help-42

[devout meteor]

okay so after i split it haw do i know what terms are "r" or "k" im a little confused

[manic vault]

r is common ratio

[plain bone]

k is the starting n value of the summation

[devout meteor]

okay so in this case n=1 so k=1

[manic vault]

This is more generalized

[plain bone]

✅

[devout meteor]

okay so i have this and now i'm supposed to plug in 1?

[manic vault]

To get the a value yes

[plain bone]

Well, you still cant directyl plug in 1

[devout meteor]

wait so what should i do?

[manic vault]

^

[manic vault]

^

[devout meteor]

i dont know what this is😩

[manic vault]

It is formula for geometric series (infinite)

$\sum_{k=0}^{\infty}ar^k=\frac{a}{1-r}$

[potent lotus]

mathisfun

[manic vault]

however this works for any lower bound k=(number)

So really it's $\sum_{i=k}^{\infty}r^i=\frac{a}{1-r}$

[potent lotus]

mathisfun

[manic vault]

For k being some arbitrary number

@devout meteor Basically just identify the first term (by plugging in the lower index) then the common ratio and plug it into formula

Red circled thing is the lower index if you did not know

[devout meteor]

so i would just do this

or is that just wrong... becasue i really have no idea what im doinf rn

[plain bone]

You need to split it up into two series

[manic vault]

No

You only evaluated it at n=1

[plain bone]

$$\sum \qty(a_n + b_n) = \sum a_n + \sum b_n$$
In this case, $a_n = \frac{3^n}{11^n}$ and $b_n = \frac{7^n}{11^n}$

[potent lotus]

King Leo

[manic vault]

Very

[devout meteor]

okay so after i split it do i still plug in 1?

[plain bone]

Still not yet- show what you got after splitting the series

[plain bone]

Now, use $\frac{x^n}{y^n} = \qty(\frac xy)^n$

[potent lotus]

King Leo

[plain bone]

This will make each series conform with the $r^n$ format

[potent lotus]

King Leo

[manic vault]

At very end

When you use fomrula

[calm coral]

@devout meteor Has your question been resolved?

[devout meteor]

sry im at work and had to step away but now i did this

[plain bone]

And now, what is r and k (r will be different for each series)

From there, you can use this formula

[devout meteor]

umm is r 3/11 and 7/11?

[plain bone]

❌ did you use [this formula](#help-42 message) properly

[devout meteor]

umm i think so

[plain bone]

So r should be the full value thats being raised to the power of n

[devout meteor]

okay ya

[plain bone]

Ie the ||full fraction||

[devout meteor]

does this look right?

[plain bone]

Once you start using the formula, you should remove the series

It should just be $\frac{\qty(\frac{3}{11})^1}{1 - \qty(\frac 3{11})} + \frac{\qty(\frac 7{11})^1}{1 - \qty(\frac 7{11})}$

[potent lotus]

King Leo

[devout meteor]

okay

@plain bone so i have 3/8 + 7/4

[devout meteor]

which is just 17/8

[plain bone]

✅

[plain bone]

✅

$$\sum_{n = 1}^\infty \frac{3^n + 7^n}{11^n} = \frac{17}8$$

[potent lotus]

King Leo

[plain bone]

,w \sum_{n = 1}^\infty \frac{3^n + 7^n}{11^n}

[potent lotus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by .king.leo.

[plain bone]

!done

[calm coral]

If you are done with this channel, please mark your problem as solved by typing .close

[devout meteor]

so in general when im doing series i just try to get it to look like r^n

[plain bone]

Series can take many different forms

This specific series is known as geometric series

But there are many more. Telescoping series comes to mind

[devout meteor]

okay thanks

.close

[calm coral]

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Closed by @devout meteor

Use .reopen if this was a mistake.

[calm coral]

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[iron ore]

Is this correct? ( I know we don’t know the exsact amount it moved , it’s not really the point of the question to make an equation I just wanted extra practice )

[swift dragon]

,rccw

[iron ore]

I’m so confused where i went wrong for odd

[swift dragon]

,rccw

[iron ore]

For odd graphs the x value stays the same and the y is opposite aka a reflection over the x axis

[inner zealot]

My only thought is that the graph is slightly off.

[iron ore]

Wdym

[inner zealot]

f(-5) = 8, therefore f(5) = -8. On your graph is closer to 10.

[iron ore]

But in the photos they are in completely different quadrants

[inner zealot]

The definition of an odd function is

f(-x) = -f(x).

So if f(-5)=8, then f(5)=-8.

[calm coral]

@iron ore Has your question been resolved?

[iron ore]

Yea I know

Wait

I’m so confused

I thought for even graphs the x values are opposite and the y are the same and for odd graphs the x values are opposite and the y are the same

I’m panicking en before if that’s wrong I’m going to fail my test

Because I’ve been using that logic for the whole unit

[inner zealot]

Nope. If you swap the x-values, the y-values also swap.

[iron ore]

I’m so confused

[inner zealot]

f(-x) = -f(x)

[iron ore]

So are they both opposite

Wait

[inner zealot]

Let's say

f(-4) = 3

then if f(x) is an odd function, you can determine that

f(4) = -3

[iron ore]

I’m so confused

[inner zealot]

So yeah, the x and y signs swap.

The easiest odd function is y=x if that helps to think about it.

[iron ore]

But I swore my teacher said we can tell if a graph is an odd graph if the y values are

Opposite

And the x stayed the same

[inner zealot]

x stays the same in the sense that

f(-x) = -f(x)

where x it technically the same, but the actual value passed to the function is different.

[iron ore]

Wait

I’m sorry

I have no idea what u mean

How is it the same if it is now negative

That doesn’t make sense

[iron ore]

I don’t unfestand those equations

Because now I can’t do the second part of the unit if what I said is untrue

Because how would I determine if a graph is even or odd

[inner zealot]

So you are probably used to seeing functions of the form f(x), correct?

[iron ore]

Yea

[inner zealot]

So that x arguments is NOT set in stone. You CAN change modify the argument passed to the function. This shows up as a horizontal shift in the graph.

f(x - 3) shifts a graph to the left three units.

If you had a function,

f(x) = x

and used the function declaration

f(x - 3) = x,

then that would show up as

f(x) = x-3.

It's what actually gets passed to the function.

So you can pass the argument -x to a function f(x) as f(-x) and all that does is flip a graph horizontally.

For even functions, this will not visually change a graph.

[iron ore]

I’m really panting

But then how would I do this now if my rule does not apply

Panicking

[inner zealot]

To add on to what I commented about passing different arguments, if

f(x) = x

and you declare the function

f(x+y)

that would equal

f(y+w) = y+w.

I merely passed the argument y+wy into the variable x.

[inner zealot]

So f(x) is declared to be an even function. That means you can apply the identity

f(x) = f(-x).

You are asked what the value of f(14) is but the value of f(x) for x=14 is not given in the table. You can use the aforementioned identity to presume that

f(-14)

is also equal to

f(14).

You know what f(-14) is, that is given by the table.

[iron ore]

Wait so I put 14 into the expression so f(14)=f(-14)

[inner zealot]

Correct.

[iron ore]

But I don’t understand what I do with that info what does it mean gh find ((14)

[inner zealot]

So you should look on the table for x = -14 and see what f(x) equals.

[iron ore]

But what happened to the 14

Im sorry I just don’t get this

I thought if SOMTHING is even it’s equal on both sides

If there is opposite values on either side that means it’s opposite

[inner zealot]

Let me rephrase what you said. If a function is even, it will have the same y-value on both sides.

[iron ore]

Yea

And opppsite x values

But there’s no postive 12 in the table

[inner zealot]

The question asked about f(14) so it is asking about x = 14.

[iron ore]

There’s no 14 either

[inner zealot]

Exactly, but because f(x) is an even function, you can also use x = -14.

That IS on the table.

[iron ore]

Ohh

So it’s 8

[inner zealot]

Correct.

[iron ore]

It says it’s wrong

[inner zealot]

It's -8.

[iron ore]

Wait

I thought we where changing only one value

[iron ore]

I’m so confused

[inner zealot]

The top row shows the values of x for which there is a given value for f(x).

There is no data for x = 14, but because f(x) is an even function, you can use the identity

f(x) = f(-x).

That infers

f(14) = f(-14).

The latter of which is given by the table.

[iron ore]

But then how would we do this for an odd

The x is opposite and the y stays the same I thought.

[inner zealot]

No, that is an even function. For an odd function, x and y both swap.

f(-x) = -f(x).

Let me make an example real quick.

[iron ore]

Wait for odd the x values r the same and the e y is opposite

So would it be 10

[inner zealot]

Ok, as I mentioned earlier, the identity for an odd function is

f(-x) = -f(x)

You are given f(-13) whose x value is not given, x = -13. It's corresponding value, howeer, is given by x = 13. You can use the aforementioned identy to determine that

f(-13) = -f(13).

f(13) is given by the table, all you need to do is negate its value.

[iron ore]

I’m so confused

I don’t understand those equations

[inner zealot]

No. You are looking at the wrong part of the table. The values for x are given in the top row and its corresponding f(x) value is given in the bottom row.

f(-10) = -13
f(-3) = -11
f(8) = -8
...

[iron ore]

But we don’t get -13 on the top of the table

[inner zealot]

Correct, but x = 13 is gven.

[iron ore]

We only get postive 13

So would it be postive 9

Since the y values are opposite

[inner zealot]

And by the odd function identity.

f(-x) = -f(x)

so

f(-13) = -f(13)

-f(13) = -(-9) = 9

[iron ore]

But that nine in the equation poppped out of no where

And then my rule applies there so how do I know when the rule applies and does not apply

[inner zealot]

It always applies for even and odd functions. It seems the difficulty lies in determining the corresponding values in the provided tables.

[iron ore]

No like in general for the graph earlier

The rule did not apply

[inner zealot]

The odd function you originally asked about?

[iron ore]

Cus I know an odd function is a reflection over both the x and y axis but then the rule is broken

[iron ore]

This

[inner zealot]

The point (-10,9) is also given. It's corresponding point will be at (10,-9) which you drew correctly.

But your graph does not go through (5,-8).

[iron ore]

Oh whoops

[inner zealot]

(-2,5) is given by the graph so the inverted graph should go through the point (2, -5) which your graph does.

[iron ore]

Also that pic is not what I drew it’s the correct answer

But I got smth wrong

But I don’t know where I went wrong

[inner zealot]

What did you draw?

[inner zealot]

For any points given, just negate the signs. (-4, 7.5) becomes (4, -7.5), etc.

[iron ore]

I’m just confused why the rule breaks there

[inner zealot]

It doesn't. That's what you do for odd functions.

[iron ore]

Like why it applies for one type of problem but not the other

[iron ore]

But then what about the table

[inner zealot]

If I asked you what f(3) is equal to, what value of x would you look for on the graph since x = 3 is not given?

[iron ore]

-3

[inner zealot]

And what is x = -3's corresponding value for f(x)?

[iron ore]

11

[inner zealot]

Well, that would be what f(3) equals.

[iron ore]

-11?

Wait

[inner zealot]

You can just literally negate the signs in the table if asked about an odd function like so

[iron ore]

I got thsi wrong

The coresponding value for -3 is -11

So the corsponding value for 3 is postive 11

[inner zealot]

Correct.

And if you were told f(x) was an even function, you could change the table like so

And then you won't need to do any mathing, you can just look at the table.

[iron ore]

Ohh

Ty

[inner zealot]

But at some point, you should learn how to solve for even and odd functions without doing that.

[iron ore]

We only learned how to solve with the even function

So f(x)=f(-x)

[inner zealot]

So here I did the Odd conversion, by negating the signs in the table.

So that way you don't need to do any math, you can just look up the answer directly for an Odd function

f(14) = 8, f(12) = -14, etc.

Even though the original table states
f(-14)=-8, f(-12) = 14, etc.

[iron ore]

Tyyy

That makes these table ones way easier

[inner zealot]

So if f(x) is an odd function, what does f(-8) equal?

[iron ore]

14?

[inner zealot]

Correct.

[iron ore]

And if it was even it would be -14 right

[inner zealot]

But say it more confidently. 😉

[iron ore]

Loll

[inner zealot]

Correct.

[iron ore]

I do overthink things 💀

Cus realistically I should not worry since I have an A+ test average in this class

[inner zealot]

Sometimes the problem is the teacher.

[iron ore]

My teacher is really good better then other years

Cus I got other parts of this unit down very well

Anything involving words is when I get mixed up

[inner zealot]

It does help to let the teacher know that you do not understand. They will always assume you do understand if you do not let them know otherwise.

[iron ore]

He tried to teach me the tables on Friday I just forget again💀 and I would go to him but my test is second block tommrow so

I got no problems asking help is more do I have the oppturnituy to ( during a study or choice block )

[inner zealot]

If I told you f(x) was an odd function and that f(-1) = -2, what would f(1) equal?

Just like the table, just negate the values.

[iron ore]

2?

[inner zealot]

Correct, but be more confident when you say it. 🙂

[iron ore]

Loll

I think I’ll do good. Cus I study like way to much. So I’m re-doing all the previous hw from the unit. Then doing the desmos. And instead of doing the five required problems for each category I do 10.

[inner zealot]

For any complex problems, it helps to write out an explanation of how you solve a problem to yourself. Come back to what you wrote a day later and see if what you wrote still makes sense. If it doesn't, rewrite the explanation so that it does. Those notes make great study notes for exams, especially when you take higher-level math classes.

This is literally what I do. I explain how I solved a problem to myself. I come back a day and a week later to reread what I wrote. If it doesn't make sense, I rewrite it so that it does.

[calm coral]

@iron ore Has your question been resolved?

[calm coral]

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[autumn pollen]

,rotate

[calm coral]

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[autumn pollen]

hello

I was wondering how I could go about proving b

i know that i can do a via a counterexample because it is false

[velvet osprey]

!msgdel

[calm coral]

The original post of this help channel has been deleted, and it will abruptly close and possibly lock. (This is irreversible.) Please claim a new channel, and don't delete the first message of any future channel you claim.

[velvet osprey]

you screwed up and now your channel will implode. gonna have to grab a new one.

[autumn pollen]

okay

[calm coral]

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[quick path]

why is maths so complicated !

[remote mural]

The thing I am struggling with is that one of the points is C(4,-1). Why did we not consider absolute value of the small area under the x axis

[marsh valley]

You want the area delimited by the curves, so what you’re integrating (or finding the area of ) is the difference of both pieces of line.

This difference is non negative even when you have a point below the x axis

[remote mural]

Can you please dumb it down a little?

[marsh valley]

Say you consider the right part of the triangle.
To find the area between two curves you subtract the one below from the one above and integrate that between 2 and 4

But between 2 and 4 we’ve already established that the difference is positive because you subtract the lower curve from the upper one

[remote mural]

Okay I’ll try to digest it

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

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[visual mural]

I dont get letter b and c, i feel like im close to getting it but not rlly

[tawdry jacinth]

b. Since 12% of the editors are already a female, we deduct the sum of 45% and 40% by 12%
45+40-12=
85 - 12=
73%
100 - 73% = 27%
Ans. 27%

c. 40% of editors minus the 12% of woman editors = 28%
100% - 28% = 72%

Are my answers correct?

[echo lance]

I was thinking of since there's 45% females let's say the entire company has 100 people so 45 people are females and 40% of the people are editors so 40

Wait I misread it

I think

12 females editors total

[tawdry jacinth]

12% of the entirety of employees

[echo lance]

So uh

[tawdry jacinth]

Not on the 40

So yeha

Yeah

[echo lance]

12 female editors, 32 females , 55 males

[tawdry jacinth]

Ym 45 females

In total

[echo lance]

So a is

Uh

[visual mural]

OHHHH SO THATS WHERE 73% came from🤯

[tawdry jacinth]

Correct me if some of my answers are wrong

[echo lance]

Wait the question doesn't say if there are male editors.

💀

[tawdry jacinth]

Yeah

But u can figure it out

It's 33%

[visual mural]

where did 100 came from?

[tawdry jacinth]

It's neither

So we exclude that total percentage sum

So not in the 73%

[visual mural]

but why 100

[tawdry jacinth]

The whole percentage

All people

Like if you drew a Venn diagram

[echo lance]

Wouldn't it be 28?

[tawdry jacinth]

Isn't it 45-12

[echo lance]

I'm no help here 💀 I'm just making it worst

[tawdry jacinth]

That means 33

[echo lance]

45 - 12 then that's the remainder of the females who do other jobs

[visual mural]

45%+40%+12%?

[echo lance]

I'm talking about male editors

[tawdry jacinth]

No that's the ditors

Editors

Editors who are male is 45

Female editors is 12

I MEAN

EDITORS are 42%

[echo lance]

Wait did it not say 40% are editors

Wait

[tawdry jacinth]

Not who are male

Oh wait nah

[echo lance]

Yea

[tawdry jacinth]

I didn't read properly

Mb

Yeah 40-12

28

[echo lance]

Lol

Anyways uh so 28 male editors

12 female editors

[tawdry jacinth]

Yes

Yuh

[echo lance]

33 females with others

And 27 males with others

So now can't we js do uh

A = 12 + 33 + 28 / 100

So 77 / 100 for a

B is 27/100

[tawdry jacinth]

Total: 100%
Male: 55%
Female: 45%
Editors: 40%
F editors: 12%
M editors: 28%

[echo lance]

C is 28/100

So a = 77%
B = 27% and
C = 28%

No?

[tawdry jacinth]

Yuh

a. she alr answered

The first question

[echo lance]

Wait uh is it 72% or 28%

Cus it says no females or editors

[tawdry jacinth]

It's says neither

[echo lance]

And only males with other jobs here

[tawdry jacinth]

Its 28% should be

Editors that are not females

[echo lance]

Wait no I got c wrong

It has to be no editors + female

55 males

And 28 are editors

So

27? For C also

Wait did I read it wrong

Oh wait no I'm stupid

28 male editors so yea

[echo lance]

Yea so this ig

[visual mural]

question, how did u get male?

[echo lance]

Basically it says 45% females

[tawdry jacinth]

So

40 - 12 is 28

Not females

Oh wait no

Yeah not females

[echo lance]

Ye so js 100-45 = 55 total males
And 40% total editors and 12% females so just minus = 28%

[tawdry jacinth]

Their asking for editors who are not females

[visual mural]

ohhh okay

[tawdry jacinth]

So 40% total editors

Minus 12% woman editors

[echo lance]

Xd I think this is so confusing if you miss read it

[tawdry jacinth]

= 28%

That are editors that aren't females

[visual mural]

uhuh

[echo lance]

Xd

[echo lance]

Wait lemme try to explain this

So basically 100% is the entire company right?

So let's say there's 100 employees

1 employee = 1%

So we have 45% females

Which is 45 females

So there's only 2 genders

Female and male

[ebon steeple]

[echo lance]

So it's 100 which is the entire company minus the females

[visual mural]

how do u know omg im gonna cry

what if its only 97

[echo lance]

So it's 100-45 which leaves 55

[visual mural]

OHHH

[echo lance]

So 55 males

[echo lance]

Huh 😭

I mean sure

[visual mural]

wait probability

[echo lance]

It's possible to mb do that

[visual mural]

…

[echo lance]

Yea so here js make it simple and say there's 100 employees

[visual mural]

uhuh

[echo lance]

So now we know there's 55 males and 45 females

[visual mural]

yes

[echo lance]

40% of the entire company are editors

[tawdry jacinth]

We assume

[echo lance]

While 12% are females

[echo lance]

So 40 employees are editors

So we want to find how many males there are

[visual mural]

OHHHH we can assume Aahhhh

[tawdry jacinth]

It's gonna be a bit complex if we only use percentages

[echo lance]

So we minus the total minus the female = male

So it's just 40 - 12 = 28 which is the amount of males

[visual mural]

male editors?

[echo lance]

So now we know
Female = 45
Male = 55
Female editors = 12
Male editors = 28

[visual mural]

yesss

[echo lance]

So cus u got a I don't need to explain it

K so b says no editors or females

So if we take out the females we get 55 left

[visual mural]

well i get it alrdy if 100 is the total number

[echo lance]

But 28 males are editors

So we take it out now

[visual mural]

but what if its not😔

[echo lance]

So it js becomes 27

[visual mural]

yes

[echo lance]

So B = 27

Now C

Ok so now we want editors

But no female

[echo lance]

So using this

We just need to take out the females

Which are 12

So it's just 40 - 12 = 28

So now
B = 27
C = 28 and since 1 employee = 1% it becomes

B= 27%
C = 28%

Oh wow I solved someone eles stuff when I need someone eles to help me

[visual mural]

what if its not 100

💃

[echo lance]

Well uh this question never says how many employees there are

And it wants probability

Not number of employees

[visual mural]

OHHHH

[echo lance]

Cus if they wanted that they would need the specify how many employees are there

Also what grade level is this 💀

[visual mural]

so if it wants probability u can assume it’s 100?

[echo lance]

Yea normally when it's a question like this and it's probability

[visual mural]

10😞

[echo lance]

And it doesn't specifying

Specify

We can assume it's 100

[echo lance]

Huh 💀 isn't this like grade 5 level

[visual mural]

FR?!

omg

[ebon steeple]

Not really

[void umbra]

I think your country just be like that tbh

[ebon steeple]

Grade 10 is fine

[echo lance]

No in my country this is also like grade 10 level

But the way they made it is quite easy ig?

[ebon steeple]

Yeah cuz not everybody does competition math? It's practically pointless anyway

[echo lance]

Cus I knew grade 12 math since grade 6 so I can't say shit 😭

[void umbra]

Idk then probably just an introductory qn then

[echo lance]

Yea I'm also quite dumb sometimes

Sometimes I mistake y² as x² and js do the quadratic formula and get the wrong anwser

[visual mural]

Thankyall

.close

[calm coral]

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[exotic cosmos]

Need help with the challenge problem

😭 no idea what to start with

[calm coral]

@exotic cosmos Has your question been resolved?

[calm coral]

@exotic cosmos Has your question been resolved?

[calm coral]

@exotic cosmos Has your question been resolved?

[calm coral]

Channel closed

Closed due to timeout

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[uncut narwhal]

is it 8

[clear delta]

show work

[dreamy stream]

feels like ans is 4

[uncut narwhal]

might not be a perfect straight line so leaning toqwards 6.4

[clear delta]

show work

the lines are assumed to be straight

[uncut narwhal]

nope, not straight

[clear delta]

which line do you think is not straight?

[uncut narwhal]

thetwo lines connecting them both

[ivory pilot]

What is the formula for the area of a triangle

[uncut narwhal]

1/2 bh

[clear delta]

you're saying you don't think this line is straight?

[uncut narwhal]

not parallel to the side length of the square

so its not 8

[ivory pilot]

Then draw the h and b of both the triangles and you will know the answer

[clear delta]

well no it's not perpendicular to the square but it doesn't need to be

also -- it doesn't say it's not perpendicular

so it might be

but it doesn't matter; you'll get the same answer

[ivory pilot]

I don't get why it would help knowing that's a straight line

Did you draw b and h @uncut narwhal ?

[uncut narwhal]

th base is 1 right

height is 8

[ivory pilot]

The height of 1 triangle is 8 ?

[uncut narwhal]

its 4

[ivory pilot]

Okay yeah but I am not sure if you understand why

So if you can draw h

[uncut narwhal]

god why did discord remove remixes

:/

but the height of one triangle is 4

so area is 2

two triangles would be 4

[ivory pilot]

It's still there?

[uncut narwhal]

not for me

[ivory pilot]

I don't have much time
So I am gonna trust that you do understand

So anything else

[uncut narwhal]

thx

ly

how to close this

[tawdry jacinth]

".close"

[uncut narwhal]

.close

[calm coral]

Channel closed

Closed by @uncut narwhal

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• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[shy wadi]

hey, im really bad with recurrence relations and im so confused, why didnt they do 2(a x 3^n) when finding p(n)?

[shy wadi]

like when they sub in un = a x 3^n i dont understand why they just said 2a x 3^n instead of also mulitiplying 3^n by 2.

.close

[calm coral]

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[exotic cosmos]

.reopen

https://cdn.discordapp.com/attachments/948770026914213909/1346080531628167238/image.png?ex=67c6e2af&is=67c5912f&hm=424b35548bd4beb61dc81da400733019b2efc066bb121df4ac341ac40690377c&

[exotic cosmos]

Need help with the challenge problem

https://cdn.discordapp.com/attachments/948770026914213909/1346080576280461364/image.png?ex=67c6e2ba&is=67c5913a&hm=1d4238518c09edc645350f4dbf41c1ba27c901152fb8a0e580182e10c252847d&

this is the diagram

[calm coral]

@exotic cosmos Has your question been resolved?

[calm coral]

@exotic cosmos Has your question been resolved?

[exotic cosmos]

😭 help

[ancient thistle]

attempt the problem where u=(0,0,0), v=(1,0,0), and w=(0,1,0)

that should've give you an idea of what's going on

[exotic cosmos]

omg you're here

and yeah i have an idea

[exotic cosmos]

btw the answer to this would be that it lies outside the dashed triangle

but okay like if they had me draw 1/3u + 1/3 v + 1/3 w and 1/2 u + 1/2 w

they probably want me to realize that they both sum to 1

so it's probably c + d + e = 1

probably because proof by example isn't a thing

😭

[ancient thistle]

stay in the triangle

oh wait no actually

[exotic cosmos]

oh

[ancient thistle]

you're

right

[exotic cosmos]

yeah both the examples

i think are in the triangle

[ancient thistle]

yes

[exotic cosmos]

😭 so what do we do to prove this

idk why it's so hard

i have like no ideas

[ancient thistle]

this

[exotic cosmos]

isn't it outside the triangle

oh wait

i totally misread what u meant

wait u,v, and w can be any vectors?

nvm it's not linearly independent

[ancient thistle]

any 3 points give you a triangle

it's the 3 vertices

[exotic cosmos]

oh i misread again

thought u meant vectors

[ancient thistle]

im giving you a really simple triangle

i do mean vectors

they're the 3 vertices of the triangle

,w plot (0,0),(1,0),(0,1)

[potent lotus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by tearfox

[exotic cosmos]

yeah this

what am i looking to do now?

wait oh

those vectors are supposed to point at these coordinates

[ancient thistle]

ultimately vectors are just points

[exotic cosmos]

so you mean u put a point at the tip of a vector?

and we disregard the tail or rather fix it to the origin

at least that's not the case in physics i guess

cuz free vectors hmm

confusing

but yeah i think in linear algebra we fix it to the origin? okay i guess then points are vectors

yeah idk 😭 what i'm supposed to be doing

[ancient thistle]

i mean

like cu + dv + ew = c(0,0,0) + d(1,0,0) + e(0,1,0) = (d, e, 0)

you can see the points are all in the xy-plane so you can just forget about the 3rd dimension

and you're looking for when (d,e) lies within the triangle whose vertices are (0,0) (1,0) (0,1)

it should be visually clear that the longest side has equation y=1-x

[exotic cosmos]

okay so

(d, 1-d) are the points on that line

and we have 0 < d < 1

but d + (1 - d) uhm is regardless still 1 anyway

[ancient thistle]

but these are not all the points in the triangle

[exotic cosmos]

mmm i'm being dumb

wait

why not?

i restricted d already

[ancient thistle]

(1/4, 1/4) is inside the triangle for instance

,w plot (0,0),(1,0),(0,1),(1/4,1/4)

[potent lotus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by tearfox

[exotic cosmos]

yeah that's true

it's like the intersection of

x,y > 0

and y < 1-x

idk why my brain isn't braining

but it's that region

[ancient thistle]

yeah

although we probably want equalities as well

[exotic cosmos]

but the problem is i can't express it in inequalities

[ancient thistle]

x,y >= 0, x + y <= 1

[exotic cosmos]

wait let me try again

😭

oh yeah true

i'm dumb

ughh

[exotic cosmos]

yeah this

[ancient thistle]

you wrote the same thing

[exotic cosmos]

idk i was trying

[ancient thistle]

i just added =

[exotic cosmos]

to do something with

d and e

[ancient thistle]

d,e >= 0, d + e <= 1

[exotic cosmos]

indeed 😭

[ancient thistle]

okay so

[exotic cosmos]

wait

[ancient thistle]

this is the model example

[exotic cosmos]

shouldn't it be d,e > 0

and d + e < 1

no equality cuz inside the triangle?

[ancient thistle]

the sides of the triangle are part of the triangle

[exotic cosmos]

don't they want inside triangle

[ancient thistle]

well id say that the sides are included in a triangle

[exotic cosmos]

oh

[ancient thistle]

this is a non issue

[exotic cosmos]

okay then

[ancient thistle]

you can just change it

the equations matter more

[exotic cosmos]

yeah that's true

not a big deal

nowww we have to bridge this

to the main question

somehow someway

[ancient thistle]

in general, you have 3 random vertices u,v,w

what you can consider is u'=0, v'=v-u, w'=w-u

translating the 3 vertices so that u is on the origin

[exotic cosmos]

do we have to translate it?

[ancient thistle]

well

[exotic cosmos]

oh you want to relate it to

our previous problem we just did

[ancient thistle]

yes

it's not really necessary but it's like

good to just have the origin be at u so you don't have to worry about it

[exotic cosmos]

now the v - u is a bit harder than what we deal with

cuz vector subtraction

😭

okay nvm it's still

a point

[ancient thistle]

well

it's just v'

[exotic cosmos]

yeah that's true

i was like "let's introduce two points n and t

but then i was stuck at

n >= 0 and t >= 0 LOL

so i'm thinking of smth else

[ancient thistle]

so it's just like this or something

[exotic cosmos]

isn't it in 3d

but yes

[ancient thistle]

well

doesn't really matter

just work in the plane that all 3 vectors lie in

3 points are always coplanar

[exotic cosmos]

ohhhh

ur right

i forgot about that

[ancient thistle]

so then like

how do you make sure you're inside the triangle

u' = 0 so we just have dv' + ew'

[exotic cosmos]

yeah true

uhhhhh

thinking

okay

[ancient thistle]

the real question is

[exotic cosmos]

yeah i have it

it's just the intersection of all the three regions

with all the three lines

[ancient thistle]

how is this any different from when we just had it on the axes

[exotic cosmos]

uhmmm

[exotic cosmos]

is this not it?

or is this bashy of sorts

[exotic cosmos]

uhmmm we kinda can rotate stuffs maybe?

area is still preserved

so that it's on the axis

[ancient thistle]

too much work

[exotic cosmos]

oh i was doing that 😭

welp

[ancient thistle]

yeah pretty much

the point is that it isn't

[exotic cosmos]

so v'' and w''

[ancient thistle]

well

let's not actually do that

but

[exotic cosmos]

😭 oh

[ancient thistle]

the idea of vectors is that you don't have to care about if they're axes aligned

axes are just a human imposition upon the problem

but intrinsically, they do not exist

[exotic cosmos]

oh it's the basis thing i guess

okay fine

[ancient thistle]

so we should expect the answer to be exactly the same

but if you want an argument for why, think about what the vectors lying on the line segment connecting v' and w' should be described as

[exotic cosmos]

w' - v'

like that vector?

[ancient thistle]

that is the vector pointing in the direction of the line segment

i want a vector which is lying on the line segment

maybe point is less confusing

so for example 1/2 v' + 1/2 w'

[exotic cosmos]

mhm

what about it

[ancient thistle]

it's on the line segment

in other words, v', 1/2 v' + 1/2 w', and w' are collinear

[exotic cosmos]

how are u doing this 😭

okay so

this reminds me of a question

i did

so you're saying

you have vector V pointing to V

you have vector W point to W

u is the linear combination of v and w

t = cu + dv

and we have to assert that the liner combination such that c + d = 1

lies on this segment?

[ancient thistle]

u = cv + dw?

if so then that's a correct statement

[exotic cosmos]

why u = cv + dw

[ancient thistle]

u, v, w are collinear if c + d = 1

im just using the notation you introduced

[exotic cosmos]

but the diagram is different right

for this diagram then i think yeah c + d = 1 then u = cw + dv lies in that line

[ancient thistle]

yes

[exotic cosmos]

idk how we argue about collinearity tho

[ancient thistle]

well

it's a true fact that can be proven

like

(1 - t)v + tw is the equation of a line

it contains v at t=0 and w at t=1

[exotic cosmos]

c + d = 1

1 - c = d

u = cv + (1 - d)w = c(v - w) +w

so u - w = c(v-w) and so u,v, and w are collinear?

[ancient thistle]

so it's that line

[exotic cosmos]

is this parametric equations?

[ancient thistle]

yes

[exotic cosmos]

i haven't learnt it yet

😭

[ancient thistle]

hm

well it's explicitly linear in t

so it can't be anything other than a line

[exotic cosmos]

doesn't this work?

[exotic cosmos]

oh so v - tv + tw

v - t(w - v)

yeah okay linear in t

v is a point

[ancient thistle]

it does

anyway

so points on the line segment v'w' are (1 - t)v' + tw' where 0 <= t <= 1

[exotic cosmos]

wait can we work with v - t(w - v) instead

the other one is confusing

maybe cuz i haven't done parametric equation

at lesat v - t(w- v) is somewhat similar to y = mx + c

[ancient thistle]

it will probably be worse

it's just dv' + ew' where d,e >= 0 and d + e = 1

[exotic cosmos]

omg i feel lik ei'm just wasting your time

okay wait

just to recap

u,w, and v are some vectors in R^3

mmmm

we translate stuffs (points)

so that we're at u' = 0, w' = w - u , and v' = v - u?

[ancient thistle]

ye

[exotic cosmos]

why do we translate stuffs

i mean other than to relate to what

we had before

can we do that?

for our final answer, this part is relevant for the "proof"?

[ancient thistle]

well, it's just to make things fall out nicer

not strictly necessary but you'll end up doing the same calculation in the end

if you don't translate, you still can't avoid saying v-u and w-u

[exotic cosmos]

mmmm

okay okay

idk if it's related

but perhaps they had that in mind?

cuz 21) says that

anyway rightttt yeah

[ancient thistle]

well

[exotic cosmos]

and all three points are coplanar

[ancient thistle]

its related but only because you calculate line segments by taking differences

[exotic cosmos]

i see

[exotic cosmos]

yeah okay we were here

i think our last part is to find conditions on

d and e right

such that dv + ew lie in the triangle

[ancient thistle]

yes

[exotic cosmos]

because c doesn't matter

[ancient thistle]

so i began with just the black line segment

[exotic cosmos]

since u = 0

[ancient thistle]

yeah

[exotic cosmos]

oh yeah

u did the black line segment

u need that d + e = 1

[ancient thistle]

yeah

[exotic cosmos]

not sure about the reasoning but i think this requires parametric stuffs right

[ancient thistle]

not really

[exotic cosmos]

i'm learning calculus from scratch and uh we didn't reach there yet

[exotic cosmos]

well i think it was this question

and someone here helped me with the proof above

[ancient thistle]

you can use your argument for collinearity to derive d + e = 1

[exotic cosmos]

yeah they gave me a hint but i'll say not the most intuitive

[ancient thistle]

hm

[exotic cosmos]

😭

[ancient thistle]

well

at least when i see that expression

im thinking that c and d are weights given to the vectors v and w

[exotic cosmos]

yes

[ancient thistle]

so i have a linear mixture of v and w

so it must lie on the line connecting them

it's like you say you want 26% v and 74% w

then okay here you go 0.26v + 0.74w

[exotic cosmos]

yes i understand that

wait nvm

[ancient thistle]

good

[exotic cosmos]

😭

[ancient thistle]

oh

[exotic cosmos]

hold on

[ancient thistle]

nvm not good then

[exotic cosmos]

i'll try not to disappoint

😭

okay nvm i don't

ugh sorry for being dumb

but yes we have a linear combination

of v and w

[exotic cosmos]

not sure where u got this from tho

[ancient thistle]

because it's linear

linear equations lead to lines

[exotic cosmos]

no i mean 0.26v is a vector fine

scaled version of v

0.74w is another vector that's scaled version of w

now what

[ancient thistle]

it lies on the line segment joining v and w

[exotic cosmos]

what exactly lies on the line seg

[ancient thistle]

dv + ew where d,e >= 0 and d+e = 1

xv + yw where x,y >= 0 and x+y = 1

you are essentially drawing the line x+y=1

but your axes are v and w instead

[exotic cosmos]

this is difficult for me

do you think i should just skip it

andmaybe come back later?

😭

maybe more context helps

or not really?

[ancient thistle]

maybe

depends on what kind of context

[exotic cosmos]

idk what kind of context is required?

this is at 1.1

[ancient thistle]

i usually teach this stuff to people doing vector geometry in highschool and not all of them have done parametric equations

[exotic cosmos]

yucks i have skill issue then

[ancient thistle]

you don't really need the full machinery of linear algebra to do vector geometry in 2D and 3D

[exotic cosmos]

yeah okay true maybe that's why it's there in

1.1

[ancient thistle]

yeah

[ancient thistle]

like if you draw out examples you can verify that this does give you the line segment vw

[exotic cosmos]

yeah i can do examples

i just can't do the proof for some reason

i don't understand the linear combination thing u said as well

wait what should i set these to?

to get what you're saying

[ancient thistle]

okay so

https://www.desmos.com/calculator/s9nwylqqin

u = (1 - t)v + tw = v + t(w - v)

[exotic cosmos]

omg your animations are crazy

okay so yeah

that is indeed true

do i just need a visual understanding of this of sorts?

this is just the c + d = 1 so if c = t then d = 1 - t

right?

[ancient thistle]

yeah

[ancient thistle]

not for doing the algebra

ofc you should know what the equations draw

[exotic cosmos]

yeah that's right

but now the justification part is particularly hard

the (1 - t)v + tw implies u,v, and w are collinear

i get the animation tho so tysm for that

now we have to tread the algebra path right

[ancient thistle]

probably

[exotic cosmos]

or what do you have in mind

[ancient thistle]

w - v is the vector whose tail is v and head is w

t(w - v) scales it down because t is between 0 and 1

so adding that to v gives you a vector on the line segment vw

[exotic cosmos]

yep

[exotic cosmos]

yep

[exotic cosmos]

oh yeah

that is indeed right

okay fine the animation explains the rest

[ancient thistle]

yeah

[exotic cosmos]

yeah yeah

omg finally

okay so

i checked the solution in the book

he didn't even explain it at all

😭

[ancient thistle]

lol

classic

[exotic cosmos]

he just did this

also skipped the challenge problem i think

[ancient thistle]

lmfao

it's a challenge for you

[exotic cosmos]

oh yeah 😭 he just gave the answer

whoops

[exotic cosmos]

certainly!

[ancient thistle]

seems about right

[exotic cosmos]

yeah it's gilbert strang's book

[ancient thistle]

writing solutions is a painful journey

having tried writing books in the past i never want to write solutions

[exotic cosmos]

omg you seem really good

like technically

like with technology

maybe u could write a super nice book

imo

[ancient thistle]

[exotic cosmos]

and well you know the math too

so even better

okay wait so

we're done with the challenge problem right

piece all of this together

i honestly would've never been able to do that

😭

shifting stuffs and everything

[ancient thistle]

well

almost

we still need to get the rest of the triangle

we only got one side

[exotic cosmos]

wdym

haven't we showed

d + e = 1

in cu + dv + ew

[ancient thistle]

no but this isn't what you want actually

because remember we had dv' + ew'

expanding this back out

[exotic cosmos]

ohh

yeah forgot about that

[ancient thistle]

dv' + ew' = dv - du + ew - eu = (-d - e)u + dv + ew

and if you translate everything back to u

you need to add it back

u + dv' + ew' = u + dv - du + ew - eu = (1 - d - e)u + dv + ew

so this explains why the coefficients need to add up to 1

it's 1 - d - e + d + e

[exotic cosmos]

ohhh

oh my god that's a crazy exercise

but okay yeah

we shifted all the vectors down by

-u

so we gotta add it back

[ancient thistle]

yeah

but this only covers the summing to 1 part

we need to justify why all the coefficients need to be >=0

[exotic cosmos]

by inspection 😭

[ancient thistle]

well

we're almost there with the line segment argument

[exotic cosmos]

wait we have to justify the >= 0?

looks like he just gives it to us for free

[exotic cosmos]

hmm okay

[ancient thistle]

for any point within the triangle, it lies on the line segment joining kv and lw for some 0 <= k,l <= 1

we already know how to describe points on line segments

so the purple point in the middle is
d*kv + e*lw for some d + e = 1, 0 <= d,e,k,l <= 1

[exotic cosmos]

k and l are the intersection points of purple with red and blue?

[ancient thistle]

they're constants which scale v and w down

[exotic cosmos]

wait we also need k + l = 1, no?

[exotic cosmos]

otherwise we get a whole parallelogram

worth of solutions

[ancient thistle]

no we don't

we just need them to be between 0 and 1

we just need them to be shorter

[exotic cosmos]

but for example

k = 1 and l = 1 satisfies 0 <= k, l <= 1

no?

[ancient thistle]

yes

that gives you the full v and w

the line connecting them is the black line segment we were working on at the start