help-23

[crude star]

thanks alot

I get it now

[solar hazel]

np!

[crude star]

.close

[safe radish]

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[safe radish]

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[ripe gale]

I need to prove the following:

[ripe gale]

if f and g are diagonalizable linear transformations from V to V, show that there exists a basis of V such that:

[flat frigate]

Vertox

[pulsar pecan]

!status

[safe radish]

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

[ripe gale]

I think those are called the "representative matrices"

I have already proven the following, I assume I can use those:

if f is diagonalizable and U is a subset of V then f restricted on U is also diagonalizable

if f and g commute then for each eigenspace of g we have: f(E) is a subset of E

oh and in the new proof I'm trying to do f and g still commute

<@&286206848099549185>

[rapid igloo]

apply this for U being an eigenspace of g

[ripe gale]

Then I know that f(U) is a subset of U

[rapid igloo]

yep

[ripe gale]

And how do I continue?

Where does the Basis of V come into play

Wait the basis of V is a subset of V

[rapid igloo]

uh

wait wait wait

Since $g$ is diagonalizable we have $$V = \bigoplus_{\lambda \in \mathrm{Sp}(g)} E_{\lambda}(g)$$

[ripe gale]

Oh god

Why

We haven't proven this yet

[flat frigate]

giovanni75

[ripe gale]

Sp stands for the span?

[rapid igloo]

it means spectrum

i.e. the set of eigenvalues

[ripe gale]

Oh I see

Is this only true if g is diagonalizable?

[rapid igloo]

yes, otherwise it's not equal to V

but the eigenspaces still have intersection equal to {0}

[rapid igloo]

i'll come back to that right after the other proof

[ripe gale]

Okay let's just assume it's true for now

[rapid igloo]

because of this, we just have to construct a basis of each E_lambda(g), made of eigenvectors of both f and g

[ripe gale]

Okay yeah makes sense

Do we actually need to construct that or can we assume it exists

[rapid igloo]

we need to construct one

well the proof is about its existence

[ripe gale]

Okay so let's say we just constructed it

[rapid igloo]

we need to

[ripe gale]

Yes but let's do that later

[rapid igloo]

well we just have to concatenate the bases then

[ripe gale]

The sum of basis will be a basis of V right

[rapid igloo]

yes

and it will contain eigenvectors of both f and g

so they will both have diagonal matrix representations in that basis

[ripe gale]

I don' get the last step

No

[rapid igloo]

the matrix stuff ?

[ripe gale]

Why they suddenly have diagonal matrix representations in that basis

[rapid igloo]

let's say B = (e_1, ..., e_n)

since e_i is an eigenvector of f

f(e_i) = lambda_i e_i

[ripe gale]

Oh right

Got it

[rapid igloo]

now about this

if E is some eigenspace of g, we know that f(E) is a subset of E, and the restriction of f to E is diagonalizable

[ripe gale]

So f(E) can be diagonalized

So each of them is diagonalizable

[rapid igloo]

yes, so there exists a basis of it made of eigenvectors of f

but the vectors of E are also, by definition, eigenvectors of g

[ripe gale]

Right since they are just multiples

[rapid igloo]

yep

so there we have a basis (of each E_lambda(g)) made of eigenvectors of both f and g

[ripe gale]

Wait 1 sec

[rapid igloo]

which is what we wanted

[ripe gale]

Wouldn't E be made of eigenvectors of just g?

Why can we use eigenvectors of f?

[rapid igloo]

because f (restricted to E) is diagonalizable

[ripe gale]

Ah yes

[rapid igloo]

so we can create a basis of E made of eigenvectors of f

[rapid igloo]

and for this, it follows from the fact that we can create a basis B = (e_1, ..., e_n) of V containing eigenvectors of g only

so this means every vector $x$ of $V$ can be written as $$x = \sum_{i=1}^n x_i e_i$$

[flat frigate]

giovanni75

[rapid igloo]

and this decomposition is unique

as B is a basis

[ripe gale]

Okay so now we have a basis for each E

[rapid igloo]

yes, and this concludes the proof

[ripe gale]

And the diagonalization of the sum still persists?

[rapid igloo]

yep because V is the direct sum of all E's

[ripe gale]

Don't we also need to prove that?

[rapid igloo]

i made some sort of proof here

[ripe gale]

I see I see

Thank you very much!

[rapid igloo]

it could be written better to be honest, but this is the idea behind it

np :)

[ripe gale]

Yeah I'll write it all down in detail now

[rapid igloo]

(note that f o g = g o f is both a necessary and sufficient condition here)

[ripe gale]

Oh yeah wait where did we use that

[rapid igloo]

when we said f(E) was a subset of E

which allowed us to consider the restriction of f to E

as a linear map

[ripe gale]

Ah right, that's the part I've already proven if they commute

I think that answers all my questions

Thank you

.close

[safe radish]

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[jade gust]

nah it's 8

[ripe gale]

Nicht 1?

[safe radish]

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[safe radish]

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

[vivid vortex]

2

[raven heart]

alright

what have you got then ?

[vivid vortex]

I have thought of proving it through induction

I have tried a few numbers to see where it leads

But nothing visible

It's sth under radical +1

And I did think of proving that form is not a square root

And assuming it is

I don't really know where to go

[raven heart]

well first we can simplify that thing a little bit

[grim plover]

consider that when u do 1*111...x=y

[raven heart]

we're only computing stuff of the form x*1 anyway

that's the only thing of interest here

$x*1 = \sqrt{(x^2+1)(1+1)-1} = \sqrt{2x^2+1}$

[flat frigate]

_aplatypus

[vivid vortex]

Yeah, that makes sense

I guess here comes the induction part

[raven heart]

therefore we have something like (x*1)^2 = 2x^2+1 for any x

that sounds a bit more manageable at least than all those square roots everywhere

[vivid vortex]

Yeah

For sure

[raven heart]

so now we can work with the sequence of squares of 1*1*....1 (n times)

we have our starting element u_1 = 1

and u_(n+1) = 2u_n^2 + 1

and we want to show u_n is not a perfect square for n>=2

[vivid vortex]

What did u mean to write on the right side?

What's for the n+1 th term?

[raven heart]

$u_{n+1} = 2u_n^2 + 1$

[flat frigate]

_aplatypus

[vivid vortex]

Oh

[raven heart]

comes from this

[vivid vortex]

Yeah yeah

[raven heart]

so for n=2 we get u_2 = 3

which not a perfect square so it works

now there's the recursive case

what would we want to show here ? @vivid vortex

[vivid vortex]

but 1 is a perfect square

[raven heart]

we're starting from n>=2

[vivid vortex]

yeah

[raven heart]

of course for n=1 it doesn't work

[vivid vortex]

i know remembered

usually ,for an induction i would notate the expression with P(n)

then i would assume a certain P(k) is true

and that would imply that P(k+1) is also true

[raven heart]

alright

so what's our P(n) here, just to see if you follow ?

[vivid vortex]

lemme just say it cus i can't type it correctly

so basically that n is the number of times that operation is done

since we have x*1

it would be n-1

is that correct?

[raven heart]

if we suppose x is (n-1) times the op, then x*1 is n times the op

i mean sure

[vivid vortex]

it's very helpful now that i think ab it

it honestly makes it more complicated

[raven heart]

that's why I defined that u_n

this operation is wack

I just tried to get rid of it + only work with integers

to get to something a bit more conventional

it's the story of math

[vivid vortex]

haha

[raven heart]

but yeah anyway I already said what my P(n) was earlier

[vivid vortex]

yeah

so P(k) would be u_k

[raven heart]

this

u_k is not a statement, it's just a number

[vivid vortex]

i meant

[raven heart]

your P(k) has to be something you can prove

[vivid vortex]

oh

2*(u_k)^2+1

[raven heart]

that's u_(k+1)

still just a number tho

[vivid vortex]

so basically

i was assuming

P(k)

which is u_k=2*u_k^2+1

is true

then i would go to P(k+1)

and i would write it in terms of P(k)

[raven heart]

uh

I think the reduction to u_k I proposed made you lose focus of the original problem

maybe I'll restart my explanation from the beginning

we want to show that 1*1*1* ... 1 is not an integer for n>=2

the idea we both had was to use induction, which sounds not too bad

[vivid vortex]

yep

[raven heart]

now the thing is this raw operation x*y is hard to deal with

[vivid vortex]

yes

[raven heart]

so we tried to simplify it a little bit

since we're only interested in the x*1 case

[vivid vortex]

mhm

and u simplified it by writing each term as a term of a sequence

[raven heart]

we saw that x*1 = sqrt(2x^2 + 1)

[vivid vortex]

which is every operation till the nth one

[raven heart]

now the thing is

our sequence u_n is not just the terms themselves

it's the terms squared

[vivid vortex]

yeah

[raven heart]

since getting rid the the square root makes things easier

[vivid vortex]

right

[raven heart]

so now our objective has changed a little bit with u_n

if x*1 = sqrt(2x^2+1) is not an integer

what does that mean for (x*1)^2 ?

[vivid vortex]

it is

[raven heart]

I mean sure it's an integer

if sqrt(2x^2+1) is not an integer

that means the thing inside the square root is not a perfect square right

[vivid vortex]

yeah

[raven heart]

and that thing inside the square root is (x*1)^2

(reminder x*1 itself doesn't have to be an integer)

[vivid vortex]

why not?

[raven heart]

1*1 is sqrt(3) right ?

[vivid vortex]

yep

[raven heart]

not an integer then

[vivid vortex]

oh

yeah

it's an irrational number

[raven heart]

so that's where the, "we need to show u_n is not a perfect square" comes from

it's because we squared everything in the equation to simplify

[vivid vortex]

yeah

right

[raven heart]

$$\begin{cases} u_n = 1 \ u_{n+1} = 2u_n^2 + 1, n\geq 1\end{cases}$$

[flat frigate]

_aplatypus

[raven heart]

reminder of u_n

[vivid vortex]

n doesn't have to be greater than 2?

or equal

[raven heart]

that's just the recursive definition

i wrote u_(n+1) in terms of u_n

[vivid vortex]

oh yeah

[raven heart]

not u_(n) in terms of u_(n-1)

[vivid vortex]

correct

[raven heart]

anyway

what's our P(n) now ?

[vivid vortex]

u_n=2u^2+1?

[raven heart]

:/

[vivid vortex]

i have never done it this way

im so sorry

[raven heart]

that's the translation of the original question using u_n

so our P(n) is "u_n is not a perfect square"

we want to show this for n>=2

[vivid vortex]

sorry, i always thought it was just how i told u above

that's how i was taught

[raven heart]

I mean yeah I guess they just make you prove equalities in high school or something

[vivid vortex]

yep

[raven heart]

that's why you're not used to that prolly

[vivid vortex]

yeah yeah

[raven heart]

but the P(n) can be whatever statement we want to prove

is the main idea here

[vivid vortex]

mhm

[raven heart]

so now

we have our P(n), that we want to show for n>=2

you know the rough sketch of how to do that I suppose

[vivid vortex]

yeah

i sometimes get stuck when it comes to writing P(k+1) in terms of P(k)

[raven heart]

yeah alright

so what do we do for our base case now ?

[vivid vortex]

ofc P(k) is not a square root

it's some sort of example

based on P(n)

and I will write what i need to demonstrate

u_k+1=2u^2_k+3

wait

im stumbling here

[raven heart]

I'm just talking about the P(2) case here

[vivid vortex]

oh

[raven heart]

base case is the start, inductive/recursive case is when you go up by 1

[vivid vortex]

it will be u_2=2*u^2_1+1

[raven heart]

yeah alright

so u_2 = ?

[vivid vortex]

3

[raven heart]

yup

is 3 a perfect square ?

[vivid vortex]

no

[raven heart]

yep

so nice ^^, we found that P(2) is true

[vivid vortex]

yep

[raven heart]

now the recursive case

can you write in plain english what we have to show now ?

(it always helps at the beginning)

[vivid vortex]

now i have to show that for P(k) I need to prove that it implies P(k+1)

[raven heart]

yeah alright

can you unfold P(k) in this sentence ?

(i.e. write it out)

I'm making sure you're following closely since you're not very used to it

[vivid vortex]

u_1=1,u_k+1=2u^2_k+1,k>=2

it's the same thing,but i just take a certain k variable

[raven heart]

that's just the definition of u_n

that's not P(n)

[vivid vortex]

oh

okay

[raven heart]

P(n) is a statement we're trying to show about u_n

[vivid vortex]

P(k),what i wrote above ,+ is not a square

[raven heart]

yeah

u_k is not a square implies u_(k+1) is not a square

that's our inductive case

[vivid vortex]

yep

[raven heart]

now maybe it's possible to do it directly like that

but to me it sounds a bit of a pain to do so

(maybe it's possible idk)

do you know about contrapositives ?

[vivid vortex]

nah,i don't think i've heard of those

ohhh

i have never heard of this name

but the concept looks familiar

[raven heart]

it's the "(A implies B) is the same as (not B implies not A)" thing

[vivid vortex]

yeah

i remember

[raven heart]

since those "not a perfect square" conditions seem painful to handle

if we do the contrapositive trick

they become "is a perfect square"

that was the idea behind thinking about the contrapositive in the first place

[vivid vortex]

oh,okay

[raven heart]

so what's the contrapositive of this ?

[vivid vortex]

u_k is a square implies u_k+1 is a square

[raven heart]

nope

you forgot to inverse the order of the statements

[raven heart]

^

[vivid vortex]

u_k+1 is a square implies u_k is a square

[raven heart]

exactly

at least we have a statement involving squares now

[vivid vortex]

mhm

and now what?

:)))

@raven heart

[raven heart]

re

sorry, I was going with the flow a bit too much

I think the statement I just presented to you is not provable

but I found another way

(which works this time don't worry)

[vivid vortex]

okay

im all ears

[raven heart]

do you know modular arithmetic ?

[vivid vortex]

a little i believe

i believe it's the one that includes %

[raven heart]

yeah

I found something interesting actually

if you look at u_n mod 4

it seems to always equal 3

[vivid vortex]

oh wow

[raven heart]

I just did on a few examples in python

[vivid vortex]

so it would be sth periodic

[raven heart]

but it's provable by induction

and this time it's a very classic thing

now the 3 mod 4 is interesting

because it turns out that (squares mod 4) can only equal 0 or 1

[vivid vortex]

yeah,correct

[raven heart]

so if we prove our sequence is always 3 mod 4 (with n>=2 ofc)

we've shown there's no squares in it

and we win

[vivid vortex]

that's interesting

[raven heart]

so yeah what we want to show is P(n) : "u_n = 3 mod 4"

for n >= 2

[vivid vortex]

i hope its quick bc i have other hw too🥲 😭

[raven heart]

nah this one is rather quick

so what about P(2) ?

[vivid vortex]

cool then

[raven heart]

is u_2 = 3 mod 4 ?

[vivid vortex]

yes

[raven heart]

yeah it's 3 that's nice

now the inductive case

u_n = 3 mod 4 -> u_(n+1) = 3 mod 4

it's just a question of applying a bit of mod arithmetic now

$u_{n+1} = 2u_n^2 + 1 \pmod{4}$

[flat frigate]

_aplatypus

[raven heart]

since they're equal in integers, they're also equal mod 4

now we supposed that u_n=3 mod 4

you can prolly finish from there

[vivid vortex]

can u finish it just for me to have it as a reference?

i have never done any exercises that require mod arithmetic in induction

[raven heart]

alright

well we suppose that u_n=3 mod 4 (that's the induction hypothesis)

therefore we have

,tex \begin{align} u_{n+1} &= 2u_n^2 + 1 \pmod{4} \ &=2*3^2+1 \pmod{4} \ &=19 \pmod{4} \&=3 \pmod{4} \end{align}

[flat frigate]

_aplatypus

[raven heart]

ignore the equation numbers at the end

and there we have it

[vivid vortex]

or i could have just proven radical 2x^2+1 is irrational

[raven heart]

if you can do it for any integer that works

but that's pretty much what I attempted at first

(the first induction I mean)

[vivid vortex]

https://youtu.be/HqpDPDbVYlM

i found this video

[raven heart]

you still have to show 2x^2 + 1 never ends up being a perfect square

and that's false anyway

2*2^2 + 1 is 9

[vivid vortex]

yeah

[raven heart]

that's the thing

I wanted to use something stronger to prove

which really relates to the function at hand

instead of some generic stuff (see the first induction I proposed)

[vivid vortex]

thanks for the help

have a great day/evening!!

[raven heart]

you too^^

.close

[safe radish]

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[vapid mica]

if I’m being asked how many 4 digit numbers are 1 mod 6, what is it actually looking for? I know 1 mod 6 is 1, but it’s obviously not asking how many 4 digit numbers are = 1

[royal pivot]

rephrase

[lean otter]

do you understand what mod 6 means?

[royal pivot]

||for how many values of n is 1000<=6n+1<=9999||

[vapid mica]

yeah, residues

[quasi bison]

we're looking for numbers which leave a residue of 1 on division by 6

[vapid mica]

but it’s a little hazy because I’m used to the cs definition, not the math definition

[royal pivot]

the same?

[vapid mica]

is it?

[royal pivot]

what is the computer science definition

oh they’re slightly different

[vapid mica]

so, I know x mod 2 is always 0 or 1, so then we’re looking for odd 4 digit numbers?

[royal pivot]

it’s not saying 1mod6

all of the numbers will be odd, but not all odd number will satisfy the conditions

[vapid mica]

darn

[quasi bison]

anything that's 1 mod 6 is odd, but not all odd numbers are 1 mod 6

[vapid mica]

ohh

[royal pivot]

mod 6 applies to the entire equation

[vapid mica]

so to rephrase, it’s looking for x mod 6 = 1?

[royal pivot]

yes

[vapid mica]

alright one second want to see if i know what to do after that

[safe radish]

@vapid mica Has your question been resolved?

[vapid mica]

so I tried to use the other form of 1 mod 6

x = na + k = 6a + 1
solved for a when x is 1003 and 9997, because those are the bounds where they have a residue of 1
which means a is in [167, 1666]
1666 - 167 + 1 = 1500

so 1500 4 digit numbers that are 1 mod 6? feels very wrong

[safe radish]

@vapid mica Has your question been resolved?

[sudden forum]

In mod 2 space, the entire integer set is divided into 2 equal parts -- evens (0 mod 2) and odds (1 mod 2).
In mod 3 space, the entire integer set is divided into three equal parts -- those that are in form 0 mod 3, those in form 1 mod 3, then those in form 2 mod 3.
Similarly,
mod n operation always divides the integers into n different "classes" -- 0 mod n, 1 mod n, 2 mod n, ..., (n-1) mod n.
So in this case of mod 6, the class of 1 mod 6 is also one sixth of the complete given set {integers between 1000 and 9999 (inclusive)}.
So number of elements in it will be just (9999-1000+1)/6 = 1500
Same logic applies for all

@vapid mica

that was my first intuition

[sudden forum]

this is also correct then

You can probably check it using a program

[vapid mica]

what an interesting way to think about it

[vapid mica]

ah you're right

thank you for the help

[sudden forum]

Indeed! They are actually called "equivalence classes," I studied a bit about them in the basic number theory course I did

[sudden forum]

You're welcome. Glad to be of help 🙂

[vapid mica]

.close

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[lean otter]

need help.... somewhere i made a mistake..
it's an exponential equation.
The solution is 3/2

[faint seal]

who said you made a mistake

write the right hand side as $$\frac{2^3}{3^{\frac{3}{2}}}=(\frac{2^2}{3})^{\frac{3}{2}}$$

[flat frigate]

kheerii

[lean otter]

i said...
oh i see
got it!
thank you @faint seal ! (:

.close

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[junior wagon]

Whats the relationship between differentials and linear approximation?

[junior wagon]

from what I understand, they both use this idea of replacing the actual change in the function with approximation using the derivative for a small amount of change

[steep pewter]

didnt take this in my classes so i dont know any fancy talk but here's the thing

[junior wagon]

Also, what is the use for differentials ?

[steep pewter]

oh

differentials give the gradient of the tangent to the curve at any point

have you been taught the first principle of differentiation?

[junior wagon]

the limit definition?

[steep pewter]

this

[junior wagon]

yeah

[steep pewter]

and did ur teach explain what it means

where it coems from

[junior wagon]

It finds the slope between x+h and x

for small h

to approximate the slope of tangent

[steep pewter]

yep

u know the formula for gradient using 2 coordinates right

in short this this takes two veryververy close coordinates on a slope

[junior wagon]

do you mean the lim definition fixing either x or y?

[steep pewter]

no the gradient between (x1,y1) and (x2,y2)

like 2 points

[junior wagon]

hm, not sure which one you are referring to

oh

[junior wagon]

yeah

[steep pewter]

my man here does the same

but with x,y and (x+h),f(x+h)

and since h limits to zero i think u can imagine what happens

[junior wagon]

so what is the differential referring to?

[steep pewter]

well teh differential here

is h

its the small change in x that we are observing

in the above formula its written as h

in ur text its called dx

well its a reference as to how small slices u want to make when differentiating or integrating

since h limits to 0 but never truly becomes 0

it need to be something

[junior wagon]

i see

[steep pewter]

the differential is that something

[junior wagon]

so in the case for two variable functions f(x,y)

there would be two differentials

one for x and one for y

[steep pewter]

not necessary

u see f(x) = y

y is defined with x

so if u tie a differntial to x

the same can be used with y

since they are related by the function

[junior wagon]

ok, so at the bottom here, they say $dx = \Delta x = x-a$ and $dy = \Delta y = y-b$

[flat frigate]

KN

[steep pewter]

well sure u can have two of those

[junior wagon]

I thought those referred to two different values of h

[steep pewter]

but ultimately dy is a function of dx

since y is a afunction of x

they do

dy and dx are different values of h

[junior wagon]

wait, I mean in the multivariable case

[steep pewter]

but f(dx) = dy

oh yeah sure sure

multivariable case would prefer two differentials

[junior wagon]

ok, make sense

Ok I think I know what differential refers to now

How does it relate to linear approximation?

[steep pewter]

well when doing linear approximation

first realize that integral and derivates are sums of instantaneous values and instantaneous rates

when u cant integrate someone(since the integral might be very hard)

[junior wagon]

I guess, would it be saying like $\Delta y = f(x + \Delta x) - f(x) = f'(x) \Delta x$? So, like saying the change $\Delta x$ (the differential?) is so close to the function that we can approximate the actual change with a linear one (ie. the slope to a very close point)?

[steep pewter]

u can take dx as say 0.00001

[flat frigate]

KN

[steep pewter]

this will most likely be specified by the question

it lets u approximate ur integral

[junior wagon]

we've only seen derivatives so far

i guess this is something ahead

[steep pewter]

i think u can find derivatives from linear approximations as well

again just take dx as say 1

for the interval of 1 starting at 0(e.g.) all the slope remains same

comparable to this in many ways

[junior wagon]

i see

makes sense

thanks

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[hallow saddle]

easy

question

[hallow saddle]

i think

just not good with vectors

I heard acceleration was the gradient of the vectors

[kind fractal]

Given the forces, you can find acceleration by
a = F/m

where F is the sum of all forces acting on the particle

[hallow saddle]

so i add the vectors

[kind fractal]

Yes

[hallow saddle]

tysm

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[lean otter]

Slog(a*b) = Sloga + Slogb
right?
(base is 10)

[lean otter]

S is a real number

[untold cobalt]

yes doesnt matter which base

[final halo]

what a slog!

[lean otter]

thanks!

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[limber drift]

need walkthrough: sin(theta) tan(theta)=6 tan (theta)

[limber drift]

pls ping when help is here.

[safe radish]

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

[limber drift]

walkthrough means everything

therefore i need help with everything

[plucky elk]

You should try it yourself first

No one's gonna do the entire problem for you

[limber drift]

sign* forget it. i got a bad feeling about how this will come out based on how u answered me.

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[safe radish]

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[fickle talon]

How do I solve this DE

$x'(f) = \frac{1-f^2x(f)^3}{f^3x(f)^2}$

[flat frigate]

roxyit

[quasi sentinel]

multiply f³x(f)²

as a matter of fact i posted this

[fickle talon]

exact form?

doesn't look like homogenous or lde

[quasi sentinel]

x'(f)f³x(f)²+f²x(f)³ = 1

(1/3f³x(f)³)' = 1

[fickle talon]

oh I see

so exact form it is

thanks

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[minor spruce]

i got the correct answer, but i don't know if i did it correctly. i basically simplified 42a + 77b = 350 to 6a + 11b = 50 and i guessed the values for a and b and repeated it for the rest. is this how im suppose to do this?

[minor spruce]

i found a bit of a pattern so it got a bit easier

[safe radish]

@minor spruce Has your question been resolved?

[hard crest]

that sounds like a great way to do it

[minor spruce]

ok ty. i thought there was some way i was missing

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[royal arch]

is this a correct trig identity?

i dont recall learning that and i cant find anythin online

what about sec(90° - θ) = 1/cos(90° - θ)

@whole acorn

<@&286206848099549185>

wait its 15 mins mb

uhm

[lean otter]

yeah dude secx is 1/cosx no matter what

here its just that sec(90-theta) is cosec(theta)

its still checks out

csc(theta) = 1/sin(theta)

we know cos(90-theta) = sin(theta)

thereby 1/sin(theta)

this isn't like a proof just making sure you get that it all checks out in the end and no matter what the angle is it stays correct

for secx = 1/cosx

[royal arch]

were both of my equations correct than

wait

i still dont understand how 1/cos(90-theta) = csc(theta)

or does it

[lean otter]

cos(90-theta) is sintheta

do you agree with that?

i think you should know that 1/sintheta = cosec theta

thats how it is...

yeah both your eqns are correct but i only see one img

[raw grail]

if you want an intuitive motivation for why this identity works, look at the graphs of cosecant and secant, then $\sec{(\pi-\theta)}$ is simply just a shift over by $\pi$ of the its original graph at the origin, which coninsides with the graph of cosecant

[flat frigate]

Blue Guilmon

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[primal gazelle]

I need help with this question

[lean otter]

hint: find the line between their centres

[primal gazelle]

oki ill try that thank you

rest is just simple computation

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[wind kite]

why do i like women

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@wind kite Has your question been resolved?

[thin bridge]

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[hallow saddle]

How to calculate tension when the object is not hanging (i.e a train)

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[scenic otter]

is this a legal move?

[scenic otter]

the thing below as an example

not the actual numbers

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[unreal kindle]

hello

[unreal kindle]

$x(x-1)y'(x) - y(x) = (x-1)^2$

[flat frigate]

lilisworld

[unreal kindle]

please i need to solve this

[glass carbon]

divide by x-1 then integrating factor?

[unreal kindle]

$y_H(x) = C*\left(\frac{x}{x-1}\right) ?$

[flat frigate]

lilisworld

[unreal kindle]

is there supposed to be an absolute value here?

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[drowsy holly]

This question could also be true right? If L1 and L2 are the same then L1 <= L2 and f hat minimizes L1 as well

[safe radish]

@drowsy holly Has your question been resolved?

[safe radish]

@drowsy holly Has your question been resolved?

[plucky elk]

yes, could but not always, which is what the question is asking about

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[boreal wharf]

hi, should this question have the equals sign as the 3 lines because of how the coefficients equal each other ? I dont understand how or why, like so root a = 7 /

[crude bear]

Hm

Your teacher guessed n = 2

there

[boreal wharf]

u sure? isnt that why it should be this ≡ sign idk

im so confused but so one side equals the other or..

[crude bear]

You are given $\left(ax^{6}\right)^{\frac{1}{n}} = 7x^3$

[flat frigate]

talestitan

[crude bear]

That equation is equivalent to $a^{\frac{1}{n}}\cdot x^{\frac{6}{n}} = 7x^3$

[flat frigate]

talestitan

[crude bear]

agree ?

[boreal wharf]

yeah

so 6/n = 3, therefore n is 2?

[crude bear]

That is one way to find a solution

Matching $x^{\frac{6}{n}}$ with $x^3$

[flat frigate]

talestitan

[boreal wharf]

yeah that makes sense, just wanted to make sure if u can do that , but then the coefficients have to be the same, so why is root a = 7 then

[crude bear]

You matched x^6/n with x^3

So the remaining a^1/n must be equal to 7

[boreal wharf]

ahhh so its because its remaining so thats the only option left

makes sense thank you

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[idle tiger]

how would i do this fast

using the unit circle

[hasty falcon]

tell me your approach first

[idle tiger]

ok my approach first

[hasty falcon]

like if you don't have to show any calculation then it is quite easy and fast

[idle tiger]

would be

since csc = 1/y

id manually look at the unit circle\

and plug every single y value

until i get -7/3.

[floral cape]

i mean in my honest opinion i think its D

[idle tiger]

well

it is d.

LOL

[floral cape]

yeah

by excluding

jus exclude

[hasty falcon]

nah hold on

[idle tiger]

but i need a fast approach

i have a test tmr

[floral cape]

fast apporach

[idle tiger]

and i just needa know how

[floral cape]

okay

[idle tiger]

can u elaborate

on excluding

[hasty falcon]

what is the unit circle you talked about?

[idle tiger]

?

just

the unit circle..

[floral cape]

yeah

no no

unit circle is much work

[idle tiger]

hold on

[floral cape]

listen

so

csc is 1/sin you know that

and sec is 1/cos

so you gotta readjust your given

1/sin = -7/3, 1/cos < 0

so now everything is much easier

[floral cape]

since the given is like this

its much easier to get sin as it wants in the querstion

first thing that caught my eye was find sin not find cos

since i already has 1/sin = -7/3 in the given

so just by cross multiplication

1/sin = -7/3

sin = 1 divides -7/3

[idle tiger]

oh

damn

[floral cape]

it gives you -3/7

and respectfully in the answers

the only answer for sin which is -3/7

is only D

thats how u exclude

[idle tiger]

what about B

b also has

-3/7

how did u exclude from b and d

i mean exclude B

and find cos that fast too

or did u just go off that the answers were placed in order in correspondence wit the question "find cos 0 and sin 0"

answer is d though haha

[floral cape]

no

look

the quesiton goes

find cos then find sin

so answers are also teh same

[idle tiger]

Oh

yea

[floral cape]

B is wrong

cuz first answer -3/7 is for cos

not sin

u get it?

[idle tiger]

makes sense

ty

yeee

[floral cape]

your welcome

good luck on your test

[idle tiger]

wait

can u elaborate on this also a little more

i think i missed the day when this was taught in school so i'm confused

cos(θ)=x
tan(θ)=y/x
csc(θ)=1/y
sec(θ)=1/x
cot(θ)=x/y```

this is what i have in my notes

i dont have the csc=1/sin shit

[floral cape]

let me make it easier for you

[idle tiger]

can u provide notes for that

ye ty

[floral cape]

sre

sure

[idle tiger]

yea my teacher was saying 1/sin and shi the other day n i was mad confused

[floral cape]

so lemme make it easier for you

cot = 1/tan no big deal

easy to memorize

right?

but these

[plucky elk]

,tex .recip trig

[floral cape]

some people might have problems

[flat frigate]

rie.mann

[floral cape]

ur teacher mean

"x" refering for it as if it will give you an angle value

like 45 degrees for example

[idle tiger]

ye

got that

[idle tiger]

didnt know tis existed ty

[floral cape]

is it simple now

same LMAO

[idle tiger]

Yup

i get it now

LMAOO

[floral cape]

LMAO

[idle tiger]

ty for ur help

[floral cape]

GOOD LUCK BRO

💪

ofc

[idle tiger]

FS

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[idle tiger]

.reopen

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✅

[idle tiger]

how would i do this one

😭

i looked on the unit circle

no sin of theta = -3/5

answer is C

[plucky elk]

use trig identities to find cos(theta)

[idle tiger]

which one should i use

i'll work from there

there's so many

cos(theta)=sin(pi/2-theta)?

[plucky elk]

$\sin^2(x) + \cos^2(x) = 1$

[flat frigate]

rie.mann

[idle tiger]

.close

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[tepid talon]

me confusion

[civic kettle]

set up the equation and solve for a

[tepid talon]

so would it be

like loga(4)

to set it up?

[lean otter]

$log_{a}4 = \frac{1}{3}$

[flat frigate]

Richard22

[tepid talon]

a = 64

so would it be log64(4)

?

[lean otter]

Your final answer should be in terms of x

There should not be a 4

Because it asks for a function

All you gave is a numeric value

[tepid talon]

but solving loga4=1/3

is equal to 64

a = 64*

[lean otter]

I know

leave a is 64

[tepid talon]

so log64(x)=1/3?

[lean otter]

But replace the 4 inside the log with x

Yes

Wait

No

[tepid talon]

O_O

[lean otter]

bro

Give a function

$y=log_{64}x$

[flat frigate]

Richard22

[tepid talon]

why does the X go away

u know what

is the 4 only to find A

OO

[lean otter]

Yes you only use the 4 to find a

[tepid talon]

gotcha, thank you sm

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[wraith prism]

Lim x tends to 0= (modx)÷x

[wraith prism]

The short method to find the limit doesn't exist. I used 0+h,0-h

[quasi bison]

1. that equals sign doesn't belong there
2. use / and not ÷ for division
3. use |x| and not "modx" for the absolute value of x
4. indeed the limit doesn't exist. that's not a reason to look for ways to make it exist.
5. you have not asked a question at all, not even a "am i doing anything wrong here?" or similar

[wraith prism]

If I use the direct method is it valid
X/x =1
-x/x =-1 both aren't equal so it doesn't exist

[quasi bison]

sure

[wraith prism]

Sorry for not including all the information

Three points are given and i have to find the area of the triangle so here if I use the determinants formula, is it the fastest way to solve ?

If there is any other method please suggest

[safe radish]

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[atomic geode]

number of ways to make a 6digit number using only 3,4,5

[frail roost]

3^6?

[atomic geode]

how

[spice grove]

Don't give out answers.

[atomic geode]

i know the procedure im just confused why 3^6

like in the last 2 spots of 6 digit number,we have only 2 choice and 1 choice

[frail roost]

look at how many numbers are possible for every digit and then you have to multiply it because for every first digit there can be every 2nd digit and so on

[atomic geode]

but in the 2nd last blank wont we have only 2 choices left?

and 1 choice left for the last blank

[spice grove]

Why is that?

[atomic geode]

we cant repeat numbers

like each digit should appear 2 times

[spice grove]

Says who?

[atomic geode]

says the question

[spice grove]

Where?

[atomic geode]

i didnt mention it

my bad

but now that i have mentioned it

[spice grove]

Are you nuts

[atomic geode]

what will be the new answer

[spice grove]

Alright then the question Changes slightly.

It is known that each digit occurs exactly 2 times, yes?

[atomic geode]

yes

[spice grove]

Then rather than doing it the way you were doing before.
You can "choose" places where these individual digits fit.

Example

You have 6 different places in your number right

You can choose any 2 of them for say 3.

2 of the remaining 4 for some other number. Leaving you with no choice for the last number.

[atomic geode]

hmm

so 3^4?

or 3^2 x 2^2

[atomic geode]

i think this

@spice grove ?

[spice grove]

How'd you get numbers of the sort 3^2

[atomic geode]

3 possible numbers for 2 blanks

2 possible numbers for the other 2 blanks

2 forced numbers in last 2 blanks

so 3^2 x 2^2

or should it be 6x5x4x3

i.e. 6!/2!

[lean otter]

can anyone help me with math

[atomic geode]

go away from my channel

@lean otter

[safe radish]

@atomic geode Has your question been resolved?

[safe radish]

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[proud vector]

im so confused how do i solve this again

[rich orbit]

is that trig ratios

[proud vector]

yes

[rich orbit]

i dont remember fully

but maybe cross multiplication

find sin 50

and multiply it by 15.2

then multiply sin H and 12

(sin 50) * 15.2 = (sin H) * 12

divide both sides by 12

and then do inverse trig ratio thingy

so sin^-1 on the calculator

idk if that makes sense

[proud vector]

thanks

.close

[safe radish]

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Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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

[stiff aurora]

any1 plz tell me how to solve for principles values in inverse trig when they doent lie in domain

[stiff aurora]

u can refer video

plzzzzzzzzzzzzzzzzzzzz

[deep matrix]

Hello

Could you show me the question

@stiff aurora

[stiff aurora]

btw bro im preparing for ISI exam u may know ig?

[deep matrix]

Which question exactly?

[stiff aurora]

this above book

[stiff aurora]

and the pic of page is my solution idk y it gives wrong answer

[deep matrix]

Bro wtf

Did you do?

How did you get 1

[stiff aurora]

rationalise

[deep matrix]

Could you please send the full question

I can't see the question

Only solution

[stiff aurora]

ok

isme (i) part hai jo ma puch rha hu

[deep matrix]

Ok

Wait let me try

[deep matrix]

Yes

[stiff aurora]

maine aise kiys

glti kahah hai