help-49

[twilit field]

I think the bot will be down for a while

I have a class now, so I'll continue from where I left in an hour

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[twilit field]

thanks a lot for the help!

[frank wolf]

np

wait

is this what people are supposed to do…?

I thought there was like

!noclopen

[midnight plank]

Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem and is confusing for helpers, making it more difficult to help you. Please be patient, even if your channel has not received much attention.

[humble torrent]

yeah maybe posting it in the help forum would of been wiser

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
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[last slate]

The graph of this and " y = 1/(x-3)" is the same

[dreamy lichen]

Almost, except for 1 point

[last slate]

huh

2 ?

[dreamy lichen]

at x = 2, this is undefined

there will be the empty circle

[last slate]

desmos doesn't show that ?

[last slate]

so is this continous at x= 2 ?

or not ?

[safe oriole]

press on the point on the graph with x=2

on desmos

it will show an empty circle

[last slate]

it says undefined

[hard umbra]

x = 2 is not even in the domain

[dreamy lichen]

it will be 0/0 when you plug in x = 2

[last slate]

nvm it shows an open circle

[dreamy lichen]

I actually drew it lol

because that's how it's commonly drawn by hand

desmos doesnt plot those 1 point discontinuities

[last slate]

so its neither continous nor discontinous at x = 2 ?

[dreamy lichen]

oh, nice, i didnt know it does

[hard umbra]

well idk what people want to say about discontinuous

its just a weird question to ask when 2 isnt even in the domain of the function

would you say that (x - 2)/(x^2 - 5x + 6) is discontinuous at x = ?

[last slate]

no

[safe oriole]

you can define a new function which is the same as y(x) that you have here when x neq 2 and to be -1 when x=2

this new function will be continuous at x=2 but the current y(x) isnt

because there is no y(2)

[last slate]

OH

so

i can't say

this and y = 1/(x-3) are the same

[dreamy lichen]

except for point x = 2...

[last slate]

Okay

i get it

thanks everyone

.close

[midnight plank]

Channel closed

Closed by @gleaming geode

Use .reopen if this was a mistake.

[twilit field]

oops , forgot about this

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[twilit field]

prove that $x-\frac{x^3}{6} \leq sin(x)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

I've reached until here

[frank wolf]

Ok so you have to prove this

[twilit field]

$\sin(t) \leq t$ and $- \sin(t) \leq t$, we can deduce $2 \sin^2(t) - 2 t^2 \leq 0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[frank wolf]

Because?

[twilit field]

No, I just wanted the bot to TeX what you just said

I find it hard to read TeX

[frank wolf]

Mhm

[twilit field]

ok so we have $sin^2(t) \leq t^2$

[grand pond]

ƒ(Why am. I here)=I don't Know

[frank wolf]

How?

[twilit field]

we square both sides of the expression here

[frank wolf]

Why are you allowed to do that?

It’s not true in general that if x <= y then x^2 <= y^2

[zealous schooner]

Hello there

[zealous schooner]

They have x>sinx and x>-sinx

[frank wolf]

Right

That’s different from squaring both sides though

[wintry galleon]

You can do it using calculus

[zealous schooner]

Yeah it is different

[wintry galleon]

easily

[frank wolf]

To be clear im not disagreeing that they can deduce sin^2(x) <= x^2 from the info they have

I’m just not convinced they’ve done it yet

[twilit field]

hmm, I don't follow

[twilit field]

I know

[dreamy lichen]

-2 < -1 but it's not true that (-2)^2 < (-1)^2

[twilit field]

true

[frank wolf]

So, what I’m trying to get you to show is that $x \geq \sin x \wedge x \geq - \sin x \implies x^2 \geq \sin^2(x)$

[grand pond]

Pseudonium

[frank wolf]

And unfortunately “squaring both sides” doesn’t work

[twilit field]

$x>sin(x)$ is always true

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

now x is always positive

[frank wolf]

When $x > 0$

[grand pond]

Pseudonium

[frank wolf]

It’s not true for 0, or for negative x

[twilit field]

so $x^2>sin^2(x)$ is always true too

[wintry galleon]

f(x) = sin(x) - (x - x^3/6)
f'(x) = cos(x) - 1 + x^2/2
f(0) = 0
f'(x) is always greater than equal to 0
therefore f(x) >= 0
sin(x) - (x - x^3/6) >= 0
sin(x) >= (x - x^3/6)

[frank wolf]

This is false unfortunately

[grand pond]

ƒ(Why am. I here)=I don't Know

[frank wolf]

!nosols

[midnight plank]

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

[wintry galleon]

^

[twilit field]

for x>geq 0

[frank wolf]

Why?

[twilit field]

thank you

[twilit field]

hmm, intution tbh

wait

[frank wolf]

Unfortunately that’s not a proof..

[twilit field]

I'll try justifying it

[frank wolf]

Ok

[twilit field]

so $x>sin(x)$ in Quadrant 1

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

with both being positive

so $x^2>sin^2(x)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[wintry galleon]

that is the proof

[frank wolf]

What’s quadrant 1?

[twilit field]

I meant [0,π/2]

[frank wolf]

I was replying to their statement about intuition, not your message

[wintry galleon]

ok

[twilit field]

then from [π/2,π] x>sin(x)

[frank wolf]

Right..

[zealous schooner]

$x^2>\sin^2 x\iff x^2-\sin^2 x>0\iff (x+\sin x)(x-\sin x)>0$

[frank wolf]

Could use iff here

[grand pond]

kheerii

[dreamy lichen]

or at least opposite implication, what you had before wouldnt help

[twilit field]

but x>1 and thus we have x^2>sin^2(x) the term on the right is bound between 0 and 1

[frank wolf]

I don’t follow

[zealous schooner]

Nvm

Why are you proving x>sinx

That’s given to you isn’t it?

[twilit field]

in the interval [0,π/2] $x^2>sin^2(x)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[frank wolf]

Because?

[zealous schooner]

Yeah okay

I think you’re going about it in too much of a roundabout way

[twilit field]

now in the interval $[π/2, infty)$ x>1, so x^2>sin^2(x) as sin^2(x) is bound between 0 and 1

[grand pond]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[zealous schooner]

There isn’t a need to split it into quadrants

[frank wolf]

I don’t follow this argument

Oh

Wait

Waiiit

Ah I see what you’re doing

[zealous schooner]

Not necessary though

[frank wolf]

Oh yeah that actually works I think

Neat

It’s not the proof I had in mind

But it works

Cool!

[dreamy lichen]

How is this proved though?

[frank wolf]

Just lmk how you deduced this

I’m convinced for x >= pi/2 now

[twilit field]

if a>b, and and b are both positive, a^2>b^2

[frank wolf]

But sufficient!

[frank wolf]

Yep!

Ok cool

Now im convinced that you’ve proven $x^2 \geq \sin^2 x$

[grand pond]

Pseudonium

[dreamy lichen]

was it done for negative x already?

[zealous schooner]

Is there a way to solve this without calculus though?

[frank wolf]

Oh well we were only worrying about x >= 0

[zealous schooner]

Both sides are even so it must be true for both

[frank wolf]

But you can use evenness to deduce for negative x

[dreamy lichen]

yeah, thats the way I'd extend the argument

[frank wolf]

I don’t think it’s worth worrying about

[dreamy lichen]

i just wanted it to be mentioned

[frank wolf]

Sure!

[twilit field]

we're only considering $x\geq 0$

[zealous schooner]

Probably not, since the expansion itself came from calculus

[dreamy lichen]

I see

[grand pond]

ƒ(Why am. I here)=I don't Know

[dreamy lichen]

I think it can actually.

oh wait its something else

nvm

[twilit field]

thanks a lot everyone!

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[frank wolf]

uh wait did we finish lol

[twilit field]

yeah

[frank wolf]

I don’t think we did

[twilit field]

.reopen

[frank wolf]

We haven’t gotten back to the original Q

[midnight plank]

✅

[twilit field]

ok

[humble torrent]

Maybe a smart use of identities is possible?

[twilit field]

ok, so what do I have to prove now?

[frank wolf]

back to the original Q?

[frank wolf]

.

[twilit field]

I have proven it though, right?

[frank wolf]

no?

[dreamy lichen]

you have proven x^2 >= sin^2(x)

[twilit field]

ok, so (x-x^3/6)-sin(x) <0

[frank wolf]

where did you get that from

[twilit field]

I differentiated the function

[frank wolf]

huh

[twilit field]

used some trig identities

so 1-x^2/2-cos(x)

[frank wolf]

i am not following what you’re doing..

[dreamy lichen]

are you assuming that if f'(x) < g'(x), then f(x) < g(x)?

[twilit field]

no, idts

[dreamy lichen]

i dont follow either then

[frank wolf]

you seem to be doing something akin to

[humble torrent]

are you integrating?

[twilit field]

no

I differentiated the function

[frank wolf]

we want to prove A. We show $A \implies B$. We show B is true. Hence A is true

[grand pond]

Pseudonium

[frank wolf]

This doesn’t work as a proof

[twilit field]

I don't follow, why can't I prove that the function is decreasing

I know it starts at 0

[dreamy lichen]

which function though?

[twilit field]

x-x^3/6-sin(x)

[humble torrent]

like, i meant i guess you could go from your differentiated thing by integrating

[twilit field]

I'm trying to prove x-x^3/6-sin(x) <0 \froall x geq 0

[dreamy lichen]

wait no

[frank wolf]

Well this is false

[dreamy lichen]

oh it does

nvm

[dreamy lichen]

do you mean it's derivative

[twilit field]

well, I aim to show it's decreasing

[twilit field]

no

[dreamy lichen]

oh I see what you're doing, but at x = 0 it's actually = 0. So it's false

[twilit field]

the inequality effectively becomes $x-x^3/6-sin(x) \leq 0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[frank wolf]

yeah

[dreamy lichen]

this one is fine

[frank wolf]

There’s a better way to phrase this btw

$x - \frac{x^3}{6} \leq \sin x \iff x - \frac{x^3}{6} - \sin x \leq 0$

[grand pond]

Pseudonium

[frank wolf]

This is how id phrase it

[twilit field]

ah, ok

thanks

[frank wolf]

You see

[frank wolf]

When you say this

To me, this reads as

[dreamy lichen]

or if you wanna use words, say "it suffices to prove that [Insert your inequation]"

[twilit field]

I see

[frank wolf]

$x - \frac{x^3}{6} \leq \sin x \implies x - \frac{x^3}{6} - \sin x \leq 0$

[grand pond]

Pseudonium

[frank wolf]

To be clear, this statement is true

But if you don’t use $\iff$, you can’t reverse the logic

[grand pond]

Pseudonium

[frank wolf]

To get back to the original thing you want to prove

[twilit field]

ah, okay

[frank wolf]

When i see “A, so B”

I read it as $A \implies B$

[grand pond]

Pseudonium

[frank wolf]

Starting from A, one can deduce B

Which is fine, except it won’t directly help you prove A

[twilit field]

so I have to write $A \iff B$

[grand pond]

ƒ(Why am. I here)=I don't Know

[frank wolf]

Yes (as long as that’s actually true)

Cause what that communicates is that you can reverse your logic

After all a proof is supposed to at least be a convincing argument

So other people are going to read it

[twilit field]

yeah, true

[frank wolf]

And so it’s useful to make sure you’re communicating well

[twilit field]

okay

[twilit field]

so if I write this, I'm good to go, right?

or is the proof missing anything else

[frank wolf]

Uh no

You seem pretty eager to claim that your proof is done..

[twilit field]

yeah, sorry

I'm just trying to cover a lot of ground before uni starts

[wind oxide]

is this done?

[frank wolf]

don’t think so, no

[twilit field]

no

[frank wolf]

I see

So, what’s the next step after $x - \frac{x^3}{6} - \sin x \leq 0$?

[grand pond]

Pseudonium

[wind oxide]

cant u just

taylor series this

[twilit field]

I differentiiate it to get $1- \frac{x^2}{2} -cos(x)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

yet to be covered

[frank wolf]

Sure

[wind oxide]

What

Isnt that covered in like

[dreamy lichen]

its an option, but you would still have to use stuff like MVT i think

[wind oxide]

HS

[twilit field]

not with proofs

[frank wolf]

Wait it is?

[twilit field]

I'm doing lang now

[frank wolf]

I don’t remember Taylor series from HS..

[twilit field]

In india yeah

but no proofs

[frank wolf]

Huh..

[twilit field]

just formula

[frank wolf]

Right

[wind oxide]

wait then like cant u still use it

[twilit field]

I'm going to be doing a UG in maths, I'd rather not until I know how and why it works

[frank wolf]

After this?

[twilit field]

$2sin^2(x/2)-2 x^2/4$

[wintry galleon]

i literally wrote it few hours ago, you are still doing the same thing

[grand pond]

ƒ(Why am. I here)=I don't Know

[frank wolf]

!nosols

[midnight plank]

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

[frank wolf]

How is this related to the previous expression?

[twilit field]

1-cos(x) -x^2/2 is this

[wind oxide]

its not that tough

but

your choice

[twilit field]

I know how to do it with taylor series, but I'd rather not

[frank wolf]

Right, so id recommend you write $1 - \cos(x) - \frac{x^2}{2} = 2 \sin^2(x/2) - 2x^2 /4$

[grand pond]

Pseudonium

[twilit field]

okay

[frank wolf]

And then?

[wind oxide]

yeah yeah ur choice

[twilit field]

so x/2=t

[frank wolf]

I would write “Let t = x/2”

Since you’re defining a new variable

[twilit field]

okay

so I then have $2sin^2(t)-2t^2<0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[frank wolf]

Ok, wait

I would recommend you write this as $2 \sin^2(x/2) - 2x^2 /4 \leq 0 \iff 2 \sin^2(t) - 2 t^2 \leq 0$

[grand pond]

Pseudonium

[quiet hinge]

You can do it with some very scary looking geometry (involutes)

[frank wolf]

Just to make clear how you’ll reverse your logic

It’s just, whenever I see the words “so then” I assume you mean $\implies$ and not $\iff$

[grand pond]

Pseudonium

[twilit field]

I mean $\iff$

[frank wolf]

I’m sure you do! It’s just good to be explicit about it

[grand pond]

ƒ(Why am. I here)=I don't Know

[frank wolf]

In general, when I read proofs and see “so then”, it gets used as $\implies$

[grand pond]

Pseudonium

[twilit field]

got it

[frank wolf]

From here, what’s next?

[twilit field]

I solve this

[frank wolf]

What do you mean by “solve”? Do you mean you prove it?

[twilit field]

yes

wait

no

[frank wolf]

Hm, then what do you mean

[twilit field]

I show that sin(t)<t

[frank wolf]

Oh sure

[twilit field]

and that -sin(t)<t

[frank wolf]

We’ve already proved that $\sin^2(t) \leq t^2$ for $t \geq 0$

[grand pond]

Pseudonium

[twilit field]

tes

yes

[frank wolf]

So I think it’s ok for you to use that fact now

I won’t make you repeat that proof

So what’s next?

[twilit field]

from that we know the function is decreasing

[frank wolf]

Which function?

[twilit field]

the OG function

[frank wolf]

Which is…?

[twilit field]

this

[frank wolf]

Where’s the function in that?

I’m looking for something that I can plug a number into, and get back a number

[twilit field]

at 0, the function is 0

[frank wolf]

Which function

[twilit field]

the OG function

[frank wolf]

You haven’t actually told me what function it is

Or maybe - here’s another way to think about it

What is the value of the OG function at 3?

If it’s a function, I should be able to plug something into it

Otherwise, I’m not convinced you have a function

[twilit field]

3-27/6-sin(3)

[frank wolf]

Aha!

How’d you get that?

What did you plug 3 into?

[twilit field]

I defined x-x^3/6-sin(x) to be f(x)

and tried to show $f(x) \leq 0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[frank wolf]

Right, so you should write that explicitly

So say

“Let $f(x) = x - x^3/6 - \sin x$”

[grand pond]

Pseudonium

[twilit field]

right

[frank wolf]

And you’re saying that this function is decreasing?

[twilit field]

yes, I have the mention that too I suppose?

[frank wolf]

Mhm

So, why is it decreasing?

[twilit field]

it's derivative is always negative

or equal to 0

[frank wolf]

Indeed!

So now I agree with you that the function is decreasing

[frank wolf]

How do you show this?

[twilit field]

the function is 0 at 0

[frank wolf]

Indeed, so $f(0) = 0$

[grand pond]

Pseudonium

[twilit field]

so if it's decreasing it can only be less than 0

[frank wolf]

Well…

-x is decreasing too

But it takes the value 1 at x = -1

[wintry galleon]

(when someone is doing the same thing though I wrote it clearly before)

[frank wolf]

So you need to be a little more careful

[frank wolf]

As I’ve said multiple times, the point is not to just provide a solution, but guide someone through it

[wintry galleon]

I provided a detailed explanation

if you see that message

[frank wolf]

You did not guide them through it

[wintry galleon]

if this is not guide, then idk what it is

I clearly wrote which point we find

[frank wolf]

Guiding means, like, actually talking to them, and letting them come up with the solution themselves

[wintry galleon]

and we proved that it is increasing

[frank wolf]

Rather than merely providing an explanation

[frank wolf]

Anyway, ideas for this?

[twilit field]

yeah, that's why I said it's 0 at 0

[frank wolf]

Right

[twilit field]

so it can only become less than 0 ow

*now

[frank wolf]

What do you mean by “it can only become less than 0”

-x is also 0 at 0

But it is not always <= 0

[twilit field]

the function is decreasing, ,so to the right of 0, it must be less than 0 from the defn of decreasing functions

[frank wolf]

Yes!

So what you mean is that for $x \geq 0$, $f(x) \leq f(0) = 0$

[grand pond]

Pseudonium

[twilit field]

I really've got to lean to include detail

yeah

makes sense

[frank wolf]

Mhm

I’m sure I must seem pedantic and I’m sorry for that

But it’s important to know how to get down to all the detail

[twilit field]

yeah

I get that

so what detail is the proof missing now

[frank wolf]

I think you’re probably done actually

So long as you string everything together in a coherent way

[twilit field]

I'll do that, thanks!

thanks a lot for all the help!

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[stone blaze]

help

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[stable mirage]

how do i find a 2nd line

[slender walrus]

howd you get the first line

[stable mirage]

derivative is 2x+1

slope of line that goes through (2,-3) would be 5

y=5x+b

-3=10+b

b=-13

[slender walrus]

methodology is already incorect

[stable mirage]

oof

[slender walrus]

you shouldn't be plugging the point into that

as that point isn't actually on the parabola

derivative gives you slopes at points on the curve

[stable mirage]

what should i do

[slender walrus]

consider a point of tangency P(p, p^2 + p)

then set
slope between (2,-3) and P = derivative at P

[stable mirage]

ok

and from there?

[slender walrus]

solve for p

you should get two cases

you'll get a point / equation for each case

[stable mirage]

so

do i set $\frac{p^2+p+3}{p-2} = 2p+1$?

[grand pond]

MWB117

[stable mirage]

?

i got p = 5 and p = 1

[slender walrus]

,w solve (p^2 + p + 3)/(p-2) = 2p + 1

p=5 looks right
seems you messed up your other value

[stable mirage]

-1 mb

[slender walrus]

yes

that'll give you your points of tangency

and you can get the respective slopes from the derivative

[stable mirage]

x and y
(5,30)
(-1,0)

x and y'
(5,15)
(-1,-3)

from there i solved for b in y=mx+b using m = y', x = x and y=y

[slender walrus]

how are you getting 15 and -3

[stable mirage]

2x+1

oh

i used 2x+x

oof

2x+1 isnt working either

[slender walrus]

how are you getting the -45 and -4

did you update your calculations

[stable mirage]

nope

havent submitted

is this right?

[slender walrus]

just type -x instead of -1x

other than that, seems fine

[stable mirage]

no way broo

i hate webassign

[slender walrus]

ah right, they've explicitly stated smaller and larger

sry missed that part

[stable mirage]

dw

new randomization is the same question lets go

alr thanks for helping

.close

[midnight plank]

Channel closed

Closed by @stable mirage

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

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[acoustic owl]

Question 23

I’m clueless

[midnight plank]

@acoustic owl Has your question been resolved?

[midnight plank]

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[iron whale]

im doing synthetic division and have reached (2x-1) as a divisor. do i just divide the binomial by 2 and continue?

[last slate]

u mean u got 2x-1 as remainder?

[iron whale]

no

im dividing by 2x-1

[noble mural]

can you be more specific id understand the question 💀

[iron whale]

$$(4x^4+3x^2-1) \div (2x-1)$$

[grand pond]

CrEpasPmkinPie

[iron whale]

i need to use synthetic division

[wary trail]

show your working

[last slate]

^

[wary trail]

and tbh don't bother latexing anything

write it and take a picture

[iron whale]

my problem is that i dont know what to do with the 2 multiplying the x

do i divide the binomial by 2?

[wary trail]

okay, and we don't know where you are ar

at

so show your working

[iron whale]

i dont have work

thats my problem

i dont know how to start it

[wary trail]

then how are you asking this question?

[mortal abyss]

then go learn

maths

[iron whale]

this is review material for pre calc

and i never use synthetic division so i dont know what to do with the 2

[noble mural]

its very hard to explain synthetic division using text

[wary trail]

$(2x-1)\cdot 2x^3 = 4x^4...$

[grand pond]

Xetrov

[wary trail]

start from here

but if you can't start it, go and read your notes

[noble mural]

u rlly should ask teachers or go watch it on yt ..

[iron whale]

i know what synthetic division is, ive used it, i just dont know how to do it with the "2"

[wary trail]

then we'll help you once you give us something more specific to work with

[noble mural]

just do what u normally do

keep dividing

[iron whale]

use synthetic with the divisor of 1?

ohh ok

thank you

thank you i got the right answer now

.close

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[umbral basalt]

Hi, did I do the calculations correctly?

[umbral basalt]

the teacher says that the correct answer is 2, because the assignment indicates to write a larger root in the answer

[noble mural]

i assume yes

cant find any faults

[umbral basalt]

Okay, I'm calm now.

.close

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[frank acorn]

I'm reading a book that's supposed to help me catch up on math.
The book illustrates "Remainder's theorem" that says the remainder of the division between p(x) and d(x)=x-a is equal to p(a).

It says we can use this concept to determine whether a polynomial is divisible by another one, since it is divisible if the remainder is 0.

Then, next page, it's showing a way to solve this equation: x^4-4x^2+3=0

It says t = x^2, so we rewrite the equation as
t^2-4t+3=0

But now he says we can use the theorems studied on the page before, to rearrange the equation as follows:

(t-1)(t-3)=0

[frank acorn]

and this is what I don't understand

How did he go from
t^2-4t+3 = 0

to
(t-1)(t-3)=0

[ornate cedar]

Factoring

[grim vector]

Factor by roots

[frank acorn]

Has it anything to do with Ruffini's rule?

[ornate cedar]

What is that

[frank acorn]

https://en.wikipedia.org/wiki/Ruffini's_rule#:~:text=In mathematics%2C Ruffini's rule is,divisor is a linear factor.

maybe you call it differently

[ornate cedar]

Thats incredibly overkill for this?

[boreal smelt]

You find roots

[ornate cedar]

It's just rewriting the quadratic with its roots

[frank acorn]

Well, i'm not getting it, how do you do that?

What are roots?

Are roots the values that replaced with t would give 0?

[grim vector]

Ok so

[grim vector]

Exact

[ornate cedar]

Yes

Do you know how to find them?

[frank acorn]

No! 😦

But still, I don't understand how he went from t^2-4t+3 = 0

to (t-1)(t-3)=0

I mean, I understand he did with roots, but how

I thought he used binomial identities

[grim vector]

When you find a root you can factor the polynomial as $(x- x_1)(x - x_2)$

[grand pond]

YakuBros

[frank acorn]

Well, he didn't explain me this

Fucking book

[grim vector]

There is another form to the quadratic too which is the canonic form

[frank acorn]

He oonly said that Ruffini's rule says that if p(x) is divisible by a binomial x-a IF and ONLY IF p(a)=0

how is that going to help me

I don't understand

[grim vector]

Its explaining with many words that you can factor by (x- root)

[frank acorn]

Hm

I guess I'll have to look this shit myself online

[grim vector]

Bprp explain it well if you want

[frank acorn]

Im watching Brian McLogan on youtube

Link me any resource if you want

I'll watch it

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[last slate]

Not sure how to solve this waterloo contest problem

[midnight plank]

@last slate Has your question been resolved?

[twilit jetty]

you here?

[last slate]

yeah

[twilit jetty]

how far are you on the problem

[boreal smelt]

I get 24

[twilit jetty]

its not 24

[boreal smelt]

Ok

Can you say what's wrong in my method?

[twilit jetty]

let pinner decide

[boreal smelt]

If I send it

Ok

[last slate]

tbh I’m lost as to where to start

you have to calculate for all possible cases like all 4 the same colour 3 the same colour etcc and add each

[twilit jetty]

it says that two circles colored by a straight lines may not be colored the same

you can count through the options that arent colored the same

theres very few of them

it helps if you assume that the top-left is "color A" and the top-right is "color B" to leave it general

theres also this dead-end here

its only connection is this other color, so its easy to choose a color for this dead-end if you leave it last

since its on its own too, once you select a color for the three-way, that allows 2 possible colors for the dead-end that cant be contested

[last slate]

That allows for the bottom left is c, then bottom right to be either b or a

vice verse for bottom right being c

and in both cars deadend can be a other a or b

Cases

[twilit jetty]

youre forgetting that the top-right is B

I think you should draw this

I also started considering cases in the bottom-left

[last slate]

Oh I see so if bottom left is c, then bottom right is a

and the deadend is b

[twilit jetty]

dead-end can be B or C

[last slate]

yeah

[twilit jetty]

remember you need to consider all cases, if you go through more of them per-search then its faster

[last slate]

Sounds good

so if bottom right is c, then bottom left is b and deadend is a or b

[twilit jetty]

there doesnt have to be a C

[last slate]

yeah

top left is a, top right is b, bottom left is b, bottom right is a and deadend is b or c

[twilit jetty]

thats also possible, yes

so heres what we've counted so far

(with color A being red and color B being green as an example)

in general, we counted this

[twilit jetty]

you can see there, A B B A BC

[twilit jetty]

then A B B C AB

[twilit jetty]

then A B C A BC

so how many cases is that so far

[last slate]

3

[twilit jetty]

each of these 3 cases leave open the dead-end

so including the dead-end, how many cases is that so far

[last slate]

oh oops 6

[twilit jetty]

dw its a separate question

now with that, how many options do we have for choosing colors for A and B

remember that A and B are connected

[last slate]

When top left is A then top right is can be b or c

[twilit jetty]

no not like that

no

[last slate]

oh

like red green

[twilit jetty]

yea

I said to choose colors

not letters

youre setting A to be a color and B to be another color

thats why we went with A B C instead of just red green blue

[last slate]

If a is red then b can be blue or green

IF a is blue then b can be red or green

[twilit jetty]

...if a is green?

[last slate]

if a is green then b can be blue or red

[twilit jetty]

yes

so in total, how many ways can you assign colors to both A and B

[last slate]

6 ways

[twilit jetty]

yep

now A and B uniquely determine the graph too

so with 6 ways to determine A and B, and then 6 ways to determine the bottom of the graph, how many ways is that

[last slate]

48 ways

[twilit jetty]

6 * 6

[last slate]

36 ways

[twilit jetty]

there you go

[last slate]

mistake oops

thank you very much

[twilit jetty]

np

[last slate]

would it be possible to add you incase there any other questions i struggle with later on

[twilit jetty]

you dont have to ask me specifically but sure

[last slate]

okay sounds good

.close

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[digital crag]

how do i do thid

[wary thorn]

Can you show the question

[digital crag]

Use the Comparison Test to determine if the following integrals converge or diverge.

[tribal tartan]

compare it to the integral of 1/x

[digital crag]

why 1/x

[tribal tartan]

ln(x)<x

[digital crag]

but isnt it

[tribal tartan]

well we know that this inequality is true

[digital crag]

improper integral > divergent integral

[tribal tartan]

hence if we take the reciprocal it’ll flip around

yep

[grand pond]

y0shi

[digital crag]

oh u already did the cross multiplication

[tribal tartan]

if you have a smaller denominator the entire thing becomes bigger

[digital crag]

wdym by that?

[marsh grove]

Like how 1/0.5 is greater than 1/1

[digital crag]

Oh yeah

[tribal tartan]

and how 1/2 is bigger than 1/3

[digital crag]

Oh oh

[tribal tartan]

and hence if we compare it to 1/x

[digital crag]

I misinterpreted that

[tribal tartan]

our improper integral would be bigger than a divergent integral

[digital crag]

makes sense

Thanks

.close

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[cunning bobcat]

how did that become that ?

[cunning bobcat]

actually nvm

I got it

.close

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[sage helm]

How would I prove that the limit doesn't exist rigorously?

[sage helm]

I mean I know that if I let m = n and m = 2n we have contradicting answers therefore it can't

But I need to be all "let epsilon > 0"

[hard shard]

there exists m >M and n>N for some arbitrary M and N in N right?

such that a_mn >0

[sage helm]

almost

[hard shard]

epsilon*

[sage helm]

N such that m, n

[subtle blaze]

less than!

[sage helm]

one big letter

[subtle blaze]

a_mn - a < epsilon

[hard shard]

you want it to be greater than epsilon to showthe limit doesnt exist

[subtle blaze]

oh

oh right we want to do it doesn't work whoops

[sage helm]

oh

yeah

[sage helm]

so if I do m = n, it's stuck at 1/2

epsilon cannot be made less than that

good enough?

@subtle blaze @hard shard

I'll get this checked in #real-complex-analysis later, thank you for your help :P

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[hard shard]

youre welcome

sorry i was watching a salt fork get made

[sage helm]

I will see the salt fork later too

[subtle blaze]

it's really good

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[twilit field]

Let h be a positive number compute the area under the curve 1/x between 1 and 1+h with the area of a suitable number of rectangles to show that

[twilit field]

$\frac{h}{1+h} < log(1+h)<h$

[zealous schooner]

To prove that?

[grand pond]

ƒ(Why am. I here)=I don't Know

[zealous schooner]

Oh this should be pretty easy

What have you tried

[twilit field]

yeah, just not sure whay they mean by a suitable number of rectangles here

[zealous schooner]

Probably two

[twilit field]

why not 1

[zealous schooner]

You can’t prove both an upper bound and a lower bound with just 1

[twilit field]

mhm

right

I first divide across by h

[zealous schooner]

$(b-a)\text{min}(f(x))\le\int_a^b f(x)\dd{x}\le (b-a)\text{max}(f(x))$

[grand pond]

kheerii

[zealous schooner]

Where the min and max are in the interval [a, b]

[twilit field]

I don't "know" integrals yet

[zealous schooner]

Could prove it directly from this

[zealous schooner]

Divide what?

[twilit field]

I think divising across by h may help more actualy

divide the entire inequality by h

then I find the limit as h tends to 0

which would give me the limit to be 1

[zealous schooner]

Oh?

Are you talking about squeeze theoremv

[twilit field]

$\frac{1}{1+h} <\frac{ log(1+h)}{h}<1$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

yeah

[zealous schooner]

Can you prove this?

[twilit field]

I mean using the squeeze theorm I can find the limit of the middle term if that's what you mean

[zealous schooner]

The squeeze theorem doesn’t work both ways

[zealous schooner]

The easiest method is just to use this

What they intend you to do is just prove this geometrically

Draw the graph of y=1/x

[twilit field]

hmm

so I draw a rectangle from 1 to 1+h/2

and 1+h/2 to h

right

[zealous schooner]

Uhh, no

Both rectangles are from 1 to 1+h

On the x axis

[twilit field]

okay

[zealous schooner]

Their heights will be different, namely the minimum and maximum value of 1/x

In the interval [1, 1+h]

[twilit field]

so the areas will be h(1) and h(1/1+h)

right

[zealous schooner]

Correct

Now can you build an inequality from this?

Compare the areas

[twilit field]

$h<(h)/(1+h)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[zealous schooner]

Uh well, no

h > h/(1+h)

[twilit field]

right , my bad

[zealous schooner]

1 is the maximum value and 1/(1+h) is the minimum

But also I meant, compare the rectangles with the curve itself

Which rectangle has an area larger than the integral and which has a smaller area

[twilit field]

ah

right

h(1) has a larger area

h/(1+h) smaller

[zealous schooner]

Indeed!

[twilit field]

then I use the squeeze therom to compute the limit

[zealous schooner]

What limit

Is this part of a different question

[twilit field]

I meant the inequality

no

[zealous schooner]

You don’t need any squeeze theorem here

You just said the result you needed

[zealous schooner]

.

[twilit field]

oh right

I'm so used to limits xD

thanks

.close

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[zealous schooner]

This is a pretty neat inequality for integrals

There’s a more general version of this as well

$g(x)\le f(x)\le h(x)\ \forall\ x\in[a, b]\implies\int_a^bg(x)\dd{x}\le\int_a^bf(x)\dd{x}\le\int_a^bh(x)\dd{x}$

[grand pond]

kheerii

[zealous schooner]

Choose g(x)=(minimum value of f(x) in [a, b] ) and h(x)=(max value…) and you’ll get the previous result

[twilit field]

hmm,so integrals preserve inequalities

thanks

[zealous schooner]

Yeah because they’re linear

Smth like that idrk

[twilit field]

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[royal mesa]

hi

[hushed spear]

hello

[midnight plank]

@royal mesa Has your question been resolved?

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[astral garnet]

I just dont quite get what this is asking

[astral garnet]

to me, C makes the most sense

if 2 non zero complex numbers are equal does that not imply their modulus and argument equal?

[fringe onyx]

not in the way you have it, no

[astral garnet]

hmm, is that because one could say have an argument that is a multiple of 2pi greater than the other one?

[fringe onyx]

yes

this is why we do branch cuts

[astral garnet]

logically i can deduce that D would also be incorrect too then

[fringe onyx]

correct

[astral garnet]

hmm idk about A and B though

[fringe onyx]

try to come up with counter examples

[astral garnet]

is the principal argument from 0 to pi/2

?

[fringe onyx]

definitely not

depends on the text book/prof

it will have at least length 2 pi

normally (-pi, pi]

[astral garnet]

yeah this is what we use in my country

[fringe onyx]

i think this is the standard

with rare deviations

[astral garnet]

then I think B would be incorrect, because we could end up with an argument outside the principal domain?

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[spice scroll]

is this the answer?

$H \in (A \rightarrow P(B)) \rightarrow (P(A) \rightarrow P(B))$

[grand pond]

Ayanokoji

[shy rampart]

What does $H(f)$ return?

[grand pond]

nameless individual

[spice scroll]

idk, what does it?

[shy rampart]

Suppose you have $f={(3,{5}),(5,{5,8}),(123,{5,8})}$, what is $H(f)$?

[spice scroll]

{<{3}, 5>, <{5}, 5>, <{123},8>}?

[shy rampart]

what is $<>$?

[grand pond]

nameless individual

[spice scroll]

the better way to represent pairs of AxB

[shy rampart]

wait im confused right here

what is the $\times$?

[grand pond]

nameless individual

[shy rampart]

nvm $f$ returns a set

[grand pond]

nameless individual

nameless individual

[spice scroll]

and it is which?

[spice scroll]

times = x in my notation i frogot to use the command \times

[spice scroll]

a set can also be a function, set of pairs

[shy rampart]

I updated the prompt

[spice scroll]

exactly what i wrote earlier just put {} for the B you changed into P(B)

thats although not neccarily true, also why im asking for help here

[shy rampart]

no, the $\times$ here is the cartessian produt

[grand pond]

nameless individual

[spice scroll]

ik

just what is

all im asking

[shy rampart]

$H \in (A \rightarrow P(B)) \rightarrow (A \rightarrow B))$

[grand pond]

nameless individual

[spice scroll]

why A->B?

[shy rampart]

e.g. in this case it gives ${(3,5),(5,5),(5,8),(123,5),(123,8)}$

[grand pond]

nameless individual

[spice scroll]

ty, ill look into it

.close

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[bleak pier]

Orthogonal means AA^-t=I

[midnight plank]

@bleak pier Has your question been resolved?

[midnight plank]

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[dark jewel]

I need help with the 19th question <@&286206848099549185>

[buoyant slate]

,rotate

[grim vector]

So y = sqrt[log(x)+y]

[fresh sparrow]

wrong brackets

should be $y = \sqrt{\log(x) + y}$

[grand pond]

artemetra

[tough spindle]

We can simplify to $y = \sqrt{\log \left(x \right) + \left(\sqrt{\log \left(x \right) + \sqrt{\log \left(x \right) + ...}} \right)} = \sqrt{\log \left(x \right) + y}$ and then write $y^2 - y = \log \left(x \right)$ and the rest is easy.

[grand pond]

PowerUp

[sinful trout]

and this is simply quadratic

solve for y

differentiate

or whatever

[grim vector]

Yeah

[dark jewel]

okay thank you guys !!

[fresh sparrow]

or just implicitly differentiate here

much much faster

[dark jewel]

okayyy

.close

[midnight plank]

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[midnight plank]

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[rose cobalt]

If $B=\bigcup^{\infty}{i=1} A_i$, then $\bar{B}\supseteq\bigcup^{\infty}{i=1} \bar{A}_i$. \
Give an example where this inclusion is proper.

[grand pond]

Ciceròn

[zealous schooner]

Are you sure it isn’t supposed to be the other way around

[subtle blaze]

what's the bar mean

[zealous schooner]

Complement I presume

[tender trench]

i would have thought closure?

[rose cobalt]

i just copied the book

[subtle blaze]

see that's why

[rose cobalt]

its closure

[subtle blaze]

we need to be sure

i wasn't sure if it was complement or closure either

[visual tiger]

it's closure yeah

did you show the inclusion?

[rose cobalt]

Yeah

[visual tiger]

for proper example:

consider the balls of radius 1- 1/n centered around 0

(for R^d)

[rose cobalt]

ohhh right

[midnight plank]

@rose cobalt Has your question been resolved?

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[hollow thicket]

Two circles, whose radii are 12 inches and 16 inches, respectively intersect. The angle between the tangents at either of the points of intersection is 29°30'. Find the distance between the centers of the circles.

[hollow thicket]

I did the math and got 8.11 inches, which is the correct answer.

But...there is another answer which is 27.1 inches.

I don't know where that came from.

I used Cosine Law.

[fresh sparrow]

show your work

[hollow thicket]

a=12 in, b=16in, C=29°30'
c=distance between the centers of the circle

by Cosine Law
c²=a²+b²-2ab cos C
= 12² + 16² - 2(12)(16)cos(29°30')
= 144 + 256 - 384cos(29°30')
= 400 - 384cos(29°30')
c = sqrt(400-384cos29°30')
c = 8.11 inches

why need to show work? i need an answer or maybe a hint on why there's another answer which is 27.1 inches

or maybe it's just an error in the book???

the problem is on number 12

[midnight plank]

@hollow thicket Has your question been resolved?

[hollow thicket]

<@&286206848099549185> ☺️

[midnight plank]

@hollow thicket Has your question been resolved?

[last slate]

hi! still confused?

[last slate]

have you considered that that angle is the one being measured?

giving you the angle in the triangle being 150.5

150°30’

[midnight plank]

@hollow thicket Has your question been resolved?

[midnight plank]

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[hollow thicket]

.reopen

[midnight plank]

✅

[hollow thicket]

how?

[hollow thicket]

so like 180-29°30'? how and why?

[inner gazelle]

is it geometry and trig?

[midnight plank]

@hollow thicket Has your question been resolved?

[midnight plank]

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[sonic nova]

Hi! Can somebody help me understand question a? I have no idea how to solve it and I am in a bit of a hurry cause I have an exam tomorrow. I would really appreciate any help. Thanks! ^^

[muted panther]

@sonic nova In order to find the matrix you want, you need to compute f(0,1,0) and f(0,0,1), as f(1,0,0) is given

But (0,1,0) and (0,0,1) both belong in Ker(f)

[sonic nova]

Can you explain in detail please? I have no idea at all T.T

I am a little bit desperate. I don't need to fully understand it cause I don't have time but I want to know how to solve it. I have 8 hours till the exam 😅

[midnight plank]

@sonic nova Has your question been resolved?

[wheat marsh]

so f(0, 1, 0) = f(0, 0, 1) = (0, 0)

[midnight plank]

@sonic nova Has your question been resolved?

[midnight plank]

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

[glacial fiber]

@midnight plank pls slove this math

[midnight plank]

Please don't occupy multiple help channels.

[glacial fiber]

Ok

[rose trout]

What have you tried? Since it's a graphical method, you should draw the lines and find their intersection.

[surreal moon]

!noans

[midnight plank]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[glacial fiber]

This math is correct?

[surreal moon]

Also, bot will not help you