusk osprey]

😭

[calm coral]

@dusk osprey Has your question been resolved?

[waxen pebble]

Do you still need a hint?

|| if so, set up a system of equations with two equations using s and x: one for the triangle angles adding up to 180 and one for the straight line spanning 180 degrees ||

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[remote mural]

I tried taking the first and second derivatives of

this function, and wasn't able to get ther ight answer

even though I thought I did everything right

the function was $\frac{x^2+4}{x^2-4}$

[potent lotus]

Remlis

[edgy raven]

for simple divisons like this I would recommend to use the product rule for less writing and less errors:
$(x^2+4)\cdot(x^2-4)^{-1}$

[potent lotus]

Jigglyproff

[remote mural]

oh that's convinenent

but what mistakes did I make though

double checked my work and everything

[edgy raven]

how did you get this

you found $(2x)\frac{(x^2-4)-(x^2+4)}{(x^2-4)}$ but how did you shorten to this

[potent lotus]

Jigglyproff

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[remote mural]

even with this im having trouble lol

[remote mural]

pretty bad bec im supposed to write an exam on this soon

$(x^2+4)\cdot(x^2-4)^{-1}$

[potent lotus]

Remlis

[remote mural]

$y'=(2x)(x^2-4)^{-1}-(x^2+4)(x^2-4)^{-2}(2x)$

[potent lotus]

Remlis

[remote mural]

$y'=\frac{2x}{x^2-4}-\frac{2x^3+8x}{(x^2-4)^2}$

[potent lotus]

Remlis

[remote mural]

$y'=\frac{4x^2-2x^3-8x}{(x^2-4)^2}$

[potent lotus]

Remlis

[remote mural]

$y'=\frac{(x-1)^2}{(x^2-4)^2}$

[potent lotus]

Remlis

[remote mural]

and obviously this isn't the answer

<@&286206848099549185>

[calm coral]

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