help-10

[eternal bloom]

You could also show try to show that a_n is bounded and monotone [didn't verify], and use the MCT to see that a_n converges. Then pass to the limit to get that the limit is 0

[obtuse pebble]

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[knotty light]

alright, will try this and see what it comes to

.reopen

[obtuse pebble]

✅

[obtuse pebble]

@knotty light Has your question been resolved?

[knotty light]

say we were tackling this

we will basically need this

and if we can say the limit is 0 then the whole thing is solved

where should i start?

<@&286206848099549185>

for the boundary this is all i managed to find

[eternal bloom]

Not sure, sorry about that. Only had some ideas on how it might go, but didn't verify if it works or not

[knotty light]

oh ok

maybe writing a_n+1 as a_n - h and a_n-1 as a_n + h ?

[knotty light]

yeap

i have no idea

this is crap, there has to be smth that im missing

[eternal bloom]

It's a challenging one. I apologise for being unable to help further and for giving ideas which did not bear fruit.

[knotty light]

thank you, i appreciate the effort

.close

[obtuse pebble]

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[carmine bear]

What are the steps to solve this?

W being a wiener process

I know the solution is this but I don't get how it got derived

[thick fog]

That's just the Ito Isometry, there isn't much to derive

[carmine bear]

Ooooh yeah

Thank you

.close

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[sacred root]

Hj

[sacred root]

Hi

$if \sin(90-\alpha) = \frac{BC}{CA}$ write down the ratio of $\sin\alpha$

[warm shale]

BASU

[sacred root]

Ok how do I solve it

I went like this

$\sin90 - \sin\alpha = \frac{BC}{CA}$

[warm shale]

BASU

[sacred root]

Then

Did the math

$\sin\alpha = \frac{CA-BC}{CA}$

[warm shale]

BASU

[sacred root]

Is this the answer?

<@&286206848099549185>

[spiral maple]

that is very much wrong

since sin(90-a) != sin(90)-sin(a)

[obtuse pebble]

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[fading birch]

so apparently this is the newton leibniz formula, how is this different from the fundamental theorem of calculus?

[fading birch]

To give some context, this was introduced to establish how integrals have a finite/infinite value depending if the original function converges/diverges, but like, why couldn't this be explained with the ftom?

[tardy epoch]

Different in what way? Looks the same

[worthy comet]

(at least at wikipedia), there are two parts of the fundamental theorem of calculus, and one of these parts is the newton leibniz formula. There is no real difference

[fading birch]

I guess it's easier to show this instead?

"discuss if the following equation diverges/converges"

like, why is this all this even necessary 🤔

So they're the same, got it o7

(I'm studying again today so I may send more questions)

Yeah, just checked the examples, the teacher only uses it when one of the bounds would cause a division by 0, so it's more of a correct notation thing

[tardy epoch]

why is what even necessary?

[tardy epoch]

did you answer your own question?

[fading birch]

I think?

[tardy epoch]

👍🏻 .close

[fading birch]

I'm specifically referring to the lim(μ->0+) and 1+μ

Would it be alright to occupy this channel until I finish studying? I may have more questions and it may be easier for staff to answer it if they know what I've been talking about rather than have questions in disparate channels

[tardy epoch]

there's no point in keeping it open for hours at a time

[fading birch]

If it's interfering with the rules then alrighty o7

[tardy epoch]

just open a new one when you have a new question

[fading birch]

.close

[obtuse pebble]

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[obtuse pebble]

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[soft temple]

Help plz

[obtuse pebble]

@soft temple Has your question been resolved?

[eternal bloom]

The integral diverges [sponsored by WolframAlpha]

[soft temple]

Yeah this formula includes the exponential integral function so I'm not too worried about it.

.close

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[neat briar]

Find the equation of the line of symmetry for the equation of the graph below.
y=12x-4x^2

[idle thunder]

in a quadratic where is the line of symmetry?

[obtuse pebble]

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[frosty nimbus]

why is set (b,b) never checked? isnt a reflexive relation supposed to have the elements (a,a) and (b.b)?

[frosty nimbus]

it sees that a,a belongs to r and automatically goes to the answer that it is reflexive, why so?

<@&286206848099549185>

[tardy epoch]

read this

[frosty nimbus]

ive done everything thats stated there

[frosty nimbus]

and for this question how is the relation reflexive if
{(1,1) (2,2)...(6,6)} are not present

[obtuse pebble]

@frosty nimbus Has your question been resolved?

[frosty nimbus]

a little help here

[frosty nimbus]

<@&286206848099549185>

.close

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[cursive coyote]

when trying to find the limit of the sequences

not sure how to visualize letter q.

also this diverges as well

[cedar lichen]

Well, what have you tried?

[cursive coyote]

well i was just thinking it would be like the other problems, cos(inf) diverges

because 2pi times infinity is infinity

[gloomy valve]

think about the unit circle and cosines period

[cedar lichen]

Also remember that n is an integer

The value of cos(2πn) is constant for integer n

[cursive coyote]

so since its 2pi it always falls at the start of the cos function?

[cedar lichen]

Yes. For integer n, cos(2npi) is always 1

[cursive coyote]

okay i think i understand

besides just visualizing it

how would i show work

[obtuse pebble]

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[timid silo]

Hi can i please get help on explaining why C is a subspace?

I understand it needs to meet the 3 axioms of closure under addition, scalar multiplication and zero vector

But I need help with proving it

[haughty coyote]

what is P3 ? Polynomials of degree 3 ? Of degree 2 ?

[timid silo]

degree 3

[spiral maple]

ok well trivially 0 is in it

cause 0=0=0

[timid silo]

Yeah i understand that, does that mean it already meets 1 of the 3 axioms as it is a zero vector?

[spiral maple]

yes

[timid silo]

ah okay ty

how would you prove it under closure under addition?

[spiral maple]

the naive test is that 0 in U, a,b in U means a+b in U, and a in U and k in K means ka in U

definition

pick 2 vectors from the set, show their sum is in the set

[haughty coyote]

you can show a linear combination of elements of C is still in C, proving both addition and multiplication at the same time

[spiral maple]

to which yes, you can prove it in 2 tests, 0 is in it, and it closed under linear combinations

[haughty coyote]

for simpler cases like that it avoids having to do twice almost the same thing

[timid silo]

Ahh okay thank you

I will try do this question again

@spiral maple thank you for your help also

.close

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[violet mulch]

hello

[violet mulch]

I do not understand what is going on in this particular part:

[haughty coyote]

You're supposed to recognise this as the Taylor series of (1+x/2)^-2 and use that to compute it

[obtuse pebble]

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[jagged edge]

idk how to move ahead

So, 2a(da/dt) + 2b(db/dt) = 2c(dc/dt ) So,i have 2 ( 12) (da/dt) + 2 (5) (0.5) = 2(13)(dc/dt)

theres 2 variables thats why

how do i find the dc/dt

[spiral maple]

how much is the length of a solid ladder changing?

[jagged edge]

thats 0.5 feet per sec

[spiral maple]

no

[jagged edge]

wouldnt that only be db/dt

umm

[spiral maple]

the length of the ladder is c

[jagged edge]

ye

[spiral maple]

how much does the length of the ladder change?

the physical length of the ladder

[jagged edge]

the physical length of ladder doesnt change?

[spiral maple]

yes

so what's dc/dt?

[jagged edge]

OH

13

[spiral maple]

....

[jagged edge]

thats the length of the ladder?

[spiral maple]

The length of the ladder does not change with time

[jagged edge]

ye

[spiral maple]

so dc/dt is clearly 0

[jagged edge]

oh

right

[spiral maple]

cause... the length doesn't change

[jagged edge]

dang the whole RHS would be 0 then?

[spiral maple]

yes.

[jagged edge]

i got -5/24

does that look right?

[spiral maple]

idk, cba to do the entire computation rn

[jagged edge]

oh lol

would the 0.5 be positive or negative?

db/dt = -0.5 or 0.5?

@spiral maple

[spiral maple]

down is negative

[obtuse pebble]

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[leaden minnow]

Rotate L and determine the coordinates of L’

[leaden minnow]

Use trig

I just couldn’t get somewhere

[obtuse pebble]

@leaden minnow Has your question been resolved?

[gloomy valve]

Let the angle between L and the x-axis be $\theta$. Let the angle between L' and the x-axis be $\theta'$. Denote the length of L and L' by $l$.

[warm shale]

Navix

[gloomy valve]

$$\sin \theta = \frac{2}{l}$$
$$\cos \theta = \frac{3}{l}$$

[warm shale]

Navix

[gloomy valve]

Now consider $\sin \theta'$ and $\cos \theta'$ and use the identities $$\sin(90 , \mathrm{deg} - \theta) = \cos \theta$$
$$\cos(90 , \mathrm{deg} - \theta) = \sin \theta$$

[warm shale]

Navix

[gloomy valve]

If that doesn't tell you much, another simple way is to look at this red triangle:

you can argue that the drawn angle is the same as the angle that L forms with the x-axis

[obtuse pebble]

@leaden minnow Has your question been resolved?

[leaden minnow]

@gloomy valve sorry didn’t check my discord

Thanks a lot for the help <33

[obtuse pebble]

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[novel rover]

Could someone please help with this

[obtuse pebble]

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[eternal nova]

can I get some help please

[junior inlet]

can you write the curved surface area in terms of r and h?

then substitute r for 12-h

[eternal nova]

can you show me how you did it

like some working

[spiral maple]

Post your working

[eternal nova]

i didnt do any

[spiral maple]

so go attempt the problem

[eternal nova]

i am just confused

i have tried for the last hour

[spiral maple]

so post your work then

[eternal nova]

i have a test coming up and I dont know what im doing

i deleted it its on my ipad

[spiral maple]

Ok, so describe what you tried then?

[eternal nova]

i searched up the surface area of clynder

and then i tried to plug in

but it was only rh not r+g

r+h i mean

[junior inlet]

well
imagine this as a hollow cylinder
cut off the top and bottom
unwrap the remaining area to form a rectangle (the area of this rectangle is the curved surface area)
find the sides of this rectangle.

[eternal nova]

ohhhhh

i get it

thanks

[obtuse pebble]

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[woven parcel]

If to find maximum, we set differential of an equation = to 0.
How to find minimum?

[forest sinew]

which diff eq?

but youre talking about the first derivative test, right?

[woven parcel]

Given an equation like 4x-x^3=y

for example

[forest sinew]

equals what

or just finding the max of that expression?

[woven parcel]

Yes, finding the max/min

[forest sinew]

well its a polynomial so

[woven parcel]

i know how to find max, but dont know to find min

[forest sinew]

odds are youll only get one or the other

but the same way

this function wont have a global minimum

only local

[woven parcel]

Global? Local?

[forest sinew]

yea

like

graph it

,w graph 4x-x^3

[warm shale]

Results provided by WolframAlpha

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

Requested by jan Niku#2119

[forest sinew]

imagine what is happening to the left and to the right

this thing is shooting off to upper or negative infinity

[woven parcel]

Ye

[forest sinew]

that means

no global max or min

[woven parcel]

Oh

[forest sinew]

but within a small region, if we restrict ourselves

we can find locally maximum places

so the first derivative test doesnt find if something is a max or a min or neither

it just finds where the rate of change is 0

it will find maxes, mins, and neithers, equally as well

second derivative test works okay

[woven parcel]

This refers to the turning point right?

[forest sinew]

yea

but it may not turn, too

consider x^3

neither max nor min at x=0

globally or locally

but first derivative test will return it

[woven parcel]

Oh yeah

What does the second derivative test does?

[forest sinew]

you should look it up

its pretty straightforward

well it deals with concavity which is harder to picture

but its more or less the same as first derivative test

take a derivative, set it equal to something, find the place

etc

[woven parcel]

Ok thx

ill try to look up for second derivative tests then

.close

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[frail anchor]

could someone explain how the equations at the bottom were obtained?

[frail anchor]

Formation of a PDE(partial differential equation) by elimination of arbitrary constants.

Let f(x,y,z,a,b) = 0
where a and b are arbitrary constants

Differentiate equation 1 partially w.r.t x and y respectively

we get:

This is what is written, in case you're having trouble reading the handwriting

[obtuse pebble]

@frail anchor Has your question been resolved?

[frail anchor]

<@&286206848099549185>

[obtuse pebble]

@frail anchor Has your question been resolved?

[frail anchor]

<@&286206848099549185>

[obtuse pebble]

@frail anchor Has your question been resolved?

[obtuse pebble]

@frail anchor Has your question been resolved?

[frail anchor]

<@&286206848099549185>

[obtuse pebble]

@frail anchor Has your question been resolved?

[frail anchor]

.close

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[nimble kindle]

how am i supposed to know the formula if its not geometric or arithmetic

the sequence just goes up by 4 each time but every other number is negative

[valid zephyr]

so every term is a multiple of 4

that's the first part right'

and now you need a way to alternate the signs

any idea on how to do that?

[nimble kindle]

wouldnt that be a negative number raised to a power

[valid zephyr]

yap

but it can't just be any negative number

because it can't interfere with the 4

so you need a number that when multiplied by the other one does not change the result

what would that bE?

[nimble kindle]

it would have to be 1

right

[valid zephyr]

right

-1 in fact

so you would get something like (-1)^n*4n

[nimble kindle]

ohhhhhhh

[obtuse pebble]

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[obtuse pebble]

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[surreal inlet]

hello

[surreal inlet]

i need help

so i have a math question

the book says i have to factorise this

[hybrid gull]

Idk if that can be factored

[surreal inlet]

ikr

[hybrid gull]

I think you have to use QF

[surreal inlet]

ok

thanks

.close

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[rich holly]

c and d are both integers, which means $\frac{3x^2+6x+C}{(x-2)(x+d)}$ is an integer. What's the value of $C+2d$

[warm shale]

hyperlix26

[rich holly]

x can't be 2 and -d

[warm shale]

hyperlix26

[rich holly]

so

[warm shale]

hyperlix26

hyperlix26

[rich holly]

now what

[tardy epoch]

where's the original question

[rich holly]

@tardy epoch

find the value of C + 2d

[tardy epoch]

that's your latex. show the original problem

[rich holly]

wdym? That's the original problem

find the value of C + 2d from this

$\frac{3x^2+6x+C}{(x-2)(x+d)}$

[warm shale]

hyperlix26

[rich holly]

where both C and d are integers

[tardy epoch]

did you invent this problem yourself? no right?

[rich holly]

no

[tardy epoch]

show the source of the problem

[rich holly]

like, a screenshot?

[tardy epoch]

sure

[rich holly]

same thing in my native language

[tardy epoch]

is this your remainder?

[rich holly]

that's just this, after I cancelled some terms

[tardy epoch]

those two are not equal

[rich holly]

oh rip

wait why not?

[tardy epoch]

show your work

[rich holly]

@tardy epoch my guess what I did wrong has something to do with the dx

[tardy epoch]

that's not how fractions work

https://studylib.net/doc/14116545/rational-expressions-common-mistakes

[rich holly]

hmmm, so it's the dx

$\frac{3+6x+c}{dx-2x-2d}$

[warm shale]

hyperlix26

[tardy epoch]

you need to learn how to manipulate fractions

you can't just cancel common terms in fractions like (a + b + c)/(a+b+d)

that is not the same as c/d

[obtuse pebble]

@rich holly Has your question been resolved?

[rich holly]

hmm

what should I do from here?

[tardy epoch]

https://youtu.be/_FSXJmESFmQ

[obtuse pebble]

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[rich holly]

holy shit I forgot it's the same as $3x^2+6x+c$ divided by $x^2+dx-2x-2d$

[warm shale]

hyperlix26

[obtuse pebble]

@rich holly Has your question been resolved?

[rich holly]

$3x^2+6x+c=(3)(x^2+dx-2x-2d)-3dx+12x+c+6d$

[warm shale]

hyperlix26

[rich holly]

hmm doesn't seem to help

$3x^2+6x+c=3x^2+6x+c$ got here

[warm shale]

hyperlix26

[rich holly]

.close

[obtuse pebble]

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[wicked holly]

Can someone please help me start this? My prof said to do column operations but I don’t know how that helps cause I can’t figure out which operation

[royal basin]

you can't figure out the operations that will take you from A to B?

[wicked holly]

Why go from A to B? I thought it was B to A?

Oh wait, i need the operations to go from A to B

[compact shadow]

det(AC)=det(A)det(C)

[wicked holly]

Thank you but how does that help in this case?

[spiral maple]

Doesn't matter what way you go, so long as you correct the det correctly

[wicked holly]

Okay thanks Mosh

Looking at B and going to A, can I do v2 - v1? Then v3 of B - v2? Etc

[royal basin]

doesn't matter

[wicked holly]

I see thank u

[royal basin]

i mean tbh like, if you use operations that preserve the determinant, i.e. no col swaps or col scales

[compact shadow]

$C=\begin{pmatrix}1&1&0&……&0\0&1&1&……&0\0&0&1&……&0\…&…&…&……&…\0&0&0&…….&1\end{pmatrix}$

[wicked holly]

That means the det doesn’t change

[warm shale]

Cogwheels of the mind

[royal basin]

then you can give no shits at all about the direction

but also instead of dumping this big ass matrix on you like CotM did

what with their love of overcomplicating things

[wicked holly]

Okay thank you. But if those operations don’t change doesn’t that mean the det will be the same?

Lol appreciate the help tho cogwheels

[royal basin]

yes, because you can do this with only col additions, you will get that det(B) = det(A)

[compact shadow]

How is det(AC)=det(A)det(C) complicated?

Nvm, everything is complicated

[wicked holly]

I’m confused

[royal basin]

no, it's the fact that you thrust this big ass matrix that is meant to encapsulate all of your row operations in one go

[wicked holly]

I think I’ll just do those operations and show it’s the same det

[compact shadow]

Column operation I think

[wicked holly]

Yes

Thank you

[royal basin]

subtract col 1 from col 2

then subtract col 2 from col 3

etc.

up to: subtract col n-1 from col n

with these n-1 operations B will become A

[wicked holly]

That’s right thank you, then I can just show that det is preserved?

[royal basin]

you've already done so.

unless your prof requires you to reprove it every time

[wicked holly]

Sorry I meant if I can make B look like A, then I can say det(B) = 10 as well?

But that just seems way too easy

[timid silo]

hello can i get help with something??

[wicked holly]

What is this user 💀💀💀💀

Lol

[idle thunder]

check out #❓how-to-get-help

[spiral maple]

Well did the operations done change the det?

[timid silo]

thank you, it says that i have to find an unoccupied channel, how do i do that?

[spiral maple]

Look at the ones that are available

[idle thunder]

#903427206980714507

[obtuse pebble]

@wicked holly Has your question been resolved?

[wicked holly]

No

Makes sense thank you

[spiral maple]

So yeah the det is the same

[timid silo]

okay thank you:)

[obtuse pebble]

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[tardy basin]

Heey sooo

[tardy basin]

I did an analasys of some data and found this corrolation

summary(mdl4)

Call:
lm(formula = y ~ poly(x, 3), data = df)

Residuals:
     Min       1Q   Median       3Q      Max 
-0.76558 -0.46323 -0.09172  0.34950  0.97851 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   9.3150     0.2435  38.252 2.13e-08 ***
poly(x, 3)1 -17.3223     0.7701 -22.495 5.05e-07 ***
poly(x, 3)2   4.8193     0.7701   6.258 0.000772 ***
poly(x, 3)3  -1.2224     0.7701  -1.587 0.163524    
---
Signif. codes:  
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.7701 on 6 degrees of freedom
Multiple R-squared:  0.9892,    Adjusted R-squared:  0.9837 
F-statistic: 182.6 on 3 and 6 DF,  p-value: 2.772e-06

And i would love to turn this data into a function

What i belive it corrolates to is something like

[warm shale]

Drillenissen

$-17.32\cdot x^3+4.81\cdot x^2-1.22\cdot x+9.315$

[tardy basin]

(looking at the estimates)

            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   9.3150     0.2435  38.252 2.13e-08 ***
poly(x, 3)1 -17.3223     0.7701 -22.495 5.05e-07 ***
poly(x, 3)2   4.8193     0.7701   6.258 0.000772 ***
poly(x, 3)3  -1.2224     0.7701  -1.587 0.163524    

This is the analasys ran

mdl4 <- lm(y ~ poly(x, 3), data = df) 

Where df is the dataframe and i belive poly(x, 3) corrolates to a third degree function

This is the plot im trying to find the formula of

[thick fog]

Your coefficients are in the wrong order

[tardy basin]

Typical, let me try to fix it

Would it be
$-1.2224x^3+4.8193x^2-17.3223*x+9.3150$

[warm shale]

Drillenissen

[tardy basin]

Because like, its still not giving a good output

(purple line on the right)

It might be that my analasys is wrong

[thick fog]

Yes, but whether it gives a good output is another story without knowing more about the data.

You are trying to fit a third degree polynomial to a curve that is approximately concave up everywhere, it's not really a surprise the result looks bad

[tardy basin]

What type of curve would you map to it?

[thick fog]

I don't know? A straight line? It really depends on the problem and what you are trying to achieve

[tardy basin]

So this is our data

And were trying to explain it with a mathmatical model

[thick fog]

That information is already in the previous two images and the problem statement is really meaningless.

If this is just a class, it probably just wants you to fit it to a straight line, a negative exponential or something and make up some conclusion about the data decaying to zero or something

The result will probably be terrible, but it's practice and it doesn't matter

[tardy basin]

Yeah

[thick fog]

If it's some specific use case beyond "we want to model it, so lets do that", the problem statement means a lot

[tardy basin]

Fair

[thick fog]

Like, our insurance team needs to model how much the company loses in top 10% extreme cases, we need to approximate the value of this particular input to tell how much to charge for insurance on this asset or if we need to refuse signing a contract at all.

That tells you a lot about what you are doing

[tardy basin]

Makes sense

[thick fog]

Honestly, I hate classes that do this.

You should have a problem you want to investigate, you formulate a hypothesis, then you test the hypothesis and write down some conclusions.

Just "Explain with a mathematical model" comes up way too much.

[obtuse pebble]

@tardy basin Has your question been resolved?

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[fading birch]

mostly a theory question today

[fading birch]

"If there are two static points A and B, and a moving point M, such that the distance of |AM| = |BM|, then the movement of M can be described as a

[spiral maple]

static as in not moving?

[warped rivet]

straight line that bisects the line segment AB

[fading birch]

a) flat plane
b) a point
c) a curve
d) a straight line

It's in 3d space so I think a plane also works?

I'm mostly wondering if A and D are both valid answers, just that A probably is more sound if I had to choose one

[warped rivet]

Yep. I was thinking 2D

[fading birch]

a plane works right? My 3d imagination is just bad

[compact shadow]

You just let A=(0,0,…,0,1) and B=(0,0,…,0,-1) and let P=(x_1,x_2,…,x_n) be a point such that |PA|=|PB| you will find that it’s equivalent to x_n being 0

[fading birch]

oh so sort of like to pyramids with their bases joined together?

So the two tips would always be equidistant to the base?

[compact shadow]

It’s just a WLOG process. Any two points in that Euclidean space you can always let the midpoint be origin, the line passing throw two points be x_n-axis

[fading birch]

got it! Thanks! o7

.close

[obtuse pebble]

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[obtuse pebble]

@gilded tree Has your question been resolved?

[gilded tree]

<@&286206848099549185>

[obtuse pebble]

@gilded tree Has your question been resolved?

[swift marsh]

He saved $4.55 on just one shirt

You have $19.95 on sale

Do you know how to calculate percentage change?

[obtuse pebble]

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[inland valve]

i set v1 + v2 + v3 + v4 + v5 = 0 and wrote it into an augmented matrix and row reduced it

[inland valve]

and found that v1 and v2 are the leading ones and pivot columns

but im not sure why the answer is wrong

😦

[forest sinew]

its kuri

[obtuse pebble]

@inland valve Has your question been resolved?

[inland valve]

[forest sinew]

trying to install stuff in matlab to see if i can do this

[inland valve]

ok np tyt~

[forest sinew]

i dont have access to python atm

@inland valve can you paste that vector code

[inland valve]

[14,-4,-4,-2,-18],[3,9,-3,-3,-9],[-11,13,1,-1,9],[17,5,-7,-5,-27],[9,5,-9,3,-15]

?

[forest sinew]

you in college? you should try matlab

 
test =
 
[ 14, -4, -4, -2, -18]
[  3,  9, -3, -3,  -9]
[-11, 13,  1, -1,   9]
[ 17,  5, -7, -5, -27]
[  9,  5, -9,  3, -15]
 
>> colspace(test)
 
ans =
 
[ 1, 0, 0]
[ 0, 1, 0]
[-1, 1, 0]
[ 1, 1, 0]
[ 0, 0, 1]

[inland valve]

yes
matlab

[forest sinew]

matrix laboratory

its free if youre in college

its purpose built for stuff like this

[inland valve]

oh ill look into it

[forest sinew]

if youre allowed to use whatever you want

if youre not then you have to use python and honestly probably using python is better experience

what were you supposed to do in python?

[inland valve]

ive been just using RREF calculators i find online XD

[forest sinew]

did you have to write your own program?

[inland valve]

i dont think you need to do anything on python

its just for copy and pasting im guessing

[forest sinew]

right but

into python

did they give you python code that finds the span?

[inland valve]

i dont think so?

these are practice problems from previous professors

that my current professor posted

so i'm not sure if im supposed to use python ig

[forest sinew]

,w column space [14,-4,-4,-2,-18],[3,9,-3,-3,-9],[-11,13,1,-1,9],[17,5,-7,-5,-27],[9,5,-9,3,-15]

[warm shale]

Results provided by WolframAlpha

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

Requested by jan Niku#2119

[forest sinew]

wew

,w span [14,-4,-4,-2,-18],[3,9,-3,-3,-9],[-11,13,1,-1,9],[17,5,-7,-5,-27],[9,5,-9,3,-15]

[warm shale]

Results provided by WolframAlpha

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

Requested by jan Niku#2119

[forest sinew]

lol

[inland valve]

idk what those are

[forest sinew]

yea i like the matlab answer

 
test =
 
[ 14, -4, -4, -2, -18]
[  3,  9, -3, -3,  -9]
[-11, 13,  1, -1,   9]
[ 17,  5, -7, -5, -27]
[  9,  5, -9,  3, -15]
 
>> colspace(test)
 
ans =
 
[ 1, 0, 0]
[ 0, 1, 0]
[-1, 1, 0]
[ 1, 1, 0]
[ 0, 0, 1]```

see if u can get your answer from this

od

oh

wait

idk i still dont trust this

i like the first one better

yea

the first one

sorry if i totally confused you

[inland valve]

np idk what those numbers are

[forest sinew]

could also try this

https://www.emathhelp.net/en/calculators/linear-algebra/column-space-calculator/

theyre the column space

[inland valve]

column space?

what is that

[forest sinew]

a hint is given in your question about this

[obtuse pebble]

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[inland valve]

.reoen

.reopen

[obtuse pebble]

✅

[inland valve]

its saying that the span is = to the col(A)?

Col(A) is the column space?

what exactly is the column space? (is it just the number of columns in A)?

[forest sinew]

so span is the collection of all places you can hit by combining a collection of vectors

column space is a type of span, it talks about a matrix

[inland valve]

This i understand

[inland valve]

idk this

its a type of span that talks about a matrixX?

[forest sinew]

https://www.youtube.com/watch?v=VLj3wun3hsc&ab_channel=DenisPotapov

im still not sure how matlab is doing this and if i need a transpose
this is why id use a calculator

https://www.emathhelp.net/en/calculators/linear-algebra/column-space-calculator/?i=[[14%2C3%2C-11%2C17%2C9]%2C[-4%2C9%2C13%2C5%2C5]%2C[-4%2C-3%2C1%2C-7%2C-9]%2C[-2%2C-2%2C-1%2C-5%2C3]%2C[-18%2C-9%2C9%2C-27%2C-15]]

[inland valve]

so column space = span?

[forest sinew]

it talks about a matrix

but yea

its a type of span

i have to go but

if you do not figure this out

[inland valve]

thi does look like the definition of span

[forest sinew]

you should post in #linear-algebra

yea

lol

[inland valve]

[forest sinew]

sorryi i just realized how late it is

i think that link i linked is good

that very last one

screw matlab its confusing me

[inland valve]

[forest sinew]

but for real theyre helpful in #linear-algebra

good luck

[inland valve]

uh okok

[inland valve]

Ty!!

.close

[obtuse pebble]

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[ivory spoke]

In this isosceles trapezoid, <ABD & <BDC are congruent because side AB || side DC?

[acoustic cobalt]

side AB isn't equal to side DC

[fallen flower]

They put parallel

[ivory spoke]

yes, they're parallel

[acoustic cobalt]

oh right sorry

[fallen flower]

They’re congruent because “If two parallel lines are cut by a transversal alternate interior angles are congruent “

[ivory spoke]

oh i see

[fallen flower]

DB is a transversal

[ivory spoke]

does that work only on isosceles trapezoid or it would also work on regular trapezoids

[fallen flower]

Since the two bases would always be parallel, it will always work for any trapazoid

[ivory spoke]

oh okay got it

suppose <C = 60 and <CBD = 80. <ABD & <BDC would be 40 deg right

[fallen flower]

Yes, but what justifies that?

(What theorem/postulate)

[ivory spoke]

well 80 + 60 = 140 and 180 - 140 = 40

oh

the supplementary thingy i forgot lol

[fallen flower]

The sun of the angles in a triangle is equal to 180 degrees

But otherwise yep!

[ivory spoke]

alr thanks

.close

[obtuse pebble]

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[ocean rampart]

/latex 4\pi \int_{h}^{h + 2r} x\sqrt{(x-h)(2r-(x-h))} ; dx

[ocean rampart]

huh

4\pi \int_{h}^{h + 2r} x\sqrt{(x-h)(2r-(x-h))} ; dx

ermmmm, sorry bout that, one sec

How would I solve this?

(this isn't homework or for a class)

it's equivalent to this, in case that's easier

[obtuse pebble]

@ocean rampart Has your question been resolved?

[ocean rampart]

<@&286206848099549185>

[leaden ibex]

$4\pi \int_{h}^{h + 2r} x\sqrt{(x-h)(2r-(x-h))} ; dx$

[warm shale]

Remavas

[obtuse pebble]

@ocean rampart Has your question been resolved?

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[serene summit]

i need some help with these questions

i really dont know what to do

<@&286206848099549185>

[astral ivy]

@serene summit do you understand this explanation here

These are pretty cool questions, is this analysis?

[obtuse pebble]

@serene summit Has your question been resolved?

[serene summit]

yeah i think so

[obsidian isle]

y = 2x^2 - 4x -5

let's solve for x

@serene summit do you know how to do this?

[astral ivy]

I would recommend graphing it and then it will be clear — do you know what an invertible function looks like graphically?

[obsidian isle]

another hint: a parabola is symmetric about its vertex.

[obtuse pebble]

@serene summit Has your question been resolved?

[serene summit]

my problem is mostly with this question

[frail fable]

@serene summit it's a complete guess

[obsidian isle]

ok. are you familiar with the concept of an "injective function"?

no complete guessing required sir

[frail fable]

But if you inverse it don't you get the same thing

[obsidian isle]

no

you have to account for the domain

[frail fable]

Mmm

[frail fable]

Is it injective because 0 -> infinity

Does infinity have a zero ???

[frail fable]

No because 0 is the smallest number and infinity can't just be 0 of else it would not equal anything

[serene summit]

not really

i cant do math

forget it

.close

[obtuse pebble]

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[astral ivy]

Dude

What

Don’t quit lol

@serene summit come back

[serene summit]

i dont understand any of this shit

im dumb as fuck

[astral ivy]

I will explain it

What class is this

I need to know what you know

[obtuse pebble]

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[serene summit]

its intermediate algebra

[serene summit]

and inverse functions

[astral ivy]

Algebra? You’re in high school?

[serene summit]

yes

[astral ivy]

Jeez

Did they teach you what sets and functions actually are

[serene summit]

im taking an online class

taking geometry at my school, learned jackshit for algebra 1

no thinking at all

all memorization in school

[astral ivy]

Dude, just don’t worry

[idle thunder]

they dont teach functions until algebra 2

and they dont even talk about sets at all in hs

[astral ivy]

This is early undergraduate work, so if you’re struggling its not your fault

Anyway

Can you explain to me what a function is

[serene summit]

given a domain, the function calculates the range

?

[serene summit]

i need to have a written solution to this by 11:59 pm tonight

[slow skiff]

Do you know what it means for a function f to be an injection?

[astral ivy]

They don’t know what a set is, or a function

[serene summit]

so if we have f(x) = x+5

the inverse would be

x-5

is that an injection?

[astral ivy]

Exactly

No

Well, it is, but that’s not what injections are about

Do you understand intuitively what an inverse function does

[slow skiff]

Ah well... This is an unfortunate question then :/

[astral ivy]

Yeah

They’re in high school for gods sake

[serene summit]

yeah

[astral ivy]

What does it do?

[serene summit]

its like reversing the function

undoing it?

i think the textbook mentioned like ctrl z

yeah so undoing

[astral ivy]

Yes exactly

So what’s the inverse of x^2

[serene summit]

+-sqrt(x)?

[astral ivy]

Yep

Those are two functions, really

+sqrt x and -sqrt x

If x is positive, the inverse of x^2 would be +sqrt x. If x is negative, the inverse would be -sqrt x.

And so that’s the issue with x^2 — it has no singular inverse function that works for every number x. You have to restrict the domain of x^2 into “positive” or “negative” to get a single inverse function.

Are you following so far? This is just what the first screenshot was saying

[serene summit]

yeah that makes sense

however the 2nd screenshot is what problem i need to tackle

[astral ivy]

Yes

So one fundamental property of inverse functions is the following: if f and g are inverse functions, when you take the output of f (which is f(x)) and put it into g, it should give you x back. In other words, g(f(x)) = x, for any number x.

@serene summit still following?

[serene summit]

yeah

i learned that

[astral ivy]

Ok cool

So now onto the first question

[serene summit]

alright

[astral ivy]

Btw dude you’re not dumb - the fact that you understand this already is great

Anyway

So I was thinking about the fact that 0 isn’t in the domain of f

[serene summit]

okay

well you can't divide by 0

[astral ivy]

So to prove that it doesn’t have an inverse, we really want to find a number in g that “should” map to 0

But that isn’t really working

Hmm

[serene summit]

well

let me get this right

a function has no inverse when there are 2 inputs that give the same output right

[astral ivy]

Yeah that works too

You could go that route, maybe it’ll lead to something

I was thinking, if f doesn’t have an inverse, there could be some number which f doesn’t map to

[serene summit]

oh

how would i go about finding that

[astral ivy]

Oh wow

@serene summit I just graphed it in Desmos

Oh wait I’m stupid

1 and -1 both map to 0

That’s much easier

Boom, you’re done

[serene summit]

oh wow

ur right

[astral ivy]

You can see the “horizontal line test” checks if a function has an inverse

[serene summit]

yeah

how do i prove it doesn't have an inverse then?

when the domain is set to 0 < inf

[astral ivy]

U mean that it does have an inverse

[serene summit]

yeah sorry

[astral ivy]

Well you can see it passes the horizontal line test — but of course that isn’t convincing enough

Hmm

Maybe we could construct the inverse function and show that it has domain R and range (0, inf)

Do you know what I mean by that

[serene summit]

yeah we could set up so its x = y - 1/y right?

and then solve for y

i need to get this done pretty soon

[astral ivy]

Yeah exactly

Sorry I’m at dinner lol

Ok here’s the trick

Square both sides

@serene summit

[serene summit]

okay

[astral ivy]

Tell me what you get

[serene summit]

when i square x = y - 1/y?

[astral ivy]

Yeah, both sides

[serene summit]

x^2 = y^2 - 1/y^2

[astral ivy]

The right side isn’t quite right

Use FOIL

[serene summit]

x^2 = (y^4-2y^2+1)/y^2

@astral ivy

[astral ivy]

Yeah

Distribute the y^2 to each term

[serene summit]

okay

i got

x^2y^2 = y^4-2y^2+1

[astral ivy]

Nah

I mean the 1/y^2 to each term on the right side

@serene summit

[serene summit]

how do u do that

idk what u get

@astral ivy

[astral ivy]

Here

[astral ivy]

From this, you have $$x^2 = y^2 - 2 + \frac 1{y^2}$$

[warm shale]

abs_0

[astral ivy]

I divided each term on the right by the y^2

[serene summit]

oh alr

i see

i thought we were multiplying

[astral ivy]

You can rewrite this as $$x^2 = \left(y^2 + \frac 1{y^2}\right) -2,$$ and then substitute $x$ for the parentheses to get $$x^2 = x - 2$$

Oh…

[warm shale]

abs_0

[serene summit]

what happened?

[astral ivy]

The y disappeared

So we can’t really solve for it any more… lol

Gimme a sec maybe I messed up smth

[serene summit]

we're running out of time

[astral ivy]

Yeah sry bruh

@slow skiff do you have any ideas

[slow skiff]

I found the potential inverse functions

But gave up trying to show that only 1 makes sense for the given domain

[astral ivy]

Hmm

[warm shale]

.nai

[serene summit]

howd u get that

[slow skiff]

Solve the quadratic you get by multiplying by y in terms of y

[serene summit]

this one @slow skiff ?

[slow skiff]

No

[slow skiff]

This one

[serene summit]

multiply y in terms of y?

[slow skiff]

multiply by y, then solve quadratic

[serene summit]

giving you that

okay

[astral ivy]

@slow skiff I have an idea. I want to prove that for any c, there is exactly one positive x such that c = x - 1/x. Playing around with this equation:
cx = x^2 - 1
x^2 - cx - 1 = 0
If this quadratic has exactly one positive root, we’re done. Right?

$$\frac{c \pm \sqrt{c^2 + 4}}{2}$$ are the roots

[warm shale]

abs_0

[slow skiff]

I think so...

[astral ivy]

The logic is “this equation has exactly one positive solution, thus the original equation has exactly one positive solution, thus it is bijective”

[slow skiff]

Yeah seems right

But if you're going in this direction, it'd be easier to just derive injectivity from the definition of an inverse

[astral ivy]

But aren’t we proving a left and right inverse? So it must be surjective as well as injective?

[serene summit]

guys im willing to put down any answer at this point

[slow skiff]

Just write it out until this

I can't think of a way that doesn't involve injectivity

[serene summit]

how does that prove there is an inverse?

[slow skiff]

It doesn't

But at least you'll get some credit

[serene summit]

somebody had posted on the message board this initially

[slow skiff]

I've never really seen calc care about surjectivity for inverses, but then again I haven't done much on this topic

[serene summit]

some of their points got redacted, but do u think this works?

[astral ivy]

Hmm

I don’t really get what they’re saying

I think they’re saying if g(a) = g(b) then a = b

a - 1/a = b - 1/b

[serene summit]

im so tired

and lost

[astral ivy]

Hmm

I don’t really know how they got to the first equation

And what this contradiction mess is

[astral ivy]

@serene summit I would write this, and then the next equation

And if you really have no time then just claim “obviously there is no way for both of these numbers to be positive. Thus this quadratic has exactly one positive root, and so our original equation has exactly one positive solution.”

[slow skiff]

a²b-b=b²a-a
a²b-b^2a-b+a=0

ba²-(b²-1)a-b=0

a=(b²-1±sqrt(b⁴-2b²+1+4b²)/2b
a=(b²-1±sqrt(b⁴+2b²+1)/2b
a=(b²-1±(b²+1)/2b

a=-1/b
2)
a=b²

Phone sucks

@astral ivy this look ok to you?

[astral ivy]

Yep

That’s how they got that first equation

Wow

But what did they do with that? Seems like they derived some sort of contradiction here

[slow skiff]

Well yeah

So... lemme reread it again

[astral ivy]

Dude take a step back and realize they’re expecting high schoolers to do this problem

Lmao

[slow skiff]

So from the domain of the function* we know that both a and b are positive

which is a contradiction with 1) because 1) shows them to be negative, let alone equal

[astral ivy]

But doesn’t that imply the second case?

[slow skiff]

How so?

[astral ivy]

I thought 1) and 2) were cases of the last equation. Ie 1) is true or 2) is true

[serene summit]

it alright guys

i figured it out

[astral ivy]

Bruh what

Seriously?

What did you do @serene summit

[slow skiff]

Fuck 5am brain

[slow skiff]

Ok no, my bad. I don't see the contradiction

[serene summit]

thanks for all ur guys help

even if its not right

im okay with it

have a great rest of ur day/night

tyvm again

.close

[obtuse pebble]

Channel closed

Closed by @serene summit

Use .reopen if this was a mistake.

[astral ivy]

Nice dude

Solid proof

Bro @slow skiff it was right in front of us lmao

[slow skiff]

No comment x)

At least we know more now for other problems

[astral ivy]

Why does calculus only care about left-invertibility (i.e. injection) when considering inverses of functions

Yeah

[obtuse pebble]

Available help channel!

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[astral ivy]

Why does calculus only care about left-invertibility (i.e. injection) when considering inverses of functions

[slow skiff]

right limit

[astral ivy]

Wait @slow skiff isn’t this problematic? He assumed there was an inverse and then showed that it didn’t lead to any contradictions

Isn’t that a circular argument?

[slow skiff]

I don't like the beginning of the proof

It has no clear list of arguments

[astral ivy]

I suppose one could argue that the function he got at the end was a valid inverse, so technically the proof “starts” with that

But yeah there’s no roadmap or anything

[slow skiff]

Either way this was kinda fun ngl x)

[astral ivy]

Yeah lol

[slow skiff]

also, just now I realised

[astral ivy]

This dude should really edit his proof tho, I don’t think enough people do that. Just because it might be technically correct doesn’t mean it’s good or even readable

[slow skiff]

x-sqrt(x^2+a)<0 for any a>0

[astral ivy]

Why

[slow skiff]

well

sqrt x^2= x
sqrt( x^2+a) > x

Even though sqrtx²=|x|, but ignore :)

[astral ivy]

lol

I guess that makes sense

[slow skiff]

Sometimes I wonder how I passed calc 1...

[astral ivy]

How do you deal with absolute value though

I know right lol

[slow skiff]

u dont think about it

[astral ivy]

Lmao they’re so annoying to deal with

[slow skiff]

wait but no

its still true

[astral ivy]

Oh it doesn’t matter

If x is positive, we’re good

If x is negative, sqrt(x^2 + a) is positive which is obviously bigger than x which is negative

[slow skiff]

yeah, exactly

[astral ivy]

Lit

[slow skiff]

[astral ivy]

Dude I just feel incompetent

This is a high school calc class, or maybe even algebra 2

[slow skiff]

The problem is that once you learn more advanced stuff, you forget to use the simpler things

[astral ivy]

Yeah

[slow skiff]

At least for me

[astral ivy]

I was excited to find some cool way to prove bijectivity

But the best answer was so much simpler

[slow skiff]

An amazing comparison: limits before and after l'hopital

[astral ivy]

Hahaha yeah

Exactly

[slow skiff]

btw about before

Why'd you say the function has to be bijection?

[astral ivy]

Wdym