help-26

[jade jungle]

do you know what an arithmetic sequence is?

[neat patrol]

a little i just cant figure what the equation was

[jade jungle]

then an easy way to remember the equation for the nth term of an AS is to just consider the first three terms.

2 is the first term, no questions asked. but how do you get 9?

[neat patrol]

add 7

[jade jungle]

excellent. and notice that all of the terms differ from their neighbours by 7.

so from 2 to 9 you add 7. from 9 to 16 you add another 7.
how many "copies" of 7 have you added, then, from 2 to 16?

[neat patrol]

2 right?

[jade jungle]

yes.

now, 2 was the first term, and we added 0 copies of 7, that is 2 = 2 + 0(7).
then, the second term 9 has us add one copy: 9 = 2 + 1(7).
16 = 2 + 2(7) for the third term.
do you see a pattern here?

[neat patrol]

ohhhh yeah yeah

[jade jungle]

specifically:
T_1 = 2 + 0(7)
T_2 = 2 + 1(7)
T_3 = 2 + 2(7)

can you now generalize this pattern?

[neat patrol]

yeahhh so the 55th would be t_55 = 2 + 55(7) right

[jade jungle]

close, but not quite!

[neat patrol]

mmhhh

[jade jungle]

look at the relationship between the bolded numbers in the same row.

[neat patrol]

ohhhh the 55(7) would actually be 54(7)?

[jade jungle]

correct.

[neat patrol]

so would that apply to most AS?

[jade jungle]

to all AS.

[neat patrol]

OHHHHH

[jade jungle]

do you think you can write down the general form of this equation?

for T_n.

[neat patrol]

maybe uhhh T_n = 2 + n(a) oh jeez idk

[jade jungle]

use d to indicate the common difference, and a to indicate the first term.

2 was the first term, and 7 was the common difference.

also, not n directly on the right. look back at the relationship we've established.

[neat patrol]

t_n = a + m(d)

rightttt????

[jade jungle]

m is?

[neat patrol]

idk 😭 uhhhhhh the

uhhh letter before n?

[jade jungle]

why are we using the letter before n?

did you intend to subtract 1 from n?

[neat patrol]

ohhhh wait so

t_n = a + (-1n(d))

oh wait i messed up

t_n = a + (n-1(d))

wait no

[jade jungle]

you're almost there!

but I'll fix the bracketing for you.

T_n = a + (n-1)d

or in proper notation, $T_n = a + (n-1)d$, or $T_n = T_1 + (n-1)d$.

[thorny flame]

Lute

[neat patrol]

ohhhh omg i was almost there

[jade jungle]

yeah, just a little bit of bracketing issues.

but yes, this formula applies to any AS. if a sequence does not respect this formula, it is not an AS.

[neat patrol]

omg thank you so much, i wasnt able to remember or actually learn AS

have a great day/night!!

[jade jungle]

you too.

[neat patrol]

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@storm orbit Has your question been resolved?

[storm orbit]

Am I on the right path?

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[dusky anvil]

hello, can someone explain how egyptians used to solve problems with unknown variables? I looked it up on google but didnt rlly get it

[placid swift]

No idea dude, maybe check out a History of Science forum for Egypt that talks about it

[dusky anvil]

oh oke, i'll let this post linger in case sm1 knows the old ways

[placid swift]

Alright sure

[jade jungle]

https://www.exeter.ac.uk/research/groups/education/pmej/pome12/article13.htm
I'm not sure if this will help your research.

[placid swift]

That should help him ig

[dusky anvil]

thats what i found on google but its a little short. thank you though

[jade jungle]

https://en.wikipedia.org/wiki/Rhind_Mathematical_Papyrus
then will this help?

[dusky anvil]

yeah this is the one im reading, its really analytic so its good

[jade jungle]

I see, sorry for resending this.

[dusky anvil]

its alright, i appreciate your help

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[wooden bison]

Hi guyseth
Is there any way to easily calculate the length of the triangle angle's sector line with given angle, length of both of its sides and the proportion between the sector and the angle itself? Gives the vibe that the compact formula already exists somewhere

[wooden bison]

The sine law does not work quite well because it depends on what angle is more than 90 degrees if theres any (it is possible in the task that requires the formula), and the cosine law leads to the knowledge of angle beta's value so far. Anyway none of those should really be used because its meant to be calculated automatically by the pc processor, and there will be thousands of iterations of this calculation and the shortest formula will be the best one

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[dusky anvil]

can a homeomorphism be defined only through the use of topology? If not, then what others are there? idrk what im talking abt so show some mercy with the use of other definitions

[quasi depot]

isn't homeomorphism already defined using topology?

[vapid lichen]

a homeomorphism must be continuous. how do you define that without topology?

[worthy flax]

i suppose if you're working in something like metric spaces then there's a continuity definition without topology

but yeah in general continuity needs topology to define

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[neon iron]

After differentiating and i found critical points, i got x=pi/4, but x=5pi/4 is also given

How is that possible?

[rough furnace]

!show

[topaz sinew]

Show your work, and if possible, explain where you are stuck.

[rough furnace]

And also the sol in your book is the better way ngl

[neon iron]

But i prefer to use derivatives tbh

[rough furnace]

Could you show us the work so we can check

[neon iron]

Ye sending

[neon iron]

The picture doesnt include sol tho

How are u saying

[rough furnace]

You should get an interval, not a value for x do you

[rough furnace]

I know they add sin and cos to become sqrt(2)sin(x-pi/4)

Or cos

I also would do that

[neon iron]

Which is faster and better

[neon iron]

How to get that

[rough furnace]

Are you familiar with a problem finding interval like that

[neon iron]

My teacher taught today

Kinda ye

[rough furnace]

Your process is to find stationary points, then check value between those value

Those "points"

[eternal sierra]

u should convert it to a singular sin or cos term before differentiating

[rough furnace]

He said I don't like that

[neon iron]

Well tbh i dont know how tho

[eternal sierra]

multiply and divide by sqrt(a^2+b^2)

[neon iron]

I was taught to find derivative and find critical points

[eternal sierra]

oh

[neon iron]

Well im open to solve using any methods

[rough furnace]

That's the sol below here

[neon iron]

So i have no problem with that

[rough furnace]

Uh...

[neon iron]

But i cant understand that

[rough furnace]

Okay...

[neon iron]

With f(x)?

[rough furnace]

I could have you understand it

[neon iron]

Please do so 🙏

[rough furnace]

Sure

do you know the sum identity

[rough furnace]

Cheat sheet ahhh xD

[neon iron]

Sin square x + cos square x?

[neon iron]

Oh

[eternal sierra]

hm?

[rough furnace]

Nothing, just that I don't have to make it from scratch in latex

[eternal sierra]

oh ok 👍

[neon iron]

Im sorry, but could you please teach me, i have to solve a lot 😭

[rough furnace]

Okay...

$\sin(x)\cos(y)+cos(x)\sin(y)=\sin(x+y)$

[thorny flame]

Alexis_Fx

[rough furnace]

Do you know this identity

[neon iron]

Yes i know this

Idk how its called, but i know this

[rough furnace]

Okay so

In general $a\sin(x)+b\cos(x)$

[thorny flame]

Alexis_Fx

[neon iron]

Ye

[rough furnace]

This is what we try to reduce to a single sin function

[neon iron]

Ok

[rough furnace]

The process looks like this

[neon iron]

Ok

[rough furnace]

You should take a look first

And tell me what confuse you

[neon iron]

Nothing

Here a and b = 1

[rough furnace]

Yeah

Apply that thing

[neon iron]

Ok did

What next?

[rough furnace]

What you got

[neon iron]

Root 2 x sin (x-pi/4)

[rough furnace]

Yeahhh

[neon iron]

We could go either way with x-pi/4 right?

Like pi/4 - x?

[rough furnace]

Nope:p

[neon iron]

Oh

[rough furnace]

cos(x)=cos(-x) but sin(x)=-sin(-x)

[neon iron]

Oh

Ye

Then ill get pi/4-x

Umm, are u here?

[rough furnace]

Yeah

I mean you can just take the differential

[neon iron]

Ok

-root 2 cos(pi/4-x)

Now expand?

[rough furnace]

=0

You should have taught how to do that....

[neon iron]

Cos(pi/4-x)=0

Cos pi/2=0

Multiplying both sides by cos inverse

Im getting x=-pi/4 😭

Is that how we do it?

[rough furnace]

No.... You find interval, not the value for f'(x)=0

[neon iron]

I cant understand, how am i supposed to find interval

[rough furnace]

Oh man I hate explaining that...

Okay

So

Let say f'(x)=0 at a,b,c and a<b<c

[neon iron]

Ok

[rough furnace]

f'(m)<0 for a<m<b

Then f(x) is decreasing in (a,b)

[neon iron]

Oh

[rough furnace]

f'(n)>0 for b<n<c

Then f(x) is increasing in (b,c)

[neon iron]

Ok how to find the other points

If we take a=-pi/4, how to find b and c

[eternal sierra]

just find where f'(x) is greater than 0 and where it is less than 0

in the interval 0 to 2pi

u can do this from the graph

[neon iron]

Dude yes, i get only 1 point

[neon iron]

But 5pi/4 is also included

[eternal sierra]

this is the graph of cos(pi/4 - x)

see in what interval it is negative

and what interval it is positive

[neon iron]

I wont be given marks drawing graphs, so how do i prove it

[neon iron]

Uh

(3pi/4,7pi/4)- negative

And the rest positive

[eternal sierra]

yes

so where it is negative, that is ur interval where f(x) is decreasing

and where it is positive, f(x) is increasing

[neon iron]

Ok how do i find it without graph

[eternal sierra]

dont know

and i dont know why u would want to

first master one method then u can go looking for others..

[neon iron]

Yare yare

Thanks for ur help, alexis and yoda

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[strange basin]

how would I solve these equations to get the unknowns

would I use elimination or sub?

[still cedar]

try moving the Fbcsin45 and the Fbccos45 to one side of both equations

i'm assuming ur trying to find Fac?

[strange basin]

Yes

[still cedar]

ok so once you've done this

can you figure out how to eliminate ur Fbc

[still cedar]

ok i mean that works too

then sub your Fbc into the 2nd equation

[strange basin]

what was your way

elimination

[still cedar]

once you move Fbcsin 45 and Fbc cos 45 to the other side

[strange basin]

did I need to know sin45=cos45

[still cedar]

its useful to know

[strange basin]

Because

[still cedar]

bc once you done this

your RHS will be equal

so then you can equate 4/5Fac = 340 -3/5 Fac

[strange basin]

from here

do I eliminate sqrt2/2

[still cedar]

yep

[strange basin]

and cos45

[still cedar]

you can solve for Fac normally from here

[strange basin]

how

[still cedar]

so your RHS=0

[strange basin]

ok

[still cedar]

and your cos45 cancels out with the sqrt(2)/2

so your equation become 340 - Fac - 3/5 x 4/5 Fac = 0

[still cedar]

yeap

[strange basin]

and then

factorise?

im trying to get FAC

By itself now

[still cedar]

so you can move all the Fac to one side

try adding up the like terms in Fac

[strange basin]

Like

-3/5*4/5

I don't know

[still cedar]

k so try moving all the Fac to one side

what is Fac + 3/5x4/5 Fac?

[strange basin]

but its a negative

[still cedar]

ye but you moved it to the other side of the equation

so now its a positive

[still cedar]

yeap

can you solve for Fac from there?

[still cedar]

ye go on

[still cedar]

yep you pretty much got it

[strange basin]

so its 100?

[still cedar]

yeap

[strange basin]

no its not

[still cedar]

so now you can find Fbc also and sub it back

[strange basin]

it says FACIS 243

[still cedar]

let me read your workings

[strange basin]

ok

[exotic dome]

I would use one linear system and solve it all

[strange basin]

which one

[still cedar]

o wait mb its 12/25

not 12/5

[strange basin]

wait

this bit?

[still cedar]

oh wait

[still cedar]

its 340 - 3/5Fac isnt it

not 340 - Fac

[strange basin]

yessss

basically we are here again

[still cedar]

your original equation was 340 - 3/5 Fac - cos45 Fbc = 0 right

[still cedar]

yeap

you can get Fac = 243 from here

[still cedar]

alr go on

[strange basin]

I don't know what to do after this

[still cedar]

so if i asked you to solve the equation 340 = 3x + 4x, would you be able to do it?

[strange basin]

I think we are multiplying tho

[still cedar]

no so you do the exact same thing here

340 = (3/5 + 4/5)Fac

[still cedar]

basically just factorise out Fac

[still cedar]

yeah they cancel out so we dont have to care about the sqrt(2)/2 anymore

basically u can treat the 4/5Fac the exact same way as the 3/5Fac term

[still cedar]

yeap

u got it

[strange basin]

thanks

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[simple canopy]

can someone explain why i^8 always = 1

[mortal steeple]

i² = -1

Take power four on both sides

[simple canopy]

elaborate please

[fallow igloo]

What is sqrt(-1)?

[simple canopy]

1 ?

[spice seal]

no

sqrt(-1) is not an element of the real numbers

[simple canopy]

oh

[spice seal]

What inspired this question?

have you heard of the complex set (or imaginary numbers) before

[simple canopy]

i was calculating the the value of i^8 and the guy said that i^8 will always be equal to 1

[simple canopy]

yes

[rigid ivy]

Do you recall how i is defined?

[simple canopy]

oh yes

its sqrt -1

my bad

[rigid ivy]

If you want to be really technical, it's actually defined to satisfy $i^2=-1$. But it's fine with how you are defining it

[thorny flame]

SWR

[rigid ivy]

Okay, so, if $i^2=-1$, then what is $i^4$?

[thorny flame]

SWR

[simple canopy]

+1 ?

[rigid ivy]

Yup

So, then what is $i^8$?

[thorny flame]

SWR

[simple canopy]

+1 ?

[rigid ivy]

yup

That's really all there is to this problem

[simple canopy]

oh

this was easy

but i suck at solving questions

[rigid ivy]

sometimes, we overthink things

The skill you need to develop over time is asking yourself the right questions that lead you to the answer you are seeking

[simple canopy]

thanks guys

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[dusky anvil]

Hello! I am studying stuff abt the pythagorean triples. Specifically, I'm in the process of determining a,b,c such that they form the components of a primitive pyth. triple. I have proven that there are only two ways for that to happen; 1) a,b are odd, and c is even
2) a,c are odd and b is even.
I think they generally mean the same thing but I proved it for the case in which a²+b²=c² without having any comparison between a and b

[dusky anvil]

oh yeah the question is whether the result of the proof is valid

[sharp pagoda]

i dont think the first way can happen

[ruby tree]

Are you only specifying that c is the hypothenuse? Or that a < b < c?

[dusky anvil]

just that c is the hypotenuse

[ruby tree]

Ok then your result is correct

[sharp pagoda]

yea i dont think the first way can happen

[dusky anvil]

how can it be disproven tho?

[sharp pagoda]

simply

[ruby tree]

https://en.wikipedia.org/wiki/Fermat's_theorem_on_sums_of_two_squares

[sharp pagoda]

the difference of 2 numbers that are odd, square and even, square is either an odd number or an odd number times an odd number

so its an odd number

and for an odd number to be square, it has to be 4k+1

[dusky anvil]

idk what 1 mod 4 means:/

[ruby tree]

4k+1 for some k

[sharp pagoda]

yea

[dusky anvil]

i had a stroke reading this sorry

[sharp pagoda]

just what i was saying

the difference of 2 squares that are 1 odd 1 even, its gonna be an odd number

and for an odd number to be square, it has to be 4k+1

[ruby tree]

Not sure how that matters since c^2 = a^2 + b^2

[sharp pagoda]

hmmm what was i thinking

lemme think

[dusky anvil]

wait so youre talking abt the case where a and c are odd, right?

[sharp pagoda]

no

thats 100% right

the one where a and b are odd

[dusky anvil]

why does it have to be 4k+1 to be square?

[sharp pagoda]

ummm

because i checked all 5 possible numbers

1, 9, 25, 49, 81

[dusky anvil]

how exactly did you disprove it tho?

[sharp pagoda]

yea thats what im thinking

i somehow proved then forgot xD

[dusky anvil]

first of all, we dont even need the difference of two squares here. only the addition of them

[dusky anvil]

happens to me too dw

[sharp pagoda]

c^2=a^2+b^2

c^2-a^2=b^2

[dusky anvil]

why would we do this tho? we can work with the first line

but if we work with the second then:

b² is even

so

c²-a² is even

[sharp pagoda]

im trying to prove mine

[dusky anvil]

and if a² is odd, then c² is even

no what

sorry

hold on hold on

[sharp pagoda]

xDDD

[dusky anvil]

,tex
$ b^2 = c^2 - a^2 $ so $ b^2 = 2n \implies c^2 - a^2 = 2n \implies c^2 - 2k = 2n \implies c^2 = 2m $

[thorny flame]

fijokazż

[dusky anvil]

so c is even and so there can exist no other divisor spare 1 that makes a primitive pyth. triple

[sharp pagoda]

yes

no

[dusky anvil]

okay why?

[sharp pagoda]

because 4k+1

[dusky anvil]

WAIT

youre right

[sharp pagoda]

right?

[dusky anvil]

the sum of two even numbers squared is ecen

even*

[sharp pagoda]

yea

but we're not there

[dusky anvil]

so only the case of a,c being odd, where c is hypotenuse is correct

[sharp pagoda]

yess

finally

[dusky anvil]

okayokay thank you for pointing that out

[sharp pagoda]

i could figure it out how to tell you but you know it

[dusky anvil]

thanks for the help broski

.solved

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[dusky anvil]

.reopen

[topaz sinew]

✅

[dusky anvil]

so

I made a stupid mistake

we had that a and b are odd

lol

ill test it here and pls someone point out whether im correct

,tex
$ a^2 + b^2 = 2n \implies c^2 = 2n $ so c is an even number, if a and b are odd

[thorny flame]

fijokazż

[dusky anvil]

is this not correct?

if it is correct, then how do we prove that we can construct a primitive pyth. triple using such numbers?

[sharp pagoda]

huh

[dusky anvil]

uhm

[sharp pagoda]

yea although this is correct

BUT

there is this thing

[dusky anvil]

if a,b are odd, then if c is even, we know that a²+b² = c² has solutions and that theyre infinitely many

[sharp pagoda]

the difference of 2 squares is always 4k + 3, if the bigger one is even

a square can only be odd if its 4k+1

this is a contrabution

you understand now?

[dusky anvil]

how do you prove this?

[sharp pagoda]

easy

$n^2-(n-(2k+1))^2=n^2-(n^2-2 \times n \times (2k+1)+(2k+1)^2)$

[dusky anvil]

what is n-2k-1?

is that just an odd number?

[sharp pagoda]

yea

[dusky anvil]

oke

[sharp pagoda]

bc n is even

[dusky anvil]

lemme check it

[thorny flame]

Roy

[sharp pagoda]

$=2 \times n \times (2k+1)-(2k+1)^2)$

[thorny flame]

Roy

[sharp pagoda]

$=4nk+2n-(4k^2+4k+1))$

[thorny flame]

Roy

[dusky anvil]

so far with you

[sharp pagoda]

$=4nk+2n-4k^2-4k-1$

[thorny flame]

Roy

[sharp pagoda]

because n is even

2n is divisible by 4

[ruby tree]

That's overly complicated but sure

[sharp pagoda]

so we get 4 * something - 1

this is just 4k-1=4(k-1)+3

[dusky anvil]

why 4? we can just get 2 outside

[sharp pagoda]

how

[dusky anvil]

2×something - 1

[sharp pagoda]

why

[dusky anvil]

hold on

[dusky anvil]

,tex
$ = 2( 2nk + n -2k^2 - 2k) -1 $

[thorny flame]

fijokazż

[dusky anvil]

which is just 2m-1

[sharp pagoda]

yes but i said

n is even

because we stated that before

[dusky anvil]

Idk man your way seems sketchy to me

[sharp pagoda]

no its not

[ruby tree]

(2n)^2 - (2m+1)^2 = 4(n^2 - m^2 - m) - 1

[sharp pagoda]

ok thats a bit simpler but its not really clear

[ruby tree]

But anyway, I think you're chasing your own tail; why would a = 4k+3 disprove the thing

[sharp pagoda]

because

i just said that earlier bruh

[sharp pagoda]

.

[dusky anvil]

but 4k+3 is 4n+1

[sharp pagoda]

how

[dusky anvil]

i think

[sharp pagoda]

ok i give up

[dusky anvil]

lol

[sharp pagoda]

ill let your teacher tell you

[ruby tree]

Ah I see, fair enough

[dusky anvil]

wait so he is right?

roy thanks for helping either way broski

[sharp pagoda]

im always right

[sharp pagoda]

you try to explain to him

[ruby tree]

Explain what? That 4k+3 cannot be 4n+1?

[dusky anvil]

ill state a question to help you answer

why does a square have to be 4k+1 to be odd?

[ruby tree]

If n^2 is odd, then n is odd, correct?

[dusky anvil]

yup

[sharp pagoda]

yes

[ruby tree]

So n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4(k^2+k) + 1

[sharp pagoda]

ok you tell him i go check other channels

bye

[dusky anvil]

okay I see

lemme state another question in a bit

[ruby tree]

Ok, I'll be back in a bit

[dusky anvil]

,tex
If a and b are odd, we have that: $ a^2 +b^2 = c^2 \implies c^2 = 2n $

[thorny flame]

fijokazż

[dusky anvil]

,tex
$ c^2 - a^2 = b^2 \implies b^2 = 4m+ 1 $

[thorny flame]

fijokazż

[dusky anvil]

I don't see anything wrong with this

[dusky anvil]

check these two thingies and validate them when you can

because if they are true, then how is there no primitive pythagorean triple with a and b being odd? (where a²+b²=c²)

[ruby tree]

c^2 - a^2 = (2n)^2 - (4k+1) = 4n^2 - 4k - 1 = 4(n^2-k) - 1

[dusky anvil]

i see

im really sorry for going in circles, but why does 4x-1 not form an odd square?

[ruby tree]

Sorry, you said c^2 = 2n, not (2n)^2, but it still holds

An even square is always divisible by 4

[dusky anvil]

true, i think

yes true

[ruby tree]

We just saw that any odd number, when squared, produces a number of the form 4k+1

And only odd numbers produce odd squares

[dusky anvil]

so we have to prove that for any case; 4x-1 =/ 4k+1

right?

[ruby tree]

I mean... sure

[dusky anvil]

=/ is not equal btw

so we have 2(x-k)=1

x-k=1/2

how do we disprove this?

[ruby tree]

x and k are integers

[dusky anvil]

very true

I see then

so only the second thingy holds

okay damn i feel stupid

thank you for the help Nel and Roy

.solved

[topaz sinew]

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[dusky anvil]

Hello. In euclid's formula for pythagorean triples, if we wanna find what conditions k and n must meet to produce any primitive triple, is it enough to say that either k is even and n is odd, or the opposite? (meaning any of the two ways is enough) k and n are connected by the formula (k²+n², 2kn, k²-n²) btw

[dusky anvil]

I have proven that (a and c) or (b and c) are odd (so b is even in the first and a is even in the second) if c is the hypotenuse btw

[topaz sinew]

@dusky anvil Has your question been resolved?

[dusky anvil]

<@&286206848099549185> please help me im desperate

[radiant thicket]

I will

How

[dusky anvil]

im not that desperate actually i just wanted attention sorry

[dusky anvil]

read the above and tell me if im crazy

[radiant thicket]

U crazy

[dusky anvil]

okay why?

i feel so helpless because our professor just whipped out a very scary statement and asked us to prove its validity, but no matter where I check, I can't find a good enough answer

if someone can explain if im right, i will appreciate it a lot

[dusky anvil]

the question is here

<@&286206848099549185>

it seems to me that we need to prove something extra which I haven't realized yet what it is

[potent crystal]

@dusky anvil what do you need help with?

[dusky anvil]

I wanna show the minimum conditions that k and n must meet in order for them to be used in euclid's formula and generate any primitive pythagorean triple

imma go for a shower and be back

[potent crystal]

You want to use Euclid’s formula to create every pythagorean triple?

[potent crystal]

i will explain to you when you come back! make sure to tag me.

[dusky anvil]

only the primitive triples

[dusky anvil]

okeoke

[dusky anvil]

@potent crystal I'm ready

[potent crystal]

What was your question again?

[dusky anvil]

so we know that euclid's formula for any pyth. triple is (k²+n², k²-n², 2kn)

and I wanna know the restrictions to be taken on k and n in order to generate only the primitive triples

(I tried proving some of them but ill confuse you if I tell you what I did)

[potent crystal]

K is bigger than N which means both are positive
But K and N are no common factors except 1 (they are comprine)

[dusky anvil]

why should the second part be true? thats what im mostly tryna derive

[potent crystal]

if you follow these rules, every triple you create with euclids formula will be a primitive pythagorean formula meaning the three numbers have no common factors except no other than 1

[dusky anvil]

Yes, but how do we know that they generate any primitive triple?

also, i wonder if theres a way to derive that (k,n)= 1 naturally, without making the assumption and proving it right

[potent crystal]

Euclid’s formula works for all primitive Pythagorean triples because, for any such triple, you can always find two numbers that fit into the formula and give you those three sides. The way squares and sums work means every primitive triple comes from this pattern.

As for why the two numbers (K and N) must not share any common factor: If they did, then all three sides you get from the formula would also share that factor, so the triple wouldn’t be primitive anymore. This shows naturally, just by looking at how the formula builds the sides.

[dusky anvil]

So if k²+n² and k²-n² have a common factor other than 1, then k and n also have that same factor?

(if im close then its good enough)

[potent crystal]

Yes

[potent crystal]

Wait, let me think

[potent crystal]

The minimum conditions for k and n to generate primitive Pythagorean triples using Euclid's formula are: k > n > 0, gcd(k, n) = 1, and k and n have opposite parity. The statement "if k2+n2 and k2-n2 have a common factor other than 1, then k and n also have that same factor?" is false. An example disproving the statement is k=3, n=1, where gcd(k2+n2, k2-n2) = gcd(10, 8) = 2, but gcd(k, n) = gcd(3, 1) = 1.

[dusky anvil]

I see. so gcd(k, n)= 1 comes from gcd(k²+n², k²-n², 2kn) = 1, right? The other two conditions I managed to prove myself

[thorny flame]

xfilin

For Euclid's formula $$(k^{2} + n^{2}, k^{2} - n^{2}, 2kn)$$ to generate only primitive Pythagorean triples, the restrictions on $$k$$ and $$n$$ are: $$gcd(k, n) = 1$$$$k$$ and $$n$$ cannot both be odd.$$k > n > 0$$If $$k^{2} + n^{2}$$ and $$k^{2} - n^{2}$$ have a common factor other than 1, it does not necessarily mean that $$k$$ and $$n$$ share that same factor. However, ensuring that $$gcd(k, n) = 1$$ is indeed derived from the requirement that $$gcd(k^{2} + n^{2}, k^{2} - n^{2}, 2kn) = 1$$, which guarantees the triple is primitive.

[potent crystal]

To make sure you only get unique Pythagorean triples (where the three numbers don't share any common factors other than one), you need to pick your base numbers, let's call them 'k' and 'n', very carefully when using the formula.First, 'k' and 'n' must be partners in the sense that they don't have any common divisors themselves, except for the number one. Think of them as being completely independent of each other in terms of shared factors.Second, they need to be a contrasting pair in terms of evenness and oddness. You can't have both 'k' and 'n' be odd numbers. One of them has to be even, and the other has to be odd. This prevents the resulting numbers from the formula from automatically having a common factor of two.Third, 'k' needs to be bigger than 'n', and both of them must be positive whole numbers. This just keeps things tidy and ensures you're dealing with actual lengths for a triangle.If you follow these simple rules for picking 'k' and 'n', the triples you create will always be the special, unique ones!

[dusky anvil]

hmmm okay. when you say that k and n can't be both odd, you prolly mean that they can't both be even, too, right?

[potent crystal]

The first requirement for generating unique triples is that 'k' and 'n' must not share any common factors other than the number one. If both 'k' and 'n' were even numbers, they would automatically share the number two as a common factor. This would break the "no common factors other than one" rule right away So, the rule about them not both being odd is important on its own, and the separate rule that they must not share any common factors (other than one) already takes care of preventing them from both being even. They work together to make sure no shared factors get into the final triple.

[dusky anvil]

okay I think you solved all my main questions so thank you very much

.solved

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[potent crystal]

[topaz sinew]

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[dapper sparrow]

What is an example of this occering? im a little confused on when the same polynomial has two different sets of coeficents

[loud oasis]

it's only a hypothetical

the conclusion is that polynomials always have the same coefficients

so it's like "if we have two polynomials which are equal as functions but may or may not have the same coefficients, then we can prove they do have the same coefficients"

for what it's worth though, this is true in fields like R and C (what your book works with) but not true in finite fields (for example the integers mod 2)

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[lament kettle]

I got a question about torque vectors

[timber jetty]

Alright

[lament kettle]

Like these are assumed to be the same

Sometimes they extend the rench or whatever and find the angle

Instead of using the one between the parts

So which angle is the angle of separation

The dotted line is the extension

[topaz sinew]

@lament kettle Has your question been resolved?

[topaz sinew]

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[lament kettle]

.reopen

[topaz sinew]

✅

[lament kettle]

No one?

[waxen summit]

The angle of separation is generally the acute one

So if you have a flat line and draw a slope from it, it'll always make two angles that add to 180, (shown by the dotted extension line)

By convention, the angle of separation is taken as the smaller of these two angles, since it directly shows how far the slope tilts away from the baseline.

Does this help @lament kettle ?

[lament kettle]

Well the point is the line ends

So you just sssume it’s continued?

[waxen summit]

right it does, but generally speaking it doesnt "have" to end

In geometry, lines can be considered infinite, so the endpoint doesn’t change the fact that there are still two angles formed at the intersection with the flat line.

the reason why they extend the line will be because the original angle, like in your example, is obtuse ( > 90), so they extend it to easier display the acute side

[lament kettle]

It’s a vector though

[waxen summit]

right, sorry yeah its not infinite, but the same convention still applies since we're looking at the angle

the angle rules still work the same way because the “tilt” of the vector is the same as the tilt of an extended line through it.

[lament kettle]

Ok thanks

[lament kettle]

The reason I’m asking is for a problem I had it said it was the larger one where you don’t extend it

But in the book they extended it

I think it still results in the same torque through tbh

[topaz sinew]

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[valid roost]

hello, i would like to get help on this problem it is precalculus i dont know how to do it could somone help please

[jade jungle]

you're asked to find rational roots. the rational root theorem should immediately jump to mind.

[valid roost]

but thats the thing idk what that is

[jade jungle]

I see. are you then allowed to use this theorem?

or do they not care what method you use?

[indigo spindle]

Well... the name of the section that this question came from is called Rational Root Theorem

So I would hazard a guess that it is the intended method

[jade jungle]

oh. I'm sorry, I missed that.

in that case, let's review the theorem.

[valid roost]

ok

[indigo spindle]

Clicking that link might also lead you to a page with an explanation of it

[valid roost]

it doesnt it just brings me to more problem i dont know how to do

[jade jungle]

the RRT says that all rational roots of a polynomial, if they exist, must take the form:
p/q
where:
p = a factor of the constant term
q = a factor of the leading coefficient (coefficient of the term with the highest power)

(a quick reminder to consider negative factors as well.)

[valid roost]

mhm

[jade jungle]

can you do it from here?

[valid roost]

yeah but how do i know if ots negetive or positive

[jade jungle]

test both.

[valid roost]

ohh

ok thank youuu

[indigo spindle]

Shove them in the equation and see if you get 0

[jade jungle]

note that the theorem does not guarantee any values found are roots.

[valid roost]

oh. .

[jade jungle]

the theorem only guarantees that if there are rational roots, they must fulfill this form.

[indigo spindle]

But if they give 0, they are

[jade jungle]

so you'll need to check any values against the original polynomial, as Greenie mentioned.

[indigo spindle]

11 has remarkably few factors too

So you're in luck

[valid roost]

ok ill try

thanks

[clear python]

The phrasing of this is taking me out

[valid roost]

im still young guys i might be a lil slow

[indigo spindle]

lol

[clear python]

It's okay, take it slow

[indigo spindle]

Did you see my DM?

[valid roost]

im scared at how mature you guys are

[indigo spindle]

Rawr

[clear python]

Discord needs to send notifications for message requests lol

[indigo spindle]

Just turn that off

You can always manually ignore someone

[clear python]

I should work on that yeah

I haven't updated my settings since 2021

[jade jungle]

so... I think I'll leave the helpee to you two then.

[indigo spindle]

I think she might be done

You did a good job Lute

Most of the work tbh

[clear python]

I didn't do anything lol

[indigo spindle]

I suggested 1 thing

Everything else was her

[jade jungle]

I did nothing. please don't attribute anything to someone who didn't even see the name of the section.

[clear python]

Lute it's okay to miss things lmao

[valid roost]

huh

guys is there an easir way to calculate if its -/+?

[jade jungle]

unfortunately not for cubics and higher.

at least, none that I know of. I apologize if there is and I just don't know of it yet.

[topaz sinew]

@valid roost Has your question been resolved?

[indigo spindle]

And yet you still got it right anyway, and provided the relevant information

[topaz sinew]

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[frozen gale]

I suppose c and d are DNE but I don't know for a, b, and e, anyone can help?

[frozen gale]

nvm I got this one

.close

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[keen wedge]

help

[clear python]

!da2a

[topaz sinew]

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/

@keen wedge Has your question been resolved?

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[frozen plank]

hey, i think ik how to prove the min (based of the fact that f(x)>=0 ig) so i would like a hint how to prove the max for f(x)

[frozen plank]

ofc f(x) is cont on [a,b]

[clear python]

You are convinced that the maximum exists right

[frozen plank]

i think the max of f is the min of g

[clear python]

That is correct, but the question is why

[frozen plank]

cuz its the min of g(x)-1

so ^2 makes it the max of f(x)

[frozen plank]

the the harder part for me to prove

[clear python]

...close but no

When is x² maximum

On an interval [a,b]

[frozen plank]

x^2 doesnt have max

only min for x=0

[clear python]

On an interval [a,b]

[frozen plank]

ohh sry

yeh right

hmm

[frozen plank]

it depends if a,b>0 or <0

i mean if they both >0

[clear python]

There is a general thing you can say still

[frozen plank]

then b is max

for example

[frozen plank]

on int [a,b] then x^2 has a max on a or b or on a critical point

[rough furnace]

g(x)max = 3 and g(x)min = -6 and notice that |g(x)max|<|g(x)min| this's why f(x) max when g(x) min and not when g(x) max

[frozen plank]

same for min

[frozen plank]

thats ik

[clear python]

Not on a critical point

The only critical point of x² is 0

And that's a min

So it has to be max at either a or b

[frozen plank]

for specifically x^2 yep

[clear python]

Yes there's a reason I'm trying to give you this example

So we just want to check which is larger between a² and b²

How do we do that

[frozen plank]

hmm

[clear python]

There is a very easy solution

As a hint, x² is an even function

[frozen plank]

i mean if 0<a<b then b is max for example cuz 0<a^2<b^2

[clear python]

Think of this

You do not need any assumptions

[frozen plank]

i am

[frozen plank]

hmm

[clear python]

Do you want another hint

[frozen plank]

lemme try few mins

thx for your patience and help @clear python

[clear python]

No worries no worries

Draw the graph of x²

That would help

Then take any interval [a,b]

This isn't the other hint btw

[frozen plank]

ok a is the max

[clear python]

Nope

[frozen plank]

bruh

[clear python]

[1,2]

As a counterexample

[frozen plank]

in this case 2 is max

b

[clear python]

[-2,1] as a counterexample to that

[frozen plank]

its where | | is larger

[clear python]

Good

[frozen plank]

so if |g(a)-1|>=|g(b)-1| then f(a)>=f(b)

i think

[clear python]

Exactly

[frozen plank]

cool

[clear python]

So you want to find the point where |g(a)-1| is as big as possible

[frozen plank]

yeh, but how do i prove that for which the absolute is larger, it is the max of f

[frozen plank]

since ik that the max would be for a which g(a)=-6

[clear python]

Because the max of x² occurs when |x| is max

The max of f(x) occurs when |g(x)-1| is max

[frozen plank]

we dont have a theorem like that :/

[clear python]

That should be smth you can just use

And even if you can't, it shouldn't be hard to prove

[topaz sinew]

@frozen plank Has your question been resolved?

[frozen plank]

ok thx!!

.close

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[hazy mulch]

can anyone help me

[left juniper]

!status

[topaz sinew]

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

@hazy mulch Has your question been resolved?

[ruby cove]

Quite easy, arc MA is 40 and CAP and CBP have same angles, thus arc BN will be same as arc MA

[hazy mulch]

so you are saying angle ACM = angle BCN

[ruby cove]

No

[hazy mulch]

so this can't be true

[timber jetty]

That cannot be true if angle ACM != angle BCN

[ruby cove]

Yeah this is sad

[left juniper]

I found the answer using the sine law

[ruby cove]

Let me try to calculate the whole thing

[ruby cove]

Is it 20?

[left juniper]

Can I DM you? The answer would be a spoiler

[timber jetty]

@left juniper Can you shoot a copy of the solution in the hlounge? I'm kinda curious

[ruby cove]

Let me solve

I’ll send it when im done

[left juniper]

Can't shafay access the hlounge?

[timber jetty]

I'll ping you there

[rough furnace]

Hmm

So ABPC is cyclic

[rough furnace]

That gives us angle ACM= ABP

So ABP=80 degree

Oh wait im not supposed to give the sol

Ig you could continue from this

[timber jetty]

Just send everything in the hlounge

[topaz sinew]

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[timber jetty]

@hazy mulch you forgot to react to the bot

Do you wanna keep the channel?

[left juniper]

Maybe, they're gone

[topaz sinew]

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[zinc pendant]

First uni linear algebra lesson and the professor is super abstract gives no examples and just writes down proofs with no explanation of what he's doing so this is gonna be fun.

I copied this proof off the board but have no I idea what is actually written.
I think P is the solution set of the non-homogeneous system Ax=b, and Q is the set of a particular solution plus all solutions of the homogeneous system Ax = 0
is that part correct
because in the end were proving P = Q by first proving that Q is a subset of P and then that P is a subset of Q
Right?

[zinc pendant]

rotated it upright

also idk where the vh is from I assume its just a variable but why isnt it just like v why vh

[valid marsh]

Yes, Q is the sum of p with the set of all solutions of Ax = 0, where p is any element of P

[valid marsh]

and it does look like the words on the graph paper could prove P = Q in the way you mentioned

[zinc pendant]

okay glad I atleast understood what we are tryin to proof

[valid marsh]

in the most recent picture, there are a few things you could say to make it clearer

[valid marsh]

like when you write x in Q implies x = p + v_h, you could explicitly write that p is the designated element of P, and v_h is a solution to the homogeneous problem

[zinc pendant]

do you want met to translate the first picture?

[valid marsh]

Sure

Is this Dutch?

[zinc pendant]

ye

Let p be a solution of the consistent system Ax = b, so Ap = b

Then the solution set of Ax = b is given by p+v_h with v_h a solution of Ax = 0

and then under it bv its
A(p + v_h) = ... although I dont know what theyre asking

[zinc pendant]

how would you write that? In highschool we barely see proofs

[valid marsh]

well since A is linear, A(p + v_h) = Ap + A(v_h) = b + 0 = b

[valid marsh]

you would write it like this:
(I’m assuming $p$ and $V$ are given)
To show that $Q \subseteq P$, let $ x \in Q$. Then, by the definition of $Q$, there exists some $v_h \in V$ such that $A(v_h) = 0$, and $x = p + v_h$. Then
$A(x) = A(p+v_h) = A(p) + A(v_h) = b + 0 = b$, where the equalities are respectively given by the construction of $v_h$, linearity of $A$, construction of $p$ and $v_h$, and a vector space axiom, so $x \in P$

[thorny flame]

soup_norm

[valid marsh]

for an intuitive example of what this slide is talking about, if you picture a plane P in 3D space (or any hyperplane in any vector space), and a plane W parallel to P but passing through the origin, then P is equal to the sum of W plus any vector p in P; that is, given any vector p in P, we know that P is the translate of W by p

In this example, P corresponds to the solution set of Ax = b, and W corresponds to the solution set of Ax = 0

[zinc pendant]

okay this I get thanks already for that

[valid marsh]

also, an example comes from differential equations

have you learned about homogeneous differential equations? the idea behind solving them is similar to what’s going on here

but you use vector spaces of functions instead

[zinc pendant]

okay I get it

[zinc pendant]

maybe? can you give an example

[valid marsh]

let me find one

$y’’ - 4y’ - 12y = 3e^{5t}$

[thorny flame]

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

[zinc pendant]

uhmm no I dont think so

[valid marsh]

let’s say you already know that y = -3/7*e^(5t) is a solution to this DE

[zinc pendant]

okya

[valid marsh]

then, if you can find all solutions of the equation y’’ - 4y’ - 12y = 0, then you’ll be able to find all solutions to the original equation

[zinc pendant]

and how would you go about that

[valid marsh]

Let W be the solution set of y’’ - 4y’ - 12y = 0

Then W + (-3/7*e^(5t)) is exactly the solution set of the original equation

[valid marsh]

This can be proven by a very similar method to what you did here

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[celest solar]

is this channel start ?

[topaz sinew]

Please don't occupy multiple help channels.

[jade jungle]

you have a channel already in #help-3.

[celest solar]

ok 🙂

[dapper storm]

Bro you got 3 channels 😅

[celest solar]

how to see ?

i cant se it !!

oh !!

[jade jungle]

I pinged the channel for you.

[true magnet]

#help-3 here

[celest solar]

ok thanks 🙂

.close

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[velvet spear]

I'm having issues with the number 0 and the zero vector, 0

[velvet spear]

V is some vector space here

[quasi depot]

What is the problem

[velvet spear]

Sorry, had to refind the page as I accidently jumped down a bunch of pages

I'm trying to prove this

Is 0 here, the number 0 or the zero vector?

I assume it is the number 0 as the book has yet to define multiplications between two elements in V

[quasi depot]

There is no general notion of multiplication on elements of vector spaces

The zero on the left is the zero of the field

And on the right its the zero vector

[jade thunder]

It may be useful to know some field properties

I hope your field has that the additive identity is not the multiplicative identity

[velvet spear]

Okay, I tried writing a proof for this based on the clarification but realzied I need to take a small step back

Yeah, no, all my missunderstandings are cleared up

[jade thunder]

What does your proof look like?

[shut obsidian]

(often written as 1 != 0)

[velvet spear]

Is it ok if I denote the zero vector as $\bar{0}$ to avoid confusion later on?

[jade thunder]

\vec

[thorny flame]

Pen

[clear python]

Yeah bar works

[velvet spear]

oh,

[clear python]

Some people use an underline

[shut obsidian]

My lecture notes denoted it as $0_V$ where $V$ was the vector space

[thorny flame]

Kepe

[jade thunder]

That’s common as well

[jade thunder]

But also, this

[velvet spear]

Was writing it and then removed it as I thought that it might be just rubbish but, $0x = 0(x) = 0(x+y+z)$, and according to the distributive law as well as V.S 4 that states that $y+z = \vec{0}$ for $y,z\in V$, gives 0(x+y+z) = 0x+0y+0z = 0x+O(y+z) = 0x+0\vec{O} = 0x$.

[thorny flame]

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

[jade thunder]

Uhh I think you’re right, it is rubbish

Lol

[velvet spear]

aghhh

[jade thunder]

What’s vs4

[velvet spear]

That for every element x, there is some number y such that x+y = 0

[jade thunder]

So you’re saying that 0x = 0(x + y + z) for some y in V, and z an additive inverse of y

[velvet spear]

yes

use the distributive property

[jade thunder]

Then you use distributivity great

[velvet spear]

then we have 0x = 0x + 0(y+z)

[jade thunder]

But the next part you have 0(y + z) = 0

How do you know that

[velvet spear]

That is why I found it to be a bit rubish as y+z = 0, but I can't show that $0\bar{0} = \bar{0}$

[thorny flame]

Pen

[velvet spear]

or really, why it holds for any element, a, from a field F

[jade thunder]

Even still if I allow that

The left side starts at 0x

The right side ends with 0x

[velvet spear]

yeah... I'm not great at this : )

[jade thunder]

So uh

[velvet spear]

still trying to get used to maths

[jade thunder]

Let me give you the first step

[velvet spear]

yes, please

[jade thunder]

0v = (0 + 0)v

(Also rationalise which vs axiom I used for this first line)

[velvet spear]

Ok so, first line you used two axioms, first that 0+0 = 0 which is defined in a field

Then you used the axiom of associtivity that 0x = (0)x = 0(x)

[jade thunder]

No

If (V, +, • ) is a vector space then 0x must only mean 0 • x

There is no other way to talk about 0x

[velvet spear]

But I am not disagreeing?

[jade thunder]

This does not exist as an axiom and neither do bracket make sense without talking about an operation

Associativity is a property of a operation not of the elements themselves, (0)x means nothing, neither does 0(x)

[velvet spear]

I'm a tad confused,

Is this not an axiom for vector spaces?

[jade thunder]

Let $(F, \mathcolor{green}{+}, \mathcolor{green}{\cdot})$ be a field, and $(V, +, \cdot)$ a vector space over $F$, then $(a\mathcolor{green}{\cdot} b)\cdot v = a \cdot (b \cdot v)$ for every $a, b \in F$ and $v \in V$

[thorny flame]

frosst

[jade thunder]

It’s not really associative

[velvet spear]

The multiplication of two scalars a and b, is a scalar, multiplied by a vector, is some other vector. $abv$, which is the same as the vector that is obtained by multiplying the scalar a with the vector bv

[thorny flame]

Pen

[jade thunder]

im being very pedantic here mind you

[velvet spear]

I understand, my previous statement about 0x = 0(x) is meaningless

[jade thunder]

but how can the white • be associative

it's not a binary operation on a set

• : F x V -> V

associative would be to say that (a • b) • c = a • (b • c) but the first one says b is in V and the second one says b is in F

if F ≠ V then you've plugged illegal things into •

and F is in general not equal to V

[velvet spear]

Ok, maybe I should clear up my current knowledge: no courses about fields or set theory, besides the basic. I'm starting this course with a background in CS

[jade thunder]

does what i just wrote make sense?

or did i lose you somewhere

[velvet spear]

No, you did not lose me

[jade thunder]

right, so it's technically wrong to say that • is associative

in practice we do just abuse the word associative and say that, but it's not quite right

[jade thunder]

but this is correct, this is the only thing i've used

[velvet spear]

(0+0)v = 0v+0v using the distributive axiom

and since 0v+0v = 0v is the same as 0v+0v=0v+0, using the cancellation law for vector addition, 0v has to be equal to 0

[jade thunder]

whats the cancellation law for vector addition

which axiom is this

[jade thunder]

did you prove that theorem already

[velvet spear]

Yes

[jade thunder]

good

[velvet spear]

I was confused about two things, firstly the notation 0 as the real number and 0 as the zero vector

[jade thunder]

ya there are many 0's in maths, there are also many different equal signs in maths as well

[jade thunder]

even here there are different multiplication and additions in maths

[velvet spear]

the second thing is this, I always thought that this was called "assocativity of scalar multiplication of a vector"

[jade thunder]

.

[velvet spear]

Maths is really hard for me. Thank you Frosst for being patient with me

I'm one step closer!

Hope you have a nice day ❤️

[jade thunder]

you too! good luck on your journey!

[velvet spear]

Thank you ❤️

.close

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[dusky ridge]

Why do a and b have to be coprime?

[jade jungle]

if they do, keep dividing until they don't.