help-10

[formal heron]

damn

[static marten]

yeah, if you times something by 1, it doesn't change

[formal heron]

wdym

[kindred valley]

4 x 1 is still 4

idk if thats what you meant

[formal heron]

we got 1/x+1 * by x+1

so we just multiply the upper part

right?

[kindred valley]

yes

[static marten]

yeah

[kindred valley]

you now multiplied the whole equation by x+1

[formal heron]

i mean it'd be a bit weird to do 1/3 * 2 = 2/6..

[kindred valley]

yeah

so now you have (x+1)^2 + 1 >= 2(x+1) can someone help texit this

[formal heron]

$(x+1)^2+\frac1{x+1}(x+1)\geq2(x+1)$

we had this

[warm shale]

General.Admiral MWstudios

[formal heron]

and now uhm

[kindred valley]

yes

change the fraction to 1

[formal heron]

$(x+1)^2+1\geq2(x+1)$

[warm shale]

General.Admiral MWstudios

[kindred valley]

yeah

now if you dont want to do the tedious expansion

turn (x+1) into y

you choose

[formal heron]

wdym

i will do the turning since it'll look cooler on paper

[kindred valley]

if you dont want to expand a binomial

oh ok then

[formal heron]

so.. y^2 + 1 >_ 2*y ?

[kindred valley]

yeah

bring the 2y over

[formal heron]

$y^2 + 1\geq2y)$

[warm shale]

General.Admiral MWstudios

[formal heron]

$y^2 + 1 - 2y\geq0$

nvm

NVM

[kindred valley]

$y^2 - 2y + 1\geq$

[warm shale]

ᨁ Dan ᨁ

[kindred valley]

idk how

oops put a 0 after that

[warm shale]

General.Admiral MWstudios

[kindred valley]

now do you know difference and sum of squares

(a + b)^2 stuff

[formal heron]

you mean (a+_b)^2

yeah

[kindred valley]

yes

you can turn the above equation into a square

[formal heron]

so it is (y-1)^2 ?

[kindred valley]

yes

[formal heron]

that was lovely

i got one from my old homework that i still havent been able to do . well since i didnt know nothing butt i'll write it if u got time

[kindred valley]

i do

[formal heron]

nvmm idk how to type that shit

i'll send a ss

[kindred valley]

yeah yea

[formal heron]

it is b

a , b >_ 0

[kindred valley]

as we discussed before, sum of squares

wait which question

[formal heron]

but we dont have a 2 there in the firs one

[kindred valley]

oh yeah

[formal heron]

so we just ()^2 it all?

[kindred valley]

the b) is not part of the equation right

[formal heron]

nono

b is the problem's notation

[kindred valley]

yes

[formal heron]

like the counting for it

[kindred valley]

square it

[formal heron]

so we get a^2 + ab + b2 >_ (3/4)(a+b)^2

[kindred valley]

looks right

expand (a+b)^2

[formal heron]

a^2 + ab + b2 >_ (3/4)(a^2+2ab+b^2)

[kindred valley]

multiply throughout by 4

[formal heron]

damn

[kindred valley]

ill be honest i do not see where it is going but im inituitioning it

[static marten]

it's pretty normal ending

[formal heron]

you could also say that you move the 4 from the fraction to the other side by reversed symbol but it's the same thing i think

[kindred valley]

mhm

[formal heron]

this way it is easier for my square dhead to understand

so

4a^2 + 4ab + 4b2 >_ 3(a^2+2ab+b^2)

[kindred valley]

4a^2 + 4ab + 4b^2 >_ 3a^2 + 6ab + 3b^2

yes

now if i can do some shifting

a^2 + b^2 >_ 2ab

see it?

[formal heron]

yeah

we just took what we could from the sides

but now the 2ab

oh

oh

oh

we got a formula

[kindred valley]

mhm

[formal heron]

a^2+2ab+b^2 >_ 2ab

we take the 2ab

[kindred valley]

wait huh

[formal heron]

what

[kindred valley]

the 2ab did not exist on the left side

[formal heron]

.

wait wtf have i done

[kindred valley]

a^2 + b^2 >_ 2ab

[formal heron]

NVMM

[kindred valley]

shift it to make a^2 - 2ab + b^2 >_ 0

and NOW you have a formula

[formal heron]

wait im blocked

a^2 + b^2 >_ 2ab

[kindred valley]

and also can you translate the question

[formal heron]

what part

[kindred valley]

the instruction

Docā something

[formal heron]

if a,b >_ 0, demonstrate that

daca means if

[kindred valley]

oh ok

a^2 + b^2 >_ 2ab

when you move a term to another sidr it changes signs

[formal heron]

it is the same as (a+b)^2 >_ 2ab , right ?

[kindred valley]

ok lesson time

(a + b)^2 is not a^2 + b^2

and (a - b)^2 is not a^2 - b^2

[formal heron]

oh since u square both of them at the same time , and here a^2 + b^2 you need to do the squaring first , then the addition

[kindred valley]

just move the 2ab to the left

and tell me what you get

[formal heron]

a^2 -2ab+ b^2 >_ 0

[kindred valley]

formula ^

[formal heron]

and i get (a-b)^2 >_ 0

[kindred valley]

yes

[formal heron]

that's greatt

thank you

i got a hold of it i think

[kindred valley]

nice

[formal heron]

i'll do a few more exercises in my note book

but with this help it should be enough since i understood the principles

[kindred valley]

cool

[formal heron]

thank you too @static marten

how do i close the channel..

[kindred valley]

.close

[formal heron]

.close

[obtuse pebble]

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[formal heron]

ty

[obtuse pebble]

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[lucid haven]

Any idea on how to solve this

[lucid haven]

<@&286206848099549185>

[static marten]

$log_2(x-2)=log_4(x^3-80)-log_4x$

this looks incomplete

[lucid haven]

What

C 103

Im sorry

Im missed the x at the end

[leaden lion]

Convert all log in same base

[warm shale]

chlamydia

[lucid haven]

So putting log_2 into log_4

?

[leaden lion]

Yes

Or log base 4 into log base 2

[lucid haven]

Okayyy

And what next

Shoot i can divide

The log maybe

[leaden lion]

Make rhs 0 and then use log properties

[lucid haven]

What is rhs?

Right side?

[leaden lion]

Yes

[lucid haven]

Ok thx

Ill manage

Oh

Can i maybe just cut the logs out on both sides

[obtuse pebble]

@lucid haven Has your question been resolved?

[obtuse pebble]

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[tame moth]

Is anyone can help me with math homework?

[static marten]

what if the answer was no

[tame moth]

😔

[static marten]

go on

[tame moth]

Express the solution set of the given inequality in terms of writing intervals and sketch its graph

[static marten]

!status

[obtuse pebble]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

@tame moth Has your question been resolved?

[tame moth]

1

[wide rose]

on which question are you now?

@tame moth

[tame moth]

All

[wide rose]

for a, make the x subject

[hollow sail]

just draw a graph for all

[wide rose]

lol

not practical

[hollow sail]

U can use the graphing calculator

It's about the same as drawing the number line

[obtuse pebble]

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[storm birch]

Can anyone help me with this?

[rich plume]

use difference of squares formula

[storm birch]

yeah I know but im not sure which one to use or how to use it exactly

[rich plume]

a^2-b^2=(a+b)(a-b)

[storm birch]

sorry I dont get it :( is it like, reversed or something? and what about the - in the middle?

[teal turret]

$a^2 - b^2 = (a+b)(a-b) \newline (x+3)^2 - (x-3)^2$

[warm shale]

Stephen

[teal turret]

Do you see what a and b correspond to

@storm birch

[storm birch]

WAIT WAIT IM ONTO SOMETHING GIVE ME A MOMENT FOR MY BRAIN TO PROCESS

.

OMG.

THANK YOU!!!!

[teal turret]

NO PROBLEM

https://tenor.com/view/pop-smoke-dance-nyc-dance-move-smile-gif-16391422

[storm birch]

<3333 IM NOT STUPID NO MORE

[teal turret]

What did u end up getting

[storm birch]

uhh wait I need to do this give me a moment

wait wait ok so

I can it like

wait

shit

omfg

ITS

(X + 3)(X - 3)

??

[crude coral]

no

[storm birch]

oh

sorry im stup[id

😭

wait so do I like

(x^2 + 2x3 + 3^2) my way into this

[vast willow]

no you are not stupid

[storm birch]

aww youre sweet

[crude coral]

yep thats (x+3)^2

now (x-3)^2

[storm birch]

okay ill do that and get back to you thank you

(x^2 - 2x3 + 3^2)

[vast willow]

nah you are not thats reality

yea exactly

[crude coral]

yeah and now subtract them together

cuz negative sign

[vast willow]

now do the subtraction

[storm birch]

awww omg okay thank you a lot you guys

[vast willow]

ok now let me tell you the way the guys before told you

(if you want)

[storm birch]

(yes please)

[vast willow]

ok sure

now we know that $a^2-b^2=(a+b)(a-b)$ right

[warm shale]

calculus is fun

[storm birch]

yessir!

[vast willow]

notice the similarity between $(x+3)^2-(x-3)^2$ and $a^2-b^2$

[warm shale]

calculus is fun

[storm birch]

YESS

yessir

[vast willow]

what can you notice from seeing both

[storm birch]

that theyre the same formula thingy uhh

sorry english isnt my first language

[vast willow]

exactly

[storm birch]

woohoo!

[vast willow]

me too dw

here a=x+3 and b=x-3

[storm birch]

yes!!

[vast willow]

so $(x+3)^2-(x-3)^2=(x+3+x-3)(x+3-(x-3))=2x(x+3-x+3)=2x*6=12x$

[warm shale]

calculus is fun

[storm birch]

ohh okay give me a moment

[vast willow]

take your time

if anything isnt clear tell me

[storm birch]

I totally understand the process but my mind is a little blown by how the formula is used in this

thank you

[vast willow]

np you will be used to these stuff later on

in which grade are you

8?

[storm birch]

a little embarrssing but no im in 10th now and we just started so were repeating things from 9th 😭

[vast willow]

not embarrassing bro there are no problems

[storm birch]

I go to a greek school if youre wondering, our system isnt great

awwww

thanks bro

[vast willow]

past few years were much worse than before in everything including learning

learning process was weakened alot in my country (lebanon)

i am in grade 12 btw

have a nice day/night

if you dont have anymore questions pls type .close

[storm birch]

okay!! thank you so much

I finally figured it out thanks to you

have a nice day/night too :)

.close

[obtuse pebble]

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[sinful eagle]

In 3D geometrically. Where goes -x or -y, -z,

[sinful eagle]

.close

[obtuse pebble]

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[timid silo]

How to solve and explanation please

[humble wind]

oi bro remove channel already taken

[timid silo]

Oi you little nub you took my spot

[humble wind]

no i didnt

[timid silo]

You did

[humble wind]

these channels arent just yours theyre for everyone

[golden dragon]

yea bro see the pinned messages for the original asker

[timid silo]

Aw man

I need help

[humble wind]

go in another channel??

[jovial moss]

Find dx in terms of du

[humble wind]

e^x dx

[golden dragon]

you can choose an unoccupiee help channel

[jovial moss]

In terms of du

[humble wind]

read how to get help

[jovial moss]

Not dx

[timid silo]

Got it thank you

[humble wind]

and delete all your messages here @timid silo

[golden dragon]

lol you dont have to

[humble wind]

yes he does

[humble wind]

1/(u-1)?

[humble wind]

^

[obtuse pebble]

@humble wind Has your question been resolved?

[obtuse pebble]

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[fluid snow]

Determine the probability of rolling an $11$ in the sum with three dice.

[fluid snow]

Would we need to determine all ways to get a sum of 11 with 3 terms?

Or is there a better way?

[amber axle]

in a way, yes

combinatorics ftw

you have 6 ways to roll the first dice

actually waaait i might be trolling

but you can count out all of the combos

but theres another way

you can find all combos of 3 dice that make 11

then * 3! for the perms of that combo

add

[fluid snow]

Well yes, that's what I originally meant with "determine all ways to get a sum of 11 with 3 terms"

But we'd have to do that by hand, right?

[amber axle]

oh yeah

but as i said

you can make ur life easier by finding all the combos

and then doing the permutations of the combos

[fluid snow]

Is there a way to find the number of combinations without listing them all out?

[amber axle]

tahts a great question

let me try something

[obtuse pebble]

@fluid snow Has your question been resolved?

[fluid snow]

Ah, imagine the 11 as 11 successive ones

Now, we need to place two plus signs somewhere

e.g. 11111+111+111

That'd be 5 + 3 + 3

So this is just 10C2 (since there are 10 in between spaces)

[amber axle]

ah interesting

stars and bars

didnt think of doing that

[fluid snow]

So is the probability $\frac{{10 \choose 2}}{6^3}$?

45/216

[amber axle]

thats... too many

my counting tells me 27 combos

but im ngl that does not fit nicely with any combinatorics i can try

https://math.stackexchange.com/questions/2396015/sum-of-three-dice-is-eleven

this is an interesting MSE question

with some complicated answers

unfortunately i am not strong enough in the combinatorics force to remember how to do this, <@&286206848099549185> ?

[obtuse pebble]

@fluid snow Has your question been resolved?

[fluid snow]

.close

[obtuse pebble]

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[tall skiff]

The distance from the top of a mountain is 5000 meters, how many staircases would you need if they are each 2 meters long and increase by 1 meter each staircase.

[wide rose]

what is the formula for the height for n stairs?

[lunar jolt]

It's like the some of x+1 each time from 2 to the value where x+1 reaches 5000

I forgot the formula I'll look it up

[wide rose]

so, is it $n + \frac{n(n+1)}{2}$

[warm shale]

TimK

[wide rose]

cause sum from 1 to n + 1*n

we get an equation $n+\frac{n(n+1)}{2}=5000$

[warm shale]

TimK

[tall skiff]

What if it increases by 3

Would i divide it by 3?

[wide rose]

no

it would be $n+\frac{3n(n+1)}{2}$

[warm shale]

TimK

[wide rose]

the general formula is $\frac{S_n+S_1}{2}*n$

[warm shale]

TimK

[tall skiff]

I dont think this one starts with 2 and increases by 3 with each staircase

[obtuse pebble]

@tall skiff Has your question been resolved?

[obtuse pebble]

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[timid silo]

I have a question, about my math problem, I'm stuck at this problem for over 1 hour

[timid silo]

So, this is the full exercise, ημ means sin and συν means cos, I will translate if you don't understand,

I solved the i) question and the iii) but I can't do the second one, the problem says that the lim of x→0 [f(X)+sin5x]/(x²+2x) is equal to 4

The answer to the first question is lim of x→0 f(X)/X equals to 3

And the last question is equal to 18, but I have no idea how to solve the second one, I already tried something but I am stuck again

( I need to solve it without using L'hopital or something, just using limits )

[shrewd crow]

translate

[timid silo]

Alright, so the

How can I solve this :

Lim f(cos(X))/(x-π/2)
X→π/2

Without l'hopital

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[timid silo]

.reopen

[obtuse pebble]

✅

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[timid silo]

.reopen

[obtuse pebble]

✅

[timid silo]

<@&286206848099549185>

[slow trench]

can you write the limit properly please ? what do you mean by function f ? I can't see how it is defined ?

[little jewel]

put x = π/2 + h

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[thorn pivot]

why without it tho

[obtuse pebble]

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[lime scaffold]

what is a single plane cut?

[thorn pivot]

you cut it with a plane:) "you rip it into 2 pieces"

[lime scaffold]

how is the answer E

does this count as a single plane cut @thorn pivot

[thorn pivot]

your shape being cut by a plane parrallel to the base of it

[lime scaffold]

.close

[obtuse pebble]

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[indigo pendant]

hi i am a bit confused. cna someone help me identity the cofractor and minors and what they represent.
https://gyazo.com/171e0f64c095050502e377fd8874a99d

[obtuse pebble]

@indigo pendant Has your question been resolved?

[obtuse pebble]

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[winter vigil]

I'm in confusion rn

[obtuse pebble]

@winter vigil Has your question been resolved?

[winter vigil]

<@&286206848099549185>

[lunar jolt]

I guess

[winter vigil]

I was debating between C and D so I just wanted to make sure

[lunar jolt]

Not it's D

[lunar jolt]

Are u undergraduate?

Just asking

[winter vigil]

Yep I am

[lunar jolt]

Good keep it going 👍

[winter vigil]

Thx, btw might I get an explanation of why the answer is D exactly pls

[lunar jolt]

Didn't that one make sense?

So the ven diagram showcases that the shaded part is located in between two regions

Outside R but yet inside S ok?

So the outside of R is R' and the it's contained by S so it's D which make more sense. I don't have the correct terminology so the language barrier is a fucking problem

[winter vigil]

No it's okay that's actually a pretty understandable explanation

My dumb self for a sec thought that S' meant that that's the only one being shaded

But thx for clearing it for me great help

[lunar jolt]

I'm here to learn and help others don't mention it

[winter vigil]

Thx, I'll prob be back again for help because I'm stupid

Anyways cyaaa

.close

[obtuse pebble]

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[obtuse pebble]

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[forest sinew]

Show that the infinite solutions $0<x_1<x_2<\dots$ to $x\sin x = 1$ are given by $$x_n = n\pi + (-1)^n z_n$$ where $z_n >0$.

[warm shale]

jan Niku

[forest sinew]

The problem also asserts what seems like correctly that $z_{n+1} < z_n$

[warm shale]

jan Niku

[obtuse pebble]

@forest sinew Has your question been resolved?

[obtuse pebble]

@forest sinew Has your question been resolved?

[warm canopy]

I would think about the intersection of sin(x) and 1/x

@forest sinew

[forest sinew]

how does this show that?

i think its intended we use an expansion

i believe the given solution is reasonable

its more finding zn

[warm canopy]

Well then I don't think you've adequately explained the problem that you want help on

[forest sinew]

i screwed around with taylor at like f(xn+eps) where f(x) = xsinx

[forest sinew]

all i can add is its a baby asymptotics class

[forest sinew]

i believe the intention is to expand xsinx or xn using taylor or some other power expansion and gather terms or so

[violet sentinel]

Have you tried turning it off and on again?

[slate zephyr]

Am I missing something or does this say almost nothing?

xsin(x) = 1 has a lot of solutions by interemdiate value theorem.

They will be around a muiltiple of ~pi assymptotically

x_n is defined as such and invesely this also defines the z_n.

Basically you only need to proof that the z_n would be positve I think?

[forest sinew]

the problem is to arrive at a solution of this form

where zn -> 0 as n -> inf

[slate zephyr]

oh that too? Seems doable though.

[forest sinew]

too?

[slate zephyr]

originally you dind't mention anything about a limit

[forest sinew]

here i meant

show solutions are npi + some alternating qty that vanishes along n

[slate zephyr]

Do you uinderstand what I mean with that the x_n and the z_n are already determined?

[forest sinew]

no

[slate zephyr]

ok

let me try explain again

We are looking at x sin(x)

[forest sinew]

we arent given that xn to work with

if its not clear

[slate zephyr]

,wolf plot x sin(x) from x=0 to x=20

[forest sinew]

we are to arrive at it, not confrm it

[warm shale]

Results provided by WolframAlpha

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Requested by m8732

[slate zephyr]

This (indirectly) defines our zeros.

[forest sinew]

zeros of xsinx-1?

[slate zephyr]

x_1 must be about 1.<something> for example

We don't get a nice formula for them

but they are already locked into place

The z_n are then inversely also set

[forest sinew]

this seems not true

[slate zephyr]

$z_n = (-1)^n (x_n - n \pi)$

[warm shale]

M8732

[slate zephyr]

why not

[forest sinew]

the xn sure

zn rely on a form of the solutions

we havent shown

[slate zephyr]

We can use this, it will work.

[forest sinew]

why

[slate zephyr]

I just rewrote your equation.

I solved for z_n.

[forest sinew]

should it be clear this is true or reasonable

[slate zephyr]

HOWVER we still need to proof that z_n is positive

and decreasing

[forest sinew]

we arent given the eqn to work with

[slate zephyr]

that is the real thing we need to do

[forest sinew]

we need to arrive at it

this is the exercise

[slate zephyr]

I simply define z_n to be this.

[forest sinew]

[slate zephyr]

If we define it like this, you arrive at your equation

[forest sinew]

of course assuming part of the solution makes the solution clear

no offense

[slate zephyr]

I am not assuming anything.

I am defining.

[forest sinew]

why is your defn reasonable

where does it come from

[slate zephyr]

if you proof anything you always need to work towards your goal using what you want to get to in th eend.

it comes from my goal to get z_n such that your equation can be satisfied

[forest sinew]

i dont think this is intended

i think my prof would dismiss this

[slate zephyr]

of course we are not allowed to USE those z_n in the proof but we can certainly work backwqards if we never need to assume the existance of the z_n to beginn with.

[slate zephyr]

qzute honestly then I have no clue what he wants

[forest sinew]

i appreciate your input

[forest sinew]

yea maybe i dont either

[slate zephyr]

I mean you can't really arrive at this

[forest sinew]

why not

weve done problems like this and arrived at similar things

[slate zephyr]

$x_n = n \pi / 2 + (-1)^n z_n$ is also possible.

[warm shale]

M8732

[slate zephyr]

Just your z_n won't be positive here

if we do $x_n = n \pi + \frac{1}{n^2} + (-1)^n z_n$ it would work still.

[warm shale]

M8732

[slate zephyr]

everything still true you want.

z_n will be creasing and converge to 0.

[forest sinew]

zn wouldnt go to zero, would it

[slate zephyr]

yeah it wouldn't here

in my second equation it would

Can you post the original question?

[forest sinew]

just producing a form of the soln ex nihilo rubs me the wrong way

sure

[slate zephyr]

It is actually really common in mathematics.

[slate zephyr]

ah

I still think my solution strategy is the only reasonable way to do this and most likely the intended way.

Maybe you can put it in different words so it bugs you less if you are not used to it.

The crucial thing really is the z_n > 0. Otherwise it is always possible with any function having zeros.

[forest sinew]

i guess im not sure what makes you believe the 'show' refers only to the 'where'

[slate zephyr]

There are no other contraitns for the z_n.

If we are not contrainting z_n as I said

I can also show that there are z_n to express the natural numbers this way.

or really anything (any infinite sequence of real numbers)

[forest sinew]

i guess sure okay

why shouldnt they be

[slate zephyr]

This is because we can DEFINE z_n like this and then verify that they do give your solutions.

[forest sinew]

how do you find zn

[slate zephyr]

^

I define them like this and verify that it works.

[forest sinew]

and how do you show zn -> 0

[slate zephyr]

That is the next step.

First we actually need to see that the z_n are positive.

[forest sinew]

can we pause for 45 minutes

[slate zephyr]

ok

[forest sinew]

sorry im just outside

[obtuse pebble]

@forest sinew Has your question been resolved?

[forest sinew]

okay so lemme see

youre saying that $x_n = n \pi + (-1)^n z_n$ with no restrictions on $z_n$ is just true once we accept that $x\sin x= 1$ has infinitely many solutions

[warm shale]

jan Niku

[forest sinew]

the fact that we can write them this way relies on nothing

the problem is actually asking, show that $z_n >0$ and $z_n \to 0$

[warm shale]

jan Niku

[slate zephyr]

yes

[slate zephyr]

yes

100%

[forest sinew]

okay

so we need to show

$sgn (-1)^n = sgn (x_n - n\pi)$

[warm shale]

jan Niku

[forest sinew]

its not immediately obvious to me how youd do this

[slate zephyr]

yeah

[forest sinew]

maybe you can take cases

lmc

[slate zephyr]

I think it is possible.

I think we need to use intermediate value theorem

f(x)=xsin(x) is continous & monotonically increasing/decreasing in known intervals

f(0)=0

f(pi/2)=pi/2>1

-> first x_n in between

somehow like this but inductively

[forest sinew]

i mean my thought was

my thought was

[slate zephyr]

f(2n pi + pi/2) is always bigger 1

f(2n pi - pi/2) is always less than 1

in between it is monotone

so it will hit 1 once each time

[forest sinew]

im not sure i follow what argument youre making

[slate zephyr]

is this clear?

$f(2n pi + pi/2)=(2n \pi + pi / 2) \underbrace{\sin(2n \pi + \pi / 2)}_{=1} = (2n \pi + \pi / 2) \geq \pi / 2 > 1$

[warm shale]

M8732

[slate zephyr]

I shoulld likely just give you more time to think.

[forest sinew]

i guess i mean what is the argument you are making overall

use ivt?

[slate zephyr]

using this

and the next statements

we can conclude

2n + pi/2 < x_{2n+1} < 2(n+1) pi - pi / 2 gfor the times f assumes 1 on the decreasing ranges.

2n - pi/2 < x_2n < {2n+2} pi + pi / 2 gfor the times f assumes 1 on the increasing ranges.

my indices in the middle may be shifted.

My point is just that we know the function comes by 1 in each interval once

Monotone and continous implies that

[forest sinew]

bleh

[slate zephyr]

Maybe I am being more precise than your course expects

It has a lot of "graphically show".

That would have never gotten accepted where I studied lol.

[forest sinew]

we are so far off of the flavor or style or method or anything weve done in this course

[slate zephyr]

I kinda figured

[forest sinew]

its this stupid prof

[slate zephyr]

unfortunately I do not know what style you preciely are looking for

[forest sinew]

he teaches these "analysis" courses

idk what I could do besides send my notes

but they arent helping me and i was there and paying attention during lectures

I cant imagine they are going to help you

[slate zephyr]

[forest sinew]

asymptotic "analysis" complex "analysis"

[slate zephyr]

I love how you put it in quotation marks.

[forest sinew]

because its not analysis lol

not that I care i couldnt pass an asymptotic analysis are you kidding me

[slate zephyr]

ye, "graphically" anything really isn't.

[forest sinew]

every time i take a proof based class i end up having a mood episode

[slate zephyr]

I am coming at this entirely from the proof side.

Becies proofs I have no clue what to put there lol.

I mean it DOES say "show".

[forest sinew]

yes

[slate zephyr]

maybe you can just say, you saw it graphically in the graph.

[forest sinew]

see this is the problem right

youd think not having to prove anything would make it easier

instead it just makes it completely unclear how you're meant to approach any problem

asking him in office hours hell always say like

oh assert this or that

or clearly

[slate zephyr]

absolutely, this is how i feel here.

[forest sinew]

you can see from the graph

blah blah

[slate zephyr]

[forest sinew]

hes very smart i get the feeling he doesnt take us seriously as students

[slate zephyr]

The thing is that it is subjective what you can see and what not.

[forest sinew]

i mean i can see that clearly there is defined behavior for odd and even n

[slate zephyr]

Sometiems maybe you see things that don'Ät generalize infinitely.

[forest sinew]

that one over and one under estimates

are we intended to assert that we know this from a graph?

that seems like such a sin lol

[slate zephyr]

that is basically what I was saying

using proof language

[forest sinew]

so here

[slate zephyr]

idk I wiol let you judge that.

[forest sinew]

how is this

oh wait

sum and difference

just give me a moment

[slate zephyr]

^

Recall f is our xsin(x)

[forest sinew]

okay fair

and opposite for odd

[forest sinew]

im not sure i track this part

sorry for the ping

you know 2n+p/2 is greater than 1

[slate zephyr]

YOU PINGED ME HOW COULD YOU AAAAAAAAAAAAAARFG

😉

[forest sinew]

ah i didnt mean to i know its annoying

[slate zephyr]

i am joking, all good

[forest sinew]

its a sin in professional communication

which we are

professionals

[slate zephyr]

well there is a sin in your question also.

[forest sinew]

yes but you missed this joke at the first opportunity

so i assumed i was now safe to use the word

[slate zephyr]

🙂

[slate zephyr]

hmm

[forest sinew]

so how do you know its x_n+1

[slate zephyr]

x_0 is before right?

in [0,pi/2]

(due to a similar argument)

x_1 is in [pi/2,pi] because of what I wrote.

it overshoots beofre and undershoots afterwards

[forest sinew]

well at x=2pi+pi/2 youre after x1 and x2

[slate zephyr]

n=0

first

[forest sinew]

oh

[slate zephyr]

then it is pi/2 to pi

[forest sinew]

forgive me too much matlab

[slate zephyr]

determining n+1=1

yay i think my indices aren't shifted

[forest sinew]

and this falls out bc f is cts

[slate zephyr]

yes

and monotone

if it wasn't monotone

it could go there many times

[forest sinew]

im not sure i follow that part

actually i dont follow the original inequality

[slate zephyr]

Look this argument

f(0) = 0

f(21783128pi) = something big > 1

=> it hits 1 one time?

no it doesn't

because it goes back and forth

[forest sinew]

i have just totally lost the skill of asking specific questions

[slate zephyr]

intermeidate value theorem means it hits it at least once

monotone means it doesn't turn around from increasing to decreasing in the middle

if it would oscillate around 1 it would be bad

[forest sinew]

so here

f(2npi + pi/2) > 1

[slate zephyr]

yes

[forest sinew]

f((2n+1)pi+pi/2) < 1

[slate zephyr]

yes

[forest sinew]

you rewrite

[slate zephyr]

s9imilar to this

but negative sins

may god forgive them

[forest sinew]

f(2npi+pi/2) = 2npi+pi/2 for even n

[slate zephyr]

yes

[forest sinew]

and 2(n+1)pi - pi/2 for n odd

we need to actually show its monotone

[slate zephyr]

no wait

[slate zephyr]

if you put 2n inside

then it is for ALL n

[forest sinew]

no

[slate zephyr]

it is only just even

if you forget the 2

[slate zephyr]

yes

[forest sinew]

hmm disagree

[slate zephyr]

why

[forest sinew]

because you pick up extra rotation in the odd case

[slate zephyr]

sin(2npi+pi/2)=1 for all n

[forest sinew]

yes

[slate zephyr]

yes

[forest sinew]

wait no

[slate zephyr]

2pi

[forest sinew]

wait yes

[slate zephyr]

2pi means full rotations

sin(x) = sin(x+2pi)

[forest sinew]

yes of course

[slate zephyr]

feel free to remove the 2

and say even and odd though

[forest sinew]

and f(2n+1)pi+pi/2 = -1

for all n

this is how i got to what i have

[slate zephyr]

wat

[forest sinew]

sorry parens

and f

$\sin \qty( (2n+1) \pi + \sfrac{\pi}{2} ) = -1$

[warm shale]

jan Niku

[slate zephyr]

yes

[forest sinew]

yes

[slate zephyr]

yes

[forest sinew]

then it follows what i said

[slate zephyr]

no because you put 2 and even

[forest sinew]

f(2npi+pi/2) = 2npi+pi/2

[slate zephyr]

yes

[forest sinew]

oh i guess i see what you mean

what i should write is this

this makes more sense right

[slate zephyr]

almost

[forest sinew]

[slate zephyr]

you need to drop the 2

on the right

[forest sinew]

[warm shale]

jan Niku

$f(n \pi + \sfrac{\pi}{2}) = \begin{cases} n\pi + \sfrac{\pi}{2} & \text{n even} \\ (n+1)\pi - \sfrac{\pi}{2} & \text{n odd} \end{cases}$

[slate zephyr]

$f(n \pi + \frac{\pi}{2}) = \begin{cases} n\pi + \frac{\pi}{2} & \text{n even} \ -n\pi - \frac{\pi}{2} & \text{n odd} \end{cases}$

[forest sinew]

hmm i dont see why we drop the 2 but ok

ah maybe i do

man i havent done real math in such a long time

[warm shale]

M8732

[forest sinew]

its like

take n=2m

take n=2m+1

then back subbing

im just dense

[slate zephyr]

you got confused with the 2 and even

[slate zephyr]

^

this

just different sign

[forest sinew]

i dont follow your second line

[slate zephyr]

even positive

odd negative

[forest sinew]

wait yes i do

hold on gimme just 45 second

[slate zephyr]

sin(npi + pi/2) = (-1)^n

f(npi + pi/2) = (-1)^n(npi + pi/2)

This is the compressed way ™️

[forest sinew]

gotcha

now i gotta show its monotone

and this is between any two n

[slate zephyr]

I mean this is clear.

[forest sinew]

not for my prof

[slate zephyr]

sin has monotonicity in these intervals

x is strictly isone

[forest sinew]

,w isone

[slate zephyr]

if you product

[warm shale]

Results provided by WolframAlpha

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

Requested by janniku

[forest sinew]

ah i am closing in on your location

[slate zephyr]

*isotone

sorry

aka monotontically increasing

[slate zephyr]

@ Moderators

[forest sinew]

yea i think i get what you mean

it seems reasonable to assert this

okay

[slate zephyr]

you could also derivativate

,wolf derivate x sin(x)

[warm shale]

Results provided by WolframAlpha

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

Requested by m8732

[slate zephyr]

waait

[forest sinew]

?

[slate zephyr]

I said nonsense.

Technically I didn't 🙂

Here is the issue though

the intervals inteded to decrease

x increases

sin(x) decreases

who wins? we don't know

this means in fact the points we looked at

are not hte minima and maxima

[forest sinew]

you know its funny

[slate zephyr]

,wolf extrema sin(x)x

[warm shale]

Results provided by WolframAlpha

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

Requested by m8732

[forest sinew]

-tanx = x is the problem we solved in class that was meant i think to be similar to this problem

like the example we were supposed to follow

[slate zephyr]

hm

I think we need to augment our argument technically

[slate zephyr]

oooh

[slate zephyr]

that is equivalent to setting this 0 🙂

we could try determine the actual maxima. I think there won't be a nice value.

I am more tempted to find more ineuqalities

[forest sinew]

im going to try some expansion stuff

[slate zephyr]

if x is big this is solely determined by x cos(x)

this is how I would argue.

it decreases becazse cos(x) < 0

[forest sinew]

man my prof made so many sign errors in this example

[slate zephyr]

and x is big enough for it not to matter. It is deifnitely a bit mire annoying but it can work.

[slate zephyr]

For some reason your professor does not sound like the most competent instructor forom what you have been saying

(even if sign errors happen to everyone)

[forest sinew]

man i hate this problem

[forest sinew]

hmm im gonna just email the prof im tired of guessing

thank you for all your time @slate zephyr

.close

[obtuse pebble]

Channel closed

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Use .reopen if this was a mistake.

[forest sinew]

if youre curious ill lyk the solution

[obtuse pebble]

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[proven olive]

hello I just closed my last chanel

[proven olive]

can I get some help

[violet sentinel]

Ok I might be completely wrong here as I haven't done these in a min, but let's consider what we have in our arsenal so far.

$|x - 4| < \delta = -\delta < x - 4 < \delta = 4 - \delta < x < 4 + \delta$.

$\medskip$

Then we have $|\sqrt{x} - 2| < 0.4$. Thus: $ -0.4 < \sqrt{x} - 2 < 0.4 = 1.6 < \sqrt{x} < 2.4 = (1.6)^2 < x < (2.4)^2 = 2.56 < x < 5.76$.

$\medskip$

Now, we know that $4 - \delta < x < 4 + \delta$ and $2.56 < x < 5.76$. So we can say that $4 - \delta = 2.56$ or $4 + \delta = 5.76$. Respectively, $\delta = 1.44$ or $\delta = 1.76$.

$\medskip$

Since we take the lesser, we can say that $\delta = 1.44$

[warm shale]

MellowDramaLlama

[violet sentinel]

That might be hot garbage and not make any sense at all

so someone correct me if I'm incorrect

but that's how I see this working out

[obtuse pebble]

@proven olive Has your question been resolved?

[proven olive]

my bad

let me read this

i was eating

yes i got it right

for this one im having trouble finding the x's

[violet sentinel]

oh so that was the correct answer? 🙂

[proven olive]

yesss\

[violet sentinel]

hell yeah

[violet sentinel]

same logic with this problem

[proven olive]

the way im doing this one is

square root

square root of 1.5

and square root of 0.5

is this logic correct?

since whatever the square root gives me if i put it to the ^2 itll give me the y output

[violet sentinel]

yep, you're on the right track. This is what you said, I'm assuming

[proven olive]

I got .2248 for this one is that correct?

[obtuse pebble]

Channel closed

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[proven olive]

.reopen

[obtuse pebble]

✅

[violet sentinel]

yeah that's the approximation, but I would put the answer down as an exact answer (in this case: sqrt(3/2) - 1)

[proven olive]

where did you get the 3 from?

[violet sentinel]

3/2 = 1.5

in your case, I'd keep it as sqrt(1.5) - 1 then

[proven olive]

what the hell wait sorry I must have got tunnel vision

where did you get 1.5?

[violet sentinel]

from you

which is correct

[proven olive]

ohhhh righttttt

fuck my bad

[violet sentinel]

np lol

[proven olive]

and the -1 comes from the y axis right

[violet sentinel]

same logic as last time. One moment

$|x - 1| < \delta \implies -\delta < x - 1 < \delta \implies 1 - \delta < x < 1 + \delta$.
$\medskip$
$|x^2 - 1| < 0.5 \implies -0.5 < x^2 - 1 < 0.5 \implies 0.5 < x^2 < 1.5 \implies \sqrt{0.5} < x < \sqrt{1.5}$

$\medskip$

Since $1 - \delta < x < 1 + \delta$ and $\sqrt{0.5} < x < \sqrt{1.5}$, then $1 - \delta = \sqrt{0.5}$ or $1 + \delta = \sqrt{1.5}$. So then we see that : $\\ \delta = 1 - \sqrt{0.5}$ or $\delta = \sqrt{1.5} - 1$. Since $\sqrt{1.5} - 1$ is smaller, we can say that $\delta = \sqrt{1.5} - 1$

[warm shale]

MellowDramaLlama

[proven olive]

ahh that makes more sense I got the overall idea

thank you top g

.close

[obtuse pebble]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[tidal ravine]

I believe the answer is yes but I do not know if I got the answer by mistake or if im even doing it right

[timid silo]

so the questions basically asking u to evalute if the lim at x=0 exists

[royal basin]

show your work for working out the limit

[tidal ravine]

is this what I was suppose to do

[royal basin]

handwriting could be better

but it looks like you got it down to h sin(1/h) ?

which is correct

[brave bramble]

f(x0) = 0, not 0^2sin(1/0)

[royal basin]

yes

[brave bramble]

But you correct that with another error haha

[royal basin]

f(0) is explicitly given as 0 and the formula x^2 sin(1/x) is inapplicable

[tidal ravine]

ohh i see

so the limit does exist even though it equals to 0

[obtuse pebble]

@tidal ravine Has your question been resolved?

[obtuse pebble]

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[spark iron]

Part b

[obtuse pebble]

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[steady knot]

I have a homework for central difference (numerical analysis)

[steady knot]

I don't know if this is correct

[timid silo]

ur finding the integral of (x-4)(x-3)(x-2)(x+4)?

what is the 2nd line? like the E^-0.5

[obtuse pebble]

@steady knot Has your question been resolved?

[sour flare]

hello

[obtuse pebble]

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[rancid tundra]

hey guys

[obtuse pebble]

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[jaunty sluice]

Is this way of proof correct?

[obtuse pebble]

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[jaunty sluice]

.close

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[hearty island]

hello, I'm trying to understand the formal definition of a limit, i keep going through it but i just dont get it, please bear with me

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[timid silo]

hi

[atomic hornet]

hi.

[timid silo]

if ax^25 + bx ^24 + ... yx + z = 0

and

zx^25 + yx^24 + ... bx + a = 0

what is the value of x

in the question x for alphabet is replaced by omega

so x is only a variable

no coefficient is x

Ωx^23

not xx^23

i literally don't understand what this problem is supposed to make us do

[tender stratus]

is that all you've been given?

[timid silo]

yes

it is not a test question

it is a question i was given by my professor to solve

as challenge

[tender stratus]

hmm

[timid silo]

i know i sound dumb

i am sorry

im not very bright in math

[tender stratus]

these two are reciprocal polynomials, but there's no way to find the roots just from what's given. I think

[high lily]

consider a simpler case with
ax^3 + bx^2 + cx +d= 0
dx^3 + cx^2 + bx + a = 0

[timid silo]

uhh

i cant solve that either sorry

[high lily]

think about any values of x where both
ax^3 + bx^2 + cx +d
and
dx^3 + cx^2 + bx + a
are the same

[tender stratus]

I think I get what you're saying, but aren't we sort of making assumptions about the coefficients if we do that? ||like x = +-1 is a solution if a + b + c + d =0, or a - b + c -d =0. but is it justified?||

[high lily]

no, no assumptions about the coefficients are being made

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

I still don't understand

im sorry

[high lily]

lets make that even simpler,

for what value of x is
ax + b the same as bx + a

(though you can, you shouldn't really need to set up an equation for this)

[timid silo]

1