help-38

[solid kiln]

SWR

[junior viper]

ohhh

i remember my teacher say something like that]

10 divided by 500

0.02

?

nah i get the same aswer for b

[kindred pier]

what are you doing for b now?

[junior viper]

0.02 x 350

which is 17500

because 10 divided by 500 is 0.02

then to find y you do x by 350

so its 17500

for b

oh my god ma

[kindred pier]

are you multiplying or dividing?

[junior viper]

oh yeah

7

multiplying

i did 350 x 0.02

it was supposed to be the other way

thats how i got 7

[kindred pier]

how did you get 7?

[kindred pier]

this is correct. 350 x 0.02 would have also been the correct

[junior viper]

thats 17500

[kindred pier]

show me

I'll bet you its not

[junior viper]

nevermind

your right

i did my calculation wrong

[kindred pier]

you were probably doing 350 / 0.02

[junior viper]

yeah

[kindred pier]

figured

[junior viper]

smart

wow

: )

your pro level

grade 9

so

[kindred pier]

it just requires patience

[junior viper]

is it only for this topic that you divide by the smaller

[kindred pier]

don't move so fast, and read carefully what you are doing

[kindred pier]

wdym

[junior viper]

like 0.02 x 350

instead of 350 x 0.02

[kindred pier]

"the smaller" of what

[kindred pier]

it doesn't matter the order

[junior viper]

oh

[kindred pier]

350 x 0.02 = 0.02 x 350

[junior viper]

ooh

what about inverse proportion

i dont get that at all

and the video was useless

[kindred pier]

this is always true btw

[junior viper]

yooo

thats epic

[kindred pier]

Inverse proportion is almost the same. Direct proportion is $y\propto x$, and inverse proportion is $y\propto \frac{1}{x}$

[solid kiln]

SWR

[junior viper]

so its just fractions

[kindred pier]

yup

[junior viper]

nooo

im cooked

are you good with ratio

sharing ratio and that stuff

something like this

what do i do

the angle of a trangle are in the ratio 1:2:9. What is the size of the largest angle

?

[kindred pier]

angles of a triangle add to 180

[junior viper]

is that the aswere

[kindred pier]

no

[junior viper]

oh

[kindred pier]

I'm just helping you set up the solution

[junior viper]

oh right

i have no clue how o solve it

to

[kindred pier]

Let the smallest angle be x, then the largest angle is 9x and the medium angles is 2x

they must add to 180

[junior viper]

9 x 2

18

and then

18 / 180

= answer

[kindred pier]

what?

[junior viper]

10

[kindred pier]

why did you do this?

[kindred pier]

why did you do this?

[kindred pier]

$\frac{18}{180}\ne 10$

[solid kiln]

SWR

[junior viper]

9 x 2 because you need to find the biggeest angle

[kindred pier]

You are mixing up 180/18 and 18/180

[kindred pier]

how does this find you the biggest angle?

[junior viper]

because it needs to add up to 180

so you just fill it up

[kindred pier]

how does 9x2 add up to 180?

what are you getting at?

[junior viper]

no it = to 18

[kindred pier]

why are you doing 9x2

it makes zero sense why you are doing it

[junior viper]

because i need to make 180 so after i find 9 x 2 i then divide by 180 by 18

9 x 2 since its the angles given

[kindred pier]

Okay

1. The angles must sum to 180
2. what does 9x2 have to do with any of that?
3. how does 180/18 mean the angles sum to 180?

[junior viper]

by doing 9 x 2 and then dividing that by 180 i get the answer. then i x all those together (the ratio thing) and i get 180

it sums to 180

?

what do i do

[kindred pier]

Okay, let's try it your way

you are saying 18 is the largest angle?

[junior viper]

no

im finding 180

do you add instead

so 9 + 2

11

+1

12

[kindred pier]

okay, and now what do we do?

[junior viper]

180 divide by 12?

oh

15

[kindred pier]

okay, now what?

[junior viper]

so 15 = something

12 parts

15 degrees is = to 12

?

[kindred pier]

No. That makes no sense

[junior viper]

oh

[kindred pier]

remember what your 9+2+1 represents

[junior viper]

sorry

[kindred pier]

your angles are in a 1:2:9 ration

we don't know what any of the angles are yet, so they're just unknowns

let's say that x is the smallest angle

Then the angle ratio from smallest to medium is 1:2

So, if x is the smallest angle, then what is the medium angle (in terms of x)?

[junior viper]

x2

[kindred pier]

It's more common to write 2x, but you are right

[junior viper]

yeah

[kindred pier]

and, what is the largest angle?

[junior viper]

9x

[kindred pier]

yup

so, your angles are x, 2x, and 9x. And they're angles of a triangles, so they must add to 180

how would you write that algebraically?

[junior viper]

add them together - 12

=12

oh

[kindred pier]

minus 12?

[junior viper]

uhh

no

i meant = 12

sorry

[kindred pier]

what equals 12?

[junior viper]

9+2+1

11 my bad

nO ITS 12#

sorry

[kindred pier]

what happened to all your x?

[junior viper]

12x?

then

[kindred pier]

yes, and this is equal to 180

[junior viper]

would you need to write x?

[kindred pier]

yes

[junior viper]

okay

[kindred pier]

because 1, 2, and 9 by themselves mean nothing

[junior viper]

12x = 180?

[kindred pier]

Your angles are x, 2x, and 9x

[kindred pier]

yes, precisely

[junior viper]

so techinically all we did was just add them up and put x in

[kindred pier]

your angles x, 2x, and 9x sum to 180. That is, x+2x+9x=180. You simplify the left, giving you 12x=180

[kindred pier]

we added x, 2x, and 9x, and set that equal to 180

[junior viper]

would you add x into every question

with ratio

[kindred pier]

x, 2x, and 9x are your angles. x+2x+9x is the sum of your angles. The sum of your angles is 180, so x+2x+9x=180

[kindred pier]

yes

[junior viper]

ohhh okay

yooo thankss so much

legend

would that question be higher level or foundation

[kindred pier]

a ratio like $1:2$ just means that one value is twice the other. We don't know what those values are, we just know what we get when we divide them. If the smaller value is some unknown $x$, then the larger value must be $2x$, because $\frac{x}{2x}=\frac12=1:2$

[solid kiln]

SWR

[junior viper]

right

thanks man

you helped me alot

thankss

.close

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[plain tusk]

help please

didnt really make a start because i got it wrong last i time tried it

[hidden dew]

plug 2 into the x

[plain tusk]

like this?

i leave the g(2) alone right?

just carry it down when solving the right side

[hidden dew]

yea

[plain tusk]

negative 2

i did 2 squared x 2

8

and then minus 10

[mild cosmos]

your answer is correct

[plain tusk]

oh ok

thx

also one more that is kinda confusing

[mild cosmos]

sure send it

[mild cosmos]

ok so in general if you are given f(x) and you are told that it has an inverse do you know the process of finding f^(-1)(x)?

[plain tusk]

sorry i lost internet for a min

[plain tusk]

yes, you change the x into a y

or at least i was taught that way from what i remember

[mild cosmos]

you replace x by y or f(x) by y ?

[plain tusk]

the x by y

wait let me show

[mild cosmos]

no actually thats not what you should do

you replace f(x) by y

then solve for x in terms of y

after that swap x and y

the result will be y in terms of x

[mild cosmos]

this y will be your f^(-1)(x)

ah wait

yes your way works too

alright

[plain tusk]

yeah

[mild cosmos]

so lets do this

[mild cosmos]

great what next

[plain tusk]

i think u kept x as in the 4x?

and made y the fx

[mild cosmos]

exactly

[plain tusk]

oh ok

mines just the opposite ig

[mild cosmos]

what i meant is to rewrite as y=4x-1 then you work to get x=... , this ... is something in terms of y after that you switch the places of x and y

what you are doing is similar

[plain tusk]

oh ok

[mild cosmos]

so lets work in your way

because you are probably used to that

[mild cosmos]

ok so from this what do you get

[plain tusk]

x+1 = 4y

[mild cosmos]

and then ?

[plain tusk]

here i worked out this bit

[mild cosmos]

so f^(-1)(x)=?

[plain tusk]

x+1 divided by 4 right?

[mild cosmos]

exactly

[plain tusk]

ah ok

that wasn't as bad as i thought it'd be tbh

[mild cosmos]

you can check that by plugging this into f

it should give you x

since f(f^(-1)(x))=x

[plain tusk]

into this?

[mild cosmos]

or you can plug f(x) into this

[plain tusk]

oh ok

anyways thanks for your help with the questions

[mild cosmos]

so $f(x)=4x-1$ which means $f(\frac{x+1}4)=4(\frac{x+1}4)-1=x+1-1=1$

[solid kiln]

pirateking0723

[plain tusk]

wait why did u divide by 4?

on the which means bi

t

[mild cosmos]

or you can do this by saying $f^{-1}(f(x))=f^{-1}(4x-1)=\frac{4x-1+1}4=x$

[plain tusk]

but why did u have to divide by 4

[solid kiln]

pirateking0723

[stable crescent]

u have y = f(x) = 4x - 1 right?

[mild cosmos]

yes

[stable crescent]

(y+1)/4 = x

[plain tusk]

us is the is the y fx-1?

[mild cosmos]

but you have $f^{-1}(x)=\frac{x+1}4$ right

[solid kiln]

pirateking0723

[plain tusk]

yes

[mild cosmos]

so $f(f^{-1}(x))=f(\frac{x+1}4)=4(\frac{x+1}4)-1=x+1-1=x$

[solid kiln]

pirateking0723

[mild cosmos]

thats where the division by 4 came from

[plain tusk]

ohh ok

[mild cosmos]

this is a method to check your work

ie to check if you got the inverse right or no

[plain tusk]

i see

[mild cosmos]

you dont need to do both f(f^(-1)(x)) and f^(-1)(f(x))

you can only do one of them

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@plain tusk Has your question been resolved?

[mild cosmos]

if you are done please type .close

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[hidden elk]

im most likely glossing over something for this problem. I'm trying to use using a non-homogeonous solution, and i've gotten to the following point:

y'-y(1+x) = 1 + x

[hidden elk]

when i apply the μ(x) portion, i get that μ(x) = e^(x+x^2/2)

but im finding that this doesn't check out

silly mistake, nevermind

.close

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[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[urban copper]

well 1 + i is a root

-1 is a repeated root

1 - i is another root

because RRT , because P in R[x]

so the polynomial is a quartic

[lilac flame]

what are its factors

[silent canyon]

Who started this trend 💀

[urban copper]

maximo

[urban copper]

P(x) = a((x-(1+i))(x-(1-i))(x+1)(x+1)

[lilac flame]

yes those are factors

[urban copper]

the hard part is the last condition

[lilac flame]

in general P(x) = ((x-(1+i))(x-(1-i))(x+1)(x+1) * f(x) for some f in R[x]

plug in x=i

[urban copper]

P(i) = a((i-(1+i))(i-(1-i))(i+1)(i+1)

[willow mortar]

What does the last condition mean ?

Aaaah got it

[lilac flame]

u can simplify this

[lilac flame]

-2a = 28

[lilac flame]

solid

haven't checked ur work but looks fine

[urban copper]

bro it takes a second to check

[lilac flame]

be thankful tf

[urban copper]

lmao, mb

[lilac flame]

u can check on wolfram alpha

[urban copper]

!volu

[trim jolt]

Helpers are just people volunteering their time to help you. Be polite and patient.

[lilac flame]

,w p(x) = x^2, what's p(2)

[solid kiln]

Results provided by WolframAlpha

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Requested by mayervietorus

[urban copper]

.solved

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[vocal cobalt]

the answer from the video and the answer from the solution is different... which one is correct? im really confused... (from ap classroom vid)

[zinc ginkgo]

What video

[vocal cobalt]

ap classroom video

[zinc ginkgo]

Which one is which

Blue is video?

[bright quarry]

the red one is a right riemann sum

[vocal cobalt]

yeah the blue one is video and the red one is the from the answer key attached

[bright quarry]

video is correct

[vocal cobalt]

so they kinda confused themselves...? when writing solution?

[bright quarry]

yes

[vocal cobalt]

ok

thanks!

.close

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[languid scaffold]

How do I set this up? Like I understand the total needs to be <3.2ppm
Would I average them like (2.7+3.42+x)/3 to get the average?

[harsh sleet]

yup

[languid scaffold]

thanks

.close

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[twin yoke]

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[twin yoke]

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[bright quarry]

area of a circle

[twin yoke]

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[bright quarry]

what

[trim musk]

its semicircle right?

[twin yoke]

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[trim musk]

y>0

[twin yoke]

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[bright quarry]

i mean yea technically you want a quarter circle

[vague walrus]

looks pretty easy

simple formula based

[bright quarry]

but just know the formula for a circle and divide it by 4

[vague walrus]

even if u dont wanna use circles u should know that

it is the general integral

[twin yoke]

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[bright quarry]

no

just find the area of a circle with radius 4 then divide it by 4

since you have a quarter circle

[restive vortex]

Yeah but how to integrate it

[twin yoke]

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[restive vortex]

We use u sub

Or sum

[bright quarry]

let x = 4sin(u) lol

[bright quarry]

you’ll just get to the same shit

[restive vortex]

I forgot trig sub

[bright quarry]

big waste of time

[twin yoke]

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[vague walrus]

i personally remember those tbh

makes it faster

[twin yoke]

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[bright quarry]

dawg turn off dark mode on desmos

[vague walrus]

area of semicircle

find it

[bright quarry]

you have reverse contrast on

[twin yoke]

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[bright quarry]

you can’t see see the grid

[twin yoke]

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[trim musk]

jeez what is this

[trim musk]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[glacial rose]

15 no. Please

[twin yoke]

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[bright quarry]

bro wtf is wrong with your desmos

[twin yoke]

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[bright quarry]

how did you manage to make it so gross

[vague walrus]

what are yall doing with desmos bruh 😭

[twin yoke]

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[restive vortex]

Desmos get obliterated

[twin yoke]

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[gray swallow]

In the quadrilateral ABCD AB 14 BC 48 CD 40 and AD 30 find the diagonal AC and the area of the quadrilateral if AC divides its area in the ratio 14:25

[gray swallow]

im having a tough time with this one

[magic eagle]

what did u try?

[safe loom]

is trigonometry allowed

[gray swallow]

i tried to find the half perimeter

the i tried a herons formula but it didnt help much

it got complicated

[gray swallow]

i dont think thats gonna help us out that much

[safe loom]

So like, AC divides the shape into 2 regions and ratio of those is 14/25?

[gray swallow]

ye

[magic eagle]

use herons formula itself

its coming

its not too complicated

[safe loom]

I mean heron's formula does do the trick here but the equation is very complicated

[magic eagle]

its just some x^2 and x^4s lmao

[safe loom]

That's the problem

In a test for example, it wont be an effective method because expanding it is very troublesome for such a small time

Though if nothing else works then ig its a last resort

[magic eagle]

can trigonometry solve this tho?

[gray swallow]

it will only get harder with trigonometry

[safe loom]

I think it can, if we focus on B and D and find its connection through the area ratio

Then use law of cosine to find AC

[magic eagle]

yea, but too many unknowns

[safe loom]

Sigh

Wait for a few sec

oh look how convenient it is

sin B = sin D

that means B + D = 180

that means cos(B) = -cos(D)

done, problem solved

[gray swallow]

oh

[safe loom]

tbh i also did not expect this

Although, something isn't right

From my observation, S_ABC is definitely smaller than S_ADC

Is the figure drawn not to scale?

If it isnt then B+D=180 will cause it to be an isosceles trapezoid

(Which obviously isnt right at all)

@gray swallow

[gray swallow]

not really

[safe loom]

huh.

lemme try the opposite way, assume S_ADC < S_ABC

yeah no that obviously doesnt work lol

@gray swallow im pretty certain something is wrong with the problem. the sine ratio should not be 1 because it will make it an isosceles trapezoid

either that or something is wrong with the figure
(yeah no the problem lies at the... problem. i have tried every possible way to draw a trapezoid but nothing satisfies the given problem)

[gray swallow]

aw man

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@gray swallow Has your question been resolved?

[gray swallow]

.close

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[jovial thorn]

$\sum_{n=0}^{\infty} \frac{n^3}{n!}$

[solid kiln]

kronium_

[jovial thorn]

I tried to do it

in my mind

but for some reason I keep failing and cant pin point the reason

everything I do just seems to make sense but it doesn't get me to the answer

$e^x$ = $\sum_{n=0}^{\infty} \frac{x^n}{n!}$

[solid kiln]

kronium_

[jovial thorn]

Now derive both sides

$e^x$ = $\sum_{n=0}^{\infty} \frac{nx^{n-1}}{n!}$

[solid kiln]

kronium_

[jovial thorn]

mutliply both sides by x

$xe^x$ = $\sum_{n=0}^{\infty} \frac{nx^n}{n!}$

[solid kiln]

kronium_

[jovial thorn]

derive once again

$e^x + xe^x$ = $\sum_{n=0}^{\infty} \frac{n^2x^{n-1}}{n!}$

[solid kiln]

kronium_

[ionic pendant]

it doesn't still start at 0 though

[jovial thorn]

what

[ionic pendant]

the n = 0 term is a constant so it goes to 0

so the first term is n = 1 after that

[jovial thorn]

yes

like

what ur saying is

$\sum_{n=0}^{\infty} \frac{n^3}{n!}$ = $\sum_{n=1}^{\infty} \frac{n^3}{n!}$

[ionic pendant]

what?

[solid kiln]

kronium_

[jovial thorn]

how is it false

[ionic pendant]

well it's trivially true i suppose

just because the n = 0 term is 0

[jovial thorn]

exactly

but whats

wrong

wutg wgat

u dud

I did

[ionic pendant]

but i don't see how it pertains to what i said

[jovial thorn]

theres a bee in my room

lemme get rid of it

[ionic pendant]

i am talking about when you took the derivative of both sides of the equation

[jovial thorn]

what then

[ionic pendant]

,, \odv*{\sum_{n=0}^{\infty} \frac{x^n}{n!}}{x} \ne \sum_{\mathcolor{red}{n=0}}^{\infty} \frac{nx^{n-1}}{n!}

[solid kiln]

cloud

[jovial thorn]

it kinda is?

equal

both of those

become e

'e'

[ionic pendant]

not the second one

[jovial thorn]

yes the second one too

[ionic pendant]

the second one isn't even a power series

[jovial thorn]

but the second one still becomes e

not an approximation

exactly 'e'

I MEAN

e^x

mb

e^x

$e^x$

[solid kiln]

kronium_

[ionic pendant]

i suppose it doesn't matter that much since the first term is 0 anyway

[jovial thorn]

right

so

what happens if

is

nx^(n-1)/n!

n! = (n-1)! * n

the n's cancel each other out

so

u get

sum from 0 to infinity

x^(n-1) / (n-1)!

and since n starts from 0

the first term will be (1/x)/(-1)!

which is

INFINITY

so it becomes 0

and then like normal

[ionic pendant]

what i'm saying is that you are applying the power rule to say [ \odv*{x^0}{x} = x^{-1} ] which makes no sense

[solid kiln]

cloud

[jovial thorn]

right

OH RIGHT

wait no

$

$\frac{d}{dx} x^0 = 0x^{-1}$

[solid kiln]

kronium_

[safe loom]

x = 0

[jovial thorn]

👀

oh shit

now I understand...

[safe loom]

i mean ||there's an easier way to evaluate this series||

but i'll let u figure it out a bit more until u give up

[jovial thorn]

how so

$e^{e^x}$ = $\sum_{n=0}^{\infty} \frac{e^{nx}}{n!}$

[solid kiln]

kronium_

[jovial thorn]

so whats ur easier way

[safe loom]

||instead of focusing on this, why not just keep differentiating?||

||also you should observe: n^3 = n(n - 1)(n - 2) + 3n(n - 1) + n||

[nova spire]

this I also like

[jovial thorn]

wait

wait

,w expand n(n-1)(n-2)

[solid kiln]

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[jovial thorn]

right

now u want me to split it up into 3 different summations?

[safe loom]

yep

[jovial thorn]

so

if I differentiate

3 times

it stays e^x

and then replace x with 0

so it becomes 1

so its

[nova spire]

uh

[safe loom]

if you do x = 0 then the series will collapse

[jovial thorn]

e

so then what

[safe loom]

x = 1

[jovial thorn]

OHH

BRUH

MB

MB

I KEPT THINKING THE WRONG WAY

damn

I see now where I went wrong

I totally forgot

I thought x was the exponent

im my head

ahhh

mb

yeah got it

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[plain storm]

4x^2+13x+2x^3-80=0

[plain storm]

how do ii solve that

i mean

find x

[summer python]

you sure you've copied the question right?

[plain storm]

yes

[summer python]

damn a cubic

no idea on this one

[plain storm]

😭

[summer python]

wait a sec

mb

try diviiding the equation with different factors

like x+1, x-1 and stuff

[plain storm]

like

x(2x(x+2)+13)-80=0

?

[summer python]

not quite

[plain storm]

huh

[summer python]

if the sum of the coefficients of the terms equal 0, (x+1) is a factor

if the sum of the coefficients of the terms with the odd powers equals 0, (x-1) is a factor

actually yeah that doesnt work here

but just in case, keep those in mind, they're helpful

[plain storm]

okay

oh i got it ok

thanky you

i did

[summer python]

can i ask how you solved it

[plain storm]

f(x)/f'(x)

[summer python]

OH THAT

damn

[plain storm]

bruh ididnt even know 😭

ok thank you

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[sweet turret]

is the sum notation a must to write

i basically moved the everything except S outside the integra

[marble wharf]

its wrong to write it twice

[sweet turret]

ok

thx

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[proven root]

Hi could I ask how do I rationalize or smth this question in order to find the limit?

[proven root]

im a bit confused because im used to rationalizing, but apparently that is not the correct term to use when trying to cancel out the numerator

and when cross multiplying, instead of -8, its positive 8..?

[restive vortex]

The limit doesn't exist

[proven root]

thats honestly so cool, can I ask how were you able to solve it

[restive vortex]

As x aproach 0

The denominator became smaller

Wait

[proven root]

the other answer I got was 1/16, is that wrong

[restive vortex]

Hmmm

[proven root]

ive double checked with calculators, AI, and manually checked it on my own but each one is different

😭

[restive vortex]

Let me solve it first

[stark bison]

,w limit of (sqrt(x^2+64)-8)/x^2

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[proven root]

this is worth like 80% of my performance task so I was desperate and had to search for a math server

[stark bison]

1/16 is right yeah

[proven root]

okay thankyou

[proven root]

oh wow

okay thankyou

im genuinely so afraid to submit this

[restive vortex]

Yeah 1/16 should be right

1/16 is right

[proven root]

okay thankyou legends ill try submitting

okay it was correct thankyou so much everyone

I appreciate it

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[leaden aurora]

This is an Undergraduate level mathematics question, would like some help as I got no clue how to covert the given equations of a great circle into an equation of a sphere.

[hasty yew]

what class is this from?

[leaden aurora]

Undergraduate, Semester 1, Maths question.

Geometry class

[main sigil]

i dont get how that circle is defined

the first equation is equivalent to
(x+1)^2 + (y-2)^2 + z^2 = 0

[main sigil]

and this defines a degenerate sphere (as in 0 radius)

(so actually a single point)

and the next equation is one of a plane

[leaden aurora]

i.e. the great circle I suppose

[main sigil]

i dont get it

the first equation defines a point, the 2nd one defines a plane

where is the circle?

[leaden aurora]

That's what I was confused about myself, I think the question might just be flawed, probably gotta ask the professor himself 🥲

[main sigil]

That'll probably be best

[naive otter]

First one is sphere

[main sigil]

show him this

[naive otter]

2nd one is plane

[main sigil]

degenerate one

so actually a point

[leaden aurora]

Alrighty then

[naive otter]

Anyways it's a circle

[main sigil]

I think that the sphere was meant to be non-degenerate and then the intersection of the plane and the sphere would define the great circle

[leaden aurora]

I think so too, I'll just ask him before class, thanks guys.

[main sigil]

(-1, 2, 0) is the point. The plane doesnt pass through this point. So the intersection is empty and we have no circle

[naive otter]

Oh lol i didn't check, ty for clarification

[leaden aurora]

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[velvet gorge]

Is 1+0.5+0.25+0.125… an integral?

[empty orchid]

No, it’s an infinite sum.

Specifically, $\sum_{n=0}^{\infty}\qty(\frac12)^n$.

[solid kiln]

;(

[velvet gorge]

Isnt that what an integral is?

[empty orchid]

No, that’s a sum.

[velvet gorge]

Can you explain the difference between an infinite sum and an integral?

Or recommend something to watch to understand it?

[empty orchid]

An infinite sum is just a sum, but has infinity as an upper bound. An integral is a “smoother form” of a sum. In other words, a sum is discrete, and an integral is continuous.

[velvet gorge]

So irrational?

Or infinite repetition in decimal places?

[empty orchid]

No.

[empty orchid]

For?

[velvet gorge]

I’m confused

[restive vortex]

Sigma is just the length

Integral is the area

[velvet gorge]

Ohh

[empty orchid]

Wouldn’t say that, necessarily.

[velvet gorge]

Why?

[restive vortex]

Yeah it's too oversimplified

[empty orchid]

Because something like the euler-mascheroni constant exists.

[restive vortex]

Euler macaroni constant

[velvet gorge]

Is that the fancy “i”

[empty orchid]

,w euler-mascheroni constant

[solid kiln]

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[restive vortex]

Is the root of -1

[empty orchid]

This comes from the sum of 1/x subtracted by the integral of 1/x to infinity.

[empty orchid]

That’s just the imaginary unit.

[velvet gorge]

😵‍💫

[restive vortex]

Diffrent number

[velvet gorge]

Yeah, but i still confused 😭

[empty orchid]

There’s no better way to put it.

$\int_a^b f(x)dx=\lim_{n\to\infty}\sum_{k=1}^n \Delta{x}f(x_k)$

[velvet gorge]

It might help you to know I’m 13 and dont have any experience in calculus other than sigma notation

[solid kiln]

;(

[restive vortex]

Read James Stewart calc book

Good book

If you really want to learn it

[velvet gorge]

👍

I get it now

[somber ginkgo]

in the context of measure theory, yeah

you can view it as the integral of 1/2^n w.r.t the counting measure

if you do more analysis, you'll see that actually sums and integrals behave quite similarly

for example, if a series is absolutely convergent, then you can sum up the series in any way you want

similarly, if you have a double integral where integrating the absolute value gives < infinity, then you can rearrange the 2 integrals

anyway that's some motivation (not the main motivation tho) behind having measure theory cus then that unites the theory between sums and integrals

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[velvet gorge]

Thanks

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[solid kiln]

938c2cc0dcc05f2b68c4287040cfcf71

[tepid prism]

if you take a vector out of S_1 how does it transform under the action of f?

[urban copper]

the output is in S1

[tepid prism]

yes but you can write it more specifically

[urban copper]

f(S1) ∈ S1

or what do you want me to say?

[tepid prism]

if you take any vector out of S1 you can represent it as a linear combination of basis vectors in your case v_1

and then you can apply the linear transformation to the linear combination

[urban copper]

λf(v1) ∈ S1

[tepid prism]

that might not be the case

in general

if u is an element out of S1, then

u = λ*v_1

[urban copper]

λf(v1) ∈ S1

this doesnt work?

[urban copper]

yeah

[tepid prism]

thats correct

[urban copper]

so?

[tepid prism]

f(u) = f(λ*v_1) = λf(v_1)

[urban copper]

yeah

linearity

scalars out

[tepid prism]

so

but f(v_1) has to be equal to some non zero number times v_1

because it is a 1-d subspace and it only has one possible direction

Im sorry I gtg

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@urban copper Has your question been resolved?

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@urban copper Has your question been resolved?

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@urban copper Has your question been resolved?

[nova spire]

If you haven't found how to continue

just remember that S1 = span(v1) and S2 = span(v2,v3)

so if f(S1) = S1

that means in particular f(v1) in span(v1)

so f(v1) = a*v1

and a is not 0, otherwise f(S1) = {0}

try the same reasoning for S2

[urban copper]

f(S2) = S2
f(bv2 + cv3) ∈ S2

[nova spire]

what are b and c

[urban copper]

non zero scalars

b,c∈ R

[nova spire]

where do they come from

isn't it easier to just investigate f(v2) and f(v3)?

since f is linear

[urban copper]

I mean

f(v2) = bv2 + dv3

f(v3) = cv2 + ev3

that's from Mb(f)

[urban copper]

problem is, is not as easy because S2 = <v2,v3>

f(v2) ∈ S2

f(v3)∈ S2

f(v2) = bv2 + dv3
for b,d ∈ R nonzero scalars

f(v3) = cv2 + ev3
for c,e ∈ R nonzero scalars

look
we know
f(S2) = S2
S2 = <v2,v3>
s2 ∈ S2
s2 = αv2 + βv3
f(S2) = αv2 + βv3

also we know

f(S2) = <f(v2),f(v3)>

[nova spire]

not necessarily

some of b,c,d,e can be zero

[urban copper]

true

[nova spire]

no, you have to use that

f(v2),f(v3) must be a basis of S2

[urban copper]

f(S2) = <f(v2),f(v3)> = S2

[nova spire]

If V = {v1,v2,...,vn} is a basis of some space E

and A is an n*n matrix

when is AV = {Av1,...,Avn} a basis of E?

[urban copper]

if A is bijective

I mean when it has only output for every single input

aka when A is invertible

[nova spire]

yup

[urban copper]

is still possible that f is bijective but one of the basis vectors in S2 map to 0

[nova spire]

if a non zero vector is mapped to 0

and the zero vector is also mapped to 0...

[urban copper]

lol

my baad

so what would be the proof?

[nova spire]

.

[urban copper]

why do we care about S1 ∩ S2 ={0}

[nova spire]

.

[nova spire]

I don't understand what you mean by that

we have two supplementary sets S1, S2

so we can create a basis of R^3 that is basically a basis of S1 + a basis of S2

[urban copper]

I dont understand why they give S1 (+) S2 = R3
is possible that for example S1 + S2 = R3
when dim S1 = 2, dim S2 = 2 and dim S1 n S2 = 1

[urban copper]

yeah

[nova spire]

dim(S1 n S2) = ...

[urban copper]

dim(S1+S2) = dim(S1)+dim(S2)-dim(S1nS2)
dim(S1+S2) = 2 + 2 - 1 = dim(R3)

[nova spire]

so you want both S1 and S2 of dim 2

but that's not what the exercise is looking for

[urban copper]

yeah I was pondering, not that it matters

[nova spire]

The question statement is generally not to be questioned with, the guys creating the exercise set the stage however they want

you can question it if what they set is false

[quartz wave]

Hi

[urban copper]

well if f(S2) = span{f(v1), f(v2)}
and f is bijective
then span{f(v1), f(v2)} is also a basis for S2

[nova spire]

you mean {f(v1),f(v2)} is also a basis

but yes

so

[urban copper]

well if f(S2) = span{f(v1), f(v2)}
and f is bijective
then {f(v1), f(v2)} is also a basis for S2

like that?

[nova spire]

so

restrict f to S2

then M_(v2,v3) (f) = some matrix

and since f is bijective from S2 to S2

that matrix that represents f must be...

[urban copper]

the matrix with columns (f(v2))_S2 and (f(v3))_S2

[nova spire]

yes

but what property does this matrix have

[urban copper]

invertible

[nova spire]

alright

and how do we characterize invertible matrices?

[urban copper]

nonzero det

[nova spire]

and there you have it!

so with the exercise's notations:

f(v_2) = bv_2 + dv_3
f(v_3) = cv_2 + ev_3

$Mat_{{v_2,v_3}} (f) = \begin{pmatrix}b&c\d&e\end{pmatrix}$

[solid kiln]

rafilou is not not born in 2003

[nova spire]

$|Mat_{{v_2,v_3}} (f)| \neq 0$

[solid kiln]

rafilou is not not born in 2003

[urban copper]

yeah, is just

yeah I am convinced is true, is just that I am struggling to prove

[nova spire]

what part

[urban copper]

how do you prove for example f is bijective

aka an isomorphism, you used it, but we assumed Mb(f) dets is nonzero

[nova spire]

a surjective function between two sets of the same finite dimension

rank/nullity theorem for example

etc...

[urban copper]

every endomorphism is surjective then?

[nova spire]

no

I said that every SURJECTIVE endomorphism is bijective in finite dimension

[urban copper]

well R3 is finite dimension and our f is an endomorphism because f: R3 -> R3

but how do we show is surjective?

[nova spire]

f is surjective, but we're not interested in that

if you want to prove it in your own time, it's because since R^3 = S1 (+) S2, every vector of R^3 is the sum of a vector of S1 and a vector of S2

but since f(S1) = S1 and f(S2) = S2

it's gonna be easy finding an inverse image for each component

and thus by linearity, we get an inverse image of the original vector

anyways

we're more interested in:

f:S2->S2 is bijective

we can call it a different name since we changed domain and codomain

say g:S2->S2 is the endomorphism such that g(v) = f(v)

[urban copper]

okay

[nova spire]

so

g is bijective

because it's a surjective endomorphism in finite dimension

(g(S2) = f(S2) = S2)

so

$Mat_{{v_2,v_3}} (g) = \begin{pmatrix}b&c\d&e\end{pmatrix}$

[solid kiln]

rafilou is not not born in 2003

[nova spire]

g is bijective

so this matrix is invertible

det(this matrix) != 0

now we can assert that M_B(f) = this

and we proved everything

in fact, here's the alternative proof of why f is bijective

now that we've shown M_B(f) = this with a !=0 and det(...) != 0

M_B(f) is invertible

i.e f is bijective

[urban copper]

I mean, proof seems okayy, is not like I can actual advice about the proof since I am still learning and I am new to linear algebra, but at least I am convinced is true when a is nonzero, about the determinant of {{b,c},{d,e}} I think we need to take that for a fact, otherwise f(v2) is a linear combination of f(v3)

[nova spire]

again, if that last part was blurry to you

I invite you to go back to the proof of:

"let f:E->F linear

then f bijective <=> f sends any basis of E to a basis of F <=> f sends at least one basis of E to a basis of F"

[urban copper]

is okay I found the link, apparently is a theorem

https://math.stackexchange.com/questions/2521291/proof-that-f-is-an-isomorphism-if-and-only-if-f-carries-a-basis-to-a-basis

[urban copper]

thank you for the help and ty for the invitation, I think is solved I think

the determinant of {{b, c},{d, e}} needs to be nonzero because

that's the only way f is an isomorphism

like if S2 = {v2,v3} is a basis

M_S2(f) = {{b,c},{d,e}}
and since f(S2) = <f(v2),(f(v3))>
we need {f(v2), f(v3)} to be a basis for S2 aswell

so if M_\S2(f) = {{b,c},{d,e}}

(f(v2))_S2 = (b,d)
(f(v3))_S2 = (c,e)
meaning we need
f(v2) = bv2 + dv3
f(v3) = cv2 + ev3 to be linearly independent otherwise
M_S2(f) is non invertible and thus
f is not an isomorphism
and if f is not an isomorphism then
{f(v2), f(v3)} is not a basis for S2

you see what I mean? I think I got it, unless I made a boo boo

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[urban copper]

.solved

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[rapid meadow]

how do you diff 1/e^(x^2)

[marble wharf]

write it as e^(-x^2)

[neon dirge]

and from there chain rule

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[sweet turret]

how does the general solution look like when 2 same eigenvalue gives one eigenvector

[sweet turret]

this seems to be wrong compared to online calculator tho

[trim jolt]

@sweet turret Has your question been resolved?

[neon dirge]

I think we're missing more context

[sweet turret]

meaning?

[neon dirge]

what the given system was

[sweet turret]

this was the question

i wanna find the actual answer

can the same eigenvector be used twice?

<@&286206848099549185>

[neon dirge]

nop

you need another vector independent of (1 0)

[sweet turret]

So there's no solution to this..?

[neon dirge]

ofc there is, just take (0 1)

[neon dirge]

and you get the result described here too

you can check by directly solving for x & y too

it gives the same result

[sweet turret]

Why is (0,1) valid

Let's say if I didn't use the online calculator how would ik (0,1) is an answer

[neon dirge]

if you have a defective eigenvalue you can solve for it with (A-lambda * I) * new_vector = existing_eigenvector

which in your case is

(A-lambda * I) * (v1 v2) = (1 0)

which yields v1 = 0 and v2 = 1

[sweet turret]

Lambda is 1 also?

[neon dirge]

so (0 1) is your new "constructed" eigenvector

[sweet turret]

Then I find the inverse?

[neon dirge]

yes

to generalize the process:

you had an eigenvalue that's a double root in the characteristic polynomial

so its arithmetic multiplicity is 2

however you found out you can only get 1 independent eigenvector

meaning the geometric multiplicity is 1

[sweet turret]

Yea

[neon dirge]

therefore you need to "construct" more eigenvectors, which aren't true eigenvectors

but they are uniquely determined by the corresponding eigenspace

[sweet turret]

Yea I def didn't learn this shi

[neon dirge]

hm but how was the procedure taught to you?

[sweet turret]

They only taught us basic of finding eigenvalues and eigenvectors, and the examples given usually has 2 eigenvectors for 1 eigenvalue

111
000
000

[neon dirge]

you mean arithmetic and geometric multiplicities always aligned?

[sweet turret]

Like this matrix I know that y and z are free variables

[sweet turret]

Not rly but mostly yes

[neon dirge]

hm k

[sweet turret]

Also I wanna ask

For eg

011
001
000

Why is (1,0,0) an eigenvector

[neon dirge]

for the eigenvalue 0?

[sweet turret]

And does that mean (0,0,1) is also an eigenvector?

[sweet turret]

Ya

[neon dirge]

I mean any multiple of (1 0 0) gets mapped to 0

[neon dirge]

why'd it be

[sweet turret]

Introduce x as free parameter?

[neon dirge]

M * (0 0 1) = (1 1 0)

not a multiple of (0 0 1)

however M * (1 0 0) = (0 0 0)

which also holds for M * x * (1 0 0) = (0 0 0)