help-10

[obtuse pebble]

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[coral owl]

i need require assistance!!

[inner sierra]

?

[coral owl]

i need help solving this

[inner sierra]

!nodpf

!nopd

!nopdf

[obtuse pebble]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

[marble estuary]

thats an entire past paper?

[coral owl]

not the entire paper

[inner sierra]

can you send a photo please?

[coral owl]

js some question

s

[coral owl]

is there a level requirement to send images

[inner sierra]

just send the image

[marble estuary]

shouldn't be

[inner sierra]

we don't encourage sending PDFs here

[coral owl]

ahh okay okay

gotcha

lemme send a pic

question 14 and its parts

[coral owl]

our teacher gives us practice past papers

[inner sierra]

which question you needed help with

[marble estuary]

think ive seen this one b4

[coral owl]

i sent a picture

[inner sierra]

which question

there's 3 questions here

[coral owl]

the whole page

[marble estuary]

which part

oh all of it

[inner sierra]

14a kinda simple

[coral owl]

yeah besides that

[inner sierra]

sub in x and find y

[coral owl]

part b and the other one

not a

i can do a

[inner sierra]

god damn

[marble estuary]

plot all the points on the table and connect them with a decently smooth curve

[inner sierra]

i don't know how to graph a cubic

😭

[coral owl]

same 💔

[inner sierra]

well

desmos it is then

[marble estuary]

cubics would have the shape
_/
/

if positive

[inner sierra]

yea

[marble estuary]

or (if the coefficient of x^3 is negative)
\_
\

[inner sierra]

you should do what someone said if you don't know how

[marble estuary]

as in, if the coefficiant of x3 is positive

[coral owl]

what would coefficiant mean in math

[marble estuary]

for c, try and rearrange the equation given into: the original cubic = something, can you try that?

[coral owl]

thats the most confusing part for me

[halcyon pewter]

the nubmer before the variable

-3x² has coefficient -3 for example

[marble estuary]

if you have a term like 5x², then 5 is the coefficient of x², basically its the amount by which smth is multiplied

[coral owl]

ohh okay

gotcha

what abt part c

im still confused on that

[marble estuary]

so you basically want to get the equation into the form cubic = line (which you can solve by drawing the line and looking at where it intersects cubic)
here cubic refers to the eqn that you plotted

so you subtract stuff until the left hand side of the equation is the same as the original equation

you'll also need to multiply by -1, because the signs are swapped it seems

[coral owl]

is thst it

[marble estuary]

yes

then when you have the cubic on the left side, whatever is on the right is the line and you plot that

[coral owl]

im still confused by it sorry

it might be cus its over msging but im trying to understand

from my previous experiences

when ive done these types of questions my teacher has said to make them into a "subject"

them as in the equation on part c

and compare it w the original equation

[marble estuary]

wdym

[coral owl]

ill try and explain best as i can w my ability

so

he says

you have to compare the original equation to the new one

which is part c

and u have to see whats similiar

between them

and afterwards u make that thing which is similiar a subject

he calls it

thats the best i can explain

theres like 1 or 2 more steps ik off but idk how to word it properly

[marble estuary]

yeah thats i think p much what i was saying

[coral owl]

ah

u js made it sound a bit confusing sorry

well ill go try it and let u know how it goes

[marble estuary]

ah mb

[obtuse pebble]

@coral owl Has your question been resolved?

[prisma dagger]

hey i have a question...

can they help seventh graders like me here?

im doing algebra

[timid silo]

welp do not mention your age if you are below 13 youwill likely get banned

[prisma dagger]

.

[timid silo]

!help but

[obtuse pebble]

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

[coral owl]

okay so i did it

but can u check if its correct

here

[obtuse pebble]

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[thorny portal]

Hi yall, i need help with this. I know how to do it the conventional way, just 1/3 + 1/4 + .... 1/8 and then find the common denominator. But is there anyway i can evaluate this summation without doing that because its rather alot of work

[halcyon grotto]

i have no idea how to do this.. pls help

[halcyon pewter]

already occupied mate

use one of the available channels

[thorny portal]

Help fr

[halcyon grotto]

ohkk

[thorny portal]

<@&286206848099549185>

[tardy epoch]

no there isn't

[thorny portal]

damn

okay then

[dark stirrup]

This is the harmonic series if you are interested, but as already stated, no simple way to do it beyond the tedious way

[trim portal]

the only suggestion i have is to do it somewhat systematically

I'd probably start by adding
1/8 + 1/4 + 1/2
then
1/3 + 1/6
and then 1/5 and 1/7

[thorny portal]

Yeah im doing that but theres no 1/2 since it runs from k=3

Thanks yall

[obtuse pebble]

@thorny portal Has your question been resolved?

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[obtuse pebble]

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[dry elm]

One message removed from a suspended account.

[wary badger]

FTC

[dry elm]

One message removed from a suspended account.

One message removed from a suspended account.

One message removed from a suspended account.

[obtuse pebble]

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[obtuse pebble]

Available help channel!

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[grizzled silo]

this question might be a little confusing but how do i intersect the lines the proper way like is there a number of units that i need to follow? because when i try to draw the lines they never intersect on the right point and im constantly redrawing to get the point thats correct.

[grizzled silo]

like if i draw the line by every one unit they dont intersect properly

[deep kraken]

Yes must correct ratio

And intersect points

[grizzled silo]

what numbers do i use for that

[deep kraken]

Find 2 x-y intersections points for each equation and connect them

[grizzled silo]

wdym

[deep kraken]

for example you have equation x-y=-2

find x when y =0, and find y when x=0

connect those 2 coordinates

[grizzled silo]

i see

[deep kraken]

similar to the other equation

then try draw again

[grizzled silo]

ima try that and be back

[deep kraken]

all good

[grizzled silo]

wait question again

do i find y when x=0 and vise versa in the orginal equation or the equation after i change it to slope intercept form

so x - y =-2 or y= x + 2 which one would i try to solve for

[deep kraken]

its the same

[grizzled silo]

alright

okay im gonna skip it and come back cuz im lowkey still confused even after doing that 😭

.close

[obtuse pebble]

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[deep kraken]

Youre cooked man

[obtuse pebble]

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[karmic hedge]

Orthogonally diagonalize the matrix giving an orthogonal matrix P and a diagonal matrix D.

[karmic hedge]

How did they get u1 = (1/sqrt(2), 2/sqrt(2))?

[restive gorge]

This seems like a typo

u_2 in P also seems to miss a 3

[karmic hedge]

Ah ok good to know

Thanks!

❤️

.close

[obtuse pebble]

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[grizzled silo]

OKAY I AM BACK. im trying to see what you are talking about, did i do this right 😭

[deep kraken]

Correct

[grizzled silo]

okay so the point is (0,1) ?

[deep kraken]

Now connect (0,2) and (-2,0)

[restive gorge]

This channel may auto close btw

[grizzled silo]

i did

[deep kraken]

The other is (1,0) and (0,-4)

Then yes

[grizzled silo]

yes but whats the final point

bcs in the original equation we need to connect both y inyercepts and slope points

[latent thunder]

2y = 5x-10

[deep kraken]

Can i see the original equations

[deep kraken]

Then the intersect point should be (2,4)

[latent thunder]

Minnh how is it not 2y = 5x - 10

[deep kraken]

Not that question

[latent thunder]

She didn't do it right tho

[grizzled silo]

yes but how are we getting to that point is where im confused

bcs ik TO intercept them but HOW is whre im confused and ik you said to get the correct ratio

but once i do the if x=0 what would y be and vise versa where do i go from there to get the singular point

[obtuse pebble]

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[crisp sleet]

bra II lol

[grizzled silo]

?

[woeful torrent]

bra II: electric boogaloo

what's your question?

[deep kraken]

I think he is just replying to old messages

[timid silo]

if you dont have question close this channel

[crisp sleet]

as its a liner equation its a straight line

the slope of a straight like is
(Y2 - Y1)/(X2 - X1) = slope
given that, 5x -2y = 10
as we know the equation of a straight line => y = mx + c (where m is slope and c is y intercept)
converting the given equation in the following equation for comparison we get
y = 5/2x - 5
as trough comparison we can say
the slope is 5/2 and the y intercept is -5

idk why I even did this

whats the point lol

[crisp sleet]

I didn't open this

oh lol why does this say my usr name
sorry mb Idk how the server works

Im a newbie

[deep kraken]

All good

[deep kraken]

Your way is correct, either way is correct since its a linear function

If more complex functions like quadratic then simplifying equation like you did wont work

[crisp sleet]

for that you can just differentiate to find the slope (dy/dx)

[deep kraken]

Yes

[obtuse pebble]

@crisp sleet Has your question been resolved?

[obtuse pebble]

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[mild sandal]

<@&268886789983436800>

[obtuse pebble]

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[mystic jetty]

@hollow flame No advertising

[hollow flame]

sorry it was a mistake

[obtuse pebble]

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[main echo]

<@&268886789983436800>

[obtuse pebble]

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[main echo]

scam begone

[hardy widget]

[high lily]

jeez, triple

[worn yoke]

it's no fun anymore because they're all mr beast again

we used to have variety

[obtuse pebble]

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[elfin heart]

I. I don't understand. What does any of this mean and how does it work. I've tried for like two days straight

[high lily]

did you try looking up the definitions of the terms in the list?

and/or have notes from class

[elfin heart]

yes I did

[elfin heart]

not technically in school this is self study

[flat mural]

u select one from the dropdown right?

[elfin heart]

yeah

i mean like. idk the terms

[high lily]

what did you get when you looked them up

[elfin heart]

well i got overwhelmed and cried (i struggle with like big blocks of information sometimes)

also the information itself just wasnt... clear. to me

[flat mural]

ok one step at a time

[high lily]

the basic definition wouldn't be a big block

Can you show what ur seeing when you search these terms

[elfin heart]

i got these for one of the searches

even if i dont get overwhelmed i just learn better when like I can talk through it tho

[high lily]

the property is summarized under that first link

[elfin heart]

yeah idk what exactly that means in context

[high lily]

draw a straight line

Actually If you click into it you should see a diagram

[elfin heart]

ik that it didnt rly help

[high lily]

show what pic ur looking at when you click that link

[high lily]

which part of that confuses you

[elfin heart]

I just get overwhelmed just like. Having information shoved in my face I don't process it well

Like ik it sounds stupid but like I'm trying

[high lily]

Point to the specific part of what ur reading here that you don't understand

actually i'll just break it down even further,
lets start with a line.
the angle on each side will be 180°, but lets just focus on one side

[elfin heart]

Okay

[high lily]

do you have any issue with that so far?

[elfin heart]

Nope

[high lily]

now lets draw a line sticking out of it

this will split that 180° angle into two smaller angles (A,B in the diagram)

[elfin heart]

okay

[high lily]

the line may extend further down, but that has no effect on A and B

here A and B are referred to as a linear pair

and the property states that their angle sum is 180°

and that's pretty much it

[elfin heart]

okay. so 2 is angles forming a linear pair

right?

[high lily]

sum to 180°, yes

[elfin heart]

now. what do the rest of the terms mean...

[high lily]

again try looking them up

[elfin heart]

i genuinely have 0 frame of reference

i have

i did try

i tried

i rly did

[high lily]

i'm more or less just repeating whats there

[elfin heart]

well it makes more sense when i ahve someone explain it to me idfk

[high lily]

show what you're getting when you look up the term
and point to the part you don't understand

[elfin heart]

its not a specific part. im telling u its not a specific part my brain just doesnt. handle just like written information like that well and stuff

like idk im sorry if this is frustrating and im sry if im irritated but like it jsut feels like no one fricking listens

[high lily]

i'm not trying to be a pain,
but i'm writing stuff too

[elfin heart]

like if u dont have time or u dont want to thats okay i get it but like just say that

like ive tried what ur asking me to do

[main echo]

you need calm down a bit and back read some stuff

[high lily]

but I tend to encourage people to try their best

if there is an issue with the language or how they deliver it, we can help clarify

[elfin heart]

i am

[high lily]

which is why I've been trying to ask you what you're reading and to point the specific part you don't understand

so I can guide you along rather than teach from scratch

[short crescent]

maybe watch some videos on the topics you're struggling with so you have someone speaking instead of just having to read

there are a lot of good youtube videos that explain a massive variety of topics well

[elfin heart]

I tried the videos too. i need to like. be able to talk to the other person idk

like I dont like having to do this I feel like an idiot but its the only thing that works most of the time

[main echo]

try talking someone irl

[elfin heart]

i dont have anyone irl

[short crescent]

^ yes, was gonna say meet up with some classmates

[elfin heart]

not in school

[high lily]

follow these steps
look up the term
show what came up
try explaining which part of what came up you don't understand

[short crescent]

are you in uni? almost all professors have office hours you can go to for help

[main echo]

teacher/professor?

[elfin heart]

im 15

i'm not in school bc no car

and mama isnt good at math and is also rly anxiety inducing to work with

[wise dust]

yo you tryna do this or the definitions

[elfin heart]

well im trying to do that

but to do that i need to know the definitions of the terms

[wise dust]

by terms, do you mean the statements

[short crescent]

what do you mean by the "definitions of the terms"

which terms?

[elfin heart]

these

idk what any of these mean

i included it in the original message

[short crescent]

this is a bisector

complementary angle means it adds to 90 degrees, supplementary means it adds to 180

[short crescent]

in this case angles A and B are supplementary

perpendicular lines means two lines that intersect at a 90 degree angle

reflexive property of congruence applies to whole shapes, and reflexive property of equality applies moreso to angle measures, variables, etc

theres every definition you need

[elfin heart]

Thank u

And then. What is the question actually asking me

Like. I don't. Understand rly.

[short crescent]

its asking you to fill in the reason why that statement is true

using one of these many definitions that i gave you

[elfin heart]

Okay

Thank u

[obtuse pebble]

@elfin heart Has your question been resolved?

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[spring orbit]

you guys i need help how to solve this exponential equation

[tawny tree]

is x real?

[spring orbit]

you have to find x

[tawny tree]

I know but is x real?

[frank coral]

Probably

[spring orbit]

probably

[tawny tree]

anyway

what have you tried

[spring orbit]

like i change to the other sides

like three on one side

tow on other side

[tawny tree]

okay good

[frank coral]

Yup

[tawny tree]

and you're stuck or?

[spring orbit]

yes im stuck

i cants solve

[tawny tree]

hint: factor

[frank coral]

(Although it’s not necessary, I’d recommend solving for u=x/6 to get rid of fractions)

[spring orbit]

i just tried so many times

i jts cant move

[sand roost]

what have you written out so far?

[obtuse pebble]

@spring orbit Has your question been resolved?

[spring orbit]

bro im just stuck at the begiging

i dont know how to do

like i put three on one side, and two on the other and im stuck

[sand roost]

okay, let's say i asked you to factor $5^5 - 5^3$

[warm shale]

حسیب ♥

[sand roost]

what's the common factor between these two?

[spring orbit]

let me see

5-3

which is 2

[sand roost]

hmm, not quite. you can do $5^5 - 5^3 = 5^3(5^2) + 5^3(1),$ right?

[warm shale]

حسیب ♥

[spring orbit]

okay

[sand roost]

so what is the common factor?

[spring orbit]

5^2 and 1

5^3(5^2+1)

[sand roost]

okay, you got the idea

[spring orbit]

okay

[sand roost]

although we would say 5^3 is the common factor, not the other ones

[spring orbit]

understood

[sand roost]

coming back to our equation, we have $$3^{\frac x2} - 3^{\frac x2 - 1}$$ on one side. can you do something similar to factor that?

[warm shale]

حسیب ♥

[spring orbit]

so i woudl do here 3(x/2 + x/2-1)

[sand roost]

you seem to be saying $3^{\frac x2 + \frac x2 - 1}$, which isn't what we did before

[warm shale]

حسیب ♥

[sand roost]

hint: out of the exponents, 5^5 and 5^3, the common factor was the smaller power, 5^3

[spring orbit]

okay

so then woudl be 3(x/2)+3(1)

[sand roost]

no, you didn't factor anything our here

in this case, you should have something like $a(b+c)$

[warm shale]

حسیب ♥

[sand roost]

a monomial times a binomial

[spring orbit]

aaa

would be (3-1)*3^x/2-1

[sand roost]

ding ding ding!

(you should probably write 3^(x/2 - 1), though)

now, try doing something similar to the other side, on your own

[spring orbit]

okay, i will try my best

[sand roost]

once you have it all factored out, you can do some nice algebra to get an easier exponential equation

[obtuse pebble]

@spring orbit Has your question been resolved?

[obtuse pebble]

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

[light dove]

$$f(n) := 1 - \frac{f(n-1)}{f(n-2)}, \quad n \in \mathbb{Z}, \quad n \ge 1$$

$$f(-1) = -1, \quad f(0) = 1$$

Prove that $f(n) \neq 0$ for all values of $n$.

[warm shale]

fall

[neon vector]

Suppose f(k) = 0 for some k ≥ 1

[light dove]

I did try that... I could only prove f(k+2) blows up

[deep kraken]

This can be done by induction i suppose

[vagrant harbor]

I don't see any other method ngl

[deep kraken]

so suppose if n>= 1 right

just take a base case, f0 = 2

then assume there exist an integer k such that k <=n

[neon vector]

What is f(k+2)

[light dove]

^

for some k>=1, f(k) = 0, a direct implication of this is f(k+2) has a division by zero and therefore blows up / is undefined

[deep kraken]

wait, i think

if f0 = 0 right

then isnt f(n-1)/f(n-2) = 1?

such that f(n-1) = f(n-2)

[neon vector]

Yes exactly

That's your contradiction

The right side must be well defined for all n ≥ 1

[light dove]

but this only works if you assume f(k) = 0 for some k >= 1

so its circular

hmm, i think im not phrasing this well, slayla made a better explanation

[zenith raft]

just look at it contrapositively. if f(n) = 0 for some n then f(n+2) \neq 1 - f(n+1)/f(n). qed

[light dove]

$$f(-1) = 1,\ f(0) = 1 \implies f(1) = 1 - \frac{1}{1} = 0$$

So $f(n) = 0$ is perfectly reachable for different initial conditions. Any argument that never mentions $f(-1) = -1$, $f(0) = 1$ cannot possibly be a valid proof, since the same argument would "prove" $f(n) \neq 0$ for the counterexample above.

[warm shale]

fall

[light dove]

or rather, i should've phrased the problem better, since its more about proving the sequence is well defined for all n>=1

and well definedness would require n=/= 0 since as we have shown above, it leads to a division by zero

[deep kraken]

i suppose for fn = 0, then f(n-1) = f(n-2)

[vagrant harbor]

That doesn't quite make sense

[light dove]

Prove that the sequence defined by $f(-1) = -1$, $f(0) = 1$, and $f(n) = 1 - \dfrac{f(n-1)}{f(n-2)}$ is well-defined for all $n \geq 1$ i.e. that $f(n-2) \neq 0$ at every step.

this would be a more precise phrasing

[warm shale]

fall

[zenith raft]

we have f(1) = 0, so f(3) doesn’t parse

the argument still holds there

but ig i see what you’re trying to ask now

[light dove]

yea, its a pretty chaotic sequence as well...no obvious patterns i could see

[zenith raft]

f ‘making sense’ is equivalent to f(n) \neq 0 for all n >= 1. but you want to know if f’s definition makes sense. the problem made it look it was being assumed f’s definition made sense

[vagrant harbor]

You kinda used f(-1) = 1 and -1 interchangeably

[light dove]

I used them deliberately and the counterexample is a different choice of initial conditions: f(-1) = 1, f(0) = 1 gives f(1) = 0.

If your proof never mentions f(-1) = -1, it would apply to this case too. But it fails here, so t needs to actually use the specific initial conditions somewhere.

[zenith raft]

it’s just miscommunication at this point

[light dove]

yea XD

[zenith raft]

your problem makes sense to me now, it just wasn’t what it looked like originally

[light dove]

but this is the problem phrased better

[vagrant harbor]

The thing is, your counterexample depends on changing conditions, but shouldn't it be assuming f(n) = 0

[light dove]

I phrased it badly initially, thats my bad though

[zenith raft]

i can also frame the problem a certain way in case anyone does not understand.

define f(-1) = -1 and f(0) = 1. is f(1), as defined in the problem, not equal to 0? if so, what about f(2)? and so on

[light dove]

If your proof is correct, it should fail when applied to a case where the conclusion is false. Your proof says: "if f(k) = 0 then f(k+2) is undefined, contradiction." This never uses f(-1) = -1. Applying it to f(-1) = 1, f(0) = 1, where f(1) = 0. Your argument would still claim f(n) ≠ 0, which is false.

[light dove]

what i meant to pose as a question in the help channel was basically that

[zenith raft]

this is a yes or no question

[vagrant harbor]

Yeah but f(n) is defined for n ≥ 1, so we can disregard the condition

[zenith raft]

well that’s the thing…..

[light dove]

XD

we need to prove f(n) is well defined for n>=1

[zenith raft]

just disregard the original question

[light dove]

also, I dont think there exists a closed form for the recurrence relation

[vagrant harbor]

I don't see exactly how the new question differs, it's just that it was indirect and became direct

[zenith raft]

define f(-1) = -1 and f(0) = 1. consider f(1), as defined in the problem. if it is 0, end the process and answer no. if it isn’t 0, what about f(2)? and so on. if it at any iteration you get a 0, answer no. if they will all be nonzero, answer yes

[vagrant harbor]

I guess that's more guided

I suppose problem solved?

[zenith raft]

i see no solution proposed

[light dove]

^

[zenith raft]

well i see several proposed but they don’t answer the question because they are answering the wrong thing

[zenith raft]

notice how here you don’t get to use f(n+2) when answering for n

[light dove]

there's not exactly a proof provided yet, and I cant find a clean pattern in the reccurence

[zenith raft]

like… if f(n) is 0 then f(n+2) doesn’t parse yes everyone knows that. it’s irrelevant to the problem. we want to know if any such n exist. there is no supposition that the formula already makes sense for all n, so f(n+2) not parsing doesn’t have any useful consequences

[vagrant harbor]

Can't you just use induction as minhh proposed?

[zenith raft]

what would the induction proof look like

[vagrant harbor]

I've yet to figure that part out

So far I could only get
$$ f(n+1) = 1 + \frac {1}{f(n-2)} - \frac {1}{f(n-1)} $$

[warm shale]

ScoobySnacks

[vagrant harbor]

Doesn't seem to help

[obtuse pebble]

@light dove Has your question been resolved?

[vagrant harbor]

Do we know if f(n) is decreasing or increasing?

[light dove]

uhm, its pretty chaotic, so I actually dont know

here is what it looks like on desmos(up to n=290):

[vagrant harbor]

I still don't understand how this works ngl

[zenith raft]

please leave ping on when replying to me

[vagrant harbor]

Well, if we prove that f(n+2) doesn't parse, can't we just prove that it is the case for f(n+1) as well? Or f(n+3)

[zenith raft]

i have to drive for a little bit and i’ll think about how to explain differently

[vagrant harbor]

Aight drive safe 🫡

[light dove]

I did get something like this but it still doesnt prove well definedness

[light dove]

drive safe

[light dove]

we need to prove that no such n exists such that f(n) = 0 given the recurrence relation and its initial conditions as defined

[vagrant harbor]

I'm not sure if that's a thing, but can't we prove by induction that if f(n) = 0 then we get a contradiction?

[light dove]

thats fair, but it only proves that given f(n) = 0, it leads to f(n+2) being undefined

[vagrant harbor]

I think I get what you're saying

[light dove]

it does not say whether a number n = 84579875439875368793 leads to f(n) = 0, aka f(84579875439875368793) = 0

ye

and there are infinitely many n, so you cant just check and prove finitely

[vagrant harbor]

Do we know for certain that f is defined?

[light dove]

this is what we are trying to prove XD

[vagrant harbor]

Yeah so it's not certain

[light dove]

because if there exists a number n such that f(n) = 0 then the recursive sequence terminates

[vagrant harbor]

Meaning it could be false

[light dove]

yes, you can try to disprove it too

basically, do the opposite, prove that there exists some n such that f(n) = 0

but i have the vibes based on gut feel that its probably well defined, aka no such n exists

[vagrant harbor]

I mean

[light dove]

could go either way

but as far as im concerned

I did the computation in C++

its well defined up to 1 trillion

[vagrant harbor]

We can't prove by induction that f(n) = 0, doesn't that imply that it can't be equal to 0?

[light dove]

well, im not sure as to whether we cant prove it, or whether we havent found a proof

[light dove]

also the entire thing blows up really fast, so i was just computing the signs of the fraction

[vagrant harbor]

The first condition doesn't stand as f(1) = 2 > 0

Even if it did, f(n+1) ≠ 0

[zenith raft]

i thought of an explanation i think. i’m not sure if it will settle it, but it’s different. let’s define the sequence in a way that is for sure well defined first. i will introduce a new symbol called X, that f(n) can be equal to (it’s just a formal symbol that doesn’t really have any meaning as a number or anything, but intuitively f(n) = X means the original formula has blown up at some point). the definition of f will be
f(-1) = -1
f(0) = 1
for n > 0:
if f(n-2) is 0, or if either f(n-1) or f(n-2) are X, then f(n) = X
otherwise (so, in the case that f(n-1) and f(n-2) are both real numbers with f(n-2) not 0), we define f(n) = 1 - f(n-1)/f(n-2), like before

just to check your understanding, confirm that you understand the following...
let n be a positive integer.

1. if f(n) = X, then for all k > n, we have f(k) = X
2. if f(n) = 0, then we have f(n + k) = X for all k > 1

now the question is, are there any X’s in the sequence?

that’s equivalent to “are there any 0s in the sequence?”

[vagrant harbor]

That's fairly basic yes

[zenith raft]

by settle i just meant settle understanding the problem btw

[vagrant harbor]

I was gonna ask about that yeah

I'm just confirming that we've been agreeing on practically the same thing. But we still lack the essence of it, which is the proof

[obtuse pebble]

@light dove Has your question been resolved?

[zenith raft]

i think i got something

actually i need to do some more work. i’m not sure if i have anything

if $f(n) = 0$, do we perhaps have $f(n-k) = \frac{f(n-1)}{(1 - f(n-1))^{k-2}}$?

[warm shale]

slayla

[zenith raft]

for all k with 1 < k < n or something

[zenith raft]

i think this would resolve the problem because it implies the sequence (after the first two terms) is geometric (until the first 0)

if there is a 0

but it is not geometric

it holds for k = 2, 3, 4 i know. still need to think about if it generalizes

[light dove]

also i just realised

the geometric pattern breaks at k=5

[zenith raft]

seriously?

[light dove]

yourr k variable is basically doing the work of my a variable

[zenith raft]

lol

[light dove]

yea XD

[zenith raft]

very sad it does not generalize. i will probably be done with the problem for the night

[light dove]

goodnight, rest well

[wet latch]

Can we not assume a k that f(k)=0 and just go back infinitely

Eventually reaching the contradiction f(-1)=f(0)

[light dove]

just go back infinitely

isnt that the problem XD

you would have to prove that after the infinite number of steps you 'go backwards', the initial conditions matches our original definition of the recurrence relation with base case f(-1) = -1 and f(0) = 1

[wet latch]

That's what I came up,I thought since we just descend infinitely at somepoint we'll reach k=1

[zenith raft]

are you claiming that all the f(k-i)’s are equal?

[wet latch]

Exactly,leading to a contradiction

I can't find another way to explain that it descends infinitely

If asked for exams or something I would just reference the well ordering principle

[obtuse pebble]

@light dove Has your question been resolved?

[hardy widget]

(Troll?) Suppose that $m$ is the smallest integer where $f(m)=0$. Then $f(m+1)=1$. But setting $n=m+2$ yields $1/0$, which is undefined. Hence we have a contradiction.

[warm shale]

Civil Service Pigeon

[light dove]

this is the problem, better stated

[light dove]

we can parameterize f(k-1) = f(k-2) = a for some a, and the recurrence will still be valid and not force f(-1) = f(0)

[light dove]

refer to this for an example of what i mean

[warm shale]

fall

[light dove]

@hardy widget

[light dove]

this would be a more precise formulation of the question I meant to pose

[light dove]

this works assuming the sequence is well defined for all n, which is what we are trying to prove, i.e. prove that there exists no such n such that f(n) = 0

[hardy widget]

[hardy widget]

(To anyone else: the original question was ill-posed. I have pinned the corrected question.)

[light dove]

<@&286206848099549185>

[deft pagoda]

wtf

stop pinging

[light dove]

this is the only time i pinged the helpers role though...

[vagrant harbor]

That was the most justified ping ever lol

[obtuse pebble]

@light dove Has your question been resolved?

[balmy mortar]

delete the role you assigned to yourself if you dont want @ helper pings

[deft pagoda]

omg stop crying

[balmy mortar]

<@&268886789983436800> someones being difficult

[balmy mortar]

even with the new pin .. this channel has a lot of history and its hard for someone new walking in to pick up where others left off. maybe its better to start a new channel with where exactly youre stuck now?

or just reask what exactly is troubling u rn

[deep kraken]

If don't want to help or receive unwanted pings then no points having helper roles though, its optional.

[upbeat plinth]

@deft pagoda the helper ping was justified in this case, and pls do not attack ppl for suggesting how to avoid getting helper pings

[light dove]

👍

.close

[obtuse pebble]

Channel closed

Closed by @light dove

Use .reopen if this was a mistake.

[deep kraken]

anyways, if you can show for all n such that f(n-1) = f(n-2) would make this contradicts

[light dove]

i have reasked it, or rather, a continuation but a different conjecture to prove, in a separate help forum post https://discord.com/channels/268882317391429632/1484101319470878770

[obtuse pebble]

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[young osprey]

Help me with quadratic equation. Completing square method

[glossy basalt]

specific please

[glossy basalt]

i mean

any example

[young osprey]

5x² - 7x + 11

[glossy basalt]

sure

First you can

factor 5 out of the first 2 terms

5(x²-7x/5) +11

next

add and minus (7/5÷2)²

inside the brackets

i.e.
5(x²-7x/5 + (7/5÷2)² - (7/5÷2)²) + 11

[young osprey]

Can u use this. ( x + b over 2) ² +c - ( b over 2 )² =0

[glossy basalt]

but your x is not with coeff 1

your example is a 5 instead of 1

[glossy basalt]

I can't use this if coefficient of x² term if 5

[young osprey]

Ohh okay

[glossy basalt]

if you wanna use this, try it on an quadratic equation like
x² + bx + c =0

where b and c are real numbers

[young osprey]

So you can also use this to complete the square in gcse o level?

[glossy basalt]

not sure about gcse o level, but if the coefficient of x² is 1, then this is good enough

[young osprey]

Alr

Kae thee boffan?

[glossy basalt]

?

[young osprey]

I meant can u factorise 18a² - 98

[glossy basalt]

oh okay , but again this coeff of the squared term is not 1

[young osprey]

I mean normal factorisation

[glossy basalt]

oh right

misread

first factor 2 out of the expression

2(9a²-49)

using identity (x+y)(x-y) congruent to x²-y²

that's the hint!

see if you know how to do it from this

[young osprey]

So 9 is 3² and 49 is 7²

So

[glossy basalt]

yea

[young osprey]

2(3a-7)(3a+7)

[glossy basalt]

correct

[main echo]

-# biscuit

[young osprey]

Alr thanks for your help

[glossy basalt]

yes?

[glossy basalt]

np!

[glossy basalt]

Cheers!

[young osprey]

Byee

[glossy basalt]

you can type .close if there's no more related questions

[young osprey]

Alr

.close

[obtuse pebble]

Channel closed

Closed by @young osprey

Use .reopen if this was a mistake.

[obtuse pebble]

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[patent spruce]

I was about to do integration by parts

[patent spruce]

Is that possible?

[patent spruce]

It’s not in the answer scheme

Part i of the same question (for extra info)

[restive gorge]

uhmm maybe but i imagine it'd be a pain

[obtuse pebble]

@patent spruce Has your question been resolved?

[obtuse pebble]

Channel closed

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[high lily]

apply double angle identity to the sin(2x) before integrating

[obtuse pebble]

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[timid silo]

Question: Find the derivative of f(x) = x²

[timid silo]

How do I?

[restive gorge]

Do u know the power rule

Also did you mean x²?

[timid silo]

Yes mb

I'm trying to solve some calculus problem that will come this academic year

I haven't started this grade yet lol

[restive gorge]

The power rule tells you to take down the current exponent, multiply that with your function, and take away 1 from the exponent

[timid silo]

Oh aightt

Next?

[restive gorge]

Thats it

Try it

[timid silo]

Ohkay lemme try now

[deep kraken]

Nice work

[timid silo]

I tried it's f '(x) = 2x ig?

[timid silo]

Oh thx

[fiery bronze]

Try it.

[fiery bronze]

Right.

[restive gorge]

Yes

[timid silo]

Oh alright this is kinda easy now

[fiery bronze]

Yeah, do you have more questions?

[timid silo]

Thanks guys

[restive gorge]

You can apply this rule on any function of the form x^k

[timid silo]

Nope thxx

[fiery bronze]

Have a nice day. You can close typing .close.

[timid silo]

Oh alright I'll note it

[timid silo]

Yea you too!

[timid silo]

.close

[obtuse pebble]

Channel closed

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[obtuse pebble]

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[patent spruce]

For part c, why is answer only 4

How do I decide that

[wild swallow]

the domain of f is x >= 0

[patent spruce]

Can I say since range of inner function must be equal to domain of outer function, so f(x) is domain of g(x) here so…

[patent spruce]

And then this

[deep kraken]

Range inside dont necessary equal to domain outside

[wild swallow]

the domain of gf(x) is the domain of f(x)

it's x >= 0

[patent spruce]

Oh, because x replaces f(x) in gf(x)?

[wild swallow]

what does that mean

[patent spruce]

But it’s supposed to lie inside the domain of outside?

[wild swallow]

to evaluate gf(x), you feed x into f

so x must live in the domain of f

[deep kraken]

Range f <= domain g

[patent spruce]

And then you feed f(x) into g(x)?

[wild swallow]

yes

[patent spruce]

Ty

[patent spruce]

I was thinking I could say that f(x) has become the domain of g(x) because that’s what you plug in g(x)

[deep kraken]

What do you mean by this

F(x) is not the domain of g(x)

The range of f lays inside domain of g

[serene elk]

intuition is close but not quite right lets say i have f(x)=x^2 if i plug in the set S={1,2,3,4,5} I don't say that the domain of f is now S. Domain is all the values of x for which f(x) is defined

[patent spruce]

Aha

Noted

[serene elk]

$g(x) \subset Dom(f)$

[warm shale]

Nyxzore

[serene elk]

for f(g(x))

[patent spruce]

What’s that sign in the middle

[serene elk]

subset

[deep kraken]

What is that sign

[serene elk]

[patent spruce]

I’ll write this in my notes, tyy

[uneven schooner]

if $A\subset B$ then $B$ contains all elements of $A$ but $A$ may or may not contain all elements of $B$.

[warm shale]

Annie Maqionde

[deep kraken]

First time i see this sign, thank you Annie for teaching me🙏☺️

[patent spruce]

My brain is loading

[serene elk]

whaaa

i must show my blud in father more things fr

[deep kraken]

Me noob

[serene elk]

its essentially this but g(x)=dom(f)

is also implied by this symbol

[deep kraken]

I think you are going opposite

This is def of fog

[patent spruce]

This is my thought process from left to right

[uneven schooner]

ohk so basically, your entire qs is why -10 is not an answer for (c)?

[patent spruce]

So what is the range of gf(x)?

[patent spruce]

Yess

[serene elk]

yeah mb

[uneven schooner]

assume it is. then $g\circ f(-10)=69$. Is $f$ defined for $-10<0$?

[warm shale]

Annie Maqionde

[wild swallow]

ran(g o f) = g(ran(f))

[patent spruce]

Nope

[patent spruce]

Could you please put it into words?

[uneven schooner]

$g$ is defined for ALL real $x$. So we don't need to worry much abt it. On the other hand, $f$ is defined ONLY for $x\ge0$. Now, If $x<0$, then $g \circ f(x)$ is not defined, since $f$ is not defined, is it? So negatives cannot be in the domain of $g \circ f$. So the domain of $g \circ f$ is just set of all $x$ such that $x\ge 0$ for all $x \in R$

[uneven schooner]

so it cannot be an answer.

[deep kraken]

Words

[wild swallow]

domain not range

[uneven schooner]

sorryyyy

[warm shale]

Annie Maqionde

[patent spruce]

The last sentence doesn’t make sense…

So for any composite function e.g fh(x), domain is domain of h(x)…and range is?

[patent spruce]

So there isn’t an answer to your question?

[vagrant harbor]

the domain becomes h(x) but it also must satisfy the conditions of the original domain of f

the range depends on both the demain of h and f and the range of f

Does this sound right to you?

[patent spruce]

…

[patent spruce]

So there isn’t a rule of thumb for range of composite functions…you just do trial and error

[vagrant harbor]

Let's take an example

[patent spruce]

But if a comprised function is defined…this applies: So for any composite function e.g fh(x), domain is domain of h(x)

[vagrant harbor]

suppose that f(x) has a range of 0 to 5, and that f(x) = 5 iff x = 6

and then suppose another function g whose range is 0 to 5 as well

what's the range of f(g(x))

[patent spruce]

0 to 5?

[vagrant harbor]

Wrong

Because f(x) can't be equal to 5

since that requires g(x) = 6

but g(x) has a range of 0 to 5

btw when I said 0 to 5 I meant [0 ; 5]

Oh I have a better example

[patent spruce]

Inclusive?

[vagrant harbor]

Yes

[patent spruce]

Why not

It’s 0 to 5

[vagrant harbor]

Because for f(x) to be equal to 5, we need x = 6

in our composite function f(g(x)), it becomes g(x) must = 6

but g(x) has a range of 0 to 5, therefore that condition cannot be true

[patent spruce]

Aha

Sounds right

[vagrant harbor]

I have a better example

• Suppose a function f such that f(x) = 3x
• And suppose a function g such that g(x) = cos(x)

What would be the range of f(g(x))

[patent spruce]

3cosx?

[vagrant harbor]

Without writing the function in terms of x

Just think of it logically

[patent spruce]

But you didn’t give me numerical values…

[vagrant harbor]

You don't need them

remember that the range of g(x) becomes the domain

what's the range of cos(x)

[patent spruce]

1 to -1

[vagrant harbor]

Very well

[patent spruce]

[vagrant harbor]

which means

f(x) for all x of [-1 ; 1]

[patent spruce]

We were just having this argument @uneven schooner

[vagrant harbor]

Think of it like this

the range of g(x) gives you the possible values of f, but not all values

For example

[patent spruce]

g(x) is a subset of f(x)?

[vagrant harbor]

In that previous example, g(x) has a range of -1 to 1, but what values can x take?

[patent spruce]

0 to 360 if it’s defined like that

[vagrant harbor]

in other words, R

so the domain of our composite function is R

So any value of x, gives us a value of g(x) which is limited to the range -1 to 1, which we then input to f

[patent spruce]

There’s a range of values of x which we feed to g(x) which we feed to f(x)?

[vagrant harbor]

So, we want x to be able to define g(x), but we also want g(x) to define f(x)

Two conditions:

1. x ∈ Dg (Dg is the domain of g)
2. g(x) ∈ Df (Df is the domain of f)

That's in term of domain

Now in terms of range we have

1. f(p) has a range for all values of p
2. g(x) has a range for all values of x
3. We know that g(x) becomes that p

So we just have to account for that change

It's basically that the range of g(x) defines the possible inputs of f(x), and then those possible inputs define the range of the function

So in our example, since the range of g(x) [ cos(x) ] was [-1;1], then we want to check the range of f(x) in the interval [-1 ;1]

So even though f has a range of R, it can't reach those values, because it can only take values from -1 to 1

Does this make sense?

[obtuse pebble]

@patent spruce Has your question been resolved?

[patent spruce]

Yes
Thank you so much!!

[vagrant harbor]

Was that it?

[patent spruce]

Yup

.close

[obtuse pebble]

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[obtuse pebble]

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[turbid locust]

problerm 2, i have no clue what they want please someone explain like im 10

[latent quiver]

do you know how to plot like

y = x - 1 and y = x^2?

[turbid locust]

i may do but probably im not getting what youre saying

[latent quiver]

the question is asking you to plot inequalities basically. A simpler case might me to visualise how something like y > x would look like

,w plot y > x

[warm shale]

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Requested by lexqa

[latent quiver]

y = x would be the diagonal line at the bottom of there that defines the (lower) boundary of the region

in your case, y = x -1 defines the lower boundary (similar to the plot above) and y = x^2 defines the upper boundary

[turbid locust]

ok i get this part

[latent quiver]

that part is basically the whole idea of the problem; is there anything else that's troubling you about it?

[turbid locust]

i cant solve it

like i need an example of how to solve it

[latent quiver]

ok sure

I'll work out an example problem in Desmos for you to connect to your problem. Imagine I am asking you to draw the region of integration of $0\le x\le1$ and $x^3 \le y \le 2x$

[latent quiver]

you can begin by drawing the vertical lines described by x = 0 and x = 1 as seen

then you can draw the functions y = x^3 and y = 2x

then try to consider the resulting region of the inequality bounds given to you in the problem one by one and intersecting them together. For example, x < 1 would be everything left of the line at x = 1

repeating the process should get you this

[turbid locust]

i kinda get it but not fully

maybe ive been studying for too long

thank you anyway for your help <3

.close

[obtuse pebble]

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[latent quiver]

u should probably take a break then yeah

[turbid locust]

ill prolly take a 30 min nap or smth

[latent quiver]

aight nap well!

[obtuse pebble]

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[clear roost]

i have considered functions where there's an f(h) < 0, a < h < b, and the previous problem (intermediate value theorem relied on finding largest f(x) such that f(x)=0) allowed to find the largest 0 from where the proof was apparent, but i am not sure how to handle edge cases where f(x)>0 for all x, yet there are multiple zeroes besides a and b. How do i handle them in my proof?

[clear roost]

I also thought of using the theorem that if f(x_0)>0, then there exists a \delta such that x_0 - delta < x < x_0+delta implies f(x) > 0, but didn't see how to properly apply it

i hope that was clear 😭

[chrome echo]

Basically, since the graph is continuous, starts at 0, goes above the x-axis at (x_0), and comes back to 0 at the end, it has to first cross up from 0 somewhere before (x_0) (that’s your (c)) and then cross back down to 0 somewhere after (x_0) (that’s your (d)); and because the graph is smooth with no jumps, once it goes above after (c), it stays positive all the way until it reaches (d), so (f(x) > 0) for every point between (c) and (d).

[clear roost]

Yes i get it intuitively, but i'm trying to formalise it

[mortal blade]

What's the previous problem being referred to

[clear roost]

(the theorem being referred to along with it)

[somber spear]

So you have to solve by using formula and properties?

[clear roost]

There are certain properties continuous functions have that should be used to prove this formally

[somber spear]

I forgot all process but you have to proof that function from a to c is -

[clear roost]

why does it have to be negative?

[mortal blade]

Oh oops I missed the ping

[clear roost]

oh, i missed another part of that problem...

yeah uh that seems useful

though it still reffers to f(a) < 0 which isn't guaranteed

[mortal blade]

God this is annoying as fuck to prove

[somber spear]

I am confused

My english hell weak

[mortal blade]

My instinct is a proof by contradiction here

You intuitively know what c and d are right

[clear roost]

mhm

the closest zeroes to x_0

[mortal blade]

Actually nvm a forward proof works with that

[somber spear]

Wait you have to proof c and d so use that intermediate or smthing theorem and find c by using a and x0

[mortal blade]

Yup

[clear roost]

as in, no need to prove by contradiction?

[mortal blade]

Now the value between c and x0 cannot take a negative value

[somber spear]

We have to do this right gang?

[mortal blade]

Cuz if it does, it would hit another zero which would be closer

[grizzled shore]

Something about considering the largest closed interval around f(x₀) such that the entire interval is nonnegative

[mortal blade]

Violating the definition we have used for c

[sly elm]

you can consider the interval [c,x0) since f(x0)>0 you can try proving c is the largest zero ofthe function in [a,x0) ?

[mortal blade]

The largest*?

Smallest would be just the point itself

[grizzled shore]

Yeah whoops

[mortal blade]

Okay I think we have too many cooks here

I'm gonna step back

[grizzled shore]

I just wanted to throw that idea out there cos it’s what I thought of

[sly elm]

sorry me step back as well

[mortal blade]

We can't all step back lmao

[sly elm]

wat th blud

off to you

[mortal blade]

Lmao sure

[sly elm]

🫡

[mortal blade]

@clear roost you still here

[clear roost]

mhm

[mortal blade]

Cool

Let's define c and d explicitly

$c = \max {p \in [a, x_0) \mid f(p) = 0}$

And similarly,

$d = \min {p \in (x_0, b] \mid f(p) = 0}$

[warm shale]

Xavier 🌺

Xavier 🌺

[mortal blade]

Claim is that a ≤ c < x_0 < d ≤ b and d is positive in (c, d)

The first part is clearly true by definition of c and d

[clear roost]

could we just consider
$c = \inf (x_0 - \delta, x_0), f(x) > 0, \forall x \in (x_0 - \delta, x_0)$
if that makes sense

that's terrible tex

[grizzled shore]

I think you need to use inf and sup btw

[warm shale]

Jolii

[mortal blade]

Right yes cuz analysis is the worst

[grizzled shore]

Yeah some weird bump function might have your head otherwise

Or some other weird thing it’s safe to use inf and sup

[mortal blade]

This is what frosst said yes

c and d are the endpoints of the largest interval containing x_0 where f is non negative

Also I think the definition you gave are slightly harder to work with

It's easier to use the ones I sent with inf and sup instead of min and max

[clear roost]

i just think that stuff like this is probably what the textbook expects me to use

[mortal blade]

But the overall proof structure would stay similar enough

What textbook is it

[clear roost]

spivaks calculus

[mortal blade]

Ah I have avoided Spivak to the best of my ability so can't comment lol

But yeah lemme modify my definition to work

[clear roost]

most of the proofs regarding continuous functions and zeroes used properties like "if f(x) is continuous and f(x0)>0 for some x0, then there exists a d suc that for all x such that x0-d < x < x0+d, f(x)>0" and etc

oh wait

no your idea was right

[mortal blade]

$c = \inf { p \mid f(p') > 0$ $ \forall$ $ p' \in (p, x_0)}$

[clear roost]

using this set we can prove that inf of that set is c

[mortal blade]

Yours wasn't wrong either lol

[clear roost]

well i mean yeah i just meant that that would probably be something to use too

[mortal blade]

Oopsie

[warm shale]

Xavier 🌺

[mortal blade]

Yeye

Either way, we don't really care as long as the c we pick works

[clear roost]

well we need to prove that the least upper bound of that set is indeed c

which seems easy

[mortal blade]

That's not what we need to prove though

Wait nvm

Yes

Right lmao this is easier

[clear roost]

alright so

[clear roost]

i'll call the set A
the set clearly isn't null because x0 is in A, and A has an lower bound a, therefore A has a highest lower bound. then let c = inf A. Thus we need to prove that f(c)=0, assume f(c)>0, because f is continuous, there is a $\delta>0$ such that $x \in (c - \delta < x < c+\delta), f(x)>0$. Because c is a highest lowest bound there exists a number x1 such that $c < x1 < c + \delta$, therefore f(x) is positive on the interval $[x_0, x_1]$, but if $x_2$ is a number between $c-\delta$ and $c$, then f(x) is positive on the interval $[x_1, x_2]$, so f(x) is positive on the interval $[x_0, x_2]$ therefore $x_2$ is in A, which contradicts the fact that c is a lowest upper bound

[warm shale]

Jolii

[clear roost]

if $f(c) < 0$, then there exists a $\delta>0$ such that $c - \delta < x < c + \delta, f(x) < 0$, similarly, there exists a number $c - \delta < x_1 <c$ such that $f(x_1)$ is positive on $[x_1, x_0]$ but thats impossible since $f(x_1) < 0$
hence f(c) has to equal to 0

[warm shale]

Jolii

[clear roost]

f(d) is basically the same i wont bother

[clear roost]

Thanks!

.close

[obtuse pebble]

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[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[hardy widget]

<@&268886789983436800>

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[cold dragon]

hello i need help understanding this problem not solving it, im confused becuase i dont get how the relationship between kinetic energy and speed is represented here

[cold dragon]

for example 4*6 is 24 how is that shown on the graph or did i just get it wrong

[median yacht]

K = 1/2(mv²)

[brazen viper]

Because KE = mv^2 / 2, if you have something going at 2v, you get (2v)^2 = 4v^2

this is where the 4 comes from in the explanation.

If we consider 11 m/s vs 22 m/s, we have at 11 our graph reads 3, and then at 22 our graph should read 4 * 3, which is 12. And it does.

@cold dragon does this make sense to you?

[cold dragon]

how did you get 4v from (2v)?

i get the rest

[brazen viper]

(2v)^2 = (2^2 v^2) = 4 v^2

@cold dragon

remember KE = 1/2 mv^2. We are replacing v with 2v, so it becomes 1/2 m (2v)^2. Then we are simplifying to get the KE after is 4 times the KE before.

\begin{align*}
T(v) &= \frac1{2} mv^2 \
T(2v) &= \frac1{2} m(2v)^2 \
&= \frac1{2} m (2^2 v^2) \
&= \frac1{2} m (4v^2) \
&= 4 \cdot \frac1{2} mv^2 \
&= 4 T(v)
\end{align*}

[warm shale]

OmnipotentEntity

[obtuse pebble]

@cold dragon Has your question been resolved?

[obtuse pebble]

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[shy vector]

hi

[shy vector]

can someone explain

step by step

given that y=4x^3 ,find the approximate percentage change in y for a 1percent change in x.

[daring ravine]

you know how when you zoom in to the edge of a circle it looks more like a straight line

you are basically using that straight line as an approximation

[shy vector]

yes

tbh like i don't get it yk

why we divide delta y by 4x cube

[daring ravine]

well do you at least agree that dy/dx is 12x^2

[shy vector]

yea i agree

[daring ravine]

ok cool

if we change x by 1%, it's like inputting 101x/100 instead of 100x/100
so our difference is x/100

[shy vector]

yes

[daring ravine]

to approximate the change, we use the derivative to find the slope of the straight line

[shy vector]

yes

[daring ravine]

the slope is 12x^2 for some x
multiply by the difference in x
12x^2 * x/100
that's the change in y due to the small change in x

[shy vector]

can i ask something like

[daring ravine]

then the new value / old value is the percent change

[shy vector]

why delta y= dy/dx into change in x

[daring ravine]

also this is a terrible way to explain it ngl its confusing

[shy vector]

its ok

[daring ravine]

dy/dx is meaning, "change in y with respect to x"

so if you multiply the rate at which y changes w.r.t x, by x, that is the rate y changes

[boreal cliff]

the way i always see this being done is taking a log on both sides and derivating.

and then pretending that $\Delta x = dx$ if x is small enough

[warm shale]

Περσυ

[shy vector]

thanks a lot 🙏 🙏

[shy vector]

why we divide delta y by 4x cube

i don't rlly understand in that part

[daring ravine]

new value / old value = percentage gained difference

[shy vector]

but isn't the old value dy/dx?

[daring ravine]

oh wait no i meant

part / whole = percentage

the part is the difference in y we just calculated

the whole is the original value

so the difference / original value is the percent gained over the original

[shy vector]

oh i understand it rn thanks a lot🙏

[deep kraken]

why @restive gorge my blud keep spamming heart emojis 😂

[obtuse pebble]

@shy vector Has your question been resolved?

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