help-39

[tropic saddle]

char poly can have multiplicites bigger than 1

[cursive wraith]

It's split with multiplicities 1 <=> T is diagonalizable

[red trail]

yeah that therom was if charestristic poly split with multiplicities 1 => T is diagonalizable
not iff

now i rememebr

[red trail]

yes

[cursive wraith]

That's an implication with characteristic poly

With minimal poly, It's an equivalence

[red trail]

ye

[cursive wraith]

This is a consequence of Cayley-Hamilton

If char poly is split with mult 1, then since min poly|char poly, min poly is also split with mult 1

Thus T is diagonalizable

[red trail]

yes

[cursive wraith]

Ok so now you have to find the minimal polynomial of T

[red trail]

but how would i find the m.p in this situation?

[cursive wraith]

How about computing T, T², etc

[red trail]

there was a therom like this but i forgot it

what does it say?

it had something to do with the minimal T-saved subspace

being {v, Tv, TTv,...}

[tropic saddle]

how about starting by computing T^2(A) for some matrix A

[red trail]

its A

[tropic saddle]

so T^2(A)-A=0

[red trail]

so I

?

[tropic saddle]

no

[red trail]

oh wait

but wait why not?

i thought i figured something

[tropic saddle]

you cant take sqrts of linear maps

also T isnt a 2x2 matrix

and that matrix doesnt even square to the identity matrix

[red trail]

ur right yes

what am i doing

[red trail]

so what can i do here?

this squared is I

[red trail]

what is its dimensions then?

[red trail]

here the m.p = x^2-1 = (x-1)(x+1) so diagonalizable?

[tropic saddle]

T goes from a 4 dim space to a 4 dim space

so a matrix representation of it would be 4x4

there are lots of matrices which square to I

you know how both 1 and -1 square to 1. its like that but even worse for matrices. you shouldnt take sqrts

[tropic saddle]

yes

[red trail]

so what is the right thing to do here im confused

[tropic saddle]

well you found the min poly

[red trail]

yes but that was by thinking of this

[tropic saddle]

from here you have T^2-I=0

so the min poly is x^2-1. (or a factor of it)

but its obvious that x+1 and x-1 arent the min poly of T

[red trail]

oh yes u mean caily hamilton

[red trail]

ye

[tropic saddle]

no

not cayley hamilton

[red trail]

why?

[tropic saddle]

cayley hamilton is about char poly

you dont need it

[red trail]

yes but u can find the min polynomial by trying "varietes" of the char poly

[tropic saddle]

if you mean factors, sure

[red trail]

yes

[tropic saddle]

but you dont need that

you just found the min poly directly

[red trail]

u mean by definition?

[tropic saddle]

yes

[red trail]

ok ty

[red trail]

my teacher did this once thats why i took a sqrt

thats what i meant

like this also

[tropic saddle]

well with that definition sqrt(I)=I

but T is clearly not the identity

[red trail]

ye thats the problem

it confused me even more

[red trail]

but here he says find A matrix not THE matrix so thats okay i guess

[red trail]

tysm for the help i think i should close this

.close

[pearl pond]

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[pearl pond]

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[hollow pike]

this is independent, right?

[glacial sequoia]

yes

[pearl pond]

@hollow pike Has your question been resolved?

[vestal tapir]

yes

[hexed stone]

Help!

[pearl pond]

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[sudden parcel]

help please

[sudden parcel]

I know you have to use product rule

[naive narwhal]

,rotate ccw

[sudden parcel]

we weren't taught how to differentiate tanx in class : /

[old marsh]

U can turn tan(2x) into sin(2x)/cos(2x) if u haven’t learned how to differentiate tan

@sudden parcel

[sudden parcel]

i see

ill do it

.close

[pearl pond]

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[vestal spade]

i do not understand how they came up with (2-c) or (-1-c)

i got to the step of creating 2x+2y=0, 0x-y=0

but do not understand how c came to be

[summer imp]

If you multiply out the left side, you get a column vector (2x+y, -y) = (cx ,cy)

Equate both coordinates and you get the system of equations

[pearl pond]

@vestal spade Has your question been resolved?

[pearl pond]

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[midnight haven]

Find the domain of $f(x)=\sin^{-1}(2x^2-1)+\cos^{-1}\left(\frac{x}{x-1}\right)$

[jolly parrot]

pun pun

[midnight haven]

im gettin closed 0 to 1/2 but the answer says closed -1 to 1/2

[sharp smelt]

!show

[pearl pond]

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

[midnight haven]

i dont have a phone

hold up lemme go get my mums 😭

[sharp smelt]

I hope you started with the sub $x=cos(u)$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

or atleast I think that should work

[midnight haven]

no?

why should i do that??

[inland ivy]

what?

it's a domain question

[sharp smelt]

ah

right

[inland ivy]

yeah the given answer is right

[cinder flower]

pun pun

[midnight haven]

hold up lemme go get the pic

[midnight haven]

hi

[inland ivy]

you solved $\frac{x}{x-1}\le 1$ incorrectly

[jolly parrot]

kheerii

[inland ivy]

ah, on the second page it's right

nvm

it seems right, the only issue is how you concluded $x\ge 0$ from $x^2\ge 0$

[jolly parrot]

kheerii

[midnight haven]

oh wait

it should

be

R

[inland ivy]

also, I hope you know that that thing you did on the first page, $x\le x-1$ is COMPLETELY incorrect

[midnight haven]

all real number

[jolly parrot]

kheerii

[inland ivy]

yeah

[midnight haven]

yea i know, i left it there forgot to cancel it, we dont know if x-1 is +ve or -ve so we cant transpose

[inland ivy]

indeed

so your work is correct except for the x>=0 part

[midnight haven]

alr i think all good now

[inland ivy]

which makes your answer the same as the one given

[midnight haven]

i got the answer now

yea

thanks

imma close now

.close

[pearl pond]

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[pearl pond]

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[heavy timber]

A spherical balloon is inflated such that its volume changes at a rate of 5 cm³/s. Determine the rate of change of the radius when the volume is 1500 cm³.

I have already do it, but I don't know if my answer is right

[tall gust]

i dont acc know but i rate the profile pic

[heavy timber]

Haikyuu fan, have a nice day :)

I know that I should use this formula

So I want to find the derivative of radious when V = 1500cm^3

Which is this

Then, I just find radious

[vestal spade]

ya

[heavy timber]

And finally I found the derivative of radious

Is it ok?

[sweet forge]

Yes

[heavy timber]

The problem is that AI gives me different result

[summer imp]

AI is notoriously bad at maths. You shouldn't trust it.

[heavy timber]

There's almost no difference but well

[heavy timber]

That's why I ask

Which would be the result for this problem?

[summer imp]

Whichever one you used your calculator for, and not AI.

[fickle schooner]

I got 0.0079

[summer imp]

The radius is wrong.

[heavy timber]

how did you do it?

[heavy timber]

I listen

[summer imp]

It's the same, just 7.1 instead of 7.4

That gives 0.0079

[heavy timber]

Perfect, thank you so much, wait me a minute until I finish it please

[summer imp]

It's nothing against you for AI, it's just that it gets hard to untangle what AI said, where it made mistakes, where you might've made some and so on.

[heavy timber]

I'll take your advice, thanks and have a nice day :)

.close

[pearl pond]

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[sharp smelt]

Simplify the following statements. Which variables are free and which are bound? If the statement has no free variables, say whether it is true or false.

$-3 \in{x \in \R | 13-2x >1}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

so this means that $6>x$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

this statment has no free variables

sorry, x is a free variable here

is the right?

[sonic patrol]

x is not a free variable

[sharp smelt]

but a free variable is just a placeholder

right?

[sonic patrol]

Not exactly

Here’s a proposition that has a free variable

Hmm wait

Ok ok I remember now

A free variable is something you can substitute

For example

“x = 3”

This is a proposition with a free variable, x

[sharp smelt]

ok

[sonic patrol]

You can think of it as a function

[sharp smelt]

and here why can I substitute x?

[sonic patrol]

You take in a number, $x$

[jolly parrot]

Pseudonium

[sonic patrol]

And you output either "TRUE" or "FALSE"

For example, you could substitute "10" for x

Then you'd get 10 = 3, which is FALSE

[sharp smelt]

ooh

ok

[sonic patrol]

Or you could substitute "8" for x

[sharp smelt]

yeah, makes sense

[sonic patrol]

Again, you'd get 8 = 3, which is FALSE

Or, you could substitute 3 for x

[sharp smelt]

but here say x=8

[sonic patrol]

And get 3 = 3, which is TRUE

[sharp smelt]

we get 6>8 which is false

[sonic patrol]

So it's really a function $p : \mathbb{R} \to {\text{TRUE}, \text{FALSE}}$

[jolly parrot]

Pseudonium

[red hedge]

sorry jus wondering what level of maths is this?

[sonic patrol]

However, here, x is not a free variable

[sharp smelt]

where I'm from just before 1st year of uni, so 11th/12th, though it's usually only done at uni

[sonic patrol]

Do you see why x is not a free variable?

[sharp smelt]

no

[sonic patrol]

Ok so

Have you met this set-constructor notation before?

[sharp smelt]

yes

[sonic patrol]

Here's one way to think about it

We have a function $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) = 13 - 2x$

[jolly parrot]

Pseudonium

[sharp smelt]

yes

[sonic patrol]

Then, we also have a function $g : \mathbb{R} \to {\text{TRUE}, \text{FALSE}}$

[jolly parrot]

Pseudonium

[sonic patrol]

Defined by $g(x) = 'x > 1'$

[jolly parrot]

Pseudonium

[sonic patrol]

So here, $g(3)$ would be TRUE, $g(0)$ would be FALSE

[jolly parrot]

Pseudonium

[sonic patrol]

Does that make sense?

[sharp smelt]

what does $='$ mean

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sonic patrol]

Ah it's like

To evaluate g

You look at the proposition "x > 1"

Substitute your input

And check whether it's true or false

[inland ivy]

that's some wild notation

[sonic patrol]

So, to evaluate $g(3)$

[jolly parrot]

Pseudonium

[sonic patrol]

We look at "3 > 1"

This is true, so we output TRUE

[sharp smelt]

okay

[sonic patrol]

This is a proposition with a free variable, actually

You can substitute x

And you get either TRUE or FALSE

[sharp smelt]

I see

[sonic patrol]

We can then compose these

We get another proposition

It takes in a real number $x$

[jolly parrot]

Pseudonium

[sonic patrol]

And checks the proposition "13 - 2x > 1"

And outputs TRUE or FALSE based on that

[sharp smelt]

I get that

but here we have 6>x

right

say I put in x=7

we get false

[sonic patrol]

Yes, so there's a way to show this on the diagram

Let's define a function $h : \mathbb{R} \to {\text{TRUE}, \text{FALSE}}$

[jolly parrot]

Pseudonium

[sonic patrol]

Via $h(x) = '6 > x'$

[jolly parrot]

Pseudonium

[sonic patrol]

Then $h = g \circ f$

[jolly parrot]

Pseudonium

[sonic patrol]

Is essentially what you're saying

[sharp smelt]

I see

[sonic patrol]

Again, h is a proposition with a free variable

You can substitute an input, a real number x

And you get an output, TRUE or FALSE

In fact, using h, we can now forget about g and f

[sharp smelt]

mhm, yea, makes sense

[sonic patrol]

Now, we want to define ${x \in \mathbb{R} | 13 - 2x > 1}$

[jolly parrot]

Pseudonium

[sonic patrol]

Which we can write ${x \in \mathbb{R} | h(x) = \text{TRUE}}$, right?

[jolly parrot]

Pseudonium

[sharp smelt]

ah

Makes a lot more sense now, thanks!

[sonic patrol]

Cool

We can also show this on our diagram

Do you want to see how?

[sharp smelt]

sure

where are you getting these diagrams from btw?

[sonic patrol]

Let's call this subset $A$, so $A \subseteq \mathbb{R}$

[jolly parrot]

Pseudonium

[sonic patrol]

https://q.uiver.app/

[sharp smelt]

👍

[sonic patrol]

Then, we can turn an element of $A$ into an element of $\mathbb{R}$, using something called the inclusion map $\iota : A \to \mathbb{R}$

[jolly parrot]

Pseudonium

[sonic patrol]

It's defined as $\iota(a) = a$

[jolly parrot]

Pseudonium

[sonic patrol]

All it does is change the type of its argument

And the point of $A$ is that the composite of those two functions should just be a constant TRUE

[jolly parrot]

Pseudonium

[sonic patrol]

Does that make sense?

[sharp smelt]

just a minute

yeah,as the common range is {TRUE} it maps to true

right?

[sonic patrol]

The idea is that we want $\forall a \in A, h(a) = \text{TRUE}$

[jolly parrot]

Pseudonium

[sonic patrol]

Because we want to make this set

[sharp smelt]

got it

thanks!

[sonic patrol]

So, at the end of the day

We're left with $A$

[jolly parrot]

Pseudonium

[sonic patrol]

But notice - $A$ is not a function! It's a set

[jolly parrot]

Pseudonium

[sonic patrol]

You can't plug anything into it

Our original proposition then reduces to $-3 \in A$

[jolly parrot]

Pseudonium

[sonic patrol]

This is either TRUE or FALSE, it's not a proposition with any free variables

Because having free variables means you're a function

[sharp smelt]

ah

I see

but If the 3 were not there, it would be free?

[sonic patrol]

So, having free variables means you're a function

So... what function are you suggesting?

[sharp smelt]

3x+1

say

[sonic patrol]

Yes, that's a function from $\mathbb{R} \to \mathbb{R}$

[jolly parrot]

Pseudonium

[sonic patrol]

But it's not a proposition

Because a proposition always outputs either TRUE or FALSE

[sharp smelt]

$a\in {x| x<3}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

so here $a$ would be a free variable ,right?

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sonic patrol]

Yes!

Because this is a function

It takes in a real number "a"

And spits out a truth value, TRUE or FALSE

[sharp smelt]

ah

got it

TYSM!

[sonic patrol]

Yayyy

I'm glad this worked

[sharp smelt]

so just to confimr

this is another example

$2\in {x|x<3}$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

here x is NOT a free variable

[sonic patrol]

Yep!

[sharp smelt]

got it!

thanks!

[sonic patrol]

Np!

[sharp smelt]

can I close this now

[sonic patrol]

Sure

[sharp smelt]

.close

[pearl pond]

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[pearl pond]

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[novel remnant]

can anyone help?

[old marsh]

uh

what do spring, summer ,autumn, and winter all have in common

@novel remnant

[pearl pond]

@novel remnant Has your question been resolved?

[midnight haven]

Seasoning

[pearl pond]

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[ruby crown]

i know continuous piecewise functions

its just creating the function

because i don't know what to do first

[regal herald]

whats the expression for the first two hours

[ruby crown]

$12

so

$0<=x<=2$

[jolly parrot]

chilly

[ruby crown]

but its not continuous

[regal herald]

it will be, but its not 12

it would be 12x or equivalent

[ruby crown]

ohh

[regal herald]

what would the total cost be for 3 to 4 5 hours as an expression?

[ruby crown]

10x?

[regal herald]

that neglects the first 2 hours

how much was the cost at 2 hours

[ruby crown]

12

24

ohh

$24+10(x-2)$

[jolly parrot]

chilly

[ruby crown]

yes?

[regal herald]

good

then after the 5th hour?

[ruby crown]

$54+9(x-5)$

[jolly parrot]

chilly

[regal herald]

si

[ruby crown]

would the expressions for those be 12x x<2

24... x<5

and 54... x>6

js to put in short

[novel remnant]

sorry i went outside

[regal herald]

uh no
ill just do set notation
[0,2]
(2,5]
(5,inf)

would be acceptable

[ruby crown]

thats interval notation i think

[regal herald]

yeah i just didnt want to write out the stuff

[ruby crown]

oh alright

i still don't get it

[regal herald]

the first is 0<=x<=2
then 2<x<=5
then 5<x

[ruby crown]

oh

but how would I make it continuous

[regal herald]

it is continuous already

[ruby crown]

because im supposed to graph it

oh

[ruby crown]

ohh

i see

[regal herald]

you achieved that

[ruby crown]

thanks

.close

[pearl pond]

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[slender dust]

we have the following functions, which are proved to have an instantaneous rate of change of 1.23 at x = 2. How would I prove this mathematically?

[slender dust]

here is my start:

[muted shale]

you need to look at the derivatives f'(x) and g'(x)

[slender dust]

how can i calculate the derivatives?

is there an equation i can use?

[muted shale]

you have to use your derivative rules

product rule, chain rule, quotient rule, etc

[slender dust]

ah okay!

.close

[pearl pond]

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[opaque compass]

What's up

yeah

sure

okay first u need to evaluate each part of the expression

what's -0.25^2017

power of 2016?

for 0.25?

[midnight haven]

longg jeez

0.25 = 1/4

which is 4^-1

btw

[opaque compass]

yeah

so yk 0.25 = 1/4

[midnight haven]

wait mb

its -1

[opaque compass]

so it equals 1/4^2017

[midnight haven]

wait

i said that

im dumb

oops

7/5

uhhh

-4 * 2 +1 -4

3

i think

wait

its 11?

im trynna do it in my head 😭

just to double check

[opaque compass]

wait the entire expression is 7/5?

[vestal spade]

bro whats difficult

[pearl pond]

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[opaque compass]

What is the area of the regular polygon? Round your answers to two decimal places.

[west sapphire]

!status

[pearl pond]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[opaque compass]

So there's a triangle within right

and you can split the triangle into two

so you have triangle CDB

[west sapphire]

yes

[opaque compass]

and you know it's a right triangle

[west sapphire]

yep

[opaque compass]

is angle DCB 72 degrees?

because 360/5 = 72

[west sapphire]

should be half that, right?

[opaque compass]

ohhh yeah

[west sapphire]

ACB angle is 72 deg

[opaque compass]

so dcb is 36

[west sapphire]

yep

[opaque compass]

which makes the last angle 54?

[west sapphire]

correct

[opaque compass]

so then we can do sohcahtoa to find the length of CB

DB i mean

[west sapphire]

yes

[opaque compass]

so 9cos(54)

which is 5.290067271

right?

[west sapphire]

correct

[opaque compass]

okay

so then we know AB is 10.58013454

[west sapphire]

oh wait hang on

oh yea ok

[opaque compass]

is it right?

[west sapphire]

sorry i lost track of which angle was which

[west sapphire]

yes

[opaque compass]

and to find the length of CD i just do 9sin(54)

[west sapphire]

yep

[opaque compass]

which is 7.281152949

so then to find the area of the triangle I multipliy 10.58013454 by 7.281152949

then divide by two

so the area of the triangle is 38.5177889

[west sapphire]

yea

[opaque compass]

and multiply that by 5 because 5 triangles fit in the polygon

[west sapphire]

yep

[opaque compass]

so 192.5889445

and round to two decimal places so

192.59

?

[west sapphire]

looks good to me

you can check using this rather nasty formula:

[opaque compass]

alright i will

tysm

for your help

ur awesome

.close

[pearl pond]

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[west sapphire]

sure, cheers

[pearl pond]

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[opaque compass]

Assume the triangle below is an equilateral triangle and the 9m segment is the radius of the circle which is perpendicular to the side of the triangle.

!status

[pearl pond]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[opaque compass]

I'm at 4

This is basically what I wrote down:

The height of the triangle is 36

(18√3 * 36)/2 = (648√3)/2 = 324√3

324√3 - 81π

<@&286206848099549185>

[worthy lance]

36?

[opaque compass]

my b

i forgot the first bit

Area of the circle: 81π (92)

Each side of the triangle is 18√3

The height of the triangle is 36

(18√3 * 36)/2 = (648√3)/2 = 324√3

324√3 - 81π

and i did this last week so idk how i got the answers

i totally forgot

[worthy lance]

How do you get height

[opaque compass]

I honestly don’t remember

[worthy lance]

How many apothems

[opaque compass]

3?

[worthy lance]

The height of an equilateral triangle is 3 times the size of the apothema

As you can see in this drawing

[opaque compass]

so 27?

[worthy lance]

That would be the height yes

[opaque compass]

oh

hm

okay

and the sides?

how would i find that

[pearl pond]

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[agile ridge]

Any hints on this?

[pearl pond]

@agile ridge Has your question been resolved?

[pearl pond]

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[spiral pivot]

@agile ridge My guess is you can use b to construct a set of equations in the a_i that form non-trivial sums of zero, meaning that you can define the kernel of the matrix, which seems like it would be a 2 dimensional kernel?

After that, perhaps you could find the eigenvalues and vectors and express b in terms of those? I feel like I'm probably overcomplicating things though. There's probably some super slick theorem I'm forgetting to make this easy, and swisher is going to put it here right after I submit this.

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[woeful harness]

unfortunately i do not come with the super slick theorem

[woeful harness]

oops now its my channel

[pearl pond]

@woeful harness Has your question been resolved?

[pearl pond]

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[warm canopy]

pls can someone help me solve this

[feral sedge]

start by calculating the area of intersection between 2 circles

[warm canopy]

i dont know how to

oh its a third

[feral sedge]

?

[warm canopy]

wait i dont get it

what is the fraction of the shaded area to one circle

[feral sedge]

Well if you knew that the problem would pretty much be solved, no?

[warm canopy]

yea

do u know or no

[feral sedge]

Not off the top of my head

[warm canopy]

thats helpful 😭

[spark cedar]

the inner white part is a equilateral triangle with side length 22cm + 3 circle segments of angle 60 deg

and if you take one of the green areas plus the inner white area its two circle segments of angle 120 deg

i guess there is an easier way

[warm canopy]

i figured it out

thanks

.close

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[sharp smelt]

$A\setminus\left(A\setminus B\right)$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

I'm trying to express this uing logic operators

I've come this far (yet to post please wait)

$x\in A\wedge\neg\left(\left(x\in A\wedge\neg\left(x\in B\right)\right)\right)$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

which further simplifes to

$x\in A\wedge\neg\left(\left(x\in A\wedge\left(x\notin B\right)\right)\right)$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

which then simplifies to

$x\in A\wedge\left(\left(x\notin A\vee\left(x\notin B\right)\right)\right)$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

which give me

[vital estuary]

what happened to negating the second statement

[sharp smelt]

ooh, that's where I was going wrong

thanks

.close

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[wet osprey]

@sharp smelt try doing these with just logic operations

without invoking x ∈ ...

[sharp smelt]

the book asks me to use logic operations

[wet osprey]

well, try not using x ∈ ...

and just manipulating the ∩ ∪ \ symbols

[sharp smelt]

[pearl pond]

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[warm charm]

how do i do this

[warm charm]

since both denominator and numerator come out positive i said one

[humble wedge]

Let's solve this step vy step

You have this question, how do you solve it?

Not just, because positive so 1. Like show your work

[warm charm]

im used to solving these with numerators and denominators whose leading term isnt inside a square root

i honestly dont know what i tried to do there

[humble wedge]

Okay let's imagine a normal case

[warm charm]

something like this i can do easily

[humble wedge]

(2x^3 + 2)/(5x^3 - 1)

[humble wedge]

Or this one

How do you solve this

[warm charm]

long division, slope of slant asymptote is the direction of the limit

[humble wedge]

There is a much easier way

[warm charm]

mmmm

[humble wedge]

Factoring

Basically what is the most amount that you can factor from both denominator and numerator?

[warm charm]

numerator it's x

denominator nothing

[humble wedge]

You skipped steps

Step by step

What do you factor from denominator and numerator?

Oh okay

Well I see

[warm charm]

x(x^3-3x+1)

over

x^3-x+8

[humble wedge]

You can factor more

Let's say you have x^2

Factor x^3 from it what does it become?

[warm charm]

i suck at factoring anything over 2nd degree

one second

[humble wedge]

Alright let's go step by step one more time

Factor x from x^2

[warm charm]

x*X

well

[humble wedge]

Now factor x^2

[warm charm]

denominator could be factored as x(x^2-1)+8

[humble wedge]

No that's not what we want

[warm charm]

ok

[humble wedge]

Let's do x^2 example for now

[warm charm]

x*x

[humble wedge]

Factor x^2 from it, what will you get?

[warm charm]

from what

denominator

[humble wedge]

No from x^2

Unrelated to your question

Just factor out x^2 from x^2

[warm charm]

factor x^2 from x*x?

oh

x^2*1/x^2

[humble wedge]

You factored x^2 from 1 here

Basically

factor x from x^2: x(x)
factor x^2 from x^2: x^2(1)
factor x^3 from x^2: x^3(1/x)

now factor x^4 from x^2

What will you get?

[warm charm]

x^4(1/x^2

[humble wedge]

Exactly

In limit to infinity questions

You will factor like this

[humble wedge]

For example here, what do you think is the most factoring you can do from denominator? (Most amount you can factor in denominator is based on the highest degree in denominator)

[warm charm]

x^3(1-x/x^3+8/x^3)

i think i see where youre going

[humble wedge]

Nice

Now factor as much as you can from numerator

What will you get?

[warm charm]

x^4(1-3x^3/x^4+x/x^4)

[humble wedge]

Nice

Now x^4 on numerator and x^3 in denominator

Cancel

[warm charm]

ok

[humble wedge]

Write remaining result

[warm charm]

x(1-3x^2/x^4+x/x^4)

over

1-x/x^3+8/x^3

[humble wedge]

Great

Now apply limit to infinity

[warm charm]

distribute?

well

[humble wedge]

You can distribute too, yes

[warm charm]

numerator gives me 0

no

[humble wedge]

Does it?

[warm charm]

gives me inf*1

[humble wedge]

Yes

Which is inf

[humble wedge]

Now solve this

[warm charm]

oh

thats the answer

mmm

[humble wedge]

What's the answer?

[warm charm]

2/5

[humble wedge]

Good

[warm charm]

ok i think i got it now

[humble wedge]

Now try this one

[warm charm]

how would u factor the numerator

brb 5 mins

[humble wedge]

Basically, we always try to factor as much as we can. It means just right before it becomes zero. Which means right before all of your values become 1/x^n
That's why we say highest degree
Like in x^5+7 we can factor right till x^5, because then it becomes 0 (as in 1/x + 7/x^6)

How much do you think you can factor out the numerator? Also note whatever you did, you can also do inside square root (aka highest degree inside square root)

sqrt(x^2 - 1) can also be written as
sqrt(x^2(1 - 1/x^2)) which is same as x*sqrt(1 - 1/x^2)

Now try solving the question

[pearl pond]

@warm charm Has your question been resolved?

[pearl pond]

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[hard locust]

having trouble figuring out what is going on with the for all n = 2,3?

[hushed oxide]

wdym?

ohhh

what they done is said

okay look, 2=2

3=3

but

3<4

3<5

3<...

to create an inequality

so for every n > 3, let the numerator=3

then uve constructed a sequence that must be smaller than the initial sequence

[hushed oxide]

does that make sense?

[hard locust]

kinda of? so every number greater than 3 is going to have its numerator be 3, infinitely

its like cos(pi) in that way right?

[hushed oxide]

uhh not really

its just more that we are trying to say

we are making a new sequence

that we know GUARANTEED is smaller

like

1 * 2 * 3 * 4 * 5 >= 1 * 2 * 3 * 3 * 3

right?

[hard locust]

i do think i get it but i dont see how this helps solve what the limit is?

[hushed oxide]

okay suppose im trying to prove that

1x2x3x4x5 >= 20

but i suck at multiplication

so i come up with an idea

let me show that something smaller than that is bigger than 20

if the smaller thing is bigger than 20, then the original thing must also be bigger than 20

right?

so i say, 1 * 2 * 3 * 4 * 5 >= 1 * 2 * 3 * 3 * 3

and 1x2x3x3x3 > 20 (54)

so then the original 1x2x3x4x5 is also > 20

similarly, we are finding the limit of a sequence we can actually solve, and since the original sequence is bigger, then it must also be greater

so if the smaller sequence goes to infinity, then the bigger sequence must also go to infinity

now, what is the limit of 2^n?

as n -> inf

[hard locust]

Inf

[hushed oxide]

yes!

[hard locust]

:D

[hushed oxide]

so then we also know. 3^n is what?

without thinking

what is its limit?

it HAS to go to inf because 3 > 2

right?

[hard locust]

Inf

[hushed oxide]

now here, we have bn which is super complicated right

supposed we had an easier sequence

that we knew is definitely smaller at every value of n

like if we could prove that bn >= 2^n

since 2^n goes to inf, where does bn go?

@hard locust where does it go?

[hard locust]

Inf

[hushed oxide]

yes!

now, we have fractions right?

like our sequence

is a fraction

do we know any sequences to do with fractions?

any limits

like a type of limit test that tells us if a sequence goes to a number or Infinity

[hard locust]

P test?

[hushed oxide]

which is what

if u have a fraction to a power

what happens

[hard locust]

If the power is greater than one it's convergent?

[hushed oxide]

uhh

thats for a series

take a^n

when does that converge

or diverge

btw i dont mean to be annoying, but do you mind staying here and ill help?

its 3:20 am

[hard locust]

I got class at 2

[hushed oxide]

so ill finish helping and sleep ahah

when is that?

[hard locust]

K, gotta leave in 15 min tho

[hushed oxide]

ah we can be done

[hushed oxide]

^

when does that converge or diverge

[hard locust]

It diverges if it's in the numerator and converges if it in the denominator?

[hushed oxide]

uhh

more that

if abs a < 1, then a^n converges

for example, (3/10)^ n

right?

but (20/13) ^ n diverges

[hard locust]

Is the power being for the entire fraction on purpose?

[hushed oxide]

yep

[hard locust]

That would diverge then right?

[hushed oxide]

because 3/10 < 1

no because think

3/10 ^ 100 is tiny

that converges

to 0

(3/10)^10000

is tiny

(10/3)^10000 is huge

[hard locust]

Oh I see

[hushed oxide]

right?

so then, if we can prove bn is bigger than a divergent sequence, then it must also be divergent

so we can see that 1/2 < 1

2/2 <= 1

3/2 > 1

4/2 > 1

... > 1

so then we have 1/2 * 2/2 * 3/2 * 3/2 ... <= bn

right?

[hard locust]

Yea

[hushed oxide]

and 1/2 * (3/2)^n

[hard locust]

Got 5 min

[hushed oxide]

and then since (3/2)^n goes to inf, then 1/2 * (3/2)^n goes to inf also

then since bn is bigger than that, it must also go to inf

and boom ur done?

!

[hard locust]

Oh so samdwitch method

[hushed oxide]

maybe think thru again if it doesnt make sense

yes similar!

its called a bound i think

sandwich

now go to class!

gooduck!

[hard locust]

Thx

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[sharp smelt]

tried multiplying and dividing by the conjugate?

[midnight haven]

yep

idk what do next though

[sharp smelt]

series expansion perhaps?

[boreal crescent]

Ye with equivalent

[midnight haven]

tan^2x has an seires expansion ?

[sharp smelt]

tan(x) does

square it

[midnight haven]

oh

Can I rule out this part?

[sharp smelt]

no

also, that's wrong

[midnight haven]

Oops

[sharp smelt]

you can't rule that part out

no

[midnight haven]

It isn't 0 or infinity,so why not?

By rule out I meant substituting x=0

[sharp smelt]

what's cos (0)

[midnight haven]

1

[sharp smelt]

so what's 1+1

[midnight haven]

2

[sharp smelt]

so how can you neglect it

,w series expansion of tan^2(x)

[midnight haven]

Mb

[jolly parrot]

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[midnight haven]

Till x^2 is fine right?

[sharp smelt]

first 2 terms

or even first term should do

[midnight haven]

This might sound dumb but can I sub x =0 here?
Because then I would have a proper 0/0 form

[sharp smelt]

use series expansions

take sin(x)=x

tan(x/2)=x/2

and cos(x)=1

for values very close to zero

these are the first terms of each of their series

should work

:D

[midnight haven]

:C

[sharp smelt]

what

tan(x)=x

[midnight haven]

This would give me 2

[sharp smelt]

not x^2

yes

is that wrong?

[midnight haven]

Yep

Answer is 3

[sharp smelt]

,w lim_x\rightarrow 0 (\sqrt{1+xsin(x)-\sqrt(cos(x))/tan^2(x/2)

[jolly parrot]

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Upgrade to WolframAlpha Pro!

Requested by physicsrocks

[sharp smelt]

wait

no idea then

sorry

[midnight haven]

Okay

btw

@sharp smelt

about the previous question,the one you told me to ask on MSE

i was wrong here

[sharp smelt]

I saw the post :D

[midnight haven]

oh okay : D

[acoustic path]

yes

the whole purpose of multiplying by the conjugate is to get a term that you can remove from the limit

[midnight haven]

How to go furthur ?

[acoustic path]

series expansion works

[midnight haven]

it gave me the answer as 2

but the correct answer is 3

[midnight haven]

is this right?

[acoustic path]

i can't read that, but it gave you 2 then it's wrong

[midnight haven]

Oh

Understandable?

[sharp smelt]

I think you maybe need to take one more term for cos?

maybe

[acoustic path]

yes you do

[midnight haven]

1-(x^2/2) ?

am i right ?

oh

that gives the answer as 3

Thanks!

-3

oh

mb

[midnight haven]

this is wrong

Thanks!

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[severe quarry]

In triangle PRS, 2a=a+ang(SPR)
ang(SPR) = a

How?

Where did they get this 2a = a + ang(SPR) from

Oh

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[visual canyon]

how do I do this kind of thing?

is there a better way than just counting across and guesstimating the two half squares make a whole 7?

sorry if that made no sense

[craggy tendon]

the circle passes through (-3,0) (1,0) nd I'm guessing (0,-6.5)

so you can "probably" find the equation => radius => diameter

[visual canyon]

wait wait

first question

"probably"

why quotes...

[craggy tendon]

since (-6.5,0)

is a guess

[craggy tendon]

this works

[visual canyon]

oh dude smart

see thats how i think its supposed to work

so that would make the correct answer

[visual canyon]

7.21

[visual canyon]

or 2 sqrt 13

thoughts

[burnt dust]

Sounds good to me

[visual canyon]

yep just checked the answer key

thank you that was so smart

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[visual canyon]

i dont fully understand what the focus of a parabola is

is it the point of the parabola touching the line?

is it slightly hovering above like all the diagrams show

wot

[craggy tendon]

um

The directrix and focus defines the parabola,
When you derive the simple parabola and get y^2 = 4ax , you start from the fact that the focus is (a,0)

It's on the axis

Of the parabola

[idle cipher]

is each axis ok?

[craggy tendon]

each axis?

[idle cipher]

idk