help-38

[uncut bough]

Yea

[valid beacon]

https://tenor.com/view/triangle-sign-lilly-singh-hand-sign-seeing-through-the-triangle-be-a-triangle-gif-26881397

triangles

[uncut bough]

But I’m thinking of just x^2 /2

[empty orchid]

now, lets take the base of the triangle (for 0<x<1) to be x

[uncut bough]

Why is it x-x^2 /2

[empty orchid]

yeah, but consider that y=1-x.

what is the area of the triangle in terms of x and y?

[uncut bough]

Can you help me making the step ?

[uncut bough]

Can you help me ?

[empty orchid]

draw it

[empty orchid]

i guess that works.

[uncut bough]

Where to get X-x^2/2

[empty orchid]

what will be the base and height of the triangle

[uncut bough]

Height is 1-x

And base is x ?

[empty orchid]

yeap

so u divide all that by 2 (A=ab/2)

[empty orchid]

yes

meknows i can integrate it normally

methinks why its not correct

[uncut bough]

What do you mean ?

[empty orchid]

methinks to myself

[uncut bough]

Ohh okeyyy

[empty orchid]

meunderstand now.

@uncut bough think of it as trapezoid rather than triangle

let me show!

[uncut bough]

Aightt thanks

[uncut bough]

So y = 0 ?

[empty orchid]

no

we are analyzing $\int_0^x f(t)dt$, so $x$ is changing from $0<x<1$

[solid kiln]

;(

[empty orchid]

therefore the area will always be a trapezoid

unless x=1

[empty orchid]

you know y=1-x, so now solve and i believe it will work?

yes it works perfectly

@uncut bough

[uncut bough]

Okey

[empty orchid]

sorry its like 2am over here

[uncut bough]

Let me try

Ohh yea

That workssss

Thank youu so muchhhh

Appreciate ittt

Thanks!!!

@empty orchid

[empty orchid]

i will be off now

[uncut bough]

Good night

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[wraith hinge]

Hi, this is a test because I am new here, and I am curious on the process works. Please help me with the following math problem: Solve for x:
\sqrt{x} = 4

[wraith hinge]

I am assuming for an easy stuff, the bot will show how solve it, and for the hard questions a human help 🙂

[quick tusk]

No bot, we're all humans

[wraith hinge]

hello are you real?

oh! wow

[mental anchor]

Humans help for all problems

[wraith hinge]

oh I didn't know this!

okay I will try and come up with some truly hard problems then 🙂

[quick tusk]

dead internet theory hitting hard huh

[wraith hinge]

lol. Well, the answer is 16

[mental anchor]

[wraith hinge]

thank you for your help 🙂

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[brave rampart]

help pls

what is the approach?

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@brave rampart Has your question been resolved?

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@brave rampart Has your question been resolved?

[past widget]

Start with bounding: for any permutation of (1, 2, ..., n), the given expression 2n > n/1 + (n-1)/2 + ... + 1/n ≥ a_1 + a_2/2 + ... + a_n/n ≥ 1 + 1 + ... + 1 = n

perhaps that might give an insight

but since you're to count number of permutations in a closed form s_n, it might not helpmuch

||induction go brrrr||

[brave rampart]

can you show how to use the induction in this case

@past widget

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[radiant portal]

I think I know how to solve it but im just stuck on 1 thing

u = x^2 - 3

du = 2x dx

[split chasm]

du = 2x dx

[radiant portal]

Thx

To get rid of the x^3 do I do

u = x^2 - 3

u + 3 = x^2

sqrt(u + 3) = x

[split chasm]

no

split x^3 into
x *** x^2**
express the x^2 in terms of u
while you use the other x with dx and convert that to du

[radiant portal]

Ah yeye

Ok I think I can solve, thanks!

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[lethal vigil]

Find the equation and length of common tangent to hyperbolas
[
\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \quad \text{and} \quad \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1
]

[solid kiln]

riddle

[lethal vigil]

,rotate

[lethal vigil]

Got slope m=±1 of tangent

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[untold sage]

Hey, im really lost about how to do these. I understand that the first one is supposed to use the stars and bars theorem but i dont know why i cant do 5^20 and i just dont know how to do b and c

[vernal palm]

is this the whole question?

@untold sage

and where did you get answers?

because it seems to be a D(n,k) question

[naive otter]

Ig in part a they applied a restriction like at least one of each type is displayed

Or smthng else !

[naive otter]

Why would we need stars and bara though ?

Part b reads, at least 4 9v, so we can leave 4 places for 9 v and do 5^16, considering the places are identical.

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@untold sage Has your question been resolved?

[vernal palm]

I don't think he's here anymore

[dense breach]

Order doesn't matter I assume

So 5^16 is too big

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[naive otter]

∞-∞ form

[half marsh]

which means?

[naive otter]

You just perform lcm

[half marsh]

lcm?

[naive otter]

I mean

(xlnx - x+1)/(x-1)lnx

[half marsh]

I did that but then i got stuck

wouldnt it be -1+x?

[neon dirge]

-(x-1) = -x + 1

[half marsh]

omg

i wrote down the wrong function

[naive otter]

0/0 form

[half marsh]

i have x/1-x instead of x/x-1

[neon dirge]

k, but with 0/0 you can derive

[half marsh]

l'hopital?

[neon dirge]

sure try

[half marsh]

lim x-> 1+ ln(x)/(ln(x)+(x-1)/x)) ?

then i can plug in 1 so it will be 0/(0+1/1) = 0/1= 0?

wait no fuck

[past widget]

which means, you substitute x = y + 1

[half marsh]

now or at the start?

[past widget]

even now would be fine

[half marsh]

aah okay

[past widget]

Let me write the expression for you

$L = \lim_{y \to 0^+} \frac{(1 + y)\ln (1 + y) - y}{y\ln (1 + y)}$

[half marsh]

huh

if x goes to 1 y will go to 2?

if x= y+1

omg

mb

i am so cooked atm

[solid kiln]

Arya

[past widget]

nevermind cookie, do you get how to proceed from here?

[half marsh]

not sure let me try please

[past widget]

btw, remember $\lim_{y \to 0} \frac{\ln(1 + y)}{y} = 1$

[solid kiln]

Arya

[half marsh]

oooh ye the special limit

i would have forgotten

[half marsh]

L'hopital?

[past widget]

do you want to do L'hopital? 🥺

[half marsh]

i dont think so since u gave me that tip

but still

i have no idea then

[past widget]

okay go ahead, but even for using L'hopital, I'll suggest simplifying the denom at least

[half marsh]

so devide by y?

in the top and the bottom

[hazy blaze]

There is no reason not to do LHôptial.

[past widget]

yes rewrite ln(1 + y) as [ln(1 + y)/y] * y, the square bracket goes to 1

[half marsh]

wait let me take a screenshot of what i haven ow

[half marsh]

oh

i can still do that and then split the limits?

[past widget]

ln(1 + y) = y - y²/2 in the num immediately gives limit = 1/2 :/

there's no reason to even know or mention lhopital

[half marsh]

how do people see this so fast

i still dont see it

💀

[past widget]

np, you'll see it someday.

I hope

[half marsh]

I hope so too

so can you show me the step i must make?

[past widget]

nuuu T-T

wait I'll show

[hazy blaze]

You need to put O(y³).

[past widget]

$= \lim_{y \to 0^+} \frac{(1 + y)\ln (1 + y) - y}{y^2 \cancelto{1}{\frac{\ln (1 + y)}{y}}}$

[solid kiln]

Arya

[past widget]

screw that :/

[half marsh]

ye i would have never ever done that in my entire life

[past widget]

why you learn limit properties T-T

lim L + lim K = lim (L + K) and stuff

[half marsh]

I knoww but i just didnt see that step

[past widget]

but you know the standard limit for ln(1 + y)/y

[half marsh]

let me write this down real quick and try to solve it then

[half marsh]

ye and the sin(x)/x and stuff

[past widget]

the only right step is to use that

and the (1 - cos x)/x² stuff

welp std. limits have their use depending on your indeterminate form

[main sigil]

wait where did cos come from

[past widget]

imagination

[half marsh]

i didnt know that one

our prof doesnt want us to memorize a lot

[main sigil]

introducing cos to this is crazy 💀

[half marsh]

so we only get little to use

[half marsh]

ye it was just an example for special limits

[past widget]

I hope your professor doesn't make you remember L'Hopital is the way for all limits

[half marsh]

no we just learned that this week

so it was more of a guess tbh

that i had to use it for these exercises

[past widget]

why do they even teach that

[main sigil]

it might be an intended way to do this exercise then

[half marsh]

anyways imma try to solve t now

hmm

[past widget]

okie then, after reducing the denom, you're free to use Lhopital

$= \lim_{y \to 0^+} \frac{\ln (1 + y) - y}{y^2} + \lim_{y \to 0^+} \frac{\ln(1+y)}{y}$

[solid kiln]

Arya

[past widget]

you can use lhopital for the left part

[half marsh]

isnt l'hopital a nice thing :-:

it seems like you hate it

[main sigil]

in my unpopular opinion, you can also use lhop right at the start to avoid excessive manipulation

[half marsh]

i am still trying to figure out this step

[main sigil]

it's just expanding (1+y)ln(1+y) to ln(1+y) + yln(1+y) and then splitting the fraction

[half marsh]

aaah

i am cooked

[past widget]

$\frac{x\ln x - x + 1}{(x - 1)\ln x}$ hmm, you can just plug an alternative solution by using lhop on this

[solid kiln]

Arya

[past widget]

no biggie

[neon dirge]

I'll put the raw l'hop one here but I thought it was also valuable you going through some limit principles c:

[half marsh]

wait

didnt u make a mistake?

[past widget]

wait, don't you mean a "xln x" on second num

[half marsh]

in the second one

[neon dirge]

oopsie

[half marsh]

for the left one

how would you do it without l'hopital?

[past widget]

👀

[half marsh]

maybe split them?

ln(y+1)/y² - 1/y?

[past widget]

nope can't do that

[half marsh]

no that probably wrong

since 1/y wil bring us nowhere

[neon dirge]

updated

[past widget]

the limit properties state that lim h + lim g = lim(h + g) given both limits exist

[half marsh]

Hmmm

ye no idea

[past widget]

You'd probably not like it but one solution to that is by using Taylor series for e^x = 1 + x + x^2/2 + ..

[half marsh]

I only know that from math memes on tiktok

never ever seen taylor series

:-:

[past widget]

wait I went 4 steps ahead

[half marsh]

i dont know taylor series

[past widget]

Lhop it ✅

[half marsh]

that the only other way?

[past widget]

you want to do it purely without lhop?

[half marsh]

well i learn from other not easy methods

so if you have any tricks that are usefull

[past widget]

scratch that.. gimme a min

[half marsh]

ye okay ill just use l'hopital this is probably too advanced for now

thank you tho!

[past widget]

not really :c I just cannot remember the easier way to do it

[half marsh]

can you maybe help with another limit?

[past widget]

see ! you get rusty looking at lhop all day :c

[half marsh]

hahaha ye

[past widget]

$\lim_{x \to 0} \frac{e^x - x - 1}{x^2} = \frac{1}{2}$ was another standard integral back in the days :p

[solid kiln]

Arya

[past widget]

did you lhop the left part, did you get 1/2 as your answer?

[half marsh]

yes

1-1/2=1/2

had to do it twice right?

[past widget]

not really but it's alr

[half marsh]

Huh

[past widget]

$\frac{\ln(1+y) - y}{y^2} = \frac{\frac{1}{1+y} - 1}{2y} = \frac{1 - y - 1}{2y(1 + y)} = \frac{-y}{2y(1+y)}$

[solid kiln]

Arya

[past widget]

and you cancel the y's get the -1/2

[half marsh]

oooh ye damn

gotta love limits

[past widget]

:p

So, ready for the alternate solution?

[half marsh]

i hope so

[half marsh]

seee

l'hop just makes the thinking part way less complicated

but tbh you wont learn lots of it

[past widget]

you had another question?

[half marsh]

yes

lemme get it

[half marsh]

can you read this?

[past widget]

yes

[half marsh]

perfect

i literally have no idea

first i thought maybe multiply by the opposite

[past widget]

won't suggest lhopping right away

[half marsh]

Noooo

[past widget]

should rationalize

[half marsh]

i wouldnt even dare to

[half marsh]

is that multiplying by root(1+tan(x)) + (root(1+sin(x)) ?

and also divide

[past widget]

$= \lim_{x \to 0} \frac{\tan x - \sin x}{2x^3}$

[solid kiln]

Arya

[half marsh]

wow wait

Huh

[past widget]

the √(1 + tan x) + √(1 + sin x) -> 2 in denom

finite limits multiply rule

[half marsh]

oooh

okay now i get it

but now

[past widget]

now change tan x to sin / cos and viola, you get your limit

[half marsh]

imma try rq

[past widget]

$= \frac{1}{2} \lim_{x \to 0} \frac{\sin x}{x} \cdot \lim_{x \to 0} \frac{1 - \cos x}{x^2}$

[solid kiln]

Arya

[half marsh]

right side l'hopital?

left side special limit

[past widget]

can do right side was special limit, but you guys didn't do it so

btw 1 - cos x = 2sin²(x/2)

[half marsh]

aaah ye

i just thought of that no way

[past widget]

you're getting smarter :p

[half marsh]

it was sarcasm :(

one day!!!

but we dont have the right side as a special limit

so i guess i must do l'hopital then

wait i am not even allowed to

wait i am

mb

[past widget]

Hmm?

[half marsh]

answer 1/4?

[past widget]

Can't you do this and it's the special limit for sin y / y

[half marsh]

well i could but wouldnt hat make it way more complicated?

[past widget]

Waito

[half marsh]

wait i might know how to do it

[past widget]

$= \frac{1}{4} \lim_{x \to 0} \frac{\sin x}{x} \cdot \lim_{x \to 0} \cancelto{1}{(\frac{\sin (x/2)}{x/2})^2}$

Yep the 4 got sent out of space

[half marsh]

ye i am not allowed to do it like that

i must make sure the limit is sin(x)/x

so using substitution and stuff until it is in the standard form

[solid kiln]

Arya

[half marsh]

i found a way

[past widget]

What about this?

Is this still not allowed?

[past widget]

Hm?

There's no worries tho, you can L'Hopital, since it's a lhop exercise

[half marsh]

it is this

and then y= x/2

x-> 0 y->0

[past widget]

✓

[half marsh]

what year are you in?

[past widget]

M graduating this yr

[half marsh]

Aaaah

makes sense

[past widget]

:p I used to be sharper before

[half marsh]

No way

[past widget]

All the algebra n stuff dulled it down

[half marsh]

nooooo

i suppose u got to go now?

[past widget]

Obv coz practice

Not really, I'm around for a bit

[half marsh]

Ln both sides?

[past widget]

mmm

[half marsh]

Ln L = lim cos(x) ln(tan(x))?

[past widget]

Step 1. indeterminate form

?

[half marsh]

yes?

cuz

[past widget]

No, I'm asking which indeterminate form it is

[half marsh]

cos will be 0

oooh

infinte to the power of 0

right?

[past widget]

✓

[half marsh]

okay luckily

[past widget]

Step 2. ln L = lim cos x ln(tan x) = 0 * inf

send cos to denom and lhop

[half marsh]

smart

we will have inf / inf

i love math

oh no

i am stuck again

how

what is the dy/dx of 1/cos(x)?

[past widget]

Don't know derivative of sec x ? :p thought so

Chain rule ✓

[half marsh]

we are not allowed to use those

only cos

sin

tan

and their inverses

[past widget]

Chain it

[half marsh]

aaah okay

so 1/sin²x?

[past widget]

-1/cos² * -sin

=_+ did you forget how to chain too

You diff wrt cos first

Then diff cos wrt x

[half marsh]

i been doing math for the entire day

my brain just stops working

[past widget]

[half marsh]

so i will be left with lim x->0 tan(x)/sin(x)?

[half marsh]

=1?

[past widget]

You should be left with cos x/sin² x

[half marsh]

I am not allowed to use sec :-:

[half marsh]

i give up

[past widget]

Btw this limit is 1

[past widget]

This is 0

Also, ln L = 0 => L = 1

[half marsh]

well at least i got that right

[past widget]

[half marsh]

but the l'hopital is just so ugly if your not allowed to use sec and stuff

so many brackets they confuse me

[past widget]

It's best to simplify as much as you can before lhop

[half marsh]

but i cant simplify ln(tan x) / 1/cos(x) right

[past widget]

✓

[half marsh]

But then i will be left with such an ugly lim

look:

[past widget]

You did have a way btw

[past widget]

yes that's not ugly

[half marsh]

`huh

[past widget]

You've not seen ugly !

[half marsh]

this is!

[past widget]

That's just cot x cosec x

[half marsh]

not for me

[past widget]

Two words, not ugly

[half marsh]

it is if your not allowed to use them

okay ill just do this tomorrow

i am too cooked

anyways Arya

you been amazing help

thank you very very much

[past widget]

[half marsh]

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[molten comet]

Hello

[molten comet]

Why is the x coming out of the integral?

[empty orchid]

its respect to y.

[boreal apex]

All other variables besides the Integration variable are constant

[empty orchid]

so in this case, x is a constant here.

[molten comet]

Yeah so xtan(x) should come out, no?

Not just tan(x)

[empty orchid]

similar to how $\frac{\partial}{\partial y}xy=x$.

[empty orchid]

they kept the x for the u-sub.

[solid kiln]

;(

[boreal apex]

You could take out both it doesn't really matter

[molten comet]

But they subbed in xy

I'm saying there's x missing in the final solution

[empty orchid]

?

[molten comet]

Or am I wrong

[empty orchid]

there is not, since it vanished in the u-sub.

[molten comet]

xtan(x)e^xy + ø_1(x) should be the final answer

Ohhh

Right

Must have cancelled with the du/dx = x

[empty orchid]

$\int xe^{xy}\partial y=\overbrace{\int e^udu}^{u=xy, \partial u=x\partial y}$

[molten comet]

Thanks!

[empty orchid]

oopsies

[molten comet]

I understand now it's fine haha

[solid kiln]

;(

[molten comet]

.close

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[molten comet]

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[craggy basin]

what is concave up vs concave down again

[craggy basin]

i refuse to accept these are the proper terms for it TT

[ionic pendant]

sometimes concave up is called convex and concave down is called concave

[craggy basin]

idk why my teacher doesnt use those

[ionic pendant]

concave up vs down is pretty common in calculus courses

[craggy basin]

this doesnt make sense

wouldnt concave up be concave and concave down be convex

[zinc ginkgo]

Nope

https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/04%3A_Differentiation/4.06%3A_Section_6-

[craggy basin]

conviced whoever created these terms never knew basic words in the dictionary

is this right?

[zinc ginkgo]

All look right yea

[craggy basin]

i cannot stand their use of concave

.close

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[knotty glacier]

I need help with this problem

[hidden dew]

are you supposed to graph it or find asymptotes or what specifically

?

[wraith hinge]

this is part b of a problem... what is the problem?

[tulip violet]

<@&268886789983436800>

[west sleet]

Calculator

[dull temple]

yeah i can help you with that. bye.

[knotty glacier]

Oh sorry

Since they were g(x) and h(x) I thought they were different

[knotty glacier]

I didn’t catch that it was two parts

[hidden dew]

the only case where this has an x-intercept is if a=0
and 0 isn't positive constant so no x-intercept

[wraith hinge]

set h(x) = 0, and it should be evident that there's no x-intercepts.

[knotty glacier]

Oh ok

So if I were to graph it, it would be a horizontal line?

[hidden dew]

the a/(x-b) where a=0 ?

[knotty glacier]

No if I set h(x) = 0

And wouldn’t no x-int mean it’s a horizontal line?

[hidden dew]

not nesseciraly (I misspell alot srry)

[knotty glacier]

It’s ok

How do I write it out then? If not h(x) = 0

[hidden dew]

you could say something like a/(x-b)=0 and the only case that it intercepts the x-axis if a=0 and because 0 is not a positive constant then there is no x intercept
might be too wordy and not sure if it could be shorter

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[limpid pumice]

Hello ,I can't find the mistake

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[limpid pumice]

<@&286206848099549185>

its divided by n and not m on the second line

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[cedar elk]

Try to resolve it with a new blank paper, idk

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[craggy basin]

what did i do wrong

[silver glade]

looks right

[craggy basin]

i only got half points

[south latch]

my guess would be that the application doesn't like the vertices at x=0 and x=1

[craggy basin]

thats so dumb

[south latch]

yes. yes it is.

tell your teacher you want to do math with pencil and paper.

[craggy basin]

i dont think hed allow that

[south latch]

outsourcing math homework to these 🦆 ing app companies is causing lethargicness and brainrot in math students

not as much as social media, but close

[craggy basin]

im really behind on math so im not really allowed to complain

[south latch]

and more anxiety for teachers than its worth

[craggy basin]

wasnt even allowed to be in the class TT

[south latch]

I think the app wanted you to draw a line segment from (-1,-2) to (2,1)

I dunno how you got those points at (0,-1) and (1,0)

[craggy basin]

i just was starting the graph at 0

they arent closed dot

[south latch]

the app doesn't know how to handle whatever you tried to do.

that's why math apps suck

[mossy island]

ah, pearson

[craggy basin]

idk im really overwhelmed with this math class and idk what to do TT

[south latch]

you'd think they of all companies have the recources to account for these contingencies

[mossy island]

they do

the whole app is bad

[south latch]

you got the answer right. Take a screenshot. Show your teacher. Get full credit.
You can move on.

[mossy island]

does your teacher/professor have office hours

probably need to discuss that 😭

[craggy basin]

i think so but i cant complain cause i wasnt allowed in this class and had to ask him to override it soooo yeah

[south latch]

all teachers like observant students

you have the right to your own credit.

your teacher will be happy you brought it up because it means you are trying and validates their existence.

[craggy basin]

yeah but i litterally said i understand all the prerecs and i dont

[south latch]

believe me or don't, man

[mossy island]

eh

ive had expections made for me before because i didnt meet the prerecs

just do okay and theyll trust you

[craggy basin]

idk if it makes sense but if i tell him im basically saying i lied to his face

[south latch]

The app did you wrong.
Next time, plot as few points as you need to

[craggy basin]

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[gray pebble]

would this graph have 3 rel extrema or 4? i thought 4 bc at x=6 its und but still changes signs

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[rigid orbit]

a + b = 1?

(y - 1 + 2 + 1) = 2(x + 1 - 1 + 2) ?

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@rigid orbit Has your question been resolved?

[past widget]

=> y = 2x is your line after all translations```

Now it is reflected over y = x. What do you think will the new line be?

[rigid orbit]

I need to swap the x y value then solve for y

[past widget]

Alr, tell me what you get on reflection

[rigid orbit]

(½)x = y

½ + 0 = ½

What happend if I were to reflect on line y = 3x + 3

[past widget]

Fortunately that is not the question

So stick to the question for now

Btw y = x/2 is correct. Now you just need to compare to get a, b and find a + b

[rigid orbit]

The result is h + 2, 2h + 4 so
2h + 4 = h + 2?

[rigid orbit]

a = ½
b = 0
½ + 0 = ½

[past widget]

HUH

y = 2h + 4 = 2(h + 2) = 2x

(h + 2, 2h + 4) is parametric for your line after all the translation

[rigid orbit]

Owh.....

[rigid orbit]

Wait, why 2(h + 2) = 2x? Bevause x = h + 2 hmmm

[past widget]

Rack your brains

[rigid orbit]

Alr thank you

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[wispy raptor]

Let ABC be triangle with incenter I. Let the incircle be tangent to AB and BC at P and Q, and reflect C over the bisector of BAC to D, how to prove the midpoint of CD is collinear with P and Q?

[past widget]

Would suffice to show PQ bisects CD

Hm, that is the question tho. I see

[wispy raptor]

yea

the midpoint of CD is kinda obscure

[past widget]

It's not, it's the foot of AI on CD

ACD being a isosceles

[wispy raptor]

oh right

[wispy raptor]

why not make P/Q the phantom point instead?

[past widget]

To show RQP' makes a cute angle 90-B/2 with AD

[vernal palm]

how do you write text like that?

[wispy raptor]

add double `

[vernal palm]

where?

[wispy raptor]

when typing

[vagrant prism]

like this

[vernal palm]

thanks

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@wispy raptor Has your question been resolved?

[wispy raptor]

<@&286206848099549185>

[past widget]

Now, join PR. Then ∆ARP ~ ∆ACI```

Can you proceed from here?

[wispy raptor]

wait a min

[wispy raptor]

sry how is R'PD=90-B/2?

[past widget]

BP = BQ

[wispy raptor]

how is ARP~ACI tho

[past widget]

Figure, that's a hint

!nosols

[trim jolt]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[wispy raptor]

PR'D=(A+B)/2

APR=...=AIC
ARP~ACI
ARP=90-C/2=(A+B)/2

like this?

[past widget]

Huh, what's the dots

[wispy raptor]

like i dont know how to fill in 😅😅

[past widget]

No that's not how

Another Hint: You'll have to use SAS criteria for similarity

[wispy raptor]

i genuinely cant see any congruence

[past widget]

Similarity not congruence

[wispy raptor]

if i figured ARP~ACI how would the proof be strucutred, if not the way i proposed?

[past widget]

Once you figure ∆ARP ~ ∆ACI => ARP = ACI = C/2 => PRC = (A+B)/2 but we have PR'C = (A+B)/2

[wispy raptor]

oh yea right it's this

so if SAS similarity then PAR=IAC and smh i gotta prove AP/AR=PI/PC?

[dreamy mango]

You can use menelaus on this i guess

I know how to solve this but I am bad at english

[past widget]

We solved it but alternate solutions are always welcome

[dreamy mango]

I am Vietnamese but I will try to write it down

I found it difficult to explain 😦

[wispy raptor]

it's nvm

[dreamy mango]

Ok

I will write it down

Wait for a sec

[wispy raptor]

prove AP/AI=AR/AC, AR/AC = cos ACR=cosADR and im stuck again

[past widget]

Lol, almost close

AP/AI = AR/AC = cos(A/2)

And included angle A/2 is equal => ∆ARP ~ ∆ACI

[wispy raptor]

bruh wait a min im fking stupid

[past widget]

Yeah that's a different question, I suppose

[wispy raptor]

shouldnt it be AP/AR=AI/AC btw

[past widget]

AP/AR is not cos(A/2)

[wispy raptor]

no we want APR~AIC tho?

[past widget]

You can rearrange the ratios though for your similarity criteria once you've derived it

AP/AI = AR/AC <=> AP/AR = AI/AC <=> AP • AC = AR • AI

[wispy raptor]

ahh sry gotit

thanks again arya 😅

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[ocean quail]

if m ∈ R and the system is incompatible, find m

[tacit loom]

add equation 1 and 2

[ocean quail]

why

it would give 2x - (m^2-m)+(m+2)z = 3

oh, you want me to solve it by saying that this and the last one are equal?

[tacit loom]

yes

[ocean quail]

no no, i need a solution that involves matrices

[tacit loom]

then use the determinant

[ocean quail]

i tried

[tacit loom]

gets messy though

[ocean quail]

wanna see what i got?

[wooden plover]

you can just start with a determinant in the first place tbh

to get the values of m for which the system ain't invertible

[ocean quail]

but -1 is NOTT the answer

so yea

[ocean quail]

yea so the determinant needs to be 0

[wooden plover]

you need the det to be 0, but that's not enough

[ocean quail]

tell me more then

[wooden plover]

the system still might be compatible even if it's not invertible, depending on the right hand side

e.g. x+y = 1, 2x+2y = 2

[ocean quail]

yea yea i know

so after you find the values of m, you just check which one works

[wooden plover]

yeah

[ocean quail]

how do i find those values

[wooden plover]

well plug in the value of m you wanna check first

[ocean quail]

cuz my method is absolute trash

[wooden plover]

then you can use gaussian elimination for example

if you get an inconsistency in the end like 0 = 3 your system ain't compatible

otherwise it is

[ocean quail]

yea we dont use that

but it doesnt matter really, i can just check the solutions after and see which one works

thats the easy part

i only know Rouche and Kronecker-capeli

which are enough imo, i dont need 40 methods to prove that a matrice is compatible or not

[wooden plover]

you asked me for something lol

[ocean quail]

oh

[ocean quail]

u wanna use this method to find the values of m?

[wooden plover]

no

[ocean quail]

i thought u meant to check the values you got for m

[wooden plover]

I'm just talking about the checking yeah

[ocean quail]

well yea, thats the easy part

[wooden plover]

but you seemed to imply your methods for checking were complete trash

[ocean quail]

i need to find the values first

[ocean quail]

the method of trying to find the values

[wooden plover]

ah ig I misread the convo

[ocean quail]

cuz none of the values i got are right

so no matter which way i check, because m is none of those

[wooden plover]

I mean actually a bit of gaussian elimination works very fine here to help you get the det

like just do some row reduction to get 1 0 0 on the first column

[ocean quail]

hmmm, thats a good idea

[wooden plover]

this won't change the det of the matrix

and you only have to compute a 2x2 det now

[ocean quail]

ill get a sum of 2 matrices, no

?

2x2 matrices

[wooden plover]

just 1

[ocean quail]

can you make all the elements 0?

im left with a 1

[wooden plover]

what

I mean keep the 1 on top yeah

then it's pretty obvious that the det of the matrix is the same as the 2x2 on bottom right after row ops

[ocean quail]

oh yea

oh oki

this worked out really smooth

thx

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[bright seal]

Given the following linear map:

[
f: \mathbb{R}^3 \to \mathbb{R}^2
]

defined as ( f(x, y, z) = (3x + 2y - z, x - z) ). Given the bases ( \mathcal{B} = { e_1, e_2, e_3 } ) and ( \mathcal{B}' = { (2,2), (-1,2) } ), determine the matrix associated with the linear map with respect to the bases ( M_{\mathcal{B}}^{\mathcal{B}'} ).

[solid kiln]

Task Bot

[bright seal]

I know how to do it with 1 base, now with 2 how to do it pls help

[sudden mortar]

do you have the definition for this sort of matrix?

[bright seal]

Given a linear transformation ( f: V \to W ), where ( V ) and ( W ) are vector spaces with bases ( \mathcal{B} = { v_1, v_2, \dots, v_n } ) and ( \mathcal{B}' = { w_1, w_2, \dots, w_m } ) respectively, the matrix of ( f ) with respect to these bases, denoted by ( M_{\mathcal{B}}^{\mathcal{B}'} ), is the unique ( m \times n ) matrix such that:

[
[f(v_1)]{\mathcal{B}'} \quad [f(v_2)]{\mathcal{B}'} \quad \dots \quad [f(v_n)]_{\mathcal{B}'}
]

where each column ( [f(v_j)]_{\mathcal{B}'} ) represents the coordinate vector of ( f(v_j) ) with respect to the basis ( \mathcal{B}' ).

Thus, the transformation satisfies:

[
[f(x)]{\mathcal{B}'} = M{\mathcal{B}}^{\mathcal{B}'} [x]_{\mathcal{B}}
]

for all ( x \in V ).

[solid kiln]

Task Bot

[bright seal]

I don't know what it means

[sudden mortar]

okay, as we can see in the definition you are asked to calculate the matrix which is defined:
for each column, take the a vector v from the base B (of the space R^3), calculate f(v) (which is in R^2) and represent it according to the base B' of R^2

and then all the columns together (next to each other, by order) are the matrix that you need to compute

[bright seal]

Step 1: Compute the images of the basis vectors

[
f(1,0,0) = (3,1), \quad f(0,1,0) = (2,0), \quad f(0,0,1) = (-1,-1)
]

Step 2: Express the results in terms of ( \mathcal{B}' )

Solving the linear systems:

[
(3,1) = 1(2,2) - \frac{1}{2}(-1,2)
]

[
(2,0) = \frac{1}{2}(2,2) - \frac{1}{2}(-1,2)
]

[
(-1,-1) = -\frac{2}{3}(2,2) - \frac{1}{3}(-1,2)
]

Step 3: Construct the matrix

[
M_{\mathcal{B}}^{\mathcal{B}'} =
\begin{bmatrix}
1 & \frac{1}{2} & -\frac{2}{3} \
-\frac{1}{2} & -\frac{1}{2} & -\frac{1}{3}
\end{bmatrix}
]

[solid kiln]

Task Bot

[bright seal]

Is this correct ?

[sudden mortar]

that's the idea, just your calculation for the linear systems is wrong

[bright seal]

Mmm

[trim jolt]

@bright seal Has your question been resolved?

[bright seal]

Is this correct now ?

@sudden mortar

[bright seal]

,rotate

[sudden mortar]

looks good

[bright seal]

thank you very much

[bright seal]

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[bright seal]

Consider the endomorphism ( f: \mathbb{R}^3 \to \mathbb{R}^3 ) such that:

[
f(v) = A v
]

with ( v \in \mathbb{R}^3 ) and

[
A =
\begin{pmatrix}
1 & 0 & 0 \
-1 & 2 & 0 \
-1 & 0 & 2
\end{pmatrix}
\in M_3(\mathbb{R})
].

[solid kiln]

Task Bot

[bright seal]

Pls help

[soft pendant]

<@&286206848099549185>

[trim jolt]

@bright seal Has your question been resolved?

[bright seal]

<@&286206848099549185>

[thorn heart]

If I got it right, you're asked to find a basis of R^3 which contains those two vectors

And then determine the expression of the given endomorphism in terms of your chosen basis, is that right?

[soft pendant]

i think we need to add a third vector but we dont know how to choose the correct one

[thorn heart]

Could you post the original exercise too?

[bright seal]

Is this

[soft pendant]

its this one

[bright seal]

With this

[soft pendant]

we are working togheter

i will let task bot write now

[thorn heart]

Alright, fairly sure it should be this then

[bright seal]

but can I add a random vector?

[thorn heart]

If you need to find the Expression of the endomorphism for some basis, it needs to be one

[thorn heart]

From my understanding, yeah

As long as it forms a basis

[bright seal]

How can i choose the correct one ?

[thorn heart]

You're not given any more information to work with

[thorn heart]

Just choose any vector that's linearly independent to the other two

Any one vector that satisfies that is “the correct one“

Though of course some are easier to work with than others, if I were you, I'd choose one from the canonical basis of R^3

[bright seal]

And the determinant must be non-zero, right?

[thorn heart]

Yeah

[bright seal]

Why are the canonical bases e1 and e3 good?

[thorn heart]

Wdym by “the canonical bases e1 and e3“?

[bright seal]

e1=(1,0,0) ,e3=(0,0,1)

[thorn heart]

(I'm gonna assume either one of those work, haven't computed the determinant)

The reason why they're good choices is basically cuz they have many zeros, they usually simplify calculations

[bright seal]

Isn't there a way to see "without doing the compute" which one is right?

[thorn heart]

What do you mean by “which one is right“?

[bright seal]

That the determinant is not 0

[thorn heart]

You've been talking about finding the “right vector“ from the start, but I'm not sure I follow what you mean

[thorn heart]

Oh, well

Depends on the case

A set of vectors is linearly dependent if there exist some scalars a, b and c such that au+bv=cz for vectors u,v,z

(I stated that for simplicity, the set can be of any length)

What the determinant does is save you that calculation

But if you can determine a linear combination by inspection or see that it's impossible, you could skip that too

In this case, though, don't think it's worth it, the determinant will reduce to a 2x2 determinant

You still there?

[bright seal]

Yes

[thorn heart]

Well, have you already determined the basis?

[bright seal]

Yes @thorn heart

[north berry]

Inbox

[thorn heart]

Alright, do you know how to proceed from here?

[bright seal]

No

[thorn heart]

Ok, well, you've got your chosen basis then, which one did you choose?

[bright seal]

(0,0,1)

[thorn heart]

So the basis is {(1,-2,1),(1,-1,1),(0,0,1)}

[bright seal]

Yes 🙂

[thorn heart]

Alright, then onto the basis change

There's a formula for this, do you know it?

[bright seal]

No

[thorn heart]

Given a matrix A associated to a linear application in terms of the canonical basis, then, the matrix A' associated to the linear application in terms of an arbitrary basis B is given by:
A'=C^(-1)AC

Where C is the base change matrix from the arbitary base B to the canonical base

[bright seal]

?????

[thorn heart]

Where did you get lost?

[bright seal]

I wouldn't use this method

[thorn heart]

And what would you do then?

[bright seal]

Like this

[thorn heart]

I haven't read it all but it can't be right

You're swapping between the basis of two different spaces

One has three elements, the other two

[bright seal]

This ?

@thorn heart

[thorn heart]

Why are you considering R^4?

[bright seal]

Its an example

[thorn heart]

Uh, yeah I guess

You've calculated the image using matrix multiplication, right?

[bright seal]

Yes

[thorn heart]

Well, alright

I mean, in case you really do not want to calculate matrix inverses

Even using the method I gave you, you have a way to work around that

C is the matrix that maps a vector in terms of the base B to the canonical base, while C^-1 will be the matrix that undoes the transformation

That is to say, C^-1 maps the canonical base to the base B

Thing is, you actually do know how these matrices will look

C will have the vectors if the base B in terms if the canonical basis (how they're shown to you, basically) and placed in columns

Meanwhile, C^-1 will have the vectors of the canonical basis in terms of the vectors of the base B(their coordinates, which you can get solving three systems)

Once you get those, the problem reduces to multiplying three matrices(I actually think this is what you were trying to do??)

[bright seal]

@thorn heart can we do the exercise ?

[thorn heart]

Welp, ok

I'll be a little bit hasty in explaining how I go about it though, hope you don't mind

[thorn heart]

Alright this is our chosen basis, which means that our first basis change matrix, C, will be like this:

C=(1,1,0)
(-2,-1,0)
(1,1,1))

As we know the vectors form a basis, the determinant will be nonzero, hence it'll be an invertible matrix

As such, the second matrix we need, C^-1, exists

That matrix can be calculated easily through the standard method of calculating inverse matrices, in this case, though, I'll just give ya the solution

C^-1=(-1,-1,0)
(2,1,0)
(-1,0,1)

Multiplying C^-1AC we get A'

A'=(4,2,0)
(-3,-1,0)
(0,0,2)

As such the endomorphism would be rewritten in the way:
f(v')=A'v', where v' is any vector in terms of the basis B snd f(v') its image, once again in terms of B

And that's all there is to it, really, I'd suggest you look at some more examples if it seems confusing

[bright seal]

@thorn heart yes

what is the result?

[bright seal]

This right ?

[thorn heart]

Yeah

[bright seal]

Is there a way without matrix inverse?

[thorn heart]

I told you before

[thorn heart]

Read from here

[bright seal]

@thorn heart I'm having trouble calculating the inverse matrix

[thorn heart]

How so?

[bright seal]

It's the first time I've tried

[thorn heart]

That's a matter of practice then

Though it's strange you're taking linear algebra without ever having inverted a matrix before

[bright seal]

Ah no its easy

[bright seal]

I got different

😭

C^-1=(-1,1,0) (-2,1,0) (-1,0,-1)

Ops

I got wrong

I fix !!!

[bright seal]

@thorn heart It came out a little different to me

Pls help

[thorn heart]

Let's see(sorry I took a while, was studying myself)

[bright seal]

Np

[thorn heart]

Don't see what you're trying to do here

[bright seal]

I'm calculating C^-1 A

[thorn heart]

Do you have C^-1?

[bright seal]

Yes

[thorn heart]

And did you get the correct result?(Asking cuz I don't see the inverse anywhere)

[bright seal]

Oh

Since the detA was 1 I left (A*)^t

So C^-1=(A*)^t

[thorn heart]

What do you mean by A*?

[bright seal]

The matrix I got with algebraic complements

[thorn heart]

Welp, alright, I'll assume you took the inverse right

(you've got the actual inverse written somewhere in this chat regardless)

[bright seal]

Yes

[thorn heart]

After that, first you need to multiply C^-1 A

[bright seal]

Here

[thorn heart]

And then multiply that resulting matrix by C

[bright seal]

Yes

I did as you said

[thorn heart]

Then it's merely a computation error

You probably made a numeric mistake

[bright seal]

🙂

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[serene crane]

can you help me walk through HOW to know which operation to do
ALSO
can someone please explain the matric rules to me
Heres what i have in my notes from my teacher

1. Row rescaling: Multiply all entries in a row by a nonzero number (Can I just pick a random number of my choosing?)
2. Row exchangement/swap: Switch the order of two rows (I understand makes sense)
3. Row repalcement: replace a row by the sum of itself and a multiple of another row (Not understanding this, is the sum the element thats at the far right end, and what does it mean a multiple of another row)
   As for the problem, how did they know to do Step1c where they multipled row 1 by 2? Also which rule is subtracting?

[ionic pendant]

1. Row rescaling: Multiply all entries in a row by a nonzero number (Can I just pick a random number of my choosing?)
   as long as it isn't 0, yes. but some numbers will be more useful than others
2. Row repalcement: replace a row by the sum of itself and a multiple of another row (Not understanding this, is the sum the element thats at the far right end, and what does it mean a multiple of another row)
   you add every element in one row with the corresponding element of the other row (1st to 1st, 2nd to 2nd, and so on)

by a multiple, they mean that row multiplied by a number as in the row scaling gase

[serene crane]

for 3), are we including the answer also?

like 1+2+9+-1

or without -1

[ionic pendant]

you don't add the row to itself

[serene crane]

oh?

please explain

OH

wait

1+2, 2+1, 9+12

[ionic pendant]

if you do r1 -> r1 + r2
you replace the first entry in row 1 with the sum of the first entry in row 1 and the first entry in row 2, and so on

[serene crane]

^ so that would be row 1 if we did r1 -> r1+r2

[ionic pendant]

yes

[serene crane]

like lets say there were 3 rows

uhh

3 1 8 | 0

(random numbers)

if I did r1->r1+r2, i can view it as theres a plus in between row 1 and 2?

Like we're adding the top variable to the variable below it

is what im asking

[ionic pendant]

sure

[serene crane]

In step2b, they divided. What rule is this.

[ionic pendant]

that's really just multiplying by 1/-3

[serene crane]

oh so rescaling

And logic wise, how did they know to multiply row 1 by 2? (Also you can combine multiple rules into one? Because step 1c is repalcement, and R1 is having row rescaling done to it

[ionic pendant]

you can only do one operation at a time

[serene crane]

then how are they multiplying r1 by two, and also subtracting r1 from r2 (Also which rule would be subtraction?)

[ionic pendant]

that's what it means to "add a multiple"

they are adding -2*row 1

[serene crane]

just saw this, not sure what row scaling gase is. But im still lost on -2 being a multiple

Im getting its row replacement

(I also realized the reason they knew to do a 2 is because they were going column by column)

[ionic pendant]

when you do row addition, instead of adding just another row, you can instead add some number multiplied by that row (we call something multiplied by a number a multiple of that thing)

[serene crane]

for that second part yourte saying

you can instead add a row multipled by a number of your choice

so I could pick -3 for all I wanted

[ionic pendant]

yes

-3 would just be less useful

[serene crane]

okay its clicking holy shit

ive learned more from you than my teacher in two days, she blazed through everything

I have one last two part question then i think i should be off

For number 2, how is the last column not a pivot if thats where its 1. Is it because theres nothing below it to make it a pivot? Also there wouldn't be 0's under it?
For number 3, Is it being its like #2, but also after the 1 theres something there (in this case 4)

Also, understanding a free variable is hard to me

[ionic pendant]

a pivot is the first nonzero entry in a row, starting from the left

so in example 3, the pivot of the second row is the "1" in column 3

[serene crane]

OH also, is the best way to go about this, going column by column and making sure all the rules are being met per column. rather than looking at it as a whole?

Like for the photo, we got the pivot points. Theres zeros under the pivots, and the last 1 (last pivot point) has only zeros in its column

[serene crane]

I see that that is the pivot, but it says "some other column is not a pivot column either"

Is the 4 a free variable?

[ionic pendant]

yes, column 4 is a free column

[serene crane]

In #2, how doesnt the last coulmn have a pivot?

Theres a 1 there

is it because theres no numbers after the 1?

[ionic pendant]

in #2 there are no "1"s in the last column?

[serene crane]

Oh not just the number 4, but the entire couln 4 is free because theres numbers after the pivot

[ionic pendant]

each column (except the last/augmented column) is either a pivot column (contains a pivot, corresponds to a pivot variable), or a free column (does not contain a pivot, corresponds to a free variable)

when they say "last column" they mean the augmented column [a; b; c]

[serene crane]

oh wait

#2 is saying they dont care about the last column

only the columns prior to it

[ionic pendant]

well they care about the augmented column only if it contains a pivot (as in #1) since that means there are no solutions. if it doesn't have a pivot then there are solutions

[serene crane]

This is clicking now, x2 and x4 arent pivots, and the last column isnt a pivot!

[serene crane]

what exactly is the augmented column?

[ionic pendant]

the column after the bar

it's "augmented" because we've basically glued two different matrices together (the coefficient matrix on the left and the constants on the right)

[serene crane]

Im still not tracking, i get how its augmented now

wait

[serene crane]

so the last coulmn, that is 0, 0, 1

is that considered to be a pivot column

because it has a non zero

[ionic pendant]

it can have a pivot, but it doesn't correspond to a variable so it doesn't really fit into the distinction about variables

[serene crane]

so x33 isnt a pivot?