help-33

[brazen jungle]

??

what is this

[jagged relic]

The first one

You should post the whole question

[ebon depot]

If I choose the first one then I’d have first 5 choices, then 4 and finally 3

543= 60 choices

But that does not include all possible combinations of the other letters

[jagged relic]

Unfortunately that's wrong because the A's are indistinguishable

[ebon depot]

How????

I have three As
A1,A2 and A3

[jagged relic]

The possible arrangements of the other letters is just 4!, you even said so yourself: #help-30 message

[ebon depot]

Why is it different every time

Why can’t it just be simple

[jagged relic]

But you don't count C A1 R A2 M B A3 and C A2 R A1 M B A3 as different solutions

[ebon depot]

I get that

[jagged relic]

That's the reason permutations and combinations are different

P(5,3) = 5! 4! 3! = 60

That's what you did

C(5,3) = [5! 4! 3!] / 3! = 10

[ebon depot]

Yes because that is only because I remember doing another where I just divide by the factorial of all letters that are the same

3As result in 3! In denominator

It has nothing with me understanding why

[jagged relic]

That's right, that's why the 3! is there

[ebon depot]

I want to be able to utilize the formula

[jagged relic]

Oh okay hold on

[ebon depot]

But I can’t distinguish between them

[jagged relic]

If you arrange the letters like CARAMBA, but count each A as different (like you said, A1, A2, A3)

then you use permutations

but since the A's aren't different

you need to divide by the number of ways you can arrange A1, A2, A3

It can be A1, A2, A3, or A2, A1, A3, or A2, A3, A1, ...

That's 3! arrangements

They all count as the same solution

[ebon depot]

Yes

[jagged relic]

So what exactly don't you understand?

[ebon depot]

Everything

Since I still can’t solve it

[jagged relic]

[ebon depot]

A1,A2,A3 A2,A1,A3 = 2
A1,A2,A3 A2,A1,A3 = 1

So the above is combinations

And bottom permutations

What does n and what does k stand for infor combinations and permutations?

[jagged relic]

The other way around, assuming you meant 6 instead of 2...

[ebon depot]

No I meant 2 and 1

[jagged relic]

Then that makes no sense

[ebon depot]

Then I don’t understand?

Like I can’t think as quick as you are professors

I need a structured way to actually understand

[jagged relic]

If you have 3 different objects, there are just 3! different ways to order them

If you have 3 identical objects, there is only one way to order them

[ebon depot]

Yes

[jagged relic]

Well, in CARAMBA, you have 4 different consonants and 3 identical vowels

You can order the consonants in 4! different ways, and you can order the vowels in only one way

The rest is determining how to place the vowels such that they don't touch each other

For that, you need to think about what it means for the vowels not to touch each other

It means there is at least one consonant in between

[ebon depot]

Yes

[jagged relic]

In other words, you can place only one vowel between two consecutive consonants

(or zero)

There are 3 places you can insert a vowel between two consonants, and also 2 places at the ends (between a consonant and nothing)

So there are 5 places where you can insert one or zero vowels

You have 3 vowels to insert

So you need to choose 3 places out of the 5

That's literally what combinations are used for: choosing k things among n

[ebon depot]

I'm so sorry Nel but i'm reading it over and over but i just don't get it

I assumed since this was a question about permutations or combinations i'd have to choose one of them

and then just utilize the formula based of n and k

[jagged relic]

Not sure what to tell you, maths is not about applying one specific formula every time

[ebon depot]

fine take over my channel

.close

[marsh citrus]

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[kind perch]

please say this isn't as complicated as it looks

[kind perch]

starting on part (a)

[past frigate]

It isn't

[kind perch]

phew which do i do first integrate or partial fraction it?

[past frigate]

What do you think, by looking at it?

Is there a way to integrate it directly?

[kind perch]

integrate

when i integrated it i got $(1/3)x^3+(3/2)x^2+4x+2$

[past frigate]

As final ans?

[elfin berry]

sealpup321

[kind perch]

I cant see how to turn this into a partial fraction?

as its not a fraction with a polynomial denominator

[past frigate]

For partial fraction you need the degree of denominator to be greater than that of numerator

So try long division first

[kind perch]

what do you mean its not over anything?

i did the long division before integrating

[past frigate]

What did you get

[kind perch]

$x^2+3x+6+(2/x-1)$

[elfin berry]

sealpup321

[past frigate]

=the numerator is it?

Or the entire question

[kind perch]

what do you mean?

[past frigate]

What is this equal to

[kind perch]

when you do polynomial long division on part (a)

$(x^3+2x^2+3x-4)/(x-1)$

[elfin berry]

sealpup321

[past frigate]

$(x^3+2x^2+3x-4)/(x-1) = \frac{{(x^2 + 3x + 6)}{(x-1)} + 2}{x-1}$

[elfin berry]

Lorentz

[kind perch]

yes

[past frigate]

Right

So you get
$$x^2 + 3x + 6 + \frac{2}{x-1}$$

[elfin berry]

Lorentz

[kind perch]

yup

[past frigate]

Now integrate it

Oh

[kind perch]

then integrated to get this

[past frigate]

One of the terms should be in form of ln tho

[kind perch]

why?

[past frigate]

The 2/(x-1) term

[kind perch]

I mean you are right but I am confused

[kind perch]

ah because i wasn't sure what to do with it i turned it into $2x^{-1}-2$

[elfin berry]

sealpup321

[past frigate]

I see

Yeah you'll need to integrate that too

[kind perch]

ye i did to get 2-2x

[past frigate]

Integration of 1/x is ln(x)

$\int \frac{2}{x-1} dx$

[elfin berry]

Lorentz

[past frigate]

And we have this

[kind perch]

oh thanks

[past frigate]

Np I guess

I assumed you knew that

[kind perch]

so how would i use that then?

[past frigate]

You have x-1 in the denominator here

[kind perch]

so $ln|1/2 * x-1|$

[elfin berry]

sealpup321

[kind perch]

?

[past frigate]

No wait that 2 is a constant, it would be multiplied with the ln thing

[past frigate]

The 2 here I mean

[kind perch]

or $2ln|x-1|$

[elfin berry]

sealpup321

[past frigate]

Yes that

[kind perch]

cool 🙂

alright so once i have this new number how do i turn it into a partial fraction?

[past frigate]

Which new number?

[kind perch]

$(1/3)x^3+(3/2)x^2+6x+2ln|x-1|$

[elfin berry]

sealpup321

[past frigate]

That's your final answer

[kind perch]

huh?

[past frigate]

The integral of the question

Yes

[kind perch]

do i not need to do the whole A/x-1 + B/x-1 +C/x-1

or whatever the numerator and denominator would be in this case

[past frigate]

I thought so too, but after using long division the denominator of 2/(x-1) was a direct integration

[kind perch]

sorry can you explain what a direct integration means?

[past frigate]

I mean like you can directly integrate 1/x to ln x

Instead of some partial fractions thing

[kind perch]

so you dont need to turn in into partial fractions...

strange... thanks 🙂

[past frigate]

I was confused too

Np

[kind perch]

yup checked and its correct thanks for all the help

[past frigate]

Coz for partial fracs the degree of denominator should be more than numerator

So I didn't know any other way

[kind perch]

🤷

.close

[marsh citrus]

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[gusty sorrel]

Where did condition 2 and 3 come from?

[marsh citrus]

@gusty sorrel Has your question been resolved?

[gusty sorrel]

<@&286206848099549185>

😭

[gusty sorrel]

.close

[marsh citrus]

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[marsh citrus]

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[muted grail]

need some help with this question

[muted grail]

Someone pointed out that we can define a function g(x) = f(x) - x and since both f(x) and x are continuous functions then so is g(x).

then we can use the fact that f(x) is a bounded function to look at two values a and b such that a > b

and g(a) and g(b) have opposite signs

this would imply the existence of a g(c) = 0, according to the bolzano's theorem.

this is the part I'm confused about.

how does the fact that g(x) is bounded (due to f(x) being bounded) and continuous (due to both f(x) and x being continuous) imply that there exists values a and b such that g(a) and g(b) have opposite signs?

<@&286206848099549185>

[spark siren]

as f is bounded there are values m and n such that m < f(x) < n for all x. now think about what does this mean for g.

[muted grail]

hmm

it would just mean that g is bounded too

right?

wait actually

if g(x) = f(x) - x

regardless of if f(x) is bounded or not, the x isn't bounded

which would mean that g is not bounded

right?

[spark siren]

take an x between m and n.

[muted grail]

ohh

at some point x = f(x)

and there will be g(x) = 0

right?

[spark siren]

for x large enough f(x)-x < n-x < 0 -> g(x) < 0
for x small enough (which means x< 0, and |x| large enough) f(x)-x > 0, which means g(x) > 0, so you know that a and b with the claimed conditions exist.

[muted grail]

ok lemme think this through

[muted grail]

x is not bounded correct? It's just like the identity function

"for x large enough f(x)-x < n-x < 0 -> g(x) < 0" so in this line, we are essentially taking the largest possible f(x) which is n, and we take a value x, and since x isn't bounded, it can be any real number

so we take x such that n - x < 0 -> g(x) < 0

[spark siren]

no, not the largest possinle. thsi would be infinity. just an x which is > n.

[muted grail]

but it wouldn't be infinity

as f(x) is bounded?

said bounds being n and m

right?

[spark siren]

if x > n then also x+1, so a largest x cant exist.

[muted grail]

i said largest possible f(x)

x is unbounded. there isn't a largest x right?

we can select x however we like

[spark siren]

we take an x which is > n.

[muted grail]

ok got it

which proves the exictence of a negative g(x)

and then do the same to prove the exictence of a positive g(x) value

[spark siren]

yes

[muted grail]

and then since it is continuous

we have proven that there exists a g(x) = 0

the we have 0 = f(x) - x => f(x) = x

which proves part i

[spark siren]

yes

[muted grail]

ok dope

just a quick question

[muted grail]

shouldn't this be m =< f(x) =< n

[spark siren]

the boundary could be asymptotic

eg. arctan x

[muted grail]

i don't know what that means

aympstotic boundries

[spark siren]

arctan is bounded, what is m and n if you think there should be an "=" in m<=f(x)<=n?

[muted grail]

f(x)

[spark siren]

take f(x) = arctan (x), say what is m and n such that m <= f(x) <= n, and for what x is f(x) = m and for what x is f(x) = n?

[muted grail]

idk how arctan behaves at different values

[spark siren]

you do not know the function arctan?

[muted grail]

i know the function

[spark siren]

is ir bounded?

[muted grail]

yes

[spark siren]

what is the upper bound?

[muted grail]

ik how it looks like

[muted grail]

I dont knw

one radian

I don't remember honestly

arctan was the function that took lengths as arguments and gave degrees right?

that's literally all I know about it

and how it looks like

[spark siren]

anyway, if a function is bounded this only means, that there are upper and lower bounds. that does not mean, that the function reaches the bounds. and ix k is an upper bound then also k+1, k+2, k+3, ---

[muted grail]

ok got it

[marsh citrus]

@muted grail Has your question been resolved?

[marsh citrus]

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[jaunty pelican]

what is the easiest way to solve a system of linear equations

[zenith python]

it depends on how many equations you have?

[jaunty pelican]

just 2

[shell ether]

Send it

i can't imagine

[zenith python]

my favorite method is reduction

[shell ether]

maybe gauss jordan method

[jaunty pelican]

how would you do that

[shell ether]

@zenith python say something

I forgot this method

[main idol]

If you don't know it, say so

[jaunty pelican]

I don't

[main idol]

You can just use algebra

There are many ways, one could be multiply one equation by a number so that when you add them you eliminate a variable

Like the first eqn by 3

[shell ether]

or replacement

[main idol]

Or second equation by -4

[jaunty pelican]

where did you get -4 from?

[main idol]

Try it all 3, none are better than the other

[main idol]

-4 + 4 = ?

[jaunty pelican]

0

[shell ether]

4x
x
4x
-4x

So we can erase this shit

Like add first equation to second

And we'll get rid of x

[jaunty pelican]

add everything?

[shell ether]

And we'll be able to solve y

Yes

But

First of all

Multiply second equation by -4

All second equation

And then addition

[jaunty pelican]

im still confused where your getting the -4 from

[shell ether]

we multiplied x by -4

And got -4x

[jaunty pelican]

but there is no -4 in the equation

[shell ether]

So we can do 4x+(-4x) and got 0

[shell ether]

But we can multiply all second equation to -4

[main idol]

-4 = (-1) * 4

[jaunty pelican]

4x+y=15

x-3y=7

you mutiply the second equation ^ correct?

by a number

[shell ether]

YES

Yes

[jaunty pelican]

i got that

so would you substite x with 4 since the first equations x is 4?

[shell ether]

Nah

don't touch first equation

[jaunty pelican]

ok

[shell ether]

multiply second by -4

[jaunty pelican]

bro

where did you get the -4 from?

[shell ether]

well as i said

[main idol]

Do you understand this

[shell ether]

We can multiply our equation by any number

[main idol]

Or this

4 + ? = 0

[jaunty pelican]

oh so we can mutiply it by and number?

any

so 4 (10)

would that work

or does it have to be 0

[shell ether]

Whaat

4(10) what it means

[jaunty pelican]

cause you said we can mutiply our equation by any number

[shell ether]

You can't replace x

[shell ether]

Yep

you can do (x-3y=7)×(-4)

[jaunty pelican]

-4x+12y=-28

?

[shell ether]

Yes

So add the first equation to second

[shell ether]

replace your second equation with this result and keep the first equation in the same place

[jaunty pelican]

12y^3=13

?

[shell ether]

no

12y

Just 12y

Why in 3 power

[jaunty pelican]

but 12y+y = 12y^2?

[shell ether]

Nah

It's 13y

Just

[jaunty pelican]

is it because theres a 1 infront of y?

[shell ether]

12y+y=12y+1y=13y

🙂

Yes

[jaunty pelican]

ok got it

so 13y=13

[shell ether]

Yes

[jaunty pelican]

y=1

[shell ether]

You found y

[jaunty pelican]

ok so whats the next steo

step

[shell ether]

4x+y=15

solve for x

[jaunty pelican]

do i replace y for 1?

[shell ether]

you can

But

you can do it after

[jaunty pelican]

x=5?

4x+1=15

[shell ether]

x=(15-y)/4

Hope you can do it

And then replace

so the answer isn't 5

Oops

14/4

Solve it

[jaunty pelican]

wait

if the y was a positive how did it turn into -

[shell ether]

well

You know that

Bro

I'm confused that you don't know basics

x+y=0

Solve for x

[jaunty pelican]

😕

[shell ether]

x+5=0

Solve it

What you'll get

[jaunty pelican]

x=-5

[shell ether]

yes

So why fucking sign changed

...

That's the same

[jaunty pelican]

cause you subtracted

[shell ether]

Sooo

[shell ether]

What happened here

[jaunty pelican]

ok lemme go back

we solved for y and got 1

so we have to plug that into our first equation 4x+y=15

?

[shell ether]

yes

We can

But we can firstly solve for x

And it will be easier

[jaunty pelican]

and to solve for x we divide?

[shell ether]

yes because it's 4x

We need x

So we divide by 4

is it understandable

[jaunty pelican]

15/4 is a decimal

[shell ether]

yes but

You have -y

(15-y)/4

because you first substract

By y

Both sides

And then divide by 4

[jaunty pelican]

ohhhhhhh i get it now i was overcomplicating it

[shell ether]

so

What is the answer

[jaunty pelican]

4?

[shell ether]

nah

(15-y)/4

Replace y by 1

[jaunty pelican]

14/4

[shell ether]

What you'll get

Yes

Solve it

[jaunty pelican]

thats another decimal

[shell ether]

What is 14/4

Can you simplify it?

[jaunty pelican]

yea into a decimal

[shell ether]

Do it

[jaunty pelican]

3.5

[shell ether]

🙂

so what is the general answer?

[jaunty pelican]

3.5 and 1?

[shell ether]

nah

Usually people say (x;y)

[jaunty pelican]

yea as a ordered pair

(3.5, 1)

[shell ether]

yes

Good

https://youtu.be/Ff9ba3PryUs?si=Q4nxqqDD_BEz8AKo

Recommend you this cheat method

Cause it's much easier

[jaunty pelican]

the method im learning is complicated

[shell ether]

Everything is complicated when you are struggling with basics

you should first learn how to do simple/medium/hard linear equations

Not even systems

Anyway I'll go

[marsh citrus]

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[dusk meadow]

A hen knows how to count. We put it in front of a packet of 2024 eggs. She counts the eggs from the first packet and places them in a second packet. Each time she has counted 4 eggs, she lays an egg which she places in the packet of those she still has to count.

How many eggs will the hen count in total?

[dusk meadow]

<@&286206848099549185>

[stoic slate]

Wasn’t this solved yesterday?

[dusk meadow]

yeah but i have a problem

im getting another answer

my original answer was 2698

but then i got 2695

help plz

<@&286206848099549185>

anyone

[marsh citrus]

@dusk meadow Has your question been resolved?

[dusk meadow]

<@&286206848099549185>

help

.close

[marsh citrus]

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[honest current]

Hello

[honest current]

Can I get help with this

<@&286206848099549185>

[marsh citrus]

@honest current Has your question been resolved?

[honest current]

<@&286206848099549185>

[lethal cairn]

gosiogiusdg

[marsh citrus]

@honest current Has your question been resolved?

[marsh citrus]

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[lyric kelp]

why is C negative?

at the end of the second line

[gleaming pecan]

this form C was chosen to make the answer look nicer = more convenient for possible further calculations

[lyric kelp]

so it doesnt matter correct?

[gleaming pecan]

but it doesn't matter at all, really

[lyric kelp]

the sign of C

ah okay tyy

.close

[marsh citrus]

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[marsh citrus]

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[flat grail]

I'm trying to compute $$\lim_{n \to \infty} {(3^n + 1)^\frac {1}{n}}$$ by the squeeze theorem. is it ok to use $$(3^n)^\frac{1}{n} \leq (3^n + 1) ^\frac{1}{n} \leq (3^{n+1})^\frac {1}{n} $$

[fathom ridge]

well you do get the right answer so

[elfin berry]

lewis_f04

[flat grail]

I'm not sure if it is ok to say that $$(3^n + 1) ^\frac{1}{n} \leq (3^{n+1})^\frac {1}{n} $$ without proof as it isn't trivial

[elfin berry]

lewis_f04

[paper raptor]

couldn't you say that for sufficiently large $n$, $3^n+1\leq 3^{n+1}\implies (3^n+1)^{\frac{1}{n}}\leq (3^{n+1})^{\frac {1}{n}} $

[elfin berry]

PajamaMamaLlama

[flat grail]

for sure, I've just never got feedback from my lecturer on how exactly this is to be done, and my exam is tomorrow

if thats how you'd do it, I'll use that

thanks for the help

[paper raptor]

I'd probably ask someone else but that seems sufficient

[marsh citrus]

@flat grail Has your question been resolved?

[marsh citrus]

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[whole jetty]

Given the following functions find the image of 0, and 2 in each case, and the antiimage of –3, 0.

[marsh citrus]

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[marsh citrus]

@whole jetty Has your question been resolved?

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[cerulean oxide]

how do i do this?

[marsh citrus]

@cerulean oxide Has your question been resolved?

[cerulean oxide]

<@&286206848099549185>

[marsh citrus]

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[cerulean oxide]

how do i do this?

[cerulean oxide]

Not sure how to procede

[deft apex]

Do you have any working for anything at all?

[cerulean oxide]

what?

what does that mean?

[stoic slate]

It means if you tried anything to solve the problem or you did nothing so far

[cerulean oxide]

i think you have to think the values of lambda

i found them to be:
"3, -1 + 2sqrt(2), -1 - 2sqrt(2)"

for A

that is what i've done so far

[marsh citrus]

@cerulean oxide Has your question been resolved?

[cerulean oxide]

so?

<@&286206848099549185>

[final skiff]

find the eigenvectors corresponding to each eigenvalue

[marsh citrus]

@cerulean oxide Has your question been resolved?

[stoic slate]

So nothing, I’m just explaining what he was asking cause you said “what does that mean?” Referring to him, and that meant you didn’t understand him

[cerulean oxide]

how do i do that

sorry i am a bit lost on this part of linalg

(A-lambdaI)*x=0, right

[cerulean oxide]

this woul be it, right

[main idol]

Did you write this

[cerulean oxide]

yes

[main idol]

Where did 3 come from

[main idol]

Oh here?

[cerulean oxide]

the first eigenvalue

is found to be 3

i hope its correct

[main idol]

Looks right

Find a vector v such that Av = 3v

You can arbitrarily pick the one of the components of v to be 1

[cerulean oxide]

i might have made a mistake

this is wolfram alphas answer

[main idol]

The order doesn't matter

Your eigenvalues look the same

[cerulean oxide]

i am not missing 1 thing

no nvm

nice

[cerulean oxide]

that is sort of what i tried here

[marsh citrus]

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[main idol]

.reopen

[marsh citrus]

✅

[main idol]

That's not how you find eigenvectors

[cerulean oxide]

i tried to do:
(A-lambda*i)*x=0

[main idol]

There's no x here

[cerulean oxide]

yeah i wasnt rly sure what x should be

[main idol]

You're solving for it

It's unknown at the moment

[cerulean oxide]

x = -y -z:
x= -y

z is a free variable

right

@main idol is this what you meant

[main idol]

I don't know what you're doing here

[main idol]

Just do this

[cerulean oxide]

(A-lambda*i)x=0

yeah, that is also what i've done

we are essentially solving (A-3I)*v=0

[hardy slate]

yeah so you have one of them

(1, -1, 0) is an eigenvector with eigenvalue 3

now get the other ones

[cerulean oxide]

ok, 1 sec

[cerulean oxide]

agree?

"So the three eigenvalues are: $\begin{bmatrix} 1 \ -1 \ 0 \end{bmatrix}$, $\begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix}$, and $\begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix}$."

[elfin berry]

// mav

[cerulean oxide]

@hardy slate after i've found these, what am i supposed to do

[hardy slate]

oh wait it's the identity matrix

wait, what

[cerulean oxide]

wdym wait, what

is it incorrect?

[hardy slate]

sorry one sec

the way you work it out is different to what i'm used to

this shouldn't be right

like ok

you're saying A (1, 1, 1)^T = (-1 - 2sqrt(2)) (1, 1, 1)^T

and A (1, 1, 1)^T = (-1 + 2sqrt(2)) (1, 1, 1)^T ?

[cerulean oxide]

mhh ok yh

both are the same

[hardy slate]

yeah that's impossible

[cerulean oxide]

ill use a comptuer to what it gets

[hardy slate]

you should probably learn how to solve systems of linear equations

that's pretty important

[cerulean oxide]

i think i can do that already

that is what we are doing right now

[hardy slate]

well, you gotta learn how to do it correctly then

[cerulean oxide]

it is correct

[hardy slate]

it's not

[cerulean oxide]

i am aware its because i did it for the wrong matrix

so my eigenvalues are wrong

everything is wrong

[hardy slate]

oh what

okay?

[cerulean oxide]

i put a -1 on a 1 that should be +1

which is why i got such weird values

[hardy slate]

that'll do it

[cerulean oxide]

give me 5 min

wow yeah

the eigenvalues are now -3,3,3

which makes it a lot easier

how do i even write up an answer for this

[cerulean oxide]

idk how to write an answer for it tho

[hardy slate]

so any vector st. x+y+z = 0 will be an eigenvector with eigenvalue 3

so you can just take any vector that isn't 0

so like, (1, 1, -2)

actually you want two eigenvectors for 3

so you want to find a second orthogonal eigenvector to that

to make the matrix orthogonal

[cerulean oxide]

wdym

[cerulean oxide]

this would be correct, right

[hardy slate]

no

for L = -3

you need x = y + z

but 2 != 1 + 0

[cerulean oxide]

why 2!?

so [2,1,1]?

[hardy slate]

!= means 'not equals'

but anyway yeah that would work

ok

so you have two eigenvectors, one for 3 and one for -3

now you just need another eigenvector for 3

it should be orthogonal to the other two

wait actually hmm

the problem now is that (1, 1, -2) isn't orthogonal to (2, 1, 1)

so

we could do (0, 1, -1) instead and that would work

and then we'd need one more

[cerulean oxide]

for L=-3?

[hardy slate]

no

(2, 1, 1) works for L = -3

[cerulean oxide]

L=3 [0,1,-1]

right

[hardy slate]

right

and then we need one more for L = 3

orthogonal to the first two

[cerulean oxide]

i though that would jsut be the same

but it isnt?=

[hardy slate]

?

ok so essentially

for L = -3, you need x = y + z and x = 2y

2 constraints

so there's only sorta 1 dimension of possibility

but for L = 2 you only need x + y + z = 0

1 constraint

so there's 2 dimensions of possibilities

a whole plane of possible eigenvectors

you want to find 2 orthogonal eigenvectors in that plane

we already have (0, 1, -1) so we need another one

[cerulean oxide]

ok, but if i insert L=3 again

do i not get the same value

[hardy slate]

wdym

there's lots of possible solutions to x + y + z = 0

[cerulean oxide]

yeah

infinite

[hardy slate]

right

so we can find 2 orthogonal eigenvectors in that plane

with eigenvalue 3

[cerulean oxide]

yeah ok

what could another one be

[hardy slate]

so you want x + y + z = 0, and it needs to be orthogonal to (0, 1, -1) as well

think about it

[cerulean oxide]

so

i need to fidn an orthogonal to (0,1,-1)

(1,1,1)

[hardy slate]

ok

but you also need x + y + z = 0

[cerulean oxide]

mhh idk then

[hardy slate]

ok

[cerulean oxide]

(1,1,1) is orthogonal to (0,1,-1) tho

[hardy slate]

it is

but you also need x + y + z = 0

but 1 + 1 + 1 = 3

[cerulean oxide]

yh

[hardy slate]

so

[cerulean oxide]

bit too high

[hardy slate]

how do you tell if something is orthogonal to (0, 1, -1)

[cerulean oxide]

dot product =0

[hardy slate]

right

so what's the dot product of (0, 1, -1) with (x, y, z)

[cerulean oxide]

y-z?

[hardy slate]

right

so you want y - z = 0

and you want x + y + z = 0

[cerulean oxide]

y=z

x=-2?

no

yh idk

[hardy slate]

??

anything that works

please

[cerulean oxide]

y-z=0 just meanst y=z

[hardy slate]

yes

[cerulean oxide]

so i hae x=-2y

[hardy slate]

yes

[cerulean oxide]

that is what i wrote

in the start

[hardy slate]

ok but you need a vector

find an eigenvector like that

[cerulean oxide]

1,-2,-2

[hardy slate]

??

x = -2y

not y = -2x

[cerulean oxide]

-2,1,-2

[hardy slate]

bruh

[cerulean oxide]

why isnt that right tho

y=z
x=-2y

y=x/-2

like what

z=x/-2

[hardy slate]

??

you need x = -2y and y = z right

[cerulean oxide]

yes

then y=x/(-2)

[hardy slate]

but you need y = z

[cerulean oxide]

you divide by -2 on both sides

then z is also just x/-2

[hardy slate]

ok

let's start over

[cerulean oxide]

no no

[hardy slate]

we need x = -2y

and y = z

[cerulean oxide]

how is thatn ot correct

[hardy slate]

pick any z

[cerulean oxide]

ok

[hardy slate]

that isn't 0

[cerulean oxide]

x =- 2y

y=z

[hardy slate]

no

pick

a number

[cerulean oxide]

48

[hardy slate]

ok

[cerulean oxide]

lets go with 1

[hardy slate]

z = 48

[cerulean oxide]

instead

z=1

[hardy slate]

ok

z = 1

then y = z so y = 1

then x = -2y so x = -2

[cerulean oxide]

yes

[hardy slate]

so it needs to be (-2, 1, 1)

[cerulean oxide]

which is what i wrote like 800 years ago

[hardy slate]

because y needs to equal z

no

[cerulean oxide]

ok now i will write it

(-2,1,1)

[hardy slate]

ok

[cerulean oxide]

and everyone is very happy

[hardy slate]

so what you've ended up with is three eigenvectors, all of which are orthogonal to each other

so now you just put them in a matrix next to each other and you have an orthogonal matrix P

done

[cerulean oxide]

does the order matter

[hardy slate]

no

[cerulean oxide]

so we have this

[hardy slate]

that isn't it

wait yes it is

ok

wait

what's happened

am i dumb

[cerulean oxide]

what?

[hardy slate]

the second and third aren't orthogonal

am i really dumb

[cerulean oxide]

what mistake did we make

[hardy slate]

god DAMN it

you solved the linear equations wrong!

i think? wait

[cerulean oxide]

i didnt

i checked with computer

[hardy slate]

ok so

[cerulean oxide]

mhh

bruh

the -1

isnt supposed to be there

😭

[hardy slate]

there we go

jesus

[cerulean oxide]

on the right, its supposed to be [1,0,-1] top

[hardy slate]

right.

[cerulean oxide]

this has to be the right one now

[hardy slate]

ok

[cerulean oxide]

x=z

[hardy slate]

so what's the eigenvector here

[cerulean oxide]

y=z

i say x=1

[hardy slate]

ok

[cerulean oxide]

x=z=y=1=1=1

[hardy slate]

ok

so let's see

yeah so all of the columns of this are orthogonal now

so it works

finally

[cerulean oxide]

phew

now i have to show that with my now P matrix

P^-1*AP is diagonal

[hardy slate]

ok so i explains this ages ago

basically alright so

you have Av = Lv for each of the eigenvectors v

like ok

Av_1 = L_1 v_1
Av_2 = L_2 v_2
Av_3 = L_3 v_3

right?

[cerulean oxide]

mate listen

the question is just find P so this is diagonal

[hardy slate]

no this is fast

look

do you see what i'm saying

this answers your question

[cerulean oxide]

my question of the assignemnt

i asked if we have found P

[hardy slate]

do you agree that this is true

[cerulean oxide]

yes

[hardy slate]

right

so then we can take like A(v_1 v_2 v_3)

AP

and this will be the same thing as PD

where D is a diagonal matrix with L_1, L_2, L_3 as the diagonal entries

so we have AP = PD

so P^-1 AP = D

so you're done, see

[cerulean oxide]

yeah idk what you've said

but this is true

[hardy slate]

$A \begin{pmatrix} v_1 & v_2 & v_3 \end{pmatrix} = \begin{pmatrix} \lambda_1 v_1 & \lambda_2v_2 & \lambda_3 v_3 \end{pmatrix} = \begin{pmatrix} v_1^T \ v_2^T \ v_3^T \end{pmatrix} \begin{pmatrix} \lambda_1 & 0 & 0 \ 0 & \lambda_2 & 0 \ 0 & 0 & \lambda_3 \end{pmatrix}$

[elfin berry]

Kaisheng21

[hardy slate]

well ok this doesn't even work

you need the eigenvectors to all be unit length

but then it works

[cerulean oxide]

yeah but it works by just computing it

[hardy slate]

oh it does? sick

anyway i gtg now

in conclusion, you gotta get more reliable at solving systems of linear equations

[cerulean oxide]

no lol

i just didnt know how to compute P

if i knew that then its somewhat just boring algebra

[hardy slate]

yeah obviously you didn't know wtf was going on

but also you need to get better at systems of linear equations because you kept making the most basic errors, like holy

that wasted a lot of time

[cerulean oxide]

it wasnt to do wit that, sadly

it is me looking at my paper

and typing it onto the computer

and then putting wrong values

like misstyping

xd

[hardy slate]

idc how you do it but you gotta get it right

cya

[cerulean oxide]

cya

enjoy ur day

ty for help

[marsh citrus]

@cerulean oxide Has your question been resolved?

[marsh citrus]

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[mystic minnow]

what have you tried

[unique basin]

You have an optimization problem at hand here

Maximize or Minimize something

Subject to certain constraint(s)

The constraint is the fact that xy = c^2

Normally to maximize something you take the first derivative and set = 0

But because you have 2 variables in

ax + by

You cant do that easily

Use the known constraint to replace one of the unknowns

if you know xy = c^2 must be true

Then you can solve for x or y

And turn it into a 1 variable

Thats what i would do to start probably

You should be able to solve for x or y in terms of a, b, c

Then solve for y in terms of a, b, c

Treat a, b, c like constants

x = (c^2)/y

From the constraint

Plug that into the other expression

a * [(c^2)/y] + by

Take derivative of that with respect to y

And set = 0

[idle ridge]

.close

[marsh citrus]

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[idle ridge]

(The user was banned)

(Sorry to waste your time)

[unique basin]

Oh np xD

[marsh citrus]

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[jovial mulch]

Weird question,
Something like
v= 2t +2t^2 +2

A. Distance of Q from origin , when t =5
B. Total distance travelled of Q from origin in the first 5 seconds

[vernal forge]

A. integrate with bounds 0 and 5

[jovial mulch]

whats the difference

[jovial mulch]

yep did that

[vernal forge]

are you sure you wrote it correctly?

in your case there is no difference

[dusky viper]

one is displacement one is distance i'd assume

[vernal forge]

but in general A is displacement and B is distance

[jovial mulch]

The distance of Q from O, when t=5
The total distance travelled by Q in the first 5 seconds

Thats the exact wording

[vernal forge]

yes

[vernal forge]

^

[jovial mulch]

Ah ok

[vernal forge]

i was asking about the equation

[jovial mulch]

Yeah

[jovial mulch]

no

very wrrong

made it up

[vernal forge]

ah okay

[jovial mulch]

https://tenor.com/view/happy-happy-dog-dog-happiest-dog-super-happy-gif-17885812

[vernal forge]

well basically

[jovial mulch]

yeah one is the displacement

Understood

[vernal forge]

$\int_{\text{start}}^{\text{end}} v(t) dt$ - displacement \
$\int_{\text{start}}^{\text{end}} |v(t)| dt$ - distance \

[elfin berry]

artemetra

[jovial mulch]

but thats weird

Whenever i did distance

I got it correct

[vernal forge]

yeah well you got lucky the object was not going backwards at any point in time

[velvet cairn]

||both will be same in this case as t²+t+1 is always >0||

[vernal forge]

yep

[jovial mulch]

so if it was -t^2

[vernal forge]

then that would've been very different

[velvet cairn]

Then u will have to find where the value is negative and integrate seperately

And then add the magnitudes

[jovial mulch]

right

thats cool you can plot it

didnt know

[vernal forge]

#desmos4life

[jovial mulch]

seriously amazed tho

welp thats it hehe

need to finish it

Thank youuu both

.close

[marsh citrus]

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[jovial mulch]

Mr cuttlefish

[jovial mulch]

Could you teach me how to use this

I understand everything you put in

How would I put in t= 5

[To find the distance]

[vernal forge]

huh?

do you know how to do definite integrals?

like the ones with the bounds?

$\int_1^3 x^2 dx$

[elfin berry]

artemetra

[vernal forge]

can you solve this?

[jovial mulch]

yes

9 - 1

[vernal forge]

you sure?

[jovial mulch]

8

[vernal forge]

bruh

[jovial mulch]

oh wait

[vernal forge]

no

[jovial mulch]

yeah

dx

[vernal forge]

integrate it first

[jovial mulch]

I write it differently

2x

6 - 2

[vernal forge]

no that's derivative

[jovial mulch]

Oh yeah

Wait

[vernal forge]

[jovial mulch]

I SWEAR

NO BRO

idk

[vernal forge]

$\int x^2 dx$

can you do this

[jovial mulch]

Yes

[elfin berry]

artemetra

[jovial mulch]

x^3/3

[vernal forge]

is this doable

[vernal forge]

correct

so

[jovial mulch]

and then whatever that gets

[vernal forge]

$\int_1^3 x^2 dx$

[elfin berry]

artemetra

[vernal forge]

$=\frac{x^3}{3} \mid_1^3$

[elfin berry]

artemetra

[jovial mulch]

Yeah

26/3

But the problem I have

[vernal forge]

yes

[jovial mulch]

So,
So far everyquestion I have done to do with distance I found out

Is that

Its always been here

Or in the exact middle, I found out

Now the question im doing right now, has situations where its on the other side