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[golden pebble]

hi how to do b)?

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[soft badge]

Hi, all i have so far is that I have written out S1, 2, 3 and 4 but cant find the pattern

[trim thicket]

isnt it an AGP

[soft badge]

sorry what does that mean

I have just started the unit on proofs

[trim thicket]

oh

[soft badge]

oh u mean arithmetic geometric progression?

yeah i think so

[trim thicket]

to find the sum u multiply this seriess by 2(common ratio) and subtract this series from the original seriess

[trim thicket]

yes

[soft badge]

wouldn't the common ratio be 4

[trim thicket]

basically each term in an AGP is product of the corresponding ap and gp term
and for an agp the common ration is just the common ratio of the gp

[soft badge]

so in this case which terms would be the ap and gp terms?

[trim thicket]

2^i-1 is the gp
i is the ap

[soft badge]

oh ok thanks

[soft badge]

wouldn't this just result in the same value as the original series

and how would i find the conjecture from the sum regardless?

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[untold turtle]

This is an example problem i made for a task, I have solved A. already, how do i solve for B (minimum) now?

You are coding a RPG game where, 1 gold = 1 defensive level/1 offensive level, both for the enemies and for the players.
Overall boss strength can be calculated by multiplying the defensive level and the offensive level.
Overall boss strength also changes depending on the difficulty the players have chosen.
The main boss located midway through the game has 56 gold to allocate to it's stats.

A. Compute for the highest possible overall boss strength for the hardest difficulty.
B. Compute for the lowest possible overall boss strength for the easiest difficulty.

[wise dagger]

try graph y=56x-x²
and if x,y≥0, try to read the y values on the curve with x,y≥0

[untold turtle]

sorry, i dont quite understand, its my first month on basic calculus

[untold turtle]

how do i go about doing that ?

[wise dagger]

for example

,w plot y=56x-x², 0<x<56

[solid kiln]

Results provided by WolframAlpha

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

Requested by biscuityxd

[wise dagger]

like this?

[untold turtle]

oh so i'll have to read the graphs?

[untold turtle]

is there a way similar to this ?

the picture i meant

[wise dagger]

yes of course, you need to set bounds to the equation

eg as i have suggested, x, y≥0 right?

i mean

eg x , y and 56x-x² must be ≥0

then since it's first increasing then decreasing

it will reach minimum before and after the max

which in this case is 0

where 0offense and 56 defense

vice versa

[untold turtle]

ooh thats a mistake on my problem

should i add a thing saying x , y ≠ 0

the prof would see how 0 offense and 56 defence adding up to 0 strength makes no sense

or does it?

[wise dagger]

aslo, must x,y must be positive integers?

or it can have negative strength too if you want

[untold turtle]

yes, i also dont know how to add that

[wise dagger]

you can just add x>0 and y>0

[untold turtle]

alright

[wise dagger]

for integers x and y

[untold turtle]

also, how do i say that x and y should be whole numbers as well in the problem

do i just say that as it is or is there some sort of symbol i can use

[wise dagger]

well, you can use

$x,y \in \bZ^+$

[solid kiln]

Biscuity

[wise dagger]

but if you find it too fancy

you can just stick with x,y are whole numbers with x,y >0

[untold turtle]

alright i think i can add this and then the fancy stuff afterwards in a parenthesis

[wise dagger]

sure

now that you have your construction done

we can go back to "since 56x-x² is increasing before max and decreasing after max"

then the number for 56x-x² will decrease when it is further away from x-value when max

hence we find the furthest away from x-value when max, which is >0 and where x is a whole number

which is when x=1 or 55

[untold turtle]

minimum numbers should be 1 and 55

[wise dagger]

the minimum numbers are 1 and 55, and the minimum strength is 1×55 which is 55

[untold turtle]

about that, my class has not been taught about anything visualising for calculus, we've been going purely on solving

[wise dagger]

interesting

[untold turtle]

odd result of teaching calculus way too early haha

[wise dagger]

i always find visualising easier 😛

but that will restrict me from understanding high dimensions

[untold turtle]

i really wanna learn visualising but the prof never encouraged it

[wise dagger]

like 4-D, 5-D coordinates systems

[wise dagger]

true, it's not really good to understand it too soon as you might miss the full picture

but for this specific question, visualisation would help a lot

[untold turtle]

do you perhaps know how i should go about solving for it without visualisation?

im required to pass it with solution on paper

which was easy enough for maximising, but i dont understand how to get the solution for minimising that much

[wise dagger]

56x-x²
=-(x-28)²+28²

vertex =(28,28²)

sorry, typo

[untold turtle]

if i basically equated for x and y with chain rule for maximise, would it be similar for minimise then?

[wise dagger]

to see, if there's only one local max in the graph, minimizing can be hard

you need to restrict your domain to suitable zone

and then find the global min on that domain

which for a "quadratic function, it's easy to see

but generally it's not

[wise dagger]

this completing square i used is to find the vertex of the quadratic

and then of course you can use differentiation to find max when 56-2x=0, x=28

but after that

you can state that
56-2x<0 when x>28
and
56-2x>0 when x<28

now with that, you can say
56x-x² is decreasing after 28
and
56x-x² is increasing until 28

[untold turtle]

ohh

[wise dagger]

next, you can construct
limit from x to +infinity 56x-x² is -infinity
and
limit from x to -infinity 56x-x² is -infinity

[untold turtle]

wouldnt x to -inf = 0?

nvm

[wise dagger]

so that we will know
56x-x² grows from -infinity to 28 and then get back to -infinity

after all the constuctions

we can safely say that further away from 28, the 56x-x² will be smaller

[untold turtle]

ok i think i can try to express that on paper ?

[wise dagger]

secondly, to find where it reaches 0 helps you too

[wise dagger]

of course

56x-x²=-x(x-56)

so 56x-x²=0 when x=0 and x=56

[untold turtle]

thats looking for familiar

[wise dagger]

factorizing

[untold turtle]

OH yeah i remember that

[wise dagger]

numbers before 0 and after 56,
56x-x²<0

so we would consider numbers between 0 and 56

[wise dagger]

please try,hope that helps!

[untold turtle]

would it be a cop out to put on paper how 0 and 56 equates to 0 so the numbers beside them should be the minimum?

[wise dagger]

first construct (for ease of writing) is let f(x)=56x-x²

[wise dagger]

yea, you'll have to make sure the numbers beside them are minimum using f'(x)

usually making a table would help

[untold turtle]

that was my first idea, but apparently, that would kind of break the point of trying to find a way to solve it for coding situations

[wise dagger]

, rotate

[wise dagger]

there are still some remaining terms can't be shown, oh well

[untold turtle]

oh wow thats a useful bot

[untold turtle]

this is for f(x) = 56x-x², yes?

[wise dagger]

yep

[untold turtle]

i think i might be hallucinating random numbers

how come i compute -56 if x=28

[restive goblet]

whats the question gusys?

[untold turtle]

and not 28²

[untold turtle]

its this

[wise dagger]

,calc 56(28)-28^2

[solid kiln]

Result:

784

[wise dagger]

,calc 28^2

[solid kiln]

Result:

784

[wise dagger]

dunno lol

[untold turtle]

huh

oh goddamn i used 26 instead of 56

[wise dagger]

lol

[untold turtle]

man this paper is so rippable right now

the initial task given to us was to make an optimisation problem and solve it using differential calculus, twist is, we were never taught either and this is a task given to us for us to be forced to learn it in our own way

[wise dagger]

that's sad

[wise dagger]

but i think you've done well

[untold turtle]

seriously, i might as well have payed myself the tuition fee

[untold turtle]

thanks, ive been at this for so long, at least im making progress

[wise dagger]

take a good rest before continuing!

Cheers!

[untold turtle]

thats a good idea, i didnt think of stopping, ill get rest for now, cheers

thanks for the help

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[limpid ore]

how would i do this?

i did x = y, and i got 1/10 for the limit, but idk how to find the other limit

[lusty delta]

the multiple paths approach is used to prove a limit doesn't converge, if this is what you mean by "find the other limit"

you can consider ||y = 2x|| or anything similar

[limpid ore]

ohh

so that'll be 2/13 != 1/10

so DNE

makes sense

what about this one then

y = x^2 gets me 1/9

[lusty delta]

you could

[limpid ore]

ok yeah

wait so

if i do 4y = x^2

numerator is gonna be y^2e^y

denominator is 16y^2 + 8y^2 = 24y^2

so it'll be 1/24

which != 1/9 so its DNE

is that reasoning right?

[lusty delta]

numerator would be 4y^2 but yes

[limpid ore]

oh yeah so 4/24

but still not equal to 1/9

ok that makese more sense now just adding ocefficients

ty!

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[modern ferry]

could someone help me with this "The matrix A10×4 has 4 linearly independent column vectors from R
10. Can one claim that any of the matrices
Are A^T A and AA^T surely invertible, or, alternatively, singular (ie, lacking an inverse)?
Hint: B^−1 does not exist if Bx = 0 for some x ̸= 0. In that case x is also x^T Bx = 0. Interpret then B = A^T A or AA^T."

[granite cove]

This below is just the question?

[modern ferry]

yes

[wraith hinge]

If $ B = A^T A$, then $x^T B x = 0 \iff x^T A^T A x = 0 \iff (Ax)^T Ax = 0$

[solid kiln]

A human being

[modern ferry]

hmm okay

what does that tell us exactly?

[wraith hinge]

Since $Ax \in \mathbb{R}^{10}$, $B^{-1}$ does not exist iff the norm of a non zero vector in $Im(A)$ is zero

[solid kiln]

A human being

[wraith hinge]

which is never the case

I think the solution is along these lines

tbh I'm doing it live, so I'm not sure

[modern ferry]

hmm okay makes sense but is there a another way to prove that it is not invereble?

[wraith hinge]

I'm talking non sense

My proof is not valid

[modern ferry]

haha oh

[wraith hinge]

Well I'd have to think about it, sorry

[modern ferry]

i asked someone else about this question another day and they started talking about the size of tha matrix that it was 10x10 and 4x4 but they never finished explaning

but i dont know where they were going with that

[wraith hinge]

What you can tell from the size is the matrix is that the rank of the matrix is less than the dimension of the end space, here being R¹⁰

But I'm not sure this implies anything on the rank of $A^T A$

[solid kiln]

A human being

[modern ferry]

hmm no maybe not

is it maybe becuse A^T A is a 4x4 and if it is less then 4 it means tht it is not inverable

[wraith hinge]

That's indeed correct, but it might be 4, and then you cannot say anything a priori about A^T A

[modern ferry]

hmm yes you are right

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[sharp thistle]

How do I find the ACD angle(AD=DO=BO=OC, && AC=BC, ACB>90)

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[pallid seal]

I think its 22.5

Its been a while since I did geometry

,rotate

[pallid seal]

@sharp thistle

[sharp thistle]

hmmm

I took a paranetric model in autodesk

and by using the dimensiosn it gives

it is clear thirty

I just gave it the similarities of sides

Your looks also good

So I will try and figure out which is the correct one

[pallid seal]

I am probably wrong to assume three equal sides would imply a rhombus

Mb

[sharp thistle]

yeah they dont seem to

but thanks for the idea

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[normal escarp]

How do I do this?

Q 2

I find derivative then it gives me 2x + 1/x = 3 but that isnt a quadratic equation so I only get one value, what u thought is that the points are opposite for second point but how would I get that?

I got (1,-1) for first set of coordinates

[wanton rune]

what happens when you multiply 2x+1/x=3 by x on both sides

[normal escarp]

Nvm i multiplied the bottom x under faction only on there forgot to do on 2x sorry lol

[wanton rune]

its good

[normal escarp]

Thanks

[simple haven]

!occupied @steel wedge

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[wraith hinge]

I need help with something

[wraith hinge]

if we have a degree such as 210 on the unit circle, the x coordinate would be negative root 3 over 2.

Would that be because since the reference angle of 210 is 30, so the if we drew a triangle at 210 degrees then when we solve, the cos of 30 is root 3 over 2 and since we are in the 3rd quadrant the x axis will be negative radius

so the root 3 over 2 value becomes negative

correct?

[spiral socket]

210 is $\pi+\frac{\pi}{6}$

[solid kiln]

dqvidutzul

[spiral socket]

do you know the cos of a sum?

the formula

[wraith hinge]

sum identity?

[spiral socket]

yes

cos(a+b)

a-b sorry

no wait

a+b

[wraith hinge]

what

[wraith hinge]

is this correct

[spiral socket]

you get $\cos(\pi)\cos(\frac{\pi}{6}) - \sin(\pi)\sin(\frac{\pi}{6})$

[solid kiln]

dqvidutzul

[wraith hinge]

can u write this in degrees so i dont have to convert it in my head

😂

[wraith hinge]

is this correct

[spiral socket]

cos(180)cos(30) - sin(180)sin(30)

[spiral socket]

yes

[wraith hinge]

like is that the reason why the x coordinate would be negative root 3 over 2

interesting

why ru explaning using the sum identity tho

[spiral socket]

i dont really understand the triangle ideology

[wraith hinge]

?

[spiral socket]

thats how you calculate the cos of bigger angles

[wraith hinge]

what ideology

[wraith hinge]

yes correct

i usually only use those identities for angles like above 360

[spiral socket]

i meant analogy, bad english

[fervent vortex]

the one you mentioned in the beginning

[wraith hinge]

nah dw bro english isnt my first language as well

[wraith hinge]

?

[fervent vortex]

the triangle one

where you talked about making a triangle with 210 deg which doesnt really work

[spiral socket]

anyways this is -1*sqrt(3)/2 - 0

which is just -sqrt(3)/2

[wraith hinge]

couldnt u drawa line from the blue 210 degree point and form a right angle triangle?

u could just do that and find out the angle inside which is the reference angle 30, from there u would already know the cos(x) and sin (y)

u would simply take into account the quadrant ur in by that point

[fervent vortex]

yeah yeah now I understand what you meant. I was sitting wondering where you'd get a triangle with a 210 deg angle lol

[wraith hinge]

ye lmao

well it wouldnt be a perfefct triangle but just an estimation so it would make more sense yk?

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[terse haven]

I need help understanding what the percentages are

[chrome root]

you need to use the inclusion-exclusion principle to figure out how many were surveyed

[terse haven]

i got 444 people

[terse haven]

is 444 correct?

[chrome root]

no, I read it again and it's not giving you all the intersections directly

so it gives you number of tablet and bluray owners who don't have cellphones

so that would be (T ∩ B ) \ C

use that

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[glossy crest]

Looking for how to start this one, seems like maybe the goal is to turn it into a double integral and then sketch?

I cant find any practice problems online or in the textbook that isnt sketching based on double integral

[neon dirge]

'sketch' as in 2D twice?

or sketch in 3D

[glossy crest]

3d

the strategy we learned in class was sketch the region in xy plane, then sketch the plane in 3d and then combine the two for the final sketch

something like this

then combined like that

i assume the goal is to get something like that for this problem but idk how to do it

[neon dirge]

careful, you want the area under the graph

so the cross section of your solid would look like

[glossy crest]

yeah so that is literally my professors work not mine

[neon dirge]

ah k, at least in this case you can view it simplified by seeing the 1/2 as the length in the Z-direction

and the above plots X & Y

[neon dirge]

meaning an object with this cross section and height 1/2 would fulfill the task

[glossy crest]

so basically

just gotta draw a solid with that cross section

height 1/2

im a bit confused

the integral is bounded from 0 to 1 in y

and also confused why the fact a 1/2 is there would mean the height is 1/2

.close

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[somber girder]

I'm not sure where to start

do I square the entire thing?

[somber girder]

I tried squaring the entire thing

and treating it as if

it was (a+b)^2

but i don't think thats correct

[orchid wagon]

try conjugate

[somber girder]

i'm not sure what that is

[orchid wagon]

$\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$

[solid kiln]

FungusDesu MSC2020 34A05

[somber girder]

so its similar to rationalising?

[orchid wagon]

yes

[somber girder]

so in my case do i do it for the entire thing

?

[orchid wagon]

yes

[somber girder]

oops

√4 - √7 = (4 - 7)/√4 + √7

is it like this?

[orchid wagon]

wrong

[somber girder]

oh wha

[orchid wagon]

you do conjugate for 2 outer roots first

not 4 and 7

[somber girder]

ohhhh

so both big roots?

[orchid wagon]

yes

[somber girder]

wait

so a is the lhs and b is the rhs

[orchid wagon]

theres no equal sign

wdym lhs rhs

[somber girder]

like a is the left part of the orignal question

and b is the right part of the original equation

[orchid wagon]

yes

[somber girder]

and one more thing

do i ignore the minus sign

?

or does it go with b

[orchid wagon]

prettysure go with

[somber girder]

okok

tysm broski

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[wraith hinge]

how is this done?

if the answer is 0, how does lim x-> -3 f(x) / lim x -> -3 (x + 3) equal 7 when 0/anything shouldn't give 7?

[delicate bobcat]

lim x-> -3 f(x) = 0 doesn't automatically mean that it's 0 in the top when doing lim x->-3 f(x)/(x+3)

consider e.g. f(x) = (x+3)(x+10)

[wraith hinge]

ohhh yeah, u're right

how do i do it ; - ;

[delicate bobcat]

A, B, D have no sense

because then limit wouldn't exist

[wraith hinge]

makes sense

how would i know that it's not e

cus it'd be a weird function?

[delicate bobcat]

mhm

It could be e.g. 1/(x+3)^2 then

as f(x)

but when divided by (x+3) would give odd power

[left oriole]

oops wait, i misread haha, ignore

i'll delete the above to avoid confusion

[wraith hinge]

thank you guys

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[cunning inlet]

I was tasked to do an equation using Cramer's rule, here's the equation
X + 3Y + 2Z = 0
2X + 7Y + 7Z = 2
2X + 5Y + 2Z = -1

[cunning inlet]

When I try to input the 1, 1, 6 it doesn't fit the equation
What did I do wrong

[lament reef]

where did you find det(A2)? when you replaced the first column with your b vector, you calculated det(A1), but I don't see det(A2) anywhere

[cunning inlet]

O damn, I forgot about A2, mb I'm tripping

In my defense it's like 6 am
Also what about the A3?

[lament reef]

det(A3) is incorrect, i think you mixed up which row you're doing cofactor expansion on in the middle of the calculation

idk why you have a zero there

[cunning inlet]

0 is the entry of a13

So it's easier

[lament reef]

yeah but then you used a21 and a31, so you should've started with a11

[cunning inlet]

Wait really

Aren't you able to pick determinants from anywhere

[lament reef]

sorta, anytime you do cofactor expansion you need to choose a row or column to do it with

you can't pick any element

[cunning inlet]

So if I'm going with row 1 it's just row 1 I can't just switch to row 3 col 2 so suddenly?

[lament reef]

yeah exactly, you gotta do it all with the same row or column

[cunning inlet]

Or collumn

Oooooo
Holy, that cleared up so much

Let me do it again and I'll update you on it

[lament reef]

sounds good lol

[cunning inlet]

YOOOOO I FOUND IT

:D
Thank you so much, you've been a big help
Cleared up a very important thing ngl

[lament reef]

niceeee >:) happy to help 🫡

[cunning inlet]

🫡

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[rich stag]

how to restrict domain

[mortal dune]

Is that desmos?

[rich stag]

Fr

[mortal dune]

Do this:

$\frac{120x}{100-x} {{0 \le x < 100}}$

[solid kiln]

Someone

[mortal dune]

bruh

ok just do

{0 ≤ x < 100}

no comma and with curly braces @rich stag

[rich stag]

So cool

Plz help with B

[rich stag]

Umm

I thought like it meant like

It couldnt get past 100% but like I think thats like about the restrictions not the asymptote

Idk wat the asymptote mean

It gets expensive when really close to 100%

We can never truly get 100%

@scarlet escarp

[scarlet escarp]

i think that is fine

it gets really expensive and we can't reach it

[rich stag]

Light work

Ty

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[echo jasper]

hi

[opal owl]

Question 16

[echo jasper]

oop

How is the domain wrong here

oh ik bracket for 6

but it will still be wrong somehow

like does 6>/=x mean it goes to infinity

[scarlet escarp]

what happens if you plug in 10

look at the numerator

[echo jasper]

oh

oh i have to do tests

wait

what happen if i plug in 10

36-100?

over 103

[scarlet escarp]

36 - 100 is under a square root

[echo jasper]

ohhhhhhh

so i always plug in 10

is a good number to plug in?

[scarlet escarp]

well no

what you should be doing is finding the domain of the numerator, denominator, and of the fraciton as a whole

the numerator has domain [-6, 6], the denominator has domain [-3, infty), the fraction requires the denominator is nonzero, so forces the denominator to have domain (-3,infty)
take the intersection of those

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[vernal fern]

3x^2+x+1=0

how do i solve this??

[scarlet escarp]

what have you tried

[vernal fern]

so i started plugging a b and c into the quadratic formula

aand then i googled it to shortcut some simplification

[scarlet escarp]

do you know the conditions to show something is a subspace?

[vernal fern]

and it simplifies to this crazy thing

a whatnow

[elder steppe]

yes the three

so there is neutral element

[elder steppe]

closed under addition

[vernal fern]

oh wait lmao he was trying to help

[scarlet escarp]

check those three

[scarlet escarp]

oh lmao

missed that

my bad dry

[vernal fern]

i think he thought you were asking

anyways what are you even sayin

[elder steppe]

i did and they obv work but i feel like ( x^2 + x -6) is wrong

i then applied a,b,c to it

[vernal fern]

i do not know those words.

[scarlet escarp]

what do you mean

where did that quadratic come from

[vernal fern]

oh maybe he was asking a question

[scarlet escarp]

oh dryforest

this channel is occupied

open one for yourself #❓how-to-get-help

[vernal fern]

oh dawg

we sent the message at the same time or smth??

[elder steppe]

well the initial condition is that p(2)=0 with p(t) being contained in p2

[scarlet escarp]

his is pinned

[vernal fern]

it was unoccupied when i started typing

lmao whoops

my b cyrus

[elder steppe]

u good

so anyway, (2)^2 +2 -6 = 0

[scarlet escarp]

im confused why you are using this arbitrary polynomial

[elder steppe]

and that would make x^2 + x - 6 work

[scarlet escarp]

go through the 3 conditions

[elder steppe]

i did

[scarlet escarp]

were they all satisfied

[elder steppe]

well obv cuz i have to show it, but i feel like im showing it wrong

im using p(t) = ax^2 + bx + (-6)c

bec that's p(t) where p(2) = 0

[scarlet escarp]

you shouldnt be doing that

at least there is no reason to

[elder steppe]

then what am i using to show that p(t) = 0

for the first condition

[scarlet escarp]

the first condition is that p(t) = 0 is in the space

[elder steppe]

neutral element containment

[scarlet escarp]

if p(t) = 0

what is p(2)

[elder steppe]

0?

[scarlet escarp]

yes

so it satisfies p(2) = 0

is the degree less than or equal to 2?

[elder steppe]

ummmm

less than?

[scarlet escarp]

haha yes

i was asking if it was <= 2

but yes

so then p(t) = 0, the 0 vector, is in the set

since it satisfies the conditions to be in V

[elder steppe]

bec p(2) = 0

?

[scarlet escarp]

and because the degree is <= 2

[elder steppe]

hmmm, thats sad cuz i did the rest of the conditions with the this arbritrary example

[scarlet escarp]

the idea for the next is to say
let p,q in V, meaning that p(2) = q(2) = 0, and the degrees are less than or equal to 2

[elder steppe]

ohhhh

[scarlet escarp]

is the polynomial (p + q)(x), defined as p(x) + q(x), in V?

[elder steppe]

so then thats the additive

[scarlet escarp]

is (p+q)(2) = 0? is the degree of (p+q) <= 2?

[elder steppe]

yes

[scarlet escarp]

show it, but yes

and finally, take p in V, c in R

is (cp)(2), defined as c * p(2), equal to 0?

[elder steppe]

YES!

[scarlet escarp]

show it, then you're done

does that make sense?

[elder steppe]

ok ok ya thank you it does

[elder steppe]

one lil question ab this

how does this make the fact that p(t) = 0

...

[scarlet escarp]

i dont understand your question

can you elaborate

[elder steppe]

so like is my reasoning supposed to be that because p(2) = 0, the neutral element is contained?

[scarlet escarp]

yes

the neutral element is p(t) = 0

the polynomial that is 0 everywhere

since it satisfies p(2) = 0, and it is of degree <= 2, it is in the set V

[elder steppe]

I put it as this:
we essentially want to show the fact that p(t) = 0. By our given set, if p(2) = 0, then we have a polynomial (0) where the degree is less than or equal to 0. Therefore the neutral element is contained in V.

[scarlet escarp]

we dont want to show the fact that p(t) = 0

the first condition is that the zero vector, neutral element, additive identity, whatever you wanna call it

is in V

the additive identity in P_3 is the polynomial
p(t) = 0

so we want to show p(t) = 0 is in V

to do that, we show p(t) = 0 satisfies:

1. p(2) = 0
2. the degree of p(t) is less than or equal to 2

[elder steppe]

so 0 is contained (as p(2) = 0), AND, p(t) is less than or equal to 2? like these are seperate ideas?

[scarlet escarp]

p(t) is not less than or equal to 2

well it is

but that isn't the important part

it is that the degree of p(t) is less than or equal to 2

and yes

review the conditions needed to be in V

they are:

1. p(2) = 0
2. the degree of p(t) is less than or equal to 2

we want to show p(t) = 0 satisfies those 2 conditions

p(t) = 0 is the polynomial defined to be 0 everywhere

[elder steppe]

oooooo

so if t = 2

, then p is 0

and thats why

wow ok i got nowhere with that thought

just bec there is an instance where p(t) = 0

that is enough?

when t = 2

?

and obv 0 has a degree less than or equal to 2

[scarlet escarp]

im not sure what your question is

again
we want to show p(t) = 0, the polynomial that is always 0, is in V

[elder steppe]

here i think im getting to it: my new explanation in a sec:

We essentially want to show the fact that p(t) = 0 is in V. P(t) = 0 is indeed a polynomial with a degree of 2 or less, and of course p(2) = 0 as it does for any t. So therefore the neutral element is contained.

sum like that?????

[scarlet escarp]

yes

but

i wouldnt use
"the fact that p(t) = 0 is in V"

i would just say

"we want to show that p(t) = 0 is in V"

dont call it a fact preemtively

[elder steppe]

i see i see

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[timid wing]

I know what the answer is but I don’t understand the process to get there, help would be appreciated.

[timid wing]

I added more numbers but I don’t know where to go from there

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[timid wing]

<@&286206848099549185>

[upbeat acorn]

@timid wing this is so alien to me, but some googling got me this. You had the right idea. Now, take the diameter to be (d-2) and the 4 in the picture to be a and b. Multiply d-2 by 2 and 4 by 4 which equal 2d-4 and 16. Set them equal to each other to get d=10. Half of d is the radius which is 5. You can then use pythag to get x!

[timid wing]

Why did you multiply d-2 by 2?

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[lyric fjord]

i integrated it, and from the original equation it = 0 when t = 4
so thats why i added on the -24

why is it wrong

[wise dagger]

i don't think it's -24, please check your caluclations on the -24 again

[lyric fjord]

why is it not 24

[vernal wraith]

Conceptual mistake.

[wise dagger]

then why it is? you didn't show your workings on the 24, i cant tell why

[lyric fjord]

i set v(t) = 0, t has to be 4

[vernal wraith]

@lyric fjord

[lyric fjord]

someone earlier helped me and said it was -24

[vernal wraith]

Then calculation mistakes

[lyric fjord]

???

[vernal wraith]

You equated v(t)=0 to find time. At that moment, s(t)=0. Substitute and find the constant C

[wise dagger]

well, since at t=4, position should be = 0
,which in such case
-3t²+24t+c=0 when t=4

[lyric fjord]

ahhhhh

i see

okok

ty

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[gusty shore]

Hi, how can I use integration to find the mean value of a function?

[crimson seal]

@gusty shore

[gusty shore]

ah cheers

so just use that formula yh?

[crimson seal]

yes

[gusty shore]

ty

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[subtle prairie]

Are these graphs G1 and G3 isomorphic?

They have same number of edges and same number of vertices and same degree at every vertex

[marble wharf]

no

the one on the left is bipartite

[subtle prairie]

actually how to verify right is not bipartite?

[marble wharf]

try splitting the vertices into two classes A and B

or use a characterisation using cycles of odd length

[subtle prairie]

kinda always stays with in sets

not disjoint

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[wraith hinge]

I don't understand how to put signs after finding the domain, can anyone explain?

is it just multiplying signs from each solution?

(I understand the ones on right, not the overall fn's)

<@&286206848099549185>

[olive tide]

Could you explain what you're getting stuck with

i mean ur solution looks correct to me

[wraith hinge]

it's not mine

[olive tide]

oh

[wraith hinge]

I don't understand how to apply signs on overall derivative fn

to figure out where it's inc, dec

[olive tide]

oh i see

this is the wavy curve method

after you factor it out, you can check when the function is becoming +ve or -ve by examining the regions between 2 zeroes of the function

cuz that's where the sign changes

[wraith hinge]

so no?

[olive tide]

as a trick, for polynomials start with +ve to the right
then mark the roots on a number line
then just switch the sign to -ve if the multiplicity of the root is odd
u can repeat this process

[olive tide]

um i personally did not understand what u meant by that

[wraith hinge]

I'll circle it out 1 sec

[olive tide]

oh

i think what they have tried to do there is check the sign of the linear factor for the given domain of x

so you are correct in that regards
the wavy curve method is derived from this idea of multiplying signs

you can read about it... it's pretty useful in many places

[wraith hinge]

Okay thank you

Can I ask another question

Or do I open new channel

[olive tide]

um i'm new here so idk about the rules tbh

[wraith hinge]

uhh

do they take the ln on both sides just to match with the mcq?

[olive tide]

nope

this is a general method

how do u wanna approach it?

[wraith hinge]

uh 1 sec

[olive tide]

uh that's not correct

[wraith hinge]

Why so

[olive tide]

what did you do to derive this?

[wraith hinge]

Just exponential derivative formula, I must have got it wrong

[olive tide]

yeah well that only works if the base is constant

[wraith hinge]

The derivative of a^b where b is form of x is a^b . dy/dx(b) . ln(a)

[olive tide]

not if it's a function

[wraith hinge]

I see

[olive tide]

this will become obvious if you use 1st principles to check

in general you need to take the log of both sides for functions like this

i remember having a similar doubt when i was just learning calc myself lol

[wraith hinge]

so for questions of type
some fn of x raise to some fn of x
I must take ln of it as a whole when performing derivative

is that correct?

[olive tide]

yup

[wraith hinge]

and what iff I was just asked the derivative of it

I mean without lhs

[olive tide]

could you explain what u mean by that

[wraith hinge]

cus we multiply y on both sides..

1 sec

ooh it's the same thing 🤦

[olive tide]

u can do a lil short cut too

if u dont like the whole y = this thing

[wraith hinge]

mhm

[olive tide]

u can write f(x) as e^ln(f(x))

then differntiate that normally using the chain rule

it's the same thing really but yeah

[wraith hinge]

Noted

Thank You So much /:D

[olive tide]

np dude o7

my pleasure

[wraith hinge]

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[wraith hinge]

.reopen

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✅

[wraith hinge]

@olive tide does something similar apply for logarithmic fns too?

sorry for the ping.

[olive tide]

can you gimme an example?

[olive tide]

it's ok

[wraith hinge]

lemme see..

derivative of log a (b) where b is some form of x and a is the base, is dy/dx(b) / (b . ln(a))

& a happens to be a function of x

[olive tide]

for this i'd personally write log a (b) as log (b)/log (a)

[wraith hinge]

gotcha

maybe

I'm over complicating now, that prolly won't be on the exam

thank you sm once again!

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[olive tide]

np o7
my pleasure

[tropic bloom]

.reopen

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[obtuse pasture]

how do you find the equation of a line if theres a fraction?

[obtuse pasture]

like (0.5 , 1/6)

[proven hound]

Is there any other info?

[obtuse pasture]

mope

[proven hound]

If you have a point then you'll need a slope to find equation of a line

[obtuse pasture]

so rise/run?

[proven hound]

But anyways
Assuming the slope to be m,
The equation is y-y1 = m(x-x1)

Here (x1,y1)= (0.5, 1/6)

So just substitute

[split chasm]

with just a single point, you can't determine the equation of a unique line

[obtuse pasture]

oh

so ould it just be undefined?

[split chasm]

not undefined, just nonsensical

[olive tide]

it's not undefined
there are infinitely many such possible lines

[split chasm]

with sufficient info
fractions are just numbers and you'd apply pretty much the same methods as you would if you had integers

[obtuse pasture]

so like

what would a proper equation be like with fractions

[split chasm]

depends on the desired form

[obtuse pasture]

y = mx+c

[split chasm]

well you'd just have fractions

instead of the nice integers you're normally used to

e.g.
$$y = \frac23 x + \frac15$$

[solid kiln]

ℝαμΩℕωⅤ

[obtuse pasture]

oh

then what if its like points?

[split chasm]

what you typed earlier

[obtuse pasture]

o

[split chasm]

they're still numbers
same rules apply

[obtuse pasture]

Ohh

okay thankss

.close

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[grizzled imp]

<@&286206848099549185>

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[grizzled imp]

<@&286206848099549185>

[grizzled imp]

<@&286206848099549185>

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[limber halo]

If I’m asked to describe Ax=b as a linear combination of the columns of A, would this be it?

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[silver kiln]

Solve the equation AX = B if:

A = |-1 0 2| B = |1 1 -1|
|-2 1 3| |0 -1 1|
|0 -1 0| |1 1 0|

[shell barn]

X = A^-1 B

Just find the inverse of A. Then multiply

[silver kiln]

so than after i find X, how can I prove it can I multiply A * X and the result should be B ? sorry if I sound dumb

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[shell barn]

First find the inverse of A

Then multiply A^-1 with B

[silver kiln]

.close

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[sharp gust]

Revisiting limits evaluated w conjugates

[sharp gust]

Where am I messing up

Ignore the 16 t part oops

I can’t see what minor mistakes I am making

[odd charm]

At expanding (2 - √x)(√x + 2)

[sharp gust]

Oh?

Ok cheers

Ohhhh

It ends up 4-x bc cancellation and it’s-*+

Then you get rid of x on top, and move the 4 out of it

Is this correct? But it feels wrong on bottom too

Nvm

Lack of sleep got me stupid mistake

Thx

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[earnest oak]

i need help with logs

[earnest oak]

i still dont under stand the rules of logs very well

[low gazelle]

Is that log_3(x)-2 or log_3(x-2)

[delicate belfry]

To make clear, is it $$2 (log_3x)^2=5 log_3(x-2)$$

[earnest oak]

second one

[solid kiln]

Good

[earnest oak]

yes it is that one

[delicate belfry]

Get them all a base of 3.

[earnest oak]

can you explaing that like i said im very bad at logs

@delicate belfry

[delicate belfry]

To be honest, I'm also uncertain.

[earnest oak]

nah its fine i understand i hate logs

[low gazelle]

U had it right I think

3 to the power of the left side = 3 to the power of the right side

[earnest oak]

but how would it look

can we meybe go into a call and you will explain??? @low gazelle

[low gazelle]

I cant I will write it in latex for u

$3^{2(\log_3x)^2}=3^{5\log_3(x-2)}$

Remember 3^log3x=x and use some exponent properties

[earnest oak]

tbh im just lost gusse i will just fail

[low gazelle]

Do you understand why this is true

Oh shit I forgor a 2

[solid kiln]

Sora Harewataru

[earnest oak]

but still with the 2 how whould it help??

?

[low gazelle]

Because this lets us cancel the logarithms and we will end up with a quadratic

Remember logs are the opposite of exponentiation

[earnest oak]

i think i understand

whats exponenriation

sorry english isnt my first language

[low gazelle]

Like a^b

a*a b times

[earnest oak]

ye i know

ok so i just do

let me write it

$2x^2=5x$

[solid kiln]

ilai

[earnest oak]

so is this how its suppouse to become or am i just stupid

oh dosnt mattar the logs arnt the same

wait its wrong

[low gazelle]

Ye

[earnest oak]

so how am i suppous to do it

[low gazelle]

First try seperating the constants and the logs

Ill do left side

[earnest oak]

ty

[low gazelle]

$3^{2(\log_3x)^2}=3^{2\cdot\log_3x\cdot\log_3x}=3^2\cdot 3^{\log_3x} \cdot 3^{\log_3x}$

3^log3x=x so you get 9x^2 on left side

[solid kiln]

Sora Harewataru

[earnest oak]

isnt it 9X9x the result?

[low gazelle]

Where is the 9x9 from

[earnest oak]

3^2

=9

wait

[low gazelle]

Where is the 2nd 9 from

[earnest oak]

bro i feel so stupid

i think im a lost cuse

[low gazelle]

Nah you are doing fine you only become a lost cause when you decide to quit

[low gazelle]

In this pic is there any part u dont get

[earnest oak]

i under stand how to get to that part but idk how to keep going

[low gazelle]

Ok try doing a similar thing for the right side of this equation

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[drowsy olive]

@pliant seal I need help

[minor urchin]

help

[drowsy olive]

The question is as follows,

[minor urchin]

how do i do this

help

but

whast your q

[drowsy olive]

Yes

I thought I claimed the channel

😓

[minor urchin]

mb

whats your q tho

ill see if im any use

[granite sedge]

you indeed did claim it

[drowsy olive]

Ok

My question goes like this

Let $R = K[x,y]$, and let $\varphi: R^2 \to R^2$ be an $R$-module endomorphism of the $R$-module, $R^2 = R \oplus R$ given by $\varphi((a_1,a_2)) = (a_1 + xa_2,ya_1)$. Find a monic polynomial $p$ with coefficients in $R$ such that $p(\phi) = 0$.

[solid kiln]

mycroftholmes1703

[drowsy olive]

I understand we have to use Generalized Cayley Hamilton Theorem, but how exactly to proceed with the computation ?

<@&286206848099549185>

[zinc ginkgo]

maybe#groups-rings-fields is a better place

[drowsy olive]

.close

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[restive thorn]

okay so this is extremely stupid
if a function f(x) whose range is (0,inf) is increasing in R

is ln(f(x)) increasing?

for example f(x) = x + sqrt(1+x^2) is increasing and so is ln(x + sqrt(1+x^2))

but can this be generalised

[wraith hinge]

whenever two functions are increasing, their composite functions are also increasing

if that's what your question is about

[restive thorn]

okay thanks

[wraith hinge]

this can be proved

[restive thorn]

okay thank you

.close

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[indigo ferry]

Its basically by definition.

If a > b then f(a) > f(b) then g(f(a)) > g(f(b))

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[frail portal]

What could a graph look like if it had a local maximum point and two local minimum points?

[indigo ferry]

,w plot x^4 - x^2

[solid kiln]

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Requested by casework

[indigo ferry]

It could look like this

[frail portal]

wow okay

IS there any other type of way that is more common?

[indigo ferry]

I think this is like the most common

[frail portal]

Okay

[indigo ferry]

I mean between 2 local min there has to be a local max

[frail portal]

oh yeah

Thanks so much!

.close

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[twilit ridge]

Hi can someone help me figure out which graphs of f(x) would work for f''(x)

I know that it will be kind of in the graph of a parabola but im not sure

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[frail portal]

The graph is a parabola with a maximum point.
The derivative is positive for x < 0 and negative for x > 0
The derivative is 0 for x = 1.
The function is increasing for all x where it is defined.
The derivative has three zeros.
The graph is linear and decreasing
The derivative is positive and decreasing for all x > 0
The function has a local minimum at x = 1.
The derivative is 0 for x = -1 and x = 1. Between -1 and 1, it is positive.
The derivative is negative and increasing for all x.
The function has a local minimum at x = -1 and a local maximum at x = 1.
The derivative is constant and negative.
The function is positive and decreasing for all x

[frail portal]

Please help much appreciated!

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