help-27

[sand dove]

that's not it

6a^13 + 1 = 6a + 1 mod 7, even if a is divisible by 7

(if a isn't, then little fermat, otherwise, 6a^13 = 0 = 6a mod 7)

[spring oasis]

yeah because for any p prime we have p | x => p | x^n

or wdym?

[spring oasis]

care to elaborate

[spring oasis]

also in the other direction, p | x^n => p | x

idk if thats what you are going for

[spring oasis]

care to explain

[near jolt]

how did u get this

[spring oasis]

oops

a^13 = a^7 × a^6 = a^7 = a

@near jolt

[near jolt]

yeah

no need to ping, cuz i dont check ping anyways

[spring oasis]

can u help dawg

[near jolt]

you should continue with what u got

[spring oasis]

how?

[near jolt]

continue simplifying?

[spring oasis]

6a^13 + 7a^5 + 4^132 = 6a^13 + 4^132 = 6a^13 + 1 = 6a + 1 (mod 7)

how?

[near jolt]

yeah this is goid enough

good enough

now you can solve mod 7

[spring oasis]

then either 7 | a or it doesn't

[spring oasis]

how

[near jolt]

sure

[spring oasis]

help dawg

im stuck

[near jolt]

a can only be 0,1,2,3,4,5,6,mod 7

[spring oasis]

correct

what about it

[near jolt]

show that only one of them satisfy 6a+1=0 mod 7

and find it

pretty literally just plug in every value

see if it works

the equation is simple enough that you can just brute force

[sand dove]

that or... 6 = -1

[near jolt]

also technically u cant just say a^{p-1}=1 mod p, cuz it onky work for non-zero values of a mod p.

But this can easily be avoided if you just check a=0 mod p, or use the fact that a^p=a mod p

[spring oasis]

too much

simpler route?

[near jolt]

this is simple enough that you should just try it

you could, like the person above suggested, use 6=-1 mod 7

[spring oasis]

then?

[near jolt]

u should, however, be comfortable with just solving it for simple cases like this

[spring oasis]

a = 1 (mod 7)

[spring oasis]

goat

[near jolt]

write down the step?

[spring oasis]

yeah so

6a + 1 = -a + 1 (mod 7)

then this is zero if and only if

a = 1 (mod 7)

[near jolt]

be a bit more detailed

why is that?

[spring oasis]

well we want that this shit is cong to 22 under mod 70

so a cant be 1 mod7

[near jolt]

I meant why is a=1 if and only if -a+1 is 0?

[spring oasis]

-a + 1 = 0 (mod 7) <=> 1 = a (mod 7)

[near jolt]

what shit?

[spring oasis]

is just simple arithmetic

[near jolt]

isn't it 28

[spring oasis]

the o original shit

[spring oasis]

right.

[near jolt]

so we have a=1 mod 7

[spring oasis]

do we?

[near jolt]

as we've just deduced didnt we

[spring oasis]

but this shit cant be cong 0 mod 7 dawg

correct

?

[ebon coyote]

wdym it can't be

[near jolt]

well how so?

[ebon coyote]

It's cong. to 28 mod 70

What does that mean mod 7 then?

[spring oasis]

ah and 28 is 0 mod 7

alright got it mate

[near jolt]

yeah that is the crucial first step

[spring oasis]

alright alright

[near jolt]

cuz 28 mod 70 means 0 mod 7

cuz 7 divides 28 and 70

[spring oasis]

alright okay

[ebon coyote]

I haven't read the rest of this convo, but assuming you haven't, similarly find some results mod 2 and 5 and then deduce what a should equal

[near jolt]

now rinse and repeat our steps for other divisors of 70

[spring oasis]

help

[near jolt]

so the ugly thing =28 mod 5

[spring oasis]

= 3

[near jolt]

well we've done one example

[spring oasis]

can u help

[near jolt]

imma leave the other two to you

lmk when u r done

[spring oasis]

is hard to use fermat

[near jolt]

u've done well for the p=7 case, same thing with p=5

gotta get them practice in

[spring oasis]

6a^13 + 7a^5 + 4^132 = a^13 + 2a^5 + (-1)^132 = a^13 + 2a^5 + 1

@near jolt

[near jolt]

so far so good

[spring oasis]

a^13 = a^5^2 × a^3 = a^5 = a

[near jolt]

continue with simplifying the powers

[near jolt]

this i donw quite get

[spring oasis]

just use fermat

[near jolt]

how did you get a^5^2x a^3=a^5

[spring oasis]

a^5 = a

a^2 × a^3 = a^5

a^5 = a

[near jolt]

good you should add that in

cuz its a bit unclear with that step skipped

otherwise perfect

[spring oasis]

no steps skipped is just fermat doing heavy lifting

anyways i get

3a + 1 = 3

[near jolt]

sure but i mean it would be more clear to the reader if you wrote a^13=a^5^2 x a^3=a^2 x a^3=a^5=a

[spring oasis]

3a = 2

[near jolt]

care to elaborate?

[spring oasis]

yes

[near jolt]

show me your steps

[spring oasis]

6a^13 + 7a^5 + 4^132 = a^13 + 2a^5 + (-1)^132 = a^13 + 2a^5 + 1 = a + 2a + 1 = 3a + 1

@near jolt

[near jolt]

cool

and this si congruent to 28 mod 5 right?

whcih is ofc 3 mod 5

[spring oasis]

3a + 1 = 3

3a = 2

gcd(3,5) = gcd(3,2) = gcd(1,2) = gcd(1,0) = 1

so inverse exists

3a = 2 <=> 6a = 4 <=> a = 4

@near jolt

@ebon coyote

do you follow?

[near jolt]

good work

now the last prime

2

[ebon coyote]

-# (fyi, if p is prime, then any non-zero a has an inverse mod p)

[spring oasis]

6a^13 + 7a^5 + 4^132 = a^5 = 0 <=> 2 | a

so we have a = 0 (mod 2)

a = 4 (mod 5)

a = 1 (mod 7)

@near jolt

[ebon coyote]

So then?

[spring oasis]

crt?

pff

[ebon coyote]

go for it then

[spring oasis]

i dont remember how

need a little bit of handholding

[near jolt]

crt just says solving mod 70 is equivalent to solving mod2,5,7

[spring oasis]

also is this system pairwise coprime

[spring oasis]

i dont understand

[ebon coyote]

-# tbh CRT in lecture notes is often mentioned in a pragmatic sense tho

That said, it should be possible to just check this by hand in any case

First, if it's 0 mod 2 and 4 mod 5, what should it be mod 10?

[near jolt]

how?

jk

but wdym by that

[ebon coyote]

-# This is from Wikipedia and not any course notes, but a course typically tries to go through the proof of the theorem, and often does the "existence" part by direct construction, like so:

[lunar harbor]

https://tenor.com/view/panda-roll-back-roll-cute-animals-gif-5302560

[spring oasis]

please dont send random gifs

[near jolt]

oh

[ebon coyote]

@spring oasis

[spring oasis]

i dont know

[spring oasis]

yeah so we need to follow this constructive algorithm

to find the solution in mod 70

[ebon coyote]

Bruh you can literally just check

CRT tells us there's a unique number mod 10 as an answer here

So we can just check - which number between 0 and 9 inclusive is even and cong. to 4 mod 5?

[spring oasis]

how?

[ebon coyote]

wdym how

That's literally what CRT says

If I have an answer mod 2 and mod 5, then I have a unique answer mod 10

[spring oasis]

i dont think I follow

[astral wind]

hey guys i’ve been staring at this projectile motion problem for a while and it’s driving me crazy i’m trying to derive the general formula for the maximum horizontal range r but here’s the catch the projectile is launched from a cliff of height h not from the ground can someone walk me through the math i’m specifically looking for the breakdown using quadratic solutions and how trigonometric identities come into play here i really want to understand how the starting height h changes the optimal angle compared to the usual 45 degrees we learn in class down for a step by step challenge if anyone is free to help thanks a ton

[ebon coyote]

@spring oasis fyi you've done much harder CRT questions than this

[ebon coyote]

!occupied - see the bot message that's also gonna reply to you:

[devout snow]

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).

[ebon coyote]

idk, wdym you don't follow? that's literally what CRT is

[spring oasis]

yeah so lets apply crt

but I dont remember how the algorithm was

[spring oasis]

I already forgot

[ebon coyote]

This is why you need to check your notes; I'm not going to help you any further on this

[spring oasis]

why are u like dat

like tsundere~ish

[lunar harbor]

Making people repeat themselves repeatedly is annoying (this also relates to you reposting the same question over and over again and starting from scratch - you’re basically telling the helpers from before that they wasted their time and were useless)

[spring oasis]

no

next time you do the exercise from scratch you have a much better grasp of the exercise

is just for clarifying things that were unobserved in detail

[ebon coyote]

But for this part of the exercise, I am literally telling you this shit is in your notes

From at least half a year ago, no less

[spring oasis]

what shit specifically

the algorithm itself?

[devout snow]

@spring oasis Has your question been resolved?

[celest lantern]

I think so.

[velvet swan]

Just re-read teresa's chapter on crt from algebra 1

[spring oasis]

are u taking algebra 1?

[lean crater]

the CRT is an existence theorem, i.e., it tells you the existence of some specific object, solution, or value exists under certain conditions

to use the algorithm, you need to find the total modulus, the partial products, the modulo inverses -- sure that works, but for the purposes of finding what number fulfils your specific mod 2 and mod 5 conditions, this would be too inefficient, which is why waes is suggesting you to just manually check!

[spring oasis]

yeah but I didnt understood their hint

[velvet swan]

No, i already passed it

May take algebra 2 in the future, algebra 1 was fun

[spring oasis]

what are the requisites for algebra 2 for cs majors

[spring oasis]

thats crazy work

[velvet swan]

Algebra 1

I think anal 1 too

[spring oasis]

for math we need algebra 1 and linear algebra (dm not dc)

[velvet swan]

Then we'll need the same pre-requisites, we don't ger special treatment for being from the dc, still, you can take any course from any other major as long as you have the pre-requisites

[spring oasis]

yes maybe you would need to take linear algebra from dm before

[velvet swan]

Yeah i may take it next semestwr if we are still on strike because only the dm is working at the moment, i took algo 2 and we haven't gotten a single lecture or class yet

[spring oasis]

dm is goated but yeah dc is suffering from the financial situation

I am taking orga 1 and we also having 0 classes whatsoever

[spring oasis]

yeah dm and df is working like almost nothing happened

dc is going crazy having the public classes and the professors leaving is fucked

at the end of the day they can go to di tella and earn 3 times more

however i am not sure the same can be said for the profs at df or dm

[spring oasis]

are you guys still having exam on algo 2 or not? is the semester completely wasted?

are they going to enable exam libre?

[spring oasis]

i ponder if you pass linear algebra dm if they give you for free linear algebra dc?

[velvet swan]

Completely wasted, they made a survey the other day asking us for our opinion of what they should do if the wages get raised and the only options were "starting over again next semester", "take classes as if nothing happened" or just "try to take as many assignments as possible". All the 3 options are pretty much starting over again since algo is the only course from the dc that has 15 weekly hours when normally a course is 10 hours

We are not going to catch up in 1 month

[spring oasis]

yeah thats illogical. the only logical possibility was to take it next semester

[spring oasis]

well one semester aint much, the problem is if next semester same thing happens

because dc doesn't even have summer courses

so you would've wasted an entire year on nothing

[velvet swan]

I mean, not for me since i still study on my own since i'm so used to it, but most students NEED to take class going to the faculty in order to study

[spring oasis]

is good that you take initiative but for example for orga 1 I tried that but at some point it becomes impossible to continue without asking to the profs some stuff (circuits, and stuff), no?

well maybe algo2 doesnt apply. since as long as you understand java you can deal with it prolly

[spring oasis]

for example I am taking this class taller de calculo avanzado where there is no book to follow, if you don't go to class you are cooked

there is no teresa fasciculo8 for example

[spring oasis]

still, one year for selfstudy? you are going to study eventually again the semesters you retake the courses

lets be honest is kind of wasted time, even if they let you take an exam libre, its not going to be easy by any means whatsoever

but what else can dc do, we need to keep the strike going, otherwise nothing will change

[velvet swan]

Yeah most dc courses have the disasvantage of having horrible documentation for self study unlike dm courses but it is what it is

[velvet swan]

Java is the least of your issues in algo 2, the hardest thing is understanding the specification language we use which is pretty much first order logic with extra steps

[spring oasis]

is it similar to specification from algo 1?

[spring oasis]

well I will see when the time happens, I enjoyed the haskell part of algo 1 aswell as the project, looking forward to taking algo 2 in the future

[velvet swan]

Not at all

[spring oasis]

maybe next semester, for algebra 1 at the start of this semester we were told that we were going to have exams and last week they told us that we will only have exams if the strike is lifted (and the financial law is fulfilled)

[velvet swan]

Let me see if i can find an exercise i've made

I'll send it to you after i'm done eating

[spring oasis]

how many times did you took it?

[spring oasis]

alrighty

[devout snow]

@spring oasis Has your question been resolved?

[devout snow]

@spring oasis Has your question been resolved?

[devout snow]

Channel closed

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[steep sapphire]

early diff calculus, unsure how to do parts c and d. i've tried finding the instantaneous rate of change at t=13 and the average rate of change from t=13 to t=14, both answers are close but not correct

[merry jetty]

change t=13 to t=12 because t=13 is at the end of the 13th day and t=12 is at the end of the 12th day, so it would be way closer to use t=12 instead of 13 since its removes the 24 hour window that the 13th one has from its start
or you can find both and then just subtract the 12th from the 13th
whichever is easier

do the same for d as well
this should fix the problem since everything else was correct

[steep sapphire]

sorry, for clarification you are saying to find dv/dt when t=12 instead?

[merry jetty]

yes

[steep sapphire]

hmm, that's also not what the textbook says

[merry jetty]

wdym
what does the textbook say?

[steep sapphire]

sorry yeah

this

for -464000+2640(12)-3(12)^2 i get -432752

[merry jetty]

one moment
let me do the math myself really quick

[steep sapphire]

i'm in no rush, thanks

[merry jetty]

Ah okay, I see why. The derivative gives a super close estimate, but the textbook wants the exact change using regular algebra instead of calc

so use the 12-13 that I said in my message as this gives the exact value

[steep sapphire]

so V(12) - V(13) instead of V'(12)?

[merry jetty]

yes

[steep sapphire]

gotcha, i did try V(13) - V(14) i assumed t=13 was the start of the 13th day

[merry jetty]

yeah, i've made that mistake before as well
so dw cause multiple people do that

[steep sapphire]

makes sense now that i think about it because t=0 is the start of the 1st day

i hate using time as the x axis value in problems, always adds another layer

yep all good

.close

[potent vector]

Is it some double angle formula thing? Idk

Isn’t this available uhh

[polar chasm]

the bot is broken, and yeah, 4sin(x)cos(x) = 2sin(2x), though thats probably not what youre supposed to use

[potent vector]

Ok

If not that what could u use?

[polar chasm]

wait so where exactly are you stuck at?

7sin^2(x) - 4sin(x)cos(x) = 4?

[potent vector]

The yellow bit

[polar chasm]

They divided both sides by cos^2(x)

[potent vector]

How do u do that step

I don’t get how sinxcosx/cos^2x is tan

[polar chasm]

cancel one cos

[potent vector]

U would get this no?

[polar chasm]

You'd get this, if it was sinx + cosx

[potent vector]

Uh

Ok

[polar chasm]

but it's sinx * cosx

[potent vector]

Alr

[polar chasm]

so you can literally just cancel it

[potent vector]

Ty

[polar chasm]

np, do you have any other questions or is that all? (ill assume that its all)

.

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[stable spire]

how to solve 3)b here [sorry it is in french but here is the main idea M(x) N(iy) 3)a prove that 3)b prove that am stuck in 3)b]

[polar chasm]

do we know anything else?

About A maybe?

[stable spire]

yes A is 2+3i

[polar chasm]

hm maybe you could use a)

if b was equal to 0, then either (zm - za) or (zn - za) are 0

so if you can prove that both zm - za and zn - za are non-zero, then b is surely non-zero as well

[stable spire]

and how can i prove it ?

[polar chasm]

The property we use here is that if a =/= 0 and b =/= 0, then ab =/= 0

this is an elementary fact, you dont have to prove it

product of non-zero numbers is non-zero

and then these 2 are the same

[left skiff]

Could someone please explain how this answer was reached in (i)

[stable spire]

first i thought about solving it until i can calculate Delta if it is < 0 then it have no solutions so it is != 0

[polar chasm]

That'd probably work too, but i think that using a) is much easier

[polar chasm]

Hi, can you go to #help-25 please?

[stable spire]

yes that is why am here searching for the second method

but i still didnt get it

[polar chasm]

do you see how those 2 are the same?

[stable spire]

yes

[polar chasm]

alright, so now we only need to prove that:

$(z_M - z_a)\overline{(z_N - z_A)} \neq 0$

[woven radish]

MathIsAlwaysRight

[polar chasm]

right?

[stable spire]

Yessssss

since zA != zM it is not equal 0 ?

[polar chasm]

But if we can prove that $(z_M - z_a) \neq 0$ and $\overline{(z_N - z_A)} \neq 0$, then we're done. Because product of non-zero numbers is non-zero

[woven radish]

MathIsAlwaysRight

[polar chasm]

Yes, exactly

if you can prove that zA != zM and zN != zA, then you're done

[stable spire]

let me think for a sec please maybe i will find it

i know that M = x it belongs to real number so Zm - Za != 0 is that enough ?

to prove the first product is non zero

[polar chasm]

Yeah, perfect

M is real, but A isnt

so zm - za != 0

what about the other one?

[stable spire]

and the opposite for the next couple ?

N is only complexe number and a is both complexe and real number ?

[polar chasm]

the proper term is "N is purely imaginary", but yeah

N has zero real part, A has non-zero real part

and so the 2nd factor is non-zero too

[stable spire]

oh that is it ?

[polar chasm]

Indeed

both factors are non-zero and so their product is non-zero as well

[stable spire]

this is much easier than the method i used thanks a lotttttt

[polar chasm]

np

[stable spire]

.close

[polar chasm]

.

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[shadow thunder]

how do i do the last one

volume of revolution

[errant wedge]

oh this one

I know the formula but there is no good way to explain it, maybe you need to memorise it before doing it

[shadow thunder]

ik

its

pi

multiplied by the integral of x dy

[errant wedge]

first, draw the y=g(x) and y=5 and find the points of intersection

you rotate along y axis means your radius is the x value

[shadow thunder]

ik

im confused of what equation do i put into the integral

[errant wedge]

draw the picture to check what is the radius

[shadow thunder]

1

[errant wedge]

it is the distance of y-axis (x=0) and y=g(x),

[shadow thunder]

5 and 1

[errant wedge]

means the radius should be $g^{-1}(y)$

this kind of question must guarantee g is invertible so no worries

[shadow thunder]

the limit is 1 and 5 right

[errant wedge]

you need to turn y=g(x) to x= h(y) , whatever function h would be

what limit

yes, but it is called bound

[shadow thunder]

isnt g^-1(x) = sqrt((y-1)/2)?

[errant wedge]

yes

oh it is $g^{-1}(y)$

[woven radish]

夜靈

[errant wedge]

sorry I made the wrong notation

[woven radish]

夜靈

[shadow thunder]

a

wait what do i put into the integral then

[errant wedge]

you still correct

$g^{-1}(y) = \sqrt{(y-1)/2}$

[woven radish]

夜靈

[shadow thunder]

yes

do i need g(y)

[shadow thunder]

???

[errant wedge]

no

[errant wedge]

integrate this $g^{-1}(y) dy$

[woven radish]

夜靈

[shadow thunder]

wait

[errant wedge]

ok

[shadow thunder]

$\pi \int \sqrt{\frac{y-1}{2}} , dy$

[woven radish]

j_stin7

[shadow thunder]

$\pi \int_{1}^{5} \sqrt{\frac{y-1}{2}} , dy$

[woven radish]

j_stin7

[shadow thunder]

i mean this

[errant wedge]

oh you need like area of circle $\pi r^2$

[woven radish]

夜靈

[shadow thunder]

Oh then i got my formula wrong

[errant wedge]

so you need $\pi \int (g^{-1}(y))^2 , dy$

[shadow thunder]

oopsss

[woven radish]

夜靈

[shadow thunder]

Omgg

[errant wedge]

goodbye if you are ok or ping me when you are back with problems

[thick lotus]

.

.

.

[sharp adder]

.close

[devout snow]

Channel closed

Closed by @sharp adder

Use .reopen if this was a mistake.

[hasty cargo]

Holy clopen

[lost laurel]

[0,1) is a clopen set

[devout snow]

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[clever sphinx]

I’ve done part i and iia i just need help with the last part

That’s my answer to i

For the last part ik the min value is when the expression equals 0

And i found theta for that

For the max value

The expression could equal 1 or -1 do i solve for both

[runic prawn]

the min is when cos(theta + alpha) = +/- 1

and max is when it's 0

you got it the wrong way

[clever sphinx]

Oh yh oops

[runic prawn]

you solve for 1 and -1 yes

[clever sphinx]

Idk if I’m doing it wrong but

I got a different answer to the markscheme

[runic prawn]

check on desmos

[clever sphinx]

Oh my equations are wrong

Im kinda confused

[devout snow]

@clever sphinx Has your question been resolved?

[devout snow]

@clever sphinx Has your question been resolved?

[ebon coyote]

you sure that's the min. value?

also fwiw you're not tasked with finding theta there
[edit: wait you are lol nvm]

[lean crater]

side note: tan alpha instead of tan theta would be correct here since theta is the unknown youre trying to solve, you always look for the reference angle when using the harmonic addition formula

[lean crater]

ah i see where youve gone wrong

thats not what youre supposed to be maximising/minimising

youre supposed to minimise $25 - (4\cos\theta - 2\sin\theta)^2$

[woven radish]

reze ♡

[lean crater]

so substituting, you should get (what you found) $25 - 20\cos^2(\theta + 26.57^\circ)$ and so on

[woven radish]

reze ♡

[lean crater]

so your greatest value would be to subtract the smallest possible value of $20\cos^2(\theta + 26.57^\circ)$, and your min would be to subtract the largest possible value of $20\cos^2(\theta + 26.57^\circ)$ accordingly

[woven radish]

reze ♡

[lean crater]

hope that helps!

[clever sphinx]

Thank you

That makes more sense now

I appreciate it

.close

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[atomic idol]

I need to figure out where this thing is continuous

My current understanding of it is that it’s not continuous for any value of x but my feeling is telling me it might be for x=0

[atomic idol]

Is it continuous for 0? How would I show that?

[sand quarry]

yeah

it is continuous at x = 0

this is like some variation of the Dirichlet function

[atomic idol]

How exactly would I prove that

[sand quarry]

epsilons and deltas

as usual

[atomic idol]

My basic idea is just
The rational number 0 lines up with our definition of y for irrational numbers

[atomic idol]

So do I just write what I already wrote and this time I say there is a ball such that.. for x=0

[sand quarry]

yeah basically

like, let eps > 0 be given, choose delta = eps and let x be in B_R(0, delta)

[devout snow]

@atomic idol Has your question been resolved?

[atomic idol]

This fine?

[devout snow]

@atomic idol Has your question been resolved?

[devout snow]

@atomic idol Has your question been resolved?

[atomic idol]

<@&286206848099549185> could someone verify my solution?

[bronze canyon]

E

[restive gust]

Idk wat this is

Im in 9th

[west ridge]

And I'm in 11th

Its not taught to me yet

Even though sets is over

[dim perch]

idk about using epsilon deltas, cant you do this by sequences?

[atomic idol]

Not sure
This is like my second problem on continuity and stuff
I have no clue how to do anything

To be clear the thing in the top right corner is my profs definition of continuity

As terrible as it is

[dim perch]

well its fine lol, just using balls instead of absolute values

[woven radish]

blanketism

[dim perch]

because im not sure if what you've written works

[woven radish]

blanketism

[sand quarry]

Epsilon delta works for proving it is continuous at x = 0

Less so for proving it is discontinuos for x != 0

[dim perch]

i think i just remember the sequences thing more

[atomic idol]

Uhh sorry where exactly

[lunar harbor]

Yeah you're right that it's only continuous at 0

epsilon delta is also good to show it

you should do what blanket said and split into cases based on whether x is rational or irrational though, this way you have an explicit formula for g

and to prove discontinuity everywhere else, ||consider density||

[dapper tiger]

you can also use the definition of continuity at 0 as lim x -> 0 f(x) = f(0) = 0
by the sequential characterization, you only need, given any sequence xn st xn -> 0, that f(xn) -> 0 too

but f(xn) is xn or 0

so if xn goes to 0, it's natural that f(xn) goes to 0, it's the same sequence with more 0

the funniest side of this is that you can reverse the problem to get a proof of discontinuity at other points

you only need to find a sequence xn that goes toward your point, but such that f(xn) doesn't

its existence is trivial thanks to civil service pigeon hint

[atomic idol]

I still don’t understand that part unfortunately

[lunar harbor]

What is $|g(x)-g(0)|$ if $x$ is rational? If $x$ is irrational? That's what we've been trying to get at.

[woven radish]

Civil Service Pigeon

[atomic idol]

For the rational numbers it’d be rational and this the number itself (or x) and for the irrational ones it should be 0 right?

[lunar harbor]

$|x|$, not just $x$. And yes it's $0$ for the irrational case.

[woven radish]

Civil Service Pigeon

[lunar harbor]

Do you see how this works very well with the epsilon delta definition

[atomic idol]

I think the differences that you mentioned like |g(x)-g(0)| could be considered as a radius

And thus a ball

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[mossy cliff]

|g(x)-g(0)| is a distance

[uncut crow]

true facts

[warped relic]

So what's the question?

[lunar harbor]

probably in reference to this ^

[devout snow]

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[flint talon]

[warped relic]

Ah

[uncut crow]

[mossy cliff]

this is it ... he asking if g continues at 0

it s clear that g continus at 0

[wicked turtle]

if it's clear, then isn't that your answer?

or are you just looking for confirmation?

you are correct, it is continuous at 0

[mossy cliff]

lim x-->0 g(x) =g(0)

[wicked turtle]

yes that is the definition of continuity at 0

have you shown that it is true?

[mossy cliff]

0 is appartient to Q so g(0) =0

and 0 not appartien to R-Q lim g =0

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[willow socket]

Hey can anyone help me with this question my eoc is 3 days away and I did nothing all year 😭 🙏

[flat ibex]

in order to combine radicals they need to have the same term inside the radical

you have a sqrt(5) and a sqrt(20)

as it stands you cant combine them

do yk how to simplify radicals

[sacred sedge]

Hint: exponent rules

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[smoky gyro]

Wrong channel

[wicked turtle]

impostor

[lunar harbor]

<@&268886789983436800>

[devout snow]

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[charred eagle]

Is this right?? Can reduction formula be in terms of I_n-2 and not I_n

[devout snow]

@charred eagle Has your question been resolved?

[slow sinew]

no. yes.

u didnt use chain rule for du/dx

only power rule

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[frosty ingot]

part c

[frosty ingot]

answer for a) 0.3446

answer for b)0.475

[lean crater]

are you stuck on part c? if so, then

!status

[devout snow]

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7. None of the above

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[mild sorrel]

Find cos(pi/11) cos(3pi/11) cos(5pi/11) cos(7pi/11) cos(9pi/11)

[hasty cargo]

Like $\prod^5_{n=1} \cos (\frac{(2n-1)\pi}{11})$ or whar

[woven radish]

P(this user is larp larp) ≥ 0

[mild sorrel]

of course

[mild sorrel]

Equivalent to finding cos(pi/11) cos(2pi/11) cos(3pi/11) cos(4pi/11) cos(5pi/11)

Because cos(4pi/11) = -cos(7pi/11) and cos(2pi/11) = -cos(9pi/11)

Like $\prod^5_{n=1} \cos (\frac{n\pi}{11})$

[woven radish]

Meolve

[mild sorrel]

Wait, the expressions are exactly equal

Because -1 x -1 = 1

So the question becomes:
Find $\prod^5_{n=1} \cos (\frac{n\pi}{11})$

[woven radish]

Meolve

[desert tree]

Hint: Multiply and divide by 2sin(pi/11)

[mild sorrel]

Oh, got it

It will becomes a chain reaction

⛓️ ⛓️ ⛓️ ⛓️ ⛓️

Thank you

.close

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[mild sorrel]

cos(pi/11) cos(2pi/11) cos(3pi/11) cos(4pi/11) cos(5pi/11) = A
2 sin(pi/11) cos(pi/11) cos(2pi/11) cos(3pi/11) cos(4pi/11) cos(5pi/11) = 2A sin(pi/11)
2sin(2pi/11) cos(2pi/11) cos(3pi/11) cos(4pi/11) cos(5pi/11) = 4A sin(pi/11)
2 sin(4pi/11) cos(3pi/11) cos(4pi/11) cos(5pi/11) = 8A sin(pi/11)
sin(8pi/11) cos(3pi/11) cos(5pi/11) = 8A sin(pi/11)
2 sin(8pi/11) cos(8pi/11) cos(5pi/11) = -16A sin(pi/11)
2 sin(5pi/11) cos(5pi/11) = 32A sin(pi/11)
sin(10pi/11) = 32A sin(pi/11)
sin(pi/11)/sin(pi/11) = 32A
A = 1/32

[devout snow]

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[robust bobcat]

Any suggestions? It is of the form 0^0 but my book only allows l'hospitals rule for 0/0 and for other forms it uses these theorems.

[robust bobcat]

I thought of taylor expanding but that doesn't seem to work

[potent dirge]

Don’t you need to do an e^ln()

[robust bobcat]

I did?

[potent dirge]

Just taking the limit of e to the power of the previous function changes the question

I don’t see the “ln()” lol

[supple knot]

$\lim e^{(...)} = e^{\lim (...)}$

[woven radish]

riemann

[robust bobcat]

why is that?

[supple knot]

you can do that because exp is a continuous function

[potent dirge]

Because of continuity, you can can bring the limit into the function argument

[robust bobcat]

I don't see how that is relevant? since is cont then lim x to something e^x = e^something

but that isn't the same as what your suggesting

[supple knot]

that would be the next step

[faint gorge]

You could approximate sin(x) with x around 0

[potent dirge]

Everyone’s fav thing to do

lol

[robust bobcat]

ye but I still have the log term which is causing all the issues

[faint gorge]

*x=1/(1/x)

[potent dirge]

You’re trying to get it into a form where you can use L’Hopitals

[robust bobcat]

I still don't understand the reasoning behind it and also why does it matter in I could just apply the limit to the x in e^x and get what I need

[faint gorge]

or rather maybe x/(1/(log))

[supple knot]

what are you even asking

[supple knot]

did you not learn what continuity gives you for limits?

[robust bobcat]

continuity means the limit at the point is the same as the function at that point

[supple knot]

what does "get what i need" even mean

[robust bobcat]

I don't understand why it is and why I need it

[potent dirge]

Yes

[primal marsh]

right, so this implies you can move the limit inside the function

[faint gorge]

you could also approximate log somehow

[primal marsh]

$\lim_{x \to a} f(x) = f(\lim_{x \to a} x)$ for a continuous function $f$

[supple knot]

you use it in the next step

[woven radish]

W*

[supple knot]

.

[robust bobcat]

oh since x is a constant

[primal marsh]

um

no it's just cuz f is continuous

[potent dirge]

No. If a function is continuous, you can bring the limit inside the function

[faint gorge]

x appraoches a as arbitrary as it wants

sin(x)~x and ln(1-x)~-x/2 might help

[primal marsh]

$\lim_{x \to a} f(g(x)) = f(\lim_{x \to a} g(x))$ for a continuous function $f$

[woven radish]

W*

[primal marsh]

still holds

the defining property of continuity is that the limit of a function is equal to the function at the limiting point

[robust bobcat]

so what am I messing up then if $\lim_{x\to a} f(x) = f(a)$ and we are saying that the function at a limit is equal to the function a the limiting point. but the limiting point is a constant and the limit of a constant is the constant

[woven radish]

BigBen

[robust bobcat]

if I sub in 2^x for x we then have an approximation at 1 though but the rest of the limit is at 0

[faint gorge]

,w taylor series of 2^x

[woven radish]

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[graceful cosmos]

<@&268886789983436800> (scam account)

[robust bobcat]

But don't we want to do it for log (1-2^x)

[faint gorge]

derive it

log is also continuous

[robust bobcat]

I mean log(1+x) =x+o(x) as x to 0 so if I let x=-2^x I have log(1-2^x)= -2^x+o(-2^x)

But -2^x then goes to 1

[faint gorge]

yea so it doesnt work

[robust bobcat]

Ok so if we approximate 2^x we have log (1-1-xlog(2)+o(x).

Then we have $e^{[x+o(x)]log(xlog(2)+o(x))$

[woven radish]

BigBen
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[robust bobcat]

$\lim_{x\to 0+} x\log x =0$ seems like it could help but the o term is in the log

[woven radish]

BigBen

[faint gorge]

you could pull it out like this from the log

[robust bobcat]

How did you go from your second to last step to last step with the o(1)?

[faint gorge]

We claim log(1+o(1))=o(1), so that means log(1+o(1))/1 -> 0, since o(1) -> 0, we end up with log(1)/1=0/1=0

[robust bobcat]

Ok so both terms go to 0 as x to 0 and then 0/1 is 0

[faint gorge]

Wdym by both terms

[robust bobcat]

I mean log1 and o(1)

[faint gorge]

o(1) goes to 0 and log(1) is 0

[robust bobcat]

Ok so then I'm left with $e^{[x+o(x)][\log(-x)+\log\log(2)+o(1)]}$

[woven radish]

BigBen

[faint gorge]

yes

basically it all comes down to xlog(-x) as x->0^- the rest goes to 0 anyway

[robust bobcat]

Ok all is clear. Thank you

.solved

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[quaint valley]

can someone help me for c please

[devout snow]

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[quaint valley]

1

i already solved it

.close

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[quaint valley]

can someone give me idea to do this pls

[lunar harbor]

constant of integration

also exponent on part a is wrong in your beginning integral

also there's no reason to expand - do a substitution if you really have to

etc

[quaint valley]

ohh

so we keep the same bracket?

[lunar harbor]

You can write the general antiderivative directly in terms of $(x-2)$ if that's what you're asking

[woven radish]

Civil Service Pigeon

[quaint valley]

so = (x-2)^4/4

[lunar harbor]

constant of integration?

[quaint valley]

+c

[lunar harbor]

mhm

now find C

[quaint valley]

by using 2,0 right

[lunar harbor]

mhm

[quaint valley]

c=0

so c = 0

[lunar harbor]

mhm

[quaint valley]

tyty

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[lunar harbor]

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[grim lion]

The question is Simplify logb(a^m)/logm(a)

[raw nebula]

what have you tried

[grim lion]

I need to simplify it as a single expression involving a logarithm of base b

i have tried:

mlogb(a)/logm(a)

nothing else comes to mind

[raw nebula]

loga^b = bloga

[grim lion]

it needs to be in base b

[timber violet]

use log_b(a) = ln(a)/ln(b)

[raw nebula]

Well if you make logb(a^m)=mlogb(a)

Then change base of logb(a)

[grim lion]

Thanks

.close

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[tawdry meteor]

How do I determine the foci of an ellipse when the c value is a whole number?

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[tawdry meteor]

EX:

[lunar harbor]

I don't follow -- why would the procedure be different just because $c$ happens to be an integer?

[woven radish]

Civil Service Pigeon

[lunar harbor]

,w foci \frac{(x+3)^2}{9}+\frac{(y+2)^2}{16}=1

[woven radish]

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[lunar harbor]

hold on is c even an integer in that case

[tawdry meteor]

Uhhh

[lunar harbor]

did you add instead of subtract

[tawdry meteor]

…..yes

[lunar harbor]

actually same for this

[tawdry meteor]

On all of them 🫩

[lunar harbor]

but in general it shouldn't be different just because c isn't an integer

[lunar harbor]

rip

[tawdry meteor]

Well fuck

I’m just gonna print out a new copy

No sense in erasing all of that

Welp

Onto hyperbola I guess

Do I add on that one 🥲

[lunar harbor]

yes

See here.

You good?

[devout snow]

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[raven forge]

Given the real numbers x, y,z satisfy the systems of equations:-

x+y+z = 6
x²+y²+z² = 26
x³+y³+z³ = 90```

Find the value of XYZ and x⁴+y⁴+z⁴
-------

[raven forge]

I can find the value of xy + yz + zx but what will be the benefit of it?

How will it be useful for me

[unkempt lichen]

so if you wanna find x^4 + y^4 + z^4 or x^3 + y^3 + z^3 or any similar looking expression, these polynomials like x+y+z, xy+yz+zx, etc tend to show up

so finding them is useful

[raven forge]

Ahm

What do you mean

[spiral raptor]

I'll give you a hint

x^3+y^3+z^3-3xyz

[spiral raptor]

When you express this in the long form

What do you get @raven forge ?

[unkempt lichen]

well this is more of a comp math thing, but for example, you can write x^3 + y^3 + z^3 as (x+y+z)((x^2+y^2+z^2)-(xy+yz+xz)) + 3xyz

[spiral raptor]

Dude i just said that...

[unkempt lichen]

so you see how xy+yz+zx shows up there?

similarly when you try to break x^4 + y^4 + z^4 down, you get it in terms of x+y+z, xy+yz+zx, and some others

it's like building a house in terms of these blocks

[unkempt lichen]

um
is there a problem?

[spiral raptor]

No no nothing...

[unkempt lichen]

okay

[unkempt lichen]

btw i dont mind if multiple people try to help

[spiral raptor]

I would try avoid giving the answer

[unkempt lichen]

idk about if you do but...i was the one helping him

[spiral raptor]

Yes mb

Also i think op is gone or smth

[unkempt lichen]

oh well it's quite a comp math-y question and they dont seem to know how to start properly
i mean this isn't the sort of thing you get from the regular theorems and such, it's just if you know it you know it

doesn't appeal to intuition too much

[unkempt lichen]

and even with this you still have like 60% of the problem left

so i didn't think it was too bad

i mean to some extent if this is the first problem of this sort that he's doing, maybe it needs to be consumed as an example

and then he knows for future

[spiral raptor]

I mean I guess. But this seems like a pointless argument to have

[unkempt lichen]

just my thoughts though

yeah im not arguing, im just saying since you mentioned

[spiral raptor]

Is that u in ur pfp?

[unkempt lichen]

no lol

it's from a movie

[spiral raptor]

Ah ok

What movie?

[unkempt lichen]

dhurandhar 2

it's an indian movie

[spiral raptor]

Oh ok

Op isn't here lol

[devout snow]

@raven forge Has your question been resolved?

[raven forge]

Pardon me, had went for dinner

[raven forge]

That's such a big and complex identity. Do I need to remember the identity for (a+b+c)³ also now?

[hasty cargo]

(actually no, you just need to remember (x+y)^3 and have (b+c) as y)

or manually derive it yourself

[spiral raptor]

Its a very common identity in Olympiad (i learnt that recently)

[raven forge]

Alright so lemme also memorize it

Well then the question is solved easily

Thank you so much all who helped!

.close

[devout snow]

Channel closed

Closed by @raven forge

Use .reopen if this was a mistake.

[spiral raptor]

I remember talking to someone who gave like 10 other identities to remember

I think you can find these types of identities online

[devout snow]

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[granite island]

hint pls idk how to start

[maiden talon]

remember that one dot product formula

[granite island]

like the one with the cos

so the normal of u is equal to the normla of v

we want to show that i mean

[graceful cosmos]

cos of the angle between u and w is equal to

u•w / |u||w|

[granite island]

yes

[graceful cosmos]

That's the dot product rearranged

[granite island]

yes

[graceful cosmos]

u•w is directly computable

Remember the dot product distributes over addition and scalar multiplication

[granite island]

yes

[graceful cosmos]

Simplify both sides with this strategy, they should turn out to be the same thing

[granite island]

so i get norm(V)uw=norm(u)vw

wait typo

ok thx for the help i gtg tho

.close

[devout snow]

Channel closed

Closed by @granite island

Use .reopen if this was a mistake.

[devout snow]

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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[harsh roost]

Okay im working on this but this part kinda confuses me

for ii

to convet this into echelon form like uh

do i make it 0 1 for the first row

then 0 0 1

then 0 0 0 1

or what?

[sudden reef]

yes

i also learned it this year, u can do it with your hand which will take long or with ur calculator

i dont know if u speak dutch

try doing that each horizontle line only has 1 unknow

and work like staircases like u said

[harsh roost]

im so confused

😭

[woven vale]

you want your first row to be [1 x x x] (so the first entry has a pivot), then the second row to be [0 1 x x], and so on
so make everything below each pivot zero

it will only look like this if you have 4 pivots, you could have some zero rows and end up with less than 4 pivots

[harsh roost]

why do i need to make it1 1 1 tho?

cuz cant it be any number?

cuz i can stil find out each x

[sudden reef]

Translate this and it has every rule u have to follow

na this guy skulled it

lock in man

[harsh roost]

issue is i got 3 rows but 4 columns so wtf do i do?

[woven vale]

it can be any number, just nonzero

[woven vale]

so i had it right the first time, sorry
you set up a 3x4 augmented matrix (so 3x5 matrix), then solve for as many pivots as you can

[harsh roost]

im confused

[harsh roost]

this sis 3x3

[harsh roost]

this is

3x4

so ....

[woven vale]

it can be rectangular
$$\left[ \begin{array}{cccc|c}
1 & * & * & * & * \
0 & 1 & * & * & * \
0 & 0 & 1 & * & * \
\end{array} \right]$$

[woven radish]

haseeb ♥

[woven vale]

row-echelon form just means that every row has a different pivot

it doesnt mean the pivots go all the way to the rightmost element

[devout snow]

@harsh roost Has your question been resolved?

[harsh roost]

but with this

i cant get x3 x4 and etc

like im confused

[woven vale]

why not? you have a 3x4 matrix, and you can put a different pivot for each of the three rows

[harsh roost]

cuz

now for 3rd row

[woven vale]

there isn't one for every column, but that's not a requirement for row echelon

[harsh roost]

we will have x3 + 3x4 = 5 (an example)

so we cant get either x 3 or x4

[woven vale]

that is correct

so what are the bound and free variables of the system?

[harsh roost]

idk what that is

@woven vale

[woven vale]

bound variables are things like 5x_3 = 2, where the variable has one numerical answer
but sometimes all we have are relations between variables, like x_2 = x_3
we call x_3 (or x_2) a free variable because it's free to be any number in R

[harsh roost]

5x_3????

whats _??

[woven vale]

$5x_3 = 2$

[woven radish]

haseeb ♥

[woven vale]

vs $x_2 = x_3$

[woven radish]

haseeb ♥

[harsh roost]

oh

[harsh roost]

stil

we will hve

x2 + 5z3 = 4

(an example

so we cant get x2 or x3

i emant x3 and x4

but ye

[woven vale]

yes, so we have a relation between x2 and x3

we call x3 a free variable, let's say t, then we solve the rest of the system in terms of t

[harsh roost]

uh okay

i see

ig

do we have to make

it 1 1 1 1

like the rows thing

or we can make it wtv

[woven vale]

no, row reduce how you normally would

[harsh roost]

just make the other parts 0

[woven vale]

yeah it doesnt have to be a 1 in front

just nonzero

[harsh roost]

like the 3 zeros i can do, but do i gott do the 1 1 1

ok

i see

ty

[woven vale]

yeah ik its not very satisfying

but you basically have this line of possible solutions, and for any fixed t, you can sub in and get all 4 variables

[harsh roost]

alr imma solve and send

[devout snow]

@harsh roost Has your question been resolved?

[harsh roost]

@woven vale is this fine?

Also so if it is easy to make it all 1's I will but if it will mak efrsctions I wont ig?

[woven vale]

yes, this is perfect

what do you mean by all 1's? if you mean the row should be [0 1 1 1] then no, because that's a different system

[woven vale]

although when you submit it properly, you should say "let t in R be free" or something

[harsh roost]

no like

y * * *

0 y * *

0 0 y *

y can be any numb except

0

and what i was saying is

[woven vale]

yes, that is fine

[harsh roost]

if i can make y = 1, by divided the whole row by 1/y i will, but i wont do it if will make the other areas in this row uneven

like for example let us sya its

0 3 5 7 : 10

i wont divided by 3

cuz it will just be horrible to do

yk wim?

[woven vale]

ah i see

yeah thats totally fine

in that example you can either scale by 1/3 earlier, or divide by 3 when you substitute later

it will happen at some point, so it doesnt matter when

at least to me, maybe your teacher cares

[harsh roost]

dividng by 3 later is easier

tbh

cuz if u divided by 3 now it's just annoiyng doing 5/3 * (3 - 4t) for example

[woven vale]

i would agree, but then when you do reduced row form later on, your pivots must be 1

[harsh roost]

then moving them to other side

etc etc

[harsh roost]

reduce row form ?

wym

[woven vale]

reduced row-echelon form (RREF) looks like
$\cv{1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 1}$

[woven radish]

haseeb ♥

[woven vale]

instead of nonzero *, you have zeroes on all non-pivot columns, and only ones in the pivots

[harsh roost]

yea making the sides = 0

is pain

😭

wait nvm meh not too much pain

[woven vale]

the benefit of this is if you have $$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 3 \ 0 & 1 & 0 & \frac 12 \ 0 & 0 & 1 & \frac 35 \end{array} \right]$$
then the equations are $$\begin{align*}
x_1 &= 3 \
x_2 &= \frac 12 \
x_3 &= \frac 35
\end{align*}$$
and you don't need to do any more sub

[woven radish]

haseeb ♥
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[harsh roost]

also why is ref
1 0 0 0
0 1 0 0
0 0 0 1

???

[harsh roost]

why is the x4 is a pivot but not x3 in row 3?

[woven vale]

the only rule is that the pivots are different columns

they don't have to be immediately after one another

in this case you might get something like $x_1 = 3, x_2 = \frac 12, x_4 = 1,$ and then you have no rule for $x_3$ so you say $x_3 = t \in \R$

[woven radish]

haseeb ♥

[harsh roost]

So for ref I have to have it like this? @woven vale

So if I asked to turn it into REF I have to make it
1 0 0 0
0 1 0 0
0 0 0 1
While its still a matrix

[woven vale]

for RREF (not REF), you have to have 0's and 1's

[harsh roost]

Ah

And for ref is what we did above?

Or am I tripping

😭

[woven vale]

it's possible it will look like
1 0 0 0
0 1 0 0
0 0 1 0
this is also in RREF, but for a different matrix

REF is the
y * * *
0 y * *
0 0 y *

but
y * *
0 0 y
is also in REF

[harsh roost]

Ah so what i did above

[solemn tapir]

using letters instead of asterisks (same),
a b c | d
0 e f | g
0 0 h | i
this is REF (RREF comes after)
but haseeb's saying that if you had a row of three 0's in the left part of your matrix,
0 0 0 | d
0 e f | g
0 0 h | i
(a, b, c = 0) and d is non-zero, this is a form of REF not RREF because

• we don't necessarily have the 1's that indicate RREF but
• even if we did, that triple row of zeroes doesn't happen in RREF.

p.s. it gives us a situation where there's no solutions idk if you've learnt that

[solemn tapir]

so this would just happen to be an REF because of that triple 0.
RREF would be
1 0 0 | a
0 1 0 | b
0 0 1 | c

importantly order doesn't matter, so i could write RREF as
0 1 0 | b
0 0 1 | c
1 0 0 | a
n that'd be allgood; same for REF, too, we don't need order

[harsh roost]

Ah ref needs order by rr doesnt m?

[solemn tapir]

that's why in REF instead of necessarily having the corner triangle, I recommend you get used to looking for that triangle but literally anywhere and in any orientation
e.g....
3 5 0 | 4
6 0 0 | 4
3 2 1 | 6

[solemn tapir]

Nuh they both don't

[harsh roost]

How so

I was told and in my notes

It has to be the 1's in order

Like
1 * * *

• 1 **
• • 1 *

Oops

The * thingy

But u get what I msan

[solemn tapir]

If you're required to put it into the proper RREF form (e.g. the question asks you specifically to use RRE elimination) then yes, but

[solemn tapir]

"don't need order" meaning you can move things around at the end, and it's still the same

I'm training you to recognise this bc it is useful when you want to do specific row transformations that you're worried won't lead you to your outcome

Js switch em round at the end

[solemn tapir]

soooo as an example

[solemn tapir]

with this one

0 1 0 | b
0 0 1 | c
1 0 0 | a

I could do three row transforms R1->R3, R2->R1, R3->R2 to get the matrix we want; im just conveniently moving them around

so when you are performing your row transforms, don't be concerned about where you're putting the new/replacement row

[woven vale]

hmm, i believe we require that pivots go down and to the right in REF/RREF, so its not truly reduced until you do the row swaps

[solemn tapir]

just focus on a triangle somewhere and if you need to you may use row transforms to put the triangle in a corner

[woven vale]

because every matrix is n row operations away from its RREF

[solemn tapir]

mmm yeah but if you were just solving a system of equations it would suffice without right?

[woven vale]

this is a great point though for when you're in the process of reducing

[solemn tapir]

Also i didn't know pivots were like that for REF that's mb

[woven vale]

it absolutely would, but we don't define that to be (R)REF exactly iirc

[solemn tapir]

Alright then

[devout snow]

@harsh roost Has your question been resolved?

Channel closed

Closed by @harsh roost

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[bronze void]

How do I find volume of a truncated cone of base radius 3cm and top radius of 2cm, with height 4 using calculus (volume of revolution) ?

[bronze void]

.close