help-10

[timid silo]

oh

[fresh slate]

Man you are genius

It worked

[timid silo]

did it

[fresh slate]

S(n) alternates between 1 and -1

P(1) =-1

[timid silo]

but this is correct

[fresh slate]

And for all next numbers it is 1

Idk how to prove it

And answer also matches accordingly

[fresh slate]

Yes it maybe correct

@timid silo Sometimes doing most basic thing is most genius thing.

Thanks man

[timid silo]

oh

[timid silo]

np

even if you compute it we still need an expression for p(n)

[timid silo]

and from this P(n) seems to be finite

[fresh slate]

Yes that's the problem and I don't think its even possible to find expression for a general value of beta

There is something special about given value of beta

[timid silo]

there maybe a closed form

[fresh slate]

I will prove it later

[fresh slate]

Highly unlikely

[timid silo]

hm.. maybe

so you canclose this for now

if you wish

[fresh slate]

Yes

.close

[obtuse pebble]

Channel closed

Closed by @fresh slate

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

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[round island]

Restarted this….can someone help me walk through it

[round island]

Welcome back to “Milo doesn’t get math”

[high lily]

are you familiar with the expansion of
(a-b)(a+b)

[round island]

Somewhat

[high lily]

what do you think that expands to?

[round island]

Foil method?

A^2 +AB -AB -B^2

Or- simplified A^2-B^2

[high lily]

(a-b)(a+b) = a^2 - b^2
is a well known / very common identity for these types of problems

Ideally you won't have to foil every time

[round island]

Right… so how do I apply that if cosx doesn’t have another part

OH WAIT THD BOTTOM

[high lily]

apply that to the denominator

[round island]

Gotcha

[high lily]

As mentioned before, leave numerator as is, in factored form

[round island]

Sinx • 1 is just… sinx right?

[high lily]

Yes

[round island]

Ok

So 1+ ~sinx -sin x~ -sinx^2

Or 1 -sin^2

[high lily]

what's that extra ~ after the sun x,
typo?

[round island]

Ignore the ~ I forgot it doesn’t cross out on discord

I forgot the cross through command on here

[high lily]

1 - sin^2(x)

[round island]

Is it…or was it your mistake?

Yeah

[high lily]

the ~ looked like a -

[round island]

And then….. 1-sinxcosx on top?

[high lily]

no

[round island]

But don’t you have to multiply

[high lily]

as mentioned before IF you want to expand,
You distribute to both trrms

[ornate sable]

Bro I suggest you to think about it once

For like 5 mins

You’ll get it trust me

[round island]

I have- I was told my original instinct wouldn’t work

[high lily]

but don't distribute here

anyway apply Pythagorean to the denominator

[ornate sable]

I meant about what @high lily is saying

[round island]

Wdym 😭😭😭

[ornate sable]

Think about what he’s trying to tell you

Instead of directly jumping

[round island]

I’m confused on what you’re trying to tell me

[ornate sable]

I’m just telling you to slow down

[round island]

I don’t have the time to slow down 🥲

[timid silo]

sin^2x + cos^2x = 1is aka pythagorean identity

[round island]

Yeah I know thank you

[ornate sable]

It’ll slow you down more if you don’t

Okay listen

[round island]

True I guess

[ornate sable]

$\dfrac{a}{b} +\dfrac{c}{d} = \dfrac{ad+bc}{bd}$

[warm shale]

robin.dabanc_

[ornate sable]

Can you prove this lemma

[round island]

It’s cross multiplying

[ornate sable]

Are you VERY familiar with this

[round island]

….no

[ornate sable]

Look at it

Get a feel for what is happening

Then it’ll sit in your mind forever

[round island]

Im familiar with it I just haven’t used it….a whole lot

[ornate sable]

I do that times that, and then that times that and boom I get the numerator

Now let’s do the same thing with your two fractions

[timid silo]

https://tenor.com/view/oppenheimer-atom-nuclear-fission-gif-8314897022799710302

[ornate sable]

Identify a,b,c,d and plugin

[round island]

Right….but why do I do that?

[ornate sable]

You’ll see

In general if you have two fractions it’s always a good idea to bring it down to one

[round island]

That’s what I was trying to do with multiplying the top?

[ornate sable]

Because the RHS you wanna prove is just one thing, and is not a sum, I.e., doesn’t have a +

[round island]

Am I not supposed to multiply the top straight across?

[ornate sable]

Yeah but ur tripping

[ornate sable]

Just follow this

Trust

[round island]

I haven’t even started the top how am I tripping 🥲

[ornate sable]

In your case can you identify a,b,c,d?

[round island]

Yeah I guess?

Im not sure what you’re getting at here

I’m trying to understand so I can apply it and do it on my own

[ornate sable]

Okay go ahead

Idk where you’re going wrong then

[neon vector]

different from yesterday it seems

[round island]

HEY

Yesterday I was still trying to understand it just wasn’t clicking

[ornate sable]

Okay bro go on don’t get distracted

[round island]

Ok ok ok

So from cross multiplying on top I have 1-sinxcosx + 1- sin^2x

[ornate sable]

That’s where your mistake is

You forgot the ()

[round island]

FUCK

Ok ok

Wait but….cosine doesn’t have an extra part

So how do I foil with nothing else

[ornate sable]

Wdym

[round island]

A

Is cosine

Just by itself

[ornate sable]

yeah but the second thing has two terms

[round island]

I can’t foil with it if there are only 3 parts can I?

[ornate sable]

You have cosx(1+sinx)

= cosx +cosxsinx

[round island]

Is that not…the same thing just written differently?

[ornate sable]

Oh mb I saw the wrong picture

Yeah first term is just cos^2 x

[round island]

Yeah

Wait

Hold on how-

[ornate sable]

Second is (1-sinx)^2

[ornate sable]

ad+bc

a=d=cosx

So ad = cos^2 x

$(cosx)^2$

[warm shale]

robin.dabanc_

[round island]

But d is 1-sinx

[ornate sable]

??

D is the thing on the bottom in the second fraction

Which, according to your question is cosx

[round island]

Yeah

I think we’re talking about two very different things

And I think I need to back up

[ornate sable]

Bro

Wait

[round island]

Because I’ve been talking about the second line- this entire time

[ornate sable]

$\dfrac{\cos x}{1+\sin x} + \dfrac{1+\sin x}{\cos x}$

Just go from here

[round island]

I’m assuming sinx is meant to be there?

[ornate sable]

Yea

[round island]

Ok I’m on the same page now

[ornate sable]

Now just do the thing

From here

First term cos^2 x agreed?

[round island]

Yes

Cos^2x + 1+sin^2x

Over 1+sinxcosx

[ornate sable]

It’s (1+sinx)(1+sinx)

You’re missing the brackets everywhere

[round island]

Yeah my teacher didn’t…foil in any of her examples

[ornate sable]

Okay now do you u nderstand the brackets

1+sinx is one big unit

[neon vector]

I won't say anything about what happened after I left the channel because it might be true that you tried, but I know that while I was there you decided to ignore my questions.

And honestly I did have a way of making this click for you, and I have done it for quite a few people in the past very easily. It seems everyone tries to explain it with this particular example, but obviously all of the other elements hide the underlying concept, which is pretty intuitive once stripped down.

Just my observation, I don't mean this in a rude way or anything and hope you figure it out and wish you the best of luck!

[ornate sable]

You’re multiplying that with itself

[round island]

I’m sorry ill do better in the future

[round island]

Gotcha

[ornate sable]

So you may distribute it

[round island]

Alright let me fix it

So 1 + 2sinx + sin^2x?

[ornate sable]

NICE

Now what’s the denominator

DONT FORGET the brackets

[round island]

She just does all the bracket shit in her head so I never remember to write it

[ornate sable]

That’s why u need a teacher like me

✨

Jk lol

[round island]

👏👏👏👏

So just multiply them…both by cosine?

[ornate sable]

Can you write it out

[final plaza]

wait

[round island]

Yeah one sec

[final plaza]

why are we doing (1+sin x)^2

[ornate sable]

Isn’t that the question

Here

[round island]

Cross multiplication?

[warm shale]

robin.dabanc_

[final plaza]

why would we cross multiply here?

[ornate sable]

ACTUALLY pro tip: don’t expand the denominator

Leave it at cosx(1+sinx)

Because you know you want something to cancel out to the get RHS

[timid silo]

to take lcm and blah blah( adding fractions)

[ornate sable]

Wait

[round island]

I’m- just gonna ask my teacher on this one I fear

[ornate sable]

Milo is cooking

[ornate sable]

Bro you’re one step away

Can you write the numerator now

[round island]

How am I one step away I have to get to secant

[ornate sable]

Just wait and watch

It’s my black magic

[round island]

I need you to keep in mind I’m very limited on the properties I can use- idk if you saw that

[ornate sable]

I saw it

[round island]

Ok

[ornate sable]

If you’re allowed secx = 1/cosx we’re good

[round island]

So cos^x + 1+2sinx+sin^2x for the numerator

[ornate sable]

Now do Pythagoras

Cos^2 +sin^2

[final plaza]

ok i see where (1+sin x)^2 comes from now, numerator

[round island]

But what about the other parts attached are they not one big thing?

[ornate sable]

Now that you’ve expanded everything out, they are free terms

[round island]

Or are they all separate parts

Gotchaaaaa

[ornate sable]

you can add in any order

[round island]

Do I need the denominator rn-

Just in general

[ornate sable]

Just keep it at cosx(1+sinx)

[round island]

Gotcha

[ornate sable]

Don’t bother

So what does num simplify to

[round island]

2 + 2sinx / cos (1+sinx)

[ornate sable]

Yes

[final plaza]

brackets

[ornate sable]

Now factor the 2 out

[ornate sable]

We got what he meant

[round island]

Factor the 2 out?

[final plaza]

still important to at least point out if distribution is something that needs work

[ornate sable]

2x+2y = 2(x+y)

[round island]

GOTCHA

[round island]

It is 😔

[ornate sable]

Okay thanks

Congrats you’re done cooking

[round island]

So… 2(sinx +sinx)?

[ornate sable]

?

How did that extra sinx come

[round island]

You said x+y

There was one thing-

I doubled it

[ornate sable]

Uh

[round island]

Is it 2(sinx + 1)?

[ornate sable]

Not quite

[ornate sable]

YES

That expands out to 2+2sinx as you see

That means you’re right

[round island]

Gotcha gotcha

[ornate sable]

Now notice 1+sinx in denominator

And voila you have it

[round island]

I’m not following

[ornate sable]

You have 1+sinx in the denominator too

[round island]

Right so cancel them?

So….2+ sinx+1 on top? Or just fully gone

[ornate sable]

Just 2

Nothing else

Beacuse you canceled the rest of it

[round island]

Where did the other 2 go

[ornate sable]

$\dfrac{2(1+\sin x)}{\cos x (1+\sin x)}$

[round island]

Right-

[warm shale]

robin.dabanc_

[ornate sable]

=

[round island]

But there’s another 2

From earlier

[ornate sable]

Where

[round island]

When you made me do pythag to get rid of sinx^2 and cosx^2

That changed to a 1

And combined with the other 1 from 1 + 2sinx + sinx^2

[ornate sable]

You agreed till here

$\dfrac{2+2\sin x}{\cos x (1+\sin x)}$

[warm shale]

robin.dabanc_

[ornate sable]

You got this right

[round island]

Yes

[ornate sable]

This is the same thing

Notice the BRACKETS

When you expand it out, you get two 2’s

[round island]

Yes

[ornate sable]

So it’s the same thing

[ornate sable]

So this is also what you have right now

Do you get that?

[round island]

Ok slightly off topic but will you show me how you get the 2 2’s

[ornate sable]

The 2sinx came from expanding (1+sinx)^2

[round island]

Right

[ornate sable]

And the other 2 from 1+sin^2 x + cos^2 x

= 2

[round island]

Mhm

[round island]

I mean from this

[ornate sable]

Oh

[round island]

I know how I got them in the first place

[ornate sable]

2(1+sinx) you can distribute

[final plaza]

robin do you mind if i interject?

[ornate sable]

Sure go on

I’m busy actually

[final plaza]

so you actually only have a single 2 factor

not two 2's

the reason is because

what you are trying to do when you cancel something from the top and bottom

is you are cancelling a multiplication and a division

[round island]

Can you explain how it’s cancelling both?

[final plaza]

when you had 2+2sin x, yes there are two 2's as written

[round island]

Because couldn’t I just rewrite it as 2 + sinx +1 +sinx+1 and still have one left over?

[final plaza]

but this isnt multiplication so you cant cancel here anyways

do you mean 1 + 1 + sin x + sin x?

because the way you have it written it gives you 3 not 2

[round island]

I don’t see how that would give me 3-

[ornate sable]

It would give 4

2+1+1

[final plaza]

you wrote 2 + sin x + sin x + 1

2+1 + sin x + sin x

3 + sin x + sin x

[ornate sable]

U missed a 1

[round island]

Oh yeah I forget that 1 isn’t part of the original term

[final plaza]

can you fix that so i know what you mean?

[round island]

The way it was explained to me- and I’m not saying this is right just trying to give some insight

Is that there are 2 terms of (sinx+1) there

By cancelling- you get rid of one of them and the one on the bottom

Which would leave you with 2 + sinx+1 / cosx

And I’ll be right back I’m being called to the office

Like 5 minutes before anything because I’m not sitting in front of my paper 😭😭😭

[final plaza]

milo, do you mind if we put this problem on hold? i believe i have identified the core difficulties you're having in math, but we need to segue and come back to the problem in particular later

if you dont mind doing this, let me know when you get back, ping me so i can see the message

[ornate sable]

Milo which grade are you in

[latent thunder]

10th or 9th most prolly

[ornate sable]

NAH

Wait possible

That’s when most people learn trig

[latent thunder]

I mean the question is a 10th ncert

[ornate sable]

I see

[latent thunder]

Or 8th icse

[ornate sable]

Are you in 10th too

[latent thunder]

No

[obtuse pebble]

@round island Has your question been resolved?

[obtuse pebble]

@round island Has your question been resolved?

[final plaza]

btw @round island you are welcome to ping me in #prealg-and-algebra too if you dont feel like opening a help thread if this should close

[round island]

I’m a junior…..

[round island]

Can I dm you? I don’t want to lose you in the abyss and you have a…very complicated user

[obtuse pebble]

Channel closed

Closed by @round island

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[spiral aurora]

Can anyone (indian) help with 11th class trigonometry?

[midnight igloo]

sure, ask your doubt

[spiral aurora]

These two

[midnight igloo]

for the (i) part have you studied something like angles in A.P

you can see all the angles have same common difference of 2π/11

[spiral aurora]

Yeah

[midnight igloo]

Jee wale ho to pdhaya gya hoga

[spiral aurora]

New hu 11th pe

[midnight igloo]

acha

[spiral aurora]

Trigonometric functions of associated angles

Iss chapter se

[midnight igloo]

Mtlb ek formula hota h pdhaya jayega chapter ke end me usse jaldi hojata hai

Jee wale hi ho na

[spiral aurora]

[midnight igloo]

yes to pdhaya jayega angles in A.P krke i would say tb tk wait krlo

[spiral aurora]

Oo

[midnight igloo]

yes fir 1 line ka solution h pehle part ka

2nd part dekhta hu

[spiral aurora]

Damn

Abhi basic concepts se nhi hoga?

[midnight igloo]

let me try mai 12th me hu to utna fresh nahi h ye

[spiral aurora]

Oh

[obtuse pebble]

@spiral aurora Has your question been resolved?

[midnight igloo]

yo

Ek aur solution h but usme bhi ek formula use hoga

[spiral aurora]

Oo

[midnight igloo]

cos(theta).cos(2theta). Cos(2² theta)...... Ye formula pdha hai ?

iska product mtlb

this

Simple sum into product lagake solve kroge to end me iska use ayega

i guess you could ask someone else i only know these 2 methods

I don't think there is any other easy method though

[latent thunder]

Sometimes we over complicate ig

tan(pi/2 - theta) = cot(theta) lagega

[midnight igloo]

this is an easy solution but he hasn't studied it

Its like 3 lines

[latent thunder]

Ye toh padha hi hoga

Tan and cot are complementary

[midnight igloo]

Ha ye nahi pdha to fir its over

[latent thunder]

2nd mai vahi lagega

[midnight igloo]

2nd wala maine try nahi kia ab tk

But product me whi lagta h mostly

Complementary

[latent thunder]

Nahi kabhi kabhi complex number bhi use ho jaate hai

[midnight igloo]

maine nahi kia us level ka lagta h

Wo qn post krke bhag gya

[latent thunder]

😭😭

[obtuse pebble]

@spiral aurora Has your question been resolved?

[obtuse pebble]

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[viscid umbra]

what did you try?

[torn flower]

ok

so by descarte rule of signs

max 4 real roots exist

then

ik that value at positive inf and negative inf is positive

also the value of derivative stays positive after 1

and value at 1 is 15

so i know it will never be zero there

[viscid umbra]

what do you get when x is negative ?

[torn flower]

a rough graph

[torn flower]

from negative inf to negative 1 again no roots

i dont get the region from -1 to 0

the derivative is $8x^7 - 7x^6 +2x -1$

[warm shale]

parthisjoking

[torn flower]

and it will prob stay negative in -1 to 1

[viscid umbra]

when 0 < |x| < 1, what property do you have about x^n

[torn flower]

that x^n is smaller than x^n-1

[viscid umbra]

so you might be able to cancel out negative terms

[torn flower]

no , i want the derivative to be negative in region before 0

and positive after 0

[viscid umbra]

you dont need the derivative

just watch your function

[torn flower]

what the helly

[viscid umbra]

if x > 0, what do you have with x^8 - x^7

?

[torn flower]

that depends

from 0 to 1

[viscid umbra]

yes

[torn flower]

i have negative result

[viscid umbra]

positive

[torn flower]

from 1 to inf

i have +ve

[viscid umbra]

since 0 < x < 1, x^8 > x^7

[torn flower]

yes

so

i get negative then

but the other terms are also there

[viscid umbra]

so x^8 - x^7 > 0

yes, and you can do the same with x^2 - x

[torn flower]

what the helly

[torn flower]

noo

[viscid umbra]

oh

uh

😭

[torn flower]

dawg

you need rest

[viscid umbra]

sorry

mayvbe

[torn flower]

<@&286206848099549185> :3

[shut lagoon]

They're not far off.

x^8 - x^7 + x^2 - x + 15 = 0 in 0<x<1 just try some pairings keeping in mind that in this interval x^m > x^n whenever m < n

[viscid umbra]

oh yes i see it

[lunar vapor]

$$P(x) = x^7(x - 1) + x(x - 1) + 15$$

[warm shale]

akeanti💕

[lunar vapor]

for $[1, +\infty[$ u will find that $P(x) \ge 15 > 0$

[warm shale]

akeanti💕

[shut lagoon]

And then for x > 1 you can do a similar pairing although you've already convinced yourself that there are no roots there

[lunar vapor]

same for $]-\infty, 0]$

[warm shale]

akeanti💕

[viscid umbra]

he already treated x > 1

[torn flower]

i did this with derivatives

[lunar vapor]

oh well

i think it might be fast to prove with no derivatives

[torn flower]

yeah i see that now

wow

thanks

.close

[obtuse pebble]

Channel closed

Closed by @torn flower

Use .reopen if this was a mistake.

[shut lagoon]

For the record a pairing that works on (0,1) is x^8 + (x^2 - x^7) + (15-x)

But their method is cleaner

[obtuse pebble]

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as this action is irreversible, and this channel may abruptly lock.

[obtuse pebble]

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[echo rover]

Yall can someone check my answers

[obtuse pebble]

Please don't occupy multiple help channels.

[dense yoke]

is it the same thing as your other channel #help-6

if it is, please stick to one channel

[obtuse pebble]

@echo rover Has your question been resolved?

[spice inlet]

@echo rover use one channel at a time

[obtuse pebble]

@echo rover Has your question been resolved?

[tardy epoch]

.close

[obtuse pebble]

Channel closed

Closed by @tardy epoch

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Remember:
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[knotty apex]

Is there a way to get a function for a= or atleast a way to get a good approximation without specific like stat functions

[flint epoch]

desmos gotchu

[tardy epoch]

You can find inverse normal distribution and use properties of the probability distribution function

[knotty apex]

So id take the intergral of the inverse? I don't really understand.

[obtuse pebble]

@knotty apex Has your question been resolved?

[knotty apex]

How would i use the properties to find it. After taking the inverse of the distribution it doesn't look helpful

[tardy epoch]

Do you know what inverse normal distribution is

[knotty apex]

I mean i got it in terms of y

Wait

[tardy epoch]

Google or find a formula in your book and show it

[knotty apex]

Frick

I did it wrong i think

Yea definitely

I got this. Does this look right? My only concern is that it doesn't go over x=.5

Ill google rq

Idk it looks right but i cant really imagine how you would go from like .68 to the a value

Cuz if u entered an b value of .99/2 it would not be defined on the intergral

Im assuming you divid the orginal answer by two

Plus if you intergral it that max value would be 1 but a can be any number

Idk maybe im stupid cuz actually the intergral would be 0

Cuz its equally on positive and negative

[obtuse pebble]

@knotty apex Has your question been resolved?

[knotty apex]

<@&286206848099549185>

[obtuse pebble]

@knotty apex Has your question been resolved?

[obtuse pebble]

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[frank arrow]

ignore the words ik its chem

but im stuck on the math part

although it probably should be easy for me to do but im lowk brain farting or something cuz idk how to multiply everything out lmao 😭 🙏

[serene elk]

what are u stuck with? Solving the quadractic?

[frank arrow]

i was stuck on distributing haha but i got it now

.close

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[cyan sequoia]

guys

[cyan sequoia]

can anyone help me with a question

[viscid bronze]

What is your question

[cyan sequoia]

this one

[torn flower]

just expand the square and compare like power terms to be equal

2nd one can be done by factorisation

[mild sandal]

or complete square for the left hand side

[viscid bronze]

hint is to expand the right side for part a

[deep kraken]

Either way

But rhs expand is easier

[cyan sequoia]

oh ok

thx ill try now

[mild sandal]

lot of green ticks being thrown around

[deep kraken]

Cuz its correct💯

[mild sandal]

cant argue against that

[cyan sequoia]

a) a=−2 and b=−10, b) x=2+1 or x=2−10

my asnwer to the question

[ionic oar]

what did you do for b)?

[torn flower]

$a^2 + b = -6$

[warm shale]

parthisjoking

[obtuse pebble]

@cyan sequoia Has your question been resolved?

[deep kraken]

Seems good

[obtuse pebble]

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[gloomy vector]

for the note, why is it that "there cannot exists a prime between (a_2,2a_2-1)"?

[gloomy vector]

or is it just wrong :(

[ionic oar]

doesnt bertrand say that there is only 1 prime in (n,2n-1)

[gloomy vector]

theres atleast afaik

[ionic oar]

what😭

[ionic oar]

oh wait its atleast

[kind hawk]

maybe the note is just about the case distinction of p<=31 and p>32. that prime would take the role of r and then maybe the proof continues the same way

I havent properly digested the proof tho

[kind hawk]

there will obviously in general be many more than just 1 prime

would be very surprising if there was only one prime between 1000 and 2000

[obtuse pebble]

@gloomy vector Has your question been resolved?

[obtuse pebble]

@gloomy vector Has your question been resolved?

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[deep kraken]

Atp just do AxB

[timid silo]

[prisma spire]

?

[deep kraken]

A cross B

[prisma spire]

Well I did a+b and then dot product with i+j+k

I guess it isn't the right approach

[rose scroll]

you want unit vector along a+b

[prisma spire]

Means a+b/|a+b|

[rose scroll]

yep

[prisma spire]

Or simply (a+b) cap

Ok got it

@rose scroll thanks

Got the answer

λ = 1

.close

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[shrewd plover]

Help me find the gradient and equation of tangent line to the graph of f of x = x Sq - 2 at the point (-1,-1)
I jst studied the topic tangent to a curve at a point help me solve the example above

[inner sierra]

picture please

[inner sierra]

you learned the derivatives yet?

[shrewd plover]

Ye I learned the first principle and some theorems

[inner sierra]

cool

so you need to find $f'(x)$ basically

[warm shale]

1 divided by 0 equals Infinity

[inner sierra]

and stuffs will get easier

[shrewd plover]

Alr let me find the derivates of the given function

[inner sierra]

the derivative is basically the gradient of the tangent line

so after you find the derivative, you need to substitute the x coords into the derivative

should get the slope

[shrewd plover]

Suppose if the point was (5,10) I would substitute 5 into the derivates of given function which is basically a slope (gradient) of tangent line

[inner sierra]

yessir

at $x = 5$ tho

[warm shale]

1 divided by 0 equals Infinity

[inner sierra]

you need to note that

[shrewd plover]

Isn't the deirvate already slope?

[inner sierra]

but the derivative still has an x in it

so that's why the question asks for the slope at P(-1, -1)

[shrewd plover]

So the deirvate of given function above is 2x

At P (-1,-1)

I substitute - 1

And I get minus 2

Which is the gradient or slope of the tangent line

?

Am I right

Then how do u get the equation of tangent line I got the idea of slope (gradient)

[inner sierra]

correcto @shrewd plover

now a tangent line

it's just a line

so it's just $y = mx + c$

[warm shale]

1 divided by 0 equals Infinity

[inner sierra]

and you already got $m$, which is your -2

[warm shale]

1 divided by 0 equals Infinity

[shrewd plover]

Ye

[inner sierra]

the tangent line is at P(-1, -1)\

so it also lies on P

so substitute in y = -1 and x = -1

so you can find c

[shrewd plover]

Alr

[shrewd plover]

It's minus 3

[inner sierra]

aight

so the line is?

[shrewd plover]

?

A man pls explain

[inner sierra]

substitute m = -2 and c = -3 into the line eq?

[shrewd plover]

In y = mx+c I substituted y =-1 x=-1 these points were given and m was minus 2

Got c as minus 3

[inner sierra]

yea

so $y = mx + c$

[warm shale]

1 divided by 0 equals Infinity

[inner sierra]

sub in $m = -2$ and $c = -3$

[warm shale]

1 divided by 0 equals Infinity

[shrewd plover]

-2x - 3

[inner sierra]

good

[shrewd plover]

So u gotta substitute into the equation again without y and x

At the end

[inner sierra]

yep

[shrewd plover]

Thanks man imma try solve some questions related to it

[shrewd plover]

Bro if the points r not given instead jst x is given

How do u get the y

?

[inner sierra]

substitute in $f(x)$

[warm shale]

1 divided by 0 equals Infinity

[shrewd plover]

Oh alr I m jst dum😭

Ty

.close

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[mighty torrent]

Would someone be able to check my proof for mistakes?

[obtuse pebble]

@mighty torrent Has your question been resolved?

[obtuse pebble]

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[eager parcel]

help me out with this integral... I don't know from where to start

[trim portal]

hmm, would subbing pi/2 - x help

[torn flower]

kings rule

yk kings rule?

[torn flower]

yea

we get 1 + tan²x in numerator

[carmine spear]

oh what?

[torn flower]

but also an extra tanx down below

[torn flower]

simplified tanx + cotx

[carmine spear]

hmm ic

[stone moon]

subsititute dx for d(ln tan x) would make your life better

[obtuse pebble]

@eager parcel Has your question been resolved?

[obtuse pebble]

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[waxen iron]

For number 28 can someone help me understand the solution for the stack exchange post:https://math.stackexchange.com/questions/649442/ax-int-ax-ft-dt-in-a-b-let-c-in-a-b-if-f-is-continuous-at-c-the

[bronze mica]

It's using two standard results: if f' exists at c then f is continuous at c, and the fundamental theorem of calculus whoch says A' = f

[waxen iron]

Ok so I'm not understanding the result of if f' exists at c then f is continuous at c

I don't understand how explanations regarding neighborhoods

[obtuse pebble]

@waxen iron Has your question been resolved?

[waxen iron]

<@&286206848099549185>

[tardy epoch]

This is a theorem that differentiable at x=c then that implies continuous at x=c too

You can prove it using limit definition of differentiable at a point

[waxen iron]

Oh I see. So $lim_{h \to 0} \frac{f(c+h)-f(c)}{h}=f'(c)= \lim_{h \to 0} f(c+h)-f(c) = hf'(c)$. Since $h=0$ limit becomes $\lim_{h\to0} f(c+h)=f(c)$

[warm shale]

BigBen

[waxen iron]

What about the neighborhoods he was talking about

[fossil ingot]

You cant just make the h dissapear

[waxen iron]

Can't I set it to 0?

[pure geode]

You can't set it to 0 because it's a limit

It likely does not have defined behaviour at 0, and even so it might have different limiting behaviour

[waxen iron]

Ok but as it approaches 0 that term should be 0

[pure geode]

Why?

h can't be 0 because it's in the denominator

[waxen iron]

I multiplied both sides by h

[pure geode]

You can't multiply both sides by 0

Also the limit is only applied to the definition and not the derivative, so having h f'(x) doesn't make sense

[waxen iron]

So h would just be a variable that is being multiplied by the f'(c)

So my setup is wrong

?

[sand roost]

it's almost there - remember that h is still enclosed in the limit, so instead of multiplying by h, multiply both sides by...

[pure geode]

Do you know the definition of continuous?

[waxen iron]

Ye we can use epsilon delta or we can say that the value of the function at c is equal to the limit as x approaches c

[pure geode]

So how can you relate that definition to the one for the derivative

Given they both contain a limit

[waxen iron]

I don't see what else to multiply by

[waxen iron]

I'm confused. Are you asking how we can relate epsilon delta to the deirvative and what do you mean they both contain a limit?

[pure geode]

The definition of continuous uses a limit

and the definition of the derivative uses a limit

[sand roost]

in fairness, it's minimally different to h. recall that
$$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \cdot h = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} \cdot \lim_{h \to 0} h$$
so what are you actually multiplying by?

[warm shale]

حسیب ♥

[pure geode]

so try and use the derivative definition to show the continuous definition

[waxen iron]

Wait is there any need to multiply by anything anymore isn't what you wrote equal to $\lim_{h\to0} f(x+h)-f(x)$

[warm shale]

BigBen

[sand roost]

it is, but i wanted you to see that we are multiplying by $\lim_{h \to 0} h$, not $h$ as a variable

[warm shale]

حسیب ♥

[sand roost]

in fact, $\lim_{h \to 0} h$ is very much constant and is equal to 0

[pure geode]

That limit is resolvent and just has value 0

[warm shale]

حسیب ♥

[sand roost]

oh lol

[pure geode]

You have to very careful if you want to combine or seperate limits

[waxen iron]

Wdym

[sand roost]

if you're asking me: you wrote $$\lim_{h \to 0} f(x+h) - f(x) = hf'(x)$$
when it should have been
$$\lim_{h \to 0} f(x+h) - f(x) = \left[ \lim_{h \to 0} h\right] f'(x)$$

[warm shale]

حسیب ♥

[sand roost]

and you do have to justify this step a bit more: how do we know that $\lim_{h \to 0} f(x+h) - f(x)$ exists?

[warm shale]

حسیب ♥

[waxen iron]

Well we end up getting that it is equal to 0. If the limit approaches a definite value it must exist

[sand roost]

this is not always true. $\lim_{x \to 0} \frac 1x$ approaches a definite value, 0, but the limit doesn't exist

[warm shale]

حسیب ♥

[sand roost]

but the reason we know $\lim_{h \to 0} f(x+h) - f(x)$ exists is because of a certain property of limits

[warm shale]

حسیب ♥

[sand roost]

any ideas which one?

[sand roost]

sorry i should clarify: the limit is absolutely equal to 0 and this is how the proof goes, but the justification isn't quite right

[waxen iron]

How can this be? If the limit doesn't exist it doesn't approach a value

[sand roost]

oh whoops x2, i thought you meant that since x approached a definite value, that the limit existed

what you said is true, but how do you know that the limit approaches a definite value?

[waxen iron]

Not sure I would just use epsilon delta to show that we have a limit

[waxen iron]

Because we have that the limit is equal to the limit of h as h approaches 0 times f'(c) which is just 0

[sand roost]

right, and how do we justify this multiplication?

[waxen iron]

We sub in f'(c) for the limit defintion of the derivative

[sand roost]

we do, and then we apply the product rule, which is important to mention

i'm being pedantic, but we do have to justify that the limits we are working with exist at every step

[waxen iron]

Right so we apply the product rule of limits and then sub in for the limit defintion of f'(c)

[sand roost]

thus we get $\lim_{h \to 0} f(x+h) - f(x) = 0 \iff $\lim_{h \to 0} f(x+h) = \lim_{h \to 0} f(x) \iff \lim_{h \to 0} f(x+h) = f(x),$ which is a definition of continuity

[warm shale]

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

[waxen iron]

https://math.stackexchange.com/questions/649442/ax-int-ax-ft-dt-in-a-b-let-c-in-a-b-if-f-is-continuous-at-c-the so in regards to the answer given in this post is he saying that if f'(c) is continuous for some interval of c+- delta and then in that interval we can apply the above logic

[sand roost]

yeah, the chain here is
f' continuous at c
=> f differentiable around c
=> f continuous*
=> A' continuous (since A' = f)

* is where our little proof comes in

[waxen iron]

Can you elborate on the point around c

[pure geode]

The implication is not two way

[sand roost]

f' being continuous at c implies that f' exists for a bunch of values around c, which means that f is differentiable for all those values

[pure geode]

From the definition of continuity we get $\lim_{x -> a} f(x)-f(a) = 0$

[warm shale]

WeAreIngram

[waxen iron]

Sorry I'm not seeing this implication

[pure geode]

multiplying by $1 = \frac{x-a}{x-a}$ we get

[warm shale]

WeAreIngram

[pure geode]

$\lim_{x->a} \frac{f(x)-f(a)}{x-a} (x-a)$

[warm shale]

WeAreIngram

[pure geode]

Then we use the multiplicative property of limits to say that

$\lim_{x \to a} \frac{f(x)-f(a)}{x-a} (x-a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a} \lim_{x \to a} (x-a)$

and the last limit is known as equal to 0, so the whole expression is 0

That is, when the left limit is defined

Thus requiring the differentiable condition

[last pilot]

use \to

[warm shale]

ren (@ if replying)

$\lim_{x\to a}$

[waxen iron]

@pure geode I'm having troubling see the interval around c

[warm shale]

WeAreIngram

[pure geode]

The interval definition is equivalent to the $\lim_{x \to a} f(x) - f(a) = 0$ definition

[warm shale]

WeAreIngram

[waxen iron]

How?

[sand roost]

fair! if f' is continuous around c, then for |x-c| < δ, we have |f'(x)-f'(c)| < ε (assuming all the necessary assumptions and what have you). this means that, for values of x that are δ away from c, f'(x) has to not only exist, but be ε close to f'(c). so we end up with f' existing in an ε neighbourhood of f'(c)

[pure geode]

Look at the similarity between the formal definition of the limit and the definition of continuity

[sand roost]

actually, you can say that, since f' is a function, f' is defined for every value on [a,b]. then you can say that, since [a,b] is sufficiently "nice" (i.e. doesnt have any holes, is connected, compact <=> closed and bounded), that f' exists in a neighbourhood of f'(c)

i.e. it follows independently from the fact that f' is a function and it's defined on the entirety of [a,b]

[waxen iron]

If you mean by epsilon delta. The are very similar just the statement inside the absolute value is slightly different

[waxen iron]

But don't we have that f' is cotnious at c

[sand roost]

sorry, im still talking about this question

[waxen iron]

I'm confused. Didn't the the guy say f' continuous at c though?

[sand roost]

yes? so we have f' continuous at c, then we have that f is differentiable, then we apply the argument above to get that f is continuous at c

[waxen iron]

I just want to clarify that our neighborhoods that we are getting is because there are x values that are at most sigma+c or sigma -c. And f' is continuous for all of those

Meaning that for each of those f' points we can show that f is continuous

By what we did earlier

[sand roost]

i'm saying that f' exists at all of those points. f' existing at a point is the definition of differentiability, so i just want to argue that f is differentiable

but i want to argue its differentiable in a neighbourhood to avoid any problems. although it might be a bit overkill

you have the right idea, though. our neighbourhoods are dictated by δ and ε, and f' exists within the ε neighbourhood (the δ neighbourhood corresponds to x-values, not f(x)-values). so we can say f' exists around f'(c)

i.e. f is differentiable in a neighbourhood of f(c)

[waxen iron]

Ok so f is continuous for all those f' in the epsilon neighborhood

[sand roost]

exactly! and then of course f(x) = A'(x) by the fundamental theorem of calculus (I), so we are done

the original question is a really good foundational question to understanding all the interactions between differentiability, continuity, integrability, etc. etc.

in fact i've seen a similar question on a university final exam

[waxen iron]

Ye. It's a challenging question

.solved

[obtuse pebble]

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[outer cipher]

$C(n)$ is the number of reachable configurations of an $n \times n \times n$ Rubik’s Cube.

[warm shale]

rozehip

[outer cipher]

Is this correct? $$
C(n) =
\begin{cases}
1 & \text{if } n = 1 \
\displaystyle \frac{(8! \cdot 3^7) \cdot \left( \frac{24!}{(4!)^6} \right)^{\lfloor \frac{(n-2)^2}{4} \rfloor} \cdot (24!)^{\lfloor \frac{n-2}{2} \rfloor} \cdot \left( 12! \cdot 2^{11} \right)^{n \bmod 2}}{(24 - 23(n \bmod 2)) \cdot 2^{n \bmod 2}} & \text{if } n \ge 2
\end{cases}
$$

[warm shale]

rozehip

[outer cipher]

It matches the first few examples on OEIS: ```
1
3674160
43252003274489856000
7401196841564901869874093974498574336000000000
282870942277741856536180333107150328293127731985672134721536000000000000000

[trim portal]

https://hlavolam.maweb.eu/number-of-combinations-for-rubiks-cube

[outer cipher]

oh that is even better 🙂

thx

.close

[obtuse pebble]

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[deep haven]

<@&268886789983436800>

[obtuse pebble]

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[simple nova]

Is this the correct proof for13?

[trim portal]

|f(x)-L| = |h(x)-L| isnt necessarily true

but other than that, its mostly correct

[simple nova]

Would I just need to separate them out into their own inequalities to fix it?

[trim portal]

yeah

|f(x) - L| < epsilon for 0 < |x-a| < delta1 and similarly for g(x)

also you should mention what L is somewhere (maybe add = L to the 2nd line with lim f = lim g)

[simple nova]

Ahh alright, got it. Thank you.

.close

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

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[novel silo]

Can someone explain to me the mistake i did. I learned the proof for formula partial integration. I did it in an other way that should be correct but i get the formula but with extra stuff?

[pure geode]

What was the original question

[shut lagoon]

It's just messy to cancel dx's like that

[novel silo]

It’s a proof we started from d(uv) and the proof followed from there

[pure geode]

Yes but what are you trying to prove

[novel silo]

Partial intzgration formula

[pure geode]

integration by parts?

[novel silo]

Yes

[novel silo]

Well i cancelled the dx’s on the first one and it still gave me the formula 🤷‍♂️

[shut lagoon]

Yes but you're collapsing du/dx dx to du, which is fine if you think about u-substitution

If you do it with your second method you're integrating 1-forms which you do without appending dx's, but that's beyond calculus

[pure geode]

What is the final formula you're trying to show

[pure geode]

Have you tried integrating both sides of the product rule

[shut lagoon]

They did that twice just with two different ways

[pure geode]

I'm confused about what precisely is happening because it reads like youve already done it

[shut lagoon]

They have, but they're wondering why their second way of doing it has extra dx's appearing after integrating both sides wrt dx

[novel silo]

Im confused sorry english isnt my first language i dont know what u mean

[shut lagoon]

In short, avoid stuff like df/dx = dg/dx -> df=dg because you're dealing with differentials and that's harder to justify at this level

[pure geode]

What is this second method attempting to do

Beside reordering the steps of the first

[novel silo]

Well i just tried it in an other way it shouldve given me the same thing but it didnt

[shut lagoon]

You can do $\int \frac{\dd{f}}{\dd{x}} \dd{x} = \int \dd{f}$ because while it looks like cancelling $\dd{x}$, you're really using the substitution $u = f(x)$.

[warm shale]

Azyrashacorki

[novel silo]

I have no clue what ur talking abt

What does substitution have to do with this

Can u maybe add extra steps 🥲

[shut lagoon]

If you have $\int \frac{\dd{f}}{\dd{x}} \dd{x}$, by setting $u = f(x)$ you get $\dd{u} = \frac{\dd{f}}{\dd{x}}\dd{x}$, so $\int \frac{\dd{f}}{\dd{x}} \dd{x} = \int \dd{u} = \int \dd{f}$.

That's just using substitution

[novel silo]

So du=df/dx but dont u get integral (du dx ) ? How do u get integral du after substituting

[warm shale]

Azyrashacorki

[shut lagoon]

In your second method you end up with $d(uv) = d(u) v + d(v)u$, which is fine in some sense, but then you'd have to integrate in a different sense than "integrating w.r.t. x".

[warm shale]

Azyrashacorki

[novel silo]

Yea i dont get it

This is confusing

I just dont see how that substition correlates to my problem

Or how else i’d have to integrate it instead of wrt x

[shut lagoon]

What I'm saying is that the first method is easier to justify through substitution

[novel silo]

Well is there anything i can do to make the 2nd method work too

Both are the same no?

All i did was cancel dx’s in first method before integrating both sides and 2nd method i cancelled the dx’s after integrating both sides

[shut lagoon]

You could salvage it by writing $u=u(x)$ and $v = v(x)$, so that $\dd{u} = \frac{\dd{u}}{\dd{x}}\dd{x}$ and $\dd{v} = \frac{\dd{v}}{\dd{x}}\dd{x}$.\

Then $$u(x)v(x) = \int \frac{\dd{(u(x)v(x))}}{\dd{x}} \dd{x} = \int v \frac{\dd{u}}{\dd{x}} \dd{x} + \int u \frac{\dd{v}}{\dd{x}} \dd{x}.$$

[warm shale]

Azyrashacorki

[shut lagoon]

The point is it's messy if you argue by treating derivatives as fractions.

And while your first method does that too, it's easier to justify it.

[novel silo]

So the 2nd one is technically wrong?

[shut lagoon]

Kinda. It's just clumsy

Essentially you sort of just have have to integrate both sides (as in just stick an integral sign out front without appending dx's) when you deal with differentials like that

[novel silo]

Ok thanks

.close

[obtuse pebble]

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[obtuse pebble]

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[hard galleon]

The sum of the radii of the inscribed and circumscribed circles of a right-angled triangle is equal to one of the legs. Find the acute angles of the triangle.

[deep kraken]

what have you tried

[hard galleon]

The center of the inscribed circle lies at the intersection of the bisectors

And described on the hypotenuse, dividing it in half

[deep kraken]

can you draw it out

[hard galleon]

But then I came to the absurdity

1 angle + 2 angle equals 90 degrees

[deep kraken]

so you try to find the 2 acute angles?

[hard galleon]

Is there something wrong with my construction?

Oh my god, it's 45 degrees there

I just miscalculated

[deep kraken]

its a square isoceles triangle

[hard galleon]

Is that right?

Since this is a tangent, it is perpendicular, but it is a median, so the triangle is isosceles. The rest is just calculations

[deep kraken]

correct

or you can just show the two triangles are similar then show isoceles

then find angles

[hard galleon]

Thanks

.close

[obtuse pebble]

Channel closed

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[deep kraken]

no worries

[last pilot]

<@&268886789983436800>

[thorny fiber]

[last pilot]

thanks mods

[obtuse pebble]

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[noble venture]

Help pls

[inner sierra]

<@&268886789983436800>

previous history: #help-11 message

[pure geode]

Moderators beat this guy up

[brazen viper]

It sounds as if this is a new user who is confused about a particular meme question

[pure geode]

They have posted lots of AI stuff and multiple troll questions

[brazen viper]

they posted AI stuff in the previous thread that was linked, but I haven't yet seen evidence of multiple troll questions

[solar elbow]

also to the moderators, this user tried to help in 23 using AI as well.

[timid silo]

they shall feel the my wrath

[brazen viper]

and it's not against the rules to post AI shit as "this is stuff that I've tried, I don't understand it, can someone else help explain."

[pure geode]

They posted about riemann zeta in 3 channels

[inner sierra]

#help-23 message

[solar elbow]

just thought it would be useful context.

[pure geode]

They posted AI to try and help someone

[inner sierra]

no that's the other guy

[polar fossil]

they appear to have been banned for 1 + 2 + 3 + 4 + ... days

[pure geode]

Wait

[inner sierra]

good one

[worn yoke]

waow i didn't know you can ban for negative days

[karmic bronze]

try using the riemann zeta function

[brazen viper]

that is indeed a problem

[inner sierra]

and even argued when i tried to warn them

[polar fossil]

i suppose they will have come back 2 hours ago

[pure geode]

This guy knows analytic continuation

[karmic bronze]

what are they like owner now

[inner sierra]

thanks for backing me up 🔥

[pure geode]

https://giphy.com/gifs/bombaysoftwares-bro-bros-forever-bhNGEPUeypOaEKDvXu

[inner sierra]

accept my fq pls

[pure geode]

?

[inner sierra]

@
Moderators

[torn flower]

its a matter of belief really

[inner sierra]

click Report Spam

[brazen viper]

.close

[obtuse pebble]

Channel closed

Closed by @brazen viper

Use .reopen if this was a mistake.

[wide coyote]

what r u guys doin here

[chrome cedar]

I was just looking for help but ok

[pure geode]

Chatting

[chrome cedar]

Sorry

[torn flower]

math

[brazen viper]

check out #❓how-to-get-help

[inner sierra]

send your question in #help-47

[wide coyote]

shoudlnt u only help here and not chat???

[pure geode]

No-one to help

[inner sierra]

it's already been handled

[chrome cedar]

That was unnecessary asf

[brazen viper]

people can get a little bit spammy when something interesting happens.

this channel will lock or refresh soon enough

[pure geode]

Sorry queen, why you direct to dm advertising like that tho?

[polar fossil]

.and then someone will be left with their name on it because they were silly

[inner sierra]

ill definitely remember those words to counter hanako's argument

[wide coyote]

hmm

[chrome cedar]

Cuz I'm trying to beat this toxic person who i considered a friend as a kind of achievement

[inner sierra]

i sent a suggestion about it and no one accepted 🥀

[chrome cedar]

She is spreading misinformation about me and is taking my insecurity to her advantage

@inner sierra what is so funny

[inner sierra]

yo

[obtuse pebble]

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[shadow atlas]

Any idea on how to dolce this expression?

[shut lagoon]

Consider log rules with the fact that 2^3 = 8 and 2^5 = 32.

[shadow atlas]

Like this?

[shut lagoon]

Yeah

[shadow atlas]

so now I use the power rule correct?

[karmic bronze]

<@&268886789983436800>

[shadow atlas]

ok solved

[shadow atlas]

.close

[obtuse pebble]

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[obtuse pebble]

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[blissful sequoia]

Hi idk how to get the pr length i think it's gonna need the phytagoras formula but i couldn't find the ps nor the pq

[old lily]

what is OQ

[blissful sequoia]

It's 5 cm

[halcyon pewter]

what's QP

[old lily]

yes so SO

[blissful sequoia]

It's 10 cm

[old lily]

SO = OQ - SQ

good idea to draw on this circle if u can

to keep track of all the lengths

[blissful sequoia]

Okayy

So it's gonna be 2 cm right?

[old lily]

yurr

PS = 5+2 = 7

[blissful sequoia]

Yes

[obtuse pebble]

@blissful sequoia Has your question been resolved?

[sand snow]

yes you need pythagoras

first draw OR and consider RSO right angle triangle and use pythagoras to get RS as sqrt(21)

OR is radius btw = 5cm

now in RSP right angle triangle SP=7cm,RS=sqrt(21) you can find RP using pythagoras

[blissful sequoia]

Sorry but could u re explain the RS part? Idk what sqrt is

[sand snow]

Square root

[obtuse pebble]

@blissful sequoia Has your question been resolved?

[obtuse pebble]

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