help-10

[plucky trail]

that doesn't sound like linear algebra

but then again I know nothing about linear algebra

[static furnace]

its linear algebra

[plucky trail]

okay yes sorry what is a system

[static furnace]

at least the basics of it

[plucky trail]

in linear algebra

[static furnace]

its a set of equations

with a set of unknown variables

[plucky trail]

i thought matrices was linear algebra

[static furnace]

you can solve it using matrices or adding/subtracting the equations from each other

[plucky trail]

hmm

okay so 2h+c = 30, 2w+h=20, c+w=15, h+c+w=x

for matrices would it be better not to add them together

[static furnace]

for this system id leave the last equation out until you find the value of each variable first

[plucky trail]

if i wanted to do this with matrices what would be the point of doing the system

dont i just get an answer from the systems

[static furnace]

to set up the matrix

[plucky trail]

but if i had the value of each variable

i could just add h+c+w

i thought i could use a matrix to get to x

is that wrong

like without having to use the systems

[static furnace]

the matrix helps you find the value of each variable

[plucky trail]

yes yes

how do i do that

without systems

unless a matrix counts as a system

[high lily]

matrix is used to represent the system

[plucky trail]

mmm mmm okay

[high lily]

row reduction operations is essentially elimination

[plucky trail]

2h+c = 30, 2w+h=20, c+w=15

so could i put these into rows or something

im sorry i am a total stranger to matrixes

or linear algebra

equation systems i get and can occasionally do but i am not very good at them

[high lily]

you're pretty much representing the same thing with a matrix here, just without explicitly writing stuff like the variables

keeps stuff more organised and makes the next step of elimination clearer

[plucky trail]

i think maybe i have been unclear in how unfamiliar i am with a matrix

i know there are rows in a matrix

and columns

i have seen a matrix

i have no clue how it works or what it is used for: I guess it is related to systems as you said earlier

once I have turned the picture into different equations

how would i go about putting it in a matrix

like what goes where, what part goes into what row/column combination

if anyone wants to explain in laymans terms what a matrix is used for it would be greatly apprciated

[surreal forge]

i disagree with this claim (or at the very least, find it misleading)

minimal knowledge of linear algebra is required to effectively solve a linear system of 3 equations in 3 variables

[plucky trail]

Okay

Yes I do not have minimal knowledge beyond there are rows and columns

I heard you could multiply them

That's all I know

[heavy ermine]

so you have to add the terms in the firs second and third row

like 2h+c=30

[plucky trail]

yes okay

i got these

2h+c = 30, 2w+h=20, c+w=15

how do i determine which number goes where

or variable i mean

[surreal forge]

to clarify, by "minimal knowledge" i mean "zero knowledge" *

[heavy ermine]

You can write them vertically aligning the same variables on the same line

[plucky trail]

i already solved this both intuitively and by using systems

i wanted to explore matrixes

i havent done anything with matrixes before

[plucky trail]

okay yes

[heavy ermine]

$0w+2h+c=30$\newline
$2w+1h+0c=20$\newline
$1w+0h+1c=15$

[surreal forge]

this is how https://tutorial.math.lamar.edu/classes/alg/augmentedmatrix.aspx

[warm shale]

Bendover

[surreal forge]

you can scroll to the very bottom to see examples on how it's done

[plucky trail]

that make sense

[heavy ermine]

and then youll divide the terms to matrices and vectors

$\begin{bmatrix}
0&2&1\
2&1&0\
1&0&1
\end{bmatrix}$

[warm shale]

Bendover

[plucky trail]

the variables dissapear

[heavy ermine]

and you can make the varible vector and multiply both

[static furnace]

we dont need them here because each column represents a variable now

[plucky trail]

oh yeah

oh okay

wait

so a matrice could be used in a 3d coordinate system

like to get a vector going in any direction

i have only used vectors in 2d planes so far

mainly in physics

[heavy ermine]

$\begin{bmatrix}
0&2&1\
2&1&0\
1&0&1
\end{bmatrix}$
$\begin{bmatrix}
w\
h\
c
\end{bmatrix}$

[warm shale]

Bendover

[heavy ermine]

then set this equal to the solution vector

[plucky trail]

let me google what a solution matrix is

[heavy ermine]

$\begin{bmatrix}
0&2&1\
2&1&0\
1&0&1
\end{bmatrix}$
$\begin{bmatrix}
w\
h\
c
\end{bmatrix}$=
$\begin{bmatrix}
30\
20\
15
\end{bmatrix}$

[warm shale]

Bendover

[plucky trail]

oh

[heavy ermine]

then you can use any method you want

[plucky trail]

30
20
15

is the solutions matrix?

[heavy ermine]

I meant vector(it doesntt have a foemal name, i just like to call it that)

[plucky trail]

it makes sense

to call it what you like to call it

[heavy ermine]

can you finish it by your own or you need help

[plucky trail]

i dont think i can solve it by myself from here but at least i know more about matrixes now

*using matrixes i cant solve it

like i said i already solved it through other means

[heavy ermine]

okay well go through it, i love the elimination mathod so we will use that

[plucky trail]

just to be clear

[plucky trail]

the rows are the variables here

or is it the columns

i assume rows

[heavy ermine]

no its just a representation of the system of equations, when you multiply it you get the first thing we started with,

[plucky trail]

when you say multiply it

what is it

multiplying the matrix with the numbers with the matrix including the variables?

[heavy ermine]

when we multiply it we get
$\begin{bmatrix}
0w+2h+c\
2w+1h+0c\
1w+0h+1c
\end{bmatrix}$=
$\begin{bmatrix}
30\
20\
15
\end{bmatrix}$

[warm shale]

Bendover

[heavy ermine]

and thats the equation we started with, no?

[plucky trail]

yes that's how i meant

yes

yeah i wasnt disagreeing i was just confused about what you meant by "it"

[heavy ermine]

so do you know how to reduce the matrix

[plucky trail]

i do not know how to reduce the matrix or what that means

also i noticed that that once you multiplied "it" that the variables went from being sorted in rows to being sorted in columns

that confuses me a little bit

[plucky trail]

here the variables are aligned nicely in columns

[plucky trail]

here they are aligned in rows?

i dont understand why that is

or if it is important or not

what are the prerequisites of using matrixes

maybe i am missing something more fundamental

[heavy ermine]

wait im trying to show you how its done

[plucky trail]

ok yep

first we wrote the givens like this 2h+c = 30, 2w+h=20, c+w=15

and then you showed me that yes there is 0w in the first one, 0c in the second and 0h in the last one

2h+c+0w = 30, 2w+h+0c =20, c+w+0h=15

seeing that makes it easier to understand where to put/align everything within a matrix

you aligned them in columns instead of rows and i was just not sure if that makes a big difference or not

2h c 0w
h 0c 2w
0h c w

but kind of pointless with the variables

2 1 0
1 0 2
0 1 1

but now i think i got the order wrong

so i am lost here

[heavy ermine]

This is the reduced form step by step$\begin{bmatrix}
0&2&1\
2&1&0\
1&0&1
\end{bmatrix}$\rightarrow
$\begin{bmatrix}
2&1&0\
0&2&1\
1&0&1
\end{bmatrix}$\rightarrow
$\begin{bmatrix}
1&\frac{1}{2}&0\
0&2&1\
1&0&1
\end{bmatrix}$\rightarrow
$\begin{bmatrix}
1&\frac{1}{2}&0\
0&2&1\
0&-\frac{1}{2}&1
\end{bmatrix}$\rightarrow
$\begin{bmatrix}
1&\frac{1}{2}&0\
0&1&\frac{1}{2}\
0&-\frac{1}{2}&1
\end{bmatrix}$\rightarrow
$\begin{bmatrix}
1&0&-\frac{1}{4}\
0&1&\frac{1}{2}\
0&-\frac{1}{2}&1
\end{bmatrix}$\rightarrow
$\begin{bmatrix}
1&0&-\frac{1}{4}\
0&1&\frac{1}{2}\
0&0&\frac{5}{4}
\end{bmatrix}$\rightarrow
$\begin{bmatrix}
1&0&-\frac{1}{4}\
0&1&\frac{1}{2}\
0&0&1
\end{bmatrix}$\rightarrow
$\begin{bmatrix}
1&0&0\
0&1&\frac{1}{2}\
0&0&1
\end{bmatrix}$\rightarrow
$\begin{bmatrix}
1&0&0\
0&1&0\
0&0&1
\end{bmatrix}$

[warm shale]

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

[heavy ermine]

do you understand whats going on

[plucky trail]

mmm no sorry, i see that the numbers switch places after the first step

but i cant determine why

[heavy ermine]

if you dont understand this you should probably take a linear algebra class

[plucky trail]

i will after summer in uni

i just wanted to have a good grasp beforehand

[heavy ermine]

$\begin{bmatrix}
1&0&0\
0&1&0\
0&0&1
\end{bmatrix}$
$\begin{bmatrix}
w\
h\
c
\end{bmatrix}$=
$\begin{bmatrix}
5\
10\
10
\end{bmatrix}$

[warm shale]

Bendover

[heavy ermine]

so this is the simplified form, they are equivallent

[plucky trail]

yeah that makes sense

to be clear

[heavy ermine]

and at last to find the solution you just multiply rhe matrix and the vector

[plucky trail]

it makes sense becauase i know the answer

w h c is the vector

?

[heavy ermine]

Then this is the last answer, the matrix is an identity matrix so its like multiplying by 1;;;;$\begin{bmatrix}
w\
h\
c
\end{bmatrix}$=
$\begin{bmatrix}
5\
10\
10
\end{bmatrix}$

[warm shale]

Bendover

[heavy ermine]

so w=5, h=10 and c=10

[plucky trail]

yes yes

that much makes sense

[plucky trail]

you mentioned a vector

that is the matrix that has w h c in it

right?

[heavy ermine]

I meant the variable vector

[heavy ermine]

yes

[plucky trail]

OK

[heavy ermine]

so was that helpful

[plucky trail]

i wont ask you to explain this entire thing

yes it has been

but would you mind explaining the first step

how you went from

0 2 1
2 1 0
1 0 1

to

2 1 0
0 2 1
1 0 1

or is that still too complicated

[heavy ermine]

i just exchanged the first and the second row

that is allowed

[plucky trail]

oh

yes okay

was there any reason

[heavy ermine]

as long as i do it to the solution vector

[plucky trail]

for doing that

like does it simplify the proble

m

[heavy ermine]

its just to make things simpler

[plucky trail]

mm ok

what made you determine that

[heavy ermine]

and when i transformed the matrix, the solution vector also got transformed the same way, it was not shown but you notice it was changed at last right

its like adding, multiplying, and stuff, you will understand it when you take linear algebra

[plucky trail]

ok

yes this has all been helpful anyway

[heavy ermine]

its a pleasure helping you, thanks

[plucky trail]

no thank you, sry for being a pain

i blame a nonexistant diagnosis

[heavy ermine]

np, i just woke up and wanted to help people, you really made my day, gotta go to school

[plucky trail]

cool thanks again have fun in school

.close

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[gloomy vector]

!da2a

[obtuse pebble]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

@timid silo Has your question been resolved?

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[ionic oar]

let ABE be x degress and solve for x

(sum of angles on a straight line is 180)

[obtuse pebble]

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[mystic tulip]

Is it okay if I didn’t arrange the variables with the exponents?

[safe haven]

yes

[heavy ermine]

All u did is right mb

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[uncut lodge]

can anyone help me understand this problem

[uncut lodge]

A rectangle has a length that is 4 times its width. If the area of the rectangle is increased by 200 square units by increasing its length by 5 units and its width by 2 units, what are the original dimensions of the rectangle?

[obtuse pebble]

@uncut lodge Has your question been resolved?

[void swift]

construct two equations

[uncut lodge]

oh

[void swift]

actually no sorry

before it increases its units

the length is 4w and its width is w right

so what should the area of this rectangle be?

before it increases its size

[obtuse pebble]

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[obtuse pebble]

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[lament heart]

I need help with part b

[safe ravine]

Waves and motion?

[obtuse pebble]

@lament heart Has your question been resolved?

[lament heart]

No it’s about hook’s law

[lament heart]

.close

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[silent spear]

is y=|x| continuous at x=0

[north cradle]

yes

its defined at x = 0

[silent spear]

thanks

.close

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[grizzled orbit]

How would one use these facts to show that the sequences converge to pi?

I skipped this question previously while working through this book but it looks interesting.

[ruby path]

lol i've done a similar problem but it was for AM and GM instead

[fossil crag]

I think I would show "bn <= an, an+1 <= an and bn+1 >= bn" by induction

[thorn marsh]

just use one of the series tests

[thorn marsh]

but what the heck is an-1

ive never seen that

[fossil crag]

it's not series convergence

it's sequence convergence

[fossil crag]

a_n and b_n are defined recursively, meaning they depend on the previous values

as an example

fibonacci is defined by "first numbers are 0 and 1, then the next number is the sum of the two previous ones"

if the next number is a_n

then the two previous ones are a_(n-1) and a_(n-2)

so a_n = a_(n-1) + a_(n-2)

@grizzled orbit if you don't see how to prove induction I can give you more hints, just ping

[obtuse pebble]

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[thorn marsh]

oh okay sorry about that

[obtuse pebble]

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[spare needle]

Hi can anyone help me with question related to backpropagation?

Lets say in the case of a simple network like above, where:

L = Loss
a = activation of a cell
z = sum of linear operation at the cell (without activation function)
w = weight

and i want to calculate the partial derivative dL/dw2 (how much L changes with respect to w2). I can use of course the chain rule like so:

dL/dw2 = dL/da3 * da3/dz3 * dz3/dw2

that makes sense to me, but now i want to calculate dL/dw1,

Based on the computation graph, i would need dz3/da2, da2/dz2 and dz2/dw1 to finally get dL/dw1 by chain ruling with my previously computed derivatives when calculating dL/dw2.

My question now is do i need to compute dz3/da2, da2/dz2 and dz2/dw1 manually or is there some way i can use the chain rule to get these as well without manually evaluating the derivative?

[obtuse pebble]

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[obtuse pebble]

@spare needle Has your question been resolved?

[obtuse pebble]

@spare needle Has your question been resolved?

[obtuse pebble]

@spare needle Has your question been resolved?

[obtuse pebble]

@spare needle Has your question been resolved?

[obtuse pebble]

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[slim surge]

Hi. Can anyone tell me is this enough to conclude f(x) is an odd function?

[obtuse pebble]

@slim surge Has your question been resolved?

[zenith raft]

if for every t in R there is an x such that xf(x) = t, then (combined with the above) f is odd
can you prove it now, or come up with a counterexample now, knowing that a counterexample needs to NOT have that property?

[zenith raft]

on second thought this might not be helpful and you should consider ignoring me

on third thought i have solved the problem and can confirm it is helpful and you should not ignore me

[rocky goblet]

i have also solved the problem and can confirm this is helpful

[obtuse pebble]

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[obtuse pebble]

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[ionic parcel]

How many elements in this powerset

[stoic yacht]

do you know the formula for the cardinality of the power set of a finite set?

[ionic parcel]

2^n

[stoic yacht]

right

[ionic parcel]

2^2

?

[stoic yacht]

right

technically the set only has two elements

1, and {2, {3}}

[ionic parcel]

Yeah

Okay

Thank you

.close

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[spare needle]

Hi can anyone help me with question related to backpropagation?

Lets say in the case of a simple network like above, where:

L = Loss
a = activation of a cell
z = sum of linear operation at the cell (without activation function)
w = weight

and i want to calculate the partial derivative dL/dw2 (how much L changes with respect to w2). I can use of course the chain rule like so:

dL/dw2 = dL/da3 * da3/dz3 * dz3/dw2

that makes sense to me, but now i want to calculate dL/dw1,

Based on the computation graph, i would need dz3/da2, da2/dz2 and dz2/dw1 to finally get dL/dw1 by chain ruling with my previously computed derivatives when calculating dL/dw2.

My question now is do i need to compute dz3/da2, da2/dz2 and dz2/dw1 manually or is there some way i can use the chain rule to get these as well without manually evaluating the derivative?

[obtuse pebble]

@spare needle Has your question been resolved?

[obtuse pebble]

@spare needle Has your question been resolved?

[obtuse pebble]

@spare needle Has your question been resolved?

[obtuse pebble]

@spare needle Has your question been resolved?

[obtuse pebble]

@spare needle Has your question been resolved?

[obtuse pebble]

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[mystic cypress]

Why is this wrong?

[mystic cypress]

The matrix in the bottom left corner is the correct one

<@&286206848099549185>

[remote temple]

Yes.

[mystic cypress]

Why is this incorrect? I did everything by the rules

[obtuse pebble]

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[mystic cypress]

<@&286206848099549185> Is something wrong with my question?

[mystic cypress]

Hello?

[obtuse pebble]

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@mystic cypress Has your question been resolved?

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[signal fractal]

Can someone help me out with the solution to this problem

[obtuse pebble]

@signal fractal Has your question been resolved?

[signal fractal]

<@&286206848099549185> please

I have a test tomorrow

[keen quarry]

use ratio test to find interval of convergence

[signal fractal]

I have

[keen quarry]

what did you get?

[signal fractal]

x(n+1)

If I take the limit of that

I get infinity

[keen quarry]

and what does the ratio test say if the result is greater than 1?

[signal fractal]

It diverges

[keen quarry]

there you go

it converges nowhere

[signal fractal]

But what about the interval

[keen quarry]

there is no interval

[signal fractal]

My book says (-1,1)

And centre 0

Sorry R = 0

[signal fractal]

🤨🤨🤨

[keen quarry]

that shouldn't be correct, since n! grows faster than x^n

which means that no matter what value of x, the series should always diverge, with the exception of x=0

[signal fractal]

So there's no radius

There's no interval

[keen quarry]

yep

there shouldn't be at least

[signal fractal]

So how would you write it in an exam

[keen quarry]

i would DNE

the interval DNE because yada yada yada

[signal fractal]

Fair enough bro

Thanks

[keen quarry]

np!

[signal fractal]

Appreciate you man you've helped out a lot

.close

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[grand relic]

on the integral -inf to inf of cos x/(x²+1) why dont we take the residue of lim z->- i (z+i) f(z)?

[surreal forge]

assuming you're integrating over a semicircular arc (positive imaginary components), the complex number -i falls outside of the contour hence its residue is not needed

[grand relic]

I kind of missed the concept of this contour thingy. Could you please brief it just enough so i can solve the questions(i dont need proof)

Id be very grateful 🙏

[surreal forge]

do you have class notes/textbook?

[grand relic]

I missed the class 😦

And i dont have any friends in class

To take it from

[obtuse pebble]

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[valid temple]

Hey there I'm doing some precalc problems, im with the one on the image:

Doing my calculations I get 18 minutes, chatgpt says its 57 minutes, and the book does not have this exercise resolve.

Thanks in advance!

[fallen pine]

don't trust chatgpt for math

oh but it looks like chatGPT is right in this case

[valid temple]

Why tho?

[fallen pine]

Because if you start with 2 instead of 1, you need one fewer split to get to the same amount

1 fewer split = 3 minutes

[valid temple]

But, I thinked of this problem as, If i got 2 amoebas, I got the double of duplication

I mean, if I start with two, at the first 3 minutes I'll have 4 amoebas, instead of 2 amoebas if i had started with one

Or not?

[jovial flame]

just a guess

[fallen pine]

that's right

[jovial flame]

might go horribly wrong

but is it 57?

[fallen pine]

yes

[jovial flame]

haha

[fallen pine]

The idea is you're one step ahead if you start with 2 instead of 1

and one step takes 3 minutes

[jovial flame]

yess

[fallen pine]

so you only need 60-3=57 minutes

[jovial flame]

did it in my head

[fallen pine]

https://tenor.com/view/me-looking-for-who-asked-looking-for-who-asked-who-asked-me-looking-gif-20318322

[jovial flame]

:(

[valid temple]

Mmmm

still don't understand

I mean

[fallen pine]

buddy I'm just saying if someone's asking for help on a math problem you don't need to be gloating that you solved it mentally 💀

[valid temple]

If i need 2^20 amoebas for filling the jar, If i had 2 amoebas, i just needed 2^10

[fallen pine]

no you'd need 2^19

half of 2^20 is 2^19

[jovial flame]

mb

[fallen pine]

not 2^10

[valid temple]

lol

thats true

Okay

That was my problem

[jovial flame]

did not mean to

[valid temple]

thinking 2^10 its the half of 2^20 💀

[valid temple]

np

[valid temple]

thank you so much!

❤️

[fallen pine]

[valid temple]

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[wraith lake]

help

[supple cargo]

chatgpt is ryt here , i am also getting 57 min
@fallen pine @valid temple
Soln
2 * (2^n) = 2 ^ 20
n=19 splits
for every split , it takes 3 mins so 19 spilts means 57 mins. 👍
ANSWER

[obtuse pebble]

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[west delta]

any hint

[obtuse pebble]

Please don't occupy multiple help channels.

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[west delta]

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[wraith hornet]

How do you prove that, if f:R -> R is continuous and invertible, then it is strictly monotone?

[obtuse pebble]

@wraith hornet Has your question been resolved?

[violet oar]

Hint: Invertible implies it's injective.

[wraith hornet]

yeah that's true

[wraith hornet]

https://homepages.math.uic.edu/~marker/mtht430/c11.pdf
took me a while to find a proof of this online but this is the only one i found that makes sense to me

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[silk maple]

whats the parametric equation for the surface of a torus?

[silk maple]

nvm

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[night idol]

Problem 58. Let p_{n} be the nth prime number. Prove that p_{n} < 2 ^ (2 ^ n)

[foggy quail]

!status

[obtuse pebble]

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

[foggy quail]

are you familiar with bertrand's postulate

[obtuse pebble]

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[rose ferry]

\

[rose ferry]

Hey I am sketching this graph

and I found the end points and y int

im just very confused how do I know how many x intercepts there are?

[brazen viper]

@rose ferry the function is periodic, there are either infinity or exactly 0

[rose ferry]

sorry I forgot to include this

x is an element of [-pi,2pi]

[brazen viper]

Ah, first you need to find the period

[rose ferry]

2pi/3

[brazen viper]

Next, find the x offset

[rose ferry]

what is that?

epsilon?

in this it is -pi/4 right

[brazen viper]

That is, the offset from the normal cosine function

[rose ferry]

yep

[brazen viper]

3π/4

[rose ferry]

where is that from?

[brazen viper]

cos(3(x -π/4)) = cos(3x - 3π/4)

[rose ferry]

ah ok

got it

so the offset is 3pi/4

[brazen viper]

So where you'd normally have cos(0) = 1, instead you have f(3π/4) = 1

And instead of a period taking 2π it takes 2π/3

Where do the zeroes normally fall?

[rose ferry]

zeroes?

[brazen viper]

Of the cosine function

The x-intercepts

[rose ferry]

the x intercepts?

ah

they fall on the x line?

[brazen viper]

As in what are the x values of the x-intercepts?

[rose ferry]

i dont know

[brazen viper]

For a normal cosine function?

[rose ferry]

oh

i should probably know...

i dont though sorry

[brazen viper]

It's ok

One of them is at π/2

[rose ferry]

oh wait

sorry

pi/2, pi, 3pi/2

[brazen viper]

And the rest are offset by kπ where k is an integer

[rose ferry]

right?

[brazen viper]

Not quite

cos(π) = -1

[rose ferry]

yeah i understand that

how do we know how many x intercepts there are in total

[brazen viper]

So, we know the x-intercepts for cos(u)

And we know that u = 3(x-π/4)

We can solve for x

And plug in values for u that we know are zeroes

And see if they're in the range

[rose ferry]

ahh i think someone told me something like this in the server before

like 2pi would = 6.blahblaj

and pi = -3.1415

[brazen viper]

Well, not quite

[rose ferry]

so when we find the points we keep addind a period until it is out of the range

right?

[brazen viper]

Maybe? It's a little muddled, and I'm not sure if I'm understanding what you're conveying

You don't need the decimal expansion of pi

[rose ferry]

example

if we find the point when y = 0 x = 5pi/12

we can find 5pi/12 as a decimal

and see if it is between -pi and 2pi

[brazen viper]

Oh, no need for that

[rose ferry]

that makes sense for me

like seeing it in numbers

i dont really understand the range not in a number

[brazen viper]

If 5π/12 is in the range -pi to 2pi, then 5/12 is in the range of -1 to 2, right?

[rose ferry]

probably lol

[brazen viper]

Yup it is

It's a bidirectional implication

[rose ferry]

the only problem is my test is tech free and i will have no access to a calculator

[brazen viper]

Yes, but you don't need a calculator

To check if 5/12 is between -1 and 2

[rose ferry]

ok

[brazen viper]

And if 5/12 is then 5π/12 is between -π and 2π

[rose ferry]

can you show me how you would find all the points

just explain your working out and ill watch

[brazen viper]

I'm trying to walk you through it

!nosols

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[rose ferry]

i wont copy this

it helps me more to just watch what you do

then i can apply it to another question like how they do in examples

[brazen viper]

Then I'll do a slightly different problem, changing the numbers and function

[rose ferry]

its okay please do this one

[brazen viper]

I cannot, again

!nosols

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[brazen viper]

I can walk you through an example

[rose ferry]

alright i guess you can change the numbers

[brazen viper]

But I can't walk you through this problem

So let f(x) = 3sin(5x - π/2)

From here I can tell the period is 2π/5 and the offset is π/2

So we have u = 5x - π/2, and I want zeroes in the range [-π/2, π/2]

Solve for x, x = u/5 + π/10

[rose ferry]

what is U?

like y?

[brazen viper]

A variable yes

[rose ferry]

when you do u = 5x - pi/2 what happens to 3sin

[brazen viper]

This is why I used u

f(x) = 3sin(5x - π/2)
f(u) = 3sin(u)

I want f(u) = 0

So 0 = 3 sin(u), 0 = sin(u)

Then I solve for u, I know that sin(u) is 0 whenever u = kπ for k in Integers

So, x = kπ/5 + π/10

Or x = (2k + 1)π/10

We want x between -π/2 and π/2, which is -5π/10 and 5π/10

We can see this is satisfied for k = -3, -2, -1, 0, 1, 2

So our zeroes in the range are -5π/10, -3π/10, -π/10, π/10, 3π/10, 5π/10

The end

@rose ferry

[rose ferry]

👍

i am watching

lol

the u is really confusing me

[brazen viper]

It's just a placeholder variable

It doesn't correspond to an axis

On the graph

[rose ferry]

back to the start

where does 3 sin go

when you do u = 5x - pi/2

[brazen viper]

I explained from here #help-10 message

[rose ferry]

sorry that doesnt make sense

ok wait i get it

[brazen viper]

I'm asserting that u is equal to this. This is my definition of u

[rose ferry]

why do you want f(u) to = 0

[brazen viper]

Because we're looking for x-intercepts right?

[rose ferry]

so is u the y value

[brazen viper]

No

f(u) is the y value

f(x) is as well

I'm kinda punning the function f a little

[rose ferry]

my head is spinning

im very lost

sorry

[brazen viper]

Using it to mean two different but very related things

[rose ferry]

i understand if you want to move on to someone else

[brazen viper]

It's fine

So

[rose ferry]

my teacher has never explained it like this

[brazen viper]

This u stuff is just me trying to be very structured with how I approach the problem

[rose ferry]

using different variable and taking out the bracket into a new function

[brazen viper]

Think of u as a convenient variable that means what I want it to mean

I want f(u) to be easy to solve

Because f(x) is slightly tricky

And then I solve f(u), and then convert the solutions in u to solutions in x

And this way I solve f(x)

It's like

[rose ferry]

and to find the endpoints you just substitute in the values they give you

[brazen viper]

You go up to your house, but your door is locked.

Now you know there's a key, but the key is under a flower pot in the backyard.

So you jump the fence, and check under the flower pot, then go in through the backdoor to open the front door

You got the front door open, a hard problem, by solving a series of easier ones

[rose ferry]

ok

[rose ferry]

how did you know when to stop adding a period

[brazen viper]

Because I know that -π/2 is -5π/10

And π/2 is 5π/10

[rose ferry]

oh ok

would this kind of question be on a tech free test

with just your own knowledge

[brazen viper]

It can be

You don't need a calculator to solve this

[rose ferry]

fck

[obtuse pebble]

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[obtuse pebble]

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[shrewd comet]

need help with iv

yes

[rose scroll]

you obtained the decay equation. Can you tell what is m in the left hand side?

[shrewd comet]

i already got the answer 🙂 thanks anyway!

.close

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[inland lantern]

question 6 (c) chapter 1 principles of mathematical analysis by walter rudin

[inland lantern]

i am a bit confused whether i am on the right track or not

i first defined the set of all numbers b^t where t is rational and t <=r. Im not sure if this requires proof but since b>1 i said that b^r>=b^t for all b^t in the set. i then i assumed that b^r was not the supremum, then there exists another rational r where s<r and b^s=supB(r). but then there exists a rational q st s<q<r and hence b^q>b^s a contradiction as b^q is an element of B(r). therefore b^r is the least upper bound

is this a valid sketch of a proof for this question?

[obtuse pebble]

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[inland lantern]

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[obtuse pebble]

Please don't occupy multiple help channels.

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185> please help

[clear cosmos]

Find det R and the transpose of R

Mutiply 1/det R with the transposed R with the correct signs changed and you will the the inverse of R

@timid silo

I would use a b c and d instead of all those numbers

I’m not sure on what to do next so, Good luck!

[timid silo]

after that i think i have to find the number of possible arrangement for replacing a b c d with the given elements

[clear cosmos]

Then the det shouldn’t be zero

You know how to find the det of R?

@timid silo

[timid silo]

yeah

[timid silo]

for an invertible matrix det R can not be zero

[timid silo]

the adj R should not also be a null matrix

any other possibilty? @clear cosmos

@clear cosmos

[obtuse pebble]

@timid silo Has your question been resolved?

[sudden sage]

you can use the characteristic equation with the coefficients redefined as tr and det A

[timid silo]

yeah thanks

i tried it

[obtuse pebble]

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[obtuse pebble]

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[quaint bronze]

how is this wrong

[quaint bronze]

uh so i just did rise over run and then opposite reciprocal (idk if thats how its spelt)

cuz I'm pretty sure my teacher just put in the wrong answer or something

[nocturne bronze]

can this conclusion be made?

[wraith hornet]

which line is the "original line"? only thing I can think of is maybe your answers should be switched

[quaint bronze]

the blue one

i made the green one (the one with a dot, if you're color blind, idk)

[wraith hornet]

then your answers are correct. maybe there’s like a space after the 2 you entered in or something idk. could be your teacher putting the wrong answer in

[quaint bronze]

also ive never seen any of that before so idk

[quaint bronze]

alr thanks

shit had me concerned

[wraith hornet]

technically yes because
$|xy|\geq xy$

$\implies |xy|-xy\geq0$

$\implies 2|xy|-xy\geq|xy|$

but i’m not sure why you’d want to make that conclusion

[warm shale]

Ryan

[wraith hornet]

are the x-scale y-scales different? like on the axes i mean. can’t see the values with those white boxes in the way lol

[wraith hornet]

personally i’d just add xy to both sides to cancel them out and be left with x^2 + y^2 >= 2|xy|

but i’d have to see the rest of the problem

[quaint bronze]

can someone tell me what any of that means

[wraith hornet]

it’s just algebra but slightly more higher level than what you’re doing

[obtuse pebble]

@quaint bronze Has your question been resolved?

[quaint bronze]

aight

honestly this is review from last year so it might be what im doing this year

[obtuse pebble]

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[civic briar]

Is this alright?

I guess I should have |r|^n

< ep at first

[median dome]

Yes

[civic briar]

Everything else is fine?

[median dome]

You should specify r=0 as a special case

Because the log isn't defined there

[civic briar]

Alright

Done

[median dome]

The rest looks good yeah

[civic briar]

Okay, thank you

.close

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[median dome]

Maybe mention that the inequality here flips because |r| < 1, but I'm just nitpicking here

[civic briar]

Eh

That's common sense

At this level, doesn't make sense to justify algebraic manipulation

Unless it's particularly hard to follow

[median dome]

Yeah lmao makes sense

[obtuse pebble]

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[stone holly]

tried taking log both sides but it turns to something horrible

[rigid skiff]

wth is this mess

goddam even taking the derivative will take to long

[obtuse pebble]

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[obtuse pebble]

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[wary meadow]

could someone help with qn11 c the answer is 2215

[wary meadow]

nvm got it

.close

[obtuse pebble]

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[obtuse pebble]

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[forest shell]

I need to find the domain and range

[modern idol]

Okay so the domain is pretty straightforward. You just have to ask yourself “are there any values I can’t plug in to this function?” (Hint, you can’t plug in a value that would cause you to divide by 0)

[forest shell]

wait so would it be sinx=1/2 then

and then i find x

[modern idol]

Yeah so that would tell you the x values that are outside the domain of this function

[forest shell]

so 30

[modern idol]

Right, but is that the only value of sin that yields 1/2?

[forest shell]

also 120 ??

wai t no

150

[modern idol]

Right there’s another one. But remember sin is periodic so you can do another circle and keep looking for more

Are you comfortable with switching to radians? It’ll make explaining this a bit easier

[forest shell]

yep

[modern idol]

So those two values in radians are pi/6 and 5pi/6 right?

[forest shell]

yeah

[modern idol]

And a full circle is 2pi. So what about 2pi + pi/6?

That would also get 0 after taking the sin right?

[forest shell]

yepp

[modern idol]

As well as 2pi + 5pi/6.

And we could do that again with adding 4pi, 6pi, 8pi, and so on forever. AND with -2pi, -4pi and so on

[forest shell]

then what would be the domain then

[modern idol]

So you can express those values in general like so:

Let k be any arbitrary integer. sinx is 1/2 whenever x = pi/6 + 2pik, and whenever x = 5pi/6 + 2pik.

So the domain of f(x) is all real numbers except for those I just mentioned.

[forest shell]

ohhh

okay that makes sense

but what about range

[modern idol]

The range is more complicated. You have to ask yourself what values you will never see using this function. For starters, graphing it can give you an idea of what to look for here.

So clearly the range, being all possible y values, is every number except that strip in the middle that the function never touches right?

[forest shell]

wait so is the range just -1/2, 2

[modern idol]

The question in this case is how do you identify the top and bottom of that strip

So that’s clear to see from the graph, but now what if you didn’t have the graph to look at? Haha

The bottom looks a little lower than -1/2 though from the graph. That’s another thing to figure out analytically

[forest shell]

im kind of confused

[modern idol]

Yeah I haven’t explained how to find the range without looking at the graph. I just wanted you to get an idea for starters

So there’s multiple ways. What level class are you in? Is this calculus or something earlier?

[forest shell]

calc ab

but we havent started school yet

like these r just practice problems

[modern idol]

Ahh okay then we can’t use any calculus techniques to get there. Let me remind myself how to solve this without calculus

Ahhhh yes that’s right. Write the function down as an equation, y=all that stuff

Just replace f(x) with y in other words.

Then, what you want to do is solve the equation for x instead of y.

[forest shell]

ummm im still kind of confused 😭

[modern idol]

One moment I’ll show ya

[forest shell]

okok thank youuu btw

[modern idol]

(Sorry for bad handwriting)

Let me know if these steps make sense. If so, I’ll explain why I did it.

[forest shell]

wait so is the final thing the range then? but why do you have to solve for x again im so sorry im just rlly frazzled rn

[modern idol]

You’re getting ahead. This is not the range but it’ll help us find it. For starters does all of that algebra I just did make sense to you?

[forest shell]

yep it makes sense algebraically

[modern idol]

Why solve for x:
This is called taking the inverse of the function. If we now, after that algebra, rename x to y and vice versa, we will have the inverse function of our original function. This helps us find the range because, crucially, the domain of the inverse of a function is the range of the original function (and, less usefully in this case, the opposite is also true that the range of the inverse is the domain of the original)

[forest shell]

ohhh so do you just make the x into y and y into x after you set f(x) to y

[modern idol]

Here, I’ve added that final step to avoid any confusion

So that function on the bottom is the inverse of the function on the top.

[modern idol]

And knowing what I just said here

You can now use the same technique you used to find the domain, but on this inverse function, and that will tell you the original function’s range.

So ask yourself with this new function what values you can’t plug in

[forest shell]

ohhh

[modern idol]

This one is trickier because sin inverse, unlike sin, is not defined everywhere

[forest shell]

wait so for this one

i cant set x to 0 right

[modern idol]

That’s true but there’s a whole bunch of other values you can’t do because that stuff is inside of sin inverse

[forest shell]

okay i get it up to this step then

bc how do i find the values that i cant do

bc of sin invers

[modern idol]

Right, so sin inverse (on its own) is obviously the inverse or sin. Whats the range of regular old sin?

sinx, what’s the highest and lowest value that outputs?

(You can think about this in terms of the unit circle or the graph of sinx)

[forest shell]

-1<=sinx<=1

[modern idol]

Bingo. And because of this

We know that we can only plug in numbers between negative 1 and 1 to sin inverse

So then the domain of this whole inverse function is going to be all of the values x that make it so we take the inverse sin of a number between (and including) -1 and 1.

So to do that, solve for x in this compound inequality:

[forest shell]

ohh okay this makes sense

so like 2?

and -1/3 i tihk

[modern idol]

Very close

[forest shell]

oh wait 2/3

but yeah i get it finally yay

tysm u clarified a lot of things

[modern idol]

-2/3 to be clear

So taking a step back, that tells us that the domain of the inverse, and thus the range of our original function, is all values except those between -2/3 and 2

[obtuse pebble]

@forest shell Has your question been resolved?

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[timid silo]

is the inner product just the dot product?

[sage geode]

The dot product is a special case of inner products

[worn yoke]

inner products generalize the properties of dot products to more types of vectors, similar to how vector spaces generalize the properties of R^n

[timid silo]

hmm okayy

so if i have two vectors a, b the inner product is just the dot product?

[worn yoke]

there are many different inner products, but the dot product is the most common type

the dot product is also called the "euclidian inner product" and is usually the "default" inner product for vectors in R^n

[timid silo]

Okayy what other inner products are there?

[sage geode]

For example in the vector space of real functions continuous on some closed interval you could define the inner product of two vectors to be the integral of their product over the interval in question

[timid silo]

<f,g> = ∫[a,b] f*g dt right but what does it exactly tell me? it kinda confuses me..

[worn yoke]

similar to the dot product, it gives us a way to define an angle between two vectors, what it means for two vectors to be perpendicular, and a way to measure a vector's "length" with a norm

[timid silo]

hmm

[timid silo]

does this apply to functions aswell? like this?

like does this give me an angle?

[obtuse pebble]

@timid silo Has your question been resolved?

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[patent socket]

So im starting to learn mathematical symbols such as N natural numbers (1...10), Z for whole integers (-10...10). I only struggle to understand how rational and irrational numbers work? Such as Q, R, I

[trim portal]

(1...10)?

natural numbers extend to infinity

[patent socket]

from 1 to 10 and further

[trim portal]

right

[rotund bloom]

me when i dont include 0 in natural number set 😠

[patent socket]

funny how that is, my book told me not to include it as a standard

[trim portal]

rational numbers are just all numbers that can be written as a fraction of 2 whole integers

like 3/6

or (-7)/2

[rotund bloom]

irrational numbers r basically numbers that cant be expressed as p/q where p and q r coprime integers

[trim portal]

or even 6/1

[patent socket]

so rational numbers are basically numbers that can be divided?

1/2, 4/3, 5/3 etc?

[worn yoke]

a rational number is a number which can be expressed as a ratio/fraction of two integers, yes

[trim portal]

Yes, those numbers

[worn yoke]

then an irrational number is any real number which is not rational

[trim portal]

such as sqrt(2)

[patent socket]

real numbers are?

[trim portal]

or pi

[trim portal]

all rationals + irrationals combined

[patent socket]

ah i see

[rotund bloom]

the set symbol is R

for reals

[patent socket]

so basically real numbers but filter out the ones that are rational

[rotund bloom]

yes

[worn yoke]

all the numbers you normally think of when we say "number", but excluding other number systems like complex numbers, etc.

[trim portal]

all the real numbers have (sometimes infinite) decimal expansion.
Like
5.35235624512050713509531.......
3.141592653589.....
3
3.25
3.333333333333333333333333333
-2.513513415136416
etc

[patent socket]

i might say something dumb now, but sqrt(3)/1 would be irrational as sqrt(3) isn't a whole integer

[worn yoke]

yes

[rotund bloom]

reminds me of the joke where they say π is rational cuz π/1

[worn yoke]

of course something like 1.463/1 is also not a ratio of whole integers but we can easily write that as 1463/1000. another property of rational numbers is that their decimal expansion either ends or repeats itself infinitely

[patent socket]

right, so to sum up

real numbers (R) are rational + irrational numbers (sqrt(3), 3/2)
rational numbers (Q) are numbers that can be written as a fraction as whole integers where the denominator != 0 (3/2, 1/2, 5/4)
irrational numbers (I) are real numbers that are not rational numbers (sqrt(3), pi, the golden ratio??)

[trim portal]

yeah

although the definition is kinda cyclical

definining irrational numbers through real numbers and real numbers through irrational numbers

[patent socket]

so we can say that

sqrt(3) is not in Q

i don't know how to write the e with a stripe

[trim portal]

yep

[worn yoke]

I is somewhat rare for denoting the irrational numbers, more common is R\Q (real numbers excluding rationals)

[patent socket]

oh right, \ is used for filtering correct?

[trim portal]

basically

it's set difference

[patent socket]

N\Z = all positive numbers

[trim portal]

almost

[patent socket]

Z\N

[trim portal]

This is better

but dont forget about 0

[patent socket]

right

[trim portal]

0 is in this set but not in N

[trim portal]

so these are all the non-negative numbers

which is positive + 0

[worn yoke]

that would give all negative integers

[trim portal]

you're right

how did I miss that

Z \ N gives the set of all whole numbers that are not in N

[patent socket]

so in other words Z\N means take the set of all negative whole numbers and exclude all numbers that are positive

[trim portal]

take the set of all whole numbers (Z) and exclude all the positive whole numbers (N)

[patent socket]

ah

[trim portal]

and you get all the negative numbers and 0

sorry for my earlier mistake btw

[patent socket]

so that would mean that R\Q = I

[trim portal]

correct

and R\I = Q

[patent socket]

alright nice

I think I'm starting to understand, thank you guys a lot!

I appreciate it

.close

[obtuse pebble]

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[obtuse pebble]

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[forest shell]

how can i determine x^2y-x^2+4y =0 that y is a function of x

[past sand]

Do you mean "how to determine y as a function of x"?

[forest shell]

uhh the question says determine whether y is a function of x

[past sand]

That makes little sense

[low vigil]

will making this equation y=f(x) work?

[past sand]

You can make it a function of x, but whether it is is basically up to you

[forest shell]

wait wdym

[past sand]

You can manipulate that equation x^2y-x^2+4y=0 into y = ..., with some x and no y on the right

(with an added constraint on x)

But x and y are usually just called variables

[rocky glen]

i think it just wants you to make

[past sand]

y(x) would be called a function (of x)

[rocky glen]

y the subject

[past sand]

Anyway, that's probably what the question is actually about, it's just phrased weird

[forest shell]

do i just put y in terms of x then?

[past sand]

Yes, we call that "solving for y"

[forest shell]

okay tyy

but then it says determine whtether it is

so whats an example of it not being a function of x

because the examples it gives you could put everything in terms of x

[rocky glen]

if it can be written

so that y can be separated

and be made a subject

then it can be a function

i imagine that is what it means

[past sand]

Can you screenshot the question?

[forest shell]

its on paper and i dont have my phoneee

its just a series of equations and it asks if y can be function of x

[rocky glen]

it has to be in this case

can u give us

[forest shell]

wait but one of them is y^2=x^2-1 can that be a function of x

[rocky glen]

other examples

[forest shell]

you could just root both sides tho

[rocky glen]

i mean it can be

[past sand]

Well again, the phrasing is just weird

[forest shell]

yeah the other ones y is already singled out so you could easily put it in terms of x

sooo

[past sand]

If you do y = sqrt(x^2-1), you lose information compared to y^2 = x^2-1

But it's still a function, in a way...

For x^2y-x^2+4y=0, if you transform this into y=<something> and nothing else, then you also lose information, though not quite as much

[rocky glen]

i think

[past sand]

But again it would still be a function

[rocky glen]

y^2 wont be

because it is

one to many

for any given x there's two values of y

[forest shell]

oh so like if

theres an x and y^2

[rocky glen]

if for any x

u have a definite single answer

then it is a function

[rocky glen]

wait continue

ohhh now i get the question

yh go on

[rocky glen]

i hope

[forest shell]

nono i c ur point cuz i googled it

and like its just if x theres only one value of y

right?

i think idk

[rocky glen]

yes

if it is a function of x

ur x value shld have a single definite

output value

[past sand]

yeah I suppose you're right

[forest shell]

also i have another question

not related

but given that f(x)=sinx and g(x)=pi(x) what is f(g(x))

like is it just sin(pix)

idk how else to write it

like i thought it would be 0 but like what if x is a fraction

so liek do i just leave it like that

[past sand]

Yes it's just sin(pix)

[obtuse pebble]

@forest shell Has your question been resolved?

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[cunning rock]

how is this a valid form of differential equation?
P(x) = k, isnt it required that it must be a function of x? both P(x) and Q(x). These 2 rules don't seem to be working here.

[restive gorge]

,, P(x) = k = kx^0

[warm shale]

𝔸dωn𝓲²s

[fossil crag]

constant functions exist

[obtuse pebble]

@cunning rock Has your question been resolved?

[obtuse pebble]

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[tender cypress]

let x,y in ]-1, 1[ use this last step to complete the proof for |yx² - y + x| < 5/4: $$|y+1/(4y) - y(x+(1/2y))^2| < ? < 5/4$$

[warm shale]

Goëtia

[obtuse pebble]

@tender cypress Has your question been resolved?

[obtuse pebble]

@tender cypress Has your question been resolved?

[obtuse pebble]

@tender cypress Has your question been resolved?

[obtuse pebble]

@tender cypress Has your question been resolved?

[obtuse pebble]

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