help-19

[clever fjord]

..,,

[glad wadi]

okay

I think I got it now

thanks

[olive sparrow]

you can also put the 4 as 2^2

[glad wadi]

ahh

okay

I got it now

[olive sparrow]

$((2x+1)^2+1)$

[clever fjord]

clonesolopros

[glad wadi]

how does it get to that

i understand the

4 * 1/4 turning into 1

[olive sparrow]

$2^2(x+\frac{1}{2})^2 = (2(x+\frac{1}{2}))^2$

[clever fjord]

clonesolopros

[glad wadi]

goated

thanks

[olive sparrow]

fr

well idk if it helps with the integral

[glad wadi]

it does doesnt it?

because I can put it into the arctan

and that will be my term

for u

[olive sparrow]

well you can do u = 2x+1?

and its about the same thing

[glad wadi]

did I do anything illegal

[dawn tiger]

everything is right up until you evaluated the integral at the last step

from this formula, your u is (2x+1) and your a is 1

[glad wadi]

oh so its just flipped?

[dawn tiger]

the problem is the 1/(2x+1) out front

[glad wadi]

how do I know which is the u and which is a?

[dawn tiger]

when you plug those into the formula, you should get 1/(1) * arctan((2x+1) / (1)) + C

the u is the thing that you're integrating with respect to

if you let u = 2x + 1

then du = 2dx, so this integral becomes 1/2 * 1/(u^2 + 1) du

then you know that the antiderivative of 1/(u^2 + 1) = arctan(u)

so your answer would be 1/2 arctan(2x + 1) + C

[glad wadi]

ahh

okay

where does the 1/2 come from though?

nevermind

I got it

lets go

.close

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[west pecan]

I just want to verify that I'm understanding this correctly

[west pecan]

is this a proper proof that the set is closed under addition?

I feel like I don't understand my own justification even if it's right lol

to verify that it's closed under addition, that'd be the same as verifying that f+g (where g is a member of the set) is also a member of the set for all g right?

[odd edge]

@west pecan Has your question been resolved?

[west pecan]

wait I think I have to add this to the end:

is this the right thought process? did I do it right?

[granite slate]

In my humble opinion, you need to show that for every f, g satisfying f'(-1) = 3f'(2) and g'(-1) = 3g(2) and for all real number r, you have f + g and r*f in your set.
(Furthermore, I feel like you show nothing actually, you are simply writing what you want to show without explaining why this is true :/)

[west pecan]

okay wait let me try again

that makes sense

is this correct?

[granite slate]

Yes, this is a lot better !

[west pecan]

I still feel don't feel confident that there isn't a counterexample for g

[granite slate]

Such as ? f and g play the same role here

(Actually, you also have to check that your set is not empty, by proving that the constant null function is in your set)

[west pecan]

well it's assumed that it's on the interval (-4,4) so it can't be empty I think

oh yeah I'm just concerned about closure under addition rn

[granite slate]

The set of integers is not empty, but if I define S the subset of the integers which are both odd and even, it will be empty

What do I want to say with this stupid example is that you still need to check that your set is not such an example too

[west pecan]

oh wait you're saying I have to show that the set of differentiable real-valued functions on the interval (-4,4) such that that property holds isn't empty?

[granite slate]

You have to keep in mind that both f and g are in your set. Does it help ?

[granite slate]

Yes

[west pecan]

oh

well f(x)=0 satisfies that property as one of the conditions for a subspace anyway

[granite slate]

Absolutely

[west pecan]

and then for scalar multiplication I wrote

[granite slate]

But in order to have a complete proof, you need to mention it

[west pecan]

yeah my question here was just about verifying the closure under addition

I feel confident with the other conditions

[granite slate]

Okey, sorry

[west pecan]

tyyy though

ur good

[granite slate]

No problem

[west pecan]

.close

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[limber willow]

for n(c_n), should the lower bound not be 0? if not im not really sure how to do this

[odd edge]

@limber willow Has your question been resolved?

[limber willow]

Actually I just kept as is and and voided using c_0 does this work

[olive sparrow]

I have one question

isnt $\frac{d}{dx}(x^n) = nx^{n-1}$

[clever fjord]

clonesolopros

[limber willow]

yes but i multiplied by x

hence nx^n instead

[olive sparrow]

but why

[limber willow]

to express (nc_n) in terms of f(x)

[olive sparrow]

btw do you know what is f(x)?

[limber willow]

yea the basic egf of c_n

its in the question

[olive sparrow]

well its the function e^x

[limber willow]

only if c_n = 1

i dont really see where this is going

[olive sparrow]

wait isnt c_n supposed to be the result?

[limber willow]

wdym

i think it wants me to use c_n and c_0 to express (nc_n)

just cant find a use for c_0

[olive sparrow]

like this?:$c_n = \frac{x^n}{n!}$

[clever fjord]

clonesolopros

[limber willow]

yea

we need $\frac{n * c_n * x^n}{n!}$

wait no

[clever fjord]

NUT

[limber willow]

given

$\frac{c_n * x^n}{n!}$

[clever fjord]

NUT

[olive sparrow]

but doesnt it tell you to find the sequence: $c_n$?

[clever fjord]

clonesolopros

[limber willow]

no just the egf for nc_n

for n >= 1

[olive sparrow]

well its the same thing

[limber willow]

how

multiplying by n shifts the factorials in the dominator down by 1

[olive sparrow]

and is there a problem?

[limber willow]

yea its not the same

[olive sparrow]

its like saying: $\frac{n}{n^2}$ is not equal to $\frac{1}{n}$

[clever fjord]

clonesolopros

[limber willow]

not the same at all lmao

thanks for tryin pal

[olive sparrow]

step1: calculate c_0

$c_0 = \frac{x^0}{0!}$

[clever fjord]

clonesolopros

[olive sparrow]

$f(n) = \frac{x^n}{n!} =c_n$

[clever fjord]

clonesolopros

[olive sparrow]

$n \cdot f(n) = \frac{x^n}{(n-1)!} = n\cdot c_n$

[clever fjord]

clonesolopros

[odd edge]

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[mystic saffron]

Im learning about cylindrical coordinates and I’m wondering if there id a reason that there is a z axis but no x or y axis or why whoever made the coordinate system chose to include z instead of x or y

[mystic saffron]

Oh wait I’m dumb

It’s because the x and y axis are a part of the polar coordinate system?

[green elm]

yes, the x and y get converted to polar representation, and z is left alone

the choice of leaving z alone is kind of arbitrary

[mystic saffron]

Ok

.close

[green elm]

you can equally well do cylindrical coordinates with y and z converted to polar, or x and z convered to polar

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[mystic saffron]

Thanks

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[chrome zodiac]

Hi

[chrome zodiac]

Where do I start?

[dawn tiger]

it might be useful to notice the change of base formula

log_x(y) is the same as ln(y) / ln(x)

you can use that to rewrite log_p(a) as ln(a) / ln(p) and log_p(b) as ln(b) / ln(p)

now try using that same change of base formula for log_ab(p) and see if you get something that looks similar to your known information, perhaps by using some other log properties

[odd edge]

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[mystic saffron]

i acquired the aos but im still stuck on the rest

[shut wedge]

The axis of symmetry passes through the vertex yes?

[mystic saffron]

yeah so i'd have my first coordinate but i don't know where i'd go from there

[shut wedge]

If you have your x coordinate, how would you find your y coordinate

[shy bramble]

can you help me with geometry?

[mystic saffron]

isnt there like a formula (i lost all my notes)

[shut wedge]

!occupied

[odd edge]

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

[shy bramble]

oh sorry

[shut wedge]

Look at your original equation

Its made up of x and y isnt it?

[mystic saffron]

yeah

[shut wedge]

So cant you use that to find your y coordinate?

[mystic saffron]

wait do i just plug it in

[shut wedge]

Yep

:\

[mystic saffron]

oh my god its right in front of me

im sorry man im tired as hell

[shut wedge]

Lmao allg

[mystic saffron]

i cant even think critically

okay let me just

figure that out

assignment aint takin it for some reason, i dont know where i messed up

[shut wedge]

Bruh

You need rest

-9(-1)^2 aint 9^2

[mystic saffron]

because of the parentheses right

[shut wedge]

Yea

Its not in the parenthesis with the square

[mystic saffron]

my bad man this is thekast thing i need to knock out then im hitting the hay

i got 8

[shut wedge]

Seems right

[mystic saffron]

yeah assignment took it

oh BROTHER one last thing man

thatll be it

let me check if my tescher had a video for it (he didnt for this one)

okay yeah he didnt

this is like the last foreign concept and thatll be over

yeah never mind im good

off to bed now! thanks man

.close

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[dry plank]

Hello I asked for help with this question over the weekend however I’m sort of confused.

[dry plank]

This is what I was told.

Since I can’t use sine can I instead use cos?

Or is that also wrong?

Wait I figured it out, we have to use the cos function

[chrome zodiac]

Whether u use sine cosine or tangent is dependent on what values u were given and what u r trying to find

[dry plank]

Using sine is incorrect in this case because that will help me find the value of RS and not ST.

Tan is out of the picture.

But cos helps me determine what the value of ST is.

[dry plank]

Yeah.

.close

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[main musk]

Translation: Solve the following inequalities I am confused about question C, how do I factorize it properly like A and B

[frigid canopy]

yes?

have you tried graphing it?

it doesn't have any real roots

notice that

[noble forge]

you need to learn to cook meth

math*

[main musk]

huh😭

It difficult to me despite me enjoying doing random calculation in games ;')

[frigid canopy]

try completing the square

[noble forge]

or maybe the quadratic formula

[main musk]

I did it where I do ac and the sum of b

but it gave me -3 and 2

well specifically -6 and -1

[frigid canopy]

there aren't any real roots though

right?

ooh

my bad

yeah

quadratic formula should work here

[main musk]

when is the quadratic formula applied btw?

when it ax^2 + bx + c?

[noble forge]

that possess no conditions

the quadratic you mentioned has no restrictions and the quadratic formula has no restrictions

[main musk]

oh so aslong as there is a,b and c I could use the quadratic formula to find x?

[noble forge]

(talking about complex number system)

[noble forge]

sure you could

[main musk]

seems like something very important to rmber forever lol

so far I have -x + √-2x -24 / 12, if I were to do -24 - 2x it would be -26 but √26 is like 5.1, idk if normal or I am messing up somewhere

[ember oak]

When $a\ne 0$

[clever fjord]

SWR

[main musk]

oh alright thank you

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[ocean hamlet]

can somebody please explain? i got them all wrong lol

[odd edge]

Please don't occupy multiple help channels.

[ocean hamlet]

Consider the relation R on the set of natural numbers greater than 10 defined by xRy if x^2 > y. Which properties does it satisfy?

[frigid canopy]

mind translating the options?

[ocean hamlet]

symmetric, transitive, irreflexive, antisymmetric, reflexive

[waxen frigate]

Any germans in here it is about stochastik and analytische Geometrie

[ocean hamlet]

bro open another channel please

[frigid canopy]

!occupied

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[ocean hamlet]

it's not reflexive because x^2 is not greater than x^2

it's not symmetric because if x^2 > y then y is not greater than x^2

hence it is antisymmetric and antireflexive

and it is transitive because if x^2 > y and y > z then x^2 > z

whats wrong?

[odd edge]

@ocean hamlet Has your question been resolved?

[odd edge]

@ocean hamlet Has your question been resolved?

[odd edge]

@ocean hamlet Has your question been resolved?

[fickle coral]

But x²>x for given conditions on x and that relation is xRx

[marble bridge]

dont understand italian

[ocean hamlet]

i translated

[marble bridge]

ping me there

[ocean hamlet]

.

[ocean hamlet]

.

@marble bridge

[marble bridge]

thanks

now that i think it may be irreflexive:

the relation is irreflexive because it doesnt not have the property of reflexivity

@ocean hamlet

[ocean hamlet]

its not irreflexive

its marked wrong

[ocean hamlet]

.

[marble bridge]

wait what?

[marble bridge]

Any number squared is always greater than or equal to itself

so no element can be related to itself under this relation,and that makes it irreflexive

@ocean hamlet

[ocean hamlet]

. close

.close

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[merry finch]

What have u tried

[tropic gulch]

i dont know what goes inside the integral

i have the bounds though

outer integral is from 0 to 2π

inner integral from 0 to √theta

[merry finch]

If you want the area

Then you just want the integrand to be 1

[tropic gulch]

oh shit i forgot

[merry finch]

Remember that a 2d integral is doing the volume under the surface

And the cross section * 1 (height) would be the volume

Which is also the cross section aka the area

[tropic gulch]

so its not this

[merry finch]

Well idk I can’t do a double integral ins my head

Can you show your working?

[tropic gulch]

wait i misclicked

yeah sure

here

apparently its wrong

[tropic gulch]

@merry finch

[merry finch]

Huh

[tropic gulch]

heres my wokr

<@&286206848099549185>

pleas

@merry finch

@dawn tiger could you help me with this

got it nvvmmm

.close

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[mystic saffron]

HALP

[mystic saffron]

ok this question wants to trick me

[dawn tiger]

what do you know about the angles in a triangle?

[mystic saffron]

they add up to 180

the thing is

x+59 and x+51 dont have te degree

so should I ignore the angle

[dawn tiger]

?

[mystic saffron]

since it's only asking for the measurement

it doesn't have a degree simble

[dawn tiger]

that isn't really important it's assumed that they're degrees

[mystic saffron]

and even then it would give me a - number

[dawn tiger]

since the 84 is in degrees

[clever fjord]

Pro_Hecker

[dawn tiger]

should be x + 51 not x + 41

but yeah

why not just solve the equation?

[mystic saffron]

ye ik that

it gives me a negative tho

is that possible?

[dawn tiger]

(x+59) + (x+51) + 84 = 180

[undone basin]

yes

[dawn tiger]

you get a negative for x

the actual angle is positive

[mystic saffron]

oh ok

[dawn tiger]

x isn't your angle

[mystic saffron]

then ye that was my question

also already solved

but how do u know which letter is which

i'm still having that thought

like is d the right angle

and be is both y-5 and 13?

because if u used y-5=13

y=18

then u plug it in

18-5=13

so since it shows on both sides

and it is congruent

fr?

thenwhat r the measurements for

so AD is 5x-5?

dc is x+12 or 5x-5?

ah

5x-5=x+12

so would u solve 5x-5=90?

since its a right triangle

and u r trying to solve for x

[odd edge]

@mystic saffron Has your question been resolved?

[olive sparrow]

isnt (5x-5) = 90?

[odd edge]

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[mystic saffron]

yes

[olive sparrow]

$5x-5=90 \implies x=19$

[mystic saffron]

yes

wait

huh

95/5

[olive sparrow]

then we have $(x+12) \implies (19+12) = 31$

[clever fjord]

clonesolopros

clonesolopros

[olive sparrow]

31º is the angle its asking

@mystic saffron

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[grand hound]

just having personal wonders, does this formula work if I choose a number z=a+bi and then I want to take it to the (c+di) complex power like
z^(c+di)

[grand hound]

it makes a lot of sense to me

but I may be wrong in some point

you basically start at z = r * e^iθ

z = r * e^iθ

r = sqrt(a^2+b^2)

so that's a real number and therefore you can turn it into e^ln(r)*e^2kπi

with any integeer k I guess

so that's how you get
(e^ln(r) * e^2kπi)^(c+di) * (e^iθ)^(c+di)

so then you do lots of distributive

pretty trivial

[spice geyser]

$e^{2k\pi i}$ is just 1

[clever fjord]

qwertytrewq

[grand hound]

yes I know

[spice geyser]

ah but for c+di

[grand hound]

ya

so I can turn |z| into e^ln(|z|)

[spice geyser]

usually ud just put it with theta

[grand hound]

yes but theta is the one for the first complex

theta is the argument for z

and I turn |z| into polar coords to work with it as a power of e

[spice geyser]

which depends on branch cut, which would already determine your power without the need for a 2ki pi factor

[grand hound]

yeah but that part is for r, r being |z|

[spice geyser]

im saying you don't need it for r. the brach cut fixes unique value of complex power

better put it should be (e^ln(r) )^(c+di) * (e^iθ* e^2kπi)^(c+di)

where e^i(theta+2kpi ) is determined by branch

[grand hound]

so I just take e^2πki as 1

and then times e^ln(r) just the same thing

oh makes sense

[spice geyser]

as long as you specify your theta from specifying a branch

it should be good.

the branch is where you pick you 2kpi factor

everything else is defined around it

[grand hound]

theta is for z tho

right?

[spice geyser]

complex logarithm given branch gives you ln(z)=ln(|z|)+arg(z)

z^c= e^{cln(z)}

which is identical to your definition

[grand hound]

yes

what's a branch tho

me no english

[spice geyser]

a brach cut helps you determine what theta is

say if my number is i and our brach is (0,2pi]

[grand hound]

the branch is theta?

oh

ok I get it

no no I get in now

[spice geyser]

Ah the branch is an interval of length 2pi

[grand hound]

is like

ok

yes

go on

[spice geyser]

so log(i) would be pi/2

[grand hound]

ya

[spice geyser]

if my branch were (pi, 3pi]

$\log(i)$ would be $\frac{5\pi}{2}$

[clever fjord]

qwertytrewq

[grand hound]

also that

yes

[spice geyser]

so essentially it is making the argument function be in that range

so, you could imagine that your e^{2kpi} is chosen already by choosing the branch

[grand hound]

ok but the theta for r would always be 0

cuz it's a real number

so I just take e^ln(r)(c+di)

which is r^c*e^ln(r)di

[spice geyser]

how do i put it, if you have r, given the same branch, then r^{c+di} might have a 2ki multiple

but, the angle 2kpi is added to the imaginary part of ln(specific number)

[grand hound]

but r is the modulus of z

[spice geyser]

since you separated into reals and imaginary, you dont need to do it twice for real and imaginary

however, if you view r not as the modulus of z, rather as a complex number, there might be a 2kpi multiple

[spice geyser]

which is not what you did ofc, but useful to distinguish those

[grand hound]

I'm confused

if I think about r as a complex number

what now?

[spice geyser]

then r would have a modulus of r, and a extra angle (when taking log) of possibly 2kpi

sorta like that

[grand hound]

yes

[spice geyser]

for z, the modulus is r (nothing needs to be done for the modulus, no angle needs to be added), and the argument is theta

[grand hound]

yes

for z

[spice geyser]

👍 so is this clearer?

[grand hound]

what's the conclusion then?

[spice geyser]

if you view r as the modulus of z, you dont need to add angles of 2kpi.
if you view r as a complex number, then r^(c+di) might have an argument of 2kpi.

[grand hound]

wdym by might have an argument

[spice geyser]

simply put, a modulus does not have an argument, while a complex number does

by argument i mean arg(z)

the angle

[grand hound]

like the theta

but of r

ok but like

it's ok if I take r^(c+di) = r^c *e^ln(r)di

[spice geyser]

it should be e^ln(r)(c+di)

[grand hound]

isn't it the same?

[spice geyser]

pretty sure the 2kpi will affect c

for example (-1)^(1/2) could be i or -i

[grand hound]

it does cuz I just had to add this

[spice geyser]

(1)^(1/4) could be i, -i, 1, -1

but in your formula 1^(1/4+0i)=1^(1/4)* e^0=1

[grand hound]

well idk about that one

wasn't the square root of a number always positive by convention?

[spice geyser]

"positivity" becomes meaningless in complex numbers

[grand hound]

oh ok

then

[spice geyser]

is -1+i more positive than 1-i?

so, thats why we do branch cut

which, if you do different kinds of branch cut, you will have all the possible solutions

it would be a weird generalization if we single out real number and force them to satisfy some positivity clause

[grand hound]

ok I get it

[spice geyser]

so that said this would be incorrect

[grand hound]

isn't 1^(1/4) all of those you listed?

[spice geyser]

yeah im just showing that your formula would only produce one outcome, that is 1

but not the other outcomes

[grand hound]

wdym one outcome

[spice geyser]

i mean if you plug in r=1 here, and c+di=1/4+0i

you would get 1 on the right side no matter what branch you pick

[grand hound]

oh cuz

e^ln(1) is 1

or

cuz d = 0

one of those I'm guessing

not sure which one

[spice geyser]

d=0

so no matter what ln(r)di=0

so you get e^0=1

[grand hound]

but you also have r^c

[spice geyser]

r^c is just 1^(1/4)

i thought u r using the real definition here

[grand hound]

and isn't that like the 4 possible numbers

[grand hound]

didn't you say that doesn't work cuz complex

T.T

[spice geyser]

if that is the case your formula didnt achieve anything, you just listed (1)^(1/4)=(1)^(1/4)

[grand hound]

I should be studying algebra and not this

[grand hound]

yes but what if you choose a diff complex

that's why I want the formula for

[spice geyser]

no, i thought u meant that r^c is powers in reals

[grand hound]

oh no

I mean like

[spice geyser]

r^c as powers in complex would yield different result

[grand hound]

that's half of the formula I'm using

[mystic saffron]

oki thx

[grand hound]

yw

[mystic saffron]

??

[grand hound]

just joking

[mystic saffron]

hehehehe

[spice geyser]

then in you original formula is |z|^c a real power or a complex power?

i was pretty sure it is real power

[mystic saffron]

z^5

[grand hound]

depends on c

ig

[spice geyser]

c is a real number though from your assumption

[grand hound]

yes but

still could give complex

if you consider

em

[spice geyser]

you are using e^ln|z|c =|z|^c

[grand hound]

what you said

[grand hound]

maybe? idk

[spice geyser]

note that the left hand side is always real

e^(real number) is real

[spice geyser]

so if you are not using real power the formula e^ln|z|c =|z|^c is wrong

because the left hand side is always real

but the right hand side could be complex

[grand hound]

wait

[spice geyser]

Ah also forgot to mention that, e^z is defined without branch cut

[grand hound]

but that's not the formula, you have to multiply by e^ln|z|di

[spice geyser]

on the left hand side of the image you posted there is a |z|^c

[grand hound]

yes

[spice geyser]

it was derived from e^ln|z|c =|z|^c

[grand hound]

I have to change that to e^ln(|z|c)?

[spice geyser]

you could just say its a real power

because |z|^c as power of real number does equal to e^ln(|z|)c

[grand hound]

but like you see how I got to |z|^c right?

[spice geyser]

yeah

it works if you assume that you are doing exponentiation in reals

which is fine

you can't say that |z|^c is a complex power tho

[grand hound]

I'm trying to get exponentiation in complexs

[spice geyser]

cuz in that case e^ln(|z|)c does not equal to |z|^c

[spice geyser]

however you have reduced it to exponentiation in the reals

BTW

$$z^{a+b}\neq z^az^b$$

[clever fjord]

qwertytrewq

[spice geyser]

for complex numbers

[grand hound]

for z complex

?

amazing

[spice geyser]

for z,a,b complex

[grand hound]

wdym for a, b complex T.T

[spice geyser]

so, if your derivation is of complex powers, it would be wrong

[grand hound]

I think it is

completely wrong

[spice geyser]

im taking a and b as complex numbers

[grand hound]

but like

if they're both complex you can add them

and get just another complex

ok makes sense tho

cuz

yeah yeah

i'm dumb

so like

it was all just my schizo

sad moment

I'll go back to studying

[spice geyser]

there is light tho

[grand hound]

I'll bother another day with why this is true but for today, enough depression

.close

[odd edge]

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[odd edge]

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[late terrace]

Could someone help explain their approach to this question? I don't need an actual final solution since it's so tedious, I'm just curious as to how someone else might apporach to solve it

[late terrace]

My orignal method was factor out 100, bring it out the integral, and integrate (10+t)e^(whatever), but left me with quite a long integration by parts process

[dawn tiger]

this seems like something they'd seriously expect you to do with a calculator

especially since they talk about rounding to the nearest dollar

[olive sparrow]

integrate e^(1-t/20), and t·e^(1-t/20)

[late terrace]

man i hate this question

I just did it my original way, a seperate way where I split the (10+t) into the e^(whatever), so that I have an integral of 10e+te, and I can just evalaute them seperately

and both are different answers from the online calculator

[olive sparrow]

$1000e\int_{0}^{20} \frac{1}{e^{t/20}}dt + 100e\int_{0}^{20} \frac{t}{e^{t/20}}dt$

[clever fjord]

clonesolopros

[olive sparrow]

$100e\int_{0}^{20} \frac{10+t}{e^{t/20}}dt$

[clever fjord]

clonesolopros

[odd edge]

@late terrace Has your question been resolved?

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[eternal glade]

Hey guys

[eternal glade]

Let A be a symmetrical 3x3 Matrix with eigenvalues a1 < a2 < a3. I shall show that F has a saddle point at v2 (eigenvector to a2)

Is there an easy way that gets around calculating every second derivative?

25?

wdym?

nice

[lethal spoke]

<@&268886789983436800>

[mild elk]

Knock it off

If you're gonna contribute to these channels, then do so

If not then get out

Idk maybe

This has nothing to do with the topic of this channel

So leave

[eternal glade]

😵‍💫

[lethal spoke]

<@&268886789983436800>

Spam

[green elm]

doesn't this essentially come down to showing that A has both positive and negative eigenvalues?

[mild elk]

Thanks for making this easy NotVoid

Oh wait he can't hear me

Because he's dead

Ripperoni pepperoni

[green elm]

bye bye whoever you were haha

[raw palm]

imagine pinging moderators when the issue was directly with the moderators

[eternal glade]

I dont think so, the eigenvalues dont have to be negative.

[lethal spoke]

I wanted him banned 🚫

[mild elk]

Yeah Bungo keep in mind the ||x||^2 term in the denominator

[eternal glade]

but i dont know how to prove it without the Hesse matrix

[mild elk]

That can cause some shenanigans

[eternal glade]

and i dont want to calculate the hesse matrix

because even one of the derivatives would take up over on line of my paper

[mild elk]

Probably try to compute directional derivatives at v_2?

[eternal glade]

thats a bon idea actually

[mild elk]

But at the other eigenvectors

[eternal glade]

yeah

that would do it

[green elm]

oh hmm, good point

[mild elk]

The intuition is that if you approach it along the eigenvector of the larger eigenvalue the function is probably going up

And along the smaller it's probably going down

[mortal trench]

hint: maybe use the fact that this matrix is symmetrical somehow.

[mild elk]

(For reference I literally had to Google what a saddle point was so I might be royally talking out of my ass)

[eternal glade]

I think i needed this to show that the critical points are just the eigenvectors

[mild elk]

(But this smells like a thing that might make sense)

[pastel orbit]

Dami helps in the help channels now?!

awesome!

[raw palm]

i found this wierd too

its the first time i saw him anywhere outside #book-recommendations

[mild elk]

Lol I popped in to handle a moderation matter

[raw palm]

[mild elk]

And was like you know what while I'm here this is more interesting than guiding middle schoolers through systems of two linear equations

[mild elk]

Wait usually I live in #discussion lol

[eternal glade]

Mhhm but why does a positive and negative directional derivate imply a saddle point?

[mild elk]

Apparently a saddle point means not a max or a min

[raw palm]

i dont even go there thats why the only place i ever saw you was in #book-recommendations

[mild elk]

So I'm guessing that if you pick a direction and say f is going up

And if you pick another direction and say f is going down

That feels pretty saddle-y

[pastel orbit]

oh you never use discussion

[green elm]

doesn't the derivative also have to be zero in order to qualify as a saddle point?
think f(x) = x^3 at x=0

[pastel orbit]

that's why

[mild elk]

🤷 again I might be making no sense at all since idk multivariable calculus

[mild elk]

Yeah I mean among critical points

[eternal glade]

wait im stupid, i completely failed at imagining it, obv. your right

[mild elk]

But I'm guessing proving that it's critical shouldn't be too hard

Overall the point is, you don't want to compute partial derivatives in xyz

But since A is symmetric

[eternal glade]

i already proved that only eigenvectors are critical points

[mild elk]

Use an orthonornal eigenbasis

[green elm]

ah got it

[raw palm]

yes thats the reason but you dont need to go to discussion to find this incident wierd

[mild elk]

That feels like the better way to go about life. Even if it boils down to computing the Hessian, compute it "in the eigenbasis"

(Probably)

(Maybe)

(Idk)

[eternal glade]

nah, your right, the directionals already do the job

[mild elk]

Oh nice

Talking shit really does work more than it has any right to

[eternal glade]

hahaha

[valid tapir]

Hello

I don't suppose anybody wants to help me with my maths revision please

I have an exam tomorrow

[odd edge]

@eternal glade Has your question been resolved?

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[surreal mountain]

im not sure where to start with this surface integral

[surreal mountain]

https://i.imgur.com/c06sSvn.png

im used to having a given integrand, f(x,y,z) and surface(s) for the bounds

[green elm]

f is constant on the sphere, yes?

[surreal mountain]

idk?

i dont get what f(x,y,z)=g(x,y,z) is doing

seems like it does nothing

[green elm]

where do you see f(x,y,z) = g(x,y,z)

[surreal mountain]

i mean f(x,y,z) = g(sqrt(x^2+y^2+z^2))

[green elm]

well what is the value of x^2 + y^2 + z^2 on the sphere?

[surreal mountain]

4

[green elm]

ok

so on the sphere, you have f(x,y,z) = g(sqrt(4))

which is what?

[surreal mountain]

2

[green elm]

certainly not

[surreal mountain]

sqrt4 = 2?

[green elm]

yes

so then you have g(2)

which is what?

[surreal mountain]

-5

[green elm]

right

so what is f(x,y,z) on the sphere?

[surreal mountain]

-5

[green elm]

yes

which is just a constant

so what does the integral simplify to?

[surreal mountain]

ah

so all that information just to tell you the integrand is -5

[green elm]

yep

[surreal mountain]

do i still need the sqrt of partials + 1

[green elm]

kind of convoluted but in the end it's simple

[green elm]

what do you mean

[surreal mountain]

https://i.imgur.com/4G7Pafu.png

[green elm]

no, that's if you're parameterizing the sphere i believe

here it's simpler

you have $$\int_S (-5),dS$$

[clever fjord]

Bungo

[green elm]

you can pull out the -5 because it's constant:

$$-5\int_S dS$$

[clever fjord]

Bungo

[green elm]

what is $\int_S dS$?

[clever fjord]

Bungo

[surreal mountain]

S

but what about bounds

[green elm]

no, that's not what the notation means

it means "integrate the surface area of the sphere, over the entire sphere"

the result is just the total surface area of the sphere

[surreal mountain]

ik but how do you get the surface area if there is no bounds

[green elm]

there's a simple formula for the surface area of a sphere

[surreal mountain]

4pir^2?

[green elm]

yes

and what is r in your case?

[surreal mountain]

4

[green elm]

no..

[surreal mountain]

2

[green elm]

that's r^2

yes r = 2

[surreal mountain]

so 16pi is the answer?

-5 didnt contribute?

[green elm]

for $\int_S dS$ the answer is $16\pi$

[clever fjord]

Bungo

[green elm]

you still have to multiply that by -5

[surreal mountain]

i see

thank you

[green elm]

sure

cheers

[odd edge]

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[tiny inlet]

is this right?

also having trouble with this one,

[latent scaffold]

This doesn't really look like it's in a "completely factored form", but it depends on what they're looking for.

For the complex one, what is i^2?

[tiny inlet]

uhh im not sure i cant remember if i^2 is -1?

or is considered -1

[dawn tiger]

i^2 = -1

[latent scaffold]

Indeed, so what is 1 + i^2?

[tiny inlet]

0?

[latent scaffold]

Yes

[tiny inlet]

i^7(0)

i dont know what i^7 is

[dawn tiger]

doesn't matter what it is

what's it being multiplied by?

[tiny inlet]

unsure

[dawn tiger]

you got that 1 + i^2 = 0
so you got i^7 * 0

[dawn tiger]

you said that here

so if i^7 is being multiplied by 0, can you know what that answer is without needing to know what i^7 is?

[tiny inlet]

0?

so the answer is 0?

i guess the a + bi threw me off

[odd edge]

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[sturdy plume]

What does it mean by "The set is the line defined by the vector"

[dawn tiger]

every point in that solution set is of the form r * (2, -1, 1)

(2, -1, 1) is a particular vector in R^3, you can imagine scaling this vector by r which would lengthen it, shrink it, flip it backwards, etc. and see that every vector in that space lies on a line that contains that vector

[desert marlin]

Very poorly worded for someone who should understand linear algebra enough to teach a. Course on it

[mental ridge]

that took me like 3 readings to understand

[desert marlin]

Well and it’s even worse because infinitely many vectors work as a solution

[sturdy plume]

I'm sorry I didn't understand that, this has been kind of difficult for me

[dawn tiger]

maybe it's autograded to take any vector on that span as an answer

[desert marlin]

The solution space is spanned by:

Provide a vector

It should say soemthing like that

[dawn tiger]

does my explanation up here make sense?

[sturdy plume]

Sorry no

That's my fault

[dawn tiger]

every solution in the set is a scaled version of that vector (2, -1, 1)

you're scaling by some parameter r

when you scale a single vector by a parameter, it generates a line

for example, imagine you were in R^2 and the solution set was (r, r)

that's the same as r * (1,1)

every vector in the solution set is a scalar multiple of the vector (1,1)

if you can imagine the space containing all solutions of the form (r,r) and considering every possible value of r

this forms a diagonal line, you could think of this as y = x

so scaling a vector produces a line

in linear algebra, we say that that line is "spanned" by that vector

[sturdy plume]

what im understanding so far is that the vector 2,-1,1 which were the parameters I found are like the borders for everything within the vector?

[dawn tiger]

maybe not the borders, but every single vector in your solution set is some multiple of the vector (2,-1,1)

[sturdy plume]

gotcha gotcha

[dawn tiger]

so that solution set is spanned by the vector (2, -1, 1)

[sturdy plume]

so I would just write the vector I put down in that box because the vector defines everything

[dawn tiger]

and what your teacher very poorly worded for that last part is exactly asking for a vector that spans that space

[sturdy plume]

Its an online course and im fighting out here to understand anything

thank you very much

.close

[odd edge]

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[rose belfry]

im having difficulty with logic, on the third row of the truth table, how is it that the overall statement is true when M is false?

[rose belfry]

would M not have to be true in order for the entire statement to be true

[drowsy root]

$P \implies Q$ is always true whenever $P$ is false

[clever fjord]

Flappie

[drowsy root]

the only time when $P \implies Q$ is false, is whenever P is true but Q is false

[clever fjord]

Flappie

[rose belfry]

so does the truth value of P not matter if Q is true?

[drowsy root]

indeed

ignore my deleted messages :)

[rose belfry]

ok, so in this specific question, M is P and N is Q?

[drowsy root]

no, they swapped it

thats why i was confused

[rose belfry]

is P always the antecedent?

[drowsy root]

\begin{table}
\begin{tabular}{|c|c|c|}\hline
P & Q & $P\implies Q$\\hline
T & T & T\\hline
T & F & F\\hline
F & T & T\\hline
F & F & T\\hline
\end{tabular}
\end{table}

[clever fjord]

Flappie

[rose belfry]

does the biconditional q supercede the p -> q conditional in any way?

[drowsy root]

P is the antecedent and Q is the consequent

[drowsy root]

wdym?

[rose belfry]

3rd row, Q is F and P is T yet the overall statement is T

(talking about the top table which is correct)

[drowsy root]

in the Q is F and P is T, and the statement is F

[rose belfry]

$P \implies Q$ is false but the overall statement is true even though it is biconditional

[clever fjord]

stwizzy

[drowsy root]

this line right?

[rose belfry]

yes i meant the compound statement on the right

what makes that T

because the consequent is F and the antecedent is T

[drowsy root]

you have M which is F, and n->m which is F, they are both false, so iff is true

[serene terrace]

A <-> B is true if (both A and B is true) or (both A and B is false)

[drowsy root]

\begin{table}
\begin{tabular}{|c|c|c|}\hline
P & Q & $P\iff Q$\\hline
T & T & T\\hline
T & F & F\\hline
F & T & F\\hline
F & F & T\\hline
\end{tabular}
\end{table}

[clever fjord]

Flappie

[drowsy root]

\begin{table}
\begin{tabular}{|c|c|c|}\hline
P & Q & $P\wedge Q$\\hline
T & T & T\\hline
T & F & F\\hline
F & T & F\\hline
F & F & F\\hline
\end{tabular}
\end{table}

[clever fjord]

Flappie

[drowsy root]

hope this helps

[rose belfry]

in this case n->m is A and (n->m)<->m is B?

[serene terrace]

m is B

[rose belfry]

yes sorry

ok i understand now, i was blurring the difference between the two statements

.close

[odd edge]

Channel closed

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[keen mirage]

can anyone help me with understanding transformations or these problems

[gilded shard]

which equation do you want help on

[keen mirage]

can we start with 8

[gilded shard]

so the "a" value in this is 3, that represents how much the porabola is being stretched by

[keen mirage]

vertically right?

[gilded shard]

yup

if the a value is greater then 1 then it is being verticaly stretched

if less then 1 but greater then 0 such as 0.1-0.9 then it is being compressed

[keen mirage]

okay but how do i gragh that do i just put a dot at 3

[gilded shard]

no so basically this is vertex form so you are given the vertex which ius h,k

you h value here is 2 coming from the (x-2) and the k value is 4

[keen mirage]

(-2,4) is the vertex?

[gilded shard]

when in the bracket you flip the sign so if it was (x+2) then your vertex would be -2,4

a "-" symbol in the bracket means its a positive number on the graph and a "+" symbol means its a negative number on the graph its a bit weird

so your vertex here is 2,4

[keen mirage]

oh

so thats my highest point

[gilded shard]

"h" which is 2 is your x value

[gilded shard]

thats your origin depending on which way the porabola opens which is determined by the a value being negative or positive (negative means opens down and has a max value and positive means opens up and has a min value

[keen mirage]

it opens down rihgt?

because its 3 which is positive

doesnt the a value determine whether it opens up or down

[gilded shard]

yes but positive means opens up

[keen mirage]

oh

[gilded shard]

4 which is your k value will be the y in the vertex

x=2, and y=4

thats our vertex

[keen mirage]

okay i understand that

[gilded shard]

the 3 which is a value

is telling us how much our porabola is being stretched by

[keen mirage]

yea i dont understand what does stretch mean

[gilded shard]

do you know the y values on the normal porabola

41014

a table for that would look like this

x|y
-2|4
-1|1
0|0
1|1
2|4

sorry its a bit messy

[keen mirage]

its okay

yea i remember learning that

dont i plug in the vertex

[gilded shard]

basically to transform your current equation into something like that you need to do certain things to the x/y values

for y

the equation is (a)y+k

so (3)y+4

and for x its x (+/- depending on what the h value is) h

so in our case thats x+2

so you plug the y value into that equation

for the first 4 that would look like (3)4+4

then (3)1+4

and so on

any questoins yet

[keen mirage]

ok.. kinda understanding it

[gilded shard]

ok so for our x values what do you think the equation would look like

[keen mirage]

our x value is 2 right?

[gilded shard]

yes

[keen mirage]

wait thats where im lost

how do i know what the equation is

[gilded shard]

so to find your x value equatoin you have to know what the h value is

we klnow thats 2

[keen mirage]

okayy..

[gilded shard]

then because we know 2 is positive not negative it would be plsu 2

and x is easier in that sence as you only have to know that

then you plug in the x value on the chart into x+2

so for the first one thats -2+2

and so on

[keen mirage]

oh so -1 +2

[gilded shard]

yup

[keen mirage]

0+2 and more

[gilded shard]

correct

[keen mirage]

okay and using those points i then plug it into the gragh?

[gilded shard]

Yes

and thats the transformation of that graph

[keen mirage]

but thats only one dont i need 2

like x and y charts

[gilded shard]

?

we only need one x and y chart for the equation

your teacher probably puts the origional one and the new one next to eahcother for you to see

x|y
-2|4
-1|1
0|0
1|1
2|4

this is the origional and then with what we went over you can make the new one

do you know if your teacher wants you to write out the transformations in word form?

such as translated 2 units right

[keen mirage]

probably

i think he would

[gilded shard]

ok

so for that we look at "a" "h" and "k"

you know how to find those values in an equation

[keen mirage]

yes

[gilded shard]

ok for 9 (a)

[keen mirage]

a determines the vertical stretch

[gilded shard]

correct

heres the equations

list all the transformations you know and well see if you miss any ok

or do you want me to go over them with you

[keen mirage]

ill try and tell u if im confused

[gilded shard]

sounds good

[keen mirage]

the relation has a vertical stretch by 1/2?

[gilded shard]

yesa

but say by a factor of 1/2

for full marks

[keen mirage]

ohh okok

[gilded shard]

or BAFO if your teacher has said thats allowed just shorter

[keen mirage]

and its vertex is -4,7

[keen mirage]

i dont think he does so i wont for now

[gilded shard]

yes but for the translation do you know what to write

[gilded shard]

ok so just by a factor of

[keen mirage]

um not rlly..

[gilded shard]

no worries

you got the vertex right

[keen mirage]

ik how to do the charts things