help-4

[minor spire]

lets keep x in deg only

to m,ake it bit simple

[rough tide]

i thought that u needed to convert it into radians to differneitate

arctan and stuff

[minor spire]

lets keep it aside for now

wont make much of a difference since we can corelate the result for rad to deg

[rough tide]

okok

[minor spire]

plug this into desmos

[ 144sec^2(x/2) ] / [ 81 + 256 tan^2(x/2) ]

@rough tide

[rough tide]

ok

i changed the settings to degrees and its slightly off as it goes on

(degree)

[minor spire]

ts

pmo

tf

is happening

[rough tide]

real

idek if mine is correct

[minor spire]

lemme check

do you see it

now

@rough tide

get it??

[rough tide]

ohhh

could i get the working out of how u differentied it if possible

[minor spire]

same as yours

[rough tide]

oh then why did it

[minor spire]

i just solved the squaresa in 1/1+z^2

idk about that

[rough tide]

thats rly werid

[minor spire]

,might be some error in the desmos eqn u put in

[rough tide]

wait lemme

rewrite it j in case!

[minor spire]

yeaq

remove the radian and deg stuff

assume x is in radian

do it for radian

[rough tide]

oh yea

then its the same

wait but bc i need it to be degrees

wld it not rly matter or how wld i convert this result back into degrees then

[minor spire]

lemme show u smth

[rough tide]

bc i already defined previously that x = in degrees

[minor spire]

i gtg but you understood it right??

[rough tide]

oh yes

okok

wait lemme j screenshot this and close it!

[minor spire]

alright

gn

[rough tide]

okok tysm

.close

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[glad minnow]

this is for riemann sums, i can understand adding the values for the left endpoint by excluding the right most point but the diagrams still confuse me a bit

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[toxic ridge]

!status

[vale dock]

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

[timber swan]

1

[toxic ridge]

you should get a relation for the sequences xn and yn

have you done conic sections

parabola specifically

[timber swan]

Not yet

Not in depth

Atleast

Can definitely try any idea

[toxic ridge]

first find angle OFAk in terms of xk and yk

[timber swan]

Hmm

Derivatives ig

[toxic ridge]

what

???

no need for derivatives here

draw the diagram

[timber swan]

If i take

The point

On parabola

Oh nvm

[toxic ridge]

this is strictly conic section only question, with need for derivative possibly only when calculating the limit

[timber swan]

I can directly find the slope

Okay

2t and t^2

Omg

I see it

Wow

got it

.close

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[timber swan]

Requires integration also

[toxic ridge]

huh?

oh bruh

ok nvm

yeah that works

[timber swan]

Yea

For the sum

[vale dock]

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[oak bramble]

x is tangent here

I forgot how to solve this kind of problem

These are perfect circles BTW, I'm not very good at drawing

[slate folio]

draw the pic accurately

[oak bramble]

I can't

😭

the circles are touching at one point, line x is tangent to both circles

[slate folio]

do it on a paper or on a computer

[oak bramble]

Is this OK?

[slate folio]

good enough

now what kind of shape is the quadrilateral

[oak bramble]

trapezoid?

[slate folio]

indeed

but it's a special kind of trapezoid

[oak bramble]

I don't see it

maybe the drawing is too poor

[slate folio]

do you see the right angles?

[oak bramble]

is it at the points of tangency?

[slate folio]

correct

so it is a right trapezoid

now the easiest way to solve this would be

take the center of the small circle and drop a perpendicular to the radius of the big circle

you would create a rectangle and a right triangle

[oak bramble]

smart

i think i can take it from here

.close

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[gleaming schooner]

just wondering, is there anyone here capable of coming up with a formula that matches each line of this text file with one variable, x?

[gleaming schooner]

file provides values for up to x=55

[junior jay]

A polynomial with degree 55

[exotic mortar]

u probably have to check differences in the numbers

and then differences in the differences sequence

[vale dock]

@gleaming schooner Has your question been resolved?

[gleaming schooner]

here are the differences if they can help

repetitions of preceding differences are excluded

[wraith heart]

yea there's no line that fits those exactly

[exotic mortar]

im level 0 noob idk how to find quads and ploys from sequence lol

[gleaming schooner]

is there a "good enough" approximation at least

[exotic mortar]

i mean this guy said degree 55

[gleaming schooner]

inelegant

[wraith heart]

https://www.mathsisfun.com/data/least-squares-regression.html

[gleaming schooner]

this results in a linear curve, undesirable

[exotic mortar]

oh

it literally names the channel to who is asking the question lol

cool

[wraith heart]

then you need to be more specific

[gleaming schooner]

a close approximation could be exponential with the 1st 5 entries linear

[wraith heart]

yea go do that

https://math.libretexts.org/Workbench/1250_Draft_3/06%3A_Exponential_and_Logarithmic_Functions/6.09%3A_Exponential_and_Logarithmic_Regressions

or take log of your 6-55 entries then linear regression

[timber cloud]

is the pattern here just f(x)=ceil(f(x-1)*1.2)?

as x approaches infinity, 1.2^x (with some offset) will start to be a good approximation, but it isn't going to be at the start

[gleaming schooner]

slightly less than that on the later entries

and slightly greater than that on the earlier entries

oh wait
the difference is in a way floor(1.2^x)

thanks

.close

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[obtuse knoll]

I’m studying linear algebra. I’ve gotten to the part where linear combination is brought up. I’ve seen linear dependency and independency. I wondered whether for a linear combination equation to have solution does the system need to be linearly dependent? Because in my eyes it’s practically the same concept where we attempt to find a solution by multiplying a multiplicatif factor for each term thus giving the end result. Thank you for being explicit. This is all new concepts to me so it would help a lot.

[obsidian tree]

Wdym by a "linear combination equation"?

If you mean a nonzero linear combination of variables equaling zero then yes they have to be linearly dependent - that's the definition

Otherwise no

But if it's just one equation involving multiple variables then they're automatically linearly dependent anyway

[obtuse knoll]

What’s essentially the difference between both concepts?

[obsidian tree]

They're pretty different

A linear combination of variables is a sum $\sum_{i=1}^n\alpha_ix_i$, where $\alpha_i\in\mathbb R$

[rocky lotus]

depression

[obsidian tree]

(or any other field)

You multiply each variable by a number, and you add them all together - that's a linear combination

You can also substitute 'variable' for vector

A collection of vectors is linearly dependent if there is a nontrivial linear combination of them ($\alpha_i\ne0$ for some $i$) that equals zero

[rocky lotus]

depression

[obsidian tree]

For example, $(1,1)$ and $(2,2)$ are linearly dependent because $2\cdot(1,1)-1\cdot(2,2)=0$

[turbid valve]

-1 *

[obsidian tree]

ty yeah -1

[turbid valve]

But otherwise yes

[rocky lotus]

depression

[turbid valve]

Linear _in_dependence, then, is when the only way to solve for e. g. a1 v1 + a2 v2 +... = 0 is if all the a's are 0

[obtuse knoll]

Okay quick question. Visually, Is the difference in solving it this? This would be for solving for k of the linear combination whereas for finding the dependency instead of (x,y) on that side we’d have (0,0)?

[obsidian tree]

I can't lie, I don't understand what you just said

But that's not an equation you can 'solve'

You can either solve for k in terms of x and/or y, or you can solve for x and y in terms of k

(the latter of which is already done for you)

[turbid valve]

(What would you have said in French? Je pourrais le traduire)

[turbid valve]

Because otherwise this is just a linear combination of terms

[obtuse knoll]

Ok it’s fine I could try to explain it another way but maybe I’m just confused the linear dependency and independency are both new concepts so I might need some time to understand. I was taking a look at finding what in French they call the base. They verify it by checking whether the vectors are linearly independent and then whether it’s an ensemble of generators which as I understood it it’s the possibility of writing all combinations of vectors of a vector space through linear combination. Then: in the examples that follow. It shows a bit how to solve it. And I was saying is the difference between the use of the two concepts as I wrote it in this image

And by “how To solve it” I meant how to find out whether we have a base or not

[turbid valve]

C'est "un" base, n'est-ce pas ?

[obtuse knoll]

Oui

[obsidian tree]

We call it a basis in english

[turbid valve]

On dit alors "a basis" (pl. "bases")

[obtuse knoll]

Oh welp I didn’t know lmao

[turbid valve]

Pour dire en français... "déterminer s'il y a un base" , c'est ça ?

[obtuse knoll]

Oui exactement

[turbid valve]

bah le mot "to determine" existe aussi en anglais

"prove" c'est plutôt "démontrer"

[obtuse knoll]

Oui mais je veux dire est ce que la manière dont je l’ai reformulé est correct? La manière qu’on fait pour déterminer k1 et k2 dans la partie linear combination marche vraiment comme ça?

Parce que je pensais que les deux concepts semblent vraiment se ressembler. Un utiliser (x,y) et l’autre (0,0) pour le déterminer

[obsidian tree]

Every nonempty linear space has a basis (this needs proving), so finding out if that equation has a nonzero solution (k1 and k2 not both equal to zero) is equivalent to determining if it has a basis

If that's what you mean

Because then you get a vector space of "solutions" to your equation

But I don't think that's a very good way of thinking about it, at least at first

[turbid valve]

Ah, then you're just confused with what each is for

[turbid valve]

Mostly explained here

Si on peut trouver k1, k2 (inégal à zéro) tels que la summation soit 0, donc les vectors sont lin. dependants

(ptn mon français se détruit rapidement )

[obtuse knoll]

You mentioned the way I wrote it out isn’t the right way of thinking of basis in the beginning. How do you suggest I should reformulate each criteria?

Would it be possible to write in the mathematical form because in text it’s a bit hard to get

[obsidian tree]

The only reason I said that is because you're thinking about linear combinations of elements in one vector space, but the "solution space" is a completely different, almost unrelated vector space

They're not really connected

But if $v_1$ and $v_2$ are vectors, and you have an equation (for example) $\alpha_1v_1+3\alpha_2v_2=0$, then each valid $\alpha_1$ and $\alpha_2$ gives you a vector (\alpha_1,\alpha_2)$. That is then a vector space, since you can multiply it by scalars (for example, $r\alpha_1v_1+3r\alpha_2v_2=r(\alpha_1v_1+3\alpha_2v_2)=r(0)=0$, if that makes sense)

[rocky lotus]

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

[obsidian tree]

idk where the error is but oh well

That's more technical than anything else though, so don't think of it like that

[obtuse knoll]

Ok well thank you for that I think it clarified it more

Quick question though. My linear algebra class is in about a month. How do the concepts complexify from this point on in the content? I heard about something called spam which sounds hard

[obsidian tree]

Span* probably

It's not that hard dw

If you have a collection of vectors, then the "span" of those vectors is the set of linear combinations of them

The span of the vector (1,2) is all vectors of the form (k, 2k)

[obtuse knoll]

Ok well thank u a lot have a good night

.close

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[earnest radish]

How to solve this?

[west cloud]

Where's your question?

[west cloud]

oh nvm

!status

[vale dock]

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

[west cloud]

and maybe you should look at the progress here #help-31 message

[earnest radish]

Actually I took the question from their because I like the question and the Google and chatgpt have so complex answers that's why came here for explanation

I don't know where to begin

[vale dock]

@earnest radish Has your question been resolved?

[vale dock]

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[west cloud]

<@&268886789983436800>

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[west cloud]

tysm

[cerulean smelt]

I need help with confidence intervals. The confidence interval calculated here is this answer: (95% confident between 0.44 and 0.64). Is this saying that 100% certainty that there is a 95% chance of the mean being within that interval from the sample mean? How can it 100% be the case that: (there is a 95% confident between 0.44 and 0.64) if the sample mean is whats used to calculate that interval (which is just and estimate of the mean) and not the actual mean (because it is uknown)

the image is from a kahn academy video

[tidal terrace]

in a another channel this is closed

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[small crystal]

kinda might not be math question tho but can desmos do regression for rationals? ive used it for other typa function but idk how to do it for rationals

[sharp whale]

you can try asking the unofficial desmos discord about this

[small crystal]

is there a server?

i remember there was one in reddit

imma ask reddit

[sharp whale]

https://discord.gg/sRk7gd7ftW here it is

[small crystal]

oh

[sharp whale]

its got the bernard as the server picture

[small crystal]

yea

thanks :)

.close

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[sharp whale]

np

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[whole plank]

In a intergral it is written as ∫ f(x) dx why doesnt the dx cancel out the ∫ and isnt dx aproach 0 so should the ∫ f(x) get timezed by 0 thus making it 0....

[wraith heart]

$\int dx = x + C$ yes?

[rocky lotus]

riemann

[whole plank]

Why?

[wraith heart]

you seem to have a misunderstanding of integrals so i suggest you start all over

do you know derivatives?

[whole plank]

Ofc

[wraith heart]

do you know fundamental theorem of calculus?

[whole plank]

Yes

[wraith heart]

,tex .FTC1

[rocky lotus]

riemann

[wraith heart]

Find F'(x) for F(x) = x

[whole plank]

1/2x^2

Wait

No

Opiset

Mb read it wrong

1

[stark wedge]

btw op came here with the same question 2 days ago

[whole plank]

Yeah and u didnt answer it

[wraith heart]

$F(x) = \int_a^x f(t) dt$

[rocky lotus]

riemann

[wraith heart]

apply the derivative to both sides and solve for f(x)

[wraith heart]

this either follows from the definition from antiderivative or fundamental theorem of calculus, whichever you learned

[whole plank]

Are you just having me say F'(x)=f(x)

Becuase if so idk how i got that answer its just in my brain

[glass kelp]

Then talk to ur brain

[whole plank]

Its just in my memory

I didnt solve to get that answer

I just know thats the answer

Or i am wrong idk

[glass kelp]

do u know what a definite integral is

like

the riemann sum thingy

[wraith heart]

yea what's this when F(x) = x?

[wraith heart]

once you find f(x), plug it in here

[wraith heart]

then this should follow

[whole plank]

1/2x^2=f(x)

Wiat

[wraith heart]

made this mistake again

[whole plank]

Yea i saw

[whole plank]

How am i supposed to plug f(x)=1 into here if f(x) isnt in here unless im just supposed to put F(x)=x

[wraith heart]

correct it's the same F(x) i said here

[whole plank]

What am i even solving?

[wraith heart]

i'm answering your question

[whole plank]

Well according to some math that i dont understand the d/dx of the intergral side is f(x) right?

[wraith heart]

my answer was conditioned that you understood FTC

[whole plank]

Yea i knew it. I dont understand why it works

[wraith heart]

alright memorizing is good enough

[glass kelp]

bruh

[wraith heart]

can you do this yet

[whole plank]

I still dont understand that becuase f(x) isnt in there to plug it in

[wraith heart]

did you find f(x)?

[wraith heart]

it was not this

[glass kelp]

F'(x) = f(x)

[whole plank]

is it just 1

[wraith heart]

yes

now use 1 * anything = anything

[whole plank]

Just multiply f(x) into there?

On the left side?

[wraith heart]

.

x = F(x) = ?

[whole plank]

∫ f(t) dt 😭

[wraith heart]

close

you're missing some limits and you're not substituting f(x) = 1

[whole plank]

Im sorry but your just not making sense in my brain i understand i can multiply one side by f(x) since f(x)=1 so the value doesnt change but ∫ a->x(f(t) dt)•f(x) doesnt seem like it changes anything and the left side doesn't seem like the play

[glass kelp]

wdym 'multiply one side by f(x)'

[whole plank]

I have no idea

[glass kelp]

what riemann is doing is using ftc to show why x is an antiderivative of 1

[glass kelp]

.

Also since F(x) is an antiderivative of f(x), F'(x) = f(x) by defn

[wraith heart]

Write the right side as simple as f(t) = 1

[wraith heart]

This is important

[whole plank]

Why is f(t)=1

[wraith heart]

What

[glass kelp]

.

[wraith heart]

You've found it

[whole plank]

For f(x) not f(t)

[wraith heart]

This is another misunderstand you need to correct

In f(x), x is a dummy variable

The graphs of f(x) against x and f(t) vs. t are both the same

[whole plank]

Yeah but were taking d/dx not d/dt right?

[wraith heart]

Correct. That doesn't change anything about what I'm saying

[whole plank]

Should d/dx of f(t) be d(f(t))/dx like d/dx of y is dy/dx

[wraith heart]

No one is calculating f'(x)

[whole plank]

Ok

Why are we not in respect to x?

[wraith heart]

Welp you're just avoiding my questions

[whole plank]

I dont understand what your question even is

[toxic ridge]

😭

What exactly do you not understand here

Follow what Riemann says

[whole plank]

What are we taking the derivative with respect to

[glass kelp]

f(t) = 1 and f(x) = 1 is essntially the same function

in that regard, F(t) = t and F(x) = x are also essentially the same function

from ftc

[ F(b) - F(a) = \int_a^b f(x) \dd x]

[rocky lotus]

k

[whole plank]

Now this is a fule i unde

Understand

[glass kelp]

what are u confused abt

[toxic ridge]

Did you understand why it works

[whole plank]

Yea ar

[toxic ridge]

Because this was your original question

[whole plank]

Area under the curve minus a different area equals the area that was left oput

[glass kelp]

do u know what is the relation btw antiderivative of a function and the function

[toxic ridge]

?

You know what that’s fine

If that’s how you understood it goodenough

[whole plank]

Ok

Ive just realized the intergral is like a limit so that goes to infinity so you times it by dx as dx goes to 0 to cancel out the infinity so your left with an actual thing right?

[glass kelp]

[ \int_a^b f(x) \dd x = \lim_{n \to \infty} \underbrace{\frac{b-a}{n}}{\to \dd x} \underbrace{\sum{k=0}^{n} f\left( a + \frac{(b-a)}{n}k\right)}_{\to \int_a^{b} f(x)}]

[rocky lotus]

k

[whole plank]

Yeah i shoulda realized those were connected ive been useing that form of summations to solve area problems for a while

[glass kelp]

close enough for understanding

[uncut thistle]

[ \int_a^b f(x) \dd x = \underbrace{\lim_{n \to \infty} \frac{b-a}{n}}{\to \dd x} \underbrace{\sum{k=0}^{n} f\left( a + \frac{(b-a)}{n}k\right)}_{\to \int_a^{b} f(x)}]

[rocky lotus]

Nyxzore

[vale dock]

@whole plank Has your question been resolved?

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[rare ruin]

i dont understnad what vector spaces are

[glass kelp]

Why

Do u know what a field is

[rare ruin]

vector field?

[glass kelp]

No like

[tidal swift]

number field

[rare ruin]

yeah

[glass kelp]

Vector space over a scalar field

[rare ruin]

axioms?

[glass kelp]

Things that are made from the scalar fields that satisfy the axioms for a vector space constitute a vector space

[rare ruin]

uh

[ebon glade]

made from the scalar fields
uhh

[rare ruin]

so anything that can be proven using axioms are part of a vector space?

[glass kelp]

Not the greatest choice of work ik😭

[ebon glade]

you know R^n, yes?

[rare ruin]

sorry im really new to this i have a test in a week about vectors ive basically tried to learn a lot of stuff in 2 days

[rare ruin]

is that

the cooridnate plane

cartesian

[ebon glade]

for n=2, sure

the set of all (x1,x2,..,xn)

[rare ruin]

oh so like all dimensions

[ebon glade]

dimension n, but yes

[tidal swift]

n dimensions*

[rare ruin]

kind of?

[ebon glade]

anyway, an important property is that you can add vectors and that you can multiply them with a number

[rare ruin]

yup

[ebon glade]

and these operations behave how you would expect them to

[rare ruin]

yeah

[ebon glade]

eg v+w=w+v

[rare ruin]

👍

[ebon glade]

(a+b)v=av+bv

and so on

[rare ruin]

yes

[ebon glade]

do you know matrices

[rare ruin]

yeah

[ebon glade]

we know that we can also add matrices (of the same size) and we can multiply them by numbers

[rare ruin]

yup

[ebon glade]

and those operations also behave like you would expect

[rare ruin]

agreed

[stark wedge]

big reveal: the set of m by n matrices is itself a vector space

[ebon glade]

or we can also think of functions

you can also add functions and multiply them by numbers

[rare ruin]

ok

[rare ruin]

it is?

[ebon glade]

so we see, that all these different situations share some common properties

[rare ruin]

oh so like if it follows the axioms, its a vector space?

[ebon glade]

these properties are so important that we gave all of this a name: vector spaces

[rare ruin]

so vector spaces are the axioms?

[ebon glade]

no

[rare ruin]

srry

[ebon glade]

a set is a vector space if it satisfies the properties

[tidal swift]

vector spaces are the structures obeying those axioms

[ebon glade]

the properties are also called vector space axioms

[stark wedge]

if it follows the axioms of a vector space, then it's a vector space

[tidal swift]

can i toss a pdf here

[rare ruin]

oh so axioms are rules

[tidal swift]

no not pirated

online class PDF

[rare ruin]

and if they are followed then its a vector space

[tidal swift]

yes!

[rare ruin]

WOW

[tidal swift]

more appropriately, axioms are conditions

or "checks" of sorts

if a set passes those checks, congrats, it's a vector space

[rare ruin]

but like anything can be a vector space then?

[glass kelp]

if they satisfy the axioms, yes

[tidal swift]

nope

i mean some sets are not fields

[glass kelp]

if they dont satisfy, no

[rare ruin]

yeah i meant that

[tidal swift]

then yes

but for example

the set of all integers Z

is NOT a field and therefore NOT a vector space

[rare ruin]

okok

[ebon glade]

importantly you cant for example multiply integers with 1/2 and always get an integer back

[rare ruin]

yeah

[glass kelp]

doesnt the scalar multiplied to the vector have to be inside the field (set) that the vector space is over

[ebon glade]

and every field (of characteristic not 2) contains 1/2

[tidal swift]

isn't that why the set of integers is not a field

[ebon glade]

so Z is not a vector space over any such field

[glass kelp]

i knwo that z is not a field

[rare ruin]

also what are linear transformations? 😔 im supposed to learn that too

[glass kelp]

but shouldnt the scalar multiplied to the vector be a number from the set that the vector space is supposed to be over

[ebon glade]

and?

my point is that Z isnt a vector space over any field

over Q or R or any field you wanna consider

[glass kelp]

oh ok. i interpreted it wrong

srry

i thought u meant z acting as the scalar field not as the vector space

[tidal swift]

a kind of mapping that moves a vector from one vector space to another (can be same) while preserving all axioms

[rare ruin]

oh okay

thank you guys

.close

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[uncut cape]

1. Using congruency of triangle i found the sides of the triangle in terms of x and used Pythagoras theorem to find the sides and hypotenuse. Using this, I found the answer to the first question.
2. But how do i solve the radii of smaller circles?

[glass kelp]

,rccw

[turbid summit]

hmm you shkuld just use the pythagorean theorem on ABC and solve for positive x

[slate folio]

He already solved the first part

[turbid summit]

oop

[slate folio]

@uncut cape drop the perpendicular from O1 to OM to create a rectangle

see if it works

[uncut cape]

I tried that, but the answer is so complicated and long and doesn't match the given options, that i know there has to be an elegant way

[slate folio]

wait what is x = btw

[uncut cape]

2

[slate folio]

alr thanks

[turbid summit]

id say use the fact that the OPC and MCO_1 (M on BC) are similar and both are right angled

oop

thats O_2 actually but

conceptually they're the same

[slate folio]

??????

[uncut cape]

The hint from the book says something about similarity and 1 -sin theta / 1 + sin theta.

I can solve for sin theta. But where does this 1-sin theta is coming from?

How does AO equal to 1 + sin theta?

If anyone can answer this, I'll be able solve to the entire question

[vale dock]

@uncut cape Has your question been resolved?

[runic bolt]

need help

[slate folio]

! occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[sleek nebula]

!occupied

[vale dock]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[sleek nebula]

Please dont spam

[nimble crypt]

If you want to post a question do it on the Available math help channels (the ones above the occupied)

@runic bolt

[vale dock]

@uncut cape Has your question been resolved?

[uncut cape]

.close

.close()

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[rough talon]

prove that $x^2+(x+1)^2=y^2$ has infinitely many positive integer solutions

[rocky lotus]

skissue.in.a.teacup

[rough talon]

using the pythag generating triples thing

either
$$m^2-n^2=2mn+1$$
$$m^2-n^2=2mn-1$$

[rocky lotus]

skissue.in.a.teacup

[rough talon]

and yup im alr stuck

why do i feel like i am getting worse at number theory

[vale dock]

@rough talon Has your question been resolved?

[toxic ridge]

I think treating it as a quadratic in x might be better

Then apply condition for integer roots

[slate folio]

I tried it and you just get back what you started with lol

[toxic ridge]

Interesting

But x isn’t there

Only y is now

And some integer k

2x^2 + 2x +1 - y^2 = 0

D = 4 - 8(1 - y^2)

[unborn python]

that is just the pythagorus trios isn't that?

[toxic ridge]

Op tried that but didn’t get anywhere

[slate folio]

yes

[rough talon]

$2x^2+2x-1-y^2=0

[rough talon]

i mean always the possibility im overlooking it

[slate folio]

the smallest solutions are
3 4 5
20 21 29
119 120 169
696 697 989

[rough talon]

sus

[toxic ridge]

We just want 2y^2 - 1 to be a perfect square

[slate folio]

some sort of dumb generating algorithm maybe

[toxic ridge]

Well also

We gotta divide by 4

[unborn python]

oh wait it is x^2+(x^1)^2, not minus...

[rough talon]

$x=\frac{-1\pm\sqrt{2y^2-1}}{2}$

right

[rocky lotus]

skissue.in.a.teacup

[toxic ridge]

Yea

[rough talon]

so $2y^2=z^2+1$

[rocky lotus]

skissue.in.a.teacup

[toxic ridge]

Also one more condition

-1 +- z must be divisible by 2

[slate folio]

z has to be odd

[toxic ridge]

Z has to be odd

[toxic ridge]

Yep

[slate folio]

so if we substitute z = 2k + 1 we get back what we started

[rough talon]

$$2y^2=4k^2+4k+2$$
$$y^2=2k^2+2k+1$$
$$4j^2+4j+1=2k^2+2k+1$$
$$2j^2+2j=k^2+k$$

[rocky lotus]

skissue.in.a.teacup

[rough talon]

didnt i encounter this in my prev problem

[toxic ridge]

I wonder why

[rough talon]

actually if you mod 4 it isnt lhs 0 and rhs 2

wait nvm

[slate folio]

wait

[rough talon]

implies either 4|k or 4|k+1

[slate folio]

LHS divisible by 4

[toxic ridge]

[toxic ridge]

??

Oh ok

Got it

Ok then make cases again sigh

[rough talon]

ok yea idr if i posted this last time and i only wrote it out but i also got this in the previous problem (like exactly) and i couldnt get anywhere

maybe you can transform this to the x^2+y^2=2z^2

well it is literally the same just y=1

[toxic ridge]

Ok now this goes beyond my knowledge of number theory

Idk if this is some theorem or not

[rough talon]

#help-1 message

x=-p^2+2ps+s^2
y=p^2+2ps-s^2
z=p^2+s^2

we can just assume either x or y to be 1 right

[slate folio]

y is clearly smaller

[rough talon]

$1=p^2+2ps-s^2$

[rocky lotus]

skissue.in.a.teacup

[rough talon]

why are diophantine equations always so terrible

wait

[rough talon]

this is literally the same what the fuckk

[slate folio]

tried the pyt generation approach

[slate folio]

got back to this

this is so stupid

[rough talon]

looks like both methods are the same

[rough talon]

ikr

[slate folio]

so we're stuck

[marsh forge]

have yall tried pells eqn

[slate folio]

wait a second this is just negative pell's equation

[rough talon]

well (dio) well (phan) well (tine)

oh

[marsh forge]

i guess that answers my question

💔

[rough talon]

do negative pell eqns also have infinite sols?

[slate folio]

only for some n

[marsh forge]

depends

[slate folio]

2 is one of them

2, 5, 10, 13, 17, 26, 29, 37, 41, 50, 53, 58, 61, 65, 73, 74, 82, 85, 89, 97, ...

[marsh forge]

x^2 - Dy^2 = -1 has sols iff the continued fraction expansion of sqrt(D) has odd period

i think

i forgot all my num theory

[rough talon]

ok so

[slate folio]

oh actually

[rough talon]

actually

$$m^2-n^2-2mn=1$$
$$(m-n)^2-2n^2=1$$
$$a^2-2b^2=1$$

[rocky lotus]

skissue.in.a.teacup

[rough talon]

ok bro

[slate folio]

if you do try to solve the second equation you get 2n^2 + 1 = k^2

lol

[rough talon]

the first one gets normal pell right

[slate folio]

both can turn into pell's

funnily enough

[rough talon]

alr ty guys

.solved

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[midnight pier]

Hmmmmm...

[rough talon]

what are you doing exactly?

[tidal swift]

what's the question

[tidal swift]

just that?

[midnight pier]

,tex $ \frac{2}{5} + \frac{5}{s + 2} = 1 $

[tidal swift]

is that an S or a 5

[rocky lotus]

玩家

[midnight pier]

This

[tidal swift]

the objective is to?

find s?

[midnight pier]

Yes

[glass kelp]

Bruh

[tidal swift]

four pages of proof goodness for one small problem

anyway

just cross-multiply, if you know how to?

[glass kelp]

Why so multi-layered

[midnight pier]

when the ultimate boredom happens

[tidal swift]

wait

[rough talon]

when the unfunny happens

[midnight pier]

yes

[river shale]

Sadge

[stark wedge]

hang on tho

[tidal swift]

so

[stark wedge]

is the first denominator s or 5

[midnight pier]

wait no i messed

,tex $ \frac{2}{s} + \frac{5}{s + 2} = 1 $

[rocky lotus]

玩家

[tidal swift]

ah

[stark wedge]

so what's your goal here?

maximize overcomplication?

[rough talon]

seems like to troll :/

[stark wedge]

also \textbf{nobody} handwrites $\bN$ and $\bR$ like you did lmfao

[midnight pier]

Yes

[rocky lotus]

Ann

[stark wedge]

ok in that case:

[tidal swift]

i mean in that case

you're doing it very well

[midnight pier]

lmfao

i mean

[tidal swift]

anyway

!done

[vale dock]

If you are done with this channel, please mark your problem as solved by typing .close

[midnight pier]

.close

[vale dock]

Channel closed

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[midnight pier]

OK

.close

[river shale]

It’s already closed

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[oak bramble]

finding area

what im confused on is

is this 8?

[glass kelp]

8 is the entire base?

[oak bramble]

exactly what im confused about

is it the right half or the entire base

[stark wedge]

badly drawn diagram

is this exactly as it was given to you by the teacher?

[oak bramble]

yep

[glass kelp]

Bruh

[stark wedge]

then complain to the teacher

[oak bramble]

its timed

i dont have time for that

[next vortex]

what are you even being asked to find?

[oak bramble]

area

[next vortex]

oh

[oak bramble]

i guess i will just assume the entire base is 8 for now and complain if i lose points here

[manic flower]

You are cheating in a test?

[next vortex]

we can’t help you with a test

that’s academic dishonesty

[oak bramble]

im not asking you to solve im just asking for help understanding the diagram

i can easily solve this its just unclear

[tidal swift]

i think right

[glass kelp]

https://tenor.com/view/spiderman-meme-peter-tobey-maguire-black-suit-gif-24756540

[tidal swift]

if you're raising this here

[next vortex]

I assume this is correct

[tidal swift]

you're very likely NOT the only one having problems

shoot this question straight to the teacher as soon as you can

[rough talon]

yes 8 is that length, go back to your test now

[oak bramble]

ok bye

.close

[vale dock]

Channel closed

Closed by @oak bramble

Use .reopen if this was a mistake.

[glass kelp]

😭

Wth was that

[slate folio]

no

[oak bramble]

i realize now

impossible for the right side to be 8

it has to be the whole base

[slate folio]

was about to say that

oh wait a second

.reopen

[vale dock]

✅

[oak bramble]

Do not tell me anything further it's a test 😭

[slate folio]

if 8 is the whole base then the red side has to be equal to 2.255

[manic flower]

Yeh 8 cant be the entire base

[oak bramble]

wait im so confused

is the problem impossible

[novel canyon]

lock in for your test

[slate folio]

I'm 100% sure the problem gave too much info

[oak bramble]

no situation i find makes sense

[slate folio]

which leads to this anomaly

[oak bramble]

so its the teachers fault if i got it wrong then?

[manic flower]

Yes

[marsh forge]

if this takes way too long just skip it man

[manic flower]

Facts

[slate folio]

so did the original problem give the lengths of that 7 and 5 side?

[oak bramble]

i swear this is literally impossible, i will save and contact the teacher

[oak bramble]

that is the original problem

wdym?

[manic flower]

Yeh tell them this problem is too ambiguous

[slate folio]

what the

[oak bramble]

i did not type any numbers there

if i typed something i did it in red ink

[slate folio]

if they ask for the area what's the point of 5 and 7-unit sides there??

[oak bramble]

no clue

[slate folio]

or this is a troll problem?

[oak bramble]

its not

its literally on a test

ok well im going to leave now but i will contact teacher

[slate folio]

is it multiple-choice?

[tidal swift]

yea just contact at this point

[slate folio]

I think it's better you contact your teacher

[manic flower]

I think the teacher wants them ti find the area of this trapezium using the formula but forgot to give values that make an actual trapezium

[slate folio]

this figure is ridiculous

[oak bramble]

yeah here are choices but im going to leave now, if you guys try to figure it out fine but i wont see cuz im leaving

[manic flower]

The sum of parralel sides smth smth formyual

28

Formula wise its 28

[slate folio]

if 8 is not the whole side then the area would be in decimals

[manic flower]

Thouugh this question should be deleted

[slate folio]

I should probably close this

.close

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[austere delta]

Can someone please check for me or this answer from the answerbook is correct? If have an angle of inclination of 116.6 degrees and 21.8 degrees. So is the last step then not 180-(116.6+21.8)?

[glass kelp]

Why is the last step that in ur working

[austere delta]

the sum of int angles is 180 so therefore 180-21.8-63.4 to get the acute anglewhere angle where they cross

[glass kelp]

I’m so confused lmao

[austere delta]

Lol me too

[glass kelp]

Where is the triangle

[inner mirage]

Ok one thing about your graph, both the angles of inclination seem to be obtuse, when one of them is acute in the question

[austere delta]

the x-axis

[glass kelp]

That is not the angle w’ere finding

Also

Doesn’t ab have a positive slope

[inner mirage]

Yeah thats what I said

[austere delta]

alright, i just gonna draw it again and see

[glass kelp]

,calc 180 - 63.4 - 21.8

[rocky lotus]

Result:

94.8

[glass kelp]

,calc 180 - 94.8

[rocky lotus]

Result:

85.2

[glass kelp]

Which is supplementary to the thing we should be finding

[austere delta]

Allright, I was looking at the int. angle only

[austere delta]

I missed that step

It's clear now, I was looking at the wrong angle

Thx a lot

.close

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[lyric copper]

hi i need help with this equation. i would like help solving it without half angle substitution or any method that requires trial and error if possible.

[lyric copper]

this is my progress so far, idk what to do next

this is the equation

[tidal swift]

,rccw

gdi texit

[west cloud]

,rccw

[west cloud]

@lyric copper are you solving for cos x or sin x?

[lyric copper]

solving for x in general

[west cloud]

ok

Try to make everything in terms of sin x then

[lyric copper]

how

[west cloud]

You have the pythagorean identity: cos^2 x + sin^2 x = 1

So you rearrange this equation to solve for cos^2 x, and then you plug it in

[glass kelp]

Ull get square root unfortunately ):

[lyric copper]

how do i get rid of the cosine functions tho

[glass kelp]

But such is life

[lyric copper]

im sorry im not following

[glass kelp]

..

[west cloud]

You can rearrange the pythagorean identity from cos^2x + sin^2 x = 1 to cos^2 x = 1 - sin^2 x, and then substitute

[lyric copper]

but theres no cos^2 x in the image

[glass kelp]

.

[lyric copper]

oh

that makes things much more complicated

[glass kelp]

I suggest u rearrange the equation

This way

[ 6\sin^2 x - 3\sin x = 2\sin x \cos x - \cos x]

[lyric copper]

what do i do about -2 sin x cos x

[rocky lotus]

k

[lyric copper]

ooo what then

[glass kelp]

Factor smth out

Quartic

[lyric copper]

whats that

sorry im new to this stuff

[west cloud]

A fourth degree polynomial

that means that its highest power is 4

[lyric copper]

oh

[west cloud]

no need to apologize

[lyric copper]

thanks but like im lowkey still not getting it, could someone kindly provide a more elaborate solution

[glass kelp]

[ 6\sin^2 x - 3\sin x = (2\sin x - 1)(\cos x)]

[rocky lotus]

k

[glass kelp]

Oh waittt

[inner mirage]

Something very good happens if you factor

[glass kelp]

[ 3\sin x(2\sin x - 1) = (2 \sin x - 1)(\cos x)]

[rocky lotus]

k

[glass kelp]

@lyric copper

[lyric copper]

yoooooooooooo

this is lit

[glass kelp]

Remember

[lyric copper]

so now i just work with 3 sin x = cos x

[inner mirage]

No

[glass kelp]

One must not cancel blindly

[lyric copper]

huh 😭

[lyric copper]

then?

[glass kelp]

one must instead note that ${2\sin x - 1 = 0}$ is a possible solution

[rocky lotus]

k

[lyric copper]

OH

right

ofc

im so dumb 💀

[west cloud]

don't beat yourself up like that. these things take practice

[glass kelp]

aren’t we all? 🤔

[lyric copper]

thanks for those nice words

[lyric copper]

yall definitely arent

[west cloud]

🧢

[lyric copper]

wait imma try it out

[glass kelp]

https://tenor.com/view/savingprivateryan-ww2-old-soldier-matt-damon-gif-4203998

In the age of icl

[lyric copper]

💀💀🙏

[tidal swift]

only me

[glass kelp]

That’s the third cross today

@lyric copper u got it?

[west cloud]

^^

[glass kelp]

Oh ic ic

Thx

[vale dock]

@lyric copper Has your question been resolved?

Channel closed

Closed by @lyric copper

Use .reopen if this was a mistake.

[lyric copper]

y'all thanks sm i got

[lyric copper]

yesss thank u sm

[vale dock]

Available help channel!

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[west cloud]

!status

[vale dock]

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

[harsh jolt]

These are the potential graphs in this case

According to me

But I can't seem to find any common condition

Here

Also this

The 1 in the left graph is ahead of the root

[inner mirage]

There is 1 more graph which you forgot, which passes through both 0 and 1

[inner mirage]

!occupied

[vale dock]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[dire pecan]

can u answe it

[harsh jolt]

Ye

Forgot about that

[inner mirage]

I mean intersects the axis at both 0 and 1

[harsh jolt]

Ye got it

So ye what is the common condition binding them together

I can't seem to find any

[inner mirage]

There is no common condition

[harsh jolt]

So how do I solve this

[midnight pier]

When you’re given something related to roots of a quadratic that isn’t explicitly finding them, but related to the number of roots, or number of roots satisfying a certain condition, you may want to investigate the discriminant. Also, limits may be useful here.

[dire pecan]

are u a bot or human

[harsh jolt]

Ye location of roots but how exactly will limits be useful here

[inner mirage]

I am busy trying to help someone else, please open your own help channel

[midnight pier]

As in finding the behaviour of the function as a certain variable tends to either one of the endpoints of the proposed half open/open intervals

[dire pecan]

how??

sorry bro

[inner mirage]

(https://discord.com/channels/268882317391429632/488120190538743810)

[harsh jolt]

Ye only the roots are domain restricted tho the function is not

[midnight pier]

You’re asked for the range of k, so you may want to investigate what bound on the interval may cause the condition of exactly one root to lie in [0,1) to fail, e.g. what constraints apply

[inner mirage]

My math teacher actually gave an algorithm for this specific case, but I can't really explain it

[midnight pier]

Have you found the discriminant?

[harsh jolt]

Ye lemme try that although idk if it will help

[midnight pier]

I don’t have my iPad right now, so I can’t draw things for you (and struggle is part of the process).

[harsh jolt]

Ye this an awesome question

[inner mirage]

,rccw

[harsh jolt]

Why is d a e and beta

[inner mirage]

alpha and beta are the roots and d and e are end points of the interval which in this case are 0 and 1

[harsh jolt]

Alr

[midnight pier]

I believe that’s what Sirin meant as in “algorithm”. Without more context it’s difficult to interpret, but as I said, I suggest you find the discriminant and examine the behaviour of f(x) and your value for the discriminant using limits (as k —> some value). I think this just boils down to casework at the end.

[harsh jolt]

Ic well I'll get a 4th degree equation

If I take 2^k as t

[midnight pier]

Oh, actually, you may want to use the full formula, not just the discriminant

But this calculation is helpful anyway

Do you know the formula for the solution to a quadratic?

[harsh jolt]

Yes

[midnight pier]

Try finding that in terms of 2^k and then analysing the behaviour as k —> 0, inf, -inf, etc.

[harsh jolt]

Will this work tho as 0 is included here

[harsh jolt]

Alr

[inner mirage]

Ye it should

[harsh jolt]

Alr

[harsh jolt]

I tried it I just get

[inner mirage]

It should be less than 0

[harsh jolt]

Ye f(0) is -1

I skipped that part

Oh wait

Sorry

Alr I get this

[inner mirage]

Ok so we have $k \in (0,\infty)$

[rocky lotus]

Sarin

[harsh jolt]

Ye

[inner mirage]

Now we have to put the extreme values of this interval in the original eq. and check its roots

Obv we can't do infinity

[harsh jolt]

Alr so we put 0

[inner mirage]

Yes

[harsh jolt]

I get $x^2 -x -1$

[rocky lotus]

playboi69

[inner mirage]

What are its roots?

[harsh jolt]

$1 \frac {1\pm \sqrt 5}{2}$

[rocky lotus]

playboi69

[harsh jolt]

Ignore that one on the side

[zinc fulcrum]

teachnalliy it is irantional

[inner mirage]

Ok so now we need to check if there is exactly one root in the interval [0,1)

[harsh jolt]

Ye

There is

[zinc fulcrum]

no root from 0-1

[inner mirage]

,w (1 + sqrt 5)/2

[rocky lotus]

Results provided by WolframAlpha

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Requested by sarin.5063

[harsh jolt]

No wait

[inner mirage]

Ok so there isn't any root

[zinc fulcrum]

yea it is all iranntonal

[harsh jolt]

It is not there

[inner mirage]

which means the answer should be $k \in (0,\infty)$

[rocky lotus]

Sarin

[harsh jolt]

0 is included

In the answer

[inner mirage]

ah I see why

We got the quadratic wrong

[harsh jolt]

Oh ok

[inner mirage]

it should be $x^2 - x$

[rocky lotus]

Sarin

[harsh jolt]

Ah shit

Alr

So roots are 0,1 and that satisfies it

[inner mirage]

So there is exactly 1 root which is 0

[harsh jolt]

Ye

[inner mirage]

In this case, instead of (0,infinity), we change it to [0,infty)

[harsh jolt]

Got it

Is this algorithm applicable for any question which has bounded roots

[inner mirage]

wdym by bounded?

[harsh jolt]

Like you know exactly one root from [0,1)