help-23

[safe radish]

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[fierce sky]

Can guys help me w this question ?

[blazing palm]

This question is so poorly writte

But i'll assume its:

$$\qty(\log_a \qty(\qty(\frac xy)^2))^2$$

[severe pond]

\left(\right)

[flat frigate]

Unban il798li

[blazing palm]

Gtg

[pseudo scroll]

what are you meant to do?

[fierce sky]

This topic is whether this answer is right or wrong ?

[uncut magnet]

those two are not equal in general

the log^2(u) ≠ 2log(u)

however, log(u^2) = 2log(u) is indeed true

[fierce sky]

Can you give me explanation for the question ?

[uncut magnet]

it most likely asks if both sides are true for all x and y

but using log identity shows that the LHS is:
2log^2(x/y)

whereas the RHS is
log(x/y)^4

[safe radish]

@fierce sky Has your question been resolved?

[safe radish]

@fierce sky Has your question been resolved?

[safe radish]

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[worn shell]

how do i tell algebraically which doesnt have an inverse? i can tell by eye that abde have inverses by process of elimination, but how do i show c doesnt have an inverse non graphically?

[fathom jewel]

You can bring up a counter example where e.g two x values map to the same y value

that would make injectivity fail and thus not invertible

[worn shell]

basically just plug and chug?

[fathom jewel]

well you can't plug in whol R

you kinda should have an intuitive idea which function wouldnt be invertible

you can actually derive the inverse algebraically like if you solve x = y² yet y = x² is not invertible if it's defined on R

[worn shell]

oh gotcha!

ty!

.close

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[final jungle]

guys

[final jungle]

i have a question

i have a math test in 2 hours

so basically

[wet wigeon]

[final jungle]

wait

ill draw a diagram first

[final jungle]

do u know trig ratios

like without calc

help me find x please

[hidden gyro]

special angles ? 30,45,60,90

[final jungle]

yeah

[final jungle]

can someone help me find X

[plucky elk]

,tex .sohcahtoa

[flat frigate]

riemann

[plucky elk]

you just need to know tan(60 deg)

[final jungle]

so what i did was tan60=x/4

where do i go from there

[plucky elk]

if you haven't memorized tan(60 deg), then memorize sin(60 deg) and cos(60 deg) and use tan(angle) = sin(angle) / cos(angle)

[final jungle]

hmm

[safe radish]

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[shy ferry]

How do I find the lengths of the sides of a quadrilateral if I know the angles and bimedians?

[shy ferry]

I guess even just a single bimedian and 3 angles technically right?

[pseudo flax]

Divide it into right triangles and use trigonometry

[shy ferry]

Yes but I am confused how to divide it

[pseudo flax]

Show you've done

[shy ferry]

like this

i guess I am supposed to find the height and with that I can find the length of the sides?

but I do not know how

[pseudo flax]

Are those triangles?

They don't look like two triangles

[shy ferry]

sorry yes

[pseudo flax]

Wrong

Connect points a and c

Wait

No

[pseudo flax]

It's correct

[shy ferry]

so I need to find the hypotenuse of those triangles

but the part i am confused about is what the opposite and adjacents would

i have the angle but without the length of the opposite or adjacent i would not be able to caluate the hypotenuse then right

and how does the bimedian factor in to this?

[pseudo flax]

BC and AD are parallel

[shy ferry]

yes

okay wait

[pseudo flax]

Bimedian equals EF + 0.5 AE + 0.5 DF

[shy ferry]

the cut offs of the trapezoid combined equal the length of the rectangle in the center?

[pseudo flax]

Wdym the cut offs

And what rectangle in the center

[shy ferry]

sorry nvm

4 = EF + 0.5 AE + 0.5 FD

4 - 0.5 AE = EF + 0.5 FD

-0.5 AE = EF + 0.5 FD - 4

AE = -2 EF - FD + 8

I guess that is the adjacent?

of angle A

sec 40 * (-2 EF - FD + 8) = AB

sec 50 * (-2 EF - AE + 8) = CD

also I guess BC = EF

[safe radish]

@shy ferry Has your question been resolved?

[shy ferry]

ok system of equations

[safe radish]

@shy ferry Has your question been resolved?

[safe radish]

@shy ferry Has your question been resolved?

[shy ferry]

okay im back i thought about it and sorry i realized you also need more

information

because it can be stretched

[safe radish]

@shy ferry Has your question been resolved?

[safe radish]

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[twin oracle]

For question 3 I don’t get it how do you know when is a expansion or compression

Like how do u know when it is affect for x or y?

[pure agate]

,rotate ccw

[pure agate]

If it compresses or expands horizontally, then it affects x. It it compresses or expands vertically, it affects y.

[twin oracle]

It reflect on y axis

So there’s negative on X

But did u plug in5?

[twin oracle]

What if it just give u the point ?

The 3 is wrong so is like change in y right but the answer key said change in x

Idk

[pure agate]

Which one? a or b?

Nevermind, it's b.

You know it goes through the origin so you have two points to make an equation.

[pure agate]

.

[twin oracle]

I mean so it had a point at -5,3 right the answer key said is -5/3x I don’t get it why

Because I thought only change - and + and the y change it so it should be like 5/3(-x)?

[pure agate]

So the standard equation that is similar would be

y = -x

That obviously does not go through the point (-5, 3).

[twin oracle]

Yeah

[pure agate]

For x=-5, it goes through y=3 when it "should" go through y=5.

[twin oracle]

Yeah

[pure agate]

So that means you should compress y.

[twin oracle]

Yeah

[pure agate]

What is the ratio that compresses 5 to 3?

[twin oracle]

But the answer said beside X so is expansion X right?

[twin oracle]

3/5?

[pure agate]

Yes, so you have 3/5 y = -x.

And if you put that in y = mx form, what do you get?

[twin oracle]

Yeah I mean sure you can use the formula y2-y1/x2-x1, but what I try to figure out is I think it change y right so it should be like 3/5(-x) but the answer key is -3/5x

If is -3/5 x shouldn’t it change x instead of y?

[pure agate]

Make sense?

[twin oracle]

But the answer key is -3/5 x though

Y=-3/5 x

[pure agate]

My bad.

y = f(x)

y = 3/5 f(x)

y = 3/5 (-x)

Have you learned about functions yet?

[twin oracle]

Oh 3/5(-x)=-3/5x

The answer key just simply it

Tyty

[pure agate]

yw

[safe radish]

@twin oracle Has your question been resolved?

[safe radish]

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[candid ocean]

Im confused did i do a mistake

i just stopped going after

because i think im not doing it right

[severe pond]

🤔

[severe pond]

yo this is like 360p

[blazing palm]

Ya can we pls have a better pic

[candid ocean]

lol sorry i took it on my macbook

[blazing palm]

This is impossible to read

[severe pond]

and do you not know derivative rules?

[candid ocean]

umm well I just learned it lol

thats why im asking for help

[broken yew]

let's confirm what it's asking for

dt/ds?

[severe pond]

have you learned derivative rules or do you have to use the limit definition

[severe pond]

much better

[candid ocean]

limit definition

[severe pond]

cry

[hidden gyro]

that's rough tbh

[candid ocean]

like we learned the most basic derivative. thing

[blazing palm]

Smth messed up here

[candid ocean]

oh

[blazing palm]

Seems like you multiplied by the conjugate- thats good

[candid ocean]

OHHH

its -25

[blazing palm]

But idt you did it properly for the numerator

[broken yew]

you can defo make life easier by factoring out the constant from the function

before doing anything

[candid ocean]

what about after that

[blazing palm]

Can you show your updated work

[broken yew]

$t = \sqrt{\frac15}\sqrt{s}$ and make sure you take that $\frac15$ outside in the 1st line of work

[blazing palm]

Is this a multiplication of fractions? @candid ocean

[candid ocean]

ya

[blazing palm]

$\frac{125 + h - 125}5 \cdot \frac 1{h \qty(\sqrt{\frac{125 + h}5} + 5)}$

[flat frigate]

King Leo

[blazing palm]

Is that what you wrote

[candid ocean]

yep

[blazing palm]

Ok, now cancel out the 125s in the left fraction

Than cancel out the h

[candid ocean]

yeah so im left with h in the numurator

and 5h(huge thing)

[blazing palm]

h/h can cancel out

[candid ocean]

but isnt it 5h

[blazing palm]

?

Can you show me what you have now

[candid ocean]

so there would be 4h left

lol idk im tired

[blazing palm]

Cancelling out the hs just leaves you with 1/(denom)

[candid ocean]

ok

oh ya

oh

and then you get 1/50

yay

thank you so much for your help!!

.close

[safe radish]

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[lime canyon]

One message removed from a suspended account.

[neon summit]

You have to use FTC and also chain rule

[lime canyon]

One message removed from a suspended account.

[neon summit]

Fundamental thm of calc

[hearty prawn]

Fundamental theorem of calculus

[neon summit]

So let's define a func G(x) which is the same integral, but instead of x^3 it's just x

[lime canyon]

One message removed from a suspended account.

[neon summit]

Then, by FTC, what is G'(x)

?

[lime canyon]

One message removed from a suspended account.

[neon summit]

@lime canyon

[lime canyon]

One message removed from a suspended account.

[neon summit]

No

[hearty prawn]

There’s a rule that states the derivative of an integral on the interval [a(x),b(x)], the answer is f(b(x))*b’(x) - f(a(x)) * a’x

[lime canyon]

One message removed from a suspended account.

[neon summit]

It's ok

If we have a function defined as an integral

Then, by FTC, it's derivative is the function being integrated

[lime canyon]

One message removed from a suspended account.

[neon summit]

So in this case, if we have the integral of (sin(t))^2 from 0 to x

[neon summit]

K np I'll write it down to make it easier goo

[lime canyon]

One message removed from a suspended account.

[neon summit]

Too

[lime canyon]

One message removed from a suspended account.

[neon summit]

i'll try using latex but i kinda forgot how lol

[hearty prawn]

By this, answer is 3x^2sin(x^6) correct if I’m wrong

[neon summit]

yeah it could be right but there's a method to doing this without knowing a formula

[lime canyon]

One message removed from a suspended account.

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[hearty prawn]

Is the top bound x^3

[neon summit]

$F(x) = int_{0_x^3} sin^2(t) dt}$

this is def wrong lol

[lime canyon]

One message removed from a suspended account.

[hearty prawn]

Probably a typo

[lime canyon]

One message removed from a suspended account.

[flat frigate]

haveaniceday12
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[neon summit]

k i give up

[hearty prawn]

It should be 3x^2

[neon summit]

ok so that's F

[lime canyon]

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One message removed from a suspended account.

[neon summit]

ok so define a functino like this, right

what is G'(x) ?

[lime canyon]

One message removed from a suspended account.

[hearty prawn]

You’re good don’t worry

[lime canyon]

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One message removed from a suspended account.

One message removed from a suspended account.

[hearty prawn]

I say just memorize the formula

It will get you the right answer all the time

[lime canyon]

One message removed from a suspended account.

[hearty prawn]

Yea

[lime canyon]

One message removed from a suspended account.

[hearty prawn]

It’s true for all integrals

There’s a rule that states the derivative of an integral on the interval [a(x),b(x)], the answer is f(b(x))*b’(x) - f(a(x)) * a’x

So whatever the integrate is

Plug in the top bound in the function

Find the derivative and multiply it

And then do the same for the lower bound

And subtract

[lime canyon]

One message removed from a suspended account.

[hearty prawn]

Sure

[lime canyon]

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One message removed from a suspended account.

One message removed from a suspended account.

[hearty prawn]

Sorry I’m in bed rn so imam use notes

[lime canyon]

One message removed from a suspended account.

[native raptor]

nave can u help me after? no one wants to help me, I just need a check

[hearty prawn]

The derivative of a constant is 0 so the 2nd half evaluates to 0

So

You just multiply the top part

[lime canyon]

One message removed from a suspended account.

One message removed from a suspended account.

[hearty prawn]

And get 3x^2sin(x^6)

[hearty prawn]

Sure I’ll try

[native raptor]

ok

[lime canyon]

One message removed from a suspended account.

[hearty prawn]

Yep

That’s right

But not plus C

[lime canyon]

One message removed from a suspended account.

[hearty prawn]

Alright perfect

[lime canyon]

One message removed from a suspended account.

[hearty prawn]

Try that with other problems

[lime canyon]

One message removed from a suspended account.

[hearty prawn]

There isn’t a C

It just is 0

Because sin 0 is 0

And derivative of 0 is 0

So that part is 0

[safe radish]

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[steel plover]

im not sure where to start

[mossy remnant]

try drawing it out

[mossy remnant]

^^

[steel plover]

i did

[mossy remnant]

what did u get?

[steel plover]

i drew this on my paper

[mossy remnant]

ok what unit are you in rn

itll be rlly to explain knowing what method i should use to tell you

[mossy remnant]

^^^

[steel plover]

wdym what unit

im like

[mossy remnant]

like trig or geometry

[steel plover]

geometry

[mossy remnant]

ok

do you know

a^2 + b^2 = c^2

?

[steel plover]

yes

[mossy remnant]

ok great

now

do you know the definition of the radius?

[steel plover]

yeha

[mossy remnant]

it is?

[steel plover]

half diametre

[mossy remnant]

right

[steel plover]

diameter

[mossy remnant]

also

its the midpoint fo the circle

[steel plover]

yeah

[mossy remnant]

to any part of it

great

now

why dont you see when the square touches the midpoint

and put a dot or smth there

[steel plover]

okay

[mossy remnant]

then another dot where the square touches the circle

then draw a line between those two dots

[steel plover]

for both points where it touches the circle?

[mossy remnant]

no

[mossy remnant]

this is point one

[mossy remnant]

this is point two

[steel plover]

oh okay

[mossy remnant]

so one at midpoint

one at touching circle

[steel plover]

okay

i got it

so now what

[mossy remnant]

thats your radius

[steel plover]

oh yeah

that is

[mossy remnant]

so how do you wanna try to solve it

[steel plover]

so like

x^2+ (x/2)^2=radius

liek that

[mossy remnant]

ur missing thing

a^2 + b^2 = c^2

[steel plover]

radius^2

[mossy remnant]

yup

[steel plover]

oh okay

[mossy remnant]

so you good now?

[steel plover]

wait

[mossy remnant]

?

[steel plover]

so i git (5(x^2))/4=radius^2

is that right

[mossy remnant]

yup

[steel plover]

and it simplyfy

[mossy remnant]

yea somewhat

[steel plover]

r=(x root 5)/2

[mossy remnant]

yup

[steel plover]

oh okay

thank you

[mossy remnant]

np

you can do .close to close this :)

[steel plover]

.close

[safe radish]

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[mossy remnant]

cant see the question

i just have the graph

[foggy cloud]

is this correct?

[calm peak]

duh

cant see ques

[foggy cloud]

sorry

[pure agate]

,rotate ccw

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Type .reopen

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[deep dagger]

I'm not sure what formula they're using to get vapor pressure in this table? I tried to follow the formula at the top, but I got the wrong answer. The original formula will probs help more. 😅 Could be one of the formulas in the 2nd photo?

[safe radish]

@deep dagger Has your question been resolved?

[crude lintel]

the equation that relates vapor pressure and temperature is the Clausius-Clapeyron equation which is not on your formula sheet

[slim plinth]

Won’t you need the change in enthalpy for that?

[crude lintel]

yes i am also confused now that i look harder at this 😭 sorry lmao give me a minute

oh was there no methanol added yet for the first data point?

[deep dagger]

ah sorry for the late response, didn't see y'all D:

the 22.8 and .9919 were room temps, so yes :>

there's 5 trials

I'm like 70% sure he used a mixture of the last two formulas, but I have NO clue how @~@

[crude lintel]

ohh okay then yeah so he is calculating what the new atmospheric pressure would be due to the change in temperature using the relationship P2 = P1T2/T1, and then subtracting that from the actual total pressure to find what partial pressure is due to the methanol

does that make sense

[deep dagger]

oooh right cause 1 would be the initial 🤔

[mossy remnant]

damn, they be asking all types of questions here, i gotta refine my other subject knowledge

[deep dagger]

does that come from (P_2/P_1)=(T_2/T_1)?

xD haha yeahh

the topicsd

change a lot

topics*

it'sscary

[slim plinth]

Yeah I don’t think it would have been Clausius Clapeyron since the pressure is a function of e to the power something there

[crude lintel]

lol i think my excessive amount of chem knowledge made me look for zebras instead of horses

[deep dagger]

gosh I love horses

anyways

OH

I CAN CHECK THE LAB PREP

maybe that has answers

I hope

the fact that the base equation isn't on the test sheet is concerning. 🤔 He puts everything we need there.

[crude lintel]

if u look at the bottom left equation which is the combined gas law, you can keep one variable constant and just vary the other 2 to get Charles law, Boyles law, or the secret unnamed pressure-temperature law ahaha

[deep dagger]

ooooh true :O

my teacher is normally really good, I SPECIFICALLY chose his class, but this specific thing is very confusing 😅

Idk why I said normally. He IS really good

srry checking something rq and then I think that's it

ahhh okay yeah, this makes sense

ty guys <3333

[crude lintel]

np 👍

[deep dagger]

.close

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[manic bane]

Sketch an algorithm (pseudo code) to find the support for all frequent itemsets, using only the set ofclosed frequent itemsets (and their support) as information

I'm a bit lost on where to even start, if someone could guide me in the right direction?
Could i use the inclusion/exclusion principle perhaps?

[safe radish]

@manic bane Has your question been resolved?

[manic bane]

.close

[safe radish]

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[plush salmon]

can someone explain what just happened

[plucky elk]

Looks like they just found an arbitrary degree 4 polynomial with 3 free variables a and b and k which is enough for the constraint requirements.

k isn't free per se, but just a representative of the constraint equation

[plush salmon]

yeah but how did they decide that x-3 is a factor

[plucky elk]

it's not a factor of the general polynomial, just the first summand

if you let f(x) equal the entire polynomial, then f(3) = k which is one of the constraints

same for 5 and 8

[brave wolf]

(there is also k, so 3 free variables)

[plush salmon]

so how do i proceed from here tho

to find the basis

[plucky elk]

oh yes you're right it is a free variable

a, b, k are the free variables

[plush salmon]

this is what they did and im so confused

OHH

wait

okay i got what they did

idk why they did what they did tho

is it just to split it into the free variables to show that its span?

[brave wolf]

Basis allows us to write every vector as a linear combination of basis vectors. Note that they wrote w = a(something) + b(something) + k*1. They managed to write w as a linear combination of some vectors (with the free variables being a, b and k). And those vectors therefore constitute a basis (because w can be arbitrary)

[plush salmon]

ohhhhhh i see

okay that makes sense

thank youuuu

.close

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[paper shell]

also note that v1,v2,v3 must be linearly independent.

here you can say that because they all have a different degree

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[visual pond]

I'm not really sure how to start

I've been working on mins and maxes

I think distance formula might be needed

[heady igloo]

what method have you learnt? projection or multivariable minimisation?

[visual pond]

we did projection in a past unit

but I think we are supposed to use a different method here

[heady igloo]

you can use the set gradient to 0 method here

[visual pond]

is the shortest path always the normal vector?

[heady igloo]

this is multivariable minimisation

yes

[visual pond]

I don't know what that means sorry
We have been finding min,max of multivariable functions using
F_x F_y
F_xx F_yy
F_xy
D (x,y) = F_xx*F_yy - F_xy^2

[heady igloo]

yes thats the gradient method, where you set the partials to 0

and use the hessian determinant to check if its a max or a min

D (x,y) = F_xx*F_yy - F_xy^2
is the hessian determinant

[visual pond]

Gotcha

I wasn't told the key words sorry

[visual pond]

is it faster to use projection though?

[heady igloo]

tbh i dont think the difference is noticeable enough to matter

if you are allowed to use any then go with the one you prefer

[visual pond]

gotcha thank you

.close

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[distant bronze]

Hey guys! I took a linear algebra course last year, and now just as self study, I want to get into some less surface-level linalg topics that weren't covered as part of the course I took.

Currently, I'm trying to understand singular value decomp, and I've got just a couple questions, mainly about the geometric intuition bout things.

1. I get that the matrices A^T A and A A^T describe how the rows and columns of A correlate with one another, and that finding the axes of maximum variance gives you the most dominant features in this space. I don't understand, however, what the eigenvectors of these two matrices mean geometrically/intuitively, and why using these eigenvectors/values (ranked by the eigenvalue size) gives you these principal directions of variation.
2. Why do the singular values of matrix A happen to be the square roots of the eigenvalues of A^T A and A A^T?
3. The idea of eigenvectors seems to be to find the axes within which direction is unchanged, and the point of singular vectors seems to be to extend this idea to non-square matrices. But singular vectors don't preserve direction like eigenvectors do, as a result of non-square matrices having different input/output space dimensions, so what do they preserve? I hear that they preserve orthogonality, but I don't really understand what that means geometrically/practically.
4. This is more of a general question about transposes in general. Say a matrix applies some sort of transformation, what will the transpose of that matrix apply as a transformation? What about the A^T A and A A^T matrices - what type of transformations will they apply? Or do transposes not really have a geometric meaning with respect to their original matrix's geometric transformation?

Thanks!

[untold cipher]

Eigenvector v of A^T A is the input direction that maximizes ‖A v‖/‖v‖; a scales it by sigma = sqrt(lambda) and sends it to unit u so Av = sigma u. collecting all such pairs gives A = U Sigma V^T meaning A turns the unit sphere into an orthogonal ellipsoid whose axes are the u’s and lengths are the sigmas while A^T is just the dot product reverse of A

[distant bronze]

okay wait sorry im tryna dissect this

[distant bronze]

So in relation to my 3rd question, the right/left singular vector doesn't really preserve anything, and right singular vectors are just those that happen to be the most important directions that A stretches space in and the left singular vectors are just the resulting vectors after A is applied to the right singular vectors?

Also a bit confused of the last piece - A^T is the dot product reverse of A, so what would that mean geometrically for a transformation A and a transformation A^T? A^T doesn't undo the transformation (unless A is orthonormal) right

And for #2, u said that a scales it by sigma = sqrt(lambda) but why do we know that's the case, that it's specifically the sqrt(lambda)?

[untold cipher]

Right vecs v_i are mutually perpendicular; A just scales each by sigma_i and spits out perpendicular u_i so those angles stay 90 deg while lengths change. A^T is the unique map with dot(Ax,y)=dot(x,A^T y); it applies the opposite rotation but the same stretch so it equals A^{-1} only when A is orthogonal (pure rotation/reflection)

[untold cipher]

Because (A^T*A)v = lam v,taking lengths gives ||A v||^2 = lam||v||^2 so A stretches v by sig = sqrt(lam); writing u = (1/sig)A v yields A v = sig u which shows right singular vecs stay perpendicular and map to equally perpendicular left vecs scaled by sig while A^T is the adjoint that reverses only the rotation part of A and equals A^-1 exactly when A is orthogonal

[distant bronze]

so it spits out perpendicular u_i so the angles stay 90 deg but not necessarily within the same direction of the right vecs v_i?

also, bit of a different question, but is there a guarantee that for a NxN symmetric matrix there's gonna be N eigenvalues, or is it jus =< N

[untold cipher]

correct

90 degree relationships are preeserved between corresponding axes

[distant bronze]

and that's only the case with those right singular vectors - no other set of vectors will maintain the same angles over the transformation?

[untold cipher]

yes, any real N x N symmetric matrix has exactly N real eigenvalues counting repeats and you can pick N mutually perp eigenvectors forming a basis; some eigenvalues may be equal but the total count (with multiplicity) is still N

[untold cipher]

basically

[distant bronze]

ohh interesting, so A^T always opposes rotation (bc A is orthogonal) but applies the same stretches

[untold cipher]

an orthonormal set v_i keeps its 90 deg angles after A only when each v_i lines up with a right singular vector (up to rotations inside any group that shares the same singular value). any other choice gets sheared or its angles distorted

[untold cipher]

write A = R*S with R orthogonal (pure rotation/reflection) and S symmetric pos def (pure stretch); then A^T = S*R^T so the same stretches act but in reverse order and the rotation flips sign. when S = I A is orthogonal and A^T = R^T = A^-1, the exact opposite rotation

[distant bronze]

hmm okay im still working thru this

[distant bronze]

okay i'm confused with how you got to the coded section in the second line by taking lengths. is that jus like how the transpose algebra works out? also i'm still refreshing myself on linalg in general so this might jus be me being incredibly rusty and having forgotten lots

[untold cipher]

start with the eigen eqn A^T A v = lam v. now dot both sides with v: v^T A^T A v = lam v^T v. because (A v)^T = v^T A^T the left side becomes (A v)^T (A v) which is just ||A v||^2, while v^T v is ||v||^2. so we get ||A v||^2 = lam ||v||^2; that is all taking lengths means. from there sig = sqrt(lam) is the stretch factor since ||A v|| = sig ||v||

[distant bronze]

wait lemme work thru it a lil more n ill get back

[distant bronze]

is it just 90 degrees that are preserved/not preserved? or will all other angles not be preserved for non eigenvectors?

[untold cipher]

apart from 0 or 180 deg cases the only angles A keeps rigid are the 90 deg ones between right singular vectors that share the same stretch factor

any other pair of directions is generally skewed

[distant bronze]

How does A^T = S*R^T? Wouldn't it equal S^T (or S^-1)*R^T?

[untold cipher]

because (R S)^T = S^T R^T and S is symmetric (so S^T = S) you get A^T = S R^T

no inverse appears since only the order flips when you take a transpose

[distant bronze]

So I guess back to my original 1st question, it's still unclear as to how the eigenvectors of A^T A and A A^T correspond to the principle directions of variation?

[untold cipher]

A unit vector v that maximizes the variance sent through A solves max v^T(A^T A)v with ||v||=1; the maximizer is exactly the eigen-vector of A^T A for the largest eigen value, the next eigen vectors (kept orth) give the next largest variances so the whole eigen basis ranks the principal directions of variation (same story in the output space with A A^T)

[distant bronze]

I might be missing something - I don't have an intuitive understanding of why it is that the maximizer for principal directions of variation is specifically the eigenvectors?

[untold cipher]

think of M = A^T A as a set of orthonormal axes (its eigen basis) where each axis is stretched by its own scalar lambda

any unit v is a mix of those axes so its output length squared is the weighted sum Sum(lambda_i * coeff_i^2)

that sum is largest when all weight sits on the axis with the biggest lambda, i.e.

when v itself equals that eigen vector and the same logic cascades for the next axes once you force the directions to stay orthogonal

[distant bronze]

what does the coeff_i represent?

[untold cipher]

coeff_i is e_i*v, the scalar projection of v onto eigenvector e_i, measuring how much of v’s unit length lies along that eigen axis

[distant bronze]

Okay this makes sense for why the highest eigenvalue'd eigenvector will have the highest output length squared. But why wouldn't the second highest one be the vector v which is like 99.999% of e1, rather than the second highest being a whole new e2?

ohhhh bc you want them to be orthogonal?

so that they're independent

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[fleet sun]

In the axiomatics of reals, there are axioms of addition, multiplication and well-ordering. And I have confusion as to why we don’t write set of reals as 4-tuple (R, +, *, >=).

[quasi bison]

bc it's cumbersome

but if you specifically want to keep writing 9 symbols where everyone else gets by with 1, go right ahead.

[median vigil]

it's very convenient to use the notation for the set as shorthand for the set and its associated structure so it's common

[fleet sun]

So, I wouldn’t need anything new excluding bilinear form to construct Euclidean Space?

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[coral summit]

For real world scenarios should I use rational numbers or irrational numbers. E.g. 22/7 to represent pi.

[velvet pier]

@coral summit what kind of real world scenarios can ya give example

[coral summit]

I want to design a packaging option to hold 10 to 16 pign pong balls

including volume & surface area

[edgy ridge]

I think you can use any and just ceil at the end, does that work?

[quasi bison]

"use rational vs. irrational numbers" is not really a question that makes sense

[coral summit]

oops

[quasi bison]

generally if you can do calculations in exact form you should, but then for subsequent manufacture you'll need to round them to some number of decimal places

(and then 22/7 might turn out to be too crude as an approximation of pi for your purposes)

also you should consider making your nickname less edgy

[coral summit]

mb

[quasi bison]

and maybe do not use the word "female" as a noun like that. it's stinky.

[coral summit]

sorry bout that

thank you for helping tho

.close

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[vague flame]

hi guys, i have an quick question so here there is an lemma and before the corollary is written it is being obtained by using an simple counting argument what is the simple counting argument shall be ? or what is that ?

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[zinc rover]

Sup everyone. So, I need a bit of help here. This one is a sequence and I wanted to check first, if it converges to 0 so I can go on with the next criteria, but then I've found out that this sequence converges to 1. My question is, is 1 the correct answer?

[cedar widget]

Its not 1

[winged cloak]

If u find that the sequence converges to a non zero, the series diverges

I think

[cedar widget]

Yeah, just the sequence does converge to 0

[winged cloak]

Yeah but that's inconclusive

[zinc rover]

My thought was, to present the denominator as "(n!) power 1/n" and since I know that 1/n converges to 0 and (n!) power 0, then I get 1 in the denominator

[zinc rover]

In this case, I don't know how to formulate it, how do we come to this conclusion?

[cedar widget]

Yeah but both n in the exponent and base get very large, so you need to account for that

If n! was n^n then (n^n)^(1/n) was just n

And n^n and n! are close in growth

[main mural]

in fact n^n grows ||faster|| than n!

[cedar widget]

Yeah, but ||striling||

[main mural]

this is a consequence of Stirling approximation formula for n!

[main mural]

yes xd

[zinc rover]

then, how do you guys propose that I find the Lim an = 0 then?

[last wren]

can you use the root test or have you not learned that yet

[cedar widget]

My point is that we should expect, due to striling a similar result to the case of n^n

[last wren]

oh wait it's 1/n nvm

carry on lol

[zinc rover]

Oh yeah, we already had that, but I wanna leave this step for after I've proven that this sequence -> 0 because this is "Trivial/necessary criteria"

[zinc rover]

I just presented the root as 1/n cause it was easier to work with

[main mural]

i think you should first prove as a lemma that n^n > n! for all n > 2

then the rest will follow much easier

[main mural]

you can do it by induction

[zinc rover]

like, I should use inequality?

[main mural]

wdym

you should prove the inequality that n^n > n!

and then you will be able to do comparison tests on your series

[zinc rover]

I mean: 1/(n power n) 1/n < 1/(n!) power 1/n

Ooooorrrr, I'm understanding it wrong

[main mural]

you can use ^ for power

[zinc rover]

it's easier for me to type it as "power", but thanks for the tip

[main mural]

do you mean $$\frac{1}{(n^n)^{(1/n)]} < \frac{1}{(n!)^{(1/n)]}$$ ?

[zinc rover]

yup

but I don't know if this way of thought is correct....

[prisma wren]

$\frac{1}{(n^n)^{1/n}}<\frac{1}{(n!)^{1/n}}$

[flat frigate]

Stitches

[prisma wren]

Prove this, then you can recognize that the LHS of this inequality is just 1/n

[zinc rover]

OH, yeah, this makes sense. And of course 1/n converges to 0 then.

Thank you guys!

[prisma wren]

does it?

The sum of 1/n converges?

Well yes, the limit of the sequence converges to 0, but the sum does not necessarily.

[zinc rover]

this is just one of the criterias I need

while 1/n does converge to 0 this doesn't mean, that the sequence converges

[prisma wren]

You're aware that this criteria is inconclusive though, right?

yes

[zinc rover]

that's why I need another criteria

I know

[prisma wren]

yes

[prisma wren]

which follows from this

[zinc rover]

I mean, I can just show that 1/n is a harmonic series and per Axiom a harmonic series diverges

[zinc rover]

yeah, that's what I mean, it's a bit diffictult to switch between languages cause I don't learn math in english

but thank you for confirming btw!

.close

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[cobalt saddle]

10.2

[cobalt saddle]

Need to see if i did the right thing

[warm warren]

Isn't the first one 1/4?

[prisma wren]

Yeah first one is wrong.

[cobalt saddle]

They can study on the other days of the week

?

Wait i see😕

[prisma wren]

Anna and Ben both studied in the school library one day this week. + School library is open Monday to Thursday -> Anna and Ben both studied one day between Monday to Thursday

[cobalt saddle]

Anyway how do you do 10.2

[prisma wren]

I would just go by counting. 4^2 possibilities in total, count the number of cases where they're consecutive.

[cobalt saddle]

?

[prisma wren]

?

[cobalt saddle]

?

[warm warren]

FInd the number of possible combos where 10.2 holds, like, either anna monday and ben tuesday or ben monday and anna tuesday, so its 8.

[prisma wren]

You wanna tell me what you don't understand or am I supposed to read your mind?

[cobalt saddle]

Can you have possibilities adove 1

[cobalt saddle]

I do not understand

[prisma wren]

I'm not asking if you understand or not, I'm asking what part of my statement you don't understand.

[cobalt saddle]

All of it

[prisma wren]

Consider two die rolls. You can get (1,1), (1,2) ... (6,6). There are thus 6^2 possible combinations.

Does that make sense?

[cobalt saddle]

Yes ,so far

[prisma wren]

Now, we can apply the same logic to the days the two people study. You could have (A: Mon, B: Mon), (A: Mon, B: Tues) and so on until (A: Thurs, B: Thurs)

How many possible combinations are there?

[cobalt saddle]

4×4

So would it be 6/16?

<@&286206848099549185>

.close

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[south lynx]

this cannot be solved by standard linear-recurrence methods, right? how do we start with it?

[south lynx]

please ping me if responding

[astral glacier]

Well the first step is to just write out a few terms

Just to get a feel for what's happening

@south lynx

[south lynx]

a1 = 17, a2 = 291, a3 would be 84684; i do not get how we have to move forward with it

[open wedge]

what do you mean by

solve the recurrence relation

[astral glacier]

Well that looks somewhat exponential

Which makes sense, every term is "roughly" the square of the previous term

[south lynx]

right

[astral glacier]

The contribution of the + n is.. quite low

[open wedge]

?

i need more context of this

[astral glacier]

Now do you just want your solution is a big O form

Cuz if so, this much suffices.

[south lynx]

well, i do not know what my professor wants. does this not mean that this recurrence admits no simple closed form in elementary functions

[astral glacier]

You can technically calculate a closed form

It will be quite messy

But it's doable I think

Firstly let's do an approximation

We know it's exponential, let's find in what way

[south lynx]

all right

[astral glacier]

Also if I disappear randomly read my nickname

And feel free to tag me

[south lynx]

does it have anything to do with log an?

get well soon, xavier

[south lynx]

never mind

no, wait

[south lynx]

doesn't it double approximately at each step?

@astral glacier

[peak estuary]

funnily enough this is on the oeis

but even that doesnt list an explicit formula

I do not know what your prof expects

[restive niche]

n+(n-1+(n-2 + (...(17+(4)^2)^2...)^2)^2)^2

[peak estuary]

thats worse than just writing out the recursion

[south lynx]

i wonder if it is just a typo

[restive niche]

Yeah lol

Idk what else you can expect here that isn't just the recursion or that

[south lynx]

.close

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[thorn epoch]

consider an equilateral triangle, draw the largest possible circle that is completely inside this triangle , given the radius of the formed circle is 16 cm what is the height of the equilateral triangle in cm

[thorn epoch]

i found a formula online for it which is

r = h/3

but why?

like why does it work

[nimble wagon]

I don't know the proof but it is the incircle

[glacial cairn]

That circle is the incircle, and its center is also the centroid, which divides the medians in a 1/3 to 2/3 ratio

[thorn epoch]

.-.

uhhhhh

wut

[glacial cairn]

The medians of an equilateral triangle are of course also the heights and the angle bisectors

[thorn epoch]

i do not understand that TvT

[nimble wagon]

Are you aware of what a median is

[thorn epoch]

middle?

[nimble wagon]

A like joining the vertex and the opposite side such that the side is bisected

[glacial cairn]

Here's an equilateral triangle

[nimble wagon]

Have you done congruency in triangles

[glacial cairn]

Those lines that go through O are the angle bisectors, the heights, and the medians

Do you agree so far?

[thorn epoch]

yea

[thorn epoch]

ye

[nimble wagon]

So you can easy prove how median, angle bisectors are the same

[glacial cairn]

So O is, at the same time, the incenter, the orthocenter, and the centroid of this triangle

Yes?

[thorn epoch]

yes

[glacial cairn]

And the centroid divides each median into a 1/3 to 2/3 ratio

That's just a property of triangles

[nimble wagon]

Sorry for the disturbance i thought you weren't aware of this @thorn epoch

[glacial cairn]

So the radius of this circle is 1/3 of a median, which is also 1/3 of a height

[thorn epoch]

wut

im trying to process this but cant

median line is the line in the middle right

then the angle bisector line are the one on the left and right side right?

js to make sure im understanding this correctly

then O is the centroid the absolute middle of the circle

then?

[glacial cairn]

No

[thorn epoch]

?

[glacial cairn]

All three lines are angle bisectors, heights, and medians

[thorn epoch]

which part i got wrong? ;V

[glacial cairn]

They're the same in an equilateral triangle

[thorn epoch]

ok

but then? how u got 1/3

[glacial cairn]

This is a triangle with two of its medians

I is their intersection

I divides each one into two parts, and for each median, the length of one part is double the length of the other

[thorn epoch]

ok

then?

;v

[nimble wagon]

We are done

Since centroid is incentre and it lies on h

h/3 is the distance of incentre from side which is the inradius

[thorn epoch]

uhhhh now i js kidda see but how do we know like the lenght outside the circle is the same as the radius?

i alr understand everything but now am stuck on how can we like confirm

the upper part = r

?

[glacial cairn]

Not sure what you're having trouble with

[thorn epoch]

im lowkey js a bit dumb

[glacial cairn]

AM is an angle bisector, a height, and a median of this equilateral triangle

O is the incenter, the orthocenter, and the centroid

[thorn epoch]

okok

[glacial cairn]

By the centroid property I mentioned, OA (red) = 2*OM (green)

OM is a radius of the incircle

Hence AM, the height, is 3 times OM, the radius

[warm warren]

The incentre is where the internal angle bisectors meet. And in an equilateral triangle, the angle bisectors are the same as the altitudes and medians, So, the incentre is the centroid. Now, the inradius is where it touches the base, i.e. OM. A centroid divides it in the ratio 2:1, so OM = AM/3. And AM is the height also because altitudes=median in equilateral triangle. So the inradius is 1/3rd the height

[thorn epoch]

ok but can i ask how do we know that the OA = 2OM like uh

[glacial cairn]

https://en.wikipedia.org/wiki/Centroid#Of_a_triangle

[warm warren]

uh thats a property that can be derived.

the section formula will help

the centroid 2:1 ratio is actually true for any triangle, not just equilateral.

[thorn epoch]

ye but like main reason im confused is cus for like OA i understand like the radius of the incircle = r ofc so then rn im asking like the

OA - (radius of the incircle) = radius
how can we like know this?

[thorn epoch]

😮 huh

[warm warren]

no,

OA

is not the radius

OM is the radius

OA = 2 OM

cuz the sides are tangent to the incircle, so OM is the radius

[thorn epoch]

yes i understand that so since OA = 2 OM and OM is the radius

that means the red line is 2 radius ye?

[warm warren]

yes?

[thorn epoch]

then since the red line includes the radius from the incricle

[warm warren]

(its twice the radius)

[thorn epoch]

so it makes it so like the part from outside the circle = the radius in the incircle

like how do we know the part outside the circle = the radius in the incricle

thats like the core reason i got confused

[warm warren]

No, we don't do that. Its the property of the centroid itself that it divides the median in the ratio 2:1

here the angle bisector is the median

so the angle bisectors meet at the centroid

so the centroid divides the angle bisector in the ratio 2:1

and the place where the angle bisectors meet is the incentre

so the centroid is the incentre

[warm warren]

And the inradius is made with the base, so OM is the inradius

and OA = 2OM

BUT AM is the height since the altitude is also the median in an equilateral triangle

So, OA = 1/3rd of the height

[thorn epoch]

ohhh

okok

i think i understand it now

tyty

.close

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[umbral haven]

Can anyone find n?

[magic junco]

yes, I can.

[quasi bison]

a nonzero amount of people can

[normal moss]

negative too

[umbral haven]

oh

yo guys

help

[normal moss]

It's next to the degree symbol

But no, have you got any ideas?

[umbral haven]

ABO is 90 degrees

we just gotta find OAC

[normal moss]

Why is $\angle ABO = 90^{\circ}$?

[flat frigate]

USS-Enterprise

[umbral haven]

cause since the tangent contacts the circle at B where B is the radius

it makes a 90 degree angle

[normal moss]

Well, B is not the radius.

OB is the radius

[umbral haven]

yh well BO

[normal moss]

Correct

Well

What's then different to OCA?

[umbral haven]

im not too sure

ah its fine

.close

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[plucky elk]

alpha and beta are not labeled

[fathom quartz]

it's not in the preview, it's in the link
preview image is not completely loaded

[plucky elk]

then

!ss

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[fathom quartz]

then

https://ibb.co/qMNxZgsJ

[safe radish]

@fathom quartz Has your question been resolved?

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@fathom quartz Has your question been resolved?

[fathom quartz]

no one ?

[cedar charm]

okay

so hint 1. draw a new point on the arc between E and F away from the points CBA to get a quadrilateral

call this new point G

Then you know what the angle EGF is

now this quadrilateral also has a special property (opposite angles add up to 180°) so you know EBF

okay wait

got okay this is gonna be annoying but I have a better solution

so |OB| = 1

okay wait lol this is actually pretty difficult

[stoic saddle]

you may use the extended sine law

just find angle EBF

[fathom quartz]

can you explain a little more ? (i'm really not good at geometry)

[tardy mango]

[https://en.wikipedia.org/wiki/Law_of_sines#HeroSection]

this implies that

,texsp ||$\frac{EF}{\sin (\angle EBF)}=2R$||

[flat frigate]

Civil Service Pigeon

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@fathom quartz Has your question been resolved?

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[wicked bone]

do all convergent series pass the divergence test / is the limit of the nth term of any convergent series always 0?

[severe pond]

yes this is the divergence test

[wicked bone]

i mean, do all convergent series pass it? for example i know not all convergent series pass the root test or ratio test

[severe pond]

i think you should reread the statement of the theorem

if $\lim a_n \neq 0$ then $\sum a_n$ diverges. this is the same thing as saying if $\sum a_n$ converges then $\lim a_n = 0$

[flat frigate]

knief

[dull sequoia]

Yes

You can prove this by contradiction

[severe pond]

wut

[wicked bone]

okay thank you thats all i needed lol

perfect

[severe pond]

its just the contrapositive of the usual statement

unless you mean prove the divergence test

[dull sequoia]

Suppose there is a lower bound r for the sequence

[wicked bone]

i getcha, makes sense

tyvm

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[fickle mantle]

Can you help me with the b

[fickle mantle]

Idk ify answer isneight

[magic junco]

That is correct

[fickle mantle]

ty >3

do you know how to prove something is congruent

[magic junco]

ye

Which one are you having problem w?

[fickle mantle]

both

i only know how to identify that it is congruent

idk what exactly I’m suppose to write

[magic junco]

What properties of congruent triangles have you learnt?

[fickle mantle]

sss

sas

Sara

Rhs

[magic junco]

What is this?

[fickle mantle]

saa

autocorrect

[magic junco]

anything else?

[fickle mantle]

no

[magic junco]

how about asa

[fickle mantle]

for similarity, ik aaa

[fickle mantle]

isn’t that saa

just turned

[magic junco]

yeah sure.

[magic junco]

What property can you apply here?

Question a.

[fickle mantle]

for a) rhs

also i gotta go do something rq bc its almost my turn too

sorry y i ireply late

[magic junco]

Alright

[fickle mantle]

question- for proving the congruent and similarity, do i need to also know the types of shape

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@fickle mantle Has your question been resolved?

[fickle mantle]

Back

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@fickle mantle Has your question been resolved?

[gritty sable]

What's the matter?

[safe radish]

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[fickle mantle]

idk how to tell if a quadrilateral or triangle is congruent

[fickle mantle]

like knowing what to write

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[fickle mantle]

.close

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[feral ravine]

can someone explain to me how imaginary numbers wor

[feral ravine]

because like the def of a square root is a nonegative number right

someone HELP I have a test tmrw and im lowk cooked

[quasi bison]

of a negative number, you mean.

anyway, the way they "work" can be summarized as:

there's a newcomer to the scene, a funny little letter called i, full name "Imaginary Unit", which behaves as if it were a variable but also has the special property that i^2 = -1

a pure imaginary number is a real multiple of i, ie 2i, 7i, -69i, sqrt(5)i, (pi-7)i and so on

a complex number is anything you can get in the form of a real number plus a pure imaginary number (both optional)

[safe radish]

@feral ravine Has your question been resolved?

[feral ravine]

ok...

but why is that

@quasi bison

and the def of a sqrt is to give the nonnegative output

so how can the sqrt of i be negative

it doesn't logically follow the definitions

[quasi bison]

that's for real numbers, but you have to let some of that go to extend yourself

[quasi bison]

confusing yourself here. nobody said anything of sqrt(i)

[feral ravine]

yea but like if i^2=-1 then i = sqrt -1 right

isn't that how it works

or am i doing this wrong

do I have to check for extraneous solutions when i sqrt both sides

how can there be a sqrt of a negative number

it just doesn't logically make sense to me

[fierce nimbus]

sqrt(i) ≠ sqrt(-1)

[feral ravine]

i mean i = sqrt = -1

bc i^2=-1

[fierce nimbus]

do u mean i= sqrt(-1)

i feel like that’s where ur getting mixed up

[feral ravine]

yes

thats what i mean

[fierce nimbus]

concept of imaginary numbers basically lol

[feral ravine]

whats the point of doing math of something that doesn't exist

[fierce nimbus]

it does exist it’s what helps us visualize problems that we cannot do with real numbers

i’m not the right person for this cause i’ve not studied it intensively but imaginary numbers play a big part in modeling real life things

[feral ravine]

can u expand on this?

[fierce nimbus]

the substitution of sqrt -1 = i just makes it easier to work with algebraic equations

[feral ravine]

so then are we just ignoring the def now

that like it outputs nonnegative number

for the rest of my math do i just ignore it

or like how does that work

does that rule just go BYE!

[fierce nimbus]

are u talking about the definition of a square root

[feral ravine]

yes

[fierce nimbus]

i think (and this is purely just my understanding) that the “imaginary numbers” exist just so that the -ves under square roots don’t violate the definition of it
by having for ex sqrt(-16) you can break it down to sqrt(-1) and sqrt(16) which gives u a 4 and the i where u can multiply it

sqrt16 itself does not violate the nonnegative output and the i substitution makes it easier to deal with

idk though don’t take my words too seriously, i hope someone else can explain it better 😭😭

[quasi bison]

yeah that part kinda goes out the window when dealing with complex numbers. in the real case its role was to ensure sqrt(x) can consistently refer to one single value

[safe radish]

@feral ravine Has your question been resolved?

[split kayak]

Depends on what you are working with, but generally, youll be ok just by:

1. Knowing your general arithmetic rules
2. Remembering that i^2 = -1

complex valued functions and the complex numbers themselves have some niche restrictions compared to the reals sometimes