help-39

[craggy torrent]

i see

oaky

[open rivet]

well this is just because i chose the x that way

you can also choose x=5748, x=837459873495, x=3874598374598734957345

etc

and you will see that it gets closer and closer to 0

[craggy torrent]

yes

[open rivet]

so, $\lim_{x\to\infty}\frac1{x}=0$

[jolly parrot]

Bonk

[craggy torrent]

yes

i can understand that

[open rivet]

okay, now what about x approaching 0?

[arctic walrus]

Hello I need help

[open rivet]

*from the right hand side

[open rivet]

!help

[pearl pond]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[arctic walrus]

!help

[pearl pond]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[open rivet]

so, im asking about $\lim_{x\to 0^+}\frac1{x}=?$

[jolly parrot]

Bonk

[open rivet]

any ideas?

[craggy torrent]

it keeps going up

?

[open rivet]

yes

what does it approach?

[craggy torrent]

a very big number

[open rivet]

what does it get close and closer to

[open rivet]

which is....?

[craggy torrent]

i have no clue

[open rivet]

its infinity

[craggy torrent]

oho ka

okay

so it never approaches a number

[open rivet]

$\lim_{x\to 0^+}\frac1{x}=\infty$

[jolly parrot]

Bonk

[open rivet]

we can get arbitrarily close to any number

[craggy torrent]

but it instead is constanly going infinately

[open rivet]

if you look at the graph, its also quite clear

it gets cut off quite early

[craggy torrent]

yes

[open rivet]

but its clear that it just goes up and up

[craggy torrent]

yes

i see that now

[open rivet]

lets look at something a bit more complicated

consider $f(x)=\frac{x^2+2x-8}{x-2}$

[jolly parrot]

Bonk

[craggy torrent]

i wanna cry

[open rivet]

its okay

❤️

[craggy torrent]

xd

[open rivet]

we can do this

so, lets analyse the limit of this function as it approaches x=2

at first glance, what do you think happens?

[craggy torrent]

its a connected thing

both left and right are connected?

[open rivet]

wdym?

[craggy torrent]

idk

[open rivet]

you mean the left and right limit?

[craggy torrent]

ye

[open rivet]

did you graph it?

[craggy torrent]

no

i just thought because its a number it is connected

[open rivet]

not quite

here, lets fill in x=2

$f(2)=\frac{2^2+2\cdot 2-8}{2-2}=\frac{4+4-8}{0}=\frac00$

[jolly parrot]

Bonk

[open rivet]

seems like we have a problem right?

[craggy torrent]

my face went from :( to :C

[open rivet]

because obviously we cant divide by 0

[craggy torrent]

ye

[open rivet]

so, how can we solve this problem?

and determine what the limit is

any ideas on where to start?

[craggy torrent]

just remove it?

[open rivet]

maybe the top term?

x^2+2x-8

try factoring this

[craggy torrent]

its all 0

[open rivet]

nono, dont take the limit yet

factor out the quadratic

x^2+2x-8 can be written as (x-a)(x-b)

and you need to find a and b

have you had that?

[craggy torrent]

are they the two x's?

[open rivet]

wdym?

[craggy torrent]

idk:(

[open rivet]

okay ill tell you

we can write x^2+2x-8 as (x+4)(x-2)

do you see where im trying to go with this?

$f(x)=\frac{x^2+2x-8}{x-2}=\frac{(x+4)(x-2)}{x-2}$

[jolly parrot]

Bonk

[craggy torrent]

i dont really

[open rivet]

okay thats fine

[open rivet]

do you see what we can simplify this to?

hint: something with the x-2's

[craggy torrent]

are we able to move it all into one equation

[open rivet]

wdym?

[craggy torrent]

since both the bottoms are x - 2

idk

[open rivet]

we can do $\frac{(x+4)(x-2)}{x-2}=(x+4)\frac{x-2}{x-2}=(x+4)\cdot 1=x+4$

[jolly parrot]

Bonk

[open rivet]

do you see why?

[craggy torrent]

kinda

[open rivet]

okay, so lets go back to our limit

$\lim_{x\to 2}\frac{x^2+2x-8}{x-2}=\lim_{x\to 2}\frac{(x+4)(x-2)}{x-2}=\lim{x\to 2}x+4=2+4=6$

[jolly parrot]

Bonk

[open rivet]

still following?

[craggy torrent]

no

[open rivet]

okay, what part speficially do you not follow

[craggy torrent]

are we figuring out the limit?

[open rivet]

yes

we have already determined that if we fill in x=2, we get a 0/0 form

to get rid of that, we factor out the top quadratic

find out that they cancel out from the x-2 in the denominator

then we just left with x+4

and there we can plug in x=2

to get our limit value of 6

@craggy torrent so you kinda know what a limit is?

[craggy torrent]

yea

[open rivet]

i suppose actually using it in problems like this is not super necessary

but i feel like you grasp the overall concept

now, we will look at derivatives

ready?

[craggy torrent]

maybe

:D

[open rivet]

okay lets go

in broad terms, the derivative is the slope at a certain point

for example this

[craggy torrent]

yes this looks more like what im doing

[open rivet]

perfect

now, we place two points on the line

(one sec im finding a photo)

[maiden plover]

HI somone understand integral has parametre??

[open rivet]

i think this one is pretty good

[open rivet]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[craggy torrent]

!help

[pearl pond]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[open rivet]

xd

[open rivet]

so lets look at this

we place one point x with corresponding f(x) on the line

then we define some distance h to the next point

that we put a bit further on

and that point on the line is f(x+h)

still following?

[craggy torrent]

a little yea

[open rivet]

okay

we get the slope by doing rise/run

seen that before?

[craggy torrent]

no

[open rivet]

okay, do you know how to get the slope between two points?

[craggy torrent]

i think so

the gradient?

[open rivet]

like if i give you A(x1,y1) and B(x2,y2) you will be able to determine the line y=ax+b that goes through both points?

[maiden plover]

how can i get my own chat

[maiden plover]

idk ow

[open rivet]

!help

[pearl pond]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[craggy torrent]

read the thing in the cahnnel

[open rivet]

click on #❓how-to-get-help

[open rivet]

@craggy torrent

[craggy torrent]

this

[open rivet]

exactly

[craggy torrent]

yea i know how to do that

[open rivet]

so from here

what are x1, x2, y1, and y2?

[craggy torrent]

f(x), f(x + h)
x, x + h

wait

nbm

thats fine

[open rivet]

no, thats correct

gj

now, whats the slope?

[craggy torrent]

f(x) - f(x + h)

---

x - x + h

it would be that

if it had numbers

[open rivet]

can you write it as a fraction please

not some weird discord formatting

[craggy torrent]

argh

[open rivet]

......

[craggy torrent]

my bad...

[open rivet]

just something like (stuff)/(stuff)

[craggy torrent]

(f(x)) - (f(x + h))/(x) - (x + h)

[open rivet]

or if you wanna be fancy and use $\LaTeX$ you can do $\frac{\text{stuff}}{\text{stuff}}$

[jolly parrot]

Bonk

[open rivet]

and dont forget brackets

[craggy torrent]

like that

or do they need to be more grouped

[open rivet]

thats still not quite correct....

but i understand what you mean

its okay

however, you filled in the variables wrong

its y2-y1

not y1-y2

y1=f(x), y(2)=f(x+h)

[craggy torrent]

ya

i noticed it earleir

thats why i said wait

nvm

[open rivet]

so we get the gradient/slope/whatever: $\frac{f(x+h)-f(x)}{(x+h)-x}$

[jolly parrot]

Bonk

[open rivet]

agree?

[craggy torrent]

yup

i get that

[open rivet]

great

but notice how we can simplify the denominator a bit more

(x+h)-x=?

[craggy torrent]

so f(h)/x

?

or am i going to far

[open rivet]

nonono

you cant simplify the numerator

f(x+h)-f(x)!=f(h)

thats something completely different

only the denominator

[craggy torrent]

okay

(x+h)-x= (x)

?

[open rivet]

no

[craggy torrent]

wait

im silly

opos

(x+h)-x= h

[open rivet]

maybe you need to revise some of your algebra haha

yes, thats correct

[craggy torrent]

thats just me being silly xd

[open rivet]

so we get: $\frac{f(x+h)-f(x)}{h}$

[jolly parrot]

Bonk

[open rivet]

agreed?

[craggy torrent]

my brain is running on low after doing all of this

yes

i get that

[open rivet]

its okay

[craggy torrent]

i understand

[open rivet]

we are almost there

so, we have the slope

in terms of some distance h between the points

now, to get the slope at a single point

we let the h get smaller and smaller

such that the distance between the two points get smaller and smaller

and do you remember how we make one approach another?

[craggy torrent]

is that the let and righ thingy

[open rivet]

kinda

we take the limit

as h approaches 0

$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

[jolly parrot]

Bonk

[open rivet]

this is the definition of the derivative

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

[jolly parrot]

Bonk

[open rivet]

see it now?

[craggy torrent]

kind of

[open rivet]

does it look familiar?

or is this completely new

[craggy torrent]

im new to this

i only knew the basics

[open rivet]

do you have a textbook you learn out of?

[craggy torrent]

just resource my tutor puts online

[open rivet]

or is it all online courses or smth

[craggy torrent]

i struggle with transport so sometimes i stay home

and watchi his online lectures

[open rivet]

i get it

[craggy torrent]

but hed oesnt record white board

[open rivet]

its okay

[craggy torrent]

and hes explainign things on it that i cant see

[open rivet]

so, now we have the definition of the derivative

[craggy torrent]

all i see is his slides

[open rivet]

so, lets use the derivative

now we have the definition

shall we try it on a few functions?

[craggy torrent]

we coudl

[open rivet]

so, lets start simple

f(x)=5

,w graph y=5

[jolly parrot]

Results provided by WolframAlpha

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

Requested by .bonk__

[open rivet]

so, how would we fill that into our definiton?

[craggy torrent]

isnt that just y = 5

[open rivet]

(btw you can also write it down on paper and take a photo if thats easier)

[open rivet]

yes, we start simple

[craggy torrent]

okok

making sure

[open rivet]

so, what are f(x) and f(x+h)?

using f(x)=5

[craggy torrent]

5,0

no

0,5?

[open rivet]

what?

lets go step by step

f(x)=?

[craggy torrent]

is it 0

[open rivet]

the derivative you mean?

[craggy torrent]

i dont know

[open rivet]

wherea re you getting 0 from

you are not telling me what you are doing

it might be right it might be wrong

i cant tell

[craggy torrent]

no im just silly

[open rivet]

because you are just saying numbers without saying where from

so, where do you get the 0 from?

[craggy torrent]

the x

[open rivet]

but x is just x

f(x)=5

and f(x+h)=5

yes?

[craggy torrent]

does the f()

mean along the y

axis

and without along the x

axis

[open rivet]

yes?

[craggy torrent]

okay

ty

that makes so much more sense now

both are 5

i see

[open rivet]

yes

so then what is $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

[jolly parrot]

Bonk

[open rivet]

when we fill in f(x)=5 and f(x+h)=5?

[craggy torrent]

5 - 5 / 0

?

so its 0?

[open rivet]

woah woah woah woah

you cant divide by 0

[craggy torrent]

one moment i need the toilet

[open rivet]

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{5-5}{h}=\lim_{h\to 0}\frac{0}{h}=\lim_{h\to 0} 0=0$

[jolly parrot]

Bonk

[open rivet]

lmk when youre back 😄

[craggy torrent]

i am back

[craggy torrent]

is this an example of what i just did

[open rivet]

thats fast....

[open rivet]

this is finding the derivative of f(x)=5 from the definition

make sense?

or still very vague

[craggy torrent]

uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuhh

its 0 tho

[open rivet]

yes

[open rivet]

do you see what happens here?

[craggy torrent]

yes

i understand

[open rivet]

great

now lets go to a bit more complicated function

lets take f(x)=x^2+5x+7

[craggy torrent]

can you format it

[open rivet]

wdym?

like this:

[craggy torrent]

like with the thingy

[open rivet]

$f(x)=x^2+5x+7$

[jolly parrot]

Bonk

[craggy torrent]

yea

[open rivet]

okay np

now what is $f(x+h)$?

[jolly parrot]

Bonk

[open rivet]

hint: replace the x with (x+h)

[craggy torrent]

i dont understand what to do with the x

s

[craggy torrent]

how am i able to remove xs

[open rivet]

wdym?

[craggy torrent]

do you want me to find a signular number for f(x)

[open rivet]

$f(x)=x^2+5x+7\implies f(x+h)=(x+h)^2+5(x+h)+7$

[jolly parrot]

Bonk

[open rivet]

do you see this?

do you see what happened?

[craggy torrent]

you added

h

[open rivet]

yes

but do you see that i replace all instances of x

with x+h?

[craggy torrent]

yes i see that

[open rivet]

before and after

that formatting 😭

[craggy torrent]

yes

[open rivet]

now, what is f(x+h)-f(x)?

$f(x+h)-f(x)$

[jolly parrot]

Bonk

[craggy torrent]

would it be 0

since theyre both the same?

[open rivet]

how ar ethey the same??

[craggy torrent]

idk im sorry

[open rivet]

i honestly dont understand how you can think theyre the same

sorry

these are not the same

[craggy torrent]

i agree

[open rivet]

okay, so lets find f(x+h)-f(x)

first off, we fill in the equations

$f(x+h)-f(x)=(x+h)^2+5(x+h)+7-(x^2+5x+7)$

[jolly parrot]

Bonk

[open rivet]

yes?

[craggy torrent]

yes

[open rivet]

can you simplify this further?

so that w edont have any more brackets

or do i need to do that aswell

[craggy torrent]

i dont know

[open rivet]

you need to relearn your algebra

tbh

because this should be like second nature

[craggy torrent]

im sorry

can you try giving me an example with numbers

[open rivet]

wdym numbers?

we are in the variable world now

here we generalise

so that we dont need to calculate every point

we get it all in one go

ill show you the proof of the product rule

,tex .diff rules

[jolly parrot]

Bonk

[open rivet]

sorry, power rule

actually, nvm

thats too much to write out

[craggy torrent]

okay are you able to show me the simple version?

of

[open rivet]

of what?

[craggy torrent]

this

[open rivet]

ill try

[craggy torrent]

so i can try understand what youve done

[open rivet]

but its quite a lot of writing

so it might take awhile

[craggy torrent]

nono

i mena

[open rivet]

just fyi

[craggy torrent]

like

with algebra

[open rivet]

but i will do it

yes

but its a lot of writing

gimme like a min or two

\begin{align*}&f(x+h)-f(x)\=&(x+h)^2+5(x+h)+7-(x^2+5x+7)\=&(x+h)^2-x^2+5x+5h-5x+7-7\=&x^2+2xh+h^2-x^2+5h\=&2xh+5h+h^2\=&h(2x+5+h)\end{align*}

[jolly parrot]

Bonk

[open rivet]

now, i will fill it into $f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

[jolly parrot]

Bonk

[open rivet]

$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{h(2x+5+h)}{h}=\lim_{h'to 0}2x+5+h=2x+5+0=2x+5$

[jolly parrot]

Bonk

[open rivet]

Thus $f(x)=x^2+5x+7\implies f'(x)=2x+5$

[jolly parrot]

Bonk

[open rivet]

its a lot of steps

can you follow them all?

any part you are having trouble with?

[craggy torrent]

im cooked

[open rivet]

this is not whats expected from you btw

you are simply expected to be able to use the rules

,tex .diff rules

[craggy torrent]

how has the top changed into the bottom

[jolly parrot]

Bonk

[open rivet]

its rearranging and distributing

[craggy torrent]

i understand the 5x and 5h

[open rivet]

$5(x+h)=5x+5h

[craggy torrent]

but wherd the x2 come from

[open rivet]

from the x^2+5x+7 part

[craggy torrent]

oh okay

how come its unaffected by the 7 before it

[open rivet]

im not sure if this helps 😭

[craggy torrent]

is it cos its not x

rather its -

[open rivet]

wdym?

do you know the distributive property?

$a(b+c)=ab+ac$

[jolly parrot]

Bonk

[craggy torrent]

yes

i know that

[open rivet]

okay, thats whats happening here

but with a=-1

[craggy torrent]

gosh my brain is so fried

[open rivet]

you are in university right?

[craggy torrent]

yes

[open rivet]

1st year?

[craggy torrent]

yes

[open rivet]

what study?

[craggy torrent]

computer science

[open rivet]

and youre only now getting analysis?

[craggy torrent]

i live in the uk

[open rivet]

at least, im assuming northern hemisphere

what course is this for?

analysis right?

or perhaps just calculus 1

[craggy torrent]

its computer science with games design

[open rivet]

yes but whats the course

you are being asked this for

[craggy torrent]

the course is called mathematics for computing

[open rivet]

ah... lol

very on the nose

do you have an example problem?

that perhaps could be on a test

[craggy torrent]

i have revision resource

[open rivet]

or would it perhaps better to call it a night and continue tomorrow?

[craggy torrent]

this is close enough to what we were just doing

[open rivet]

i suppose

but this is much simpler

[craggy torrent]

this is another thing similar

[open rivet]

yeah thats mroe like it

there we need derivatives

here is the power rule: $\dv{}{x}x^n=nx^{n-1}$

[jolly parrot]

Bonk

[open rivet]

do you see how they used that?

[craggy torrent]

i havent looked into it

and im getting really tired

i appreciate it a lot

do i just close the thing myself

[open rivet]

!done

[pearl pond]

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

[open rivet]

you can come back tomorrow if you like

[craggy torrent]

i might

ty a lot

sorry if i was too much

.close

[pearl pond]

Channel closed

Closed by @craggy torrent

Use .reopen if this was a mistake.

[open rivet]

no its alright

happy to help

😄

[craggy torrent]

is it ok if i add u

[open rivet]

yeah idm

but if you have maths questions its better to ask here

since i might not always be there

[craggy torrent]

yes

okay

[open rivet]

and here i can do the

$\LaTeX$

[jolly parrot]

Bonk

[open rivet]

things

and also others might be able to help aswell

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[quartz yoke]

could someone please help me on 18

oh wait give me a second

yeah nvm

.close

[pearl pond]

Channel closed

Closed by @quartz yoke

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[wind moss]

Is my answer correct? Am I cooked?

[wind moss]

Question 4b

[pearl pond]

@wind moss Has your question been resolved?

[pearl pond]

@wind moss Has your question been resolved?

[pearl pond]

@wind moss Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lime river]

So I had attempted the b part like this but now I am stuck how can I find for a given n the first decreasing type seq or last increasing type seq?

[lime river]

I have defined what a decreasing or increasing type seq in my attempt

[lime river]

<@&286206848099549185>

Pls help

[pearl pond]

@lime river Has your question been resolved?

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lime river]

No

[lime river]

.

<@&286206848099549185>

[feral sedge]

My guess is this probably has to do with the zeckendorf decomposition, or you can try rounding n*-1/phi

[lime river]

Well my question still remains unchanged that how would I know for a certain n for what a would it have it's first decreasing seq or last increasing seq
Or how would I know for certain n and a how many terms do they alternate for

[feral sedge]

<@&268886789983436800> I'm told this question is from some summer school application for India and the last person to ask it was banned because external help isn't allowed

[pearl pond]

@lime river Has your question been resolved?

[opal lantern]

yeah uh

please dont

.close

[pearl pond]

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[rapid wagon]

Given subspace $U={p(x) \in V | p(1)=p''(1)=0}$ of vectorspace $V=\mathbb{R}_4[x]$ \
Find a base B for U such that every coefficent of $x^3$ in every polynomial in the basis is 100

[rapid wagon]

What am I doing wrong ?

[jolly parrot]

prograce

[pearl pond]

@rapid wagon Has your question been resolved?

[pearl pond]

@rapid wagon Has your question been resolved?

[rapid wagon]

<@&286206848099549185>

[pearl pond]

@rapid wagon Has your question been resolved?

[agile terrace]

Wanna play roblox with me?

[pearl pond]

Channel closed

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[rapid wagon]

.reopen

[pearl pond]

✅

[pearl pond]

@rapid wagon Has your question been resolved?

[rapid wagon]

<@&286206848099549185>

[maiden plover]

is there somone who understand frensh

[robust kraken]

jtai dit envoie ton exercice

[pearl pond]

@rapid wagon Has your question been resolved?

[weary orbit]

me

[pearl pond]

@rapid wagon Has your question been resolved?

[wheat elbow]

did you forget to put coefficient of x^4?

[rapid wagon]

No it's on purpose this is my answer

But it's wrong

Maybe I should plug in coefficents of x^4 yea

[rapid wagon]

OMG it worked

[rapid wagon]

THANK YOU

THAT WASSP STUPID

.close

[pearl pond]

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[pearl pond]

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[remote willow]

Show that there are no nonzero natural numbers m and n that satisfy the equation √m + √n = √253

[remote willow]

I've reached the relation 2√253m = 253 + m - n

But don't get it how I can actually demonstrate that there are no nonzero natural numbers m and n

[tardy reef]

LHS is irrational, RHS is integers

[tardy reef]

for this

[dapper kraken]

overkill method:
prove that sqrt(n)+sqrt(m)=sqrt(a) has positive integer solutions iff a can be divided by a square thats not 1

[remote willow]

Makes sense, square root of 253 is irrational and because m and n need to be natural numbers, there is ~0% probability that the sum of the square roots of two natural numbers could equal to square root of 253

if they are perfect squares, the condition does not hold

But mathematically, can it be proved?

[tardy reef]

I was talking about this condition
2√253m = 253 + m - n
If such m and n exist, then you would need addition of three integers to be an irrational number.

which is ofc never possible

[pearl pond]

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[vagrant trout]

Why does the absolute value disappear for n² when we compare it to 1/n² but it doesn't disappear for sin2n? Is it because the series starts at n=1 and it's always positive, unlike the periodic sin function?

[hot canyon]

Isn't |x²| = x² for all real "x"? What do you mean by "the absolute value disappear for n² when we compare it to 1/n² but it doesn't disappear for sin2n"

[vagrant trout]

|sin2n/n^2| = |sin2n|/n^2 <= 1/n^2

This part

I was asking if the absolute value only stays on sin2n because the series starts at n = 1 therefore n^2 is always positive, whereas sin(2n) is periodic and can have negative values, so it remains with the absolute value.

[hot canyon]

That is correct, sin 2 is +ve and sin 4 is -ve

[jolly parrot]

The following error occured while calculating:
Error: Unexpected type of argument in function multiplyScalar (expected: number or Complex or BigNumber or bigint or Fraction or Unit or string or boolean, actual: function, index: 0)

[vagrant trout]

Alright, ty

[hot canyon]

,calc sin (4 rad)

[jolly parrot]

Result:

-0.75680249530793

[vagrant trout]

I see

Tyty

.close

[pearl pond]

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[pearl pond]

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[final rampart]

Hey

[final rampart]

an exercise in a book provided us with the graphs of two functions $f(x)$ and $g(x)$ and they also said that $f(x)$ is a 1st degree polynomial while $g(x)$ is a 3rd degree polynomial:

[jolly parrot]

stoicindiehacker369

[final rampart]

This is all the information I have.

[spare lark]

<@&268886789983436800> money for assignement

They are in multiple channel

Also

[final rampart]

$\text{Then, the problem asked to show that}$ $f(x)=\frac{3}{8}x+\frac{3}{4}$ . $\text{This was quite easy, I just found the slope of}$ $f$ ($\frac{0,75-0}{0-(-2)}=\frac{\frac{3}{4}}{2}=\frac{3}{4}\times \frac{1}{2}=\frac{3}{8}$ .
As it is a linear function, so, has the form $y=ax+b$ :
$a=\frac{3}{8}$ and $b$ is equal to the $y$ value in the point of the function that intercepts the $y$ axis, so $b=\frac{3}{4}=0,75$ .
Therefore: $f(x)=\frac{3}{8}x+\frac{3}{4}$ . Ok, sure, linear binomial easy.

[jolly parrot]

stoicindiehacker369

[final rampart]

But then it asked me to find out the algebraic expression for $g(x)$. I actually arrived at the correct result, but I want to know why my thought process worked and in what instances it might not work.

What I did was:

$g$ is a 3rd degree polynomial, therefore, it must have 3 roots. I can see that $-2$ and $2$ are both roots of $g(x)$ , therefore $x+2$ and $x-2$ are linear factors of $g(x)$ . I need to find a 3rd linear factor.

I assume that either $2$ or $-2$ are repeated roots.
So, $g(x)$ is either equal to

1. $a(x+2)(x-2)(x-2)\iff a(x^{3}-2x^{2}-4x+8)$
   or it's equal to
2. $a(x+2)(x+{2})(x-2)\iff a(x^{3}+2x^{2}-4x-8)$

As I know that the $y ;\text{intercept}$ is equal to 3, I just gave a name to each function (one of them had to equal $g(x)$) and I did:

$m(x)=a(x^{3}-2x^{2}-4x+8)\iff m(0)=3\iff m(0)=a(0-0-0+8)=8a=3 \iff a=\frac{3}{8}$

$n(x)=a(x^{3}+2x^{2}-4x-8)\iff n(0)=3\iff n(0)=a(0+0-0- 8)=a\times(-8) =3\iff a=\frac{3}{-8}=-\frac{3}{8}$

The coefficient of the $x^{3}$ term cannot be negative, otherwise the graph of the function would have suffered a reflection by the $x$ axis (both curves would be switched in the graph), so $a$ must equal $\frac{3}{8}$.

So, $g(x)=\frac{3}{8}(x^{3}-2x^{2}-4x+8)\iff g(x)=\frac{3}{8}x^{3}-\frac{3}{4}x^{2}-\frac{3}{2}x+3$

Then I went up to the solutions and it indeed was correct.
The thing is: why is "I assume that either $2$ or $-2$ are repeated roots." a valid assumption?
Can't a 3rd degree polynomial function have imaginary roots? Couldn't $2$ and $-2$ be just two roots of the function and the 3rd root be imaginary?

(I heard somewhere that there's a theorem -- I think by Euler -- that a polynomial of degree $n$ always has $n$ roots but they can either be real or imaginary)

[jolly parrot]

stoicindiehacker369

[final rampart]

Like, was it correct to assume that $g(x)$ can only be equal to $a(x+2)(x-2)(x-2)$ or $a(x+2)(x+2)(x-2)$?

Of course that, after I found out that $\frac{3}{8}x^{3}-\frac{3}{4}x^{2}-\frac{3}{2}x+3$ , it's easy to just execute polynomial division (or synthetic division - which is just an algorithm to make polynomial division quicker) and divide $g(x)$ by $x+2$ and $x-2$

After that it's very easy to discover the multiplicity of each root.

[jolly parrot]

stoicindiehacker369

[final rampart]

Anyway, my question is:
Can I use this thought process again? Isn't it kinda flawed? If so, in what situations might it fail?

And also:
As $f(x)=\frac{3}{8}x+\frac{3}{4}$ , the slope of this linear function is the same as the coefficient of the $x^{3}$ term in $g(x)$ , why? Is there a relation between these two?

[jolly parrot]

stoicindiehacker369

[brazen vector]

you cant just assume that

but instead you can look carefully at the graph

[calm wing]

notice that the graph of g just touches the x axis and doesn't cross it at x=2

[brazen vector]

and notice that indeed there is a double root at x=2

if you want split the curve into 2 parts

[calm wing]

this indicates a root of multiplicity higher than 1, which in your case of a cubic must necessarily mean that it's a root of multiplicity 2

[final rampart]

How do I know that it is a double root?

[brazen vector]

the first part ends when the graph touches the x-axis at the first time

and the second is when it touches it again

[final rampart]

Wait, that's great, but why is it that "if the graph touchesthe x axis at a point the multiplicity of that root is higher than 1"

[calm wing]

the more formal way is to notice that right before x=2, the function is decreasing, and then after it is increasing, meaning f'(x) changed sign at x=2, meaning f'(2) = 0

[final rampart]

I never heard about this^^^haha

[calm wing]

if b is a root of multiplicity n of f(x), then b is a root of multiplicity (n-1) of f'(x)

[final rampart]

Uhh, what exactly is f'(x) related to f(x) in this scenario?

[calm wing]

derivative of f

i don't know if you covered calculus tho

[final rampart]

I never learend derivatives and stuff

I have an idea of what it is

But just read about it for fun

[calm wing]

mm okay

[final rampart]

Idk anything haha

[tropic saddle]

you know that g has roots at 2 and -2, so if you use polynomial division then you have g(x)/((x+2)(x-2)) = ax+b for some real numbers a,b

in particular, ax+b has another real root

[final rampart]

But, from what you said, could a derivative of

(x-a)(x-b)(x-b)

Be something like (x-a)(x-b) ?

[tropic saddle]

which would also be a root of g

but from the picture it then has to be 2 or -2 again

[calm wing]

maybe not (x-a) but it would definitely have a factor of (x-b)

[tropic saddle]

so the problem that the third root could be imaginary cant happen

[final rampart]

I don't really get it

[tropic saddle]

(or more generally, imaginary roots would have to come in pairs which also cant happen here)

[calm wing]

but he doesn't have the equation of g, no?

[final rampart]

I cant't divide g(x) by ((x-2)(x+2)) because I didn't know the expression of g(x) yet

[final rampart]

Yeah exactly.

[tropic saddle]

what I mean is that you know that g is a third degree poly and you know that in principle you could divide it

[final rampart]

I did prove that it was the only way afterwards

After discovering the expression of g

[tropic saddle]

and you know that if you divide out two roots from a third degree poly then you necessarily are left with a linear poly

[final rampart]

Yes

[calm wing]

oh yeah, g has real coefficients so all complex roots must come in pairs

[tropic saddle]

you dont know what that is yet, but you know what form it has to have

[calm wing]

so having two real roots implies having a third real root as well

[final rampart]

Yeah exactly idk what any of that means yet hahah

Are you saying that:

If a 3rd degree polynomial has roots a and b, then, if i can divide it by ((x-b)(x-b)) the quotient must be another linear binomial?

Is that what you're saying?

[tropic saddle]

yes

[final rampart]

Oh

Wait I think I'm getting there

[final rampart]

This^^^^

When a polynomial function touches the x axis instead of crossing it

It means that it has a root with multiplicity greater than 1

Never heard that stated, but let me wrap my head around it

Oh wait

Got it:

If we have, for example, two functions m(x)=ax-b and n(x)=ax-b

If we do $m\times n(x)$

[jolly parrot]

stoicindiehacker369

[final rampart]

The answer would be a quadratic with a root b that has multiplicity of 2

Right?

And the quadratic function would only touch the x axis and not cross it

Since it has only one root

Right @calm wing ?

[tropic saddle]

m*n has a double root at -b/a, yes

and it only touches there and doesnt cross

[final rampart]

Well

[final rampart]

How does this generalize to assuming that when a graph of a polynomial function touches the x axis at x=b , b is a root of that polynomial with a multiplicity greater than one?

[tropic saddle]

well it doesnt give a rigorous proof. or at least none that I can think of

but it gives good intuition

maybe another perspective for some intuition: imagine the graph touches the axis from above but doesnt cross. now move the graph a tiny bit down. now you have two roots. as you move the graph back up again, those roots move together and then "merge" meaning you now have two roots at the same point, aka a double root

(again, not a proper proof)

[final rampart]

Well, yeah, I get the intuition

Where can I find a proof for this

[tropic saddle]

the only proof I can think of right now uses derivatives

[final rampart]

Oh k

[final rampart]

Then... I'll probably write down your username somewhere and ask you about it when I learn about derivatives

Can I do that?

I really want a proof for this hhaha

[tropic saddle]

I can tell you the proof now and you can look back on it. if the graph only touches the axis, then it needs to have a min/max at that point, meaning its derivative needs to be zero. but then its a double root. done

[final rampart]

Well that makes sense. The derivative is the instantaneous rate of change of a function right?

And at a min/max it makes sense that it must be 0

[tropic saddle]

yes

[final rampart]

Only part I don't understand is "If a derivative of a function f(x) is zero at a point where it touches the x axis, then the x coordinate of that point is a root with multiplicity n>1"

[final rampart]

Is that what you're saying?

[final rampart]

This^^

[tropic saddle]

if f has a root at a, then f(x)=(x-a)^k*g(x) for some integer k which we dont know yet and g(x) a polynomial with g(a) not zero

[final rampart]

Edited^^

[tropic saddle]

this follows from polynomial division essentially

now, you can compute the derivative of that

with the product rule

[final rampart]

$f(x)=(x-a)^{k}\times g(x)}$

I'm juat tryna get this

[tropic saddle]

no, g down

[final rampart]

Ohhhh

Ok

[jolly parrot]

stoicindiehacker369
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[final rampart]

Ok makes more sense now

It's difficult for me to get the maths in just latex code

[final rampart]

I need to see the output haha^^

[tropic saddle]

ok I can tex it

so, we can compute the derivative

[final rampart]

Well, I did it^^

[tropic saddle]

$f'(x)=k(x-a)^{k-1}g(x)+(x-a)^kg'(x)$

[jolly parrot]

Denascite

[tropic saddle]

from the product rule

[final rampart]

Uhhhh, I know nothing about derivatives😅

[tropic saddle]

well...

[final rampart]

But well, at least you gave me an intuition for it hahah

[final rampart]

Is this ALWAYS true?

That the point at wich the function touches the x axiss

at x=a

then a is a root of multiplicity either 2 or higher

(depending on the polynomial. in my case it could only have 3 roots)

[tropic saddle]

yes

[final rampart]

Ok then

[final rampart]

Then I can find its multiplicity by polynomial division

So I guess I'll do that from now on when I have a graph. No guesswork needed.

[tropic saddle]

yes

[final rampart]

This time I just guessed that either 2 or -2 were double roots

[final rampart]

Also, when I make a function like f(x)=(x-2)(x-2)

It equals x^2-4x+4

And, in vertex form:

y=(x-2)^2-0

So, yeah, in vertex form I can see that 2 is necessarily a double root

Anyway

Thank you both

I'll come back to this one day late when I learn about derivatives

Thank you Denascite and artemetra

.close

[pearl pond]

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[marsh moss]

Each edge of a right parallelepiped is 10 cm, and the acute angle of the base is 45°. Find the volume of the parallelepiped.

[marsh moss]

as far as I'm aware

I'm not sure if h would also be 10?

if that were the case I would just take 10^3
but I don't understand why I'm given an angle

[rancid depot]

the base is a rhombus

right parallelepiped is a parallelogram prism

[marsh moss]

still don't rly get it, I'm given that each edge is 10

[rancid depot]

the base is a diamond shape with length 10

one of the angles is a 45 degree angle

<> looks like this

[marsh moss]

so those would be equal?

[rancid depot]

all sides are 10

the one crossed out is also 10

the area of a rhombus depends on how thin it is

based on the angle

[marsh moss]

ait sure I can find the area of the rhombus but I still dont have h

[rancid depot]

the word right means that the height is the side length

you can think of it like the rhombus is the base and you build walls straight up

[marsh moss]

the height is side length?

side length being?

[rancid depot]

10

each edge of a right parallelepiped is 10

[marsh moss]

yea im very lost

im given a b and c why am I given an agle

back to what I said

can't I just take 10^3

[rancid depot]

the parallelepiped is not a cube

start with a square

keeping the top and bottom parallel, move the top edge to the right and down a little

you get a rhombus such that the angle between the sides is 45 degrees

let me draw it

[marsh moss]

ok

[marsh moss]

how did u conclude it's a rhombus anyway? wouldn't the base shape be a parallelogram

equal sides

ok

[rancid depot]

each edge is length 10

[marsh moss]

yea forgot

[marsh moss]

how can i find h from having this?

[rancid depot]

it says its a right parallelepiped

i.e. not skew

so the parallelepiped, if placed on the rhombic face would be vertical

no tilt

[marsh moss]

ok

[rancid depot]

so measuring the height would be equivalent to measuring a side length

[marsh moss]

we are going in circles
could you set some example please

side length is 10

that means that all the sides are 10

i dont get it

[rancid depot]

lets say you have a table with a surface that is a rhombus

the legs of said table are vertical

now you want to measure the volume that it takes up

[marsh moss]

the answer i got is 50sqrt2

is that correct

[rancid depot]

close

thats the area of the rhombus

but you need to multiply by the height

[marsh moss]

500sqrt2

[rancid depot]

yes

[marsh moss]

so basically it's always best to find the area of the base shape

and then multiply by height

ait thanks man

[rancid depot]

yes

this only works with right prisms

if it is a skew prism, you need to deal with another angle

[pearl pond]

@marsh moss Has your question been resolved?

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[fossil badge]

The fish catch of a ship was 100 tons one year and 200 tons the next year. This increase was illustrated with the following images:

a)
[Bar graph illustration]

b)
[Cylinder illustration]

Discuss briefly the statistical validity of these representations.

[fossil badge]

a) is good, but we would need numbers below the bar?
b) the B shape is not 2x larger and it´s in 3d form?

[plush bramble]

,rotate

[fossil badge]

annnnd? @plush bramble

[pearl pond]

@fossil badge Has your question been resolved?

[fossil badge]

<@&286206848099549185>

[pearl pond]

@fossil badge Has your question been resolved?

[pearl pond]

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[orchid cobalt]

guys im a little stuck on c!

the correct answer is 322

but i got 331 ;-;

[regal heath]

note that all the arrows pointing North are parallel

what can you do with this information?

[orchid cobalt]

the z patter-?

[regal heath]

oh well i think the way you did it also works?

[orchid cobalt]

oh

[regal heath]

oh no nvm

[orchid cobalt]

but erm im like 9 degrees off

[regal heath]

ignore that

[orchid cobalt]

ok!!

[regal heath]

consider these two green lines that are parallel because they all point north

you have that angle 142

[orchid cobalt]

mhmm!

[regal heath]

btw nvm doesnt work because you don't know if it makes a 90 degrees angle but you can work it around in a similar way using 270 degrees

[orchid cobalt]

OH

180 + 142?

[orchid cobalt]

oh ok

[regal heath]

yes

[orchid cobalt]

alright i think i get it now!!

tysm!!

[regal heath]

np!

[orchid cobalt]

.close

[pearl pond]

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[brave sluice]

would anyone be willing to check my answers?

a) F
b) T
c) F
d) T
e) F
f) T
g) T
h) F
i) T
j) T

[ocean hornet]

What’s a g set

[ocean hornet]

Oh I see

I think I agree with all of them

[brave sluice]

thank you! 🙏

i can finally put this chapter to rest

.close

[pearl pond]

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[cursive wraith]

for c) X could be a single element-set

the action could also be the 'do nothing' action

[brave sluice]

yeah

that means it's false right?

[cursive wraith]

in general yeah

for d

[brave sluice]

.reopen

[pearl pond]

✅

[cursive wraith]

think about X = {0,1}

and the action such that ex = x and gx = 1 otherwise

[brave sluice]

then let g in G not equal to e
g0 = g1 = 1

[cursive wraith]

and wait

might not be an action

mb

sorry not an action

[brave sluice]

because you can apply g^-1

[cursive wraith]

d is true

[pearl pond]

@brave sluice Has your question been resolved?

[pearl pond]

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[frank shale]

when making a table for a log function to help you sketch a graph, especially when given a transformed log equation, is there a rule of thumb to go by when choosing x-values?

I'm sure it depends on the base you're dealing but it seems you prioritize whole number powers with 1 or 2 negative powers and that tends to work better when dealing with smaller bases like 2,3,4,6 but base 10 feels pretty annoying to see much of the curve

[frank shale]

Just trying to find a way to be a little more efficient when sketching graphs for upcoming tests/exams, since we're not using graphing apps like desmos. I tend to waste a lot of time getting out an appropriate sketch

[sharp quest]

I would just choose powers of the base, so i easily know the y-value

[frank shale]

ahh fair, thanks

.close

[pearl pond]

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[pearl pond]

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[small cape]

((x+4)(x+4)(2x-2))/(2(x+4)+4(x+4)(2x-2))

[small cape]

How do I simplify this?

I tried, but got different answer when I checked with calculator

To view steps, I have to pay

I'm not paying. Anyone hlep?

,help

[jolly parrot]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[small cape]

,help cmd

[jolly parrot]

You really shouldn't take it literally :upside_down:. Please type ,help ping, for example!
The full command list may be found using ,list.

[small cape]

,list

[jolly parrot]

My commands!

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[small cape]

,list

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[small cape]

((x+4)(x+4)(2x-2))/(2(x+4)+4(x+4)(2x-2))

Ping me!!!!

[pulsar lark]

Which question

[pearl pond]

@small cape Has your question been resolved?

[small cape]

This

The only one I sent

[small cape]

This it the answer

I need to know ze steps

[pearl pond]

Channel closed

Closed by @small cape

Use .reopen if this was a mistake.

[small cape]

No

No close

.reopen

[covert whale]

cancel out like terms from numerator and denominator

im guessing you need to simply this equation?

[small cape]

Yes

[small cape]

That’s what I did. Could you do it and show the steps please?

[covert whale]

yes so first step is to cancel like terms

sure

[small cape]

Thanks

[covert whale]

so you see i’ve cancelled the like terms

sorry it took so long i had to get a paper

@small cape

after this step, what would you be left with?

[mystic crypt]

1/6 ?

[covert whale]

yes 🙌

[mystic crypt]

what is this website by the way ?

[pearl pond]

@small cape Has your question been resolved?

[pearl pond]

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[small cape]

My bad

[small cape]

I was at piano

[small cape]

I got the same answer

How did Mathway get that then?

I thought I was doing something wrong

[pearl pond]

@small cape Has your question been resolved?

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[pearl pond]

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[sharp smelt]

why does length matter here

[sharp smelt]

I am able to solve it without length too

[sharp vigil]

not all information included is necessarily relevant. it may be a bit of a trick question in that regard

[sharp smelt]

True

right

thanks

.close

[pearl pond]

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[pearl pond]

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[heady sigil]

Tried integrating to find the volume of a donut (torus) on my own and am left stuck here (2nd image)

[heady sigil]

For context am using the shell method

[half tendon]

Is that like the disc method?

Pi times int f^2(x) dx

[heady sigil]

The disk method would be treating the donut like the culmination of annuluses

Then you'd integrate up

[half tendon]

Annuluses?

[heady sigil]

Rings

You'd itegrate the area of the rings that make up the height of the donut

Here's the approach and the second image shows the shell method on the torus, taking tiny cylinders with dx thickness with radius x and height y integrating longtidually from one end to the other