help-36

[tired walrus]

no way around it and no excuse to omit them except when the thing wrapped in them is a single number or term.

[hollow hedge]

[brackets?]

[tired walrus]

(brackets)

[hollow hedge]

uh

okay

[tired walrus]

(parentheses), if you want to be formal precise and technical

doesnt fucking matter you can use {these} for all i care

anyway

i think i've seen you come here with a question of exactly this type

vertical asymptotes happen when the denom is zero

so when is x(x-1) = 0?

[hollow hedge]

yea tahts what i dont undestand

to get the VA you have to equal the denom to 0

[tired walrus]

earlier you said you "don't need the logic" and "just want the procedure to get from problem to answer".

i am confused. do you want to know the why behind it all, or not?

make up your mind once and for all.

https://www.youtube.com/watch?v=cjXseEPP9vc

[hollow hedge]

i do not care about the why at all

why are you lowkey so passive aggressive😭

its a tutoring server chill out

[tired walrus]

i am not being passive-aggressive at all.

[hollow hedge]

lmaoo okay wtv

[tired walrus]

if you don't give a shit about the why, then why this?

those are not the words of somebody who gives no shits about the why.

[hollow hedge]

i do not care about the why i just need to be able to solve the equations

math theory is a waste of my time personally

[tired walrus]

if you truly didn't give a shit about the why, you would just accept "find VAs => simplify then find zeros of denom" as a given

[hollow hedge]

i shouldnt have to argue with my tutor

go somewhere else pls

[tired walrus]

we are not tutors

[iron dagger]

you will not be able to reliably solve any of these problems if all you care about is the procedure that may or may not be filled with gaps in logic or explanations

[tired walrus]

!vol

[final saddle]

Helpers are just people volunteering their time to help you. Be polite and patient.

[hollow hedge]

go away politely

[tired walrus]

ok off i go then

[hollow hedge]

thank you

[muted glacier]

I think this conversation will take longer than just 11 minutes of a video, that was made for this purpose

[iron dagger]

example, in the last ticket you made I literally gave you the most dumbed down procedure you could make and you were still struggling to do the exact same type of problem

[hollow hedge]

holy fuck

im just asking where 0 is applicable like hop off

if you think im dumb and you dont want to help then dont say anything

this shit is so toxic

why are you even typing rn

[iron dagger]

I gave you the explanation right here

you're getting heated at somebody that's been trying to help you for 2 hours

and no I don't think you're stupid

because I used to be almost exactly like you and only wanted to know how to do a problem and not why it works dawg

[hollow hedge]

nah your right i wasted 2 hours with people who just cant help me unfortunately

would have been more effective to watch the video 4 times

not saying its your fault just saying this doesnt work for me apparently

.close

[final saddle]

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[opaque ember]

i suggest a break before trying the problem again

[hollow hedge]

nah the problem is easy asf im just confused as to why sometimes 0 is an answer and sometimes its not

the video will answer

[opaque ember]

its not easy if youre struggling with it

[hollow hedge]

ive done way harder math than this

in comparison its easy

its just confusing bc everyone here wants to give 20 minute explanations for what should be 2 sentences

but again thats not on them this is just not the answers im looking for

i took stats and ts was SA

[opaque ember]

you want easy short answers but many times it blocks understanding so you get confused by similar problems

[final saddle]

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[mint orbit]

I have an algebra question

[mint orbit]

I am trying to work out the 3-term shanks transformation.

[soft zealot]

jan Niku

jan Niku

[mint orbit]

I have the solution, but to be honest I'm not sure I could solve this on my own. How do we go about eliminating the dependence on a and q?

[soft zealot]

jan Niku
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[mint orbit]

I tried back-substitution or experimenting with some of these products but its not really clear what is supposed to happen, or what process gave this answer

[faint edge]

If you sub that in to any of your given equations I don't see how you'd be able to work your way to a true equality in order to verify it is a solution.

Are you sure there isn't more info on what a and q are?

[mint orbit]

they are general

The idea is that some sum $\sum ^n a_j =: A_n$ has some transient component described by $aq^n$

[soft zealot]

jan Niku

[mint orbit]

which vanishes in the limit to infinity along n

if its important

but i dont think it is for the algebra

which is what i dont understand

[gleaming niche]

If u move A and Aj to one side u get $A_j-A = aq^j$, which tells u the sequence $(A_j-A)_j$ is geometric

[soft zealot]

Bob the Builder

[mint orbit]

how does that help for obtaining the given solution for A?

or does it give us access to partial sums

[gleaming niche]

It gets rid of the dependence of a and q

[faint edge]

Are they doing that geometric sum trick?

Where you sum all of the equations and do geometric sum memes?

[mint orbit]

yea but its not a solution in terms of A

[mint orbit]

i dont know, i got curious because my book doesnt explain

i checked wikipedia

but they cite the page in my book

[mint orbit]

for some (A_j - A)_j sure

but we want to solve for A

ito not a or q

i guess

maybe im missing your point

[gleaming niche]

This does give A in terms of A_j

Which is what u want right?

[mint orbit]

i dont understand

we want A in terms of $A_{n+1}$, and $A_n$, and $A_{n-1}$ only

[soft zealot]

jan Niku

[gleaming niche]

yeah thats what i mean by A_j

ig

[mint orbit]

[gleaming niche]

my point is that a geometric sequence is only defined by the fact that the quotient btwn consecutive terms is constant

so $\frac{A_{n+1}-A}{A_n-A} = \frac{A_{n}-A}{A_{n-1}-A}$

[soft zealot]

Bob the Builder

[gleaming niche]

dunno if that gets the solution u want in the end tho

but i think it might

[mint orbit]

oh, interesting

[faint edge]

So, you'd solve for A from this?

[bleak granite]

,w (a-x)/(b-x)=(b-x)/(c-x), solve for x

[soft zealot]

Results provided by WolframAlpha

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

Requested by xxmrfancyu2xx

[mint orbit]

huh

@gleaming niche thats fascinating

i didnt even think to use the fact that the transient part was geometric

[faint edge]

Huh wolfram formula looks pretty close.

[mint orbit]

everyone here is so much more clever than me

its the formula

for sure

thats really cool

new trick in the toolbelt

thank you all

.close

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[jagged flare]

$ABC$ is a triangle on a plane, with $G$ being its centeroid, if $M$ is a random point on the plane prove that $AM^2+BM^2+CM^2=AG^2+BG^2+CG^2+3MG^2$

[soft zealot]

skissue.in.a.teacup

[jagged flare]

i tried cos rule on AGM, BGM, CGM

that gets $AG\cdot MG\cos(AGM)+BG\cdot MG\cos(BGM)+CG\cdot MG\cos(CGM)=0$

[soft zealot]

skissue.in.a.teacup

[cyan kayak]

Let A, B, C, G, M be vectors. |AG|^2 = |(A-G)|^2 and so on. G = (A+B+C)/3

You can probably do something with this?

[jagged flare]

vectors

maybe you can set one of em to be the origin so its cleaner?

M maybe or G?

lets try G

$|a-m|^2+|b-m|^2+|c-m|^2=|a|^2+|b|^2+|c|^2+3|m|^2$

how do you deal with squares of magnitudes

[soft zealot]

skissue.in.a.teacup

[jagged flare]

|a-m|^2=|a|^2+|m|^2-2|a||m|

is this a thing

[final saddle]

@jagged flare Has your question been resolved?

[final saddle]

@jagged flare Has your question been resolved?

[severe verge]

@jagged flare any problems you're having here?

[jagged flare]

dunno how to continue

[severe verge]

you're on the right track

[jagged flare]

got to this but isnt this basically the same as cos rule

[severe verge]

do the same for the other two

[jagged flare]

wdym other two

[severe verge]

since you pick G as the origin, a + b + c = 0

[jagged flare]

why

[severe verge]

G is the centroid

[jagged flare]

i dont get it

is it a theorem?

[severe verge]

yes

G is the centroid if and only if GA + GB + GC = 0

@jagged flare do you want me to prove it?

[jagged flare]

ill just find a proof online

[severe verge]

it's quite an easy proof

[jagged flare]

ok sure

[severe verge]

let M be the midpoint of AB

from the parallelogram rule we have GA + GB = 2GM

but since |GC| = 2|GM| and GM and GC are in opposite directions, 2GM + GC = 0

hence, GA + GB + GC = 0

[jagged flare]

oh i was thinkung about that but didnt think it will actually work

uhh what do you do with that

[severe verge]

repeat this with b-m and c-m

[jagged flare]

i thought like (a+b+c)-2(ab+ac+bc) but this uses magnitude

[severe verge]

see what you get

[jagged flare]

it should all cancel out to be $|a| |m|+|b| |m|+|c| |m|=0$

[soft zealot]

skissue.in.a.teacup

[severe verge]

yes

and since we know a + b + c = 0...

[jagged flare]

wait

isnt mgnituse always positive

[severe verge]

wait you're using magnitude?

[jagged flare]

how can adding 3 of them be 0 unless all three are 0 (which implies degenerate triangle)

[severe verge]

it's wrong then

[jagged flare]

is that not what |a| means?

[severe verge]

you don't use magnitude in this case

[jagged flare]

directional magnitude update when

[severe verge]

what you want to use is vec(AM) = vec(GM) - vec(GA)

since you pick G to be the origin then we write vector AM shortly as m - a

and also |AM|^2 = (vec(AM))^2

so |m-a|^2 and (m-a)^2 are the same thing

[jagged flare]

oh so you can just drop em?

[severe verge]

for squares then yes

just drop them

[jagged flare]

|vector|^2 is a number but vector^2 is another vector is it not?

oh is it some dot product stuff

[severe verge]

Dot product

[jagged flare]

i see

so $am+bm+cm=(a+b+c)m=0$

[soft zealot]

skissue.in.a.teacup

[severe verge]

you can think of it as multiplying 2 vectors with same length and angle between them being 0

[jagged flare]

i see

alright thanks!

.solbed

.solved

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[vital surge]

<@&268886789983436800>

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[timber plume]

just had a question

[timber plume]

does it always have to be sinC

[dusty quarry]

no

[timber plume]

can it be sinA or sinB, given that they are between two sides

[dusty quarry]

half times two sides, then the angle in between those two sides

[timber plume]

ok thanks

.close

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[lofty sinew]

It would have to be c*sinA

And sinB would require finding a different height h

[timber plume]

.reopen

wdym?

[final saddle]

✅

[dusty quarry]

look at how h is formed

[lofty sinew]

Or... sorry that latter part wasn't quite right

The (a*sinC) in the equation comes from h=(a*sinC)

So you could replace it with h=(c*sinA)

But to use sinB you would have to pick a different base value

[dusty quarry]

in other words, you'll need a different height

[lofty sinew]

The three valid combinations are
b*a*sinC, b*c*sinA and *a*c*sinB

Basically, sin of the angle times the lengths of the two neighbouring sides

[timber plume]

different height?

[lofty sinew]

So this equation is often read as "Half times base times height"

The "h" in the equation would be the height, and the line it's perpendicular to (90 degree angle) is the base, side "b"

[dusty quarry]

these two lines can be heights too

sorry, bad drawing

[tired walrus]

a better way to remember the formula which doesn't rely on specific letters is:

you involve two sides and the angle between them

[dusty quarry]

^

[tired walrus]

this way you don't have to tie yourself to specific letters anywhere

[timber plume]

but essentially they follow the same concept

the perpendicular height

[timber plume]

but they are in an inclne

[tired walrus]

"perpendicular height" is a bit of a pleonasm

[timber plume]

one dude was saying it cant be perpendicular if its in an incline

[tired walrus]

and no, seia's diagram is not to scale, but she did mark these things as perpendicular

like you see those little boxes at the ends

those boxes are how you mark a right angle in a diagram

[timber plume]

yea ik that

here is what im talking about

[tired walrus]

this looks like it's from a completely unrelated context

[timber plume]

how so?

[tired walrus]

well you're talking about an object on an incline

rather than a geometric diagram with lines making angles of 90° or not

AND trollstar even said "doesn't have to be" but you misread that as "can't"

which there's a big difference between these too...

one dude was saying it cant be perpendicular if its in an incline
this is gross misremembrance

or rather it's an extreme over-generalization that only leads to confusion on your part

[timber plume]

so in thhis case, it has to perpendicular

the height from the apex

[dusty quarry]

yes. otherwise your sines won't work

[dusty quarry]

if that bothers you, just turn the triangle around

[timber plume]

yup ok

[dusty quarry]

anything else?

[final saddle]

@timber plume Has your question been resolved?

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[still ruin]

Im having difficulties with this question it reads, angle A is 90°, ADEF is a square, AB=6 and AC=10, what's the length of AD?

[still ruin]

I LITERALLY don't know where to start

I mean is this question even possible with pre highschool knowledge or is it just flat out impossible.

[dapper hull]

,rccw

[still ruin]

Num. 13

Thx

[still ruin]

Bruh help me😭

[severe verge]

similar triangles

[still ruin]

Hold on bot don't autoclose it yet

[still ruin]

This is what i end up with

It's a fucking anomaly

[severe verge]

?

they are not conguent

[dusty quarry]

,rccw

[still ruin]

DE=EF?

[severe verge]

yes

[still ruin]

when 2 triangles are similar and one of each their lengths are the same that means they are congruent am i wrong?

[severe verge]

but in those two similar triangles DE does NOT correspond to EF

[still ruin]

Correspond?? What does that mean? Are you sayin that DE corresponds to FC but we have the info that ADEF is a square (maybe a language barrier but in Indonesia square means that all 4 sides have the same lengths)

[severe verge]

yes DE corresponds to FC

but

correspond in the context of similar triangles

[still ruin]

But wha bruh🥀

[still ruin]

DE/FC=BD/EF?

[severe verge]

correct

[still ruin]

DE is still equal to EF btw

[severe verge]

so?

DE is in the numerator while EF is in the denominator

you can not say FC = BD from there

[still ruin]

Ok it makes DE²

[still ruin]

Yeah true😔

Bru🥀

DE=AB btw so BA=AD🥀🥀🥀🥀

[still ruin]

Look at that bs😭😭😭

[severe verge]

set up an equation

let AD = x

[still ruin]

ok

x=AD?

We already what BA is

[severe verge]

what is BD and CF?

[covert tree]

there are 2 ways to calculate BD, one is to use BA-BD which gives 6-x. And hint for the other way: look at the triangle BDE

[still ruin]

BA-x?

[severe verge]

yes

[still ruin]

dude we already know BA=AD
And we KNOW what BA is and it's 6🥀

[severe verge]

so 6-x

???

how is BA = AD?

[still ruin]

Literally the last image we just saw

[covert tree]

BA includes the segment AD

its cant be equal

[severe verge]

you missed the term DE*BA

[still ruin]

My ass is somehow autistic to day bruh🥀

Alr i got it

Also how do you fight against the fact that BDE and EFC is congruent
I mean they're similar and they one side with same length?

Idk ts is fucking with me

[still ruin]

☝️

[severe verge]

here triangle BHA and BAC are similar

and they share the side BA

but they are NOT congruent

[final saddle]

@still ruin Has your question been resolved?

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[still ruin]

Ok fair

.close

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[nocturne lagoon]

Q 22 I have no idea what I'm doing

[nocturne lagoon]

I got p=qtanA

[dapper hull]

in Q22 you should draw a right angled triangle it would help

[nocturne lagoon]

Ok

Does p^2 + q^2 equal 1

[dapper hull]

no but sin^2A+cos^2A would

[nocturne lagoon]

22a?

[dapper hull]

what did you find for the hypotenuse ?

[nocturne lagoon]

1

[dapper hull]

no

[nocturne lagoon]

I got the answer

.close

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[nocturne lagoon]

.reopen

[final saddle]

✅

[nocturne lagoon]

Q22b I'm not sure the triangle is

[dapper hull]

still the same triangle just use the double angle identities to write sin(2A) and cos(2A) in terms of sin(A) and cos(A)

[nocturne lagoon]

Ok ty

Did I do something wrong

[dapper hull]

you didn't tell what was the hypotenuse still cuz it is not 1

[nocturne lagoon]

O oops

That was the other question

Wait but y did I get the answer

[dapper hull]

what answer did you get ?

[nocturne lagoon]

P for 22a

I got q for 22b

[dapper hull]

seems right

[nocturne lagoon]

What was ur hyp

[dapper hull]

sqrt(p^2+q^2)

[nocturne lagoon]

Wanna know how I got 1

[dapper hull]

i don't know about that :3

[nocturne lagoon]

Well

Since tanA is sinA over cosA

SinA equals p and cosA equals q

[dapper hull]

not really

sinA might be kp and cosA=kq

[nocturne lagoon]

But there is no k

In this case

[dapper hull]

at least there isn't but sinA and cosA could be any multiple of p and q

[nocturne lagoon]

So how'd u get the answer

[dapper hull]

same idea just now i know the hypotenuse is sqrt(p^2+q^2) i'd write it as R for simplicity
sinA=p/R and cosA=q/R and then things would work out after adding the fractions and some factoring

[nocturne lagoon]

Wouldn't u get to P(q^2 + p^2)

[dapper hull]

/p^2+q^2
which would simplify to p for 22a

[nocturne lagoon]

But how can u just divide

[dapper hull]

$q({\frac{p}{R})(\frac{q}{R}})+p(\frac{p}{R})^2$

[soft zealot]

qimmah

[dapper hull]

$\frac{pq^2+p^3}{R}$

[soft zealot]

qimmah

[nocturne lagoon]

Ye

[dapper hull]

Factor p and then it’s just R/R

[nocturne lagoon]

O ic

[dapper hull]

Same idea for the other question

[nocturne lagoon]

Alr thanks

[dapper hull]

Btw @nocturne lagoon
The reason p^2+q^2=1 happens to work it’s because it simplifies in both questions leaving you with just the other expression

[nocturne lagoon]

.close

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[nocturne lagoon]

Q24 I'm not sure what to do

[tired walrus]

do you know any trig identities?

such as angle sum, or maybe double-angle...

[nocturne lagoon]

Ye

[tired walrus]

ok, can you state all the double-angle identities you know

[nocturne lagoon]

Sin2A=2sinAcosA

Cos2A=Cos^2(A)-Sin^2(A)

Tan2A=(2TanA)/(1-tan^2A) tan(A) cannot equal to + or - 1

[tired walrus]

missing brackets on that one...

also lots of missing brackets around trig functions but that one's more of a whatever

anyway

have you seen this one? $\cos(2A) = 1 - 2\sin^2(A)$

[soft zealot]

Ann

[nocturne lagoon]

Ye

[tired walrus]

ok

do you see what to do now?

[nocturne lagoon]

O

I did 1-sinA

Oops

For cosa

[tired walrus]

???

[nocturne lagoon]

I did Cos(A)=1-Sin(A) idk what I was thinking

[tired walrus]

yeah me neither

that's blatantly false

[nocturne lagoon]

I think I was thinking complementary but don’t know where the 1 came from

Thanks for helping sometimes I can’t spot where to use some identities

.close

[final saddle]

Channel closed

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[nocturne lagoon]

.reopen

[final saddle]

✅

[nocturne lagoon]

Now I’m stuck on 25

,rotate

[nocturne lagoon]

Here is what I tried

[dapper hull]

how about starting with expanding cos(theta+x) and the negative one too

[nocturne lagoon]

Ye middle right

[dapper hull]

yes
divide the whole numerator and denom. by costheta cosx

[nocturne lagoon]

I have a question

How do u find that?

Ye I did that

[dapper hull]

what'd ya get ?

[nocturne lagoon]

(1-tan theta tanx)/(1+tan theta tanx)

[dapper hull]

to make our lives easy substitute tantheta and tanx with any variable and try to solve for tanx

[nocturne lagoon]

Ok

Wait combined or separate?

So like X and y

[dapper hull]

seperate ofc

[nocturne lagoon]

Ok

[dapper hull]

i reccomend other variables than X

[nocturne lagoon]

Ok

Can I have a hint

[dapper hull]

first of tell me your substitution
Or send your work and let me see

[nocturne lagoon]

,rotate

[dapper hull]

Remember that it’s equal to A

[nocturne lagoon]

I got it

[dapper hull]

Also you are trying to make
B= some expression in A

[nocturne lagoon]

Good idea to make it more simpler

[dapper hull]

What'd you got ?

[nocturne lagoon]

,rotateccw

[dapper hull]

Write it in terms of B and A cuz it’ll take some time to write

[nocturne lagoon]

,rotateccw

[dapper hull]

I saw it dw

[nocturne lagoon]

Ok

Thanks again

[dapper hull]

Looks right

Just remember to make your B back to tanx

[nocturne lagoon]

O ye

What’s a horizontal plane

[dapper hull]

The x axis
left to right

[nocturne lagoon]

So how do u plot angles?

It’s q26 btw

[nocturne lagoon]

Here

[dapper hull]

As in plane
Could you send the whole question once more

[dapper hull]

Yeah a plane we could draw a triangle With points P Q R

[final saddle]

@nocturne lagoon Has your question been resolved?

[nocturne lagoon]

.close

[final saddle]

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[dusky wasp]

This is just same as a1 divides b1 and a2 divides b2 right

[terse crypt]

,rccw

[dusky wasp]

,help

[soft zealot]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[dusky wasp]

,list

[solemn igloo]

!status

[final saddle]

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

[dusky wasp]

2

[solemn igloo]

Alright, what have you done so far?

[dusky wasp]

Ive just listed out the pairs i could find

,rccw

[dusky wasp]

The relation is symmetric

So 20 elements so far

I'd like to find an efficient way to solve this problem other than counting every possible solution

!status

[final saddle]

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

[dusky wasp]

!help

[final saddle]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[dusky wasp]

!

!commands

!status

[final saddle]

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

[dusky wasp]

6

[tribal jackal]

!done

[final saddle]

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

[dusky wasp]

.close

[final saddle]

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[tranquil pine]

,,\sqrt{1+x^2+y^2+x^2 y^2}+xy\frac{dy}{dx}=0

[soft zealot]

!

[solar crest]

do you know how to factorise the expression inside the sqrt

[tranquil pine]

I sent the xy dy/dx to rhs then squared on both sides

On factoring lhs, we get (1+x^2)(1+y^2)

,,\sqrt{1+x^2+y^2+x^2 y^2}=-xy\frac{dy}{dx}

[soft zealot]

!

[tranquil pine]

,,1+x^2+y^2+x^2 y^2=(xy\frac{dy}{dx})^2

[soft zealot]

!

[scarlet sequoia]

this might not help a lot

[scarlet sequoia]

it is better to keep it in this form

as suggested, try to factor what's inside the square root

and it will inspire you a function sub

[tranquil pine]

Oh whats that

[scarlet sequoia]

you just sub something like u = .... in terms of y (possibly x)

[tranquil pine]

,,\sqrt{(1+x^2)(y^2+1)}=-xy\frac{dy}{dx}

[soft zealot]

!

[tranquil pine]

What now

[scarlet sequoia]

yes

so now we can technically distribute the square root

$\sqrt{1+x^2}\sqrt{y^2+1}=-xy\frac{dy}{dx}$

[soft zealot]

Raphaelisius Maximus MMIII

[scarlet sequoia]

this is just to make the substitution more obvious

[tranquil pine]

Oh

[scarlet sequoia]

make a new function out of y, that will be more interesting to focus on

[tranquil pine]

Like y2+1=v?

[scarlet sequoia]

you're very close

but then we'll have some sqrt(v) involved

[tranquil pine]

Then diff v then get dy/dx?

[stone kettle]

i think collect dx and x terms and y and dy terms then integrate

[scarlet sequoia]

and we would like to make the equation as easy as possible

[scarlet sequoia]

that's another possibility

some f(y)dy = g(x)dx

[tranquil pine]

I did i couldnt the final in proper form given by book

[stone kettle]

root(1+x^2)/x dx= - y/root(1+y^2) dy

[tranquil pine]

Idk even if mines right

,,\frac{\sqrt{1+x^2}}{x},dx= \frac{y}{\sqrt{y^2+1}}, dy

[stone kettle]

whats ur

[soft zealot]

!

[tranquil pine]

I got smth like this

[stone kettle]

yep thats it

but negative sign

[stone kettle]

this

[tranquil pine]

I didnt get negative cuz i squared here

[stone kettle]

noo

root alr there

[scarlet sequoia]

we didn't square in the end remember

[stone kettle]

^

[scarlet sequoia]

we kept the sqrts

[tranquil pine]

But i got back the roots

[scarlet sequoia]

then that's maybe why you got an incorrect answer xdd

[tranquil pine]

How😭

[scarlet sequoia]

you went from a = -b to a^2 = b^2 to a = b

[tranquil pine]

Wanna listen to how i approached this?

[scarlet sequoia]

do you get what I'm saying

[stone kettle]

yes

[tranquil pine]

Idk whay u r saying

[stone kettle]

u wrote "root of something = negative" - complex ????

[scarlet sequoia]

when you sqrt, you have to consider that there are TWO possible stuff

[stone kettle]

so thats actually wrong

[tranquil pine]

Oh

[stone kettle]

this

[scarlet sequoia]

so get rid of the "squaring then taking the root", which just adds possibly unwanted solutions

and just stick with the original roots

[tranquil pine]

,,\frac{\sqrt{1+x^2}}{x},dx=- \frac{y}{\sqrt{y^2+1}}, dy

[soft zealot]

!

[tranquil pine]

Ok well on integrating b/s

[stone kettle]

yes

now its ez

[tranquil pine]

$LHS=-\frac{1}{2}, \log (y^2+1)$

[soft zealot]

!

[tranquil pine]

Les add c later

[stone kettle]

yeah

[tranquil pine]

I used x= tan theta for x terms

On integrating x

[scarlet sequoia]

not quite?

[tranquil pine]

,,\sec \theta + \log (\csc \theta - \cot \theta)

[soft zealot]

!

[tranquil pine]

Anything weong?

Wrong*

[scarlet sequoia]

well take the derivative of that

what do you get

[tranquil pine]

-y/(y^2+1)

[scarlet sequoia]

so

[tranquil pine]

Oh i forgot root 💀🤡

Then LHS $=-\sqrt{y^2+1}$

[scarlet sequoia]

that's better

[soft zealot]

!

[tranquil pine]

Ok now for x

I used $x = \tan \theta$

[soft zealot]

!

[tranquil pine]

Can u see the img

?

K nvm

Then on integrating i got

,,\sec \theta + \log (\csc \theta - \cot \theta)

[soft zealot]

!

[tranquil pine]

After subbing $\theta$ in terms of x

[soft zealot]

!

[scarlet sequoia]

that seems like a tiresome sub

[tranquil pine]

I asked the server to help

Few told to sub x= tan theta

[scarlet sequoia]

yeah, so try sqrt(1+x^2) = u

[tranquil pine]

Oh k

[scarlet sequoia]

mmmmh I don't see the point

[tranquil pine]

Oh bruh

Why am i so dumb 😭🤡

[scarlet sequoia]

udu = xdx

and x^2 = u^2-1

[tranquil pine]

RHS =$\frac{(\sqrt{1+x^2})^3}{3}$

[soft zealot]

!

[scarlet sequoia]

no?

[stone kettle]

lol

[tranquil pine]

How

[scarlet sequoia]

show me your u sub

[stone kettle]

put x=tan theta

[scarlet sequoia]

not sure this works

[tranquil pine]

,,u=\sqrt{1+x^2}

[soft zealot]

!

[stone kettle]

it does

[scarlet sequoia]

I'll show mine and then you can show yours

[stone kettle]

i find trig substitution easier

[tranquil pine]

Then $du=\frac{x}{\sqrt{1+x^2}}, dx$

[stone kettle]

for some others it may vary

[scarlet sequoia]

close the frac

[soft zealot]

!

[tranquil pine]

Ye mb

[scarlet sequoia]

yes that's correct

[tranquil pine]

Then x cancels then u x u = u^2

Then on integrating u^2 u get u^3/3

[scarlet sequoia]

uh no

xddd

x doesn't cancel

[tranquil pine]

Oh

Yes, bruh 😭

Ok so what now

[scarlet sequoia]

well x doesn't cancel, so what do you get instead

[tranquil pine]

1/x^2 du

[scarlet sequoia]

no

u doesn't cancel either

[tranquil pine]

U^2/x^2 du

[scarlet sequoia]

yes

now write x^2 in terms of u

[tranquil pine]

I shud go to 1st grade to learn fractions tbh

U^2-1

[scarlet sequoia]

alright

[tranquil pine]

Divide u^2/u^2-1

Oh i can plus or minus 1

I can add*

[scarlet sequoia]

yes

[tranquil pine]

U-1/2 log blah blah 😭

[scarlet sequoia]

we're now at $\int \left(1-\frac{1}{u^2-1}\right),du$

[soft zealot]

Raphaelisius Maximus MMIII

[tranquil pine]

Ye

[scarlet sequoia]

well "log blah blah" is a bit more complicated than that

you gotta split fractions first

and then you can blah blah

[tranquil pine]

Split what

[scarlet sequoia]

you know

[tranquil pine]

U+a/u-a

[scarlet sequoia]

1/(u^2-1) = A/(u-1) + B/(u+1)

[tranquil pine]

Use direct formula bruh

Why use partial fractions

[scarlet sequoia]

what would that be?

[tranquil pine]

1 min

,rccw

[scarlet sequoia]

alright

[tranquil pine]

We have direct formula instead of actually expanding and solving

[scarlet sequoia]

if you've been taught about that formula then yes use it

[tranquil pine]

Ok thanks got the ans

Ye got it

[scarlet sequoia]

idk what is the idea behind tan sub

but knowing the answer it wouldn't have been pretty

[tranquil pine]

What

Ye but

[scarlet sequoia]

.

[tranquil pine]

What does ur name mean

[scarlet sequoia]

how much can a tan sub can help us if there's no tan in the final answer xd

[scarlet sequoia]

just pompous form of my original nickname

[tranquil pine]

Forget abt the sub, but we can convert the theta terms of x using right angle triangle

[tranquil pine]

Ye sure sounds

Ok thanks

.close

[final saddle]

Channel closed

Closed by @gray mica

Use .reopen if this was a mistake.

[stone kettle]

lol 12th class?

[tranquil pine]

Yes 😭

[stone kettle]

lmao same

jee 😭

[tranquil pine]

Me just boards🗿

[stone kettle]

🫡

[solar crest]

some places dont have direct formula

i use a sec sub here, some use partial frac

[final saddle]

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[exotic shale]

what does he mean by rearrangement

[terse crypt]

moving numbers around

https://en.wikipedia.org/wiki/Riemann_series_theorem

[exotic shale]

wdym moving numbers around

[tired walrus]

adding up the same sequence of numbers but in a different order.

[exotic shale]

how can that be different

here it says this

but this isnt really a rearrangement its a completely new sequence

[vital crag]

You're taking about two different things

[vital crag]

This is a definition of conditional convergence

[exotic shale]

ok

but how can rearranging make a different result

if its the same terms

[vital crag]

Read alternating harmonic series

[terse crypt]

oh

[final saddle]

@exotic shale Has your question been resolved?

[final saddle]

@exotic shale Has your question been resolved?

Channel closed

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[final saddle]

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• Show your work, and if possible, explain where you are stuck.
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[drowsy epoch]

. @dim robin

[dim robin]

need to find the exact value , and where do i get help there are like MANY channels

[drowsy epoch]

This is your channel now

[dim robin]

oh okay

[drowsy epoch]

You pick the ones, where there is no name, else it's occupied already

[dim robin]

okay

how do i solve this

i tried to write the first term which is 1/sin(pie/4) . sin (5pie/12)

i tried multiplying 2 numerator and denominator

[drowsy epoch]

I was thinking of applying maybe sine addition theorem first

,, \sin(a+b) = \sin(a)\cos(b)+\cos(a)\sin(b)

[dim robin]

i dont think it would work

will be tooo much messy

[strange coral]

This is jee advanced 2016

Hint:- B-A= pi/6

[drowsy epoch]

,, \sum_{k=1}^{13} \textstyle \f{1}{\sin(\pi/4-\pi/6+k\pi /6) \sin(\pi/4 + k \pi/6)}

[dim robin]

?

[strange coral]

A question from the paper

Jee Advanced

Of 2016

9 years ago

[dim robin]

okay let me search it up

[dim robin]

hmm

ya it is applicable for every term till 13th

[vital surge]

you will need to make use of this

sin(x - y) = sin(x)cos(y) - cos(x)sin(y)

[strange coral]

Multiply and divide by sin(pi/6)

[strange coral]

B is the latter angle and A is former

[dim robin]

okay so you just multiplied and divided by sin(A-B)

so

[strange coral]

Yes

[strange coral]

B-A*

[dim robin]

sin(A-B)/ sin A sin B , and does it matter B-A because if we take A first it can come minus but we are also dividing by the same , so minus is gone

and everytime it is -pie/6 or pie/6

the difference*

[strange coral]

What?

Oh

[strange coral]

Yeah it works

I was telling you wrt my assumption

[dim robin]

whats sin(A-B) /sina sinb simplify to

like

cot?

ya

[strange coral]

Cotangent

Yes

[dim robin]

hmm

its not in multiplication anymore

[strange coral]

Yes

[dim robin]

so its eaier to sum

[strange coral]

Just solve and use angle minimization

[dim robin]

easier*

[strange coral]

And see the magic

[dim robin]

oh there is symmetry

i tried it on k=1 and k=2

i took 2 out

and the terms cancelling

[strange coral]

Yes

On the right path!

[dim robin]

yup same for k=3

feels like its almost solved

[dim robin]

would it continue till 13th?

[strange coral]

All terms would cancel out except

[dim robin]

THE LAST

TERM

OF 13TH

yes

[strange coral]

So,

It would be 2(cotpi/4-cot58pi/24)

[dim robin]

i got 2(cot pie/4 - cot 29pie/12)

reduce 29pie/12

so

[strange coral]

Mb

[strange coral]

This is the correct sum

[dim robin]

29pie/12 = 2pie + 5pie/12

[dim robin]

ya then im right

58pie/24 is 29pie/12

[strange coral]

It’s done

Just basic simplification

Now

[dim robin]

cot 5pie/12 = pie/4 + pie/6 , so use cotangent addition formula

im doing it

[strange coral]

It’s done

[dim robin]

LETS GO

ans is 6-2root3

[strange coral]

No

[dim robin]

thanks

[dim robin]

huh?

[strange coral]

Cot 5pi/12 is 2-sqrt(3)

Other wise it’s all correct

[dim robin]

ya mb

its 2-root3

[strange coral]

It’s a standard submultiple

[dim robin]

I MESSED UP AT FINAL PART 💀

[strange coral]

2(root(3)-1)

Check again

[dim robin]

yes 2root3 -2 is 100% ans

[strange coral]

Yes

Correct

[dim robin]

lets goo thanks

[strange coral]

@dim robin

Close the channel using ‘.close’

[dim robin]

.close

[final saddle]

Channel closed

Closed by @dim robin

Use .reopen if this was a mistake.

[final saddle]

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• Show your work, and if possible, explain where you are stuck.
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[old quarry]

The number of distinct real roots of x4 - 4x3 + 12x2 + x - 1 = 0 is

[old quarry]

i started by differentiating

but i didnt get anything useful

[tranquil pine]

Is this class 11 - 12 standard math?

[old quarry]

12th yes

[tranquil pine]

Oh, I am from class 10 nevermind

[ivory vessel]

||rolles theorem?||

[old quarry]

i havent learnt that properly yet..is that the way to do this? if yes then ill leave it for later

[versed crater]

What did differentiating tell you

[finite hinge]

Descartes's rule of signs

[old quarry]

nothing

[magic coyote]

Dyk cubic determinant

[versed crater]

No it always tells you something

[old quarry]

i dont know that

[old quarry]

nope

[finite hinge]

That's the easiest way to get roots

[old quarry]

4x^3-12x^2+24x+1=0 what does this tell

[old quarry]

huh i need to get the roots?

[versed crater]

That tells you where the critical points of the original quartic are

[old quarry]

but i cant find any values of x for which the eqn satisfies

[versed crater]

In particular, since the coefficient of x⁴ is positive, the starts at inf and ends at inf (on the -inf and +inf part of the x axis)

[old quarry]

yeah i noticed that

[versed crater]

Okay well, you just wanna find out how many local minimas there are

[old quarry]

yeah

how do we do that

that +1 is making my life hell

[versed crater]

Well, look at the 2nd derivative and see where the inflection points are

[old quarry]

oh

okay one second

[versed crater]

The inflection points give you some more info on where the critical values of the cubic might be

Which in turn tells you where the roots of the cubic might be

Which in turn tells you if there are local minimas and where they are

[old quarry]

second derivative is
x^2-2x+2=0

[versed crater]

b² - 4ac seems negative

No roots

No inflection points for the quartic

What does that mean

[old quarry]

no points in which concavity changes?

[versed crater]

Yeah so it just goes down and up

Only 1 real root to the first derivative

If you wanna be fancy you can even try newton’s method

[old quarry]

oh so because it has no roots weve concluded that theres only 1 local minima?

[versed crater]

So there’s only 1 local minima and you need to find it

Well if there were other critical points

Namely more than 1 minima

[old quarry]

alr alr how do we find it tho? thats an ugly ass cubic

[versed crater]

You’d need to change concavity to do that

I’m sure there are tricks for it that I don’t know but you could try newtons method

It works pretty well on a set up like this I believe

[old quarry]

i dont know that one sorry 😔

[finite hinge]

Use graphing tool

Demos

[versed crater]

That’s cheating

[finite hinge]

Ik

[old quarry]

yeah

if we were allowed to use desmos then why would i even ask the question

[finite hinge]

You can't do rational root theorem and derivative 🤷🏿

[finite hinge]

I'm just suggesting 💔💔

To see how many roots cuh 💔💔

[versed crater]

You should know newtons method

It’s very clever

https://youtube.com/shorts/LjtA3D3Gizg?si=9ZKL0jEK2TqqAz8i

Oh beautiful a yt shorts to teach maths

[old quarry]

😭 i did not understand that

"initial guess"?

what does this mean

[versed crater]

You guess some number

Anything

Really, anything you’d like

And you iteratively get closer to the answer

[old quarry]

okay lets go with 1

[versed crater]

I’m pretty sure this isn’t how you’re supposed to do this though

[stone wagon]

(assuming the function is nice enough 🤓)

[versed crater]

But I’m stuck and I haven’t got another idea

[old quarry]

but in their case they already knew how the curve looks

[versed crater]

We have a cubic with 1 root here

[stone wagon]

yeah alr

[versed crater]

You need only to know the tangent line at the initial guess

Which you can find by putting your initial guess into f’

[old quarry]

oh ok

and then what do we do

[versed crater]

Then find the x intercept of this tangent line

That’s your next guess

Then repeat

[old quarry]

huh

is this a particular method to find the minima?

[versed crater]

Can you imagine that with some initial guess x = 1 and the given function f(x) you can draw a tangent line at (1, f(1)) with slope f’(1)

[old quarry]

wait

[stone wagon]

newton's method but on the derivative xd

[versed crater]

For finding roots of equations

[old quarry]

@versed crater f(0) =-1 so we know that the minima will be below the x axis somewhere and graph goes till infinity so from there cant we conclude 2 roots?

so we dont have to actually find the minima

[versed crater]

So y - f(1) = f’(1)(x - 1)

[old quarry]

ohh

[versed crater]

You can also do that

But then you’re guessing where on ℝ will f be negative if at all?

It’s great if you can immediately guess it but it mightn’t be obvious

[old quarry]

yeah i agree with ur point..this is actually why i didnt get the answer

[versed crater]

But you could argue that looking at the coefficient the minima can’t be that far away from 0

But say you try 0 and it doesn’t work

What might you try next

The newton’s method gives you a way to determine what’s a good next guess

[old quarry]

ohh alr alr got it

this is the minima

wont it be way too ugly to solve for

[versed crater]

I suppose you can also just use the derivative to tell you where to go

The derivative will go from negative to positive

So pick a number to start at

Then find f’(number)

If it’s positive pick a smaller number

If it’s negative pick a bigger number

[marsh glacier]

u can differentiate and set equal to zero

[versed crater]

If your derivative changes sign you know the minima is somewhere in the middle of your last 2 guesses

[versed crater]

Well we don’t need to solve for it

We just need to know if the minima is negative or positive

[marsh glacier]

after that use rolle theorem

[versed crater]

If it’s negative you have 2 real roots

[old quarry]

so we are basically finding where the nature of the curve changes and that is our point?

[versed crater]

If it’s positive there are no real roots

If it’s 0 there is exactly 1 real root

[old quarry]

yeah okay

[versed crater]

Yeah you guess a point then use the derivative at that point to help you pick a new guess

[old quarry]

i havent learnt that yet..will try once i do tho

[versed crater]

Since you know the general shape is like U shape

[marsh glacier]

its pretty simple

u just need to find the local maxima and minima

[versed crater]

New theorems are always simple if you already know them

[old quarry]

okok i think i got it now thanks a lot

[old quarry]

yes that is our problem brother 😭

[versed crater]

That’s crazy because this wasn’t easy to do either

[marsh glacier]

i mean

cubic formula?

[versed crater]

You’re just being ridiculous now

[marsh glacier]

what

we were taught

[old quarry]

cubic formula is insane

[marsh glacier]

its actually not

[versed crater]

@old quarry despite this probably not being the solution I suppose it works, I’d say ask ur teacher how you were meant to do it and then ping me and tell me as well 😆

[marsh glacier]

depressed cubics can be slved easily

[old quarry]

alr alr tomorrows the discussion ill lyk

[marsh glacier]

and x-b/3a can be used to eliminate the x^2 term

[versed crater]

Mate if you let someone use cubic formula

You might as well just send the fucking quartic formula nuke on this problem

Like what

I’m pretty sure that’s not the intended solution

[marsh glacier]

i dont know the quartic formula

pretty sure it doesnt exist

[old quarry]

idk the cubic formula

[old quarry]

it does

[versed crater]

.

[versed crater]

It’s actually not

[marsh glacier]

that was for cubics

cubics aint TOO hard

like they may be a lil hard

[versed crater]

We have formulas for all polynomials up to but not including quintics

So there is a quartic formula

[marsh glacier]

hmm

i didnt know that

[versed crater]

Now you do

[marsh glacier]

thx for telling