help-39

[short drum]

Cross multiplying we get 15cx=100

So cx=100/15=20/3

[rigid geyser]

okay okay

[short drum]

After that we realize bx+xc = bc

Bc=15 because bac is isosceles

[rigid geyser]

wait hang on

i did BA/AC = 10/CX

so then i cross mutliplied

then i got 15CX = 10x10

CX = 6.6666

what did i do wrong

[short drum]

I think that’s right assuming u rounded but 20/3=6.666…..

With infinite 6

[rigid geyser]

oh wait i js didnt put into s>d

haha lol sorry

[short drum]

Yeah it’s alright do u understand it from here?

[rigid geyser]

yes!

[short drum]

That’s great 😃

[rigid geyser]

wait how is BC isos

i thought just BA was 15

and only AX and AX were isos

[short drum]

Bac is isosceles

[rigid geyser]

AC*

oh is it cuz theyre similar

[short drum]

Because both angles a and c are 80

[rigid geyser]

yep yep sorry

[short drum]

U have to prove that they’re both isosceles until u cnan know they are similar

Because u can then use aaa similarity

[rigid geyser]

so now i do BX = BC - XC

yuh i got 25/3

[short drum]

That’s right I think

[rigid geyser]

yess it is

i checked with textbook

thank you so much orange!!!

i understand it so much now

[short drum]

Alright that’s great good luck 😊

[rigid geyser]

.close

[pearl pond]

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[lunar cobalt]

So, I've just written down this rule, but I don't understand it at all. What is it supposed to mean?

||Sub-question. Please, rate the comprehensibility of my hand writing from 1 to 10. Didn't try my best, just wondering||

[toxic lichen]

to factorize an expression, number, or whatever else means to write it as a product of two or more things.

[oak ivy]

does it mean like

[toxic lichen]

and your handwriting is like an 8. solid but could use a bit of aesthetic improvement

[oak ivy]

When we factorise x^2-5x+6

We get (x-3)(x-2)

So if we mutiply it we get the original expression?

[toxic lichen]

@lunar cobalt this isn't really a rule btw more of an explanation for what the word "factor" means

[oak ivy]

Ann what I said is right?

[lunar cobalt]

Could you give me an equation to this?

Because word explanations don't seem to penetrate me 😅

[oak ivy]

I wanna see if what I said was correct or wrong

[oak ivy]

This

[toxic lichen]

???

sorry, but in math you WILL have to deal with words

[oak ivy]

Bruh ann ignoring me

[toxic lichen]

one way or another

[toxic lichen]

yes what you said is right

[oak ivy]

That's al

Just see what equation I sent above @lunar cobalt

[oak ivy]

This

[plush moss]

Wow amazing frfr

[lunar cobalt]

So, it essentially says that

6/3 = 2
2 × 3 = 6

?

Is this what the thing's referring to?

[toxic lichen]

more like just the 2*3=6 bit

the 2 and the 3 are the factors

[plush moss]

It's rewriting the expression as the multiplication of two or more other Expressions

[toxic lichen]

wym "like"

that's exactly what it is

[plush moss]

Mb

I like using like

[oak ivy]

Thanks

[lunar cobalt]

Ah

[plush moss]

@lunar cobalt ask here

[lunar cobalt]

I think I understand

[plush moss]

Don't make another help channel

[lunar cobalt]

So it essentially says "so, you have the number 10. Did you know you could view it as 2 times 5? Well, now you know it"

[vernal jasper]

😂 we always get caught for the improper wording! It’s how we explain things

[plush moss]

Yes

Anyway since you've created https://discord.com/channels/268882317391429632/1018700472988749904

You can either close this channel and use that

Or vice versa

[lunar cobalt]

.close

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[midnight haven]

can someone help me complete this induction proof?

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

@midnight haven Has your question been resolved?

[placid geyser]

If $d = l$, \
$\Sigma_{i=0}^l\frac{i+l-l}{i!}\cdot l! \ =\Sigma_{i=0}^l\frac{i!}{i!}\cdot l! \ =\Sigma_{i=0}^ll! \ = l\ l!$ \
How did you get $\frac{(2l)!}{l}$?

[jolly parrot]

Suika

[midnight haven]

wolfram alpha told me

but why would you substitute d with l?

you know the last one is a 1 and not l right?

[placid geyser]

Didn't you write d = l?

[midnight haven]

no

[midnight haven]

d=1

[placid geyser]

Oh, 1

Lol, the handwriting

So that all is 1?

[midnight haven]

yea

[placid geyser]

Now it makes more sense

[midnight haven]

I am struggling what to do after applying induction hypothesis

[midnight haven]

Or maybe proving this would be simpler? This I got from factoring out d! and then i basically do an induction proof over just the sum to later substituting it back

[placid geyser]

I'm not sure

I'm not experienced in proofs using induction

[midnight haven]

hmm

What does a man have to do to prove this induction

[placid geyser]

The same thing a woman has to do to prove it

[midnight haven]

crazy bro

[vital bronze]

@midnight haven has your problem been resolved if so type .close

[midnight haven]

it hasnt

[vital bronze]

<@&286206848099549185>

[pearl pond]

@midnight haven Has your question been resolved?

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[simple eagle]

The circled part

How can we jus write it like the thing writen with the arrow part

[pearl pond]

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[simple eagle]

.close

[rose cloud]

hello

@rose cloud new here

[pearl pond]

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[simple eagle]

Ive been on this for the last 10 mins can't figure put what im doing wrong

[plush sundial]

whats the orig ques?

[pearl pond]

@simple eagle Has your question been resolved?

[simple eagle]

24

[plush sundial]

,rccw

[oak ivy]

!status

[pearl pond]

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

[compact bison]

the formula seems correct to me, prolly some calculation mistake

[simple eagle]

4

[simple eagle]

Okayu

.close

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[oak ivy]

Is the answer option4 @simple eagle

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[quiet reef]

What is the diffrence of Homoegnous/Non Homogenouse and Linear/Nonlinear Boundary conditions? If im writing a math papper on how fourier series and seperation of variables can be used for the 1d heat equation, what should i pick? Im still very new to this topic

[shrewd mist]

Homogenous B.C.s are equal to zero. So if u(0,t) = 0, u(L,t) = 0, the B.C.s are homogenous. You can solve inhomogenous BCs but its not as straightforward as directly applying a seperation of variables

linear/nonlinear is a vague term.

I would start with solving the (homomogenous) neumann heat equation with no heat generation or absorption. homogenous neumann means the ends are at 0 degrees

[quiet reef]

What would linear and non linear look like ? And could it be applied to the Dirchlet, Neumann and Robin boundaries??

[shrewd mist]

you might have a nonlinear/linear PDE, not BCs

"nonlinear/linear" BCs are a vague term that isn't really important/common

linear PDE means that the function and its derivatives only appear in the first degree, and are not multiplied by each other. so no u^2, (u_x)^2, u_x*u_t, etc

[quiet reef]

Ya its probably that applied to the Neumann Dirchlet and Robin etc

[shrewd mist]

a nonlinearr PDE can have homogenous/inhomogenous bcs, those are two seperate subjects

the heat equation is linear, for example

nonlinear PDEs are very hard to solve analytically, or to prove E&U

Idk what your paper's on but unless you want a fields medal just stick to linear, homogenous PDEs with homogenous BCs. You can add some spice to it by including heat generation/inhomogenous BCs but you should stick to your standard linear PDEs (heat, wave, lapalce, etc(

[pearl pond]

@quiet reef Has your question been resolved?

[pearl pond]

@quiet reef Has your question been resolved?

[quiet reef]

i was thinking of something like this

[quiet reef]

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[pure sail]

hi

[pure sail]

not really a questions specific

but I wanna know, how discrete mathematics is related to computer science

how can discrete maths be useful

[tropic saddle]

computers are all about 1s and 0s

[pure sail]

hmm

and?

isn't computer related to digital logic?

oh

digital logic is from discrete maths

but the only few things that are useful in computers are just like, and, or, not, and basically anything can be made

how does more in depth discrete maths correlate to computer science.

[tropic saddle]

discrete math is such a huge topic. basically anything to do with finite amounts of things is in some way discrete math

[pure sail]

hmm

is discrete maths useful anywhere else

like except for solving riddles

[tropic saddle]

discrete math is not solving riddles

most things computers do is discrete math. from sorting stuff to planning your travel route to encrypting your data

[pure sail]

🤔

sorting stuff

how

is related to discrete maths

cuz till now I only know that there is something called big O for efficiency of sorting algorithms and different type of sorts there is.

how does discrete maths relate

[tropic saddle]

that is discrete math

part of it, anyway

discrete math is such a huge topic

[pure sail]

so basically in discrete maths term

big o is belong to Discrete maths

alright, how will knowing proofs n stuff, help with improving programming in the future

[tropic saddle]

for example it will be good to know that the program actually does what you hope it does

there are ways to prove that it does

[pure sail]

🤔

alright

gotcha

thanks for the ideas

.close

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[dire swift]

Need to simplify this to just a voltage source in series with a resistor using source transformation

I was able to simplifiy a bit but I’m stuck

Please ping if you reply

[pearl pond]

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[pearl pond]

@dire swift Has your question been resolved?

[dire swift]

!close

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[peak prawn]

$\frac{dy}{dx} = -6sin(2x)cos^2(2x)$

[jolly parrot]

Yousif

[peak prawn]

do i use these trig identities now?

[toxic lichen]

\sin and \cos

but also

show the entire question please?

like uncrop

[peak prawn]

b

also

i just realised

im not supposed to derive

💀

[toxic lichen]

ok a and b dont have anything to do w each other but at least you did give the full thing for part b

you need to find the integral of cos^3(2x) and they nudge you in the right direction

$\int (1-\sin^2(2x)) \cos(2x) \dd{x}$ can be done with the substitution $u := \sin(2x)$

[peak prawn]

ya ik but like

[jolly parrot]

Ann

[peak prawn]

u see how it says

'write cos^3(2x) as (1-sin^2(2x))cos(2x)'

how does that work

[jovial kiln]

you just did it

cos^2(2x)*cos(2x)

the first term will just be 1-sin^2(2x)

[peak prawn]

how so?

[jovial kiln]

for a moment, lets consider 2x = k

you get cos^3(k)

so cos^2(k)*cos(k)

hence (1-sin^2(k))*cos(k)

now plug back 2x

[peak prawn]

ahh

okok

that makes sense

now uh

in terms of the integral

do we have to differentiate

and go from there

[jovial kiln]

this

[peak prawn]

$\int (1-u^2)cos(2x) ,dx$

[jolly parrot]

Yousif

[toxic lichen]

\cos

also incomplete substitution

you didnt work out du

[peak prawn]

$\frac{du}{dx} = 2\sin^2(2x)\cos(2x) \ du=2\sin^2(2x)\cos(2x) \cdot dx \ dx = \frac{du}{2\sin^2(2x)\cos(2x)}$

correct?

[jolly parrot]

Yousif

[jovial kiln]

if u = sin(2x) then
du / dx = 2cos(2x)

[peak prawn]

yeah i fixed it now dw

[jovial kiln]

your dx expression is wrong

dx = du/2cos(2x)

[peak prawn]

oh right lol

$\int(1-u^2)cos(2x) ,\frac{du}{2cos(2x)}$

[jolly parrot]

Yousif

[peak prawn]

@jovial kiln

[jovial kiln]

yes

[peak prawn]

now i integrate or?

[jovial kiln]

integrate

[peak prawn]

how

😭

[jovial kiln]

cancel out cos2x

then apply power rule for integrals

[peak prawn]

$\int (u-\frac{u^3}{3}) ,\frac{du}{2}$

[jolly parrot]

Yousif

[peak prawn]

.close

[pearl pond]

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[daring bay]

So i got A(2/1) and B(1/3)

[daring bay]

So AB vector is (-1/2)

but if i wanna calculate the coordinates of the middle of the vector AB

can i just do (-1/2) * 0,5

[rough forge]

sorry bruder

[daring bay]

sorry bro du musst mir nt helfen 😭

ich bin so eine last

[spare lark]

Idt so

You should just do the mean between A and B components

[daring bay]

what so 1,5/2

[spare lark]

Ye

[daring bay]

man weird

and the same if u have 3 axis?

[fathom star]

alternatively OA+1/2*AB works too

[spare lark]

Yes

[daring bay]

but what am i calculating when i do 1/2 * AB

[spare lark]

The vector from A to midpoint of AB

But since A isn't the origin

You have to add OA

Where O is the origin

[daring bay]

hmmm

how is A not the origin

[spare lark]

Definition

A(2,1)

[fathom star]

the origin is always (0,0)

[daring bay]

yeah

but the vector starts at (2,1)

ok wait let me try to imagine it

but how is the origin relevant

u can place vectors anywhere right

[spare lark]

Cuz you're in a space given two basis vector that allow you to have coordinates

[daring bay]

yeah

[spare lark]

I don't get what you mean by u can place vector anywhere

[daring bay]

yeah and when u do that u start at A right

[spare lark]

No "start" is (0,0)

To get your midpoint

You do half of the vector

On this i agree

But since you start at 0,0

You first need to go to A

Which is represented by adding OA

And so you have OA + 1/2AB

[daring bay]

my teacher did

1/2(A+B)

yeah

[spare lark]

In coordinates

[daring bay]

shit

fuck

[spare lark]

Which is the easier way to do

[daring bay]

ill just memorize this

[spare lark]

Think of it as mean of coordinates

[daring bay]

yeah

ngl vectors have been confusing me alot

[fathom star]

3b1b "essence of linear algebra" playlist is really good

[wispy hornet]

to find the coordinates of the midpoint of a line passing through A and B, i would have just done it like this - M (the midpoint) =((x1+x2)/2, (y1+y2)/2) where x1 is the x-coordinate of point A and x2 is the x-coordinate of point B, and similarly, y1 would be the y-coordinate of point A and y2 the y-coordinate of point B.
so vector AB has the endpoints A and B and to find the midpoint you can simply use the coordinates of A and B to find it. however i am not sure if i understood your question correctly, but this would be my guess to find the coordinates of a midpoint.

[daring bay]

ok tysm another question

[rough forge]

was ist deine frage

[spare lark]

Spezi

[daring bay]

punkte sind A(1/-2/1) B(4/-1/6) C(7/-2/k) D(4/-3/3)

ich k so haben

dass sich ein parallelogram bildet

also AB = DC

[rough forge]

ok

wie ist AB definiert und DC

[daring bay]

AB ist 3,1,5

und DC muss dasselbe sein

oder

[rough forge]

ja schreibs mal runter was DC ist

also C-D

[daring bay]

(7,-2,k)-(4,-3,3)

[rough forge]

was ergibt das in einem

[daring bay]

K MUSS 8 sein

gg

[rough forge]

ja

einfach mal sachen runterschreiben

[daring bay]

wie

meinst du allgemein

ja

[spare lark]

System of eqn and equality of vector

[daring bay]

sorry ill talk in english from now on

[rough forge]

your law

[spare lark]

I understand german dw

[daring bay]

😭

[rough forge]

er ist ein franzose

[daring bay]

CD ist der gegenvektor von DC oder

könnte man sich nicht einbilden dass man bei B anfängt

und dann halt BA und CD hat

[spare lark]

Putting a - both sides would change nothing to result yeah

[daring bay]

stellen sie die beiden diagonalen des parallelogramms ABCD als linearkombination der vektoren a und b dar

[rough forge]

ok

[daring bay]

wait man fuck

im so confused

please wait a second

this is how the parallelogrma looks right

[rough forge]

,rccw

[rough forge]

yes

[daring bay]

but

lets say we swtich D and C

then my calculation wouldve gone wrong

[spare lark]

Ofc its not the same figure

[daring bay]

no

like

Here its AB = CD

but that doesnt make sense right

[spare lark]

No

[daring bay]

how

we go from A to B

and drom C to D

[spare lark]

No

From d to c

Look arrow

Oh you switch or not ?

[rough forge]

this goes from C to D

[daring bay]

Yes

so

It does matter where imagine the points ?

[fathom star]

the points have a clear place in space

that is specified by the coordinates

[daring bay]

so how do i know in what corners the points are

[fathom star]

that depends on the specific coordinates

[rough forge]

you know AB = CD wo what you have works

[daring bay]

wdym

[rough forge]

kannst du nochmal die aufgabe senden

[rough forge]

,rccw

[rough forge]

das hast du vorher iwie hergeleitet

[daring bay]

also ich check nicht

woher weiß ich an welchen ecken die jeweiligen punkte sind

[rough forge]

mach eine ungefähre skizze

so sieht das ding ungefähr im raum aus

[daring bay]

aber woher weißt du das

hab ich zeit überhaupt für die skizze

oder egal

[rough forge]

ja ich zähl halt

beschrifte halt die achsen mit zahlen

A heißt 1 entlang x, -2 entlang y und 1 entlang z

dann wird doch B entlang x-Achse und z-Achse weiter weg sein

usw

[daring bay]

ok ok ok

danke

ok nächste aufgabe

[rough forge]

hat 20sek oder so gedauer

[daring bay]

wie mache ich die b)

[rough forge]

übringes da stand a und b nicht A und B sportsfreund

[rough forge]

mal mal eine diagonale

a = AB und b = AD

[rough forge]

a und b spannen diese ebene auf

also existiert eine linearkombination für den roten vektor aus grün und blau

du weißt btw dass AB = DC ist

es ist nur vektoraddition

[daring bay]

ja

[rough forge]

du willst ja entlang blau gehen und dann entlang grün

[daring bay]

ja

[rough forge]

a war blau und b grün

[daring bay]

also ist die linear kombination a+b

[rough forge]

ja genau

[daring bay]

hä ok

[rough forge]

ja hä

überleg doch

[daring bay]

aber wie ist a+b aufeinmal in der mitte

😭

[rough forge]

mitte?

[pearl pond]

@daring bay Has your question been resolved?

[daring bay]

also

wie ensteht der vektor in der mitte

zwischen a und b

[rough forge]

meinst du die c)

den mittelpunkt?

@daring bay

[daring bay]

also die diagonale

[rough forge]

ja was jetzt diagonale oder mittelpunkt

[daring bay]

diagonale

also warum ist a+b die diagonale

wie leitet man das her

[rough forge]

ich habs erklärt gehe jetzt pennen

[daring bay]

aller

ok

ich such

danke

[pearl pond]

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[pearl pond]

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[manic cape]

(5a - 6) - (9a+8)

[manic cape]

My answer was = 4a - 14

Somehow this answer was incorrect on the test and the teacher marked it but didn't tell me what I got wrong

[orchid heath]

Firstly, open the brackets.

[ocean flower]

5a-9a=-4a

[manic cape]

can someone explain

[orchid heath]

$(5a - 6) - (9a + 8) = 5a - 6 - 9a - 8 = 5a - 9a - 6 - 8$

[jolly parrot]

SELVATOR

[manic cape]

I removed the brackets

what did I get wronge

I got = 5a -9a =4a

=-6-8=14

= 4a -14

[orchid heath]

It is 5a - 9a, not 9a - 5a

And 9>5

[manic cape]

yes

[orchid heath]

$0 - 1 = -1$

[jolly parrot]

SELVATOR

[orchid heath]

$1 - 2 = -1$

[jolly parrot]

SELVATOR

[orchid heath]

$5 - 9 = -4$

[jolly parrot]

SELVATOR

[orchid heath]

$5a - 9a = -4a$

[jolly parrot]

SELVATOR

[manic cape]

I got 5a -9a =4a

o i see I didn't put -4a

wair

wait

thats the answer ?

@orchid heath

[orchid heath]

$-4a - 14$

[jolly parrot]

SELVATOR

[orchid heath]

Remember to open brackets and be careful about -

[manic cape]

dam the teacher gave me 0/4

[orchid heath]

-( 5 - 3) = -5 + 3

[manic cape]

its only the sign

I hate this

[orchid heath]

Ehm... sign changes a lot

[manic cape]

wow

[orchid heath]

-5 apples is not 5 apples

[manic cape]

yes

[ocean flower]

what if you had a million dollars vs negative 1 million dollars

[manic cape]

I forgot the sign

[orchid heath]

Be careful next time!

Anymore questions?

[manic cape]

well I know the difference but I forgot

nope thx

[pearl pond]

@manic cape Has your question been resolved?

[pearl pond]

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[torpid schooner]

Problem 28

[torpid schooner]

Am I able to do DCT for both the absolute value part and the non absolute value part? For example I’m comparing the original thing against 1/2n and the both converge to 0

Absolute value of it and the

Non abs

Does that makes sense idk

[versed mica]

this doesn’t matter

if a series converges absolutely then it converges

$\sum a_n \leq \sum |a_n|$

[jolly parrot]

knief

[versed mica]

so if |a_n| converges then so does a_n

[torpid schooner]

Ah I see I’m looking back at the rules

Thank you

.close

[pearl pond]

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[torpid schooner]

.reopen

[pearl pond]

✅

[torpid schooner]

Nvm

.close

[pearl pond]

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[torpid schooner]

Nvm I have a question

.reopen

[pearl pond]

✅

[torpid schooner]

For problem 31, shouldn’t this be absolutely convergent?

Because you’re essentially taking the limit of 1/ln(x) to infinity

Which is 0

[oak sand]

lim -> 0 is not a sufficient condition, but a necessary one

[torpid schooner]

Could you explain?

[oak sand]

assume we have sum(an) and we want to study its convergence

[torpid schooner]

Ok

[oak sand]

if an -> 0, that doesn't mean sum(an) is convergent

but

if sum(an) is convergent, then necessarily an -> 0

[torpid schooner]

Okay, what would be sufficient in this case then? How would I check for its convergence

[oak sand]

there are many ways to do this, do you understand french ?

[torpid schooner]

No sorry

[pearl pond]

@torpid schooner Has your question been resolved?

[plush bramble]

This only proves the terms go to zero, not that the sum of the terms go to zero. They are two different things

[pearl pond]

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[torpid schooner]

Well, for the alternating series test, the series needs to decrease and the limit needs to tend to 0 in order to converge correct? 1/ln(x) decreases and it goes to 0 if the limit gies to inf

[torpid schooner]

But then again 1/n would be divergent

So idk what I’m talking about

Im just saying my thoughts

.close

[pearl pond]

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[spiral pivot]

Because it alternates

That's what you're missing here

[torpid schooner]

Oh I see now

Thank you

Sorry

🙏

[pearl pond]

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[full pine]

help

[full pine]

i do not understand a single thing

all i know is we use hook's law and the fact that the sum of all forces is equal to mass * acceleration

[pearl pond]

@full pine Has your question been resolved?

[novel dagger]

t = 0.224s?

[pearl pond]

@full pine Has your question been resolved?

[full pine]

how

[pearl pond]

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[queen ice]

for question 1 a (i), why is -2<x<-1 a solution?

[vital crescent]

I don't quite understand the stuff you have written on the right. Can you explain what you did there?

[toxic lichen]

well, it looks like he wrote out the sign of each factor separately

but then came up short in putting all that together

[queen ice]

so for each x value that made f’(x) = 12(x+1)(x-2)(x+2) equal to 0, i made three separate piecewise thingos to show like the increasing, decreasing, and stationary points of f(x)

[vital crescent]

yeah i suspect thats where the error arose

[vital crescent]

right, but the casework is showing the behavior of (x-2), (x+2), (x+1) separately, you would have to put those cases together in order to obtain the behavious of f'(x)

[queen ice]

ohh

[vital crescent]

I suggest looking at intervals where behavior changes: say from -infinity to -2, -2 to -1, -1 to 2, and 2 to infinity

[toxic lichen]

Interval | x<-2 | -2<x<-1 | -1<x<2 | 2<x
---------+------+---------+--------+------
x+2      |  -   |    +    |    +   |   +
x+1      |  -   |    -    |    +   |   +
x-2      |  -   |    -    |    -   |   +
---------+------+---------+--------+------
Product  |      |         |        |

make a sign table like this

ive deliberately left the "product" line empty for you to fill

[vital crescent]

that was typed up so fast

[queen ice]

so it’s negative, positive, negative, and positive

ohhhh

[toxic lichen]

it's a very basic ascii table

[queen ice]

so it’s monotonically increasing at -2<x<-1 and x>2

does monotonically increasing mean only increasing between intervals?

[vital crescent]

I don't know if you were instructed to include endpoints like -2<=x<=-1 but that looks right

[vital crescent]

honestly "values of x where f(x) is monotonically increasing" is something I have never heard

I only ever used monotonically increasing as a property on all of the domain

[queen ice]

nah i wasn’t

[vital crescent]

Do you perhaps have the definition for that?

[queen ice]

nah i don’t think so

if u really want, i could show u my notes i took in class and u could see if anything answers ur questions?

[vital crescent]

yeah sure

that particular notation is uncommon i think

[queen ice]

lmk if u can’t read anything

[vital crescent]

yeah that looks fine

I guess your teacher might have meant "increasing" instead of "monotonically increasing"

[queen ice]

what does monotonically increasing mean

[vital crescent]

Adhering to your definition of increasing: Monotonically increasing means that whenever $x<y$, $f(x)<f(y)$

[jolly parrot]

qwertytrewq

[vital crescent]

so you could imagine such a function to always be going upwards

[queen ice]

then here would they mean that at, for example, x = 2, then the function would continue to increase beyond that point?

so basicallt just wherever the function is increasing

[vital crescent]

Well monotone increasing is really a property of the function at every point. It's hard to separate the points out.

But I guess you could say that... but if you use this, notice that when -2<x<-1, the function in your original question does not always increase beyond x.

[queen ice]

oh yea true

so they just mistyped it or smth

[vital crescent]

So i think you should just interpret it as increasing at x

in particular, by your definition, this is just saying "find all point where f'(x)>0"

[queen ice]

yea that makes sense

also

while i have my notes attached

the top right hand image

the one where it talks abt graphs of a function…

when f(x) is increasing, then f’(x) > 0

does that when the function is increasing, the gradient is icnreasing?

or actually

the gradient is positive?

[vital crescent]

it should really be "f(x) is increasing at x, then f'(x)>0"

because in the line above you pointed out "at a particular value x"

[queen ice]

but is it the same thing

like what u said compared to what i wrote in my notes

[vital crescent]

Oh are you asking the following:
if f(x) is monotone increasing, does that mean f'(x)>0 at every point?

[queen ice]

no i’m asking:

if at a particular x value where f(x) is increasing, is the gradient at that x value positive?

i’m just trying to clarify what the sentence actually means

[vital crescent]

I believe that is how you defined it

[queen ice]

ok just making sure

[vital crescent]

you should ask your teacher to be sure

[queen ice]

yea i will

and also

Q1b

can u help me with that

[vital crescent]

For example: you can have $f(x)=\begin{cases} x-1 & x<0\ 0 & x=0\ x+1 &x>0\end{cases}$ and ask whether $f$ is increasing at $0$.

[jolly parrot]

qwertytrewq

[vital crescent]

ok let me take a look

[vital crescent]

Well I suggest you to think about how the function behaves at each point x

pick a couple test points, say -2,-1,0,1,2,3 and try to answer how f(x) behaves

[queen ice]

but i dont have the equation of f(x)

so how do i test

[vital crescent]

but by behaviour I don't mean figuring out what f is

but rather, think about the slope

For example: at x=-2, f'(x) is very very negative, that means that my function must be rapidly decreasing at x=-2

[queen ice]

i thought the tangent would be positive at x=-2

since it’s increasing until x = 1

and then decreasing after that

[vital crescent]

I think you might be getting confused.

increasing at x means that f'(x)>0, but as you see, f'(x) is always negative

[queen ice]

im so confused

[vital crescent]

In general, the more negative f'(x) is, the more the function dips, and conversely the more positive f'(x) is, the more the function increases

[queen ice]

when u say f’(x) > 0, are u saying the gradient is positive or the whole function f’(x) is positive

[vital crescent]

when I say f'(x)>0 I am saying that at some value x, f'(x)>0

[queen ice]

so the gradient is positive at some value x?

[vital crescent]

When the whole function f' is positive, it is usually denoted "f'>0", or "f'(x)>0 for all x"

[queen ice]

ohhh

[vital crescent]

yeah

Let me give a clear and formal definition:

Given a function $f$,
\begin{itemize}
\item we say that $f$ is increasing at some point $x$ when $f'(x)$ exists and $f'(x)>0$
\item we say $f$ is decreasing at some point $x$ when $f'(x)$ exists and $f'(x)<0$
\item we say $f$ is stationary at some point $x$ when $f'(x)$ exists and $f'(x)=0$
\end{itemize}

[queen ice]

ohhh

[jolly parrot]

qwertytrewq

[vital crescent]

you should confirm with your teacher if this is correct, but let us work with this definition for now

[queen ice]

alr sure

so since f’(x) is always less than 0, then f(x) is decreasing?

[vital crescent]

If you want some practice:

Practice: Define the function $f(x)=\begin{cases} x-1 & x<0\ 0 & x=0\ x+1 &x>0\end{cases}$. For which $x$ is $f$ increasing, for which $x$ if $f$ decreasing, and for which $x$ is $f$ stationary?

[vital crescent]

yeah! however, remember that f'(x) is 0 somewhere

[queen ice]

when x=1

[vital crescent]

so it is decreasing except at one point

[jolly parrot]

qwertytrewq

[queen ice]

oh wait so the graph is literally always decreasing excpet (1,-3)

[queen ice]

when x = 0, f is stationary
when x is larger than 0, f is increasing
when x is less than 0, f is decreasing

[vital crescent]

Also note that there is a couple nice things, for nice functions like this:

\begin{itemize}
\item If $f'(x)>0$, then the function will look like it is ``increasing" near $x$ (this is an informal notion of increasing which I have not defined yet)

\item If $f'(x)<0$, then the function will look like it is ``decreasing" near $x$

\item Lastly if $f'(x)=0$, then the function will look ``flat" near $x$.
\end{itemize}

[queen ice]

so if the gradient is less than 0, then the function at a particular x value will be decreasing AND flat?

or at least look like it’s decreasing and flat?

[vital crescent]

well I think you need to try and compute the derivative

[vital crescent]

oh wait mb mistyped

[jolly parrot]

qwertytrewq

[vital crescent]

^this is the correct one

f'(x)=0 -> function will look flat near x

[queen ice]

ah okay

[vital crescent]

here is an example: suppose that the red curve is the function. look at point A and point B

you can see that, if you zooom in, the function looks very flat at A, and B

[queen ice]

yes

[vital crescent]

https://www.desmos.com/calculator/4m3kh4xvbx

you can also check out this function, zoom in at wherever

[vital crescent]

you will see that this seems to be the case

[queen ice]

yes yes

[vital crescent]

when you zoom in at x=0, you realize that the function is increasing. And if you compute the derivative at 0, you will see that the derivative is 1, which is indeed >0

[queen ice]

and so if the derivative is larger than 0, then the function at x=0 is increasing

[vital crescent]

yup, and it looks like it is increasing (intuitively), when you zoom in

[queen ice]

yes

[vital crescent]

Now another fact that we need to know is that how ``big" f'(x) also matters

Do you know how f would behave at x if, say f'(x)=1? How about if f'(x)=6?

[queen ice]

the gradient would be much steeper when f’(x) = 6 compared to f’(x) = 1

[vital crescent]

yeah!

in fact, if f'(x)=6, then when you zoom in very very close on the graph at x=6, it will look like a straight line with slope of 6

[queen ice]

mmmm

[vital crescent]

https://www.desmos.com/calculator/n4gdaym8ar here is an example, try zooming in at (3,9)

[queen ice]

yes

bc the derivative of x^2 is 2x

[vital crescent]

in general, when you zoom into f at point x, it will look like a line with slope f'(x)

[queen ice]

so 3 x 2 = 6

mmmmm

[vital crescent]

you should try graphing a couple of functions and see this for yourself

[queen ice]

i will when i finish this

[vital crescent]

right so given these intuitions, and given the graph of f'(x)

can you figure out how f behaves at some test points?

say, how does f behave at x=-2, or x=2?

[queen ice]

at x = -2, wouln’t the gradient be negative there?

os positive

or*

[vital crescent]

your graph here says its negative

just to be clear we are looking at the original question

tryna apply these key intuitions

[queen ice]

oh wait

correct me if im wrong but

at x = -2

y would be very very negative right?

so that would mean the gradient is also ver very negative?

like y would have a very negative value

and so the gradient would be very steep and negative

[vital crescent]

precisely!

and the y value when x=-2 is precisely the value of f'(x)
(this is the definition of a graph)

[queen ice]

yes

[vital crescent]

Ok

now maybe do a couple other points, once you have a rough idea of how the function behaves, you can sketch a graph

[queen ice]

wouldn’t it just be the same graph as the derivative

but the stationary point is at 1,-3

cus even at x = 2

y is still negative

just less negative compared to x = -2

[vital crescent]

not really, you know that f' is decreasing at almost all the possible x's

but the derivative certainly is increasing from -infinity to 1

[queen ice]

wait huh

so it’s increasing then decreasing?

[vital crescent]

the derivative function, f', is increasing then decreasing.

but the function f, as we previously deduced, is almost always decreasing

[queen ice]

ohh

isn’t it just a straight line down

or like an oblique line

[vital crescent]

but there is a stationary point at x=1, as you discovered

if you have computed a couple values you will figure that ||f is decreasing less and less rapidly, until it becomes flat at x=1, then after 1, it starts to decrease more and more rapidly||

[queen ice]

wait so like a cubic???

[vital crescent]

you could think of it that way, and in this case it is precisely a cubic.

However, there are functions that behaves like what I said, but is not a cubic

[queen ice]

wait but if it’s a cubic, what would its y-int be?

[vital crescent]

you aren't supposed to know that (maybe later you will learn techniques on how to deduce what f is but don't worry about that for now). the question asks only for a sketch of the graph, so as long as you match the general behavior of the graph it's fine

[queen ice]

alrlr

and it doesn’t intercept the x-axis again?

[vital crescent]

if it is always decreasing then yeah, it will intersect x axis exactly once. In this special case, f is not decreasing at x=1. However, you can still prove that f intersect x axis exactly once.

[vital crescent]

looks good

[queen ice]

alrlr

thanks man

learnt a lot

.close

[pearl pond]

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[pure sail]

#help-43 message

[pure sail]

sorry I went inactive in here

I'd like to continue

[toxic lichen]

well, for everybody else's convenience

we left off on the question of whether f_3 is surjective

@pure sail do you think it is or do you think it isnt

[pure sail]

it is surjective

[toxic lichen]

why do you think so

[pure sail]

cuz the input R+ cross R+ always have the output R+ and not complex

[toxic lichen]

what does it mean for a function to be surjective?

[pure sail]

I just know if I proved that it's true

wait

there's meaning

[toxic lichen]

no forget about proof

you need to be able to say the general definition of what it means for a function to be surjective

[pure sail]

every element in R+ output, is associated with at least one R+ cross R+

I think

every element in codomain is associate with at least one element in domain

[toxic lichen]

your wording is still kind of sloppy but better than before

so,

for your function to be surjective, each value in R+ has to be returned at least once by it.

but now think about the fact that f_3(a,b) is ceil(something).

[pure sail]

ye

[toxic lichen]

what do you know about the ceiling function? what kinds of values can it produce as output?

[pure sail]

integers

[toxic lichen]

right

[pure sail]

rounded up

[toxic lichen]

are there real numbers which aren't integers?

[pure sail]

no

[toxic lichen]

oh that's news

[pure sail]

wait

yes

[toxic lichen]

are there positive real numbers which aren't integers?

[pure sail]

0.5

[toxic lichen]

indeed

does your f_3 ever return 0.5

for some input

[pure sail]

no

[toxic lichen]

so is it surjective

[pure sail]

yes

wait

can you explain once more, I'm a little not catching on

[toxic lichen]

for your function to be surjective, each value in R+ has to be returned at least once by it.

but we have just found that 0.5 is never returned, despite being in the codomain.

[pure sail]

ohh

that makes sense

since R+ x R+, it will never return a value of decimal numbers which are included in R+ due to ceiling function

if I understand correctly, this^?

[toxic lichen]

you're trying to over-generalize it

and say more than necessary

[pure sail]

hm

I see

I just had to have found one counter example like 0.5 to prove that the function is not surjective?

ok thx

[pearl pond]

@pure sail Has your question been resolved?

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[queen ice]

for Q2, why is y=0 and x=1 an asymptote

[queen ice]

oh i see why x = 1 is an asymptote

wait y=0 isn’t an asymptote

then how would i draw this??

[pearl pond]

@queen ice Has your question been resolved?

[nimble lily]

(0, 0) is a solution

[queen ice]

yea

[pearl pond]

@queen ice Has your question been resolved?

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[cunning veldt]

assuming they dont switch midway (intersect so the upper becomes the lower and vice versa) you could just plug a value of x in the range of x values we're integrating over for both functions and see which function yields a larger y, thatll be the upper function

[tropic saddle]

you could even just integrate it however you want and if the result is negative just flip the sign

[cunning veldt]

you could do it if they intersected midway too but ypud have to get these intersection points (sey them equal to each other and solve for x) then test out the x values of each region post being split up(split jp by the intersection)

[warm crest]

hi, i wanna know where can i get help for maths

[cunning veldt]

oh right

[cunning veldt]

see #❓how-to-get-help

[pearl pond]

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[urban gale]

number

19. D
20. D

[urban gale]

could anyone verify if my answer choices are correct?

@unborn abyss

[toxic lichen]

!noping

[pearl pond]

Please do not ping individual helpers unprompted.

[urban gale]

oops sorry

[pearl pond]

@urban gale Has your question been resolved?

[pearl pond]

@urban gale Has your question been resolved?

[pearl pond]

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[cursive wraith]

well yes

the inverse of an invertible map remains invertible

functions can be one to one and onto without same domain and codomain

the function f:{1,2,3}->{2,3,4} given by f(x) = x+1

is invertible

F^-1 is from V to U btw

[rare burrow]

$x \neq y$ implies $f(x)\neq f(y)$

[jolly parrot]

00100000

[cursive wraith]

onto: every element of the codomain is the image of someone
one-to-one: the image of a function and its domain are in one to one correspondance: same image means same input

[cursive wraith]

equivalently this is also one-to-one

[rare burrow]

and for all y in the codomain, there exists some x s.t. $f(x)=y$

[jolly parrot]

00100000

[cursive wraith]

same image means same input (different input means different image)

if there is an invertible function in between a domain and codomain

they are indeed seen as the same size

that's how we define "same size" in set theory in fact for the cardinal

oh and also

[rare burrow]

funnily enough, for infinitely large sets, their "size" is actually characterized by bijective functions between them and other sets

[cursive wraith]

I should add that "size" can have different meanings

two vector spaces can have same cardinal size, but not same dimension size

cardinal size is characterized by simple bijections

dimension size is characterized by isomorphisms (linear bijections)

[rare burrow]

haha u have very interesting equivalent definitions. i always think of it as "bijective means same cardinality," but of course the existence of a true inverse implies bijectivity

yeah

[cursive wraith]

yeah in cardinal size that's an example of functions that don't have the same size

one is bigger because you can draw an injection from the first to the second

but there is no bijection

R_n[X] and R^(n+1)

[rare burrow]

the diagonalization argument is just a demonstration that the reals aren't of the same cardinality as the naturals

[cursive wraith]

$\bR_n[X] =$ polynomials of degree $\le n$

[jolly parrot]

rafilou is not not born in 2003

[cursive wraith]

$\bR^{n+1} = {(x_1,...,x_{n+1})\mid x_i\in \bR}$

[jolly parrot]

rafilou is not not born in 2003

[rare burrow]

woah u type latex fast

im starting to think that I'm just slow asf at latex lmaooo

[cursive wraith]

when domain = codomain, a linear map is then called an "endomorphism"

we have a lot of theory regarding endomorphisms

[rare burrow]

love that word 😁

[cursive wraith]

that we don't have with usual linear maps

first interesting thing you can do when domain = codomain

is composition

so for endomorphisms, you can give meaning to T^2, T^n,...

second interesting thing: in finite dimension

if you have a left inverse

then it will also be the right inverse

and vice versa

[rare burrow]

category theory territory 🤤🤤

[rare burrow]

oh, i never knew it was the case for finite dimensions only

very cool

[pearl pond]

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[cursive wraith]

yeah, in the general case, left inverse <=> injective and right inverse <=> surjective or swap I don't remember

so in infinite dimensions, endomorphisms can be injective but not surjective, and vice versa

simple examples in $\bR [X]$ (set of all real polynomials)

[jolly parrot]

rafilou is not not born in 2003

[cursive wraith]

the derivative endomorphism is surjective but not injective

the integral endomorphism $f(P) = \int_0^xP(t)dt$ is injective but not surjective

[jolly parrot]

rafilou is not not born in 2003

[cursive wraith]

anyways channel close Imma stop yappin

[rare burrow]

lmao i appreciate the yap

[pearl pond]

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[swift pilot]

Hey

[swift pilot]

not sur how to find the convergence

do I have to factor out t²?

[cursive wraith]

you want to prove convergence? I guess you can use convergence theorems

$\lim_{x\to \infty} \frac{(\ln x)^\alpha}{x^\beta} = 0$, $\alpha,\beta > 0$

[jolly parrot]

rafilou is not not born in 2003

[cursive wraith]

use this property to find a comparison with a known convergent integral

[swift pilot]

I already try

but I just get that ln(x)^b/x^a =< 1/x

[cursive wraith]

so the problem is

[cursive wraith]

in this property, you can use any value of alpha, beta > 0

what you want to do is completely get rid of the ln(...) thing

but you want to use a beta small enough

so that what remains is enough to get convergence

so say x = 1+t^2 to save trouble

[swift pilot]

okay I see

[cursive wraith]

$\frac{\ln (1+t^2)}{1+t^2} = \frac{\ln(1+t^2)}{(1+t^2)^\beta} \times \frac{1}{(1+t^2)^?}$

[jolly parrot]

rafilou is not not born in 2003

[cursive wraith]

pick some appropriate beta

and everything will be fine

[swift pilot]

we can take beta = 1/4?

[cursive wraith]

sure

any beta < 1/2 works

[swift pilot]

yes

okay and it exists a moment where ln(1+t²)/(1+t²)^beta =< 1

and I just have to multiply by 1/(1+t²)^smth and it proves the convergence

right?

[cursive wraith]

yeah it proves the convergence of the integral on [that moment,+infinity)

[swift pilot]

yes but there's no problem at 0 right?

so it proves the convergence

thank you!

.close

[pearl pond]

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[pearl pond]

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[plucky seal]

Since the function f(x) = Sqrt(x^2 - (a + 1) - x + 9) defined on real numbers, what is the largest range of values of a?

[plucky seal]

the solution said since the inside is positive the delta must be lower than 0 nd i dont understand why?

[oak ivy]

$f(x)=\sqrt{x^2-(a+1)x+9}$

[jolly parrot]

devthemasked

[oak ivy]

Is this the question?

[plucky seal]

yes

that is

[autumn fossil]

They said that the function should be defined on all real numbers

but square roots are only defined for non-negative real numbers

e.g. sqrt(-1) is undefined

so we want that inside of the square root to be non-negative, otherwise the square root would be undefined

that means we want x^2 - (a+1)x + 9 to be positive

and it has to be positive for all values of x

*non-negative

[oak ivy]

positive or 0

[autumn fossil]

yeah, same thing as non-negative

that means that it has to have at most one root

[plucky seal]

I understand that what i dont understand is why b^2-4ac must be lower than 0

[autumn fossil]

because if it had 2 roots, it would be negative between those 2 roots

[plucky seal]

ah ic

[autumn fossil]

yeah, just moving towards that

so it must have either 0 or 1 root

[plucky seal]

yeah makes perfect sense

[autumn fossil]

and that happens when the discriminant is <= 0

[oak ivy]

When D<0 aren't the roots imaginary?

[autumn fossil]

Yeah, they are, but presumably we are working in reals

and if we are working in reals, we might as well say there are no roots

[plucky seal]

What would happen if we werent working in reels and instead in complex what would change from this answer?

[autumn fossil]

in complex numbers, square root is always defined

so a could be anything

but the question itself mentions we're working in reals

[plucky seal]

even values which makes the b^2-4ac positive?

[autumn fossil]

Yeah, sure

sqrt(25) = 5

that wouldnt change

complex numbers dont exclude reals

[plucky seal]

Gotcha