help-13

[buoyant latch]

What

What the heck is an x axis in ℝ²

[shrewd violet]

you can only stay in one axis

so it spans the entire x axis

[buoyant latch]

But what is the x axis

[buoyant latch]

Ok I get what your prof is trying to say

But my point is

The x axis is arbitrary

I could make that the y axis if I want

I could make it slanted if I want

I can rotate it however I like

As long as the 2 axis aren’t the same line, it’s fine

[shrewd violet]

for x and z

or my example

[buoyant latch]

Also obviously since every component can be non-zero the subspace spanned by B is not just axis’es

That’s bad English but I wanted to emphasise it

This spans an “axis” only because the second component is always 0

[shrewd violet]

but isnt that the same thing in our case

wait no

x and z have to be equal

for the span to be r^3

so what would we say the span is here?

[shrewd violet]

you still here?

[buoyant latch]

Sorry friend bothered me

[buoyant latch]

No

[shrewd violet]

how ?

[buoyant latch]

That sentence makes no sense

Span is a function that takes a set and returns a set

[shrewd violet]

returns a set??

[shrewd violet]

is span not a set of lienar combination of vectors?

[buoyant latch]

What vectors

You didn’t specify

[shrewd violet]

in r^2 or R^3

[buoyant latch]

Idk what are you trying to say

[shrewd violet]

the definition of a span

If v1 v2 v3 are vectors in R2 and R3 the set of all linear combinations of these vectors is the span

[buoyant latch]

What the hell is vectors in “ℝ² and ℝ³”

Things are only vectors if they are an element of a vector space

[shrewd violet]

anyways

this is offtopic from my question

what would be the span of this question

x[2,1,2],2y-x-2c[1,0,1],x[0,1,0]=[x,y,z]

this is the span?

[buoyant latch]

It says or

[shrewd violet]

oh myb

[buoyant latch]

The span is a set

[buoyant latch]

This is not a set

[shrewd violet]

Span {x[2,1,2],2y-x-2c[1,0,1],x[0,1,0]}=[x,y,z]

[buoyant latch]

What are the square brackets for

[shrewd violet]

which part

{} or the []

[ ] are the vectors

{} is the set of vectors

also this is linearly dependent

since x and z are arbitrary

can you please tell me what the span is?

[buoyant latch]

I don’t get it

What’s the coefficients doing

[shrewd violet]

this

[shrewd violet]

.

[shrewd violet]

hello

@buoyant latch are you still here 💀

i just want to know what the span is

[cedar kiln]

@shrewd violet Has your question been resolved?

[buoyant latch]

$\text{span}((2, 1, 2), (1, 0, 1), (0, 1, 0)) = {x \in \mathbb{R}^3 | x = a(2, 1, 2) + b(1, 0, 1) + c(0, 1, 0), a, b, c \in \mathbb{R}}$

[wraith dagger]

Frosst

[buoyant latch]

Sorry I’m a bit busy atm

[shrewd violet]

but i did do that?

[shrewd violet]

here

did i not?

[buoyant latch]

What is this x

Or 2y-x-2c

[shrewd violet]

=b

[shrewd violet]

x is a and c

since x = z

[buoyant latch]

but that’s not what I wrote

But I never said anything about x = z

[buoyant latch]

Read what this says

[shrewd violet]

there is no a and c

they are arbitrary

[buoyant latch]

But that’s the point of a span

[shrewd violet]

so what do i say??

for a and c

that is what i masking

if there doesnt exist an a or c in which you can get any x y z

[cedar kiln]

@shrewd violet Has your question been resolved?

[cedar kiln]

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[umbral pawn]

I got 2/xy but not sure if its correct

[rare vault]

,rccw

[rare vault]

looks right to me

[umbral pawn]

Oh, ok good thanks

.close

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[kind magnet]

Having some trouble with Difference of quotient.

I learned in class that I should just replace f(x) with x^2 + 3, which cannot be factored, but in my example I don't have f(x), I only have f(8), would I then plug 8 into (8)^2 + 3? which would give me 67?

[obsidian coral]

Yes

[kind magnet]

but if f(8) = 67, can I just replace all f(8) by 67?

[obsidian coral]

What do you mean by all?

You only have one f(8)

[kind magnet]

alI have f(8 + h)

is there something I can do with that?

like 67 + h

[obsidian coral]

f(8 + h) and f(8) aren't the same

[kind magnet]

ok

[obsidian coral]

You plug in 8 + h

[kind magnet]

so it would f(8+h) - 67

and then I distribute and then I can plug it in I see

wait no

[kind magnet]

how do you just remove the f of f(8+h)

[lost mason]

f(8) is 67 in all circumstances

since f(x)=x^2+3 and 8^2 = 64 +3 is 67

[kind magnet]

correct

[lost mason]

Isnt the answer for this just 0

67+h-67=0

or no

[kind magnet]

I don't think I can plug in 67 at f(8 + h)

I'm trying to figure out how to get rid of it

wait I can't even distribute

[obsidian coral]

f(8 + h) means to plug in 8 + h for all instances of x in the function f

It's the same logic as f(8)

[kind magnet]

so it would become 8+h -67

[obsidian coral]

Where you plug in 8 for all instances of x in the function f

[obsidian coral]

No

You have f(x) = x^2 + 3

If you plug in 8 + h for x, what do you have now?

[kind magnet]

(8+h)^2 + 3

[obsidian coral]

Exactly

Now you just expand and simplify

[kind magnet]

(8+h)(8+h)+3

[obsidian coral]

Continue

[kind magnet]

h^2 + 16h + 67

I'm doubting everything I do, because this feels so odd

[obsidian coral]

It's all correct

[kind magnet]

ok so can I take that and plug it into the original function now?

so all that + 67

[obsidian coral]

Original function being difference quotient?

[kind magnet]

h^2 + 16h + 67 + 67 / h

that's what I am guessing

so h^2 + 16h + 134 / h

[obsidian coral]

Add parenthese for the numerator

[kind magnet]

(h^2 + 16h + 134)/h

[obsidian coral]

And isn't difference quotient something like f(8 + h) - f(8) in the numerator?

[kind magnet]

yes

but I plugged in my answers

so I'm at (h^2 + 16h + 134)/h

[obsidian coral]

If it's minus f(8) why did you add here?

[kind magnet]

hold on let me check

ah indeed

(h^2 + 16h)/h

so basically the 67 disappears

I can factor out an h

h(h+16)/h

and I'm guessing h cancels giving me..

wait no can I cancel here?

h+16/1

which is just h+16

[obsidian coral]

Yep

[kind magnet]

and now I guess I just plug in 0 for h

0+16 = 16

[obsidian coral]

Yep

That's right

[kind magnet]

oh wow that was a lot

ok I understand now, but it was tricky

thanks a lot!

.close

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[torn pendant]

This algebra problem is causing me grief...Everything except x is constant, it's just my actual values are quite tedious to write. I'm having trouble separating this and solving for x. I've been trying to use laws of exponents, natural logarithms, a bunch of random other things and can't seem to isolate x. In particular, it feels like having x by itself as a variable on one side as well as being in the exponent on the other side is making it difficult for me. Can anybody help?

[cedar kiln]

@torn pendant Has your question been resolved?

[torn pendant]

<@&286206848099549185>

[cedar kiln]

@torn pendant Has your question been resolved?

[torn pendant]

i gave up on educational benefit and just matlab'd the solution like a champ

.close

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[lime wagon]

Can soneone help me understand the meaning behind this integration constant

[royal loom]

are you asking why there are arbitrary constants of integration in general?

[lime wagon]

the calculations make sense to me, but I do not understand what the integration constant actually means. Like what is c=-1/10

[royal loom]

that is because of the initial conditions

[lime wagon]

I know, but what does it mean in the cotext of this exercise

and when can we neglect integration constants? Clearly, we integrate both sides of the equation but we keep only one integration constant. Where is the other one?

[royal loom]

that is because

x+c=y+d

is the same as saying

x=y+d-c

and d-c are arbitrary constants

so we can combine them

x=y+C

it is thus only necessary to keep the constant on one side of the equation

x+c=y+d was only an example, but x and y can be replaced with more complicated functions and the reasoning holds true

[lime wagon]

arbitary constants? the one constant comes from integrating the velocity, and the other one comes from integrating time

[royal loom]

yes but they are arbitrary without initial conditions

[lime wagon]

so its a constant that helps us solve equations, but what is the meaning behnd it when we find a value like -1/10 for it

[royal loom]

the meaning is that based on the initial conditions of our differential equation, it is required that the constant in our solution is equal to -1/10 for these initial conditions to be true

[lime wagon]

oh I guess the meaning behind it in this context is just -1/v -kt

[royal loom]

our differential equation in general is true with the constant left as C

but if we want to know about specific cases

like oh the speed starts at 5 mph and the blah starts at blah

then we have to solve for a specific value of C

that makes it such that our specific cases are true

[lime wagon]

when do we need to keep integration constants seperately, and not combine them? Can I also combine integration constants when integrating twice?

[royal loom]

we don't need to keep them seperately if they are just being added like that

no, because when integrating twice you will get C and then you will get Cx+D

or Ct+D

whatever your variable is

one of the constants will be attached to the independent variable

so you can't combine it

[lime wagon]

oh ye ofc

thx

.close

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[fast relic]

I am sooo confused

[cedar kiln]

Please don't occupy multiple help channels.

Please stick to your channel.

[junior dome]

#help-17

[cedar kiln]

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[hexed sluice]

3 )It is known that the number of fish in a given lake will decrease by 7% each year
unless some new fish are added. At the end of each year, 250 new fish are added to
the lake. At the start of 2018, there were 2500 fish in the lake.
(a) Show that there will be approximately 2645 fish in the lake at the start of 2020. [3
marks]
(b) Find the approximate number of fish in the lake at the start of 2042. [5 marks]

[hexed sluice]

I need help with this

I can get an answer, but I can’t put it into a sequence/series like I’m supposed to

[mental trail]

Have you tried finding a recurrence relation ?

[hexed sluice]

What is that?

Here’s my work of what I’ve done so far

<@&286206848099549185>

[faint rain]

What is the subject?

[hexed sluice]

Sequences and series

[faint rain]

and what's the problem?

[hexed sluice]

3 )It is known that the number of fish in a given lake will decrease by 7% each year
unless some new fish are added. At the end of each year, 250 new fish are added to
the lake. At the start of 2018, there were 2500 fish in the lake.
(a) Show that there will be approximately 2645 fish in the lake at the start of 2020. [3
marks]
(b) Find the approximate number of fish in the lake at the start of 2042. [5 marks]

Here’s my work

https://cdn.discordapp.com/attachments/903430334681608232/1152195758414647306/71643516898__2729639D-430D-4915-85D0-E63CCA066DFD.jpg

[faint rain]

yes but what is the problem in b

I just don't get what it is...

[hexed sluice]

You have to figure out how many fish are in the lake in 2042

[faint rain]

and?

[hexed sluice]

You’re supposed to make a sequence or series to solve it

But I just kinda calculated each year individually

And I need to figure out how to do it the right way

[faint rain]

so you mean make a function for it?

[hexed sluice]

If that works, sure

[faint rain]

b:
  f(x) = 2500 / pow(1.07, x-2018) + (x-2018) * 250```

[faint rain]

I think this is what you need

btw pow means power

[hexed sluice]

How to I find the number of fish in 2042 then?

[faint rain]

put x as 2042

and calculate the function

[hexed sluice]

Is that 2500 being divided?

[faint rain]

by the percentage

[faint rain]

although

[faint rain]

  f(x) = 2500 * pow(0.07, x-2018) + (x-2018) * 250
``` this works too

[hexed sluice]

Can you explain the pow thing

I don’t get the ways it’s weird

Written

[faint rain]

do you know what "power by" means in math?

[faint rain]

f(x) = 2500 * 0.07^(x-2018) + (x-2018) * 250

[hexed sluice]

Oh okay I get it now

[faint rain]

is also a way

[faint rain]

in what grade even are you?

[cedar kiln]

@hexed sluice Has your question been resolved?

[cedar kiln]

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[soft mist]

can someone help me make sense of this problem?

idk if its worded weirdly or if im just stupid

[cedar kiln]

@soft mist Has your question been resolved?

[blazing dune]

@soft mist
What does the word "washer" mean here

Oh I understand it now

Are you still here @soft mist?

[soft mist]

yea

[blazing dune]

Ok so the first thing we want to know is what are the values of A and B

On the graph

I think that is a bit easy to know
@soft mist

[soft mist]

yeah, a=1 b=7

[blazing dune]

Yeah
right now we want to devide the distance between them to three intervals
I mean we want to set 2 new points while the distance between each point is equal to the other

I think that is also a bit easy

Do you know what will be the points

@soft mist

[soft mist]

going off the graph shown

im guess 3 and 5?

guessing

[blazing dune]

Yes
It can be done
Doing this
(7-1)/ 3=2
Then add this value to 1
Then add it again until you reach 7

So you will get
1, 3, 5, 7

Now let's choose the type of approximation we will use to calculate the volume of the shape

There are 3 choices of how to do this
The first one is left point appximation which in this case will yield less volume for the shape

The second one is the midpoint approximation which will yield almost the exact volume

The third one is the right point appximation which would yield higher volume for the shape
@soft mist

Depending on the graph he went for the third one

But according to me I think the best way to do it is using the midpoint approximation

But it depends on what you want

Still here @soft mist ?

[soft mist]

yeah

im guessing midpoint is the way to go here

[blazing dune]

Ok

So now let's take the midpoint of each interval

You know the midpoint of each interval right ?@soft mist

[soft mist]

2 4 6

[blazing dune]

Right

Now how we gonna know the 2 radiuses of each washer knowing these points

Do you know @soft mist ?

[soft mist]

you mean 3 radiuses?

[blazing dune]

Umm
The washer has
2 radii
One which starts from the axis of rotation to the end of the washer
The other one starts form the axis of rotation and end in the end of the hole

[soft mist]

oh you right

[blazing dune]

I mean one for the hole in it
And the other for the whole washer

[blazing dune]

So do you know how to do it?

[soft mist]

i think i got the rest of the problem figured out

ty

[blazing dune]

No problem

Don't forget to close the channel using
.close if you wanted to assure the answer later then you can open another help channel

[soft mist]

.close

[cedar kiln]

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[crimson sedge]

Last thing I got to solve and I'm stuck 😭😭

[crimson sedge]

I got to find y

[spice kraken]

,rotate

[tropic oxide]

,rcw

[spice kraken]

ah

[crimson sedge]

In the previous question I proved that f is monotonusly decreasing and

[tropic oxide]

$\mathcal{L}(y^2 - 3y) - \mathcal{L}(2y-6) = y^2 - 3y + 6$

[wraith dagger]

Ann

[tropic oxide]

this is how that reads to me

[crimson sedge]

It's -5y at the end not -3y

F(y²-3y)-f(2y-6)=y²-5y+6

[tropic oxide]

are F and f two different functions?

[crimson sedge]

No the same they both f

[tropic oxide]

and... ok wait, what even is f

can you maybe show the entire question

cause this looks like not enough info

[crimson sedge]

I don't have f but I have g(x) = f(X)-x

That's all

[tropic oxide]

can you please just show the entire problem

[wet fossil]

What is g(x) ?

[crimson sedge]

It's in greek I'm sorry you won't understand 😭

[tropic oxide]

just send it

i'll ask whatever needs to be clarified

[crimson sedge]

Number 9

I just change the greek later λ to y so it's more understandable

It just says f is monotonously decreasing and so is g

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

Has anyone got a clue..?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[regal pewter]

okay im supposed to be good at math but i cant seem to isolate b here, equation solvers online gave me errors when i plugged it in

[regal pewter]

that whole (2t√bg/√m) is an exponent btw

[cedar kiln]

@regal pewter Has your question been resolved?

[regal pewter]

okay im legally allowed to <@&286206848099549185>

[spice kraken]

I'm not sure if it's possible

[crimson delta]

this is cursed

maybe possible in terms of the lambert W function

but very cursed having it in the exponent, as a scalar and as an added term

[finite vessel]

<@&286206848099549185>

[feral coyote]

!15m

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[feral coyote]

• open your own channel

[cedar kiln]

@regal pewter Has your question been resolved?

[cerulean harness]

show ur work @finite vessel

[cerulean harness]

this can be done in various methods, so plz consider showing ur work if possible

[cedar kiln]

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[remote iron]

I have a doubt

[remote iron]

What is fractional part of x definition

<@&286206848099549185>

[feral coyote]

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[outer carbon]

Your question is not clear

[remote iron]

Fractional part of x definition

{x}

This

[feral coyote]

the number after a .

{1.2} = 0.2

[remote iron]

Oh

Like [x] -x

[feral coyote]

about pinging

[remote iron]

Yeah

.close

[cedar kiln]

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[remote iron]

Ty

[junior dome]

actually x - [x]

[remote iron]

Oh

[cedar kiln]

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[lofty oasis]

can somoene help give me the solution?

[neat dune]

well do you recall the equation for friction force in terms of mu

Genuine question cus I don't lel if I had to guess I would say mu times the Normal force tho

Bet just googled it I was right

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@lofty oasis Has your question been resolved?

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[gilded hull]

5a - 5d

i think for 5a, the answer is b

[lunar lynx]

Why do you think that is ?

[gilded hull]

x is blairs age, x +6 represents ryan being 6 years older than blair

24 is the age combined

[lunar lynx]

And x and (x+6) combined will be (x+6) ?

[gilded hull]

oh its d then ?

[lunar lynx]

Yes.

[gilded hull]

ohh

makes sense now

[lunar lynx]

x + (x+6) = 24

Like you explained

I believe you can answer rest of them, now.

[gilded hull]

what about 6x?

@lunar lynx

[lunar lynx]

Well, 6x represents kind of nothing significant. You can call it something like - it represents 6 times the current age of Blair.

[gilded hull]

ok

thank you

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[graceful island]

we had a practice quiz
i ended up getting F(x)=0.4x
putting none of the other listed answers wasn’t it either
am i missing something??

[lunar lynx]

How did you get that?

[dark cedar]

Is anyone open minded?

[cedar kiln]

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

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[graceful island]

F(x)=KX
F(x)=20
X=50
K=F(x)/X
K=0.4
F(x)=0.4X

[lunar lynx]

X is not 50.

Look at the unit.

You are supposed to put values in their SI unit.

[graceful island]

si unit?

[cerulean star]

They gave you centimeters but the formula uses meters.

[graceful island]

got it

[cerulean star]

I am wondering about the sign though

Is that important?

[graceful island]

no clue…

[lunar lynx]

Sign is not important. F(x) just gives magnitude. Sign is used with discretion.

[cedar kiln]

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[old monolith]

What is the difference between a cross product's vector form and its component form

[old monolith]

.close

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[crimson sedge]

This problem is kinda confusing me

[crimson sedge]

Im not sure how to add / subtract the vectors so i can solve for $\theta_2$, the angle the boat would have to move in to reach the clearning

[wraith dagger]

Replaced by new brandon H

[crimson sedge]

so far I've gotten

$$4 \cos(\theta_2) +1.1 = 4 cos(\theta_1)$$

[wraith dagger]

Replaced by new brandon H

[crimson sedge]

But i think the right side is wrong

idk

The resulant vector without course correction wold be $$\vec{v} = ( 4cos(\theta_1) + 1.1, 4sin(\theta_1))$$ i think. But the y component doesn't look right

[wraith dagger]

Replaced by new brandon H

[crimson sedge]

oh

i guess it's right

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

😞

I hate

.close

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[limber snow]

I dont understand this proof

I get it wit induction for k+1 they got 2k + 2

but why did they remove an edge

to get 2k

and then

add an edge back

to get 2k + 2

it doesnt make sense to me

[cedar kiln]

@limber snow Has your question been resolved?

[untold torrent]

You know that for every graph with k edges the sum of degrees is 2k.

You want to prove this is true for k+1 (Which is, that a graph with k+1 edges will have a sum of degrees 2(k+1))

So, in general, if you have a graph with k + 1 edges, you can remove an edge.
You get a graph with k edges, which you know will have a sum of degrees 2k (by your assumption).

Then, add the edge back, which adds a degree of 2, which means that the sum of degrees in your graph is 2k + 2.

Which is 2(k + 1), thus proving what you wanted.

[limber snow]

why does it add a degree of 2?

because each edge has 2 sides?

[untold torrent]

Yes

It connects 2 vertices, and adds 1 to the degree of each

So in total it adds a degree of 2.

[limber snow]

got it now

I understand

[untold torrent]

This is true even if it connects a vertex to itself.

[limber snow]

tysm

[untold torrent]

No problem!

[limber snow]

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[hard ember]

How to integrate sin2x/(sin^4x+cos^4x) dx

[violet flume]

factor

you need to rewrite the denominator

[hard ember]

1-1/2 sin^2 2x

[languid vector]

Just know that the bottom is a trig identity

[dire geode]

Show how you got this

[cedar kiln]

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[shrewd sparrow]

How’s it looking

[shrewd sparrow]

Not good

Thats whay

I ended up with a fraction

What went wrong

<@&286206848099549185> i tried again , more organized. Still went wrong , what’s up here

Wait

[languid vector]

Um look at last step

[shrewd sparrow]

No that was still incorrect there

[languid vector]

-1 + 7 = 6

And -1 + 19 = 18

6y = 18

[shrewd sparrow]

Yeah I saw that

It’s still wrong

I know the answers.

That ISNT one of the answer options I have

The answer would be y=3

Which isn’t an option I have here

So I’m aware

My answer is 1,2,1

And 3 fits into none of those answers lol

[languid vector]

If you get (3,6,3) it could be simplified into 1,2,1

[shrewd sparrow]

I don’t think that’s a possibility here

I might have found it

Let me plug some things in and see

[nimble veldt]

how can this be an solution for the first equation? x+y-z=4?

[shrewd sparrow]

My y is supposed to be a 2

[nimble veldt]

and x? and z?

[languid vector]

Yeah that’s true, lol I didn’t check

0 != 4

Either you copied the equation wrong or you’re looking at the wrong answers

[shrewd sparrow]

X 1 Z 1

My teacher GAVE me the answer to these

[nimble veldt]

again: how can this be an solution for the first equation? x+y-z=4?

[shrewd sparrow]

My teacher gave me the answer sheet

[nimble veldt]

just do it. x = 1, y = 2, z = 1 and then calc x+y-z?

[shrewd sparrow]

1, 2 , -1 , sorry

But

This is problem # 2

1 + 2 - (-1)
1 + 2 + 1

[languid vector]

1,2,-1 would work

[shrewd sparrow]

So there is that

Yeah

But y is supposed to be 2

[languid vector]

Look at this

[shrewd sparrow]

Im awareX when you mentioned it I looked at my paper and then look at my picture and noticed I didn’t have the - sign

My mistake

Still does not change my issue here lol

[nimble veldt]

the solution doesnt match

[languid vector]

Make sure you didn’t copy down the equations wrong

[nimble veldt]

so you have an error before.

where does the 4z come from?

[shrewd sparrow]

I noticed that error and fixed it. I multiplied everything by 2 and must have mistaked my Z for a 2

This whole page is full of mistakes LOL

There is so much going on

As you can tell

I have NO idea what I’m doing

Im going in circles but not the kind that lead me to the answer

I put dashes at the top by the original equations , they aren’t negative signs, just to clarify that

I marked them to remember which ones I used

Not that it helped any

I tried using different colors so it wouldn’t all mesh into confusion and I could see the visible changes I made at each step

I put it all into steps to help , didn’t work 💀

I promsie you guys in really trying for this I’m legit so confused

Ugh you know what

Im

Lmao

Help

Im dying here guys

[nimble veldt]

again here is a mistake.

but: you have the solution. you can check every line for yourself.

[cedar kiln]

@shrewd sparrow Has your question been resolved?

[languid vector]

The signs XD

[shrewd sparrow]

Yeah we aren’t

It’s not

It’s a -18 but

[languid vector]

Why doesn’t your first term half a negative on the top right

[shrewd sparrow]

What do you mean

[languid vector]

Wasn’t it -x + 2y - z

[shrewd sparrow]

No

I kept doing it over and over so I marked the equations I was using so I could keep track of the order I did it

[shrewd sparrow]

^

[still swan]

How did you get x = 13 + 3y

Check that one more time

There shouldnt be a 13

[shrewd sparrow]

Well now I have

X=-5 +3y

I plugged it in and got 4y = 9

Ugh I have so many errors

[languid vector]

Then your teacher’s solution of 1, 2, -1 wouldn’t work

1 + 2 - 1 != 4

[shrewd sparrow]

-(-1)

1+2-(-1)

[languid vector]

Oh wait, it’s a negative lol

[shrewd sparrow]

Yeah

Im so

UGUHH

this is literally the first problem I did at home and I have been trying to hours kown

This says it’s imposible

As far as my algebra standards go

Can anyone please explain what’s going on

Well it works in a calculator

So I’m confused. 😟

[random crater]

Are you allowed to put that in a matrix?

[shrewd sparrow]

A what

It already is the matrix dawg

[languid vector]

Your teacher’s answer is wrong

[shrewd sparrow]

Are you sure

I plugged the answers into the equations. I got the right answers.

It equaled 4,-6 and 7

[shrewd sparrow]

How did we come to this conclusion

And either way unless it IS a fraction answer, my answer is STILL wrong lol

[cedar kiln]

@shrewd sparrow Has your question been resolved?

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[wise dome]

limit of sin theta

[modern compass]

???

[dire geode]

,tex .original

[wraith dagger]

riemann

[tropic oxide]

,calc sin(42069)

[wraith dagger]

Result:

0.067173600243978

[tropic oxide]

$\lim_{\theta \to 42069} \sin(\theta)$ is about $0.06717$ @wise dome

[wraith dagger]

Ann

[tropic oxide]

deleted user?

oh they left lmfao

.close

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[raw wren]

if i multiply by 2 to abs value equation will it become thin? just like in case of parabola?

[tropic oxide]

this doesn't look like "becoming fatter" to me

[cedar kiln]

@raw wren Has your question been resolved?

[raw wren]

soory i mean thinner

[tropic oxide]

well, i showed you exactly what'll happen.

[raw wren]

yeah got it ann

ty

.close

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[kindred crag]

Hey, I have a question regarding determinants. Our lecture notes say, that the determinant of A is linear for all rows and columns of A, where A is a quadratic matrix of the size nxn. We have an equation like this: [det([\vec{a}_1\cdots\lambda\vec{a}_j\cdots\vec{a}_n]) = \lambda det([\vec{a}_1\cdots\vec{a}_j\cdots\vec{a}_n])]. Does it mean I can write [det(\begin{bmatrix}1 & 2 & 3 \ 1 & 2 & 3 \ 1 & 2 & 3\end{bmatrix}) = 1\cdot2\cdot3\cdot det(\begin{bmatrix}1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1\end{bmatrix})]

[wraith dagger]

ballcrunch

[cedar kiln]

@kindred crag Has your question been resolved?

[kindred crag]

<@&286206848099549185>

[cedar kiln]

@kindred crag Has your question been resolved?

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[wintry pier]

Let $(a_n){n\geq 0}$ defined by $a_0=0$ and $a_1=4$ and $a{n+2}=2a_{n+1}+a_n$ for all $n\geq 0$. Find the reminder of the Euclidean division of $a_{202021^{202021}}$ by $202021$ ($202021$ is a prime number).

Found that $a_n=\sqrt{2}\Big((1+\sqrt{2})^n-(1-\sqrt{2})^n\Big)$ but I don’t know what to do next

[wraith dagger]

Joseph.P

[cedar kiln]

@wintry pier Has your question been resolved?

[wintry pier]

<@&286206848099549185>

<@&268886789983436800> he’s using chatgpt I think

[celest seal]

bizarre, it seems they sent that message and immediately left the server lmao

but yeah thats AI generated

dealt with

[wintry pier]

Thanks

Can someone give me a hint ?

[gritty galleon]

You can use binomial theorem to simplify further

@wintry pier

[wintry pier]

Like $$a_n=\sqrt{2}\Big(\sum_{k=0}^n\binom{n}{k}(\sqrt{2}^{n-k}-\sqrt{2}^k\Big)$$

[wraith dagger]

Joseph.P

[wintry pier]

I think my formula is wrong

[cedar kiln]

@wintry pier Has your question been resolved?

[wintry pier]

<@&286206848099549185>

[pulsar pier]

This formula is correct.

But I don't think it will be helpful

What are they asking for?

[wintry pier]

The reminder of the Euclidean division of $a_{202021^{202021}}$ by $202021$

[wraith dagger]

Joseph.P

[wintry pier]

So like $a_{202021^{202021}}\equiv x\pmod{202021}$

[wraith dagger]

Joseph.P

[wintry pier]

And I want to find x

[pulsar pier]

I wanted to check some theorems here

It is either 0

Or $a_{202021}$

[wintry pier]

I tried and it’s not 0

[wraith dagger]

theGearless

[wintry pier]

It would be a very very big number

[pulsar pier]

There is a theorem called Little Fermat.

Yes

But just reduce the power of the LHS first then try to find someway to reduce the problem

[wintry pier]

$a^{p-1}\equiv 1\pmod{p}$ right

[wraith dagger]

Joseph.P

[pulsar pier]

This is Euler-Fermat

Fermat's little theorem: $a^{p}\equiv a\pmod{p}$

[wraith dagger]

theGearless

[wintry pier]

Oh yes

[wintry pier]

How do I do that ?

[pulsar pier]

Here

Now we now that the value of x is $a_{202021}$

[wraith dagger]

theGearless

[pulsar pier]

We just need to find it.

[wintry pier]

Just know that $a_9$ is approximately a million

[wraith dagger]

Joseph.P

[wintry pier]

No I’m saying shit

[glad dagger]

Hi

[wintry pier]

, w calculate sqrt(2)((1+sqrt(2))^{202021}-(1-sqrt(2))^{202021}))

[glad dagger]

Hi

[wintry pier]

Hi sorry

[wraith dagger]

Failed to get a response from Wolfram Alpha.
If the problem persists, please contact support.

Sep 16, 2023, 12:34 PM

[pulsar pier]

I think it is not doable by hand

[wintry pier]

So what can I do ?

[pulsar pier]

I tried to find a pattern but to no vail.

A simple code can solve it really fast but still.

[wintry pier]

Ty

.close

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[woven flare]

not many thoughts so far

[crimson delta]

try c=0 first

[woven flare]

oh ok

sorry i might need a little more help

[cedar kiln]

@woven flare Has your question been resolved?

[woven flare]

<@&286206848099549185>

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[near jolt]

How to get

[near jolt]

Basically, this was instructed in class. They showed us that the combination of translation first, followed by rotatiion is R * T

Then you need to multiply that total transformation by each of the points of the original triangle, in order to achieve the desired triangle

I wanted to confirm if this is true, and also I am confused about the rotation matrix. It was originally taught with a vector form the origin as an example. But if we rotate about the origin here, it wouldn't properly rotate the triangle would it?

Thanks

[cedar kiln]

@near jolt Has your question been resolved?

[cedar kiln]

@near jolt Has your question been resolved?

[near jolt]

In summary, is the total transformation matrix (rotation and translation) applied to each vertex of the triangle or what?

For vectors attached to the origin, rotation matrix makes more sense, how does it work for triangles floating around the cartesian?

[cedar kiln]

@near jolt Has your question been resolved?

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[barren island]

Hi, can anyone help me with this? Thanks!!

[cedar kiln]

Please don't occupy multiple help channels.

[wintry pier]

Close this channel

[dire geode]

!onechannel

[cedar kiln]

Please stick to your channel.

[dire geode]

.close

[cedar kiln]

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[marsh sleet]

hm

[cedar kiln]

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[grand totem]

What are the steps for solving

No need actually

@cedar kiln

Consider it closed

[dim gulch]

do .close

[grand totem]

@cedar kiln .close

🥸

.close

[dim gulch]

no need to ping lol

[cedar kiln]

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[grand totem]

What are the steps for solving this

[cedar kiln]

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[dim gulch]

Let p>3 be a prime number.
Let N be the smallest natural exponent such
that 2 and 3 have the same remainder
after division by p.
Show that N > 2023^2023 for infinitely
many primes p.

[dim gulch]

I wrote it as 2^N ≡ 3^N (mod p)

annd plugged in some values and found that for p = 5, N = 2 works

which violates N > 2023^2023 so have i written the modulo wrong or?

[spice kraken]

?

N>2023^2023 is not for all prime p

you just need to show that there exist infinitely many primes p such that N>2023^2023

[dim gulch]

oh so assuming N > 2023^2023 show that there exist infinitely many primes that satisfy this?

[spice kraken]

yes

[crimson sedge]

No not quite

[dim gulch]

but wouldnt that violate N is the smallest natural number?

[crimson sedge]

That would be implying that for a single given N there are infinitely many primes p such that 2^N == 3^N (mod p)

Which is not what they're saying

[crimson sedge]

They never said it's the smallest natural number, they said it's the smallest natural exponent such that...

a.k.a. the smallest natural exponent of those which satisfy the given properties

[dim gulch]

so it would be the smallest natural exponent greater than 2023^2023 that satisfies 2^N == 3^N (mod p)

[crimson sedge]

Basically, they're saying that for infinitely many primes p, the smallest natural number N of those for which 2^N == 3^N (mod p) is greater than 2023^2023

[crimson sedge]

No, the "greater than 2023^2023" part is not part of the definition of N

N is just "the smallest natural exponent that satisfies 2^N == 3^N (mod p)"

[dim gulch]

yeah so wouldnt this violate what were supposed to prove?

[crimson sedge]

No

All they're asking is if the statement is true for infinitely many primes p

Not all primes p

[dim gulch]

ohhhhhhhh

sorry it just hit me

[crimson sedge]

Infinitely many integers are prime, but not all integers are prime

[dim gulch]

so doesnt matter that it doesnt work for some of them as long as infinitely many other primes work

[crimson sedge]

Mhm

By the way all the parts you found confusing can be rephrased

"for infinitely many primes p" can be rephrased as "There exists an infinite set of primes such that for every prime p in the set we have..."

Which is arguably quite a bit more complicated but it makes it harder to think this means anything else

[dim gulch]

yes thats more clear to me for some reason

its harder to accidentally make assumptions

[crimson sedge]

Oh, and also you can make the 2023^2023 condition part of the definition in this way

Actually no

Still not

[dim gulch]

yeah i was gonna say wouldnt that immediatly prove this

[crimson sedge]

No, I just rephrased the theorem

We still have to prove the rephrased version

[dim gulch]

oh

[crimson sedge]

Though you can rephrase the "The smallest N satisfying ... is greater than 2023^2023" to "There is no N satisfying ... smaller than or equal to 2023^2023"

By the way there's a simple way to prove this theorem

The only hint I'll give you is be lazy

[dim gulch]

lmao

Assume that it isnt,
Well it is
QED

[spice kraken]

ah

I see the solution now

[crimson sedge]

I know, right?

bruh theorem

[dim gulch]

i feel so left out sad moment

[crimson sedge]

there's only one way to remedy this boy

@dim gulch Any leads...?

[dim gulch]

i

seem to be just plain stupid

[crimson sedge]

right

Any leads?

[dim gulch]

iddk im trying to reduce the exponent but

[crimson sedge]

do not

just

think of a brute force way to force this to be true

[spice kraken]

the answer involves no calculation lol

[dim gulch]

is it that you can scale N

[spice kraken]

no

[crimson sedge]

Okay, fine, here's another hint: 🅱️ery 🅱️ig

[dim gulch]

okay im probably overthinking. this but if you multiply by 2's and 3's itl eventually. go around the. mod at different rates and will be equal at some point

[crimson sedge]

Hmmm yeaaaah nooo

[spice kraken]

it could work

[dim gulch]

also sorry my keyboard randomly types .'s sometimes

[crimson sedge]

The goal is actually to stop thinking about modular arithmetic

I kid you not

[spice kraken]

they go around the mod and be equal

[crimson sedge]

The solution is that lazy

[dim gulch]

lmfao actually

[spice kraken]

what if they don't go around?

[dim gulch]

this entire time ive been. staring at. a mod equation

[crimson sedge]

well stop

[dim gulch]

then it doesnt matter bc there exist infititely other ones that you can prove this. somehow

[dim gulch]

okay ;-;

[crimson sedge]

Think of the question this way

The solution becomes much clearer like this

Instead of trying to make them equal but only past a point, you're trying to keep them unequal until a certain point

By "past a point" I mean past a certain value of N

I'm considering N to start at 1 and then increase until the condition is satisfied

[dim gulch]

yeah ill be thinking about that now i guess

[crimson sedge]

Now the question becomes: What's the easiest way to make sure N doesn't fill the condition for as long as possible?

[dim gulch]

make p be somethingn i dont know what yet

p be bigger. than N

[crimson sedge]

And don't forget that mods are annoying ||I'm talking about modulos not moderators pls don't ban me <3|| so you should try to avoid them like the plague

[spice kraken]

lol

[crimson sedge]

Hmmm okay... Not quite

But you got the spirit

[dim gulch]

aaaaa

[crimson sedge]

It seems you got the idea to make p so big that the modulo can be forgotten up to a point

That's a good idea but you didn't make p big enough for that to work

[dim gulch]

yeah so make p bigger than 2023^2023

[crimson sedge]

Almost!

It actually has to be even bigger

[dim gulch]

2023^2023 + 1?

[crimson sedge]

No

It's not an off-by-one thing

Not even close

[spice kraken]

think of why you wanna make p big

[dim gulch]

make p big so that 2^N = 3^N doesnt matter until N > 2023^2023

[solid juniper]

you mean moderators

[dim gulch]

isnt equal i should say

[spice kraken]

very close

[crimson sedge]

I'm pretty sure I said the exact opposite

Layla is trying to frame me!

Ban her

[dim gulch]

so make p > 3^(2023^2023) ?

[crimson sedge]

Your phrasing is very wrong but you seem to have the spirit yet again

[dim gulch]

lmao i

[crimson sedge]

It's not 2^N == 3^N that won't matter until N > 2023^2023

[dim gulch]

its that the remainder wont be equal until p > than both

right

please

[spice kraken]

yes

cause there's no solution for 2^N = 3^N without mod

[dim gulch]

WOOOOOO

yea since you would get ln2 = ln3

[dim gulch]

so wouldnt that mean you would need to make p this?

[spice kraken]

correct

[dim gulch]

*greater than this

[dim gulch]

:DDD

[crimson sedge]

Until p < than either

If p > both then the remainders won't be equal

[dim gulch]

mm yeah that makes sense so it would actually be 2^2023^2023

[crimson sedge]

So close!!

[dim gulch]

i hate it

[crimson sedge]

It's actually bigger

And it's still not an off-by-one mistake

If p is greater than both then the remainders won't be equal

But if p isn't greater than both then who knows what happens

[spice kraken]

wait what

N always exist

by the euler's theorem

[dim gulch]

okay now you guys have an argument while i continue to fail at this :p

[crimson sedge]

Couldn't it be that 2^2023^2023 < p < 3^2023^2023 and 2^N == 3^N (mod p)?

Edit: for N <= 2023^2023*

[spice kraken]

N<2023^2023 is still possible

I think

[crimson sedge]

Right, and we want to guarantee that that's not possible

[spice kraken]

also then p won't be infinite

[crimson sedge]

We want a family of primes p where we know for a fact the statement holds

Which is why I would just say p > 3^2023^2023

@dim gulch I basically just gave away the answer this time

[spice kraken]

yeah the answer is just that

[crimson sedge]

sowwy uwu

[spice kraken]

and N = p-1 is always a solution

[dim gulch]

anyways im back

[crimson sedge]

For p > 3^2023^2023, we know that if N <= 2023^2023, then 2^N < 3^N < p, hence 2^N == 3^N (mod p) would imply that 2^N=3^N which is a contradiction

So the smallest N satisfying the mod equation must be > 2023^2023

Or in short, for any p > 3^2023^2023 (of which there are infinitely many), the smallest N satisfying the condition must be greater than 2023^2023

@spice kraken Do you think I gave away too much?

[dim gulch]

so again why wouldnt p > 2^2023^2023 work

[crimson sedge]

Or is it okay to put everything together myself after having put Laito through this mental torture exercise?

[crimson sedge]

[spice kraken]

actually I gotta go

[dim gulch]

i mean i did say the annswer at 1 poinnt just not with full logic

[dim gulch]

see you, thank you!

[crimson sedge]

This logic requires that p is greater than both sides of the equations

This is necessary to turn the modular equation into a regular equation

And the biggest of the 2 sides of the equation is 3^N

[dim gulch]

yeah i get it now

this has been a hell of journey

[crimson sedge]

One hell of a journey to find the laziest answer possible

Just be lazy and make p very big

[dim gulch]

beeg p

thats what ima send my teacher

solution: yes, make p big QED

well thank you so much

[crimson sedge]

Don't forget to .close

[dim gulch]

i wont, see you sometime again!

.close

[cedar kiln]

Channel closed

Closed by @dim gulch

Use .reopen if this was a mistake.

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[modest nebula]

hello I am reading about hilberts axioms in geometry and I am having trouble understanding an example in this book

[modest nebula]

It is for the axioms of betweenness

it is to show that R^2 satisfies the betweenness axioms

[jolly remnant]

What's your question about it?

[modest nebula]

I dont understand the part "if l meets the segment AB then 0 will lie between phi(A) and phi(B)" why is this true?

[jolly remnant]

If you suppose they have the same sign

You should get that phi (intersection) also has the same sign (not zero)

[jolly remnant]

You can also parametrize the line AB, plug the parametrization into phi and see it's increasing or decreasing

[modest nebula]

what is phi intersection?

[jolly remnant]

Phi takes in points on the plane

Intersection is a point on l also on the segment AB

[modest nebula]

oh i think i understand it. Because phi intersects AB, the intersection point D satisfies phi(D) = 0 so that 0 lies between phi(A) and phi(B)

[jolly remnant]

Phi is a function

It doesn't intersect anything

Phi just gives a value to every point on the plane in such a way that the entire line l gets mapped to zero

So any point on line l has phi(point)=0

Because phi(intersection)=0 then phi A and phi B must have opposite signs (parametrize and intermediate value theorem by way of contradiction)

And then since C isnt on the line phi C is not zero. So it's either positive or negative

Intermediate value theorem again with BC and AC

[modest nebula]

ah right intermediate value theorem

if both phi(A) and phi(B) had the same sign, then 0 wouldnt be in the interval [phi(A), phi(B)] so it could not have an intersection point with the line l, contradiction

[jolly remnant]

Because after parametrizing you're left with a linear function

You have to use linearity too

IVT is for the later steps in logic

This step you use linearity

For example x^2-1 is positive at x=+-2

But it's still negative in between

[modest nebula]

oh yeah, it would need to apply to a function R -> R so parameterize by f(x) = phi(x, mx+b)

[jolly remnant]

Probably better to do x(t) = at+c, y(t)= bt+d

You have vertical lines and whatnot

[modest nebula]

hmm but then y is not a function of x?

[jolly remnant]

No

Lines on the xy plane dont have to be functions of x

It's entirely valid to have vertical line segments

If it's not vertical you can write y as a function of x anyways with the parametrization I gave you

[modest nebula]

wait i think im applying ivt wrong. If phi(A) and phi(B) are both positive, then how do we arrive at the contradiction?

ivt says that it will take on values between [phi(A), phi(B)] but it doesnt tell us that it doesnt have a root

f(t) = phi(at+c, bt+d), with f(t_0) = phi(A)>0, f(t_1) = phi(B)>0

f(t') = phi(intersection) = 0 with t_0 < t' < t_1

@jolly remnant

[cedar kiln]

@modest nebula Has your question been resolved?

[jolly remnant]

as mentioned before, you don't apply IVT in this step

use linearity

[jolly remnant]

you have a definition of phi something like (kx+vy+c)