help-38

[polar ocean]

Well I suppose you could move on and check it later

There are some other easier values to fill in

[wraith hinge]

Sigh

So for the third one

For cos

[polar ocean]

Yep

[wraith hinge]

Would it be 180

[polar ocean]

Yeah

[wraith hinge]

What abt the tan one

[polar ocean]

Hmm

I don’t really remember my trig values but

I think it comes out to some nice value

[wraith hinge]

How do I figure it out

If tan is y/x it doesn’t even show up on the unit circle

Like idk how to get it

[polar ocean]

Hmm

Well I’m not sure if you’re allowed to use a calculator but if you are

You could do that

Or maybe

Oh wait

-tan(-x) = tan(x)

So just find tan(x) = 1

That’s 45 right

Ok so that means -tan(-45) = 1
or tan(-45) =-1

[wraith hinge]

Huh

So I write -45

[polar ocean]

Yeah

Or since tan is periodic you could shift it to the positives

Add 180 I think

But -45 should work I think

[wraith hinge]

Okieee

What abt the last one

Can tan or sin be undefined

[polar ocean]

Ok so tangent can be

But sine cannot

Because of the range I mentioned earlier

-1 to 1, it can’t be any greater or any less

[wraith hinge]

So the last one is one of the two that don’t work

[polar ocean]

Yes

[wraith hinge]

Which one is the other one that doesn’t work sgain

I feel like we did all 3

Oh is it the first one

[polar ocean]

Well since we know 2 and 3 definitely work

Yeah

First one probably

Technically you can find a value for tangent but there isn’t a nice one

[wraith hinge]

Hmm

What would it be

Like how would it not be nice

[polar ocean]

It’s transcendental

I think

Maybe possibly don’t quote me on that

But it’s definitely not an integer and definitely not a fraction

[wraith hinge]

Okay so for my explanation

[wraith hinge]

For the last one do I say this

Or do I need to expand on that

[polar ocean]

You can just say there is no value of sine that results in an undefined value

[wraith hinge]

Okk

And for the

1st one

Do I say for tangent it would be in decimals

[polar ocean]

Hmm

How about saying it cannot be represented as an integer

[wraith hinge]

Wdym by that

As a radian/degree?

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[grim sparrow]

I'm trying to prove a) at the moment

[grim sparrow]

the way I'm planning on doing this is going to be
a) use the fact that S is embedded to find a slice chart around each of its points
b) extend the function f to each of the slice chart seperately
c) glue them all together using a partition of unity

my trouble at the moment is with b)

here's everything I've written down so far

I need to define f_p to be an extension of f to U_p

but I'm struggling to figure out how to formally do this

if S and U_p were subsets of R^n, I could do this easily: set f_p(x_1, ..., x_k, x_k+1, ..., x_n) to be equal to f(x_1, ..., x_k)

for example, I could extend any function f defined on the x-axis of R^2 to all of R^2 by setting f_p(x, y) = f(x)

but we don't have coordinates on the manifold itself so I'm confused on how to do this now

any suggestions?

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[grim sparrow]

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[tired badger]

why was x1 added

[tired badger]

<@&286206848099549185>

[trim lichen]

we start from the point (x1, y1) and go from it by (x2-x1)/2 horizontally

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@tired badger Has your question been resolved?

[tired badger]

ohh ty

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[bitter prairie]

help for 5🥲🥲

im so confused and atp i give up solving it haha

[thorn bay]

you didnt use my advice from last time

#help-5 message

[bitter prairie]

i forgor

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[broken cave]

Did I make any mistakes?

[broken cave]

It is a orthogonal matrix so the answer should be like this

The magnitude of vectors should be equal to 1

So there are 2×2×2 matrices

Did I make any kind of mistake?

[marsh forum]

1 or -1

[marsh forum]

the answer is ||16||

[broken cave]

@marsh forum I don't really care about the answer

But where did I make a mistake?

1 or -1

It's 2×2×2 fundamental principles of counting

[marsh forum]

Understandable, yea

[broken cave]

Right?

[marsh forum]

one minute

[broken cave]

Ha

[marsh forum]

I think you should recheck your matrix multiplication

[broken cave]

Ohkk

Oh fk

[marsh forum]

also, you get 2a_1=2a_1

[broken cave]

Huge mistake

I won't get any conclusions from multiplication

Hm

[marsh forum]

Hmm, what matrix did you get

[broken cave]

By checking by multiplication I got the same matrix 🥹

PQ and QP commute

So this must be valid

[marsh forum]

I think what you want to use is $P = Q^{-1} PQ$

[solid kiln]

What a wonderful world !

[broken cave]

Okay

P=Q'PQ

[marsh forum]

well, P=Q^TPQ

actually

[broken cave]

Yeah Q^T can be written as Q'

Let me try again

[marsh forum]

Also,why are you doing this at 11 pm 😭

[broken cave]

🥹🥹🥹 it's 11 Am here

I solved the problem till 2 pm

Am*

[marsh forum]

ah

I was concerned that someone was doing JEE adv problems for fun

lol

[broken cave]

There's one more jee advanced problem I solved and I got it right

[marsh forum]

That;s nice

[broken cave]

No I'm going for jee advanced after a year

[marsh forum]

You do know the papers have been released?

[broken cave]

Yes

[marsh forum]

Also notice $|det (Q)|=1$

[solid kiln]

What a wonderful world !

[broken cave]

Yeah orthogonal matrix

Do I have to multiply 3 matrices to reach a conclusion 💀

[marsh forum]

wait, i ahd asked this earlier, let me search for the soln I got

just realised

[broken cave]

Ha

So is there a trick ?

[marsh forum]

#help-46 message

[broken cave]

Holy

Moly

I knew this kind of shit can be done but I didn't do it ,it came into my mind at 1am but didn't apply it

Instead I chose to do wrong multiplication 💀🥹

[marsh forum]

Is this a different question

[broken cave]

Yes

[marsh forum]

lemme check

[broken cave]

It was blurry now it's fine

[marsh forum]

give me 5 minutes to go though it

[broken cave]

This is -2

[broken cave]

Ok

[marsh forum]

How did you get Tr(Q)=5

[broken cave]

Clayey Hamilton theorem

Characteristics equation of matrix

[marsh forum]

You do realise Q^2-5Q+6I = (Q-2I)(Q-3I)=0

you can expand it to verify

where 0 is the null matrix

should have used O my bad

[broken cave]

The determinant value of Q-2I is 0 and Q-3I is 0 so Q is not a null matrix but determinant value of Q-2I and Q-3I is 0

[marsh forum]

yes

[broken cave]

Np=Not possible

And ignore the things written in a cloud shape

That's the same thing written in my language

[broken cave]

Situation for AB=0

[broken cave]

So it is not gonna affect anything I think

[marsh forum]

yea, it won't

[broken cave]

Yes

[marsh forum]

I don't get why you sent this note though?

anyway, now to find det{Q-6I}

[broken cave]

Yes

[broken cave]

I did it here

(x-6)(4-6)-yz=(1-6)(-2)-(-2)=12

[marsh forum]

that seems to check out

[broken cave]

Ah okay

[marsh forum]

I think we're done, no?

[broken cave]

Yes

Thanks my friend

🫂🫂🫂🫂🫂🫂@marsh forum

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[fair tangle]

please let me know if this is correct

[marsh forum]

Is this a test

[fair tangle]

its a test review but if i get it wrong i have to do 3 more of the same questions and my hands hurttt 💀

it stacks up the more you get wrong so you practice but ive gotten 6 wrong and I dont think im getting it

if its wrong can you help me figure out where i went wrong I have my work but my handwriting sucks

ok thank you

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[torn shale]

Number of ways to put things in a circle

[torn shale]

Can someone help me figure that formula out

Like- number of ways to put 7 children in a circle

Pretty sure this has smth to do with permutations and combinations but idk where to start

[trim jolt]

@torn shale Has your question been resolved?

[dreamy path]

theres a simpler way of thinking about this ||cutting the circle into a line||

youre right it has to do with permutations

[torn shale]

Oh so its seven factorial

[torn shale]

This

[dreamy path]

almost.. keep in mind you can cut the same circle in seven different ways and get different lines

so you need to work backwards to account for how youve basically counted the same thing 7 times

[torn shale]

But how does the order matter here

Its just seven children on a line

How can we get different lines from the same children (arguing that its a combination not a permutation)

[dreamy path]

i guess it depends on whether the direction of your circle matters

[dreamy path]

so if i have a circle of 4 things

b
a c
d

i can have abcd, bcda, etc depending on where i cut

[torn shale]

Mhm

[dreamy path]

does that help?

[torn shale]

I dont think its relevant to the question im asking and its rather relevant to when order matters which i dont think it does when it comes to the children problem

The options are 1 , 7 , 720, and 5040

Its a choose question

[torn shale]

Oh is it six factorial

Because for the first child it wouldnt matter where we put him at all. The order will only matter starting with the second kid

So it will be (n-1)!

[dreamy path]

yeah!

[torn shale]

Thank you!

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[obsidian fjord]

Not sure where to start on this one, just need a little push, thanks.

[dapper swift]

$\frac{dv}{dx} = \frac{dv/dt}{dx/dt} = \frac{a}{v}$

[solid kiln]

south

[obsidian fjord]

I follow through and get -4v=dv/dx

[dapper swift]

yeah and you can check from the answer that dv/dx = -8e^(-4x), which is indeed -4v

[dapper swift]

great all that's left to do is to rearrange and integrate both sides

[obsidian fjord]

alright thanks

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[manic lagoon]

We have f(x,a,b) = 9x^2 - 9(a+b)x + (2a^2 + 5ab + 2b^2)
If we are told that x = (2a+b)/3 is a root of this, that is, for any value of a and b, if x = (2a + b)/3 is put in the function, the output is 0, cannot we conclude immediately that (2b + a)/3 is also a root of this function?

[carmine spade]

yes

[manic lagoon]

So, if f(x,a,b) = f(x,b,a) and f(g(a,b), a,b) = 0, does this imply that f(g(b,a), a, b) = 0?

[carmine spade]

yeah think so

[marble wharf]

f(g(a,b),a,b)=f(g(a,b),b,a)=f(g(b,a),a,b)

where the second equality is just because names dont matter

[manic lagoon]

Thanks.

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[manic lagoon]

Wait, isn't the second equality not always true?

.reopen

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✅

[marble wharf]

I am switching the names

[manic lagoon]

Are you saying that f(g(a,b),a,b) is always equal to f(g(b,a),a,b)?

[marble wharf]

lets do a middle step

you give me f(g(a,b),b,a) and I dont like your names so I switch them to f(g(x,s),s,x)

importantly, relative to each other the variables have the same names

its a,b,b,a in that order before and now its x,s,s,x

so now I give this to a friend again and he doesnt like my names either and renames it to f(g(b,a),a,b)

its still the same expression but with the names switched

anyway, I have to go

[manic lagoon]

Okay, thank you for your time

Consider f(x,a,b) = x - a - b

Let g(a,b) = 2a + b
Then f(g(a,b),a,b) = (2a+b) - a - b = a
f(g(b,a),b,a) = (2b+a) - b - a = b

<@&286206848099549185>

[marble wharf]

well ok a key point is that it was zero so the resulting expression didnt depend on a and b anymore

so we were just allowrd to switch names

the name switching is a bit weird in that regard

[manic lagoon]

Okay. Thanks.

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[silver void]

how do i solve for all x such cos(x) = 1/4

[silver void]

oh nvm

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[lavish cosmos]

i have been asked for my real analysis class to bring in an example of "A sequence of functions fn : [0, 1] → R which converge uniformly on [a, 1] for all a > 0, but do not converge uniformly on [0, 1].

Im gonna be honest i dont reall understand the topic and how i would i find an exampl eof this

[lavish cosmos]

to show this do i just need to find a f(n,x) that when i take the lim n appraoch infinite of x = 0 its different to every other limit from (0,1]

[nova spire]

do you understand at least what uniform convergence means or is supposed to represent?

[lavish cosmos]

its like an epsilon band that all the fn(x) conervge inside?

[nova spire]

it's kinda like that

just taking the example of uniform convergence to the 0 function

Imagine you have two straight horizontal bands below and above the functions f_n

[lavish cosmos]

yes

[nova spire]

they rest precisely on the maximum and minimum value of f_n for every n

you have uniform convergence if both those horizontal bands manage to get to 0

[lavish cosmos]

what so the limit of each should be the same for all x values, when n appraches infintiy

[nova spire]

it's not just every f_n(x) goes to 0 for every x

they all go to 0 at a "similar rate"

or at least they have a minimum required speed to converge to 0

[lavish cosmos]

they explanatnio they gave us was only 2 slides and didnt use the idea of rate i dont think

[nova spire]

you've seen the sup norm for functions right

[lavish cosmos]

yeah ive seen the supreme of functions

[nova spire]

$|f|\infty = \sup{x\in D} |f(x)|$

[lavish cosmos]

no i havent

[solid kiln]

rafilou is not not born in 2003

[nova spire]

well that's weird but ok

basically that norm is "how far away can this function be from 0"

[lavish cosmos]

hmm ive seen supremumjust in the context of a max value

[nova spire]

for continuous functions, it's kinda like the max

[lavish cosmos]

yeah okay

[nova spire]

this "norm" stuff goes through all the points on the curve of f

[lavish cosmos]

okay

[nova spire]

and it finds the point the furthest away from the x axis

[lavish cosmos]

yeah makes sense

[nova spire]

and it gives its distance

[lavish cosmos]

yeah

[nova spire]

so

if we can manage to get that "maximum distance" to converge to 0

then every point must converge to 0 AT LEAST at that speed

[lavish cosmos]

okay

[nova spire]

because you can't be further than the furthest point

just one thing to know

is that when you look at multiple functions f_n(x)

even if for every fixed x, f_n(x) goes to 0

the maximum distance can sometimes not decrease at all

this can seem weird, because every point individually goes to 0

but if you think about the "point of maximum distance" changing for each function

it's kinda like a whack-a-mole

you try to make some point go to 0 because they're the furthest away

but then some other point becomes the furthest away

[lavish cosmos]

hmm okay

[nova spire]

it's those types of functions

that won't converge uniformly

[lavish cosmos]

was what i was describing before converging pointwise

[nova spire]

Now, have you seen examples of functions that don't converge uniformly

brb

[lavish cosmos]

all good

[nova spire]

back

[nova spire]

so?

[lavish cosmos]

we saw the example fn(x) = x^n on domain [0,1]

all we saw was that lim n appraoch infitnity fn(x) for [0,1) was 0 and 1 for x = 1

[nova spire]

that's a very good example

now it's not exactly a whack-a-mole in this case

[lavish cosmos]

hmm

[nova spire]

because the functions we deal with aren't continuous

and so the "point of maximum distance" doesn't really exist

but there are some points that are always above a certain distance

[lavish cosmos]

why arent the function contious

[nova spire]

well the limit function to be precise

f(x) = 0 if x < 1 and f(1) = 1

[lavish cosmos]

hmm

[nova spire]

that's not exactly continuous now, is it

[lavish cosmos]

oh okay yes

[nova spire]

before that I talked about uniform convergence to 0

but in general

if f_n converges uniformly to f

that means f_n - f converges uniformly to 0

so now when we look at f_n - f

we're not exactly dealing with continuous functions

[lavish cosmos]

okay

[nova spire]

so there's no exact "point of maximum distance"

but we can always find points above a certain distance

so if there are always points above a certain line

[nova spire]

our "horizontal bands" can never hope to get to 0

[lavish cosmos]

yeah okay

[nova spire]

but now, let's say we investigated those functions on [0,a] where a < 1

what happens now?

[lavish cosmos]

dont know if im getting this right but for my example to find a non unfirom conerging thing

i would need to find a functino that fn(x) - f(x) cant equal 0 at x= 0 but does equal 0 at (0,1]

[lavish cosmos]

what function are you reffering to?

sorry im pretty lost

[nova spire]

those f_n(x) = x^n

[lavish cosmos]

oh okay

[nova spire]

that we were investigating at [0,1]

[lavish cosmos]

yeah

[nova spire]

what happens if we look at them on [0,a] instead

a < 1

like it seems the problem of uniform convergence was around 1

so what if we cut off a chunk next to 1

[lavish cosmos]

then it woudl be uniformly conervgent for [0,a]

[nova spire]

alright, yes, it would be the case

can you prove it?

with the definition you have in your textbook

or your lessons

[lavish cosmos]

so i would need to find an M, such that N. > M, fn(x)-f(x) less than epsilon

[nova spire]

yes

|fn(x) - f(x)| < epsilon

[lavish cosmos]

yeah

[nova spire]

and M doesn't depend on x

[lavish cosmos]

no they depend on epsilopn

M does right

[nova spire]

yes

for all n > M, for all x, ...

that's what the "uniform" is about

[lavish cosmos]

is the reason we look at n to infinitry because its the largest n can be so if it doesnt work then it wont ever work?

[nova spire]

that constant M is "uniform" on the values of x

[lavish cosmos]

okay

[nova spire]

uh I didn't get that

[lavish cosmos]

all good

nonsense

so uniform convergence is finding a M that works for all the pointwise versions of x?

[nova spire]

yeah, that's what uniform means

[lavish cosmos]

okay

[nova spire]

M doesn't depend on the "part of the space" chosen

[lavish cosmos]

makes sense

[nova spire]

compared to pointwise

which is, for each point, I can find an M

that may depend on x

[lavish cosmos]

yeah okay

[nova spire]

alright so notice that in our example, f should be 0

so we need |fn(x)| < ...

and now notice that the maximum f_n(x) can ever be is a^n

|fn(x)| <= a^n

[lavish cosmos]

yeah

[nova spire]

and that doesn't depend on x

but a^n goes to 0

because a < 1

so using the definition of "a^n -> 0"

for every epsilon

[lavish cosmos]

yeah

[nova spire]

there is M

such that for all n > M

a^n < epsilon

and that's it

[lavish cosmos]

okay

we cant find a M for x=1 so thats why it isnt uniformly convergence there?

[nova spire]

that's why the proof we had for [0,a] doesn't work for [0,1]

[lavish cosmos]

yeah

[nova spire]

because saying |fn(x) - f(x)| <= 1^n

isn't very enticing

[lavish cosmos]

so we choose to look at a because it would be the largest distance from 0? Then looked at its limit as n apporach infincity as see its 0

[nova spire]

yeah, in that case

[lavish cosmos]

okay

[nova spire]

x = a is always the point of maximum distance from 0

[lavish cosmos]

yeah

we wanted its liimit to equal 0 because f(x) = 0?

[nova spire]

yeah, f_n(x) - f(x) must uniformly converge to 0

and since f(x) = 0 everywhere on [0,a]

[lavish cosmos]

okay

[nova spire]

we're good

now, to show more rigorously why x^n doesn't converge uniformly to "f(x) = 0, f(1) = 1" on [0,1]

as stated

we would like to show "for all epsilon, there is M, for all n > M, ..." is false

so we could find an epsilon that doesn't verify that statement

so for example epsilon = 1/2

because

f_n(x) = x^n has f_n(0) = 0 and f_n(1) = 1

so I can find some point

where f_n(x) = 1/2 right?

since it's continuous

[lavish cosmos]

yeah

[nova spire]

so

that point may depend on n

let's call it x_n

[lavish cosmos]

yeah

[nova spire]

but now we have f_n(x_n) = 1/2

so we're NEVER gonna have "|f_n(x) - f(x)| < 1/2 for all x"

[lavish cosmos]

yeah makes sense

[nova spire]

alright

so that was a very good example

of f_n converging uniformly to f on [0,a] for every a < 1

but not on [0,1]

[nova spire]

we're not that far off from what you wanted right?

[lavish cosmos]

haha yeah

[nova spire]

I'll let you find out how to adapt what we did

to get exactly what you want

[lavish cosmos]

okay

[nova spire]

I have to go so ping helpers if you're stuck again

alr cya

[lavish cosmos]

can i just ask when trying to think of an example is there any easy way to tell if it will uniformly conervge

[lavish cosmos]

thank u

[nova spire]

uh

there are a few tricks

ofc if you can't find the sup of |f_n - f| on the spot

there's this nice property:

"let f_n: I -> R and f:I -> R. If all f_n are continuous and f_n converges uniformly to f, then f is continuous"

[lavish cosmos]

okay

[nova spire]

so "uniform convergence preserves continuity"

meaning

if you have a sequence of continuous functions

and the pointwise limit is not continuous

do you think it can converge uniformly?

[lavish cosmos]

no i hope

[nova spire]

exactly, it is impossible

[lavish cosmos]

hooray

[nova spire]

so

f_n continuous and f not continuous

[lavish cosmos]

so use that check but then need to prove for the domains

[nova spire]

means DEFINITELY NO

f_n continuous and f continuous means we don't know

[lavish cosmos]

okay great

thank you for all ur help really apprecaite it

[trim jolt]

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[forest stone]

For 10 I am trying to make my inductive hypothesis. I assume that n=qb+r, 0≤r<b. Next I will make n+1= q'b+r' 0≤r'<b. How do I know that both q and r need to change. Couldn't it be only one to have a +1 increase in n?

[marsh forum]

You have $n=qb+r$
$n+1 = qb+(r+1)$, if 0≤r<b-1, you. have, r'=r+1

[solid kiln]

skye ( aka wai)

[marsh forum]

can you do it from here

[forest stone]

Why b-1?

[forest stone]

?

[marble wharf]

there are two cases

either the remainder is less than b-1 or the remainder just happens to equal b-1

what happens in each of those cases for n+1

[forest stone]

I'm confused why we're doing b-1 in the first place

[marble wharf]

lets do a few examples

lets say b=5

[forest stone]

I thought the for the equation to be true r must always be less than b so if we make a new equation with r' then b still must be greater?

[marble wharf]

can you write down the q and r for all numbers up to 10 ?

1 = 0*5 + 1

2 = 0*5 + 2

and so on

[forest stone]

How did you decide to keep q constant and increment r by 1 is that because of adding 1 to the original equation?

[marble wharf]

tell me other q's and r's which work

I didnt really decide to do anything

those are just the only q's and r's that work for these cases

[forest stone]

Well that was my first part of the question do we need to change both q and r or can we only change one for the implication?

[marble wharf]

can you please write out the list I asked of you

[forest stone]

Just want to make sure we're keeping q constant and just incrementing r by 1 correct?

[marble wharf]

stop thinking abstractly

just do the examples

[forest stone]

Ok 3=05+3, 4=05+4 5=05+5, 6=05+6, 7=05+7 8=0 5+8, 9=05+9 10=05+10

[marble wharf]

no

5=0*5+5 does not satisfy that 0<=r<5

[forest stone]

So it breaks if you go past b

[marble wharf]

yes

do the list again

[forest stone]

1=0×5+1, 2=0×5+2, 3=0×5+3, 4=0×5+4

[marble wharf]

whole list

[forest stone]

But you said it is wrong to go after this since it breaks the inequality no?

[marble wharf]

5 = 1*5+0

those are valid q and r

[forest stone]

Oh ok

[marble wharf]

the pattern breaks. but not the whole thing

[forest stone]

Cant you do q for infinity then since r will always be less than b then?

[marble wharf]

but also 0<= r

0<=r and r < b

[forest stone]

Yes but if we are keeping r and b constant then we can just keep increasing q since it is not affected by the inequality

[marble wharf]

but the equality n=qb+r still has to be true

I do not understand what you mean

can you keep writing the list?

[forest stone]

yes that is what I mean that we can keep writing our listed of numbers with changing q such as 15 = 3 * 5 + 0 , 20 = 4 * 5 + 0 ..... since q does not play a role in our inequality

[marble wharf]

yes for multiples of 5 that works

but not all numbers are like that

[forest stone]

woudln't this work for all q

also going back to my orignal question how can I make my inductive hypthesis with new q and r or is it new q or new r?

[marble wharf]

can you please just keep writing the list instead of trying to guess how it works abstractly

6=?

7=?

8=?

9=?

10=?

[forest stone]

would you like me to do this changing q correct?

[marble wharf]

if you can manage it by changing q, sure

[forest stone]

ok so 6 = 1*5 + 1 7 = 1 * 5 + 2, 8 = 1*5+3, 9 = 1*5+4, 10 = 2*5+0

[marble wharf]

yes

so

we can now look at the complete list

when did we change q? when did we change r? how did we change them?

[forest stone]

we changed q when r was about to become equal to b

[marble wharf]

in other words when r was b-1

[forest stone]

yes

[marble wharf]

and then we are back to this

[forest stone]

ok, but I am still confused how this lets us figure out how we should write out our inductive hypothesis?

[marble wharf]

the induction hypothesis is just that for n there exist q,r with n=qb+r and 0<=r<b

in the induction step you have to do two cases

depending on r<b-1 or r=b-1

[forest stone]

ins't that what we assume

[marble wharf]

the induction hypothesis is what we assume

[forest stone]

don't we want to find it show that with that assumption we can find n+1 = qb+r

[marble wharf]

for n+1 we want to find q' and r'

which will depend on q and r

[forest stone]

yes that my question when we write this out how do we know if it is both q' and r' because that would imply that they both change but from our list we see that only one changes at a time

[marble wharf]

just because its called q' does not mean that it cant equal q

[forest stone]

oh ok

so we go back to our assumption which is n = qb+r such that 0<= r<b and now add 1 we have n+1 = qb+r+1. Can we now say that r' = r+1. But then how can we prove that r' <b?

[marble wharf]

.

[forest stone]

so I would need to prove that r' is equal to b-1 or less than b-1

also for induction woulld't I need to transfomr my assumption into the n+1

[marble wharf]

you do two cases depending on this

[forest stone]

so you are saying that after adding 1 to both sides I need to examine these both cases and show that r' can only fit into those two cases

[marble wharf]

r fits into only one of those two cases

[forest stone]

but aren't we looking at r'?

[marble wharf]

and depending on the case you have to define r' differently

[forest stone]

isn't r' just defined to be r+1 though?

[marble wharf]

not when r=b-1

[forest stone]

since then r = b but how can we execute our inductive hypothesis then if we are adding 1 to our assumptino?>

[marble wharf]

look at the list we wrote

what are q' and r' when r was b-1

[forest stone]

q = 1 and r = 4

[marble wharf]

so n=9

[forest stone]

yes

[marble wharf]

and for n+1=10 we have?

[forest stone]

q =2 r = 0

[marble wharf]

bad notation

q'=2 and r'=0

[forest stone]

yes

[marble wharf]

so q'=q+1 and r'=0

if r=b-1

[forest stone]

ok so I want to just make sure I am following everything. We will show that n = qb+r | 0<=r<b impliles n+1 = q'b+r' | 0<=r'<b. Now we add 1 to both sides of n= qb+r and then we have either r = b-1 or r < b- 1

[marble wharf]

yes

[forest stone]

So if we prove that it is either case we have proven that our inductive hypothesis work

[trim jolt]

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[fierce rain]

If i had some polynomial degree 3 divided by another polynomial degree 3 but that polynomial is int eh form (x+2)(x²+1) where i cant simplify it more

[fierce rain]

If i wrote it as A + B/x+2. +(cx+d)/x²+1

And then cross multiplied

Sub valeus of x

And form 4 equations to solve a b c d

Would that be a correct method

[trim jolt]

@fierce rain Has your question been resolved?

[fierce rain]

<@&286206848099549185>

[obsidian fjord]

Just A/x+2 then the second fraction

[fierce rain]

Is A wrong?

[dusky sleet]

No, the constant term still exists because the degrees of the numerator and denominator are equal

The form you put is valid, as you have first done polynomial division to get a remainder, then do partial fractions as normal

[fierce rain]

Yh i thought that

I just had a exam and i feel loke if i dod remainder theorem it would of been better

[obsidian fjord]

I learnt it a different way, where we subtract and then continue, my bad.

[fierce rain]

Yh subtraction is a good way

I should of done that

[obsidian fjord]

are you talking about fm?

[fierce rain]

Yh

[obsidian fjord]

ah

[fierce rain]

💀

Im so annoyed

[obsidian fjord]

There wasn't enough time I feel

too many questions

[fierce rain]

I did that method yh

[obsidian fjord]

this time

[fierce rain]

And when i sovled for a b c d

I got it wrong

When i checked w mu calculator

[obsidian fjord]

i got 2- A/x+2 and then bx-c

[fierce rain]

Yh thats correct

The way i did it was lomgwinded

[obsidian fjord]

How'd you find it though?

[fierce rain]

57/75 i think

I made too much mistkaea

Im cooked

I need to pattern paper 2

[obsidian fjord]

Yeah same

[trim jolt]

@fierce rain Has your question been resolved?

[fierce rain]

Csn i get a second op

Opinion

[fierce rain]

<@&286206848099549185>

[fierce rain]

.

[fierce rain]

<@&286206848099549185>

[trim jolt]

@fierce rain Has your question been resolved?

[fierce rain]

No

[trim jolt]

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[trim jolt]

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[bold anchor]

close

.close

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[lament rose]

how do i do the theorem

[stoic garden]

Do you know the conditions and the conclusion of the Intermediate Value Theorem?

[stoic garden]

If so, restate them here, and I'll help you apply it to this problem.

[lament rose]

i dont really know the rules

[stoic garden]

Condition: $f(x)$ is continuous on $x \in [a, b]$. \
Conclusion: $f(x)$ takes on every value between $f(a)$ and $f(b)$ on $x \in [a, b]$.

[solid kiln]

@stoic garden

[red loom]

the IVT states for a continuous function on a closed interval ([a,b]) and let (N) be any real number such that (f(a)<N<f(b)) then there exists a point (c\in[a,b]) such that (f(c)=N)

[solid kiln]

PajamaMamaLlama

[stoic garden]

In this case, we're trying to prove that the function is equal to 0 at some point on the interval (-1, 0). This means that we want to prove f(x) crosses y = 0 between (-1, 0). Do you understand this?

[lament rose]

so it has to be true for everything between -1 and 0

[stoic garden]

No, it has to be true for at least one point between -1 and 0.

[lament rose]

but also continuous?

[stoic garden]

Sorry- can you explain what you mean by true?

[stoic garden]

f(x) being continuous is the condition we must satisfy.

[lament rose]

ok

when i plug in the 0 and -1 i get -2 and 1 so that means there is at least one possible solution in between

is that right?

@stoic garden

[stoic garden]

You must explicitly state that this is partially because -2 < 0 < 1.

[lament rose]

i knew it was between those numbers but i guess it wouldnt hurt to put that down

so it is true because -2 < 0 < 1

[stoic garden]

It’s not necessarily about what you know. All of this must be specified in order to get credit on a free-response question.

[lament rose]

That tells that there is something between it, but it also has to be continuous?

How exactly do we tell its continuous

[summer haven]

All polynomials are continuous

[south star]

It is continuous if the graph can be drawn in one stroke

[summer haven]

If it wasn't, there would be a gap or asymptote

Which there isn't

[simple rock]

Did you get your answer? @lament rose

[lament rose]

What would be an example of something that doesn't work

[simple rock]

Wdym?

[lament rose]

Like if there was no point in between

[simple rock]

That's only possible if the function wasn't continuous

[lament rose]

As long as they don't equal?

[simple rock]

Oh you want example of a function that isn't continuous in that interval

[lament rose]

Ye

[simple rock]

f(x)=1/x

f(-2) = -0.5
f(1)=1

-0.5 < 0 < 1
So there should exists and x satisfying f(x)=0 if the function is continuous.

But try and you won't find any x for which 1/x = 0.

Because the function is discontinuous at 0.

You understand? @lament rose

[lament rose]

So since there is a single discontinuity it fails?

At 0

[simple rock]

Yes you can't use IVT if there is atleast a single discontinuity

[lament rose]

Oh

[simple rock]

But if you prove the function is continuous, you can use IVT

[lament rose]

But if f(x) = x^2
F(1)=1
F(-1)=1
It won't work because there is nothing in between, so 1≠1

Correct?

[south star]

well f(a) can't equal f(b) according to the definition

[south star]

It doesn't work because f(a) = f(b)

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[primal scaffold]

Basic question in probability theory about the expectation of discrete random variables. I simply don't get this equation in the book I'm reading.

[primal scaffold]

It's part of a simple derivation for a variance expression

[primal scaffold]

All I'm getting is that the expectation of a real number is the real number itself? Which doesn't make sense to me. Though I understand the rest of the basics like the linearity of expectations.

I'd also mention I understand the formal definition of a random variable and what a probability distribution is, etc etc

[ionic pendant]

the expectation of a real number can be interpreted as the expectation of a "random variable" except it's not random at all it just always returns that number

[primal scaffold]

How would that work with respect to this definition?

[primal scaffold]

This does clarify a bit but I'm just curious

[ionic pendant]

define the probability as 1 for that number and 0 for everything else, then it would just be summing up 0 + 0 + 0 + 0 + x + 0 + 0 + ...

[primal scaffold]

Aha okay, that clears it all up actually. Thanks

.close

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[keen pewter]

can someone please explain this to me?

[trim jolt]

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[eternal mica]

i think this problem has some sort of error because even the solution doesn't make complete sense
perhaps they meant AB = AC

[eternal mica]

because they said that angle B = angle C, but based on the way it should be drawn according to the prompt, angle A = angle C

@keen pewter

[south star]

sup

[eternal mica]

not you changing your name 😭

i thought i pinged the wrong person for a sec

[south star]

my bad

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[west birch]

can I factor this x^2-x-6=0?

[south star]

let me write it out:

$x^2-x-6=0$

[solid kiln]

hello

[hidden dew]

yes

[south star]

so what two number multiply to -6 and add to -1?

it's -3 and 2

$\left(x-3\right)\left(x+2\right)=0$

[solid kiln]

hello

[west birch]

oooh

okay

ah i wrote this out and still couldn't figure it out (x+a)(x+b)=x^2+x(a+b)+a*b

thx

[south star]

$\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab$

[solid kiln]

hello

[south star]

yes

[west birch]

how to you write it so fast in that code

do*

[south star]

use desmos graphing calculator then copy it over

just put in between dollar signs

[hidden dew]

practice

[south star]

type in the stuff here:https://www.desmos.com/calculator/ then copy it here

[west birch]

(x+a)(x+b)=x^{2}+x(a+b)+a*b

(x+a)(x+b)=x^{2}+x(a+b)+a*b

[south star]

put it in dollar signs

[south star]

$(x+a)(x+b)=x^{2}+x(a+b)+a*b$

[solid kiln]

hello

[south star]

use dollar symbol

[simple rock]

$(x+a)(x+b)=x^2+x(a+b)+ab$

[solid kiln]

Shubham1029

[west birch]

$(x+a)(x+b)=x^{2}+x(a+b)+a*b$

[solid kiln]

AllNamesAreTaken2342329

[west birch]

ooooh thx

$(x+a)(x+b)=x^2+x(a+b)+a*b$

[solid kiln]

AllNamesAreTaken2342329

[hidden dew]

you could just ab

[west birch]

ab?

[hidden dew]

instead of a*b

[west birch]

sure

merci

[hidden dew]

if you want to test latex you could go to #latex-testing

and close this room if you are done

[west birch]

.close

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[inland bear]

can someone help me with this question?

why can i not leave it as squareroot 1/y the whole thing but have to turn it into 1/squareroot y?

[south star]

Hi. I can help

Let me read the image first

[inland bear]

okay

[main sigil]

you actually dont really have to do that

as long as you can go to y^(-1/2), you should be fine

[inland bear]

but if i did this, the answer would be different

[south star]

$\sqrt{\frac{1}{y}}=\frac{\sqrt{1}}{\sqrt{y}}=\frac{1}{\sqrt{y}}$

[solid kiln]

hello

[main sigil]

$\sqrt{\frac{1}{y}}=\sqrt{y^{-1}}=y^{-\frac{1}{2}}\$
$\frac{1}{\sqrt{y}}=\left(y^{\frac{1}{2}}\right)^{- 1}=y^{-\frac{1}{2}}$

[solid kiln]

MathIsAlwaysRight

[main sigil]

both of these work equally

doesnt really matter which one you do

[inland bear]

so i just have to simplify it

[inland bear]

so from this i have to simplify it into 1/rootY.

[south star]

Yes. And you antiderivative doesn't work.

[inland bear]

oh alright,

[south star]

That's why you can't go the route you took.

[inland bear]

im abit curious why do i have to simplify it for it to work though?

[south star]

It's because it accounts for any negative exponents.

Allow me to write it out:

You solution can also be written as:

$\frac{2}{3}\left(y^{-1}\right)^{\frac{3}{2}}$

[solid kiln]

hello

[south star]

Your antiderivative

[inland bear]

oh so that would work too right?

[south star]

But try deriving this:

You don't get $\sqrt{\frac{1}{y}}$ again.

[solid kiln]

hello

[inland bear]

oh yeah right

[south star]

So for that reason, it's best practice to always convert things to exponents.

[inland bear]

okay, thank you

[south star]

Otherwise you'd still have to account for the y^-1 somehow.

Is there any other questions you want help with?

[inland bear]

no thats it thank you

[south star]

Bye

[inland bear]

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[keen pewter]

why did we square RHS?

[keen pewter]

ah

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[earnest tundra]

the figure shows the graf to a function f

determine then the function

i thought the answer was -3

[pseudo goblet]

(graph* in English)

But here you don't need to determine f

You're asked for a definite integral

i.e. the "area under the curve"

(where if the curve is below the x-axis we take the area as negative)

!occupied

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[pseudo goblet]

Ask in #geometry-and-trigonometry, and don't just ping people you don't know

[earnest tundra]

so calculate the negativa area from -3 to -2 then take that - the area from -2 to 0?

[pseudo goblet]

yh

The integral is the green area minus the red area

[earnest tundra]

my phone is in swedish so it autocorrects when i type in english

[pseudo goblet]

ah fairs

(this is why I disable autocorrect lmao)

[earnest tundra]

yeah like i said well then i get it

thank you

have a nice day

[pseudo goblet]

np

[earnest tundra]

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[wheat mango]

Yo guys

I have an issue I can't identify a series

So essentially

[wheat mango]

$f(x) = x^{l+1}\sum_{n=0}^{+\infty} c_{2n} (-k)^n x^{2n}$

[solid kiln]

Mr. Rotor

[wheat mango]

$c_{2(n+1)} = c_{2n} \frac{1}{(2n+l+3)(2n+l+2)-l(l+1)}$

[solid kiln]

Mr. Rotor

[wheat mango]

I graphed it and it looks to be something like

$f(x) = (1-e^{-ax})\sin(bx)$

[solid kiln]

Mr. Rotor

[wheat mango]

But i just can't do the bridge

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[urban grotto]

Anyone have any advice for integrating this thing?

Try as I might, I cannot get it into a useful form

E is some arbitrary constant

[trim lichen]

this looks quite ugly 💀

i mean maybe u := e^(-ax) is gonna help

somehow

[urban grotto]

Astute observation my good friend. This integral is hurting my feelings

yeah i thought about that

[trim lichen]

and then you'll end up with a quadratic in u under the root

[urban grotto]

yep

[trim lichen]

so maybe some trig sub will follow

[urban grotto]

been playing with that for a bit, wasn't able to get it into a recognizable form. close, but nothing i could find an established identity for

[fierce lake]

Sub e^ax = t it's solvable

A bit lengthy but definitely solvable

Factorise the denominator into a perfect square form

[trim lichen]

complete the square

is what i was gonna suggest

[urban grotto]

Tried that, but I'll give it another shot

[trim lichen]

you'll have uhh... -(t-1)^2 + E + 1 under the root if i didnt mess up arithmetic

so $t-1 = \sqrt{E+1} \sin(\theta)$ maybe

[solid kiln]

Ann

[trim jolt]

@urban grotto Has your question been resolved?

[urban grotto]

shoot me

ive been playing around with it and gotten it into the following form

I believe that the algebra to get here was fine, but that u in front of the square root is annoying me'

(I dropped some prefactors to focus on the form of the integral)

Here i have
-Done a u substitution as suggested earlier
-Completed the square

The square root seems like it boils down to a secant, but again, that u in front is killing me

That square should not be on the u, it should be on the u-1 term

[bright oracle]

You want the last bracket to be ()^2

[urban grotto]

yes, sorry. typo

i should have paid more attention 7 years ago in Calc II

[bright oracle]

Then you can make another substitution

I'm pretty sure the answer to this is gonna be truly horrible

But it's doable

[urban grotto]

truly horrible sounds par for the course

the prof for this course is an evil man

[bright oracle]

You wanna do a trig sub

And then you need a 3rd sub after that I think

[urban grotto]

what the actual fuck

thanks for the advice

[bright oracle]

Yeah I think we need a sin(θ) style sub

And then a tan(1/2 θ) sub

You know it's serious when the tan(1/2 θ) sub comes out

[bright oracle]

I haven't done all the working so this might not be exactly correct but I think you're aiming for something along the lines of √((E + 1) - (u - 1)^2)

Then sub u - 1 = √(E + 1) sin(θ)

That should simplify it quite a bit

[urban grotto]

i think this is the direction

im going to close this channel and think about this problem indepdently, see if there's other ways to go about it to avoid this integra

thankks

.close

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[lament rose]

Did I do this right

[lament rose]

i probably did that very wroong

[stoic garden]

When you applied the Quotient Rule, you messed up what must be subtracted. Also, you didn't use the Product Rule properly for x g(x).

[lament rose]

dont i need to do the product rule first

[stoic garden]

No. You should use the Quotient Rule. And within that, you should be using the Product Rule for d/dx (x g(x)).

[lament rose]

so put in quotient rule form then inside use product rule

so for the quotient rule is (dirivitave of the numerator times denominator)-(numerator times dirivative of deniminator then divide by denominator squared

and when the numerator is prime do i write it like x'g'(x)?

or is it like xg'(x)

this one looks correct, right?

@stoic garden

[stoic garden]

I’m not sure what you mean, but I can definitely say you’re still not using the Product Rule.

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@lament rose Has your question been resolved?

[stoic garden]

Until here, it’s correct. I can’t check the rest of it.

[lament rose]

I only use the product rule if it's like (xg(x))'

?

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[simple haven]

This is probably just an open problem, but I couldn't find much online about it. But it is a reasonably well known fact that the (mn)th Fibonacci number $F_{mn}$ is divisible by both $F_n$ and $F_m$. (It is, however, not necessarily divisible by $F_m F_n$.) If we consider all of the factors of $k = (k_i)$ and take the LCM of $(F_{k_i})$, then this number divides $F_k$. If we perform this division, because the Fibonacci numbers are exponential, there will be other, leftover factors. We call this the "primative part" of $F_k$. We can notate this as $p(F_k)$. These are given by A061446.

Other than $p(F_6) = 4$, these values seem to be square free. Are they?

[solid kiln]

OmnipotentEntity

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@simple haven Has your question been resolved?

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@simple haven Has your question been resolved?

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@simple haven Has your question been resolved?

[placid radish]

Scrolling through some sources, your conjecture, if true, is at least as hard as proving the non-existence of Wall-Sun-Sun primes: https://en.wikipedia.org/wiki/Wall–Sun–Sun_prime

but it looks like ppl are saying that from heuristic, it looks like there are infinitely many wall-sun-sun primes. Exhibiting any of those primes would be a counterexample to your conjecture (by the equivalent definition listed under the Wall-sun-sun prime wiki).

[placid radish]

and that is at least as hard as proving fermat's last theorem as i just found out

[glossy root]

What do you mean it looks like there are infinitely many wall-sun-sun primes?

[placid radish]

I mean that using some sketchy probability calculation, assuming that primes are random.

you sort of assume that a number around N has a change of being prime with probability 1/log(N)

[glossy root]

Well we've checked well over a dozen primes and none of them are wall-sun-sun primes so I think there's a pretty strong heuristic the other way.

but yeah, that's a fair point. I haven't run the numbers on that so I can't say

[placid radish]

"checked finitely many primes" means basically nothing considering that primes are infinite though .

its hard to quantify how strong of an evidence these finitary checks are

[simple haven]

Either I'm too thick to understand it, or I simply missed it, but can you point out where the wiki says that these two things are related?

[glossy root]

https://math.stackexchange.com/questions/4624851/primitive-divisors-of-fibonacci-numbers

[simple haven]

Thanks!

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[placid radish]

pick F_{\alpha(p)}, where alpha(p) is the smallest number n such that p divides F_n, but we know that p^2|F_{alpha(p)} since it is a wall-sun-sun prime (listed in the wiki, it is equivalent), but this gives us that p(F_{alpha(p)}) divides p^2.

sry for the late response i had something going on

oh the stack post explains it exactly the same, lol

[simple haven]

Thanks! @placid radish

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[frail heron]

why is my answer incorrect?

for 16 a

is it just imprecise or complete nonsense

[trim lichen]

nonsense

[ionic pendant]

it consists of every linear combination of the columns of A (or the span of the columns), not just the columns themselves

[trim lichen]

16b is also nonsense

[frail heron]

yeah, I made that up