help-19

[peak heath]

oh

[vivid wyvern]

Try factorizing it using (x-2) as a factor first

[peak heath]

okay i found 3/4

[vivid wyvern]

how?

[peak heath]

rational root theorem

i feel like there is a better way to do this

than just bashing

[vivid wyvern]

Could be

U can wait for other helpers to give u the better way ig

[peak heath]

.close

[odd edge]

Channel closed

Closed by @peak heath

Use .reopen if this was a mistake.

[odd edge]

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[next briar]

Hello

[next briar]

#help-30 message

Continue

[atomic hornet]

I am trying to derive conic section equations from plane and cone intersection and failing… does anyone know where its already done

[next briar]

The method is explained in

#help-30 message

It doesn’t yield the 2D equation of conic

@edgy axle

[odd edge]

@next briar Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

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[odd edge]

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[eternal fog]

Help

[vernal yacht]

Hiiii

What are you looking for?

,rcw

[eternal fog]

19th question

[eternal fog]

Pls tell how to do 19th question

[eternal fog]

19

Pls help anyone

@everyone

[unkempt lichen]

do not use this ping in a server with this many people (even if it's disabled, thankfully).

[eternal fog]

Ok sorry

[unkempt lichen]

also

!15m before helpers ping

[odd edge]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[eternal fog]

But can u help me

Ok

[brittle plinth]

what's the minimum value for a^2 - 4a + 6

[eternal fog]

2

[brittle plinth]

now we need to choose the minimum value between 1 and a^2 - 4a + 6

[eternal fog]

Yes

[brittle plinth]

but a^2 - 4a + 6 >= 2

so what number do you choose

[eternal fog]

No we need most general values of x

[eternal fog]

2?

[brittle plinth]

do you understand what the min {1, a^2 - 4a + 6} is?

[eternal fog]

Yes it's 1 i think

[silent oxide]

Hello I am new. My name is askeladd

[unkempt lichen]

!redir

[odd edge]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[brittle plinth]

!occupied

[odd edge]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[brittle plinth]

alr so now we need to find x so that sin x + cos x = 1

[eternal fog]

Yeah

[brittle plinth]

I think you can do this part by yourself

[eternal fog]

Yes I did it but

I got answer 2npi which is wrong

[brittle plinth]

how did you get that result

[eternal fog]

Sinx+cosx=1

[silent oxide]

Nope

[eternal fog]

So any value of n in 2npi satisfy this equation

[eternal fog]

Then?

[brittle plinth]

that's totally incorrect

[eternal fog]

Bruh how

[silent oxide]

What question are you doing?

[eternal fog]

This 19th

[brittle plinth]

you're missing solutions

you just cannot blindly conclude that 2nπ is the set of all solutions

[eternal fog]

Hm

What can I do then

[brittle plinth]

alright so let's try to solve it properly

[eternal fog]

Ok

Sinx+cosx lies b/w 1 or greater then 2 which is not possible

[brittle plinth]

do you know the identity $sin x + cos x = \sqrt {2} \cdot \sin {(x + \frac{\pi}{4})}$ ?

[silent oxide]

This is tricky

[eternal fog]

No

[clever fjord]

Desert Storm

[eternal fog]

So max value of sinx+cosx is √2

[brittle plinth]

or another idea is to simply square the whole thing up

[eternal fog]

Ok

[brittle plinth]

which way do you wanna pick

[eternal fog]

Square

[brittle plinth]

alr

so what do you get from squaring it up

[eternal fog]

x=npi/2

[brittle plinth]

no no

I'm not asking for a solution

[eternal fog]

Sin2x=0

[brittle plinth]

alr good

so what does 2x have to be?

[eternal fog]

2sinxcosx

2sinxcosx=sin2x

[brittle plinth]

I'm not asking about sin 2x Identity

[brittle plinth]

you already have this

[eternal fog]

General solution?

[brittle plinth]

yes

[eternal fog]

Sinx=siny then x=npi+(-1)^n.y

Now?

[brittle plinth]

what the

do you know the general solution for sin 2x = 0?

[eternal fog]

Yes I should be 2x=npi+(-1)^n.y

[brittle plinth]

what is the y at the end?

[eternal fog]

0

It should be npi/2

[brittle plinth]

ok good

now we do have a little problem

sin x + cos x has a cycle of 2π while sin 2x has a cycle of π

so we do need to check our solutions on the unit circle

[eternal fog]

Ok

[brittle plinth]

now we found out that the solutions must have the form nπ/2

[eternal fog]

Wait I have an idea

[brittle plinth]

when we map it on the unit circle we get {0; π/2; π; 3π/2}

[eternal fog]

We can also use other formula which u told

[eternal fog]

This

[brittle plinth]

yea that formula will be quicker

but since we're almost done with this method so let's continue l

[eternal fog]

Ok

[brittle plinth]

point out the correct solutions from this set

[fringe cradle]

c?

[brittle plinth]

C is wrong

[eternal fog]

C is correct option

[brittle plinth]

wait what

[eternal fog]

In the book

[fringe cradle]

its pretty simple

[eternal fog]

How

[fringe cradle]

sinx + cosx = min of the two functions

the 2nd function is (a-2)^2 + 2

hence the min value of 2nd function = 2 at a = 2

so min of the two functions would always be = 1

[eternal fog]

Ok

[fringe cradle]

frm this we get sinx + cosx = 1

[eternal fog]

Yes

[fringe cradle]

root2(s(x+pi/4) = 1

s(x+pi/4) = 1/root2

[brittle plinth]

nvm I blundered they wrote the option in a terrible way

[fringe cradle]

now if u are doing jee which i suppose u are

they probably told u abt

general solutions for trigo equations

[eternal fog]

Yes but how u k that

[eternal fog]

Yes

[fringe cradle]

yeah then use the result for sinx

[eternal fog]

Ok

Wait lemme try

Yes bro ur correct

I got the answer thanks

[eternal fog]

U also doing jee?

[fringe cradle]

yea im an year above u though

i gotta give in 26

[eternal fog]

Me in 27

Jee mains or advance?

[fringe cradle]

?

both

[eternal fog]

Same

[brittle plinth]

suppose I can close the channel now?

[eternal fog]

Ok

[fringe cradle]

yea u may

[brittle plinth]

.close

[odd edge]

Channel closed

Closed by @brittle plinth

Use .reopen if this was a mistake.

[eternal fog]

Can we talk in dm pls

[odd edge]

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[manic grail]

this a mistake?

[manic grail]

shouldnt it be (e^c)/x

[brittle plinth]

e^c is c

[manic grail]

ah ok

so new c

[brittle plinth]

it's still a constant

[manic grail]

alr

ty

.close

[odd edge]

Channel closed

Closed by @manic grail

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Remember:
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[atomic hornet]

uhh I need some explanation about Exclusive-OR operator, here's a ss regarding the truth table.

[short terrace]

Well for two operands it's just "is exactly one of these true"

[devout hedge]

the exclusive OR operator is true when either input is true, but not both.

if there are more than two inputs, this operator is true if an odd number of inputs is true.

[short terrace]

However, for multiple operands it becomes weird

[atomic hornet]

Alright

I'll keep this thread opened for more questions 👍

lemme show my work

[devout hedge]

I think you can truth table this.

[short terrace]

In classical logic that works

[atomic hornet]

I'm not sure if there's any other method to show this other than truth table

[short terrace]

I'm wondering if this is even true in non classical logic

Hmm it's false in minimal logic

And in intuitionistic logic

@atomic hornet what are the axioms of the logic system you're working with

[atomic hornet]

uhhh I'm not sure about that

[short terrace]

There is

[short terrace]

Okay hmm tell me one thing

Is A ∨ ¬A always true for you

[short terrace]

Yeah that's basically the proof you'd do using truth tables

[atomic hornet]

Well, haven't dug into it to far but imo it's correct

[short terrace]

Hmm okay classical logic then

In that case there is a way to prove

[atomic hornet]

bro, I'm not a postgrad 😭

[short terrace]

The proof isn't that bad lol

Like you can genuinely follow it

[atomic hornet]

is this sufficient to show the result?

[short terrace]

Yes

[atomic hornet]

Should I add some explanation?

[short terrace]

I imagine they expect you to use truth tables anyway

[short terrace]

You can yeah

[atomic hornet]

Alright

[atomic hornet]

how would you describe it?

is it necessary to describe every result in the table?

[short terrace]

Firstly I'd replace L and R with the corresponding expressions

[short terrace]

Nahhh

[atomic hornet]

makes sense

[short terrace]

And I'd add a column for ¬p

That's all

[atomic hornet]

Alright bet

I'll move on to the next question

lemme try first

[short terrace]

Try to make a column for every expression you end up using

[atomic hornet]

is there any way to simplify the expression in the first palce?

[atomic hornet]

yeah, this is correct

[short terrace]

Not really, this equivalence is trying to show you that ∨ distributes over ∧

That is a simplification you can use eventually

[short terrace]

It is, but I'd suggest showing the values for q ∧ r, p ∨ q and p ∨ r

At least on exams

[atomic hornet]

Alright

[devout hedge]

sorry for barging in, but I'd recommend showing the truth value of anything in parentheses separately.

[atomic hornet]

I'd like to, but I'm using whiteboard rn

[short terrace]

Or anything that's being negated

[atomic hornet]

probably have to use excel for this

[short terrace]

Ehh just keep it in mind for assignments and stuff

[atomic hornet]

So basically this one needs simplifying. First thing I come up with is de morgan law

[short terrace]

Good idea

What did that get you

[atomic hornet]

-p ^ -(-p ^ q)

[short terrace]

Looks good yeah

[atomic hornet]

$\neg p \wedge \neg(\neg p \wedge q)$

[clever fjord]

This is sad 😢

[short terrace]

Now you can use de-morgans again

[atomic hornet]

alright

$\neg p \wedge (p \vee \neg q)$

[short terrace]

Would have a negation on q

Negation on the q

[clever fjord]

This is sad 😢

[short terrace]

Yup

Now you can use the previous thing

[atomic hornet]

hmmm, looks like it's fully simplified

[short terrace]

Use this

[atomic hornet]

is there anyway to simplify it further?

[short terrace]

Yes

Distribute the ∧ over the ∨

I know this sounds like you're actually complicating it but trust me

[atomic hornet]

Alright

$(\neg p \wedge p) \vee (\neg p \wedge \neg q)$

[clever fjord]

This is sad 😢

[short terrace]

Good

[atomic hornet]

ah I see smth

[short terrace]

What do you see

[atomic hornet]

The first part can be simply omitted

because the decisive factor is the second part

[short terrace]

Well eventually yes

It's a consequence of two things

¬p ∧ p is always false

And false acts as identity for ∨

[short terrace]

In classical logic again

[atomic hornet]

$\neg p \wedge \neg q$

[clever fjord]

This is sad 😢

[short terrace]

That's what you wanted to show it equivalent to

So you're done

[atomic hornet]

So this expression simply means If p and q are both false, the result would be true

[atomic hornet]

tysm

[short terrace]

You can see that easier using de-morgans once again

[clever fjord]

Xavier 🌺

$\lnot(p \lor q)$

[short terrace]

(this operation is often called nor)

[atomic hornet]

Alright, Imma work on this one myself for 3 minutes

[short terrace]

Good luck

The good thing about classical logic is that you can always just truth table things

[atomic hornet]

how about smth not classical?

[short terrace]

As a last resort

[short terrace]

Then it's harder cuz you have to work with axioms

[atomic hornet]

That seems to be true

regardless

[short terrace]

It's an axiom of classical logic lol

The consequence of which is that variables are either true or false

This is called the law of excluded middle

[atomic hornet]

I see, makes sense

alr, I need to work on that

[short terrace]

Good luck

You can either truth table it or

[atomic hornet]

nah, Imma try simplifying it

[short terrace]

You can use the things you've done in this channel so far

Yeye, good luck

[atomic hornet]

$(p \to r) \wedge (q\to r)$\
$= (p\wedge q)\to r$

[clever fjord]

This is sad 😢

[atomic hornet]

Alright, I'm a bit stuck

I have no idea how to turn this into LHS

[short terrace]

Well let's start with rewriting p → r

And q → r

[atomic hornet]

before that, could you tell me why this transformation is incorrect?

[short terrace]

Sure show

[atomic hornet]

My logic is:
Both cases should be true to yield a true eventual result
The cases are: (if p is true then r is tru) and (if q is true then r is true)

OH wait

[short terrace]

p → r is also true if p is false

[atomic hornet]

p doesn't have to be true

yeah..

[atomic hornet]

I'll try this out

[short terrace]

Good luck

[atomic hornet]

$(p \to r) \wedge (q\to r)$\
$(\neg p \vee r) \wedge (\neg q \vee r)$

[clever fjord]

This is sad 😢

[short terrace]

Good

Now do you see what you can do

[atomic hornet]

lemme check

[short terrace]

It's using smth you've done in this channel

[atomic hornet]

ah, distribution

[short terrace]

Yes

[atomic hornet]

Alr

$r \vee (\neg p \wedge \neg q)$

[clever fjord]

This is sad 😢

[short terrace]

Good

[atomic hornet]

mhm, looks good

[short terrace]

Now do you see what to do

[atomic hornet]

I would take out neg first

[short terrace]

Yes

De-morgans

[atomic hornet]

$r \vee \neg (p \vee q)$

[clever fjord]

This is sad 😢

[short terrace]

Yup

Now what

[atomic hornet]

lemme think

[short terrace]

As a hint, ∨ and ∧ are commutative operators

[atomic hornet]

damn.....I have a strong impulse to truth table this

[short terrace]

You're very close

[atomic hornet]

Can I manipulate the result expression?

[short terrace]

You have r ∨ ¬(smth)

[atomic hornet]

I legit have no idea how to move forward

[short terrace]

Yes you can

[short terrace]

Let's flip this

[atomic hornet]

Oh

[short terrace]

Does flipping it give you something familiar

[atomic hornet]

yeah, it's like doing it reversely

[short terrace]

Ye

Or and and are commutative

[atomic hornet]

Alright, I'll try memorizing the entire cheat sheet lol

[short terrace]

Lololol

[atomic hornet]

I actually have this, but apparently I'm not familiar with it

[short terrace]

Yeah you'd need to remember those

Thankfully with enough practice you'll just start observing them naturally

[nocturne brook]

big fan of "domination laws"

[atomic hornet]

Alright, Imma bring up the next question

[atomic hornet]

gangster law 🔥

"Reverse E" is called the existential quantifier

what does it do specifically?

[short terrace]

Oh ∃

[atomic hornet]

I'm kinda confused about the term "existential quantifier"

[short terrace]

I was very confused at what reverse E meant

Essentially it just says there exists an x such that a property P(x) holds

It says that smth exists

As an example

[clever fjord]

Xavier 🌺

$\exists n \in \mathbb{N}$ | n is even

[atomic hornet]

So basically n is a natural number

and there exist n that is even

[short terrace]

For this case yes

[atomic hornet]

Can I say "and n can be an even number"?

Are they the same?

[short terrace]

Your property here is P(n): n is even

Properties are usually hard yes/no statements

[atomic hornet]

I see

So my statment would be imprecise in this case

[short terrace]

Well sorta

[atomic hornet]

hmmm

[short terrace]

P(n): n may be an even number

Just adds room for confusion

[atomic hornet]

I see

[short terrace]

Achieves the same thing, just adds english confusion lol

[atomic hornet]

Not really, my statement is a bit uncertain

[short terrace]

Yeah

[atomic hornet]

precisely speaking, they're different stuff

[short terrace]

Ye

[atomic hornet]

Does the thing on the right mean "for all x, P(x) is true"?

left*

[short terrace]

Yes

Wait no

That's an or

[atomic hornet]

Alright, it starts making some sense

[short terrace]

For all x, either P(x) or Q(x) is true

(or both)

[atomic hornet]

wait, but there is no

[short terrace]

Yeah parentheses in logic are.. weird

[atomic hornet]

So basically that for all sign include both of the functions right?

[short terrace]

Yes

As an example

[clever fjord]

Xavier 🌺

$\forall n \in \mathbb{N},$ even($n$) $\lor$ odd($n$)

[atomic hornet]

wait, but there is no comma

[atomic hornet]

I'm referring to this one

[short terrace]

Oh yeah that doesn't matter

Parentheses are.. less used in logic than they should be

It would add clarity to use them, but it would also make expressions much much worde

So they use this concept of "binding"

[atomic hornet]

Alright

[short terrace]

Think of ∀x P(x) as x → P(x)

[atomic hornet]

hmm...I'll think about it

appreciate it so much, to you and Lute. I'll call it a night for now since it's time for dinner.

Have a good one 🙂

@short terrace

[short terrace]

You too! Enjoy dinner!

[vernal yacht]

.solved

[odd edge]

Channel closed

Closed by @vernal yacht

Use .reopen if this was a mistake.

[atomic hornet]

gotta be fancy, closing with my green role acc

[short terrace]

(you could have closed it from the same account)

[atomic hornet]

LOL

[short terrace]

Lmaoooooo fair enough

[odd edge]

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!

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[weary tartan]

I was preparing for the 2nd level of the NMTC (a maths comp), and I was solving a question from the 2016 paper

[weary tartan]

what theorems would I need to do this?

here's what I know and tried to use:

Equal chords subtend equal angles at the centre (and converse)
the perpendicular from the centre to a chord bisects the chord (and converse)

equaal chords are euidistant from the center (and converse)

the angle subtended by an arc t the center is twice the angle subtended at any point on the circumference

andles in the same segment are equal

<@&286206848099549185>

[errant dew]

Have you drawn a diagram/sketch?

[weary tartan]

Yes

give me amintue

I was tgrying to find it but gave up

there is an unecessary amount of shit going on

but this is it

B is where the lne AC intersects the circle centered at q

y is ab's midpoint

[errant dew]

Tbh I’m not good at geometry so someone else will need to help sorry :/

[weary tartan]

the rest is just pythagoras

[weary tartan]

dw abt it

thanks for trying

[errant dew]

You can ping helpers again if you want, it was a busy time earlier

[weary tartan]

<@&286206848099549185>

\text{Two circles with centres at P and Q and radii √2 and 1 respectively intersect each other at A and D and PQ= 2 units. Chord AC is drawn to the bigger circle to cut it at C and the smaller circle at B such that B is the midpoint of AC. Find the length of AC.}

hopefully bigletters make the question more approachable

<@&286206848099549185>

[errant dew]

Put $ around text

[golden sonnet]

,tex Two circles with centres at P and Q and radii $\sqrt2$ and 1 respectively intersect each other at A and D and PQ= 2 units. Chord AC is drawn to the bigger circle to cut it at C and the smaller circle at B such that B is the midpoint of AC. Find the length of AC.

[clever fjord]

Dhairya

[weary tartan]

thank you

question:

diagram:

[atomic sierra]

U killed dolphin

[weary tartan]

uh

it was a mistake

whatever dolhpin is

[atomic sierra]

Uhh sry wrong server ig, strange

[weary tartan]

.resolved

im going to sleep now I cant wait

.solved

[odd edge]

Channel closed

Closed by @weary tartan

Use .reopen if this was a mistake.

[wild glacier]

If you want to I found something like this you can just search this question on google.

[odd edge]

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>.
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[viral robin]

yo
2 S's,
3 C's,
4 R's

how many arrangements are there is none of the c are next to each other

[viral robin]

can someone explain why my working is wrong

S_S_R_R_R_R_R_

7 gaps for the C to be in

7C3 * arrangement of the S and R which is 6!/(2!*4!)

7C3 * 6!/(2!*4!)

[late dust]

S and R?

!original

[odd edge]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[viral robin]

hey nel

how are you

sorry nto A

not a

sorry A --> is S

[wild glacier]

Exact question

[late dust]

So why 7C3?

[viral robin]

For another competition, a team of 9 people consists of 2 swimmers, 3 cyclists and 4 runners. The team members stand in line for a photo

[viral robin]

im choosing 7 gaps for the 3 to stand in

oh wait no

its 7P3

[wild glacier]

That's it?

[late dust]

Are you going to post the entire question or what

[viral robin]

How many different arrangements are there of the 9 people if none of the cyclists stand next to each other

[wild glacier]

Much better

[late dust]

Ok, so it is 2 S, 3 C, 4 R

[viral robin]

yo

[late dust]

That should be fine

[viral robin]

is it

mark scheme says its wrong

[viral robin]

but they are all different

[late dust]

Show

[late dust]

Why do I even bother...

[viral robin]

nel are u a statistic man

[errant dew]

We don’t need the original question it’s still the same problem

[late dust]

There is a good reason we prefer pictures

[errant dew]

What is that reason?

[late dust]

OP failed to provide the actual complete question with all the details 4 (four) times

[errant dew]

We don’t need the details though? The abstraction is fine

[viral robin]

hello

how are you guys

[late dust]

Suit yourself then

[viral robin]

nel this isnt you

[errant dew]

Honesty don’t like the xy thing because it’s overused

In this case it’s fine uh oh

[viral robin]

apologies i was trying to save ya some reading

bbmaths do you know how to do this question

hello? did everyone block me or something

[errant dew]

No there’s just lots of help going on rn

[errant dew]

Aren’t there 8 gaps?

[viral robin]

no sorry this is wrong

at the end there is one _

discord wont let me write it

[errant dew]

Still 8 _

Count them

Off by 1 error

[viral robin]

wtf

wait why do i count 7 in my book

oh no i wrote anoother r

[errant dew]

7 letters gives 8 _

[viral robin]

sorry

S_S_R_R_R_R_

[errant dew]

Okay I’m starting to understand now .-.

[viral robin]

bro it was a mistake we already established there are 4 R's

[errant dew]

They don’t count C’s as the same thing

There’s C1 C2 C3

Same with R and S

[viral robin]

oh yes

your right

ok thanks mate

have a good day

.close

[odd edge]

Channel closed

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[odd edge]

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[left bridge]

Example 13 -

Question: I firstly just want to know the logic behind how a²+b² = (a+b)² -2ab ?
(Pardon me if my question is dumb! My basics are weak so leveling them up)

And second question: from where did variable 'x' came in the 3rd step

[wooden python]

(a+b)^2 = a^2 + b^2 + 2ab.

subtract 2ab off both sides and you get that

and in step 3, if you read to the end of it, it says

where x = a + b
they put (a+b) as a new letter

[proven harbor]

bhai came nahi come

[left bridge]

Ok so they let 'x' = a+b

And put it

[wooden python]

yes and that is what they write

[left bridge]

Are you english teacher assigned 😅

[left bridge]

Ok, Thanks

.close

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Channel closed

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[wooden python]

he's right, if your verb has "did" in front then the verb itself is not put in past tense again

"You came" but "Did you come?" and not "*Did you came"

[left bridge]

Did is used for past and came is also used for past so how am I wrong?

[wooden python]

in english past tense is marked once only

that is just how the language works

[left bridge]

So I whether had to write "did" instead did came or I had to write come instead of "did"

[left bridge]

Understand

[wooden python]

we should move to #discussion

[left bridge]

I understand the mistake

So will do someother time

.close

[wooden python]

(it's already closed)

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[tame mesa]

Can we say f(x) is differentiable in this?

[woeful briar]

no

you can say f(x+y)-f(y) is integrable

[weary pelican]

"for every t in R" 💀

[woeful briar]

t=0 💀

[tame mesa]

Yea I thought we can't say that

I solved for f(x) being differentiable idk how to proceed for non differentiable

[quasi falcon]

$-\int_0^t f(y) dy = F(0)-F(t)\$
$\int_0^t f(x+y) dy = \int_{-x}^{t-x} f(y) dy = F(t-x)-F(-x)$

Thats the furthest ive got rn, should try to work this out

[clever fjord]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[tame mesa]

Ok I think I got it

take f(x+y) - f(y) = g(x,y)
now take x as a constant for now, we can see that the lhs is independent of t, hence the integral must be of the form kt, hence g(x,y) is constant for y

[fringe cradle]

is a linear polynomial one of the functions

[tame mesa]

Yea

[fringe cradle]

i just get f(x) = cx

no other functions

[tame mesa]

Yea I had got that too

And that is the only solution I think

[fringe cradle]

hmm

[fringe cradle]

u prepping for jee asw?

[tame mesa]

Yea

This is not jee question though it's from iit bombay bs in mathematics entrance exam

[fringe cradle]

oh

i didnt know

can yu help me w / one q

its frm differential eqns

[tame mesa]

Um okay

Dm me

.close

[odd edge]

Channel closed

Closed by @tame mesa

Use .reopen if this was a mistake.

[weary pelican]

you should open your own channel

[tame mesa]

Yea that can work as well

[weary pelican]

I don't follow that reasoning

[tame mesa]

Um so we got that f(x+y) - f(x) = h(x), now we put that in the eqn and we get f(x) = h(x), hence f(x) + f(y) = f(x+y)

And this is just cauchy's functional eqn

[weary pelican]

no, how do you arrive to the conclusion that g(x,y) is independent from y

that I absolutely don't get

[tame mesa]

The lhs is independent of t, so the integral in the rhs has to be of the form k*t

[weary pelican]

you're skipping a lot of steps but ok

[tame mesa]

Can't we say that directly?

[weary pelican]

well for it to be rigourous:

Let G(x,t) = that integral

for t > 0, we can write G(x,t) = th(x,t)

and then we have f(x) = h(x,t)

so h(x,t) is actually h(x)

but now

[tame mesa]

Bro you can't do that, you can't substitute y inside the integral

[weary pelican]

??

G(x,t) is the result of the integral

not the integrand

[tame mesa]

Oh sorry mb

Continue

[weary pelican]

well now

the problem is what is h(x)

[tame mesa]

Wait I'm confused now lemme think

Maybe the problem is not done

[fringe cradle]

u can solve it with partial differentiation i believe

[tame mesa]

.reopen

[odd edge]

✅

[fringe cradle]

thats how i did it

[tame mesa]

You differentiated and then got f(x) = x right?

[weary pelican]

you don't really need differentiation

[tame mesa]

I did that initially but I need f(x) to be differentiable for that

[fringe cradle]

differentiated once w respect to t , once w.r.t x and subtracted

[weary pelican]

the way to do it is to let F(x) = int(f(y)dy) from 0 to x

then tf(x) = [3 terms in F]

[fringe cradle]

yea

[tame mesa]

Can't we say this?

[weary pelican]

now, "swap" t and x to get xf(t) = ...

and witness the magic

[weary pelican]

you never showed the integrand is necessarily y-independent

[tame mesa]

Ok yea ig we need to show what I said was kinda hand wavey

[weary pelican]

(the fact that LHS doesn't depend on y means nothing, y is an integration variable)

[tame mesa]

No I thought that we got the integral as k*t so that is enough

Wait why isn't that enough

Okay yea ig this is enough

[weary pelican]

you never showed that?

[tame mesa]

We got the integral as f(x) * t, and we are treating f(x) as constant wrt t

[weary pelican]

HOW

how did you get that integral as f(x)*t

what's your justification

since the beginning I'm telling you there's none

[tame mesa]

Just cross multiply the t bruh

[weary pelican]

ok wait sorry on this one but

$tf(x) = \int_0^t(f(x+y)-f(y))dy$

[clever fjord]

Raphaelisius Maximus MMIII

[weary pelican]

you can treat the integral as tf(x)

but that says nothing about the integrand

[tame mesa]

Ok now differentiate this wrt to t

and I think we can do that cuz both sides are differentiable

[weary pelican]

for that to be true we need at least f to be continuous

[tame mesa]

You can't integrate a discontinuous function

Can you?

[weary pelican]

you can

[tame mesa]

Wait what

[weary pelican]

even with the classic riemann integral

I can integrate f(x) = 0 if x < 0, f(x) = 1 else

and that gives me $\int_0^x f(y)dy = \max(0,x)$

[clever fjord]

Raphaelisius Maximus MMIII

[tame mesa]

But isn't integrating just antiderivatives?

Like the opposite of differentiation

[weary pelican]

that's true for continuous functions

[tame mesa]

Oh

[weary pelican]

but "antiderivatives" a lot of times make no sense when you integrate non continuous ones

[tame mesa]

Damn I'm speechless

[weary pelican]

so, to solve this question

you take the """antiderivative of f"""

$F(x) = \int_0^xf(y)dy$

[clever fjord]

Raphaelisius Maximus MMIII

[weary pelican]

and now, with simple integration rules

you rewrite this as tf(x) = ...

and then you swap x and t to write xf(t) = ...

[tame mesa]

You get tf(x) = xf(t) ?

And then we put t = 1 and get a linear polynomial

Damn that was nice

Thanks a lot

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[wide oasis]

how do I solve this?

for inverse we switch x and y right? but how is it possible in the case of modulus?

[silk stream]

take cases

[wide oasis]

ok for x>0 and x<0 right?

[silk stream]

yes

[wide oasis]

alright lemme try it

[silk stream]

is it option A

?

[glossy basin]

I don't think it's A

[silk stream]

what u got?

[glossy basin]

I think you got ±that expression so it will be B or D

[silk stream]

oh yes u are right

so it should be B

[wide oasis]

it is B actually

yk when I made the cases I got negative sign under the square root

[silk stream]

yes

[glossy basin]

This is so ez if given in mcq cuz by observing, if you take +ve x, you should get -ve y and -ve x gives +ve. So f^-1 can be +ve as well as -ve and its sign is opposite to that of x. So B

only for mcqs else ofc you should know how to solve

[wide oasis]

how is this defined?

[glossy basin]

You will eliminate that case if you got negative sign under root

but can you show your work to confirm it

[silk stream]

when u use x>0 y will be -ve

[wide oasis]

that case is when x>0

so do I eliminate it?

[silk stream]

so -ve will come inside the root

no

[wide oasis]

yea

[silk stream]

when u open x with +ve function become -ve there fore y will be -ve

[wide oasis]

sure just a min

[silk stream]

so to compensate that -ve there will be -ve sign in the root

[wide oasis]

@glossy basin

[silk stream]

this is not the way to inverse a funtion

[wide oasis]

you just interchange x and y and then solve for y right?

[silk stream]

no

[wide oasis]

then?

[silk stream]

like in this y=x^2/1+x^2

[wide oasis]

yea

[silk stream]

so u cross multiply

then take x^2 common

then write x^2= whatever u got

then root them

now u have to interchange the x into y and y into x

[wide oasis]

ohhh

[silk stream]

ur first expression is correct but second is not

[mystic saffron]

Are we talking about imaginary numbers?

[silk stream]

where imaginary number come from?

[mystic saffron]

You said negative will come inside the root

[silk stream]

see when u open x>0

then y will be -ve

correct see the expression

so there will be -ve sign inside the root to compensate the ive value of y

thats what i am explaining

[mystic saffron]

Of which class is this 😭

[silk stream]

not of ur class i guess

[mystic saffron]

JEE?

[silk stream]

yes u can say that

[wide oasis]

sorry it's 1-x in the denominator

[silk stream]

well this is basic

[silk stream]

yes

[wide oasis]

brother, this is the same thing.

[fringe cradle]

bc itni lambi discuss iss bakwas pe

nice

[glossy basin]

This is correct. Notice in 1st case, when x>0, y<0 then then you exchanged x and y which made x<0, y>0 in new equation so the minus inside the root is cancelled by this x<0

[fringe cradle]

aise niklega jee

[mystic saffron]

Best of luck then phadi bhaji 🤝🏼

[silk stream]

phadi bhaji?

[mystic saffron]

Bisht from Uk

[silk stream]

ohh pahadi

[mystic saffron]

I used to think Shitij is a girls name

Lol

[wide oasis]

damnnn yes this is true

i get it now

[silk stream]

😂

[mystic saffron]

From which district of UK you are?

[wide oasis]

and what about the modulus and signum?

[silk stream]

i am from rishikesh

district dehradun

[mystic saffron]

Ohh nice

[wide oasis]

guys you can chat in dm please

[mystic saffron]

Sry

[wide oasis]

no problem

[silk stream]

u get the que?

[wide oasis]

i got it what the negative sign has to do in the root

in the square root i mean

but what about the modulus?

and signum?

ok I even understood the modulus

[silk stream]

u cant do that yourself

[mystic saffron]

But roots are generally for a positive function

[glossy basin]

you got
f^-1(x) = √... for x<0
f^-1(x) = -√... for x>0

so f^-1(x) = sig(-x)√...

y = √(whatever in x)
here "y" is the f^-1(x)
so you can write your answer as
f^-1 (x) = √(whatever in x)

this is what i wrote above

[glossy basin]

I can restate from here everything because I think it's getting confusing if you want

[wide oasis]

i just wanna know when x>0 wouldn't it be f^-1(x) = +sqrt?

[glossy basin]

from here
do you understand that

x>0 → y=-x²/(1+x²) → y=±√(-x/1+x)
Here, at the start we had x>0, y<0, ok? Then we exchanged x and y for which we got x<0, y>0 so in final expression
y=√(-x/1+x) where y>0, x<0

Then we say since y=f^-1,
f^-1(x) = √(-x/1+x)

is it ok till now?

[wide oasis]

okkkk so when x<0 y>0 in the inverse function right?

okkk so we can write it as sgn(-x) outside now

yes

i got it finally

[glossy basin]

yes

perfect

same for the other

then you can combine both equations

[wide oasis]

yyeaaa

thanks mann

thanks for your help

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[south dirge]

How would I start this lol

[coarse wyvern]

What's the question?

[odd edge]

@south dirge Has your question been resolved?

[south dirge]

That’s what I’m saying

Idek what it wants me to do tbh

[coarse wyvern]

What about that text on the upper left, it's cropped

[south dirge]

Says solve the following example problem using the convolution sum method

[coarse wyvern]

What about slide 8, this is slide 8 or what

[south dirge]

I’m guessing it’s a old problem it’s the only picture on the page

[coarse wyvern]

My guess is slide 8 mentions the problem and this pic shows the circuit (idk tho)

[odd edge]

@south dirge Has your question been resolved?

[south dirge]

I sent it to Chegg so hopefully it’ll get done lol

[steep pine]

check if theres anything on slide 8 (ques or smth)

[odd edge]

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[eternal sky]

help

please

[odd edge]

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[odd edge]

Channel closed

Channel closed due to the original message being deleted.
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as this action is irreversible, and this channel may abruptly lock.

[queen plover]

.close

close

uhhh

.close

help

[atomic hornet]

It's already closed

[queen plover]

oh ok

[atomic hornet]

claim another one

[odd edge]

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[jagged hemlock]

how do I find the phase shift

.close

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[versed siren]

i don't know where to begin... i guess graphically, to minimize the area of R i pick a point closest to x=0

[versed siren]

like...

the area isn't the smallest when a of P is 1/2, right?

it'd technically be zero or next to zero

[short terrace]

Normal line through P

At 0 the bounded area is infinite

Cuz the normal line is the axis

[versed siren]

where does Q factor in???

[short terrace]

Q is just the intersection of the normal with the parabola

[versed siren]

oh hold on

the blue line is the tangent line, right?

[short terrace]

Yes

[versed siren]

it'd be generated by using the first derivative to find the slope... wouldn't you have to use point-slope formula at every point?...

my calc 1 knowledge is kinda sparse, i only managed to retain some of the derivative stuff

[short terrace]

Yes that would work

[versed siren]

how do i take f(x) and move it to the other side of the equation

(g is just a select x value)

like

surely there's a way to actually get to the fact that the area is minimal at x=1/2

[short terrace]

You could find a way to express the area in terms of a

And then differentiate wrt a

[versed siren]

okay so i guess that'd involve a (technically) indefinite integral

i'd have to actually pick C so it aligns with whatever it should be

so if it's based on two equations

then it'd be like

[versed siren]

f(x)

and g(x)=x^2

so int f(x)-g(x)

from wherever they intersect?..

i'm lost here, idk what i'm doing anymore

[versed siren]

i'm pretty sure i'm getting to this but i'm also pretty sure i lack the knowledge surrounding the techniques to do this?

[short terrace]

I mean what's the area of the bounded region

[versed siren]

you find area by taking the integral

[short terrace]

Yes

[versed siren]

i defined this process here

but i don't know how to take my generalized equation and turn it into what i'd recognize as valid for g(x)

this would be my g(x)

...welll

[short terrace]

We have something helpful thankfully

The parabola is explicitly y = x²

[versed siren]

sorry, that's the bottom function

what am i missing?

[short terrace]

Hold on I'm gonna do this myself

Okay

You find the formula of the line through P normal to the parabola

Then you find where it intersects the parabola again

Then you integrate the difference

This gives you the area in terms of a

Now you differentiate wrt a to find the maximum

I can confirm that this works

[versed siren]

-# what the fuuuuck

[quasi falcon]

Yeah, you can do that, but i feel theres a geometric proof for this, lmao

[versed siren]

i barely remember my discrete maths course

like

[short terrace]

There probably is yeah

[versed siren]

idk if this is how it works, but this is MTH 264 (calc II) and proofs were covered in MTH 288 (discrete math)

would it stand to reason that it might not have been covered how to write a formal proof? e_e

[short terrace]

If it's calc2 you're definitely supposed to do what I said

[versed siren]

so basically

i just actually don't get it

you're asking me to find two points using the normal line, where it intersects x^2

[short terrace]

Yes

You have the equation y = x²

And you're gonna get another equation from the line

Solving these two will give you the intersection points

[versed siren]

yeah

but that's just a line at one point

[short terrace]

It will intersect the parabola at a second point

[versed siren]

right, so these make up the bounds of the area equation

[short terrace]

Yes

[quasi falcon]

Ill try to just put that into words youre probably most comfortable with and provide visual proof:

1. Find the equation of the normal line, you already did, right?

[versed siren]

yeah

like, i can find the area just fine

i don't know how to derive the equation for Rarea

[short terrace]

Well it's the area under the line

Minus the area under the parabola

Within those bounds

[versed siren]

but we're finding the area of just one point

[short terrace]

We're finding the area of the region

Read the question properly

[versed siren]

show that the area of R is smallest when a=1/2

[quasi falcon]

2. Find the intersection points. For all values that correspond to x > 0, theres two points. one is the one you found the line, the other is at the opposite side "somewhere"

[unkempt lichen]

R is a region though.

[short terrace]

R is the region bounded by the normal line and the parabola

[versed siren]

but it's variable

a could be 0.0001

[quasi falcon]

We will care for that later

Just use a as a random letter / variable / number

[short terrace]

You find the area of the region in terms of a

[quasi falcon]

Going back to 2. This is more of a common property of parabolas that you should know generally but:

a line and a parabola intersect at most two times, for our case, its always 2 times because of it being perpendicular to the curve.

_we should probably have to prove this is the case but we will ignore this as being trivial for now 💧 _

[versed siren]

is this proper or should i reframe this?

[quasi falcon]

Thats good

[versed siren]

so to get the intersection points

it'd become x^2=(wahtever im lazy)

[quasi falcon]

btw, you should probably put "2a" and "a^2" instead of the general function notation

Do you remember how to find the intersection points of two functions?

[versed siren]

you just set them equal to eachother

[quasi falcon]

yeah

[versed siren]

x^2=-1/2a(x-a)+a^2

[quasi falcon]

yep, well, try to simplify that

Consider that its basically a quadratic

[short terrace]

And that you know one root

[quasi falcon]

One of the roots will always be a, since we defined that bound as our starting point.

What we are setting here is basically this

Where the two purple lines are set at the intersection points, so they bound the region we care

[versed siren]

right

[quasi falcon]

Do you know how to find those or you want help with it?

[clever fjord]

Weathers

[versed siren]

i feel like i've taken a wrong step somewhere

[short terrace]

As a hint

x = a is a root

So (x-a) is a factor of your quadratic

[quasi falcon]

you should end up with sum like this, close enough:
$-x^2-\frac{x}{2a}+\left(a^2+\frac{1}{2}\right) = 0$

But yeah, as xavier just said, (x-a) is a root, aka, an intersection point.

[clever fjord]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[quasi falcon]

And because how polynomials work, you could divide this expression by (x-a) or just eyeballing it by multypling with another binomial.

If youre not too sure about polynomial division, you can use quadratic resolution

[versed siren]

it's been so long since i've just used a random variable as just a constant that i just don't grasp this

that vaguely looks like a quadratic

[quasi falcon]

a^2 + 1/2 is "just a number"

to make it easier

[versed siren]

that entire thing is just a number??

if i divide it by (x-a) it just blows up because the first term (-x^2) doesn't have an a

[quasi falcon]

$a = -1\ b=-\frac{1}{2a} \ c = a^2 + \frac{1}{2}$

[clever fjord]

∫ᴄ 𝐅·𝑑𝑟 = ∬ʀ ∇⨯𝐅 𝑑𝐴

[versed siren]

OH.

[quasi falcon]

as in, ax^2 + bx + c

[versed siren]

the last are c because they lack any x components (or rather have x^0)

[quasi falcon]

try to use resolution and simplify it as much as you can

[versed siren]

i need to take a shower and go to sleep, but just for additional clarification

uhhh

the resulting root expression should become an input to the bounds of the area integral for Rarea right?

[quasi falcon]

Youll keep the "a" all the way through, and then derivate respect to it

[versed siren]

what the fuuuuuck

[quasi falcon]

why?, remember how when you derivate using f(x) you can use the f'(x) = 0 to find the maxima and minima of x

[versed siren]

of course

wait

[quasi falcon]

Well, you can do that to find the maxima or minima by terms of a

[versed siren]

so when i resolve the "definite" integral

and i plug in a of whatever for x

it just gives me an equation with respect to a

and then i differentiate that?

[quasi falcon]

Without plugging a random x

[versed siren]

right

i mean like

uhh

[quasi falcon]

The definite integral will only have a and no x.

[versed siren]

i mean with int_0^1 x^2 dx for an example

you get the equation x^3/3

and you plug in 1 and 0

[quasi falcon]

yeah

[versed siren]

i just plug in a and whatever instead