help-39

[raven spade]

ah okay noted

i'll try it out thanks

.close

[pearl pond]

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[sonic cradle]

Let ( f ), ( g ), and ( h ) be three arithmetic functions, and ( f^{\sim} ), ( g^{\sim} ) be the inverses of ( f ) and ( g ) respectively. Which of the following statements are correct? Check each correct proposition.

\begin{itemize}
\item If ( f ) is totally multiplicative, then ( f^{\sim} ) is also totally multiplicative.
\item If ( f ) takes values in ( \mathbb{Z} ), then ( f^{\sim} ) also takes values in ( \mathbb{Z} ).
\item If ( f ) takes values in ( \mathbb{Q} ), then ( f^{\sim} ) also takes values in ( \mathbb{Q} ).
\item The inverse of ( fg ) is ( f^{\sim}g^{\sim} ).
\item If ( f ) and ( g ) are multiplicative, then ( f \cdot g ) is multiplicative (where ( f \cdot g ) is the function associating to each ( n \in \mathbb{N}_0 ) the number ( f(n) \cdot g(n) )).
\item The inverse of ( f \cdot g ) is ( f^{\sim} \cdot g^{\sim} ).
\item If ( f ) is totally multiplicative, then ( f \cdot (gh) = (f \cdot g)(f \cdot h) ).
\end{itemize}

[jolly parrot]

habibjr.

[pearl pond]

@sonic cradle Has your question been resolved?

[sonic cradle]

no.

<@&286206848099549185>

[sonic cradle]

precision: ( f^{\sim} ) and ( g^{\sim} )
are a notation for the inverse for dirichlet convolution

[pearl pond]

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[drowsy edge]

hi

[drowsy edge]

im having a hard time w this

[pearl pond]

@drowsy edge Has your question been resolved?

[pearl pond]

@drowsy edge Has your question been resolved?

[pearl pond]

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[limber sierra]

The communicating vats

[limber sierra]

I have a question

the Px in the left side is the same as Px in the right side, right?

But how can they be the same if the height for the Px on the left is twice that of the second Px?

And if they are equal to eachother, then would it be correct to consider that Py is equal to P2?

[marble sigil]

they're equal but water pressure changes with height, so P1 and P2 are different

like the deeper you go in the ocean the higher pressure you get

[limber sierra]

I see

and what about Py and P2? are they also equal to eachother?

[marble sigil]

right

[winged blade]

yo

Someone help me with this

[limber sierra]

Alright thank you

[limber sierra]

go to #help-21｜아리스킨충1

the unoccupied channels please

[winged blade]

I got one

[limber sierra]

.close

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[crimson burrow]

Where is 91.25 coming from here?

[tepid ibex]

365/4

[pearl pond]

@crimson burrow Has your question been resolved?

[crimson burrow]

oh.

[pearl pond]

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[spring crystal]

How do u write this in exponent form?

Like sqrt(x) for example is x^1/2

[unborn abyss]

yeah just like that

just with a bunch of parentheses

whenever you see a √ replace it with (...)^½

[spring crystal]

But how do I solve that

[unborn abyss]

wdym solve

[spring crystal]

Well doesn’t the like sqrt1-t get sqrt like 3 times

[spring crystal]

Find derivative

[unborn abyss]

oh

you should start with that

anyway

it's just the chain rule

[spring crystal]

Yeah but idk how to write it in exponent form

[rugged elm]

u dont have to

[spring crystal]

Then how I solve it

[rugged elm]

this is how u would continue

do you know how to calculate the derivative of square root of x?

[spring crystal]

Yeah

But idk the exponent for the sqrt in the back

So idk how to solve those

[rugged elm]

you dont need to do that

[spring crystal]

Ok but I only know how to find deriv of sqrtx cus ik it’s exponent

And I use power rule

So without it idk how to solve

[unborn abyss]

ok

take your original thing

and every time you see √stuff, replace that with (stuff)^½

you will have some nested parentheses. that's okay.

[spring crystal]

(3t)^1/2 + (2)^1/2 +(1-t)^1/2

Well that’s wrong

But idk how to write it

@unborn abyss

[unborn abyss]

ok

see how there's a biiiiiiiiiig square root symbol over everything?

take the stuff inside that square root symbol

and rewrite it on your piece of paper

then

put parentheses around it

and write ½ to the top right

note: you will have other square root symbols inside of there. That's okay! we'll get to those

and let me know when you've done that and what it looks like

@spring crystal ^

[spring crystal]

Idk how to make the 1-t go to 1/2 for a third time

[unborn abyss]

this is what i wanted you to write

although you might find it easier to start from the inside but still

[spring crystal]

what about the inside

[unborn abyss]

you'll end up with $$(3t+(2+(1-t)^{\sfrac12})^{\sfrac12})^{\sfrac12}$$

[jolly parrot]

hay²le

[unborn abyss]

do you see how each layer of square root became a (...)^1/2?

compare what i wrote to your original problem

[spring crystal]

Yeah

It’s very confusing to look at

[unborn abyss]

well... yes

if you're writing it on paper you can make the inner stuff smaller

to help you keep track

you do have physical paper, right?

and a pencil?

[spring crystal]

Not right now

[spring crystal]

What inner stuff

[unborn abyss]

see how it's got layers like i mentioned

layers of parentheses

[spring crystal]

Yeah

[unborn abyss]

if you were writing this in person with the pencil and paper that you should always have when doing math

then you could make the inner layers smaller

[spring crystal]

I don’t understand what that means

Can u show me

[spring crystal]

Ok

So how do I solve that

[unborn abyss]

apply the chain rule

[spring crystal]

@unborn abyss am I suppose to apply this like 4 times?

[unborn abyss]

should be 3 times i think

[spring crystal]

How am I supposed to apply the second time

With the 3t in front

[unborn abyss]

remember how the chain rule works

you should find a way to make your t's not look the same as +'s

[spring crystal]

but there r other numbers in front of the f’ this time?

@unborn abyss

[unborn abyss]

i don't understand what you mean

to me it looks like you accurately wrote down the first part, the f ' ( g (t) ) term, and then wrote down something that wasn't really g ' (t) at all

[spring crystal]

Yeah I’m stuck on that part

The g’(t)

[unborn abyss]

what's g (t)?

[spring crystal]

That?

[unborn abyss]

what does that say?

is that a + or a t

[spring crystal]

Plus sign only after 2

[unborn abyss]

you should write your t's with hooks on them

[spring crystal]

Ok I’ll change it to x later

[unborn abyss]

ok well if you do that then don't use x for multiplying

anyway

look at the original expression, i think you miscopied something

[spring crystal]

I derived the 3t

So it became 3

[unborn abyss]

i think you need to review and practice simpler cases of the chain rule

[spring crystal]

Ik how to do basic ones

This one is confusing cus it has 3 sections

[unborn abyss]

why did you do that?

the inner function doesn't get differentiated in that part

[spring crystal]

That was the g’(x) part

[unborn abyss]

no it isn't? isn't that the f ' ( g(t) ) part?

neatness and practice

and paper for the love of god

get some paper

i think you missed that in the original post it had a 3t + portion

somewhere along the line you dropped either the t or the + and i honestly cannot tell which one

[spring crystal]

Oh

@unborn abyss this what I got for f’(g(t))

[unborn abyss]

ok good

now find g ' (t)

hint: you'll have to use the chain rule again

[spring crystal]

Idk how to for this

It’s too confusing

[unborn abyss]

this is simpler than the original one

what's g (t)?

you need to practice derivatives in general

[spring crystal]

This

[unborn abyss]

ok good

$3t + (2+ (1-x)^{\sfrac12})^{\sfrac12}$

[jolly parrot]

hay²le

[unborn abyss]

so what's d/dt of that?

you already seemed to know what to do with that 3t

[spring crystal]

Am I supposed to bring 1/2 to the front again?

@unborn abyss

[unborn abyss]

you will need to use the chain rule again yes

and the power rule

and the sum rule

,tex .diff rules

[jolly parrot]

hay²le

[pearl pond]

@spring crystal Has your question been resolved?

[pearl pond]

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[terse stone]

How do I figure out this?

[last summit]

how old is both sexes

[terse stone]

I'm not sure

How would I find that out

[tall flint]

,calc 10034+37101+7647

[jolly parrot]

Result:

54782

[tall flint]

so it's the column total

anyway, you focus on the northeast column and then use whatever method you were taught to find the child dependency ratio to find that ratio

[terse stone]

I didn't understand the way the teacher explained it thats why I reached out to this chat because I didn't get it, there method was too in depth

[tall flint]

we have no way of answering this, I reckon none of the helpers are in your class

[terse stone]

how? I thought I could ask questions here its not like the questions sacred to the class

[tall flint]

because we don't know what definitions you're using

[compact ridge]

The child dependency ratio is number of people under 15 / total population

[tall flint]

and we can't access the course materials with them

[terse stone]

how would I find that

[compact ridge]

Luckily I did geo

[tall flint]

you would find that in the course materials

[compact ridge]

So 10034/54782

[tall flint]

!nosols

[pearl pond]

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

[compact ridge]

Not relevant here as they're not struggling with the mathematics

[tall flint]

[compact ridge]

It's a geography question that happens to use maths

[tall flint]

that's not up for you to decide

[compact ridge]

As you said it requires outside knowledge

I fail to see how giving them the answer would stunt their problem solving in this situation

[compact ridge]

They just need to remember the definition for next time

[tall flint]

[terse stone]

so would the answer be 0.18312?

[compact ridge]

Looks about right

It should be around 10/50 after cancelling 1000 from top and bottom

[compact ridge]

That's how you know your answer makes sense

[terse stone]

I can keep trying again

[compact ridge]

Oh well look in your course notes

The definition is probably different

[terse stone]

course notes?

I just started college a month ago I'm still new

[compact ridge]

The materials your classes are using

Like slides

[terse stone]

we sadly don't have slides only sometimes pdf on zoom

[compact ridge]

Like those pdfs I mean

[tall flint]

check the question again. It says round to the nearest thousandth

[terse stone]

Ima try to look at his pdfs rq also so would it be 0.183?

[tall flint]

is 0.183 rounded to the nearest thousandth?

[compact ridge]

Mhm

Yeah I think that's the issue, totally missed that

[unkempt sedge]

What I saw , the formula says the ratio is (population of under15+above 65)/(population of 15-64)

So it's 10034+7647/37101

[compact ridge]

Ahhh right that makes more sense

[unkempt sedge]

,calc 17681÷37101

[jolly parrot]

The following error occured while calculating:
Error: Undefined symbol ÷37101

[compact ridge]

Yeah it's the child dependents / working age population

[unkempt sedge]

,calc 17681/37101

[jolly parrot]

Result:

0.47656397401687

[pearl pond]

@terse stone Has your question been resolved?

[pearl pond]

@terse stone Has your question been resolved?

[buoyant stump]

Try 0.476. You don't take 5 as 10 if the number preceding it happens to be even.

[pearl pond]

@terse stone Has your question been resolved?

[pearl pond]

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[rain ore]

yo

[rain ore]

can someone help me with my math

I can't figure out what 5x8 is

I just recently learned how to do addition and subtraction

I was wondering if anyone could help me learn math

[long dirge]

5x8 is the same as you add 8 number of 5

a.k.a 5+5+5+5+5+5+5+5

[rain ore]

oh that's 40 right?

[unreal thistle]

Yes correct

[rain ore]

yay

I've never been to school so I don't know any math

I've had people on discord help me with some though

[long dirge]

oh really?

[rain ore]

yeah

[unreal thistle]

There are some free courses on Khan academy

[rain ore]

my uncle comes over and helps me with spelling and typing sometimes

[unreal thistle]

Check it out if you have time

[rain ore]

ok

I haven't gone to school because my parents never have time to bring me

[long dirge]

...

[rain ore]

and the school is pretty far away

[unreal thistle]

it's ok man

[long dirge]

how far is it?

[rain ore]

I don't know

[unreal thistle]

you can learn pretty much everything in your home of what they teach in school

[rain ore]

and I'm not home schooled because they're always at work

well thanks for the help bye

.close

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[hybrid haven]

is this derivative wrong or something

[sterile tusk]

it looks correct

[hybrid haven]

wait which equation are they originally deriving from

isnt it P=8n-nsqrtn

from the question

@sterile tusk

[sterile tusk]

yep

they took the derivative of P

[hybrid haven]

where did the -3/2 come from

[sterile tusk]

well $n\sqrt{n}=n^{\frac{3}{2}}$

[jolly parrot]

y0shi

[sterile tusk]

and they just used power rule

[hybrid haven]

how does nsqrtn =n3/2

[sterile tusk]

the negative was originally there so it stays

well what’s sqrt n with a fractional power?

[hybrid haven]

n^1/2

[sterile tusk]

yep

and by exponent rules

how do we combine them?

$n^1n^{\frac{1}{2}}=?$

[jolly parrot]

y0shi

[hybrid haven]

ohh

adding

[sterile tusk]

yep

[hybrid haven]

that makes sense

ty

[sterile tusk]

yw!

[hybrid haven]

i got one more question

why is the x 10?

in this question

[sterile tusk]

dC/dx is the rate of change in the marginal cost

so if i plug in x=11 i would get how much the cost increases to get the next item

which is why we plug in x=10 to get the 11th item

[hybrid haven]

is taht just always the case

[sterile tusk]

well not really

it’s just in the context of this problem

[hybrid haven]

whats the context that makes it like that

i dont entirely understand it

[sterile tusk]

C is the amount it costs

so dC/dx is the cost per item

so if i want to find how much the next item would cost

i just have to find that using the equation given

this question is a little weird in my opinion

[hybrid haven]

whys it weird

[sterile tusk]

actually never mind

it’s with respect to x

[hybrid haven]

so i only minus one from the x when its derivative of the cost right

@sterile tusk

.close

[pearl pond]

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[tulip cradle]

Is this vector symmetrical because the diagonals are the same

[merry carbon]

The entries on “opposite sides” of the diagonal are the same, but yep it’s symmetric

[tulip cradle]

Oh Yh

Ok thanks

.close

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[sour bobcat]

could someone help me get the correct answer, I dont understand why mine is wrong

[pearl pond]

@sour bobcat Has your question been resolved?

[spiral pivot]

Remember that matrix multiplication is not commutative, so left vs right multiplication is important.

AX(D+BX)^-1 = C

AX = C (D + BX)
AX = CD + CBX
AX - CBX = CD
(A - CB)X = CD

You've gotten this far correctly

However, now you must left multiply by (A-CB)^-1

You erroneously left multiplied on the left hand side and right multiplied on the right.

@sour bobcat ^

[sour bobcat]

thnx

[pearl pond]

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[main musk]

Question 5 pls

[pearl pond]

@main musk Has your question been resolved?

[pearl pond]

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[unreal flume]

I’m trying to prove the derivative power rule using limits then prove that limit with epsilon delta but I’m stuck while on the epsilon delta part

[rough stream]

It's a bad strategy. We prove the limit laws with ε-δ, then use the more powerful limit laws to handle the power rule

[unreal flume]

Like what

[rough stream]

Like where

[warm current]

!show

[pearl pond]

Show your work, and if possible, explain where you are stuck.

[unreal flume]

Here

[pearl pond]

@unreal flume Has your question been resolved?

[pearl pond]

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[wet osprey]

What’s with the $dxdy$ in $\iint_D f(x, y),dxdy$

[jolly parrot]

Frosst

[wet osprey]

Shouldnt it be more like $d(x, y)$

[jolly parrot]

Frosst

[sterile crystal]

Because one integral is basically smooshed inside another integral

So the dx Integral is inside of the dy one

[wet osprey]

But that implies there’s an order that dx happens then dy

Shouldn’t $\iint_D$ and $\int_A\int_B$ be different things?

[jolly parrot]

Frosst

[sterile crystal]

The region D encompass the bounds of A and B

Just a shorthand

[warm current]

There's a theorem that proves there is no difference

[grizzled dust]

when you think of it as a riemann sum that dA is like a little piece of area thats dx by dy and the sum can be interpreted / broken up to be doing two regular single variable integrals in a row and it can be shown that the order doesn't matter

[west sapphire]

except when there is 😁

[wet osprey]

Well fubini tells us if they exist then it’s the same?

[west sapphire]

fubini says if the abs val is integrable then you can integrate using iterated integrals, in either order

[wet osprey]

So iterated integrals aren’t the same thing as this double integral then

[west sapphire]

they give you the same answer under fairly general conditions but no they're not the same thing

[wet osprey]

If $D = A \times B$, then $\iint_D$ is the same as $\int_{A\times B}$?

[jolly parrot]

Frosst

[west sapphire]

what's the significance here of using two integral symbols for the first and one for the second

[wet osprey]

Well I’m not too sure what this 2 integral notation means

Is it the single one by definition?

[west sapphire]

it's a sort of unfortunate classical notation, people (like in calculus classes) use it for area integrals even though it clashes a bit with the iterated integral notation

[wet osprey]

So what’s an area integral

[west sapphire]

once you get to the level of studying lebesgue integration, people typically use a single integral symbol regardless of the dimensionality of the space

[west sapphire]

this guy here

[wet osprey]

So the area integral is the Riemann integral in multidimensions

[west sapphire]

i said "area" integral but i really meant that the differential element represents area

i.e. integrating a 2-dimensional function over some 2-dimensional region

[wet osprey]

Right, people use dA rather than dxdy?

[west sapphire]

yea that's common

[wet osprey]

A for area? To represent a part of R?

[west sapphire]

in calculus books they often use the convention of not switching to dx dy until they are doing iterated integrals

after all, you can do dA in polar coordinates too for example

[west sapphire]

yea

[wet osprey]

So there’s a difference here

The double integral is riemann where the single one is lebesgue

[west sapphire]

wellll i don't know that i would automatically assume that's true, i'm sure you can find different conventions among different authors

[wet osprey]

Oh

[west sapphire]

but i'd say the use of a single integral symbol seems more prevalent once you get into the lebesgue theory

because they want to deal with integrating over all sorts of spaces

or stating theorems that apply regardless of how many dimensions, for example

[wet osprey]

Do you also then say that if it works with the lebesgue integral then it surely also works with the Riemann integral

[west sapphire]

so now you're integrating with respect to some abstract measure

[wet osprey]

Cos fubini’s theorem is for lebesgue integrals

[west sapphire]

the lebesgue integral can handle situations where the one dimensional "slices" (the ones you integrate using the one dimensional iterated integrals) wouldn't be riemann integrable

so just because you can use iterated integrals and swap their order in the lebesgue world, doesn't mean it holds for the same function using riemann integrals

for most "reasonable" functions it'll work fine with riemann though

[wet osprey]

Oh it might still not be Riemann integrable

[west sapphire]

right

even in the lebesgue situation, you can have a 2-d measurable function which has some 1-d slices that aren't 1-d measurable or are measurable but integrate to infinity

[wet osprey]

Ok suppose you had a function that was Riemann double-integrable in the sense that this is well defined (has the same value for any partitioning)

[west sapphire]

but if the 2-d function is absolutely integrable, such badness only occurs on a set of measure zero so it's fine

[wet osprey]

Then you can equate it to its iterated integral version?

[west sapphire]

(that's also part of the fubini theorem)

[west sapphire]

hmm, if it's absolutely riemann double-integrable then it's also absolutely lebesgue double-integrable, so you can definitely do the iterated thing and the results will be equal if you're using lebesgue 1-d integrals

[wet osprey]

Right because on the individual measure they’re non-zero but from the perspective of the product measure it’s 0

[west sapphire]

i bet there are degenerate situations where the riemann 1-d integrals can still be problematic though

but offhand i'm not sure

[west sapphire]

yea imagine a function of 2 variables that's zero everywhere except along the x axis, where you make it the characteristic function of some 1-d nonmeasurable set

[wet osprey]

If it’s Riemann double integrable then surely each part is also Riemann integrable?

Otherwise it wouldn’t equal for any partitioning no?

[west sapphire]

check this

so things can get screwed up in either direction

iterated integrals ok but double integral not ok

or

double integral ok but iterated integrals not ok

yet another example of why the riemann integral sucks haha

[wet osprey]

He was cooking

A feast I must say

[west sapphire]

fo sho

[wet osprey]

Ok so this is not true

Oooh

Perhaps it’s because it’s just the partitioning of areas

Not individual partitioning of each dimension

[west sapphire]

even the statement of fubini is a mess when you're using riemann integrals, check this as well

[wet osprey]

It seems they can’t swap the order here either?

[west sapphire]

not automatically, you need stronger hypotheses

[wet osprey]

It’s just saying the integral of the 2nd part of Q exists

[pearl pond]

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[chilly ermine]

how would i start solving this?

would this as a start be correct?

[grizzled dust]

yeah so that's one vector parallel to the plane, now just use the direction vector of the line (i.e. <1, 2, -3> for the second and take their cross product)

[chilly ermine]

then what do i do with the normal?

[grizzled dust]

you can write an equation of a plane using any normal and any point in the plane

[chilly ermine]

and that would be the answer?

[grizzled dust]

so the logic is that if you think of any vector formed by some point P_0 in the plane and any other point P in the plane, the vector formed by those 2 points would be perpendicualr to a normal vector right?

so dot product between a normal and vector from P_0 to P would = 0, this can be used to write an equation

notice how the vector (x - x_0, y - y_0, z_z_0) is just a vector between two points in the plane (x, y, z) and (x_0, y_0, z_0) and then that equation is just writing a dot product that equals 0

with the normal vector (a, b, c)

[chilly ermine]

So this would be the answer?

[grizzled dust]

im too lazy to do actual calculations, but as long as you understand the concepts im sure its fine 😄

[chilly ermine]

so basically i just cross producted 2 direction vectors to get the normal then used point P to create the vector equation would that be right?

[grizzled dust]

yeah basically if you have any 2 vectors that are parallel to the plane you can get then a normal

as long as they arent colinear

and from that just use that equation of the plane, its easy to remember if you understand intuitively that its basically a dot product set equal to 0

[pearl pond]

@chilly ermine Has your question been resolved?

[chilly ermine]

alr thanks

[pearl pond]

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[dusk hound]

Prove that the set of the real numbers is not bounded inferiorly using the definition of infimum, axiom of completeness, and other theorems.

It is of real analysis or mathematical thinking

[dense rover]

aoc is an axiom of ℝ

[pearl pond]

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[dusk hound]

What is aoc?

[pearl pond]

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[pearl pond]

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[storm saffron]

Hello. Can I get a hint on where to start from with this one? I can't think of a simple way to simplify it...

Find all such values of a, for each of which the inequality is true for all real values of x.

[hallow remnant]

um so let y=x(a-cos2x)

so in terms of bounds on y, when is sin(y) between -1 and 1 inclusive?

[storm saffron]

I honestly feel dumb rn. I can only think of when |x| <= 1/|(a-cos2x)|...

or well, if something equals zero

[grizzled dust]

think about possible inputs and outputs of the sin function

you can stick anything in, but you only ever get values out between -1 and 1 (inclusive)

[storm saffron]

true..

[grizzled dust]

sin(theta) represents values of y coordinates of points on the unit circle

so they can only be between -1 and 1

so the output of that expression in the middle is always going to satisfy the inequalities...the only other potential problems would be if something is undefined for a subset of the real numbers, but there's no division by 0, no functions with domain restrictions etc. so the inequality just holds for all real numbers

and for all values of a

[storm saffron]

looking at the function, it gets out of the restricted boundaries pretty quickly...

and the suitable values of a seem to be around [-1.4; 0]

ouch

okay, I got it right this time...

the answer

And I still don't understand how to get it..

[storm saffron]

I guess I might have made the right point though.. I'll try to solve it this way and see if I can get something interesting..

wait, what would I even get from that?..

..

[pearl pond]

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[waxen sundial]

"True or False: If A is a positive definite matrix, then A^3 is a positive definite matrix"

[waxen sundial]

I'm not exactly sure how to approach this

how do matrix operations affect its positive definiteness (or whatever the word is supposed to be)

[west sapphire]

do you have a characterization of positive definiteness in terms of eigenvalues?

[waxen sundial]

what do you mean by characterization

like the definition from my instructor?

[west sapphire]

a theorem that tells you "a symmetric matrix is positive definite if and only if..."

[waxen sundial]

I have the following
"A self-adjoint matrix A: V->V is called positive definite if <Av, v> > 0 for all v =/= 0 (v vector =/= 0 vector)"

[abstract dirge]

What can you say about the eigenvalues of such a matrix

[waxen sundial]

its not an observation I made but I believe our instructor said something like, the eigenvalues will be greater than zero

and if positive semi-definite, then the eigenvalues are >= 0

[west sapphire]

ok

suppose the eigenvalues of A are all positive

what can you say about the eigenvalues of A^3?

[waxen sundial]

im not sure,
How do A and A^3 relate to each other

i know its the same matrix A but just with a different power

[west sapphire]

well one way to find out is to diagonalize A

[waxen sundial]

diagonalizing A to obtain SDS^-1 representation and then taking D^n (in this case D^3) ?

[west sapphire]

yep

[waxen sundial]

hmm

[west sapphire]

A^3 will be S(D^3)S^-1

[waxen sundial]

but then in that case D^3 is just the eigenvalues cubed

wait no

im dumb

uhh

wait no im right

[west sapphire]

that is correct

and if you cube a positive number...

[waxen sundial]

result is a positive number

[west sapphire]

yep

therefore...?

[waxen sundial]

T

[abstract dirge]

T

[west sapphire]

T

[waxen sundial]

similarly i have this one

[west sapphire]

why are we typing T haha

oh true/false

[abstract dirge]

$\top$

[west sapphire]

yes

[waxen sundial]

but i think i can use what we just did

[jolly parrot]

Philka

[abstract dirge]

Obviously

[west sapphire]

yea use the same type of argument

[waxen sundial]

so lemme just make sure i got this right

negative definite means every single eigenvalue is strictly negative

[west sapphire]

yes

[waxen sundial]

A^8 thus cannot be negative definite because its an even power

[west sapphire]

right

in fact it'll be positive definite

[waxen sundial]

strictly greater than 0

so if we had negative semi-definite A, A^8 could only guarantee positive semi-definite but not necessarily positive definite

[west sapphire]

yep that's right

semi-definite but not definite means zero is an eigenvalue

and zero will still be an eigenvalue if you take powers of A

[waxen sundial]

it has to be an eigenvalue of the matrix, if its semi-definite?

like if we know A is semi definite

[west sapphire]

if it's semi-definite and not definite then yes

[waxen sundial]

then at least one eigenvalue is necessarily zero?

or i guess, not at least one, but one

[west sapphire]

at least one if you count repetitions

in the extreme case you have the zero matrix

[waxen sundial]

but zero matrix has no eigenvalues right

[west sapphire]

it has zero as an eigenvalue, with multiplicity n

(assuming it's an nxn matrix)

[waxen sundial]

oh ok

[west sapphire]

(it's diagonal, so the eigenvalues are on the diagonal)

[waxen sundial]

sounds good. i might be back for more questions cuz my assignments are brutal 💀

But i super appreciate you breaking down the steps to answer this T/F man

[west sapphire]

sure, if you get stuck or want to check your answers, just open a new channel

cheers

[waxen sundial]

.close

[pearl pond]

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[waxen sundial]

.reopen

[pearl pond]

✅

[waxen sundial]

Actually, I'm back 💀
I just wanted to check how can we know A is diagonalizable

[west sapphire]

it's symmetric, right?

[waxen sundial]

hmm

[west sapphire]

usually that's part of the definition

of positive definiteness

[waxen sundial]

the definition includes self-adjoint matrices and, those are symmetric if real entries

[west sapphire]

ah yea good

self-adjoint if complex

symmetric if real

then the spectral theorem tells you that it's diagonalizable

[waxen sundial]

Damn

that is convenient

Ok that completes my justification, ty 🫡

.close

[pearl pond]

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[midnight haven]

hello

[midnight haven]

we have the following problem:

f:R\ {0} -> R , f(x) = (x+2)e^{1/x} . the equation of the oblique asimptote to +infinite is:?

so first we have to find m

which is defined by this:

which is 1

<@&286206848099549185>

now i have to calculate the n

which is defined by this

bcz y = mx+n

asimptote

but i dont know how to calculate this

e^1/x is 1

x+2 - x as x -> inf is theoretically 2

right?

so the final answer 2?

so y = x + 2?

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

<@&286206848099549185>

<@&286206848099549185>

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@midnight haven Has your question been resolved?

[pearl pond]

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[lyric helm]

Can someone walk me thru how to get angle theta?

[pale haven]

you can get BC with cos theorem

AB and AC with pythagor theorem

finally cos theorem again for the theta angle after you get all the required sides

[lyric helm]

okk tyy

[pearl pond]

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[red geyser]

right okay

,w graph x=cos(t)+1, y=sin(2t)+1

[jolly parrot]

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Requested by akishbrabus

[red geyser]

perhaps the slopes are the same at different places

what are "both" tangent lines. Is is asking for the equation of the tangent line at t=pi/6 and t=5pi/6

Then, they may have the same slope

no, because the y-intercepts may be different

you must compute that using point-slope form, etc.

the point of intersection?

no, because the tangent lines may not go through that point? I don't know where on the graph it is at t=pi/6 or 5pi/6

you need a point on the tangent line. Plug t=pi/6 into (x(t), y(t)) to get a point on the tangent line at t=pi/6

oh are you trying to find the slope of the tangent lines at the intersection?

Then yes, 1,1 works

the slopes should be different okay I understand

those are the wrong numbers for t I believe

t should be like pi/2 and 3pi/2

well, you can check that 3pi/2 works

manually

right

i basically just solved for x(t)=1

yes

did that work?

no

if I plug 1,1 into the second equation it doesn;t work

oh its probably a type

typo

it should be y=-2x+3

that sounds about righr

that seems right

You too!

[pearl pond]

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[red geyser]

bro you're back again 💀

lmaooo

the first bounds are from pi/2 to 5pi/6

the second are from 5pi/6 to pi

yes, I think so

its pretty easy to botch these integrals so you can probably just check on wolfram or something

yes, I'm afraid I primarily do algebra and the consistency of my integrals is in question

[pearl pond]

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[wild star]

yo

[wild star]

I was just studying about transforming from Cartesian coordinations to spherical coordinations

and between my notes i found this law :

$\frac{d\vec{U}{\phi}}{dt} = \frac{\partial \vec{U}{\phi}}{\partial \phi} \frac{\partial \phi}{\partial t} + \frac{\partial \vec{U}_{\phi}}{\partial \theta} \frac{\partial \theta}{\partial t}$

[jolly parrot]

mino

[wild star]

can someone tell me the name of it so that i can look for it more ?

[fossil jewel]

Gradient

I think

[unkempt yacht]

chain rule?

[red geyser]

https://tutorial.math.lamar.edu/classes/calciii/chainrule.aspx

[wild star]

Hmm

i'll look into these suggestions

thnx guys !

.close

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[patent quail]

im stuck at findind d and b

[patent quail]

i just want the values that make the function continuous

[lavish portal]

Do you know how to find the values?

[patent quail]

find the limit from the right and left

and see if they are equal right

[lavish portal]

yes and for them to be continuous they need to be equal

so find the values that make the limits equal

[patent quail]

so after finding the limits make them equal to each other

[lavish portal]

You can also just find the value that makes the equations equal exactly at the point where they connect

and solve yeah

[patent quail]

so what happens to the x in there

i subsitute the x at where they meet right

kindly try it out please im stuck at b -d = 2

i cant move forward

<@&286206848099549185> please

[pearl pond]

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[pearl pond]

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[patent quail]

.reopen

[pearl pond]

✅

[patent quail]

no

[pearl pond]

@patent quail Has your question been resolved?

[patent quail]

no

<@&286206848099549185> please take a look at this problem

[lavish portal]

Ok sorry

@patent quail I was at church so I had to stop so

can you show you work

[patent quail]

at the moment no

my phone is off so i cant take the photo

[lavish portal]

Ohhhh I see where

[lavish portal]

The full function is continuous

so you can find b by finding where it is continuous with y=2

does that make sense

So find b first with the other equation

[patent quail]

not really

i got a to be 2 and c to be 2 but for b and d all i have is a relation

b-d = 2

[lavish portal]

The reason you ended up with this equation was because

There are multiple b and d that will make it continuous

To find the b value, we need to have an “anchor”

so instead of comparing it with another unknown equation find the value of b that is continuous with y=2 (the other equation next to it)

oh I see why it’s weird…

I think finding d and b is related to the mean value theorem which I don’t know @patent quail

[patent quail]

forget the mean value

lets just deal with the one that makes the function continuous

[lavish portal]

you can’t find b and d

that’s because there are infinite combinations of b and d

[patent quail]

oh okay

[lavish portal]

try it out yourself

[patent quail]

i already did

[lavish portal]

b = d+2

and d = b-2

[pearl pond]

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[lavish portal]

I think you need to apply mean value theorem

[patent quail]

okay ill apply that later

what about this one then

[pearl pond]

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[mint parrot]

Can someone help me with question 33 and 34

[dawn nimbus]

show us ur work

[mint parrot]

Im trying to work out the questions with variables but it asks the verify the identity for the thinks so idk

Do I just work it out with the variables?

[dawn nimbus]

yes

[mint parrot]

That would actually kill me 💀

[dawn nimbus]

i mean, not a lot of work

[mint parrot]

Really 😢?

Ok

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[pearl pond]

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[vital halo]

can anyone convert this into a cosh function

[vital halo]

i don't get what a is

[toxic fractal]

a would be the lowest point above x axis

[pearl pond]

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[pearl pond]

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[sage walrus]

What do you do when the number is below the variable?
so here:
tan55 = a/6
we know what to do: 5 * tan55 = a

but then in this example:
tan40 = 12/a, how do I find a?

[sage walrus]

@lunar nacelle

[wild fable]

wrong ping lol

u have to do the opposite operation of what has been done to solve for the variable

[sage walrus]

uh....

[wild fable]

multiply both sides by a

what happens

$a\tan40=12$

[jolly parrot]

water beam

[sage walrus]

tan40 * 2 is..

[wild fable]

huh?

[sage walrus]

1.678

wait- No

I don't get it

[wild fable]

where are you getting 2 from

[sage walrus]

it's a mistake I meant 12

[wild fable]

look

3x = 1

how do you solve for x

[sage walrus]

divide 1 by 3?

tan40*12 is 10.069

[wild fable]

you divide both sides by 3

[wild fable]

we arent doing tan40 * 12 and dont use the decimal value

using exact values

keep it as tan40

[sage walrus]

so what do I do ?

[wild fable]

listen

if you have 3x = 1

you divide both sides by 3 to isolate x

[sage walrus]

ok, so 3/3 is 1

and then 1/1 is 1

so 1x = 1

x = 1

[wild fable]

no...

[sage walrus]

what am I doing wrong?

[wild fable]

$\frac{3x}{3}=\frac{1}{3}$

[jolly parrot]

water beam

[wild fable]

do you see this

you divide both sides by 3

[sage walrus]

oh

That's what I've done

[wild fable]

no

thats not what youve done

[sage walrus]

oh, right

cuz 1/3 is 0.3 🤦‍♂️

ok, so x = 0.3

what does that mean?

[wild fable]

dont use decimals

keep it in exact value

[sage walrus]

ok

3/3x = 1

3/1...

I have a different idea.

3 * 3x = 1x
3*1 = 1

[wild fable]

you are getting the wrong idea

i dont know why youre trying to do things differently

[wild fable]

none of what you wrote here is right

https://youtu.be/kWOTmyoaWJg?si=1Yr7rP-kJlTUk-Ej

watch this video

[sage walrus]

ok- Like I said

1 divided by 3 is 0.3

and 3 divided by 3 is one

[sage walrus]

just like he says at 5:40, same method

[wild fable]

no its not the same thing as what ur doing

you are saying 3 times 1 is 1

and 3 * 3x = 1x

you are dividing both sides of the equation by 3

[sage walrus]

oh

ok, so In here we need to do:

[sage walrus]

tan40 = 12/a
so we divide both sides by the upper number? (12)

10 = 12x

x = 1.2 ?

[wild fable]

i dont think you understand what youre doing here, you are getting all these random numbers from nowhere

[wild fable]

i suggest you watch the video in full

[sage walrus]

a is 12

alpha is 40

I need to solve for b

[wild fable]

and you wont be able to solve for b if you dont know the general approach to these types of questions

like i said watch the video

once you have watched the video

come back to this question

and see if you can do it

[sage walrus]

I think I've got it

[wild fable]

tell me

[sage walrus]

something like this then? (

[wild fable]

not quite

[sage walrus]

yeah I watched the video

just now

[wild fable]

$\tan40=\frac{12}{a}$ yes?

[jolly parrot]

water beam

[sage walrus]

yep ^

[wild fable]

you want to solve for a yes?

[sage walrus]

I want to solve for a

[wild fable]

what happens if you multiply both sides by a?

$a\cdot\tan40=\frac{12}{a}\cdot a$

[jolly parrot]

water beam

[sage walrus]

I get
a * tan40 = a* 12/a

[wild fable]

does this line make sense to you?

[wild fable]

this is what happens if you times both sides by a

[sage walrus]

got it 👍

[wild fable]

and what 12/a times a simplified

12a/a

[sage walrus]

it just means 12

ig

[wild fable]

yes

[sage walrus]

so we got
a * tan40 = 12 * a

[wild fable]

you just told me 12a/a is 12 simplified

how can there be another a next to the twelve

[sage walrus]

one second..

[sage walrus]

so on the right side we have
12/a * a

[wild fable]

do you understand the fact that $\frac{a}{a}=1$

[sage walrus]

but that just turns out to be 12/a

[jolly parrot]

water beam

[sage walrus]

eh?

why's that

[wild fable]

if you have 2 apples and share it among two friends, how many apples does each person get?

[unreal thistle]

a gets cancelled out

[sage walrus]

oh--- because no matter what it is, If it's divided by itself it'll always be 1

[unreal thistle]

Yes

[sage walrus]

ok ok

[sage walrus]

But here we don't do a/a

we do 12/a * a

[unreal thistle]

We do

[wild fable]

it is the same thing

[unreal thistle]

12*a is 12a

[wild fable]

$\frac{12}{a}\cdot a=\frac{12a}{a}$

[jolly parrot]

water beam

[sage walrus]

still don't get it

[wild fable]

what do you not understand here?

[unreal thistle]

You have multiplied 12 and a in the numerator

[sage walrus]

a is like x?

[unreal thistle]

Yes

Wtf

[sage walrus]

ok so
12/x = 12x/x * x

[wild fable]

why are you changing a to x?

[sage walrus]

idk i guess It's just simpler

for my goofy brain to understand

[wild fable]

you are overcomplicating everything

[sage walrus]

I know

I'm genuinely trying tho 😂

oh----

[wild fable]

https://youtu.be/AtBUQH8Tkqc?si=JUDhaZYPJK-5Tulk

[sage walrus]

so we just temporarily take 12 out

[wild fable]

this may help you with the basics

[sage walrus]

do a * a

which is 1

and then we got 12/1

[wild fable]

because it seems you are missing a lot of basic knowledge here to solve equations

[sage walrus]

Is this correct tho?

and so we got:
a * tan40 = 12/1

[wild fable]

yes and 12/1 = 12

[sage walrus]

yeah ok

so we got:
a * tan40 = 12

now.. we switch sides?

[wild fable]

no

now you divide both sides by tan40

$\frac{a\tan40}{\tan40}=\frac{12}{\tan40}$

[jolly parrot]

water beam

[wild fable]

what happens to tan40/tan40

[sage walrus]

it became 1

[wild fable]

yes

soooo.....

we get?

[sage walrus]

so
1 = 12 / tan40

[wild fable]

no not 1 = 12/tan40..

you forget the a is still there.

[sage walrus]

a * tan40 is 1
so we are left with:

1/tan40 = 12/tan40

[wild fable]

you cannot simply ignore the fact that the a exists

[sage walrus]

right?

[wild fable]

no

listen

[sage walrus]

I'm not ignoring it, But a * tan40 is just.. tan40

cuz u said a is 1 when divided...

[wild fable]

are you going to listen

[sage walrus]

oh- i see the mistake now

yeah, I'm listening

[wild fable]

a/a = 1

any number divided by the same number

will be 1

it doesnt matter if its 90000000/90000000

or 20000000000239493/20000000000239493

it will still be 1

do you understand yes or no

[sage walrus]

yes

[wild fable]

so now that you know this

what is tan40/tan40

[sage walrus]

1

[wild fable]

yes.

$a\cdot\frac{\tan40}{\tan40}=\frac{12}{\tan40}$

[jolly parrot]

water beam

[wild fable]

and 1 times a is just a

so what does a equal?

[sage walrus]

on the left side

It's
a * 1

on the right side, It's just 12/tan40

[wild fable]

yes

so what does a equal

[sage walrus]

a * 1 = 12/tan40

right? ^

[wild fable]

ok but

you dont need the times 1

anything times 1 is just anything

[sage walrus]

so
a = 12/tan40

[wild fable]

yes

[sage walrus]

oh wow ok

and now we just simply solve 12/tan40 ?

[wild fable]

put that into calculator or whatever yes

https://youtu.be/4CKDqvddhhg?si=66zT8ATm72MQpXrR
you should rlly watch this and some other videos on algebra

[sage walrus]

a = 14.301

alright, I'll watch it

thanks man

have a good one 👍

.close

[pearl pond]

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Use .reopen if this was a mistake.

[pearl pond]

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