help-23

[potent seal]

Even 360-22.6 is another value

[sonic dove]

yeah since the tan has a periodicity of 180 degree you can add negative value with that other wise your solution is right

[potent seal]

Also note that tan(180+33.7)=tan(33.7)

[ashen parcel]

wait so when do i know when to add 360 or 180

[sonic dove]

you need to know the periodicty of a particluar function

[ashen parcel]

this one is 0-360

[sonic dove]

go through this article http://mrseehafer.pbworks.com/w/file/fetch/69682432/252_SMP_SEFST_C04L06.pdf

[safe radish]

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@frosty glacier Has your question been resolved?

[safe radish]

@frosty glacier Has your question been resolved?

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[royal rune]

Find the number of permutations of the set {0, 1} in 10 places with the constraint that there cannot be two consecutive ones

[royal rune]

I know this can be done with a recursive formula but I tried another method which I thought was good but in the end it gave me the wrong answer.I took the even positions with 0 first and then do permutations of the rest. Then we take 0 on odd positions and then take permutations of the rest

so 2^5+2^5

but the correct answer is 144

[peak estuary]

well what about 1001...

fits neither into the first or second option

[royal rune]

oh right

[peak estuary]

recursive was also my first idea

[royal rune]

so the recursive approach is the only way?

[peak estuary]

not sure about only

but probably easiest

I mean in general you can convert a recursive formula into a direct formula

at least pretty often

whether you can easily interpret that formula is a different question

[royal rune]

🙂

thank you

.close

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[sinful glen]

I need to inspect the following Sequences of functions for pointwise and uniform convergence:

f_n(x) = x^n * (1 - x)

Can somebody help me or at least guide me for this exercise?

[sinful glen]

i forgot to precise that f_n is defined on [0,1]

[idle grove]

usually you look for the zeros of the expression first

like in this case when x takes 1 or 0 fn is constantly zero

[sinful glen]

For the pointwise convergence i found that for x = 1 f_n(1) -> 0

and for x = 0 too

[idle grove]

good

the next step is to examine if there is any other numbers that is easy to determine its convergence

[sinful glen]

alright

but how can i find such value?

[idle grove]

this step comes in experience

[sinful glen]

ok

[idle grove]

usually you look numbers that make the expression constant

for example sin(2xpi*n) then taking x=integer you get a constant expression

[sinful glen]

like for my function if i take x=1/2 it makes it constant right?

[idle grove]

why

[sinful glen]

idk 😅

[idle grove]

1/2 ^n * 1/2 ==1/2 ^(n+1)

[sinful glen]

i just took a value

yeah this is not constant my bad

[idle grove]

in this case, the trivial values are 1 and 0

since they make fn==0

[sinful glen]

ok like i showed in the pointwise convergence

[idle grove]

now try to bound the fn for x different than 0 and 1, is it possible?

bounding expression make convergence much easier

for example if you have sin or cos you can probably always bound it by 1 or -1

some other tricks might bound log(x) by x^a for every a>0

[sinful glen]

i'm not sure if i really understand what you mean by bounding the expression

[idle grove]

its like sandwich but having only one bread

[sinful glen]

ok i think i get it

[idle grove]

did you learn about absolute convergence?

[sinful glen]

like 0 always occures in the function between [0,1] and you want me to find another value that always comes right?

[sinful glen]

yes

[idle grove]

absolute convergence is a sandwich

bounding the original expression by its absolute value

sure, upper bounding

[sinful glen]

alright

[idle grove]

were you able to do any bounding? i cant do any

[sinful glen]

if i get it right there's no other x that you can bound in the function fn

[sinful glen]

no i dont find any neither

[idle grove]

ok now its time to think qualitatively

take an arbitrary x between 0 and 1

[sinful glen]

ok

[idle grove]

x^n goes to zero, (1-x) is a constant once fixed a x

[sinful glen]

i agree

[idle grove]

so fn(x) goes to zero

[sinful glen]

so it converges to 0

[idle grove]

pointwise

[sinful glen]

it converges pointwise to zero ok i can still follow

[idle grove]

thats all for pointwise convergence

now for unif conv, if fn converges uniformly then it must converge to its pointwise limit

[sinful glen]

since x^n -> 0 for any x with n -> inf fn converges to 0 for any x

[idle grove]

yes

mb

[sinful glen]

yeah sure since the function is defined on [0,1]

[idle grove]

so for uniform convergence what definition do you have

you want to see if the pointwise limit is suitable for uniform convergence limit

[sinful glen]

i have a definition for uniform convergence but i dont think its the one you want in this case

[idle grove]

Do you unos the difference between ppintwie and uniform?

[sinful glen]

yes

more or less

[idle grove]

Try to say it in daily life way like how would you speak to your younger brother

[idle grove]

This is a hard one

[sinful glen]

For the uniform convergence we want to analyze if the graph of fn is in a tube (defined by epsilon) around the graph f. That means if all the graphs fn are around / near the graph f

[idle grove]

But ir you manage to do it you will be able to learn the true meaning of it

[sinful glen]

Idk how to explain what the graph f is exactly

its very hard imo to explain it easily

That's best representative image i have for this definition

[idle grove]

Good definition

Try to formalize it

[sinful glen]

Ok ok

[idle grove]

Try with supremum

[sinful glen]

For each x : |fn(x) - f(x)| < epsilon with epsilon > 0

Wait with supremum?

[idle grove]

Not very clear

[sinful glen]

I have to admit we did not learn any definition with the supremum for this

[idle grove]

Think the diference between fn and f, taking their supremum

Value rangend in the dominium of fn

[sinful glen]

sup(|fn(x) - f(x)|) < epsilon ?

[idle grove]

whats is its difference to this one? Can you figure it out?

[sinful glen]

The difference is that we only are looking at one specific value that is the supremum instead of all x

[idle grove]

Yes, eventhough they ser the same concept

Are*

[sinful glen]

alright

But how can i find the supremum in the fn function?

more in the fn-f function

[idle grove]

The point is f is zero, so the problema turn into a bounding problem

[sinful glen]

i agree

[idle grove]

And the key idea is you take limit of n

[sinful glen]

|fn(x)| = x^n(1-x)

the limit of n?

[idle grove]

I mean limit of fn when n goes to inf

[sinful glen]

oh yes that i agree

[idle grove]

Taking supremum kills the "for all x"

[sinful glen]

ok

[idle grove]

And taking limit gives you the case when n =infinity

And they combined are uniform convergence

Fn converges to f uniformly when this limit of supremum goes to zero

[sinful glen]

yes but i dont know the supremum or do i?

[idle grove]

You dont know

The point is you dont hace to know

because what is inside supremum goes to zero

so its supremum must go to zero

[sinful glen]

oh

i get it

alright

Thank you

[idle grove]

you can also do it in its standard definition way

[sinful glen]

Sure

Thanks for you time and help it helped me very much to understand this subject better 😄

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[rotund gull]

could someone help me with this question ? I have tried to get a function so i can integrate it and calculate the area

[rotund gull]

not sure where i am going wrong

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@rotund gull Has your question been resolved?

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[buoyant pond]

Maximum of
sinx + sin2x

[buoyant pond]

how do i do without derivatives

derivaties gives me irrational stuff i dont like

can it be done with trigno only

[lone arch]

Without derivatives? Do we really want to avoid them here

[buoyant pond]

yes 4cos²x + cosx -2 is bad

I think

[lone arch]

You don't have -2 there

[buoyant pond]

oh

it is there i thinm

The f'x

[lone arch]

$f(x) = \sin(x) + \sin(2x)$

[buoyant pond]

Ah sorrry

Sin(2x)

really sorz

[lone arch]

Still

[crystal pilot]

are you trying to find a way to do this without differentiating?

[buoyant pond]

yes

[crystal pilot]

kk

[buoyant pond]

because it had irrational stuff

[lone arch]

Ah ok, we have 2x there

[lime dust]

I think you can only put in a range

But not getting the maximum just like this

Mb I’m wrong

[buoyant pond]

you might be right, bro i hv been trying this for too long

[crystal pilot]

gimme 2 mins

i might have something

not 100% sure if it'll work

[lone arch]

,w plot -cos(x), cos(2x) * 2

[lime dust]

Hold my beer 🍺. xd

[buoyant pond]

[crystal pilot]

lmao

yeah no idea

i have a really shoddy method

but nothing good enough

sorry :(

[buoyant pond]

Ah damn its actually 2sinx + sin(2x)

yea sorry from me too

[crystal pilot]

i dont think it will make it any easier

[buoyant pond]

That qn was fcked

[crystal pilot]

you would probably have to differentiate

[buoyant pond]

this ones better

[crystal pilot]

cuz differentiating is really fast

especially with these functions

[buoyant pond]

yee

[lone arch]

That'd give us 2cos(x) + 2cos(2x) = 2(cos(x) + cos(2x)), differentiated

So we just have to find when cos(x) = -cos(2x)

[buoyant pond]

yea

and pi/3 ig

well thNnks

[lone arch]

Yeah, one of them is pi

[crystal pilot]

pi is an inflection point

if you differentiate again

i think at least

[lone arch]

Yep

[buoyant pond]

oh

[lone arch]

pi/3 is good

Even if it's a minimum, we got this

Because 2sin(x) + sin(2x) is symmetric to the origin

(assuming pi/3 actually makes f' 0)

[safe radish]

@buoyant pond Has your question been resolved?

[safe radish]

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[outer horizon]

1

3 root 11

[outer horizon]

Rarionalise

[hard crest]

$\frac1{3\sqrt{11}}$ ?

[flat frigate]

ヘイリー

[outer horizon]

Yes

Do u x it by 11

[hard crest]

multiply by $\f{\sqrt{11}}{\sqrt{11}}$

[flat frigate]

ヘイリー

[outer horizon]

What?

[safe radish]

@outer horizon Has your question been resolved?

[final halo]

Well have you tried multiplying your original expression by it?

[safe radish]

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[rough citrus]

hello, im having some trouble understanding why when solving quadratic equations we can get 2 solutions but only one is correct (in regards to functions)
i apologize if the questions is formulated in an incomprehensible way, im not quite sure how to ask this question

example: x-1 = sqrt (6+2x)

[rough citrus]

i think what im not quite understanding is why, for example, the sqrt of 16 cant be -4

[stoic dune]

√16 is +4 always

√ returns positive numbers

Sometimes we can adjust for when we want to include the negative root by including the ± in front

But they haven't here

[final halo]

If you're trying to solve x² = 16 by square rooting both sides, you need to remember/know that sqrt(x²) = |x|

Bc like kaynex said its a convention that sqrt always gives you the positive root

[rough citrus]

i see, is there an intuitive explanation to why it gives a positive root or is this just a matter of accepting it as a fact

[stoic dune]

Once upon a time, the smart people of the time said "let √ give positive numbers"

It is just a convention that everybody follows

[rough citrus]

hahahaha

i see

i was wondering also

im not sure what the term is in english cuz i dont seem to get any relevant results, but its directly translated to false roots

i dont understand why they appear when working with functions

i can give an example if that doesnt make any sense

[thin bridge]

extraneous solutions

[rough citrus]

i see, thank you

[thin bridge]

info / restrictions canb e lost when applying certain operations

x-1 = sqrt (6+2x)
when squaring both sides of that, to get
(x-1)^2 = 6 + 2x
you get a quadratic equation that has two solutions
but that equation no longer has the restriction from the original equation that x-1 >= 0

[rough citrus]

hmm

and it had that restriction because it was equal to the sqrt of a number, which is positive?

if i understand correctly

[thin bridge]

technically non-negative (to include 0) but yes

[rough citrus]

yes, thank you

ohhhhhhh

[thin bridge]

(and 6+2x would also need to be non-negative)

[rough citrus]

why?

oh

nvm

cuz of imaginary nubmers

numbers

i presume

and all that, havent gotten into that quite yet

ok so if we square, its no longer restricted to non-negative solutions, which entails that we can get negative solutions

what is the issue with this?

[thin bridge]

because the solution to the original equation has the restriction

[rough citrus]

sry if its a stupid question hehe

[thin bridge]

that's the problem

i.e. if you get a value of x where x-1 is negative,
then that isn't actually a solution to the original equation

[rough citrus]

ah i see

isnt this all just based on the convention that sqrt x is a non negative?

[thin bridge]

yes

[rough citrus]

so are there scenarios where it wouldnt be incorrect?

[thin bridge]

but more generally ranges (and domains) of functions

you'll encounter similar stuff when dealing with logs

just need to ensure that they don't conflict with the original equation
there can be situations where all the solutions reached are valid

[rough citrus]

i see

do you know any good resources that i can read that go more in depth on this phenomenon?

[lament bluff]

If you want something higher level next would probably be to look up the fundamental theorem of algebra or pick up a textbook on complex numbers

But I think I can just give a few more clarifying examples and that should suffice for most instances

Imagine I wrote the expression:

Sqrt9 + sqrt4

If you treat square roots as both positive and negative values

Well this expression now has 4 possible values

Maybe the sqrt9 is 3 while the sqrt4 is -2

You can get -3, -1, 1, 3

This is not useful or practical

We want to at least specify one of these

If you want 3, do sqrt9, if you want -3, just do -sqrt9, easy

That's why we have this convention that square roots are always positive, it doesnt say that the other solutions don't exist, we just use it because it's useful to specify one

Now to see why extraneous or double solutions happen, watch this:

1=1
Square both sides
1=1
No problem, equation was true before, and it's still true after, what we want

-1=1
Square both sides
1=1
Oops, was false, now true

The squaring function f(x) = x^2 is what we call an injective but not bijective function

You can't reverse it. You don't know whether this came from 1=1 or -1=1

So remember, every time you square both sides of an equation, you run a risk of extraneous solutions and must check for them.

Now consider the opposite

x^2 = 4

Square root both sides

x = sqrt4
x = 2

We know squaring gives us an extra solution, so square rooting means we LOSE a solution

We forgot x= -2

Remember, every time you square root both sides of an equation, include plus/minus (or absolute value)

This should give a pretty good foundational intuition for how quadratics always give 2 (not always distinct) solutions

[rough citrus]

ahh i see, is it possible to lose the correct solution when sqrt

in the same sense that we can get an extraneous solution when squaring

[thin bridge]

ahh i see, is it possible to lose the correct solution when sqrt
only if you don't sqrt properly

[rough citrus]

properly meaning that i add +- ?

[thin bridge]

starting from
$$x^2 = 4$$
applying sqrt to both sides:
$$\sqrt{x^2} = \sqrt{4}$$
$$\blue{|x| = 2}$$
leads to
$$x = \pm2$$

[flat frigate]

ℝαμΩℕωⅤ

[rough citrus]

ahhhhhh

[thin bridge]

many skip part the blue step and jump to the last line

this is the source of the misconception that sqrt has two outputs

[rough citrus]

i see

thank you both very much, your explanations really helped! 👍🏼

[safe radish]

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[fiery quiver]

May someone help me??

[lunar rampart]

you can use law of cosnies and law of sins

youd get a quadratic you can solve for BC, and then use law of sins to find the angle

[safe radish]

@fiery quiver Has your question been resolved?

[fiery quiver]

So is this valid??

[safe radish]

@fiery quiver Has your question been resolved?

[safe radish]

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[cinder mauve]

<@&286206848099549185>

[lyric wraith]

one sec

[cinder mauve]

Np

[lyric wraith]

sorry im in a lecture rn and im drawing on a mouse

[cinder mauve]

😭😭😭

It’s fine

[lyric wraith]

but the main idea is the washer method

[cinder mauve]

Yes

[lyric wraith]

find the anti deriv and just solve for k

[cinder mauve]

I don’t undeetand

[lyric wraith]

lower and upper if doing horizontal slicing is ( f(x) - 0 ) ^ 2

[cinder mauve]

Understand

Yeah I understand that part

[lyric wraith]

is lower 0 -> 5

[cinder mauve]

Wait

So we solve for x

So we take x out all thay

That*

And zero is down and 5 is up

Am I right?

[lyric wraith]

find the intersections first by letting the functions equal to each other

so y= 3x^3 + 2 = 5

[cinder mauve]

Okay

Then

[lyric wraith]

one sec

i got u

[cinder mauve]

But it’s rotated about y -axis

Thanks fr😭 I hate this section

[flat frigate]

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

[cinder mauve]

Thank u…

But why 2 and 0?

[lyric wraith]

because i did horizontal slicing

since it's rotated around y-axis

you rearrange in terms of x

[cinder mauve]

Yes

But where is the two from?

[lyric wraith]

and then find intersections of both functions,

[cinder mauve]

I have a qs

[cinder mauve]

Do u find the intersection after taking x out ?

[lyric wraith]

you dont take x out

you rearrange in terms of x

[cinder mauve]

Yes what I mean basically

[lyric wraith]

and then solve for y

yea

[cinder mauve]

So after u rearrange in terms of x u find intersections?

[lyric wraith]

yea

[cinder mauve]

Okay lemme try solving

[lyric wraith]

i made a mistake

[cinder mauve]

Why did u take radical 3 out?

Like u didn’t use it

[lyric wraith]

it's the other function equal to each other

[cinder mauve]

Wdym

[lyric wraith]

because 0^1/3 = 0

never mind i did it well

*right

sorry i cannot multi task

[cinder mauve]

Okay

I am gonna try

[cinder mauve]

I don’t understand this part

Nvm

[lyric wraith]

Ok

[cinder mauve]

Anyone

<@&286206848099549185>

[safe radish]

@cinder mauve Has your question been resolved?

[safe radish]

@cinder mauve Has your question been resolved?

[safe radish]

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[toxic goblet]

how do i solve this

[delicate bobcat]

Did you learn l'hopital's rule

[toxic goblet]

no

[safe radish]

@toxic goblet Has your question been resolved?

[shadow glade]

you should be able to use the sin t / t limit to solve this

[safe radish]

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[mortal patrol]

How do i prove this

[obsidian oracle]

what is f?

[mortal patrol]

isnt that the function mentioned there?

the (1+x) part

[obsidian oracle]

?

you're mentioning f^(n)(x) for n >= 2

i'd like to know what f is

[mortal patrol]

no clue that was all that is mentioned

[obsidian oracle]

ok i infered that it's sqrt(1+x)

but isn't it written anywhere?

[obsidian oracle]

like I'd want the full question

this is part b)

isn't there some part a), some introduction?

[mortal patrol]

nope its just an integral unrelated to this

[obsidian oracle]

please can you at least show the full question to make sure?

[mortal patrol]

bet gimme a sec

how did u infer it was square root of 1 + x btw

im just confused cause usually im given a series

and asked to prove by induction

but this one got me baffled dont even know where to start

[safe radish]

@mortal patrol Has your question been resolved?

[safe radish]

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[shadow glade]

because you have (x-3) in a denominator

it would be division by 0

it is a bit confusing at first sight why it isnt an asymptote

[eager vault]

can I say something

[shadow glade]

oh its multiplication between those two fractions, i thought it was subtraction, then yeah that makes sense since its just a removable discontinuity

[eager vault]

it is because you dont have a "=" in ur equation

which is ur path to Y

[shadow glade]

wut?

[shadow glade]

that'

[eager vault]

ow my bad sorry

[shadow glade]

that's just how you enter stuff in desmos

so think about this simpler example:
$\frac{(x+2)(x-3)}{(x-3)}$

[severe pond]

no there are multiple type of discontinuity

[flat frigate]

Soosh

[shadow glade]

that fraction simplifes to (x+2) UNLESS x = 3 right? (in which case it is undefined)

so basically it ends up being a line with a hole in it

right because you no longer have an x - 3 denominator

those two versions are NOT equal for x = 3, if x = 3 then doing the cancelation is not valid because you would be dividing by 0

it isnt, simplifying involves dividing both numerators and denominators by (x-3), that would be division by 0 if x = 3

when you simplify something like this, you would have to make a note that the domain doesn't include x = 3, but the simplifcation is valid for all other real numbers

youre not allowed to cancel out (x - 3) with (x - 3) for x = 3

so if you DO cancel them out while working on a problem, you have to make note that $x\neq3$

[flat frigate]

Soosh

[shadow glade]

its a subtle point that many people gloss over

yeah its R - {0, 3}

the point is that 2nd representation is only valid for x not equal to 3

becaue if x were equal to 3 in the original representation then you would not be able to carry out the "simplifcation" because that is division by zero

but if you start out with the second version you don't have this problem, because you didn't get to it from somewhere else

$

not sure how else to explain it 😄

its not the same equation though 😛

yes

pretty much

ok lemme give you another example:

say i asked you to solve this problem:

$\frac{(x-3)(x+2)}{x-3}=5$

[flat frigate]

Soosh

[shadow glade]

what is x?

but does that fulfill the original equation?

right, so 3 is not a solution, because it is actually undefined

what you did was cancel out x - 3 with x - 3 right?

what?

riight

so when you solved though, did you cancel out the top and bottom x - 3 to "simplify" to solve?

right, but the problem is you didnt at the time know the value of x

so you werent aware that you are dividing by 0 by canceling out the top and bottom of the fraction

so that "simplifcation" is ONLY valid if the quantity you are canceling isn't 0

dividing by 0 causes problems and it's why you got a seemingly "Valid" solution to the equation, but that isn't in fact valid

same idea as your original problem

so when you have equations like these when some number makes the equation undefined, it's ok to do cancelations, but you would note somewhere on the side of your paper (or in your brain if you can keep track) that x = 3 is undefined for this equation, so the simplification youre making is only valid if x is not equal to 3, if you end up getting a solution like x = 5 at the end, then no harm no foul, that doesn't conflict with the domain, but if you DO get 3, then oh shit, the simplifcation wasn't valid and seems like there is no solution

$\frac{(x-3)(x+2)}{x-3}=10$
say i change the equation a bit

[flat frigate]

Soosh

[shadow glade]

now ok we make note x isn't 3 and cancel top and bottom (x-3) and we get (x + 2) = 10, then x = 8, if we go back to the original equation, it works just fine

exactly!

haha love to see that "click" moment

it's fine, actually great to see a willingness to understand and trying hard to, thats what matters

this is a subtle point and a lot of people gloss over it completely

and will get questions wrong when the solution IS that number that isn't in a domain, or just completley not notice it if it isn't 🤷‍♂️

and then be confused when they get questions wrong, and probably still not learn it

but yeah, the domain restriction carrying over is a good way to phrase it actually, you can see why if you start with this:

there was never a domain restriction to carry over

for x = 3 anyway

the technical name for these 1 point holes in the graph by the way is a "removable discontinuity"

youll hear that in calculus classes

!done

[safe radish]

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[stable sedge]

hi i am not sure if i did this problem right cuz the answer is supposed to be 14.4m: Stephanie is 117cm tall, and her eyes are 106 cm off the ground. She is using her new binoculars to look at the bird that is perched in a tree. The angle of elevation is 28 degrees, and Stephanie 25.0m from the base of the tree. What is the height of the bird in the tree?
This is what i drew and i got 13.3m...

[stable sedge]

thanks in advancee

[thin bridge]

where are 28 and 62 coming from

where's 20.6879 coming from

[stable sedge]

oh

28 is the elevation i subtracted 180-28-90 = 62, i used sine law

like sin28/11 = sin62/x = x = 11sin62/sin28 = 20.6879... cm

but i dont know if thats the right diagram

or if i even did this right lol

tested it three times and i got 13.3

[thin bridge]

why are you using 28 for elevation

question says 24

[stable sedge]

oh i wrote wrong

mb its 28

😭

[thin bridge]

diagram is also misleading

[stable sedge]

wdym?

[thin bridge]

in what you drew, you implying that the bird will be around the height of steph

in fact the height of steph (and the 11cm difference) doesn't actually matter here

[stable sedge]

oh trueee

rlly?>

how would u do it without a side length?

[thin bridge]

try redrawing the diagram

[stable sedge]

okk

[stable sedge]

like this?

[thin bridge]

yes

[stable sedge]

ok hmm

oh

[thin bridge]

sine law is also unnecesasary as you have a right triangle

[stable sedge]

oh

i kinda forgot bout those lol

lemme see

oh tan

[stable sedge]

WHAT

i still got 13.3

💀

tan62 = 2500/x

[thin bridge]

dunno why you didn't just use the given 28° angle

but this will get you the height above eye level

[stable sedge]

i am finding the height of the bird?

wouldnt that be the bird?

[thin bridge]

get you the height above eye level

height of the bird would be height above ground level

[stable sedge]

oh the height of the bird is the height of the tree???

ok icic

[thin bridge]

not necessarily

[stable sedge]

ok true

well the height to the branch

[thin bridge]

making stuff unnecessarily complicated
height of the bird is the height of the where the bird is sitting

[stable sedge]

yes

sorry for the hard timee

thank you tho

.close (sorry again)

[safe radish]

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[thorny star]

Why do I need to scale the left matrix to look like the right matrix when I can already solve for x1, x2, and x3? I know I need to do that to get it right I just dont know why

[hard crest]

as a human, you can "abort" this whenever you want, but ultimately you're going to multiply 6 by 2/3 to get your answer

[thorny star]

I know the scale, but WHY are we scaling when the main goal is to get x1, x2, and x3 which I can do without scaling

[hard crest]

how are you planning to get x3

[stoic dune]

I imagine the goal isn't necessarily to solve, but to get rref.

[thorny star]

pearson kinda fucks this but that fourth column is actually what the row equals

wouldnt x3 just be -3/2

[hard crest]

nope, sure wouldn't

[thorny star]

instructions just say "solve the system"

hmm

[stoic dune]

On the left matrix, the bottom row reads:
-3/2 x3 = 6

From there, how do you get x3?

[hard crest]

$-\f32x_3 = 6$

[flat frigate]

ヘイリー

[thorny star]

o shit

i did the same steps for another problem, without needing to do any scaling or fancy tricks and im so confused on why

but yeah that -3/2 is attached its not actually x3 rip

[hard crest]

correct

you'll note the one on the right says

$1x_3 = -4$

[flat frigate]

ヘイリー

[hard crest]

ie x3 = -4

[thorny star]

OH

[hard crest]

so when we get 1 in the appropriate spot

we can just read off the solution

[thorny star]

hence why we basically require a 1 or a 0 to get started

so we can actually know for sure wtf that variable is

[hard crest]

yeah, 1 in the variable we care about and 0 everywhere else

this is just adding and subtracting equations, but without writing down the variable names

[thorny star]

linear systems but in matrix form

[hard crest]

yeah

[thorny star]

its the format thatll take some getting used to.

since im new and all

[hard crest]

and yes pearson should probably have a vertical line in there, but it's very common to not bother writing that

[thorny star]

Wouldn't this be wrong?

wait no thats right

[hard crest]

why do you think that's wrong?

[thorny star]

thats a part of the system i did last time using the same method im doing with the new problem

but it worked that time

[hard crest]

yeah

you could have also scaled the row by 1 / -2.5

(is effectively what you did)

[thorny star]

ill be honest im just now learning scaling and i have no idea of anything over it

[thorny star]

but for this all i saw was the -2.5 was attached... so i divided

[hard crest]

it's just "multiplying both sides of an equation by a number"

[thorny star]

any number?

[hard crest]

yeah

like if you had um

2x + 7y = 14

you could divide everything by 2

and get

x + 7/2 y = 7

[thorny star]

ah, so that troublesome 2 is now a 1 we can leave

[hard crest]

it's a 1, so we can leave it for now and we'll clear out the 7/2 on the way back

it also lets us easily scale it up to match others

like if down below there was an equation with like -5x in it

we could scale that first one by 5

and add them and clear it out

[thorny star]

so we can just keep scaling it up and down to suit our needs multiple times over?

[hard crest]

yep!

[thorny star]

useful!

still a bit conflicted on the 2 matrices i posted and why they must use different methods

[thorny star]

is it because this we cant just divide by -3/2?

[hard crest]

we can divide by -3/2

or we can multiply by -2/3

they do the same thing

you can use either method for either matrix

the nice thing about row reduction (like you're doing) is that you just kind of do the same steps every single time

just with different numbers

the only time it gets weird is if you end up with a row of all 0s

[thorny star]

why the fuck do i have written down -1.5x = 6 to x3 = -12

[hard crest]

dunno, that seems wrong

[thorny star]

okay i dont know where my brain cells were 10 mins ago but im understanding you

what im doing is just straight up dividing which is ok

scaling seems to be the opposite, but still has the same result

ofc in other problems i cant do what i just did and must scale to properly be able to eliminate

does that sound right

[hard crest]

"scaling" is just multiplying so yes it'll be the opposite of dividing

ie dividing by 2 = scaling by 1/2

[thorny star]

yep!

do you happen to know why we can just scale 1 system anything we like?

[hard crest]

it's just multiplying both sides of an equation by a number

like if you have um

[thorny star]

ik how to do it

[hard crest]

$\gsq = \bsq$ then we can say that $7\gsq = 7\bsq$

[thorny star]

we cant do that with a normal equation is my point

[flat frigate]

ヘイリー

[thorny star]

oh wait, what you can do to one side you can do to the other

[hard crest]

yeah we can

yep

[thorny star]

so these systems are really entirely seperate equations

hence why we can just shuffle them around and do calculations on 1

[hard crest]

yeah! they act as separate equations and still follow all of the rules associated with those, but they also can be slammed together when convenient :3

"slammed together" meaning added together

[thorny star]

last question, can you explain replacement?

[hard crest]

yeah this is the "slam together" thing i was talking about

basically when you're solving? you start with 3 equations and then you produce many many more

(2 here, sometimes more sometimes less)

like in your original problem it was 3. anyway

you start with 3, and you can quickly produce a lot more by scaling and stuff

and adding together two equations

but it turns out that if you do that, your equations will be redundant

that is, you'll start saying the same thing in more than one way

you'll have extra equations that aren't helping

so we can throw the original ones away and keep the simple ones

so in the case of your image

we start with the equations
\begin{align*}
2x + 4y &= 8 \
1x + 0y &= 9
\end{align*}

[flat frigate]

ヘイリー

[hard crest]

and we start by scaling the second row by -2, so we add $-2x + 0y = -18$ to our pile of equations

[flat frigate]

ヘイリー

[thorny star]

why -2 specifically

[hard crest]

because of what's about to happen ;)

[thorny star]

what happened was the zero swapping sides

[hard crest]

we then add that equation to the first row, so:
\begin{align*}
2x + 4y &= 8 \
(+) -2x + 0y &= -18
\end{align*}

[flat frigate]

ヘイリー

[hard crest]

and now we can add these two equations together, adding the left side and the right side

[thorny star]

wait are we combining from top to bottom instead of side to side now

[hard crest]

yeah - this is a somewhat difficult thing to understand why we're allowed to do that

it again comes back to "what you do to one side you have to do the same thing to the other"

[thorny star]

thats why i mentioned this because i knew these systems interacted with each other but i didnt know how

[hard crest]

so instead of doing exactly the same thing, we do something that's equal to the same thing

[thorny star]

you chose the -2 due to the 2 in row 1 colm 1 correct

[hard crest]

yeah exactly

[thorny star]

do we do that with whatever we want to become 0? pick the opposite of the number above them

[hard crest]

yep! because you can see that the 2x and the -2x cancel each other out

[thorny star]

why was a 0 so critical in the bottom left here?

or is it just an example lol

[hard crest]

it's just an example, but the convention is to put 0s into the bottom left of the matrix

as much as possible

just lets you read them off in order

[thorny star]

ik you want to have an L shape of 0's in the bottom left, tho ive been seeing sometimes where you want it to be below all leading numbers

havent studied that far though

[hard crest]

L or really more of a triangle

but yeah so once we add those two equations we get yet another equation

4y = -10

so if we list out all of the equations we've seen so far

\begin{align}
2x + 4y &= 8 \
1x + 0y &= 9 \
-2x + 0y &= -18 \
0x + 4y &= -10
\end{align}

[flat frigate]

ヘイリー

[hard crest]

that's a lot of equations, we started with two!

it turns out that we could pick just about any two of those and still have the same amount of information

[thorny star]

yeah its the same just written differently.

altho from the 4 only 2 of those are unique right

[hard crest]

yep exactly

but the simplest, safest thing to do is to keep the one we created (that's why we made it)

and just throw away one of the starting points, whichever one is convenient to get rid of

[thorny star]

why?

[hard crest]

why what, why throw something away?

[thorny star]

i dont understand the utility. Why are we chucking the new one if its the same as the old one in terms of the variable we'll need?

[hard crest]

oh we keep the new one

[thorny star]

actually why do this at all lol is it to swap the 0 to another side

[hard crest]

yeah (and i hate that they gave you an example with a 0 in it to start)

we're trying to get a 0 into the bottom left of the matrix

[thorny star]

im never gonna have a 2x2 cute matrix in any exam so i dont really care for this example

okay i think i get the utility.

[hard crest]

yeah usually 3x3 or 4x4

but the same principle applies

[thorny star]

say we're like

0 4 3
2 0 3
0 0 4

our issue would be that 2 in row 2

if we use replacement that 0 in row 2 can be swapped with the 2

[hard crest]

hmm
well, in this case we want to start with the top left

yeah we'd have to swap some rows around

the top left needs to be nonzero

[thorny star]

oh?

i made that up but i didnt think it mattered

needs to be a 1 right

[hard crest]

so our goal is to get something like this

1 0 0 | 7
0 1 0 | -3
0 0 1 | 5

[thorny star]

yes!

lets say that

[hard crest]

that would be the destination

[thorny star]

1 0 0 | 7
1 0 0 | -3
0 0 1 | 5

we need to fix row 2 number 1

can we swap that over with the 0 next to it

using replacement

[hard crest]

nope, can't swap horizontally

why not? because you're mixing variables

remember what this says when we stick the variables back in

1x + 0y + 0z = 7
0x + 1y + 0z = -3
0x + 0y + 1z = 5

or, in other words, x = 7, y =-3, z = 5

it's a solution to the system of equations

[thorny star]

i think i got the wrong idea from this shitty example

[hard crest]

i agree i hate that example

[thorny star]

is the fact that the 0 got swapped to another side by design or accident

[hard crest]

swapped on purpose? no, usually there isn't one to start, but there will be one at the end

would you like me to make an example?

[thorny star]

please

[hard crest]

2x + 5y = 19
x - 2y = -4

there are going to be fractions :P

[thorny star]

yum

welp first we do -2!

wait no

[hard crest]

yeah no not -2

[thorny star]

we would do -1/2

[hard crest]

hmm i'd say just 1/2

we're trying to turn that 2x into just x

[thorny star]

i thought we were getting rid of that x

row 2

[hard crest]

oh if you want to do it in that order that works

[thorny star]

which am i meant to choose

[hard crest]

looks like your book is doing it in the order you suggested

[thorny star]

fuck the book, do you prefer starting at the top?

because that seems to be the trend

[hard crest]

yeah usually, it's nice to have 1s on the diagonal so you can scale easily

[thorny star]

thats a wonderful idea, lets do it

[hard crest]

ok :)

so let's turn that 2x into x

and the only thing we can do is scale

(we're only concerned with that first eqn for right now)

[thorny star]

yep!

[hard crest]

so what's the appropriate scale factor?

[thorny star]

to turn that into a 1 we can scale by 1/2

[hard crest]

great

so what do we end up with?

[thorny star]

x + 5/2y = 19/2
x - 2y = -4

you know i should probably keep my fractions and decimals consistent

[hard crest]

yeah probably lol

fractions are usually preferred

[thorny star]

see i always see that in calc but i never liked em

because i suck at simplifying

[hard crest]

it avoids the repeating decimal issue

[thorny star]

ah. right.

what would 9.5 be as a fraction

[hard crest]

well you started with 19 and divided by 2

so 19/2

[thorny star]

or i can just do 19/2

okay, just seems i couldve simplified but seems not

[hard crest]

great so the first equation is done for now

now we look at the second one

[thorny star]

we need something on the second to be a 0

[hard crest]

yeah, ideally the x position

we can't scale to get that

[thorny star]

i know the x would be easier, but is there any other reason thats ideal?

oh

[hard crest]

again, we want the 0s to be in the bottom left

[thorny star]

that matters for a 2x2?

[hard crest]

just means that you read off the solution top to bottom x to y

so, no? but it matters for bigger ones and i don't want to step by step go through a 4x4 with you haha

so we can't scale to get rid of that x, so our only option is to add another equation

[thorny star]

aye aye captain

wait cant we just multiply the whole equation by 2

im dumb

[hard crest]

we definitely can do that, but what would it help?

[thorny star]

sorry was thinking about

when you multply by the top and add to bottom

i still need to learn how this all meshes

anyways

[hard crest]

yeah so to get rid of that x, we really need to find or create a -x

easiest way to do that is to take that first equation and multiply by -1

[thorny star]

multiply just the x on row 1 by -1 right?

[hard crest]

the whole thing

remember, balanced equations

but, you can hold onto the row 1 that you have

[thorny star]

oh yes correct, sorry

[hard crest]

just write this new thing as row 3

[thorny star]

why when i can replace?

[hard crest]

because we worked hard to get that 1 in the top left corner

[thorny star]

waiiiit what you mentioned before makes sense ive just never kept the old equation

[hard crest]

you could scale it and then scale it back

just more effort

[thorny star]

okay ill listen do you want the new equation in the middle of the system

[hard crest]

hmm, let's put it at the end so that "row1 " and "row 2" still make sense

[thorny star]

x + 5/2y = 19/2
x - 2y = -4
-x -5/2y = -19/2

[hard crest]

wait what happened to the -x

should be
-x - 5/2 y = -19/2

[thorny star]

man i keep trying to just cancel the thing

[hard crest]

that's this step ;) we're going very slowly on purpose here, you'd do this all in one go with practice

so now we can add rows 2 and 3 together

[thorny star]

i love fractions

[hard crest]

yeah sorry

[thorny star]

x + 5/2y = 19/2

• 4.5y = 13.5

[hard crest]

you and your decimals

x + 5/2 y = 19/2
- 9/2 y = -27/2

[thorny star]

hehe

awe

[hard crest]

and it looks like you ripped that negative sign from both sides which is fine

ok so now that the x is gone from row 2

we can deal with the y column

we want that to be 1y

[thorny star]

hold on for a second

cant we get what y equals now?

[hard crest]

we can and we're about to

once we get 1y = (thing)

then, well, we've identified what y equals

[thorny star]

ah yes, solve for y in row 2

[hard crest]

yeah effectively

[thorny star]

y is 3/4

or in my language, 0.75

[hard crest]

i don't think that's right

[thorny star]

did i google the wrong

sigh

im sorry give me a minute

its 3

not using my calculator just google and forgot about parenthesis

[hard crest]

yep, and if we want to do it by scaling we'd multiply both sides by 2/9 to clear the fraction

so $\frac92y \green{\cdot \frac29} = \frac{27}{2} \green{\cdot\frac29}$

[flat frigate]

ヘイリー

[hard crest]

cross cancel and you're left with y = 3, as expected

so now what are we looking at

x + 5/2 y = 19/2
y = 3

[thorny star]

x + 5/2(3) = 19/2

[hard crest]

you could do that

or you could look at it another way

we can scale the bottom equation and add it to the top one this time

[thorny star]

you prefer that over shoving a variable in?

[hard crest]

since we have this pinpoint cannon that can eliminate y's wherever it sees them

[hard crest]

it's more structured, less fiddly

don't have to write out x's and y's all the time

ultimately i like it because this is how a computer would do it and it's always the same steps evry time

ofc you're free to do either way

this one seems like more work for a 2x2 and tbh it is, but for large ones this is much faster

[thorny star]

im asking so i get explainatory answers like youre doing rn

im not stubborn or dumb enough to keep this elementary algebra that im doing to the end

always nice to hear from someone knowledable

[hard crest]

yeah the less i have to constantly write x every line the better

[thorny star]

x + 5/2 y = 19/2
y = 3

lets scale

[hard crest]

yep! which one do you think we scale and by how much

remember our target to eliminate is that 5/2 y

[thorny star]

honestly no clue. 1/3? 2/5?

[hard crest]

-5/2 :)

[thorny star]

is that 3 ever going to be used up there

[hard crest]

yep it'll be important right now

[thorny star]

so we just flip the sign to scale

[hard crest]

oh, leave row 1 alone for rn

[thorny star]

right yes this is seperate

[hard crest]

scale row 2 (and write the scaled version as row 3)

[thorny star]

does the 5/2y go away

or just become negative

[hard crest]

it will in a sec, that's the goal

so if we take row2 and scale it by -5/2

we get
-5/2 y = -15/2

[thorny star]

thanks dont got my calculator so simplifying is a mess

[hard crest]

yeah a bit but here

x + 5/2 y = 19/2
- 5/2 y = -15/2

[thorny star]

x + 5/2 y = 19/2
y = 3
-5/2 y = -15/2

[hard crest]

the 5/2y bit cancels out entirely, and then you just have 19/2 - 15/2

[thorny star]

ah yes we add just like before.

[hard crest]

yep!

[thorny star]

do we just pull that y = 3 out?

so we can focus on adding

[hard crest]

(i'm deliberately scaling by negative numbers and adding here because i find it easier to work with. many people will subtract equations instead , but i often lose negative signs that way by mistake)

yeah that's off to the side

it's very useful and we love it but we don't need it right now (since we have its scaled version)

[thorny star]

yeah and what we're adding atm is basically the same equation just expressed differently

[hard crest]

yep

[thorny star]

y = 7

[hard crest]

that doesn't make any sense

sorry i mean like

we already know y = 3

and we're trying to get rid of the y here

[thorny star]

i subtracted 19/2 by 15/2

[hard crest]

and got?

[thorny star]

ah

x = 7

[hard crest]

19 - 15 is not 14

[thorny star]

i should really get my calculator x.x

x = 2 im so sorry

[hard crest]

gotta practice those math minutes

yeah great so now we have x = 2

and if we look at what we did in a matrix it was like

2  5   19
1 -2   4

1  5/2  19/2
1   -2  4

1  5/2  19/2
0  -9/2 -27/2

1  5/2  19/2
0   1    3

1  0   2
0  1   3

those were each of the steps

see how we ended up with the identity matrix (1s on the diagonal)

[thorny star]

do we always want every number on the matrix to be 1 or 0

[hard crest]

yeah we want one number per row to be 1, and the rest to be 0

[thorny star]

well.. for a 2x2

but yeah i get it.

[hard crest]

for a 3x3 it's the same

[thorny star]

we want the ability to solve for variable

[hard crest]

yeah

really we want it to tell us the variables

[thorny star]

for a 3x3

1 0 0
0 1 0
0 0 1

[hard crest]

yeah with the last column being the variable values

[thorny star]

short me of practicing more, thats all the questions i have

youre phenomenal thank you so much

[hard crest]

this is an 18 minute video of someone going through these steps for a 4x4. I'll be hoenst and say that I haven't watched it all the way through. It looks like he does the 1s in the diagonal but doesn't put 0s in the upper right part

which is fine, as you say it's pretty easy to solve at that point

https://www.youtube.com/watch?v=4VAXv6yRULU

[thorny star]

toss it my way anyway

once i get adept at this itll turn more fun

tho ive had a lot of fun just going thru this with you

[hard crest]

yeah practice it! and it's okay to do it by substitution as well if that's faster, or reorder the equations first if you think that'll be faster

you'll get good at planning and routing

it's like a speedrun :P

we were very mechanical, but with a little different strategy we could have avoided all the fractions

[thorny star]

what you were doing is what i needed to know, knowing how to bypass steps is only good if you realize it

thanks again, have a wonderful rest of your day

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[pseudo forum]

is this correct so far

[pseudo forum]

or do i keep the masses

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[loud ridge]

how do I prove this statement? when tiling a rectangle size a*b with 1 *2 rectangular tiles, there will always be a line (shown in pink) that divides the rectangle into two halves without crossing any 1*2 tiles in half.

[safe radish]

@loud ridge Has your question been resolved?

[loud ridge]

so far I've been trying to prove it by contradiction but to no avail
these are some of the facts I've noticed:
there are (a-1) vertical pink lines and (b-1) horizontal pink line, in total a+b-2 lines
to block a pink line we need to position a tile such that the pink line crosses it in half, so to block all of the pink lines we will need to fit a+b-2 tiles inside the rectangle which takes up (a+b-2)*2 of the area
in order to tile the rectangle the area needs to be even, that means either a or b (or both) needs to be even

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@loud ridge Has your question been resolved?

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@loud ridge Has your question been resolved?

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@loud ridge Has your question been resolved?

[jolly shard]

I think this can be proven false if you mean in exactly half

[jolly shard]

I have an 8x8 that has no pink divisors

[safe radish]

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[balmy marlin]

how arent both 3x

[tardy mango]

$e^{3\ln x}=e^{\ln x^3}=x^3$

[flat frigate]

Civil Service Pigeon

[balmy marlin]

how did 3 just become an exponent there

is it a log rule

[plucky elk]

,tex .log rules

[flat frigate]

dr. matlab plot

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[dapper ledge]

can someone help with the bounds calculate the integral of f over the region
f(x,y,z) = z(x^2+y^2+z^2)^-3/2
over the part of the ball x^2+y^2+z^2 <= 16 defined by z >= 2

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[dapper ledge]

.close

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[timid mason]

I have a question regarding asymptotes:

Let f(x) and g(x) be the two functions that are shown in the image.
Divide f(x)/g(x) and sketch the graph.
Mark all asymptotes.
I've marked the asympotes as vertical lines that pass through the points where g(x)=0

My question is:
Should x=0 be considered an asymptote if the functions are only limited to what you see on the image (i.e. the functions stop at the end of the visible coordinate system; they dont go to infinity)?