help-13

[cedar kiln]

@cobalt plover Has your question been resolved?

[flat mica]

it is

conversion to barycentric coordinates, though that's not necessarily the best way to think of it

barycentric coordinates are coordinates relative to a reference triangle of some kind

examples from wikipedia

of reference triangles

[cedar kiln]

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[gentle lintel]

I've arrived at a divergent integral

[gentle lintel]

When doing differentiation under integral sign, can it happen if I've chosen an inappropriate I(α) even after perfect working

[cedar kiln]

@gentle lintel Has your question been resolved?

[cedar kiln]

@gentle lintel Has your question been resolved?

[cedar kiln]

@gentle lintel Has your question been resolved?

[crimson sedge]

U can use the integration by parts its easier

[gentle lintel]

using duis was required

But yeah i see how

[crimson sedge]

Aight lemme see

[crimson sedge]

I found it

Well
What about changing ur alpha function to sum easier

[gentle lintel]

that's my question

[crimson sedge]

U want the answer?

Or just clue

[gentle lintel]

I'm wondering if certain I(α) functions lead to divegent answers regardless

or whether I've done something silly

[crimson sedge]

Nooope what uve done is correct

[gentle lintel]

So some I(α)s simply won't work then?

That's strange

[crimson sedge]

Lemme use latex

[crimson sedge]

Here are the 2 rules to use duis

Or u can use differential if u want
But i found the continuity is easier

Yes i checked
The differntial also is diverge

Therefore u need to check for another form

[edgy spade]

Can I ask why are you doing feynmans?

Isn't this solvable by IBP?

Yes it is solvable by IBP

@gentle lintel

Oh wait somebody already mentioned it

I see you have to solve using a separate method

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[hallow scarab]

I want to show that f(T_1) = T_2 -> f is a homeomorphism, since f is bijective, it's obvious that it has a continuous inverse, but I stuck at proving f is continuous, how to show that for any V in T_2, f^{-1}(V) is in T_1?

[cedar kiln]

@hallow scarab Has your question been resolved?

[hallow scarab]

<@&286206848099549185> XME?

[hallow scarab]

IMHO it's completely fine that there exists a V in T_2 such f^{-1}(V) is not in T_1

[cedar kiln]

@hallow scarab Has your question been resolved?

[unborn mirage]

What's your definition of a continuous map? Because what you said doesn't work

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[torn marsh]

am I correct? I said A<X<C or C<X<E

[torn marsh]

for b)

[slate lintel]

yes that's fine

[torn marsh]

but my answer sheet said something else

[slate lintel]

did it say the graph was also decreasing at point C?

[torn marsh]

and i dont understand why

[torn marsh]

u could say tha

[slate lintel]

because that is true but i didn't expect it to be what your answer sheet said

[torn marsh]

It said "Between A and E"

[slate lintel]

okay yeah

what does "decreasing" mean

[torn marsh]

when the y value decreases

[slate lintel]

um... what does that mean? like if i said the function was decreasing at C, what does that mean

[torn marsh]

its gettting lower?

[slate lintel]

actually let's look at B because it's 'obviously' decreasing there

[torn marsh]

yep but why did it say between a and e

c wasnt decreasing at the origin

[slate lintel]

what does this mean? aren't we just talking about the point B? the function just has one value at that point

[torn marsh]

yes it has one value at that point but as as it approaches c, it decreases

[slate lintel]

okay, you're getting warmer, i like the "approaches" term

[torn marsh]

appreciate it

[slate lintel]

usually what we say is that stuff on the left is higher than the point, and stuff on the right is lower than the point. does that make sense? does that seem like a reasonable definition for "decreases"?

again, as we zoom in close enough

[torn marsh]

Yep

because the line is just a bunch of

[slate lintel]

okay. so using that definition, what's going on with C?

[torn marsh]

points

the stuff befiore C is

higher

the points after C is lower

[slate lintel]

yea exactly

so the function is decreasing at C

[torn marsh]

where did u get this definition of decreasing?

I was never taught of it

[slate lintel]

what were you taught?

[torn marsh]

because it was self explanatory

like the actual meaning of decreasing

[slate lintel]

which is...?

[torn marsh]

this

[slate lintel]

yes that's the definition i was taught

what were you taught?

[torn marsh]

i was never taught

about that

[slate lintel]

did they just use words like "decreasing function" and expect you to intuitively know what that means?

tsk tsk

[torn marsh]

yep

😭

[slate lintel]

you are right that the function looks flat there

and that's important for derivative reasons (the derivative is 0 there)

but the function very much is decreasing at that point

which is why your answer sheet says it's decreasing on A < x < E

[torn marsh]

understandable

does that make my original answer correct as well

[slate lintel]

well no, because you said that it's decreasing on A < x < C union C < x < E

which misses the point C

having said that it's a very common error and i would expect others in your class to make the same error

[torn marsh]

understandable, that makes very sense now thank u

my other question is

could we say

y"(2)

cuz i found the seecond derivative

but im tryna take the second derivative of a particular input

[slate lintel]

yeah that's valid

well okay y''(2) is a little bit of abuse of notation

i'd probably say y'' at x = 2 if i was being precise

but it's fine

[torn marsh]

ill prob stick with y" at x = 2 just to avoid uneccesary lose of marks

[slate lintel]

since you have y = f(x) you can say f''(2) if you want

[torn marsh]

but i didnt write f(x)

or f'(x)

it was all y in each line

[slate lintel]

if they take off points for y''(2) they're a nerd

[torn marsh]

fax

tyy

.close

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[short hatch]

[
\lim_{x \to 0} \left( \frac{\ln(1 + x^2)}{(\sin x)^2} \cdot \cos\left( \frac{1}{x} \right) \right)
]

[wraith dagger]

Koren

[short hatch]

the sin and ln stuff is 1 and cos(1/x) dne, what justifies all of the expression dne

limits arithmetic?

[azure swift]

There isn’t really any limit rules for that

Unless you have some in your notes or textbook that are special

A good way of arguing if you think it’s dne is to find two different set of points which approach 0 but give different values at the limit

Or if it approaches -infty or infty; argue by unboundness

[short hatch]

you mean use heine?

[azure swift]

Look at say the points x = 1/(pi*n)

Just an example

[short hatch]

yeah i understand, we used that

[azure swift]

With n going to infinity

[short hatch]

it's going to 0

so we plug that in

[azure swift]

Mhm

[short hatch]

and then cos(pi*n)

its (-1)^n

[azure swift]

Mhm

[dreamy void]

You could multiply by x²/x² then you can use the limits sin(u)/u and ln(1+u)/u and cos(1/x) is bounded basically

[short hatch]

so we should do 1/(2 \pi n)?

[azure swift]

Don’t need to

[short hatch]

this part i understood

oh cause i know (-1)^n diverges

[azure swift]

You get something that behaves like (-1)^n and that’s not convergent

[short hatch]

it's a known limit

[azure swift]

Yes

[short hatch]

yeah

so i finished or i need another set of points?

to see theyr'e diffrent

[azure swift]

I think that would be enough

Maybe give more arguments for why it’s like (-1)^n near infty

[short hatch]

unit circle?

and periodicity of cos?

wait

by doing the x_n=1/(pi n)

i just prove cos(1/x) isn't convergent

i want to prove that
non convergent * convergent = non convergent

[azure swift]

Yeah good that might be easier

[short hatch]

but how do i do that :(

[azure swift]

Or atleast for something that is like (-1)^n

Okay let’s just do it concretely

Our factor besides (-1)^n is ln(1+x^2)/sin^2(x) (I’ll keep x here for a moment)

[short hatch]

i think i should do
x_n = 1/(2pi*n) and y_n = 1/(2 pi n + pi) and show this gives 1 and this gives -1 thus the limit doesnt exist

is this good?

[azure swift]

Yes

I’m attempting to show those parts here

But for when n is even or odd

sin x is x near 0

[short hatch]

same as ln(1+x) and x

[azure swift]

So ln(1+x^2)/sin^2 x is ln(1+x^2)/x^2 ner 0

Yes

[short hatch]

you can multiply by x^2/x^2

i think i got it

[azure swift]

And then u can just argue for when n is odd or when it’s even

[short hatch]

tysm

.close

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[hollow wind]

For a ti83 how do I know what to enter into a function without looking it up? For example if I want to use "binompdf(n,p,c)" or any other similar type of function, is there anyway to know what it wants me to enter without looking it up on the internet? I'm just trying to cover my basis for if I forget the order of things to enter during a test

[tropic oxide]

you should be able to look up a user manual for the TI83

[cedar kiln]

@hollow wind Has your question been resolved?

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[fervent sparrow]

I was reading the letter "COPY OF A LETTER FROM SIR WILLIAM R. HAMILTON TO JOHN T. GRAVES, ESQ" in which Sir Hamilton was explaining his thought process and how he came to discover quaternions. I read the paper in hopes of acquiring a better intuition. I have seen 3Blue1Brown's video about visualizing quaternions using stereographic projection, while it did provide a great perspective, it didn't settle with me.

From the letter, I found out that Hamilton focused initially on creating 3 dimensional numbers, he called them triplets. His aim was to make them satisfy a product rule in which the magnitude of the product is equal to the product of the magnitudes of the factors. He couldn't do that with 3 dimensional numbers so he had to add another dimension in order for things to work out.

Only after he finished sorting the algebraic properties out, did he start to concern himself with the geometrical meaning. The equations in the image offer great insight. Basically, it is the trigonometric elements of a hypersphere. Where the sine element of the of the hypersphere radii or the modulus mu is a 3d vector. Or commonly written in modern vector notation: q=cos(theta/2)+sin(theta/2)u

But again, I am not very great when it comes to thinking of 4 spatial dimensions. So instead of thinking of it as the elements of a hypersphere radii, cant we just think of 3 concentric circles? If mu sin(rho) is a vector in 3d space, then cant mu and mu cos(rho) also be represented in 3D space in a circle on the plane that is parallel to mu sin(rho)?

[cedar kiln]

@fervent sparrow Has your question been resolved?

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@fervent sparrow Has your question been resolved?

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[violet drum]

hiii not an actual problem question

[violet drum]

but how essential is trigonometry

[lone dune]

pretty essential

[slender ginkgo]

Very

[violet drum]

as in law of sines and cosines

[slender ginkgo]

VERY INPORTANT

[lone dune]

very essential

[violet drum]

double and half angles

[lone dune]

yea

[violet drum]

in calculus?

[lone dune]

well if youre thinking of skipping them dont

yes

[slender ginkgo]

Not much

These formula not much

But the idea of trig? A lot

[lone dune]

they come up every once and then in integrals

[violet drum]

Im not thinking of skipping them

I js wanna know if ill even use them in calc

[lone dune]

yes

[violet drum]

and if I should master em completely

[lone dune]

youll use a lot of trig

[violet drum]

cause Im doing a school course 💔

for precalc during summer

[lone dune]

and a good amount of the formulae

[slender ginkgo]

Ap?

👀

[violet drum]

yep Ap calc BC

[lone dune]

well it doesnt hurt to learn

yes u need trig

the idea of trig

[slender ginkgo]

I’ve played this game before

[lone dune]

helps a lot

[violet drum]

ik im js time managing so if I can get ahead with calc topics ill do that

cause yk summer more energy

ez

[lone dune]

yea

[slender ginkgo]

U need some basic trig idens

[lone dune]

i did that

but

learn trig and stuff first

before self studying calculus

trust me

[violet drum]

ok ok

[slender ginkgo]

Double angle and inverse trig

[lone dune]

i made the mistake of trying to learn calculus before even finishing trig and precalc

[slender ginkgo]

But overall, u need trig yes

[lone dune]

and everytime something trig related came up i was clueless

which was pretty often

[slender ginkgo]

Before precalc 💔

[violet drum]

also that weird circle with random roots for radians its actually so down bad

[lone dune]

dude i looked at the first few chapters and thought it was the same as alg 2

[violet drum]

its actually making me stress out ngl

[lone dune]

the unit circle?

[slender ginkgo]

Scammed

[violet drum]

yes

[lone dune]

its pretty useful

just do trig before calc and save urself the time later

[slender ginkgo]

Ap calc ain’t that much compared to uni calc imo

No need to stress out

[lone dune]

arent they almost the same

mostly

[slender ginkgo]

They didn’t include epsilon-delta my g

[lone dune]

except that

yea

[slender ginkgo]

No Cauchy mean value theorem

[lone dune]

what the hell is that

[slender ginkgo]

No rigor

[lone dune]

[slender ginkgo]

Oh dear

[lone dune]

ive always thought of it as calc 1 is analogous to calc ab and calc 2 is analogous to calc bc

but the uni courses have a bit more

[slender ginkgo]

Ye

[violet drum]

we only have calc AB and BC

I take calc Bc to get over with it and js take stats on my last year

[sand cradle]

There is a reason it's included in pre-calc

[lone dune]

yea thats how every highschool does it

[sand cradle]

They are directly related to polynomials and rational expressions

[slender ginkgo]

Huh

[versed coral]

i mean

[slender ginkgo]

I’ve only seen trig combined with polynomial in math competitions

[versed coral]

i NEVER learnt nor drew nor looked at that circle

more than once

the more u practice the more u will get used to it tho

u will learn what each value corresponds to

[violet drum]

real but this course is using it in the most random topics haha

[versed coral]

wdym combined

[violet drum]

like polar plane uses some

[sand cradle]

I should clarify, I meant the series definition of sin and cos of which the partial sums are polynomials

[slender ginkgo]

Ahh

[lone dune]

long story short just educate yourself on trig and precalc before self studying calc

[sand cradle]

And that int 1/(1 + x^2) dx = arctan(x) (+ C)

[lone dune]

[violet drum]

what do u even study in pre calc completely

[versed coral]

oh

i see now

[lone dune]

polynomials and a bunch of rational functions

trig

polars

[violet drum]

like on the first half I saw mostly algebra

[versed coral]

i love trig

[lone dune]

sequences and series

[violet drum]

and a bit of trig

[slender ginkgo]

Like show that sum of some sun angles is this

[lone dune]

matrices sometimes

[slender ginkgo]

U use vieta

[lone dune]

but that doesnt show up in highschool calc

[violet drum]

2nd Ive been js doing the sequences matrices bectors

vectors

[dusty wren]

?

[lone dune]

wait

[sand cradle]

Did I say something wrong?

[lone dune]

its +

not -

[dusty wren]

1/(1+x^2)

[lone dune]

i just realized lol

[sand cradle]

Sorry, I meant to say +

[slender ginkgo]

Where + c

-1/2 point

[lone dune]

👹

you right

[dusty wren]

This is why i dont give written exams

[versed coral]

• 🧀

[dusty wren]

Sorry i reacted too aggressively lol, meant to put one ? not three, mb

[lone dune]

@violet drum if you do end up self studying calculus go check out khan academy and the organic chem tutors videos

they explain a lot of things well

even for precalc and trig

[violet drum]

I was also thinking on buying 1 single book ngl

that has tons of problems

and some explanations

[slender ginkgo]

Like this: Show that
[ \cos\left(\frac{2\pi}{7}\right)\cos\left(\frac{4\pi}{7}\right)\cos\left(\frac{6\pi}{7}\right) = \frac{1}{8}]

[wraith dagger]

k

[lone dune]

you could

well

[slender ginkgo]

free book online

[dusty wren]

Stewart or spivak are good

[violet drum]

and doing the free resources

[slender ginkgo]

(not promoting piracy btw)

[dusty wren]

For all of calc

[violet drum]

or that

[lone dune]

if your goal is to just self study ap calculus for school

just get an ap exam review book from princeton or something

and youre set

lol

[slender ginkgo]

yes

[violet drum]

real

[lone dune]

thats what i did

they have a LOT of practice problems and tests

[slender ginkgo]

same😭

just one book

didnt even use the online classroom

[lone dune]

yea

[slender ginkgo]

my attention span aint that long

[lone dune]

funny story actually my neighbor had a garage sale and gave away some calc bc review book when i was a freshman

so i took it for shits and giggles

and then decided to self study because i thought it would make me cool

[slender ginkgo]

it does make u look cool

[dusty wren]

Atleast u stuck with it

[slender ginkgo]

like those studygram tiktok people

[dusty wren]

Most of these stunts i pull i abandon halfway

[lone dune]

yea thats like what got me into math in the first place

[lone dune]

thats going to be me with physics

[slender ginkgo]

ap physics c?

[lone dune]

yea

i "self studied" ap physics 1

but im taking it in school this year too

[slender ginkgo]

cool

[lone dune]

yea

anyway

@violet drum u can close this now

[slender ginkgo]

princeton review 🔥

[sand cradle]

I'd leave matrices to linear algebra, you'll probably have taken some form of that when you arrive at multivariable calculus

[lone dune]

yea they are my lord and saviors

[lone dune]

yea true

matrices dont show up until multivar calc

in terms of the calculus courses

[violet drum]

we have a whole unit on those 💔

and they r fun ngl

for some reason I like em more than trig

[sand cradle]

I mean, it will help you in linear algebra

[slender ginkgo]

linear algebra is fun 💔

what subspace are we living in

[sand cradle]

in the kernel of a rank-one map

[violet drum]

If calculus doesn't take my soul Ill probably self study linear algebra on my own time

[lone dune]

self studying linear algebra isnt that bad idt

[violet drum]

LIKE WHEN I FINISH CALC

[slender ginkgo]

calc wont

linear algebra will

[lone dune]

yeah

oh

i wouldnt know

[violet drum]

💔

[lone dune]

i went straight to multivar calc

khan academy is carrying me through it

[violet drum]

💀

i dont even know what that is

[slender ginkgo]

its bad for me cuz i tried to do it in 2 days and fried my brain

[lone dune]

multivariable calculus

like calculus 3

[lone dune]

do what

multivar?

[slender ginkgo]

linear alg

[lone dune]

oh

[sand cradle]

I find it a bit weird to go through the entire calculus sequence and afterwards analysis, here we take analysis right away

[violet drum]

WYM DO IT IN 2 DAYS!?

like the whole thing?

[slender ginkgo]

thats why i cant do linear alg properly

it doesnt work

[lone dune]

analysis isnt tied to calculus though is it

it is its own separate thing

[sand cradle]

analysis is calculus with proofs

[lone dune]

oh shit yea

[slender ginkgo]

stopped at transformation of space cuz my brain

[lone dune]

epsilon delta stuff

[slender ginkgo]

was gone at that point

[lone dune]

i was planning on self studying multivariable calc and linear algebra by the end of hs

[sand cradle]

It goes every essentially everything you'd do it in calculus while being rigorous

[lone dune]

and starting with analysis and other further math in uni

[violet drum]

real

[sand cradle]

That's why I find it a bit weird to do first spend so much time on calculus (sorry I was referring to my own message, not replying to you)

[slender ginkgo]

Can’t remember shi from that learning experience 💔

[lone dune]

oh i see what u mean

[violet drum]

yall follow the math sorcerer

[slender ginkgo]

Nah

[lone dune]

analysis looks way more complex than just calc though

or atleast for me

but the only gist of it ive had was epsilon delta proofs for limits

[sand cradle]

It's really the same, but rigorous, no handwaving

[slender ginkgo]

Series and sequences too, I think?

[sand cradle]

Building up everything logically

[lone dune]

but being rigorous in general seems tricky to me because you needa think

[lone dune]

nope

[slender ginkgo]

Which imo makes it feel like pain

[lone dune]

like when you write proofs you need to reason for each step

cant just say something like "trust me bro" and let it slide

[sand cradle]

You can if both you and your audience know it obviously holds

A proof is really just like a conversation

[slender ginkgo]

Sequence and series isn’t in ra??

[sand cradle]

But you should justify the large steps lol

[lone dune]

well yea you can cite theorems and stuff

but the main process is still you trying to justify something

rigorously

but it also looks really cool

and elegant

[sand cradle]

What I also find weird is the length of some of these intro to proof-writing books

I don't think it's very effective to read 400 pages on how to write proofs

[lone dune]

😭

[sand cradle]

Instead, just start real analysis or (abstract) linear algebra and you will learn that along the way

[lone dune]

dont they come with problem workthroughs and stuff

[sand cradle]

Sure but I mean, your analysis book probably has some worked out examples too

[sand cradle]

(Still it is probably a good idea to get familiar with proofs beforehand, for example by reading the 30-page document Loch gave)

[lone dune]

where can i find that

[sand cradle]

#proofs-and-logic in the pins here

[lone dune]

ah okay got it

[sand cradle]

#proofs-and-logic message

[lone dune]

but yea in general i feel like learning by just reading a book explain it then doing it yourself would be kinda tricky to visualize and stuff

[sand cradle]

All my analysis 1 books would include something like this at the beginning though anyways

[lone dune]

like the book would probably explain it intuitively

but just getting a visual understanding of it would help a lot

or atleast thats how i feel

like for me if it werent for khan academys videos with visuals on whatever multivariable calc concepts they were teaching, i definitely would not have been able to understand those topics as well as i do

[sand cradle]

True

[cedar kiln]

@violet drum Has your question been resolved?

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[spring forum]

<@&268886789983436800>

[cedar kiln]

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[quick pewter]

Hello. I have a question about K-Means clustering.

[quick pewter]

I have 26 pieces of data that make up a single record. I then have 100,000+ records in this data format.

I attempt to K-means cluster the data. After using PCA(2), the graphed result does not form "clusters".

How do I fix this problem?

I can share an image of the graph i am getting if that is helpful.

Example of graphed results:

@gleaming path

[spring forum]

!noping

[cedar kiln]

Please do not ping individual helpers unprompted.

[spring forum]

also, for this you might want to ask in #theoretical-cs , because people who are proficient in that area are more active there

(note that to access the channel, you'll need the undergraduate math role from id:customize )

[quick pewter]

Thank you for responding.

[cedar kiln]

@quick pewter Has your question been resolved?

[limber monolith]

ola

[cedar kiln]

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[glacial moth]

I got the right answer but I am not sure if I solved the problem in the correct way. Can someone please tell me if what I did is all good?

[exotic furnace]

Looks good to me

[glacial moth]

thank you

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[still thorn]

I'm trying to simplify this into a term without a summation

[still thorn]

I know that the infinite summation is equivalent to e^N, but I can't see how to apply that to this problem

[slender ginkgo]

Remainder theorem should give u a fair approximation?

[still thorn]

the problem wants an exact expression

is simplifying this impossible and I've went wrong somewhere else?

[slender ginkgo]

I wa’ed it

The result has a gamma function

[still thorn]

hmmm ok thanks

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[hallow scarab]

for a map f: U -> V to be continuous, for any open subset T in V, the set f^{-1}(T) should also be open, it's the definition from the textbook Introduction to Topological Manifolds

[cedar kiln]

@hallow scarab Has your question been resolved?

[void sand]

my suggestion is to identify a U such that f(U) = V

then you can use the hypothesis that f(T_1) = T_2 to conclude

[void sand]

nothing to do with topology, just set theory

[hallow scarab]

sorry but I didn't get it, for such a U what can I do?

[cedar kiln]

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[void sand]

.reopen

[cedar kiln]

✅

[void sand]

can you tell me what U satisfies f(U) = V?

again, just set theoretically, don't think about topology

f is a bijection

[hallow scarab]

yes, then this is automatically satisfied

[void sand]

that's not a good enough argument

f being a bijection serves a purpose here

it's going to allow us to find this U

it does not give us the result immediately

[hallow scarab]

yes, for any V there exists one and only one such a U

[void sand]

sure, what U is it?

[hallow scarab]

f^{-1}(V)?

[void sand]

there you go!

f being a bijection means that f(f^-1*(V)) = V is true

now can you use the hypothesis that f(U) is in T_2 if and only if U is in T_1 to conclude that f is continuous?

[hallow scarab]

I see! Thank you very much~😀

[hallow scarab]

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[cunning plover]

What do I do? Im stuck

[void sand]

a couple things to say

your derivative for the numerator is a bit off - you need to use the chain rule

as for your actual question, why not consider evaluating that function at t = 0?

[cunning plover]

How do you use the chain rule for this problem? And do you mean plugging in 0 for evaluating? I got 0/0

[void sand]

to question 1, e^2t is a composite function with inner function 2t and outer function e^t; you’ve got to use the chain rule when calculating its derivative

to question 2, I meant plugging in 0 for the new quotient, after you’ve L’Hopital’d it

[cunning plover]

So the chain rule would be(2t(e)-1)/sin (t) then take the derivative of it?

[void sand]

L'Hopital's rule says that $\lim_{t \to a} \frac{f(t)}{g(t)} = \lim_{t \to a} \frac{f'(t)}{g'(t)}$

[wraith dagger]

higher!

[void sand]

my point is that you've calculated f'(t) incorrectly

because you didn't use the chain rule when differentiating e^2t

[cunning plover]

oh ok

thank you

.close

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[void sand]

ah, also. welcome to the mathcord! @cunning plover

[hallow scarab]

hi, the latex equations are automatically compiled here?

[void sand]

@wraith dagger is in the server, so yes

you can test it out in #latex-testing

[hallow scarab]

gotcha

[cedar kiln]

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[vale depot]

I know I have A and B values wrong I just don't know where I went wrong and what to do

[vale depot]

,rccw

[high bridge]

x cannot be equal to 1/2

that is not in the domain of the original function

even if you sub 1/2 into the A(1-2x)^2 thing

you get 0A+0B=2

which does not make sense

what you might need to do is sub in x=0 and then x=1 to solve a system

[vale depot]

I'm on a lil caffeine crash, explain it to me like I'm 5

Did I atleast get the a/thing + .... right

[high bridge]

you got everything up to the second last line correct

A(1-2x)^2+B(1-2x)=2x+1 is correct

now you just gotta find A and B

[vale depot]

Okok

[hallow scarab]

I think the integral should be $\int \frac{A}{1-2x}dx = -\frac{A}{2}ln(1-2x)$

[wraith dagger]

奇偶奇偶

[high bridge]

奇变偶不变?

anyways

[hallow scarab]

jojo的谐音

[high bridge]

ahhh XD

it is not A/1-2x

you need a A/(1-2x)^2 term

[hallow scarab]

ah yes, sry

[vale depot]

So im subbing in x=1 and 0

[high bridge]

yeah that works

[hallow scarab]

it's $A(1-2x) + B(1-2x)^2 = 2x + 1$ right?

[wraith dagger]

奇偶奇偶

[high bridge]

OHHH WAIT

I AM STUPID

@vale depot line 4 is wrong

[vale depot]

😭

[high bridge]

since you multiply by (1-2x)^2

you are left with A(1-2x)+B=2x+1

not B(1-2x)^2

just B

[hallow scarab]

yes, I am really stupid

😅

[vale depot]

Now I'm more confused

[high bridge]

you have A/(1-2x)+B/(1-2x)^2=(2x+1)/(1-2x)^2 right?

[vale depot]

I do not

[high bridge]

???

[high bridge]

that is what you should get in partial fraction

[vale depot]

I had A/(1-2x)^2 + B/(1-2x)= 2x+1

[high bridge]

?

okok let's go back to square one

[vale depot]

Please

[high bridge]

you have (2x+1)/(1-2x)^2 right?

your goal is to make that into A/(1-2x)^2+B/(1-2x) right?

[vale depot]

Yes

Well I have the (1-2x)^2 under B but I assume it's not difference

[high bridge]

no difference

sure lets get it that way then

[vale depot]

Kk

[high bridge]

you make (2x+1)/(1-2x)^2 into A/(1-2x)+B/(1-2x)^2

therefore:

(2x+1)/(1-2x)^2=A/(1-2x)+B/(1-2x)^2

agree?

[vale depot]

Gimme 5 secs to write it down on paper

[high bridge]

k

[vale depot]

Sure

[high bridge]

k you got it

?

[vale depot]

I think so

[high bridge]

k now you multiply the equation with (1-2x)^2

you will get 2x+1=A(1-2x)+B

[vale depot]

OHHHHH

Okok

[high bridge]

and now you can sub to find out what A and B is

[vale depot]

Let me try

[high bridge]

k

[vale depot]

So to find B, A must be multiplied by 0 right?

[high bridge]

I mean

not neccesarily, but that does make it easier

[vale depot]

So can I sub x= 1/2?

[high bridge]

yeah

[vale depot]

But how do I find A if I can't sub x to make B = 0

[high bridge]

I mean

solve for B first?

and then sub in an x to NOT make 0A?

[vale depot]

So just anything that isn't 0A

But if I do x=1 then A is -1 and if I do x=2 A = -3/4

[high bridge]

that probably means you solved B wrong

XD

or like something else went wrong

what did you get as B

you should have B=2

[vale depot]

I do

[high bridge]

ok

2x+1=A(1-2x)+2

when x=1

3=-A+2
A=-1

when x=2

[vale depot]

Yup

[high bridge]

5=-3A+2
A=-1

I don't see what is wrong with that XD

[vale depot]

Fml

I wrote 5=A(-4)+2

[high bridge]

oof

[vale depot]

I'm a lost cause

Ok I can do it from here

Ty ty

Thabks for patience

[high bridge]

np :>

[vale depot]

.close

[cedar kiln]

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[calm berry]

hello, posted similar problems yesterday (im sorry) and i want to check if these are right

while im here, i want to ask if anyone has a solution manual to judson's abstract algebra theory and applications? (just to streamline the process a little)

feel free to ping btw

[cedar kiln]

@calm berry Has your question been resolved?

[cedar kiln]

@calm berry Has your question been resolved?

[humble karma]

For your second inquiry, do note that except in perhaps very specific instances, like resources dedicated to self-study, or more computational subjects, college-level (and above) maths textbooks seldom have dedicated solution manuals.
If they do, they're often hard to find, and might not even contain solutions to each and every one of the exercises.

If you can find one, and it does help you in checking your answers, that's fine. You should however be wary of "giving up" too early and peaking at solutions too fast, which doesn't really help you learn. Do note that for more entry-level textbooks (like first courses in abstract algebra), most of the exercises will have already been solved online in one way or another (ahum mathSE).

I'll take a look at your proofs for this exercise over the next hour.

[calm berry]

Appreciate the advice thanks

[wraith dagger]

Azyrashacorki

[humble karma]

As for problem specifics :

For the second part of a), in here you're missing a symbol on your third line (nitpick), but more importantly, as obvious as the step might be, you're taking a leap from line 3 to 4. It could be worth explaining why you get there with cases.
Also, you directly conclude but it would be nice to say that this x is in A1uA2, and thus that f(x) = y is in f(A1uA2) in order to close the loop for the backward inclusion.

b) looks fine except for the general issue I pointed out

[calm berry]

thanks for the help

[calm berry]

mmm alright

i see

i kind of get this

[calm berry]

so something like this would be more appropriate?

[humble karma]

Yeah that seems good

[calm berry]

ok thanks

[humble karma]

c) your last try is correct, only again the reordering stuff (just in the first inclusion, for the second you did it in the right order and the logic seems fine).

I didn't go through all of d) but the argument is similar to c), so you can just reorder and you should be good

[calm berry]

can i show u an example of the reordering just to make sure i get it

[humble karma]

Yes sure

e) looks good. You can also try to prove it without using the previous exercises for practice, but it's a good thing you used it so

[calm berry]

this is for part a)

[humble karma]

Only thing is f(x) is in f(A1) or f(x) is in f(A2)

Otherwise it's much better

[calm berry]

what about this

[humble karma]

Like this line

A1 and A2 are subsets of X, while f(x) is in Y

[calm berry]

oh wait im an idiot yes ofcourse

[humble karma]

I mean it's clear what you were thinking but it wouldn't be right. The rest seems good to me!

[calm berry]

thanks a bunch

thanks for the help

.close

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[abstract ruin]

How i do these

[slender ginkgo]

what have u done so far

[abstract ruin]

none. my teacher taught me while my classmates answered but i forgot

math js aint for me 😭

[slender ginkgo]

it aint for anyone🥀

anyhow

x-intercept -> f(x) = 0

y-intercept -> f(0)

horizontal asymptote -> the line that the function is approaching as x is very very large (positively or negatively)

vertical asymptote -> the line that the function is approaching as y is very very large (positively or negatively) - usually at the point where the denominator is 0 but the numerator is not

using this info, can u solve 1.

[abstract ruin]

i cannot 😭

can i get a vid of how to do this

like this lesson specifically

[slender ginkgo]

https://www.youtube.com/watch?v=XE-Z2-F3oWw

[frail jolt]

what is problem?

[slender ginkgo]

op

op doesnt know where to even start

[abstract ruin]

u get it

[slender ginkgo]

if u dont like this vid

u can try searching 'rational functions'

thats the topic

[abstract ruin]

Kk ty

[lusty osprey]

try and draw the functions

if you can't approach a problem break it down

and if there's something fundamental that you still don't know then you have a more specific problem to solve

here you need to first make sure you understand what asymptotes and intercepts are

[cedar kiln]

@abstract ruin Has your question been resolved?

[pastel vault]

there's three separate cases you need to learn from this

1. degree of numerator = degree of denominator, so questions 1 and 4

vertical asymptote is when denominator = 0

horizontal asymptote occurs when you take the limit to infinity, so $\lim_{x \to \infty} \frac{4 - 2/x}{5 + 12/x} = \frac{4 - 2 \cdot 0}{5 + 12 \cdot 0}$

2. degree of numerator < degree of denominator
   coincidentally that matches up with question 2 (and 5), so the vertical asymptotes are when 3x^2 + 5x - 2 = 0
   the horizontal asymptote will be just y = 0

3. degree of numerator > degree of denominator, so question 3

you still have a vertical asymptote when denominator = 0, but you have no horizontal asymptote

instead, the asymptote is that of a straight line: for example $\lim_{x \to \infty} \frac{x^2 - 3}{x + 1} = \lim_{x \to \infty} \frac{x - 3/x}{1 + 1/x} = \frac{x}{1}$ is the slant asymptote

[wraith dagger]

south

south

[cedar kiln]

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[ancient iris]

find the whole number parametars a,b,c such that (x-a)(x-6)+3=(x+b)(x+c) has infinite many real solutions

[cedar kiln]

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

[ancient iris]

i dont get this problem and the solutions are (a,b,c)=(10,-9,-7)

1

[gritty galleon]

well i am sure you know that a polynomial equaiton of finite degree n has atmost n real solutions

[ancient iris]

yes

[gritty galleon]

but ur told this has infinite solutions right?

[ancient iris]

yes,that is what the problem states

[gritty galleon]

meaning it cant have a finite number of solutions

[ancient iris]

yes

[gritty galleon]

and thus it cant be a polynomial equation

[gritty galleon]

nice, now expand

[ancient iris]

x2-6x-ax+6a+3=x2+xc+bx+bc

ok x2 cancels out

[gritty galleon]

yes

to make it into a non-polynomial equation, what else should cancel out?

[ancient iris]

idk?

the 3 i am guessing

[gritty galleon]

not quite

what makes a polynomial equation a polynomial equation?

variables right?

[ancient iris]

yes

then the x's?

[gritty galleon]

yes

good

[ancient iris]

how do i need to add something cause i cant see them canceling

[gritty galleon]

well ur supposed to find a,b and c

[gritty galleon]

so if u find the right a,b and c u can force the x on both sides to cancel out

[tropic oxide]

not sure this is a good angle of attack?

[ancient iris]

can i set an equation or I need to guess

[gritty galleon]

set an equation ofc

[gritty galleon]

what else would u do here

[tropic oxide]

i mena the idea is obviously that the two sides should be identical as polynomials

im just not sure that i like the idea of casting the goal as "reaching a non-polynomial equation"

[ancient iris]

i can just set x=6 then

ok one solution is (6,-5,-3),(a,b,c)?

[tropic oxide]

you had this equation, yes? $$x^2 + (-a-6)x + (6a + 3) = x^2 + (b+c)x + bc$$

[wraith dagger]

Ann

[tropic oxide]

yes i can see the original just fine

im just confirming that you did the work expanding the brackets on both sides

[ancient iris]

yes

and x^2 cancels out

[tropic oxide]

sure.

ok, and we want this equation to have infinitely many real solutions.

do you understand that this is only possible if both sides actually represent the same polynomial of x?

[ancient iris]

yes just like x=x

[tropic oxide]

meaning that the coefficients have to match

so you have:

• -a - 6 = b + c
• 6a + 3 = bc

you need to find integers a, b and c that satisfy both of these equations

i feel as if @gritty galleon was trying to lead you to this but then led himself and you somewhere else

do you see where to go from here?

[daring linden]

These are not whole numbers tho

[ancient iris]

i suppose adding/subtracting

[ancient iris]

what

[tropic oxide]

the original problem says integers

[daring linden]

Oh ok

[daring linden]

I saw from here

[tropic oxide]

@ancient iris celobrojni means $\bZ$, right? not $\bN$

[wraith dagger]

Ann

[ancient iris]

yes

[tropic oxide]

yeah ok

[tropic oxide]

not quite. this isn't a linear system

it's going to be a bit more complicated to solve than that

but one thing you CAN do is substitution

[ancient iris]

quadratic quation i guess is

[daring linden]

Yeah hopefully it factors

[tropic oxide]

specifically i recommend isolating a from the first equation and substituting it into the second

[ancient iris]

what i can assume is that this problems are for my next year of math competition so ig it is quadratics and vietovs formulas

[daring linden]

I think quadratic part is done

[tropic oxide]

vieta's formulas

anyway this won't be a simple quadratic, sorry

[opaque root]

Still there are more integer solutions

[tropic oxide]

yes there are, i found all of them myself just now

[ancient iris]

x1+x2=-b,x1x2=c

where x1 and x2 are the solutions

[tropic oxide]

with all due respect this is not going to be relevant here

[ancient iris]

it's better that way

[tropic oxide]

...

[daring linden]

4 sets of solutions

[ancient iris]

a^2+6a+33=b2+bc+c2?

[clear apex]

28 Deaths in Connecticut

[daring linden]

#discussion

[ancient iris]

i did something wrong prob,no way am i going to solve this

[daring linden]

See in these integer problems you usually try to get a form like (a+k)(b+l) = m where k,l,m are constants so that we can see factors of m and make cases

[daring linden]

So try to make a form like that with these equations

[tropic oxide]

how did this happen...

[ancient iris]

squared the first one and and subratracted the second haha

[tropic oxide]

....... no good, sorry

[gritty galleon]

😅

[tropic oxide]

ok maybe i do not know how to explain it without risking that you'll spend 4 hours doing useless things that both don't help you solve the question AND don't even learn anything

[opaque root]

Do not do things without a reason.

[tropic oxide]

but doing what i told you to do,

you get:

a = -b-c-6

from which

bc - 6a = 3

and thus

bc - 6(-b-c-6) = 3 ...

[daring linden]

Ts guy

[ancient iris]

thanks

[tropic oxide]

???

what

[daring linden]

No OP

Not you

[ancient iris]

thanks guys

[daring linden]

Solved?

[ancient iris]

i get it

so guessing game

i hate these

but it is okay

[daring linden]

Not really guessing

Why do you think its guessing?

[cedar kiln]

@ancient iris Has your question been resolved?

[cedar kiln]

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[crimson sedge]

Is this correct?

[pastel vault]

a few mistakes

m^2 + 9m + 14 = 0 so the product of roots is 14/1 = 14

[crimson sedge]

Noted

anything else?

[pastel vault]

similarly go fix the product of the roots for equation 5

[pastel vault]

all quadratics have two roots when you consider imaginary numbers

[crimson sedge]

I just use c/a

[pastel vault]

(the generalisation: all polynomials of degree n have n roots when you consider imaginary numbers, is called the fundamental theorem of algebra)

[pastel vault]

no this part is correcting the 'nature of the roots'

oh wait so imaginary being in the form k * i, where k is real I assume

[crimson sedge]

Yes, but

[crimson sedge]

Yes

[pastel vault]

yeah then that means that okay, the nature of the roots column is correct for q1, q2, q3, q4

q5 should be the roots are (purely) imaginary

a = sqrt(10)i, -sqrt(10)i

[pastel vault]

b^2 - 4ac < 0 does not imply 'roots are not real or imaginary'

your slide is wrong

[slender ginkgo]

weird phrasing, i think

[crimson sedge]

Yeah

[slender ginkgo]

but the intention is right that there is no real root if D < 0

[crimson sedge]

Yes

[pastel vault]

yes this is what they meant to say

no real roots

apart from that everything else is correct

[slender ginkgo]

'or imaginary' isnt necessary

[crimson sedge]

[wraith dagger]

Result:

25

[pastel vault]

okay q4 discriminant is good just panicked

[snow oxide]

hello

[slender ginkgo]

hi

[cedar kiln]

@crimson sedge Has your question been resolved?

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[gusty copper]

,, \int \frac{1}{\cos(x)\sin(x)}\dd{x}

[cedar kiln]

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

[gusty copper]

no clue where to start

1

cuz i dont want the solution ?

[hallow pelican]

ooooh yeah sorry

[gusty copper]

its alright

[dreamy void]

Maybe 1 = sin²(x)+cos²(x) helps

[karmic field]

Maybe

Multiply by cosx

[karmic field]

/cosx

[rapid mauve]

hint: 2 sin x cos x = ??

[karmic field]

And replace cos^2x with 1- sin^2x

Or that

[dreamy void]

You then get sin(x)/cos(x) and cos(x)/sin(x) which you can easily u-sub

[karmic field]

Use this

Or idk still gotta multiply

By cosx/cosx or sinx/sinx

[gusty copper]

ig this one worked for me ?

i got 2 \int tan

[dreamy void]

You would have \int tan(x) + cot(x)

[gusty copper]

ah yeah

[dreamy void]

But then just use u-sub

It will be something with ln

[gusty copper]

,w int cotanx

[wraith dagger]

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[rapid mauve]

that works too

[karmic field]

Should be ln|tanx|

[dreamy void]

You got the form f'/f whose antiderivative is ln|f|

[karmic field]

You use this

[gusty copper]

i see

[gusty copper]

didnt work for me

[karmic field]

Then multiply by sin(2x)

Buttom and top

[gusty copper]

i got 2 \int 1/sin(2x)

[karmic field]

Then replace buttom with 1- cos^2 (2x)

[dreamy void]

re ti kaneis

[karmic field]

Then sub u =-cos2x

[rapid mauve]

from here i think you can chain rule directly?

[karmic field]

And you have derivative on top

But integral of 1/sinx is standard

[gusty copper]

uh how

[karmic field]

Ti kano tou leo pos linetai

[karmic field]

No

[idle tusk]

yeah actially this might work better

[karmic field]

Multiply by sinx/sinx

[gusty copper]

,, 2 \int \frac{1}{\sin(2x)} \dd{x}

[idle tusk]

this reduces to partial fractions

[karmic field]

After subbing the 2x

[karmic field]

Sub

U=2x

Then multiply by sinu/sinu

Then make buttom 1-cos^2 (x)

Then sub z= -cosx

[dreamy void]

This is complicated asl

[karmic field]

And its solved

[karmic field]

Thats how you solve it

[dreamy void]

That's how you solve it

[karmic field]

Or if you remember then Identity int 1/sinx = ln|tan(x/2)|

But thats how the identity is proven

[dreamy void]

Nobody remembers that

[karmic field]

Thats the reason i explained how to solve it

But you should remember it

Anyone doing a lot of calculus learns it

[idle tusk]

that's how its solved

[karmic field]

Or is given it on a paper of the test

[rapid mauve]

or you can turn the 1/sin into a csc

[karmic field]

Its the same thing

Its as saying to remember it

I just gave the proof of it aswell

[idle tusk]

i don't remember (anti-)derivatives of sec/csc for my life lololol

[rapid mauve]

i mean if it helps him remember ig

[karmic field]

Cause its good to know where formula comes from

[dreamy void]

very elegant

[karmic field]

You can find it at the spot

But usually in tests its given

[idle tusk]

yeah exactly

[karmic field]

Cause there is no point doing it

[idle tusk]

lucky 😭

[karmic field]

Like if you do engineering they dont want you to do that every time

Its not the point

[karmic field]

Im not speaking of experience

[dreamy void]

lmao

[karmic field]

I just finished highschool im studying on my own im just saying from what ik of engineering courses

But yes its good to know how to figure it out if you care about math imo

[rapid mauve]

fair

[gusty copper]

.close

[cedar kiln]

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[cedar kiln]

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[wraith dagger]

Nyxzore

[cedar kiln]

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