help-23

[hoary seal]

ohhh, so were looking for the upper bound

[fathom jewel]

If you had now a sine term, good luck solving for x

[hoary seal]

and then isolate the upper bound only

[fathom jewel]

you isolate x in 1/(2x²)<eps

[fathom jewel]

yes sir

although the last line is wrong

[hoary seal]

my last line is wrong why?

[fathom jewel]

You are not defining M to be x that's silly

The whole idea is M such that x > M

and M is that lower bound you just derived

in other words, you made a wrong assignment of variables

[hoary seal]

there is no upper bound then

since the funct is cont

[fathom jewel]

which func is constant

[hoary seal]

g(x)?

[fathom jewel]

g(x) is not constant, it depends on x here obv

g(x)=1/(2x²) was an upper bound for |f(x)-L|

and M(eps) is a lower bound for x

[hoary seal]

but we used the upper bound to find for the lower bound?

[fathom jewel]

yes

[hoary seal]

we coudve used the lower bound as well

so -1/2x^2

and solved for M

[fathom jewel]

no

we are only interested in upper bounds

for |f(x)-L|

the whole idea is to show that |f(x)-L| becomes small enough, you can like zoom in and there will be still points inside that little window

working with a lower bound of |f(x)-L| doesn't yield anything useful, cause you cant control how small |f(x)-L| would get

[hoary seal]

so was that the answer

just solving for M was our goal

I think my prof wants to prove it

[fathom jewel]

That was scratch work

[hoary seal]

yeah

[fathom jewel]

the good thing is the scratch work already tells you how to start now

[hoary seal]

yes

so choosing values

for M

and epsilon

[fathom jewel]

Only M

epsilon is something arbitrary

imagine it's something given from the outside and your M(eps) has to work with that eps everytime

[hoary seal]

so were not getting the value of epsilon

[fathom jewel]

that's what your proof is about

[hoary seal]

once we prove it

[fathom jewel]

once you prove it, you are saying, give me any eps>0, my M(eps) will handle it always

[hoary seal]

so once we prove it, we can see if it works by plugging in epsilon which will give the corresponding M

[fathom jewel]

you can do that yea

that's why we prove things in the first place

it guarantees results

[hoary seal]

<x?

[fathom jewel]

yes

[hoary seal]

isn’t it x>M

[hoary seal]

Oh it’s the same thing

[fathom jewel]

yes

[hoary seal]

so first step is to prove x>M

howd o I do that

[fathom jewel]

blud

that is given

If x>M then ... is what you prove

You did the scratch work to find a suitable M

[clever tinsel]

you're not interested in the case where x < M anyway, if I may add.

[fathom jewel]

cause that's thankfully never the case with our M

[fathom jewel]

You are skipping steps

If a new reader looks at this they will be like "how?" "what just happened?"

Let eps>0 was good adding that, introducing variables that you intend to use, so the reader isn't confused

From the def. is fine

Then you write a small sentence like choose M = bla such that x > bla

then you start with |f(x)-L| and show every step until you arrive at < eps

using x > bla

[clever tinsel]

do you think a template for this kind of proof would help?

[fathom jewel]

maybe but i wouldn't do that

[clever tinsel]

alright, sorry for interrupting.

[fathom jewel]

all g

I think the philosophy of letting someone fail and thus gain understanding is of more of value than handing them blindly a template to follow

[clever tinsel]

that is a fair point, sorry for not considering that.

[fathom jewel]

Cause a machine can also just follow blindly orders but it cannot analytically think and that is the actual skill you acquire doings math

[clever tinsel]

I agree. sorry once again; I'll step back.

[fathom jewel]

It's okay lol no need to be sorry, OP seems to be gone anyway

just an exchange of ideas

[safe radish]

@hoary seal Has your question been resolved?

[hoary seal]

Oh I’m back

Let epsilon>0 and Choose M = some number > 0 such that |f(x)-L|<epsilon,
Then solve for epsilon once u sub values in
sin(x)/2(sqrt(1/2*epsilon)^2| < epsilon

[clever tinsel]

wait wait, so you're presumably using the Euro sign as epsilon, but then in your expression you have the word epsilon in it.

so is the Euro sign meant to be epsilon or some other variable?

[fathom jewel]

what the blud

[hoary seal]

Epsilon

[clever tinsel]

then what does the word epsilon itself mean in your expression in the last line?

[fathom jewel]

Let €>0 and Choose M = some number > 0 such that |f(x)-L|<€,
such that is wrong, that's what you want to show

[hoary seal]

Edited

[fathom jewel]

You choose M such that x>M cause this is what you will be using in your proof

[hoary seal]

Yes choosing a value greater than x

But we don’t know the value of M , it’s a general function

[fathom jewel]

Then solve for epsilon once u sub values in
sin(x)/2(sqrt(1/2*epsilon)^2| < epsilon
That's also terribly stated

You can't just plug in stuff, it's an inequality, not equation

[fathom jewel]

This is how you structure the proof, and in the middle of inequalities you have to apply x>M

[hoary seal]

Oh then we can plug in the function, limit and make it less than epsilon

Which we already have from our scratch

[fathom jewel]

yes and you are trying to show it's always less epsilon

yes

[hoary seal]

|Sinx/2x^2| < epsilon

[hoary seal]

But u gotta also plug in M into x

To solve for epsilon

[fathom jewel]

Two things: you dont write < epsilon out of nowhere, you derive that result

Second thing, you dont plug in M for x, you bound your expression using x>M, thus you inject epsilon automatically into the inequality and the math sorts itself out

[hoary seal]

Two confusions:

1. how do we derive the result
2. Bound your expressions using x>M? wdym?

[fathom jewel]

1. Algebra (look at your scratch work on how you started out)
2. I can answer that only if you did 1. because only then you will follow up with the why

After you start with |f(x)-L| this is how you continue

If you are at this step we may continue

[hoary seal]

Underneath the line

[fathom jewel]

remove the < eps

this is your goal not your starting point

[fathom jewel]

ok but here is the think how are |f(x)-L| and |sin(x)/(2x²)| related

just two expressions and you didnt derive them in your proof

[hoary seal]

so its all about connections

for proofs

[fathom jewel]

of course it is

else it's just a claim

a proof is just a chain of logical arguments that follow up

[hoary seal]

do we need to show the working out from our scratch

[fathom jewel]

here is the thing, you dont need to show how you got M

but in your actual proof, again you want to show |f(x)-L| < eps

so you start with |f(x)-L|

and from the scratch work you can see how to simplify this expression in the first place

so that we can apply x>M(eps)

and then your prove ends hopefully at < eps

îf you are stuck we can also do it together

in latex

[hoary seal]

so i would need to define whats f(x) and L

[fathom jewel]

no that's given

f(x)=(x²+sin(x))/x² and L=1/2

[hoary seal]

yes so do we write tht in our proof

[fathom jewel]

yes

then next step is to simplify it

[fathom jewel]

basically these are all the steps

i also realized the =< were technically unnecessary but anyway

[hoary seal]

why coudnt have we done the scratchh work in our proof because were doing the same steps

[fathom jewel]

because the scratch work is not a proper proof, you are working backwards

it would be circular since you assume what you want to prove in the first place is true

[hoary seal]

is proving when u need to go from finished product to the start product

[fathom jewel]

it's only for you to discover a possible M

no other way

but sometimes when we dont know how to start, we do backwards thinking in hopes of gaining something that might help

[hoary seal]

I see

[fathom jewel]

If you write these steps with = after you started with |f(x)-L|

[fathom jewel]

here

then this is where things get interesting for the proof, but before that i want you to be ready

[hoary seal]

good so far?

[fathom jewel]

the "which simplifies" step is cocky, it's not trivial, in the exam you should really do the extra step with 1/2=x²/(2x²) but yes good so far, dont forget the =

[hoary seal]

do I find the upper bound now

[fathom jewel]

how can you bound |sin(x)|

[hoary seal]

I mean the same proccess from the scratch work

[fathom jewel]

which means

be concrete

write it mathematically

[hoary seal]

bound sinx by 1 and -1

[fathom jewel]

yes

so |sin(x)| =< ?

[hoary seal]

1

[fathom jewel]

yes

[fathom jewel]

the chain of logical steps

but now comes the part that separates your scratch work from the real proof

from 1/x^2 we need to derive that it is smaller than eps

fortunately we have our assumption

so we know something about x and how to continue bounding it (thanks to our scratch work)

[hoary seal]

do we make thhis an equal sign

[fathom jewel]

we dont

[hoary seal]

from the previous step

[fathom jewel]

cause |sin(x)| is not always equal to 1

from the previous step it's immediate

x^2 cancels and we have |sin(x)| left and for the denom we can omit the abs values cause x^2 is always positive

[hoary seal]

how did u get this. tho

[fathom jewel]

read the msg

|sin(x)| is bounded from above by 1

at some point you cant forget that

[hoary seal]

but when u say |sin(x)| < 1

isnt that same as -1 < |sin(x)| < 1

[fathom jewel]

what

i mean

that's true but you dont need the lower bound

our focus is going up not down

[hoary seal]

if u put abs, it only looks for positive values

right

[clever tinsel]

at some point you have to remember the ultimate goal of what you're trying to do with these steps.
this is why Tairi at one point told you to write what you're given and what you are asked to do.

you cannot keep losing sight of the goal like this, OP.

[fathom jewel]

thanks Hanako Yukari

[clever tinsel]

?

[fathom jewel]

OH SRY

mb

bruh

[clever tinsel]

I would have thought Hanako would not be this cordial with this OP...

but either way, sorry for interrupting; please proceed.

[fathom jewel]

oh you know tairi

[fathom jewel]

that's not the point

we dont care at all about lower bounds right now

[hoary seal]

can u show me like a demo

[fathom jewel]

right now we are climbing up a mountain, so your focus is upwards not looking down

[hoary seal]

with another question

i want to see sample

[fathom jewel]

what sample what?

[hoary seal]

with another question

[fathom jewel]

we are about to enter the most crucial part and you are gonna forfeit lol

fine

[hoary seal]

im asking bc from the start of proof, my understanding is not there yet

[fathom jewel]

ok but then say that sooner not in the middle of something, especially that has been repeated for like 4 times now

[ionic blaze]

that does not matter, in this case it's helping figure out the limit from the fundamentalz since it is squeeze theorem, we know that |sinx| is always non-negative AND is either less than or equal to 1/2x^2, as it approaches infinity, 1/2x^2 becomes zero, so logically, what's next

[hoary seal]

Ill come back to this question

.close

[safe radish]

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[forest osprey]

Hi

[forest osprey]

What does chord refer to

Idk

Does it mean tangent at a point

Yo

[arctic raven]

a chord is just a line containing at least two points of your shape

I would've called it a secant line

[peak estuary]

in this case the points (x,f(x)) and (x+h, f(x+h))

[arctic raven]

it's like how chord is used in the context of circles also

[forest osprey]

Ook

So average gradient across that stretch

Hu

Its gradient at any point as well

[forest osprey]

But this too?

[broken forum]

"average gradient across a stretch" sounds like you are taking a bunch of points between two boundaries and averaging the gradient. It is simply just the gradient line between two points

[forest osprey]

Ok .close

.close uhh

[safe radish]

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[stoic dune]

Did they forget a ... in that sum?

[shrewd frost]

Yess

[stoic dune]

Because if so, I think rationalising the denominator of each term could go pretty hard

[shrewd frost]

Yess this is the general term but what can i do next

[primal gazelle]

Each term is $a_k = \frac{1}{\sqrt{4k-3} + \sqrt{4k-1}} $

[shrewd frost]

I tried telescoping series

[primal gazelle]

do what kaynex said

[primal gazelle]

yes thats the idea

[shrewd frost]

But terms are not cancelling

[primal gazelle]

!show

[safe radish]

Show your work, and if possible, explain where you are stuck.

[timid ridge]

u don't need exact answer for this problem

u just need to know what it's around

[timid ridge]

ok cool

[plucky elk]

,rotate

[timid ridge]

so it's going to be something like
1/2 * (sqrt(n + 2) - sqrt(n))

[stoic dune]

It's almost telescoping. It's like you're missing half the terms you need

[shrewd frost]

Yess

[cunning pasture]

Is it multiple choice?

[shrewd frost]

But how can i do it

[shrewd frost]

Nope

[cunning pasture]

Single answer out of the four options?

[shrewd frost]

Yess

[timid ridge]

that's multiple choice

[shrewd frost]

Oh ya sorry i mistkanely thought he is saying about multiple correct

[cunning pasture]

So single answer?

[shrewd frost]

Yesssss

[cunning pasture]

Then S > 24 cannot be the answer, because then S > 18 would also have to be true.
S < 24 cannot be the answer, because one of S > 18 and S < 18 must also be true.

Now we just decide between C and D.

[shrewd frost]

Yess

The answer given is B

Maybe something is incorrect

[cunning pasture]

If B is correct, C must also be correct.

[shrewd frost]

Do you know how to solve this

[shrewd frost]

Yesss you are absolutely right

[cunning pasture]

Perhaps we have the choose the tighest inequality.

[shrewd frost]

How can i do that??

[rugged arch]

i mean that's what the context suggests

[cunning pasture]

Anyway, is it intended to be done using calculus?

[shrewd frost]

No idea i have not tried that thing

[cunning pasture]

Do you know calculus?

[shrewd frost]

A little bit

[cunning pasture]

ok

because I don't

[shrewd frost]

But if i do with it i will be having (4n-1)^3/2 with the limit from 2500 to 1 with 4n-1 also with same thing

It will be so much bad

Thia question should have a other easy way

@cunning pasture any idea bro how to approach this question?

<@&286206848099549185>

[near sky]

hi

[cunning pasture]

Hi Prof.

[compact flame]

after a bit of work

i come to

$$
S = \sum_{i = 1}^{4999} \frac{\sqrt{2i-1}}{\sqrt{2i+1}} - 4999
$$

[flat frigate]

fluX

[compact flame]

oh wait im wrong

[shrewd frost]

How did you do this? @compact flame

[compact flame]

.

[near sky]

hm.. this would yield something

[compact flame]

try doing smth with the "missing series"

like 1/sqrt(3) + sqrt(5) + 1 /sqrt(7) + sqrt(9)

maybe a combined analysis of this with the series above yields a solution

[compact flame]

note that they sum to form this expression

[cunning pasture]

Maybe this sum can be compared to a telescoping series which is smaller than this sum.

Because root(10000)/4 = 25

And WA gives its value > 25.

[near sky]

a very loose bound would be due to concavity of square root function the sum of this series is less than the original.
if you add them your sum becomes telescoping

this sum will be less than twice the original sum

[cunning pasture]

Maybe use that 1/root(x) > 2(root(x) - root(x-1)).

[safe radish]

@shrewd frost Has your question been resolved?

[shrewd frost]

Can you give me a little bit idea how will i do that

Gimme a small example please

I am getting confused

[cunning pasture]

Ok, I know a solution (searched).
By the last inequality,
1/(4root(x)) > (root(x) - root(x-1))/2
|| 1/(2root(4x)) > (root(x) - root(x-1))/2||
||Take x = m + 1||
1/(2root(4m + 4) > (root(x) - root(x-1))/2
|| 1/(2root(4m+3)) > 1/(2root(4m+4) > (root(x) - root(x-1))/2 ||
So 1/(root(4m+3) + root(4m+3)) > (root(x) - root(x-1))/2
||1/(root(4m+1) + root(4m+3)) > 1/(root(4m+3) + root(4m+3)) > (root(x) - root(x-1))/2||
1/(root(4m+1) + root(4m+3)) > (root(x) - root(x-1))/2
||Now take the sum of both the expressions from m = 0 to m = 2499 or x = 1 to x = 2500||
||We get S > 1/2((root(1) - root(0) + root(2) - root(1) + root(3) - root(2) + ...... + root(2500) - root(2499)) = 1/2(root(2500) - root(0) = 1/2(50) = 25||
So S > 25

[near sky]

denote your sequence with A
general term of A $a_{k} =\frac 1 {\sqrt {4k -1}+ \sqrt{4k -3}} = \sqrt{4k-1} - \sqrt{4k -3}$

now define a new sequence for the missing terms \

$b_{k} = \frac1 {\sqrt{4k-1} + \sqrt{4k+1}} =\frac{\sqrt{4k+1} - \sqrt{4k-1}}2$

now compare kth terms of both you will find $a_{k}> b_{k}$

[flat frigate]

professor paradox

[safe radish]

@shrewd frost Has your question been resolved?

[shrewd frost]

Thank you so muchh @near sky @cunning pasture

[safe radish]

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[mossy ridge]

to guess a random number and being told higher or lower each turn the number is from 1-10000, whats the expected number of guesses to end game at picking randomly but still following higher lower criteria

[stoic saddle]

the best strategy is binary search

[mossy ridge]

Yeah but they said random picking

[stoic saddle]

oh ok

[mossy ridge]

Whilst still following if they are being told its higher

So theyll pick a higher random number

[radiant ice]

can you reword it?

cuz isn’t the answer just halving the higher/lower etc

[mossy ridge]

like can i give an example

[radiant ice]

sure

[mossy ridge]

say i choose say 2000

so you say 1354(random gueess from range 1-10000)

I say higher

You go 9456(random higher guess)

I say lower

You go 2679(random guess but it lies between 1354-9456)

Like this

How many turns is it expected before you get to the number

[radiant ice]

well i can be (n-1) since it’s random

for example

take i say the number 10,000

then you guess 1

then 2

then 3

….

9,999

10,000

[mossy ridge]

Thats not random tho

[stoic saddle]

it may depend on what the number is. like a number at an extreme can probably be guessed in a random no. of guesses faster than a number around the middle

[mossy ridge]

I mean that can happen but very unrealistically

[radiant ice]

well it may look like it isn’t random

i’m just saying like

n-1

[mossy ridge]

Yeah thats what, its like very hard cuz u have to account a general case for every pick too i feel like

[radiant ice]

try with a smaller number

1-5

oh wait it is just N guesses

that’s the most probable

[mossy ridge]

I mean in N guesses youre sure to get it

It says expected

Like as in most probable number of guesses

[radiant ice]

yes

[stoic saddle]

why not start simple, say the number is 10k. so E_10k = 1/n (1+E_0) + 1/n(1+E_1) + 1/n(1+E_2) +.... 1/n(1+E_n-1) when E_k is the expected number of guesses to get the last number for k many numbers.

please excuse me if im wrong

[mossy ridge]

Wait ur cooking

[stoic saddle]

this should be the case that E_n = 1 + 1/n + 1/(n(n-1)) + 1/(n(n-1)(n-2)) + ... + 1/n!

maybe not

[mossy ridge]

It looks like linear search

Like ur picking one and throwing it away

Each miss eliminates a lot of numbers each turn

[stoic saddle]

wait that formula isnt correct

interestingly

E_n = H_n

where H_n is the harmonic sum

[stoic saddle]

so with that I wouldnt expect anything much better for non nice values of the correct number

[mossy ridge]

Isnt linear the worst

[stoic saddle]

i mean in terms of a nice formula

[mossy ridge]

I mean yeah that formula is pretty funny

I mean why do i even have to do ts in cs

Ts pmo gng

Oh damn theres a formula for this

[stoic saddle]

great

so it is connected

[mossy ridge]

Yeah it definitelt looked like so

[stoic saddle]

wait it doesnt depend on the correct number?

or is it averaged over all designated correct numbers

[mossy ridge]

Apparently it doesnt

Its the overall expected value when youre also picking a random number

From the interval

[stoic saddle]

alr thats cool

now prove it lol

[mossy ridge]

AW HELL NA

😭

I leave ts to the math nerd

[stoic saddle]

you have some high level cs for pre uni

[mossy ridge]

Man how were humans smart enough to do all this

Ugh

[stoic saddle]

its very cool

[mossy ridge]

oh sorry for the smurf 😄

[stoic saddle]

damn

[mossy ridge]

I got into uni 2 years ago i should change it

[stoic saddle]

or just keep it ;)

[mossy ridge]

Haha true that

[safe radish]

@mossy ridge Has your question been resolved?

[safe radish]

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[fervent forum]

Hi, I have an exam tomorrow but I needed help with a question on the continued proportion lesson

[fervent forum]

What happened in the third step here? What property exactly was used and how did it change it to that?

[plucky elk]

what exactly is the third step

this?

[fervent forum]

Yes

What happened here?

[plucky elk]

do you have a table of properties of proportion in your book

[fervent forum]

I do but this one was never mentioned
The ones I have:
Interchanging the means
Interchanging the extremes
Addition and subtraction of anticedents and consequents
Inverse of the proportion
Cross multiplication and its inverse

[plucky elk]

show the formulas for all of them

[plucky elk]

yea property 3

[fervent forum]

The application? (K)

[fervent forum]

But it wasn't even used.

[fervent forum]

There's no k

[plucky elk]

yea it was here

[fervent forum]

Huh?

[plucky elk]

why do you need a k in the solution

you're already told they're in proportion

[fervent forum]

Yes it says find x and y
I understand until they were x over y = y over 4
But how did it turn to x + y over y + 4 = y over 4?

[plucky elk]

.

[fervent forum]

Oh, but where did the y over 4 come from?

[plucky elk]

if A = B and A = C then B = C

i'll use capital letters to distuinguish from property 3

[fervent forum]

I don't get it...

[plucky elk]

which one

[plucky elk]

this?

[plucky elk]

or this?

[fervent forum]

Yeah this

I still don't understand where did the 4 over y come from

[fervent forum]

We replaced the k with it?

[plucky elk]

there is no 4 over y

[plucky elk]

in this step

[fervent forum]

I mean y over 4 sorry

[plucky elk]

did you learn this property

[fervent forum]

No

[plucky elk]

i suggest reading and learning it then. it's a very basic algebra fact https://www.cuemath.com/numbers/transitive-property/

maybe your book calls transative something else

[fervent forum]

But what does it have to do with that?

[plucky elk]

they use transitivity here

[plucky elk]

and property 3

[fervent forum]

Ok

[plucky elk]

this is all you need to read to understand

[plucky elk]

this statement

[fervent forum]

OH WAIT
"= one of ratios"
y over 4 is a ratio so we replaced k with it?

It's because the sum of the anticedents over the sum of consequents would equal one of ratios and y over 4 is already one of the ratios

?

[plucky elk]

yes that's what "are in proportion" means

[fervent forum]

Oh I get it now

I feel so dumb lol 😭💔

Thank you for helping me tho

[plucky elk]

"y over 4 is a ratio" is pretty wrong tho

[fervent forum]

It's one of the ratios

?

[plucky elk]

being in proportion has a very specific meaning you should remember. not just y over 4 is a fraction

[fervent forum]

Ok

[plucky elk]

"Important Note 2"

[fervent forum]

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[midnight steeple]

I mean technically this is undefined right

[midnight steeple]

$Cov(X,a) &= E(Xa)-E(X)E(a) \ &= aE(X)-aE(X)\ &= 0$

[flat frigate]

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

[midnight steeple]

$Corr(X,a)=\frac{Cov(X,a)}{\sqrt{(VarX)(Var,a)}} = 0$

[flat frigate]

astral

[midnight steeple]

And Var(a) = 0

I don't think there's any other way to do it?

[opaque berry]

sm1 help with my statistics

[frozen veldt]

yeah I suppose that's why they're uncorrelated in a somewhat pedantic sense

I mean certainly they can't be correlated (because the correlation coefficient is not defined) so therefore one could say they're uncorrelated

[midnight steeple]

lmao cool

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[solar inlet]

Can any1 draw components in here as i cant figure it out

[hard crest]

start with the 6 N one, it'll be like 6 sin alpha north and 6 cos alpha to the right

the 5N one is similar, use the fact that it's alphaº from south (or 90º - alphaº from the right)

[safe radish]

@solar inlet Has your question been resolved?

[solar inlet]

Tysmm

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[white pebble]

guys whats an easy way to figure out what the range of an function is?
for example 2x^(2) + 3x + 1

[white pebble]

i know theres an way with the vertex form

[dull sequoia]

for a quadratic you can find the vertex and direction of the parabola

[white pebble]

but idk

is the vertex an extrempoint?

[dull sequoia]

yeah

a min or max depending on the direction of the parabola

[fathom jewel]

Servus

[white pebble]

is the direction upward open for my example because it is 2x^2 and not -2x^2?

yo

nvm

how do i convert it to the vertex form, i am stuck idk what now

[dull sequoia]

yes

[dull sequoia]

complete the square!

[white pebble]

wym by that

[dull sequoia]

completing the square is how you derive the quadratic formula

have you seen the quadratic formula before?

[white pebble]

i am not sure

maybe

[dull sequoia]

i'd say go google or look at your textbook for the completing the square section

[white pebble]

do u mean this?

[fathom jewel]

Brauchst du Help

[dull sequoia]

yes i do

[white pebble]

tf

[dull sequoia]

that's the quadratic formula

[white pebble]

bruh

[white pebble]

is there no easier way to figure out the range

i dont wanna remember an weird formula

[dull sequoia]

it's not a very weird formula

[fathom jewel]

the vertex is perfect

[dull sequoia]

the trick is to remember how to get to the formula, not the formula directly (although many people are able to remember the formula because it is used often)

[fathom jewel]

for range

[white pebble]

the thing is that its just not worth it to remember just for the range

[fathom jewel]

frosst knows their stuff

[white pebble]

thats like 0.1% of the examen

[white pebble]

where is it used to?

[dull sequoia]

if you're only trying to pass the exam you've got the wrong attitude to learning :/

[white pebble]

i mean i wanna get an good grade

[dull sequoia]

if one wanted to know the roots of a quadratic the formula will give it directly

there are many occasions when one wants to know this

if you study calculus, in integration you will again meet completing the square again

quadratics is quite the backbone to a lot of maths, it will show up all the time

[white pebble]

okay but what the hell is this

tahts not the same i sent or?

[dull sequoia]

that's called completing the square

it will turn out to be the same

[white pebble]

😔

[dull sequoia]

you can do that procedure to ax^2 + bx + c = 0

and then you will find that x = your formula

<@&268886789983436800>

[white pebble]

okay ill try

[dull sequoia]

if you get confused you can google this there are plenty of material online on this topic (completing the square)

[white pebble]

okay ill check

[fathom jewel]

Tim do you also need to study differentiation for the exam

[white pebble]

ableiten?

[fathom jewel]

yes

[white pebble]

ja klar

i mean with examen Abi gk

[fathom jewel]

Because you can also use this method to find the vertex (even though it is overkill)

[white pebble]

i always used the pq formula

[dull sequoia]

is that factorising?

[white pebble]

idk

[fathom jewel]

The pq formula is for the roots, but I guess you can also use the roots to deduce the vertex

[dull sequoia]

do you get something like a(x-b)(x-c)

[white pebble]

oh

wait no

vertex is extrema i got confused

[dull sequoia]

so to be honest there is another way

[white pebble]

for extrema i jsut do f'

[dull sequoia]

well that also works actually

i didn't know if you'd studied calculus yet because typically quadratics is taught long before calculus is introduced

there's also the method of using the axis of symmetry

[white pebble]

ye i maybe had it once but didnt needed it till yet

[fathom jewel]

This quadratic stuff is taught in middle school usually

[dull sequoia]

then you can plug in the point to know where the vertex is

yeah i learned about this in grade 8

[white pebble]

is this the vertex form of the example?

[dull sequoia]

<@&268886789983436800>

[white pebble]

<@&268886789983436800>

do u know why its only b/2 where i marked it?

shouldnt it be bx/2

[dull sequoia]

they've split it up

you see how in the left it's got a b * x rectangle

they cut that rectangle vertically into 2 pieces

1 of the halves got put horizontally under the square

[white pebble]

ye but why is it b/2 and not bx/2

that doesnt makes any sense

[dull sequoia]

The width is b/2

The area would be bx/2

[white pebble]

oh

im dumb

[white pebble]

how did they go from the top to the bottom?

how the hell did they simplified that

[dull sequoia]

Try expanding the bottom

[white pebble]

ye okay that makes sense but how did they do that

[dull sequoia]

They look at it and go oh I can factorise this

(Notice that they’ve constructed this part specifically so that they can factorise it)

[white pebble]

no

[formal grail]

if you have (x+a)^2 it's always x^2 + 2a + a^2

[formal grail]

yes

[white pebble]

wait i jsut realized

can i just cut the middle from x^2 + 2a + a^2 and take the x and a and square them?

does that makes sense?

ye i think i got it

but what do i do if i got functions like these

[fathom jewel]

The first two are just with a=0

3x+1 what is that graphically

[white pebble]

and what would be e ?

[white pebble]

okay i didnt thought abt taht one, thats just infinity

[fathom jewel]

in the first one there is nothin so naturally?
-# also dont you recall x²?

[fathom jewel]

wdym by infinity

[white pebble]

the range of y is R

[fathom jewel]

yes

but the way you said "infinity" wont get you points

[white pebble]

okay i think im stupid

i thought to comlicated, if i try to imagine the functions in an coordiante system then this isnt hard

x^2 is the range of y, R > 0 ?

how do i say that

[fathom jewel]

positive y

but what about 0?

[white pebble]

wym

oh

its the >=

[fathom jewel]

can y=0 be attained

[white pebble]

whats attained

[fathom jewel]

angenommen werden

[white pebble]

hä

[fathom jewel]

Do you know what range means in the first place for a function

[white pebble]

what value y can be?

[fathom jewel]

yes

so my question is can y=0 be?

[white pebble]

yes

[fathom jewel]

for what value?

[white pebble]

x = 0

[fathom jewel]

yes

so you can state the range as [0,oo)

[white pebble]

how exactly would i write it as an answer

W = [0,oo)

?

[fathom jewel]

for example, if you denote W as Wertebereich

[white pebble]

yes

bruh okay thanks for the help

.close

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[rich echo]

67

[rich echo]

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[median vigil]

@rich echo Don't abuse the help channels.

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[honest idol]

yo this question i s regarding critical points on a 3D multi variable surface,

i need help understanding the intuition right when solving the system of equations, i just dont understand the correlation exactly here

[honest idol]

like when i took the partial derivatives, i got infinite tangent lines going in the x direction, and infinite tangent lines going in the y direction

but how does that exactly turn into two lines intersect where the intersection is their solution

(im assuming thats whats going on there)

also how do you tell if this a local minimum or maximum or an absolute minimum / maximum without graphing

[quiet plume]

Can you clarify what you mean by infinite tangent lines?

[quiet plume]

In general one uses the determinant of the Hessian matrix as an analog of the second derivative test for 1-variable functions to classify the critical points locally.

To determine if it’s absolute you need to study the function in more detail there’s isn’t a particular all-encompassing method. (At least not for a domain which isn’t compact)

[honest idol]

okay so

say you have an infinite differentiable 3D surface

and you take its partial derivative with respect to x

what this means is that for points in that space, y stays static and x changes

so for every point (x,y,g(x,y)), youre calculating the instantanious rate of change going from (x,y,g(x,y)) to (x+h, y, (fx+h, y))

ya know

so thats just gonna be a set of infinite parallel tangent lines pointing towards the x direction

and vise versa

[quiet plume]

At every point you get one line which is tangent to the surface with direction $\begin{bmatrix} 1 \ 0 \ f_x(x,y)\end{bmatrix}$ and one line with direction $\begin{bmatrix} 0\ 1 \ f_y(x,y)\end{bmatrix}$.

I wouldn't really think of them as infinitely many tangent lines in the x direction and infinitely many many tangent lines in the y direction though.

[flat frigate]

Azyrashacorki

[safe radish]

@honest idol Has your question been resolved?

[quiet plume]

The goal for critical points is that if you have an extremum somewhere at a point (x,y,f(x,y)), it ought to be the case that those two lines are just horizontal, i.e. they lie in some plane z=constant, i.e. the partials are both 0.

This amounts to solving the system of equations $$\begin{cases} f_x(x,y) = 0\ f_y(x,y) = 0\end{cases}$$.\

In this specific case you showed above, those equations are linear. Generally they may not be, you just need both partials to be 0 simultaneously at that point.

[flat frigate]

Azyrashacorki

[honest idol]

but before this though im having trouble understanding how those tangent lines in the direction of [0 1 f(x,y)]^T and [1 0 f(x,y)]^T correlate to the system of equations

[quiet plume]

Well we set the last coordinates to 0 so that the z-direction is 0, so that those tangent lines lie in a plane z=constant.

[honest idol]

im tryna visualize it

[quiet plume]

There's lots of ways to interpret this.
It's the same as requiring that the tangent plane at that point be horizontal (like parallel to z=0).

[honest idol]

so like, when you take a derivaitve and set it = 0 youre looking at when its horizontal

so that is at least 1 tangent line if a critical point exists

[quiet plume]

Both of the partials need to be 0 for the plane to be horizontal.

[honest idol]

why would you think of a plane?

i mean its just two orthogonal tangent lines

right

[quiet plume]

They're not necessarily orthogonal, but they will be linearly independent, which means they identify a plane, which is the tangent plane.

[safe radish]

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[quiet plume]

You could also interpret it without talking about tangent lines at all.
If f attains some extremum at a point, then it should be at a point where it doesn't change in neither the x or y direction (just like a 1-variable differentiable function has derivative 0 where it has a supremum)

[safe radish]

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[twin prawn]

Hello, dumb question, but what is the value of k here?

[twin prawn]

I'm asking because when I'm to solve, I might need to find the value of K

k

[stoic torrent]

Hello there

[twin prawn]

.close

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[twin prawn]

OPS

.reopen

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✅ Original question: #help-23 message

[twin prawn]

hello

[stoic torrent]

Right so for average rate of change,

You can simply apply change in R / change in T

[twin prawn]

nonono i get how toget the average rate of change and the inst

im asking for the value of k, cause the problem states that k is a constant

[twin prawn]

(f(a) - f(b)) / (a - b)

[stoic torrent]

you cant comment on value of k sadly

[twin prawn]

so just.. retain k as is

[stoic torrent]

Yeah

[forest mountain]

searching up the law gives k as a constant of 5.67*10^(-8)

[stoic torrent]

But idts hes doing it from ohysics point of view

[forest mountain]

yeah

[twin prawn]

so what do i do? ill answer it with variable k? no need to sub k with the actual value?

[stoic torrent]

K isnt a variable and since your a math student, yeah it wouldbe fine to let k as it is

Unless the value of it has been mentioned to you

[twin prawn]

right

[forest mountain]

i would write the answer in terms of k if you don't know it like what yajant mentioned

just like answering in terms of pi or something

[twin prawn]

right

[twin prawn]

so basically this

[forest mountain]

[twin prawn]

bet

thanks guys

.close

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[upbeat swan]

A two-compartment, closed container holds a mixture of 50% distilled water and 50% neutral vegetable oil. The mixture is left undisturbed in the laboratory for several days. During this time, the vegetable oil increasingly settles in the upper part of the container. This process occurs at a decreasing rate; that is, initially a large amount of oil settles in the upper part of the container per unit of time, but over time this process slows down. At the end of the observation period, all the oil has settled in the upper part of the container, and all the distilled water in the lower part. The following diagram represents the container, serves to facilitate understanding, and acts as a legend for the possible solutions.

[nimble wyvern]

!status

[safe radish]

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

[upbeat swan]

What snapshots can be observed at the beginning of the period, at exactly

half the time, and at the end of the period?

I basically need to choose bwtween these options

The option that describes the liquids in the container

So im thinking

It has to be C or E

[nimble wyvern]

yes

[upbeat swan]

Becuase. B is in the wrong direction

A is linear

So its bad

And d is bad

And i was thibking

That

It should be E

Becuase

It says that the vegetable oil rises up

At first fast, then slower and slower

And C is basically the opposite because it rises faster and faster

This is my thought process

Am i wrong?

[slate ingot]

I feel like a mirror image of c would make the most sense

[nimble wyvern]

yeah

[upbeat swan]

Hm i dont follow

Becuase the curve of the upper half needs to increase

[slate ingot]

Thats why i said a mirror image

[upbeat swan]

Becuase more and more vegetable oil is at the top

[slate ingot]

Like, if you interchanged the position of the oil and water, then c would make sense

As it is, neither of the two make a lot of sense to me

[upbeat swan]

Yeah but the others are striaght up wrong

[slate ingot]

Id probably go with e, and then challenge this question afterwards

[upbeat swan]

So it has to be between C and E

[slate ingot]

The question can be incorrect

[upbeat swan]

Ok so youre saying its E

Same as me

Now look at the solution

[slate ingot]

E is the most appropriate answer of the given options

[slate ingot]

I believe the question is incorrect

[upbeat swan]

It makes no sense

[slate ingot]

I think they interchanged the position of oil and water in the middle observation

[upbeat swan]

How is the process slowing down if the curve from C is increasing!

[slate ingot]

It makes sense if you take water on the right, and oil on the left

[upbeat swan]

Water is down and oil is up

[slate ingot]

Yes

[upbeat swan]

And im saying

If it says that the slope of the curve should be decreasing over time( cuz there is less and less oil going up over time)

Then how can it be C if the curve of C has a slope that is increading over time!

Increasing

[slate ingot]

Im saying that C is the correct option, but if and only if The water is on the right, and the oil is on the left, ie, a mirror image
The slope will then be decresing

As it is, neither of the 5 options are correct, unless the examiner decided to troll you and view the vessel from behind as opposed to the front

[upbeat swan]

I dont understand why you think its mirrored

The starting position is 50% 50%

Then

The curve that is above

In the upper half of the container

Goes towards 100% oil, so to the right

And the one in the bottom half goes towards 100% water , so to the left

[upbeat swan]

Look here

For the top half 100% is on the right side and for the bottom half the 100% is on the left side

So it's correct

The oil goes towards the right and the water towards the left

[slate ingot]

Well then, can you explain to me why there is more water in the upper portion in option E, when water is supposed to occupy the lower portion?

[upbeat swan]

Ugh this is frustrating

So my idea was that there are 2 graphs in here

Separated by the middle line

The curve in the upper half describes the percentage of oil in the upper half

And the curve in the lower half describes the percentage of water in the lower half

At first they are both at 50%

So a straight line through the middle

Then the curve describing the oil, so the curve in the upper half goes towards the right

And the one on the bottom goes towards the left

Becuase the oil and water reach 100% over time respectively

So this is exactly what we would have in E

[upbeat swan]

So in my view there is no curvw that shows water in the top half

Only in the bottom half

So i guess we are viewing the problem in 2 fundamentally different ways

[slate ingot]

We are viewing a problem with no correct solutions imo

[upbeat swan]

Well usually these questions are solevd by eliminating the even worse options

Until the least worst remains

[slate ingot]

The curvature on E isnt correct
The curvature on C is correct, but they mirrored the graph

[slate ingot]

This is what it should look like

[slate ingot]

Yes precisely

E is the most appropriate answer, but it isnt the correct answer

[upbeat swan]

Yet the book says its C

[slate ingot]

If this is something that showed up in once of your cram school/coachings tests, you can talk to your mentor about this question
Thats what we would do. And either the question would be cancelled, or it would be stated that they observed it from behind

[slate ingot]

Either they mean that it can be observed from viewing the container from the behind, in which case C is correct
Or the question is erroneous

Not all questions are correct yaknow

[upbeat swan]

Ok so im gonna assume that the answer is E

And that the model solution is wrong

[slate ingot]

Brother the question in itself is wrong

Unless its from a test or something, just strike it out

Because E will not get you any marks if this is a option present

[upbeat swan]

But this makes no sense

[slate ingot]

How so?

[upbeat swan]

At the end the oil has to be at 100%

[slate ingot]

This isnt the end my man

This is the middle part

[upbeat swan]

So how would your end look like

[slate ingot]

Precisely what you said

Oil is upwards, water is downwards

[upbeat swan]

So your end would look like the end from E?

Or like the end from B

[slate ingot]

E

[upbeat swan]

Hm so in the bottom half, your curve is heading towards the rightmost value

And you re saying that in the end, it will reach the leftmost value

?

[slate ingot]

What?

Im saying that oil will be up, and water will be down by the end

[slate ingot]

Here, bottom part is mostly water, and upper part is mostly oil

[upbeat swan]

Ok so what does your graph represent

Whats the x axis and the y axis

[slate ingot]

Honestly, all i meant is that this will be the intermediate curvaturw

[upbeat swan]

Yeah but you gotta respect the x and y axis

They tell you that the x axis is the % of oil and of water

So you know that in the end you will get the end of E

[slate ingot]

I already told you, the book messed that up

[upbeat swan]

Nah

I explained

Why its correct

[slate ingot]

The curvature on E is incorrect

[upbeat swan]

Why

Do you understnad that there are 2 separate graphs?

Maybe it helps

If you cover the bottom part of the contained with your hand

So you just see the top half

Which represents

The oil

Which

[slate ingot]

1. how is there more water above the x axis, and less water below the axis?

[upbeat swan]

Has to go towards 100%

[upbeat swan]

Ok lets look at E

It starts with

50% water and 50% oil

[slate ingot]

Yes

[upbeat swan]

Now lets look at oil only

From the intitial 50%

In the top half of the container

It rises

Towards 100%

Towards the rightmost corner

[slate ingot]

Hang on a second

THATS WHAT THISE GRAPHS MEANT

https://tenor.com/view/sealyx-naked-gun-the-naked-gun-facepalm-facepalm-meme-gif-11743768702765397589

I thought the Area was what we were looking at 😭

Thats totally on me

But then, why wont it be c?

[upbeat swan]

🤣🤣🤣🤣🤣🤣

[upbeat swan]

Because as i said

It says in the question

That

The oil is rising

But it rises slower and slower

So i was thinking

That this mens

Means

That the slope of the curve keeps decreasing

[slate ingot]

Yeah fair enough

[upbeat swan]

At first the slope is big then it slowly oges towards 0

So then i look at C

And I see that

The slope keeps increasing

[slate ingot]

I think i see the mistake

[upbeat swan]

It doesnt decrease

[slate ingot]

Look at the graph for the water portion, the slope is decreasing in c, and increasing in e

The slope for the graph of oil in c looks like its increasing, but its actually decreasing

The reference axis for the oil portion is the top right, not the bottom left

If you get what i mean

[upbeat swan]

Hm

So you re saying

That the slope of C is decreasing

[slate ingot]

Yeah

[upbeat swan]

How

[slate ingot]

Its concave downwards

[upbeat swan]

So we are only looking at the top part

Or?

[slate ingot]

The top part cannot be judged by taking the bottom left as the reference

The reference for the top part is top right

[slate ingot]

Note the discontinuity at the midpoint line

Ygwim

[upbeat swan]

Yeah lets ignore the bottom half

[slate ingot]

Then youll need to take the top right as the origin

[upbeat swan]

The origin of what

Isnt the origin at 50%

In the middle

Thats the start

[slate ingot]

No

Not for determining the slope anyways

[upbeat swan]

🤮

I dislike this qurstion

[slate ingot]

🫂

[upbeat swan]

Ok

Lets just look at the top part

Just pretend that the bottom part doesnt exist

So the oil

[slate ingot]

Why

[upbeat swan]

We look at.

[slate ingot]

Why do you want yo make this harder for yourself

[upbeat swan]

Just like that

Nah just

Ignore the bottom

So

[slate ingot]

If youre going to look at the top part

[upbeat swan]

Oil goes from left to right, 0% to 100%

[slate ingot]

Then flip your phone around and look at thebottom part, thats whatll be more accurate

[upbeat swan]

We can take the line separating the compartments as the x axis

where x is the % of oil in the top part

[slate ingot]

Thats the fallacy

The midpoint is not the origin

This int a standard cartesian plane

[winter whale]

I like your name

[upbeat swan]

Just prwtend it is

[upbeat swan]

The this line to be the x axis

Wait

No

Omg

🤮🤮🤮🤮🤮🤮🤮🤮🤮🤮

https://tenor.com/view/destroying-door-gif-7336940

[slate ingot]

Real

[upbeat swan]

This is what i would do to that container

[terse imp]

heyyy

[slate ingot]

Sir this is #help

-# not that i have the right to say that

[upbeat swan]

Sup

[upbeat swan]

Which one

[terse imp]

How u guys doin

[slate ingot]

So, like, did we come to a conclusion?

[upbeat swan]

We are trying to understand this problem

[terse imp]

what are u guys discussing bout?

[terse imp]

send...

[upbeat swan]

Its above

[terse imp]

ok lemme see

[upbeat swan]

Check the pinned message

[slate ingot]

.

[upbeat swan]

Um

So anyways

How tf do we get a curve

If

[terse imp]

Ah shit fluids

[upbeat swan]

The middle photo

Is supposed to be

A single instant

From the midway of the process

[slate ingot]

Youre misinterpreting the curve

[upbeat swan]

Its not a function

[slate ingot]

Its not

[upbeat swan]

Then wtf is it

[upbeat swan]

Just read the whole problem, you re only supposed to choose between some pictures

[slate ingot]

The vertical distance from the base is literally the point in the vessel

And the horizontal distance is like the mole fration curve in liquid solutions

The more you go to the left, the more water
The more you go to the right, the more oil

You get my point?

[upbeat swan]

Keep in mind that the end photo shows 2 lines

Separate

[upbeat swan]

Yes, but oil is only described in the top part

And water in the bottom

I think

And look

If you look at the container from sideways

,rccw

[upbeat swan]

You literally get

0-100

And 100-0

Idk

[slate ingot]

Nooo 😭
Initially its a homogenous emulsion of oil and water

Then it starts seperating out

[upbeat swan]

I think its like this