help-23

[swift owl]

So the first 2, 12

is this problem random guesses?

[left gyro]

no

[swift owl]

or like there is an actual way to solve this kind of thing

[left gyro]

we're building up to those skills by having you try this out first

I can definitely say youre taking a bit longer than usual if you could do (f o g)(-4) immediately

[swift owl]

?

i am not sure what you mean

[left gyro]

can you calculate (f o g)(-4)?

[swift owl]

that is provided

wait

[left gyro]

(f o g)(-4) is provided youre saying?

[jolly kiln]

i have a large feeling that f(12)=12 and g 14 is 12

cause only those numbers exist here

its just a pattern tho nothing official

[swift owl]

that's how i feel about this whole thing

i just don't understand it

[left gyro]

cmon

whats g(-4)

[swift owl]

-2

that given

[left gyro]

then why did you screenshot the problem with all the boxes blank

[swift owl]

the ones blank are the ones need to be filled out

the others are provided

[left gyro]

(f o g)(-4) is not provided

[swift owl]

true

[left gyro]

you didnt even look at the option on the table I swear

please find (f o g)(-4) then place it in the table

[swift owl]

that's a composition function

[left gyro]

do you know what the o means

[swift owl]

the g is inside the f i am pretty sure

[left gyro]

(f o g)(x) is f(g(x))

and (g o f)(x) is g(f(x))

knowing that, find (f o g)(-4)

[swift owl]

ok give me a sec

[jolly kiln]

its the other way

[swift owl]

is it?

[jolly kiln]

jup

[left gyro]

no it isnt

if you search up composition of functions you can see the order

[jolly kiln]

and thats wikipidea

[left gyro]

this is khan academy

wait

this is wikipedia

[swift owl]

wouldn't that be -4 still since g(-4) is -2 and then f(-2) is -4

[left gyro]

where are you getting that from

[jolly kiln]

im on the german wikipedia

[left gyro]

then its a notation issue that depends on the country

@swift owl what country is the assignment in

[jolly kiln]

and my german text book say the same as me it seems its different dependend on location

[swift owl]

USA

[jolly kiln]

than its his order

[left gyro]

so (f o g)(-4) = f(g(-4)) = f(-2) = -4 is correct

[jolly kiln]

weird tho its different orders in different countres lol

[swift owl]

is this right tho?

[left gyro]

yea it is

[swift owl]

ok

[left gyro]

try filling in the other boxes using only the provided info

[swift owl]

i think i am understanding this a little

[left gyro]

yea

[swift owl]

my prof just gave us a textbook, no vids and 100 problems due next friday

[left gyro]

thats concerning

[swift owl]

and this is like the first 10 problem

[left gyro]

well at least you know whats going on exclusively from knowing (f o g)(x) = f(g(x)) and (g o f)(x) = g(f(x)) which is nice

this means you should be able to figure out part of this already

this is the only "outside" info youll need to solve the problem

[swift owl]

do you think this looks about right

[left gyro]

I didnt actually do the problem yet hold on a sec

recheck (g o f)(-2)

[swift owl]

ohhh -2

[left gyro]

yep

similarly recheck(f o g)(12)

[swift owl]

14

[left gyro]

yep

also recheck (g o f)(14)

[swift owl]

oh yea another 14

thank you

[left gyro]

also true

np

[left gyro]

nice

now do you want to see a more rigorous way of getting those 12s

[swift owl]

yea

[left gyro]

say we dont know what f(12) is

however they tell us (g o f)(12)

which is -4

so g(f(12)) = -4

now what do you think f(12) is, knowing g(f(12)) is -4

[swift owl]

12?

[left gyro]

yep

because g(12) = -4

so since g(f(12)) = -4, you can take a guess that f(12) = 12

now g(f(12)) = g(12) => f(12) = 12 isnt guaranteed

you can do that if you assume g is injective

[swift owl]

Intriguing

[left gyro]

"injective" means "different inputs must lead to different outputs"

so for g(12) and g(f(12)) to both lead to -4,

12 and f(12) are both the same

you see how that works?

[swift owl]

Yes, thank you

[left gyro]

np

[swift owl]

.solved

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[devout peak]

can i ask how this is wrong

final ans would be

[thin bridge]

How'd you get that first line

can you show your work for the decomposition

[thin bridge]

decomp looks fine
you didn't integrate correctly

[devout peak]

how

oh i see

the values are reversed

brb

@thin bridge

so final ans is

@thin bridge

[thin bridge]

yes, but since 7 is clearly positive, you don't need abs bars there

[devout peak]

ok

ty

jeez

it was confusing me so badly

for future ref

if i ge a -ln value

im wrong right?

[thin bridge]

depends on what you have

in this question you have
1/((x-1)(x+5))
which is > 0 for x> 1
so you'd expect the result to be positive

that might not be the case for something else

[devout peak]

hmm i see

[safe radish]

@devout peak Has your question been resolved?

[devout peak]

.close

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[devout peak]

i need help with this

[grim bison]

You forgot to set the bounds for ue^u

should be from 1 to infinity

[inner parrot]

the bounds are wrong btw

[grim bison]

well idk I didn't check

[inner parrot]

you used the substitution u=1/x

yeah but its important or else youll get the wrong answer

[devout peak]

im here now

[inner parrot]

first of all lets fix the bounds first

u=1/x

so if we go from x=-1

what would the lower bound be for u?

[grim bison]

Ah I see now

[devout peak]

u=-1

[inner parrot]

yep

and now for the upper bound

what would that be?

[devout peak]

inf

1/inf

= inf

no?

[inner parrot]

mm not quite

[devout peak]

well

[grim bison]

he mistook it

oh

[devout peak]

it approaches zero

[inner parrot]

since we are integrating from -1 to 0

[grim bison]

1/x

[devout peak]

1/inf tends to 0

[inner parrot]

we are approaching 0 from the left hand side

[grim bison]

yes

[devout peak]

do i just write 0

0^-

[inner parrot]

so whats $\lim_{x\to 0^{-}}\frac{1}{x}$

[flat frigate]

y0shi

[devout peak]

<0

[inner parrot]

well yeah its less than 0

but what does it approach

[devout peak]

-inf

[inner parrot]

yep

you also seem to be missing a negative sign

since u=1/x, du=-1/x^2 dx

so overall you should have something like $-2\int_{-1}^{-\infty}ue^u du$

[devout peak]

yes

[flat frigate]

y0shi

[inner parrot]

-2 mb

and you can swap the bounds too since we have that negative

$2\int_{-\infty}^{-1}ue^u du$

[flat frigate]

y0shi

[devout peak]

ARGHH

IM RUNNING OUT OF TIME

XDD

let me post

@inner parrot

4/e final answer?

[inner parrot]

close

its cuz you forgot the negative sign

its -4/e

[devout peak]

how come?

[inner parrot]

when you did the substitution

u=1/x

du=-1/x^2 dx

[devout peak]

isnt it -2/e

[inner parrot]

see how theres a -1 there

well cuz you factored out the 2 initially

[devout peak]

yeah

[inner parrot]

$2\int_{-1}^{0}\frac{e^{\frac{1}{x}}}{x^{3}}dx$

[flat frigate]

y0shi

[inner parrot]

this is the original right

[devout peak]

2(-2/e)

[inner parrot]

yeah

,w 2\int_{-1}^{0}\frac{e^{\frac{1}{x}}}{x^{3}}dx

[flat frigate]

Results provided by WolframAlpha

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Requested by doctormysterio9

[devout peak]

but you wrote

[inner parrot]

yeah thats -4/e

mhm

you evaluated the integral bit

and you just need to multiply it by -2

[devout peak]

• • • = +

• times - = +

neg * neg = pos

[inner parrot]

wait what

[devout peak]

you;re saying -2 got factored out from the beginning?

[inner parrot]

yeah

[devout peak]

ok

and now i have

[inner parrot]

cuz you kept the bounds like that

i assume you did the integration part correctly

[devout peak]

now its

this

[inner parrot]

cuz -4/e is the answer i checked with desmos and wolfram

[devout peak]

-1/e -1/e = -2/e

correct?

[inner parrot]

yeah

okay so look

how you went from -inf to -1

[devout peak]

-2 * -2/e

[inner parrot]

not -1 to -inf

[inner parrot]

thats different than this

you can get rid of the negative by changing the bounds from -inf to -1

[devout peak]

https://tenor.com/view/dizzy-gif-6376271150502901118

[inner parrot]

okay so

the integral we have after the substitution

[devout peak]

top is supposed to be negatice

[inner parrot]

is $-2\int_{-1}^{-\infty}ue^u du$

[flat frigate]

y0shi

[inner parrot]

you went from -inf to -1

[devout peak]

and when we swap then bottom is pos

[inner parrot]

but its -1 to -inf

yeah

[devout peak]

because

[inner parrot]

$-2\int_{-1}^{-\infty}ue^u du=2\int_{-\infty}^{-1}ue^u du$

[flat frigate]

y0shi

[devout peak]

yes

ok

i got it

thanks

[inner parrot]

yw!

[devout peak]

i really have to practice

https://tenor.com/view/last-day-of-school-gif-4872695840845483552

[inner parrot]

its alright

[devout peak]

it didnt help that i was in a rush too

[inner parrot]

practice makes perfect

youll get the hang of it eventually

[devout peak]

i hope my midterm on sunday isnt as rough as that one was for me

and thats an easier question

_>

[inner parrot]

damn

[devout peak]

im doomed

[inner parrot]

best of luck to you

[devout peak]

thanks for helping me though

[inner parrot]

just do more practice

[devout peak]

i appreciate it

[inner parrot]

yw!

ofc

[devout peak]

like that was dizzy dizzy

but when i do partial fractions integration

i get it 1st try

ARGHH

ty again

[inner parrot]

haha

[devout peak]

.close

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[stray chasm]

hi so i've understood the proof itself but don't inductions need a base case for it to be valid

(this is the group theory part of abstract algebra by dummit and foote. chapter 3 section 4)

[light shoal]

well certainly it's true if |G| = p

but they said as much

[stray chasm]

yeah but that isn't what the induction hypothesis is

[light shoal]

well what do you want to consider the base case, |G| = 1?

[stray chasm]

if you look at the third part of the proof (starting with N = <x>) |G/N| is not p

[stray chasm]

that wont work tho cuz then no prime divides |G|

[light shoal]

sure, so then it's vacuously true i guess
or you can consider |G| = any prime to be the base case(s), and there it's obviously true

[stray chasm]

yeah but isnt that more of a special case

[turbid crag]

base cases are special cases

[stray chasm]

cuz it works because |G| is prime, there's no guarantee it holds true if |G|=p+1

[light shoal]

but you're assuming that p divides |G|

and |G| = p is the smallest possible |G| satisfying this condition

[stray chasm]

oh. yeah

my bad 😭 thank you

[light shoal]

nw, they could have been clearer about that

[stray chasm]

yeah this book is good but it has sections where the proofs are a bit convoluted

thank you!

[light shoal]

which book is it?

yw

[stray chasm]

abstract algebra by dummit and foote

[light shoal]

oh yea

yea their proofs are not always the best

but it's a good book anyway

[stray chasm]

yeah!

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[rain path]

Hello everyone, I have a question. Can anyone answer it

[sour loom]

sure

[brave wolf]

What question and why did you open 3 help chanels?

[safe radish]

@rain path Has your question been resolved?

[quick crater]

.close

[magic junco]

This one should be kept, just in case

[quick crater]

oh ye

@rain path

im confident this is a raid tho

made account at the same day, have the wierd gibberish name, joined today

[turbid crag]

true

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[lean otter]

If A and B are two square matrices such that AB=A and BA=B, then

[lean otter]

only B is idempotent
only A is " "
both A and B are " "
none of these

ive concluded that both A and B have to be identity matrix

and because square of I is I, hence both A and B are idempotent

and this is the correct answer too, i just want someone to check if my reason is correct

[light shoal]

surely A=B=0 works

[lean otter]

yea that too

O^2 is 0

so that would work too

so am i right?

[kind tinsel]

ABA = A^2 and then replace BA with B, and then AB with A

and use symmetry

[lean otter]

how does ABA become equal to A^2?

[kind tinsel]

right multiply the first equation by A

[lean otter]

yea right

same we could do for BA=B

alr thanks

det(A)=0
so c option should be correct too right?

but the answer key says its only option A

need help with this one question too pls

[kind tinsel]

did you verify this fact?

[light shoal]

for 53?

det A is not 0, the columns are LI

[kind tinsel]

or are you guessing

[lean otter]

oh wait what?

lemme calculate again

oh yea lmfao 😭😭😭

its not 0

mb

[lean otter]

yea

[light shoal]

by the way, if (a) is true you know (c) is false

becuase A^2 = I implies that A is its own inverse

[lean otter]

OH yea right

that makes sense

alr thanks both of u

imma close now

.close

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[fervent edge]

hi chat ^_^, i need some help with basic maths!

[fervent edge]

i gotta find the radius of the circle!

yet i've no idea how 😔

[turbid crag]

let $OC=OD=x$, then we have, by the law of cosines, $(3+x)^2=(4+x)^2+5^2-5(4+x)$

[violet stratus]

Also this triangle is a pretty popular one

[fervent edge]

huh.

[turbid crag]

i dont quite remember it lol

||r=2||?

[flat frigate]

nameless individual

[fervent edge]

looking at the answers sheet it should be 4

[turbid crag]

i was just guessing

[fervent edge]

ooh

[turbid crag]

i remember something about 567 or something lol

[fervent edge]

ah.

[turbid crag]

oh it was 357

[fervent edge]

yup, got it

thanks y'all ^_^

how do i uh

close this

[violet stratus]

yep you can get 357 triangle from 578 btw

[violet stratus]

.solved

[fervent edge]

.solved

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[turbid crag]

how

[turbid crag]

oh nvm

me being dumb today

also nice board

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[opaque pasture]

Can someone check two of my answers please?

[turbid crag]

please, we are not problem solvers for you

[lime dust]

To know if your solutions are correct, you have to input the values you got in your original equation

In case you got more than 1 solution, so it as many times as solutions you have

[opaque pasture]

!status

[safe radish]

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

[opaque pasture]

@turbid crag then what's the point of number 4?

I've already done the manual work and am just looking for confirmation, these just happen to be multiple choice selections

[opaque pasture]

Okay, thanks

.close

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[turbid crag]

its just that it is quite spammy

[opaque pasture]

To you.

It was literally two channels in total, four questions. How is that spammy?

[turbid crag]

there's gotta be a particular reason you want us to check your work... otherwise everyone would get a help channel

anyhow, we'd like to see some work; we're not checking just the answer

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[golden loom]

If from any point on the common chord of two intersecting circles, tangents be drawn to the circles, prove that they are equal.

[thin bridge]

is this the original question

was a diagram provided to you as this doesn't make sense
the described point will be within the circles
you won't be able to draw tangents to the circle from within the circle

[golden loom]

thats what i am thinking

no diagram is provided

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@golden loom Has your question been resolved?

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[cunning pasture]

how to find the sum of Gp?

[cunning pasture]

a+ar+ar^2+ar^3...... ar^(n-1)

[magic lion]

there's a formula for it?

[cunning pasture]

Ikr.

[magic lion]

do you want to prove the formula?

[cunning pasture]

Yes

could oui give moi hints to proceed?

[magic lion]

suppose the initial sequence as Sn and mark is as equation -(1
multiply the equation - (1) by r on both sides and mark as eqn - (2
subtract eqn 2 from 1 and find put the Sn alone

[cunning pasture]

oh got it. thanks

.close

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[safe radish]

@vivid whale Has your question been resolved?

[safe radish]

@vivid whale Has your question been resolved?

[buoyant shadow]

@vivid whaleit's the same answer as if you would draw from 4 decks

linearity of expectation removes the distinction

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[lusty spear]

can someone varify that these two are equivalent ??
I transformed the term because its easier to see a bound that way, but I need to be 100% sure its correct

$\frac{k}{4} \cdot 2^{(k-1)/2} = \frac{k}{4 \cdot \sqrt{2}}\cdot 2^{k/2}$

[flat frigate]

barış

[lusty spear]

I basically just pulled the sqrt 2 down, thats valid right , or am I missing somethign here ?

[arctic raven]

looks real

[dull sequoia]

It’s good

[lusty spear]

thanks !!!!

.close

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[vernal zephyr]

Pls help me resolve into partial fractions

[glass carbon]

,rotate

[vernal zephyr]

@glass carbon ???

I got no idea ABT this

[glass carbon]

What have you tried

[vernal zephyr]

Is this correct one

[dull sequoia]

That’s not even the same denominator

[vernal zephyr]

How

It's X squared into X minus one

[glass carbon]

firstly

look at the degree

[dull sequoia]

But when you multiply them all together you have an extra x

[vernal zephyr]

Because there is an X squared

So I need to write it two times

As X and then as X²

[dull sequoia]

Hmmm

[glass carbon]

That's fine, but

[vernal zephyr]

Someone from this server told me to do this

[dull sequoia]

Maybe that works actually I haven’t done this in a while

[glass carbon]

still degree of the nominator is higher than the denom

[vernal zephyr]

Oh isee

Do I have to divide

Ok

[glass carbon]

Yes

and then do that

[vernal zephyr]

@glass carbon pls tell me if the division is correct

[glass carbon]

yeah

if there is a "+" between them

• 2/(...)

[vernal zephyr]

Sorry didn't get it

Wdym

There a minus

[vernal zephyr]

That just means there is a minus

And the numerator is positive two

Hehe

[glass carbon]

yes, and that's wrong, it should be +

(only + )

[vernal zephyr]

Ok

Now resolve into par.fracs

.

[glass carbon]

yes

[vernal zephyr]

Like the previous way

In the pic I sent

Above

[glass carbon]

yes, that's valid

[glass carbon]

,w 2/(x^2(x-1)) pfd

[flat frigate]

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[glass carbon]

all good

[vernal zephyr]

How did you do that

What the

I couldve asked a calc all along?

[glass carbon]

Well, in WolframAlpha you have to subscribe to have a step-by-step solution

[vernal zephyr]

How do you do that from a server

[glass carbon]

It's provided by a bot called TeXit

you simply write ,w then your prompt and that's it

WolframAlpha will try to solve it

[vernal zephyr]

Do you have to learn all the commands

[glass carbon]

and the view from the browser:

[vernal zephyr]

.

[glass carbon]

not really, you can just write "partial fractions" and it will work either

but generally it's enough to just paste an equation/expression

[vernal zephyr]

,w 2*7

[flat frigate]

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[vernal zephyr]

Woah

[glass carbon]

,w tangent line at x = 3 to y = x^2 - x + 3

[vernal zephyr]

,w integrate 1

[flat frigate]

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[vernal zephyr]

,w derivative of x2

[flat frigate]

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[vernal zephyr]

Yooooo

That's a smart bot

Thanks very much

[glass carbon]

yeah, WolframAlpha is really good (much better than AI)

[vernal zephyr]

,w integrate -2/x^2

[flat frigate]

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[vernal zephyr]

@modus

@glass carbon

I was meant to integrate the partial fractions

[glass carbon]

That should be easy now?

[vernal zephyr]

This is the answer

The middle one

It doesn't align with my partial fractions

[glass carbon]

don't forget about (x+1)

which you obtained previously

[vernal zephyr]

Yea that's where the x2/2 came from

And also the x

[glass carbon]

yep

[vernal zephyr]

Butt there are only two terms after it

But partial fractions consists of 3 terns

[glass carbon]

2/(x-1) gives 2ln(x-1)

[vernal zephyr]

And 2/x

Gives 2lnx

But what ABT 2/x^2

[glass carbon]

well it would give 2/x (taking the minus sign into account)

finally:
x^2/2 + x + 2ln(x-1) - 2ln(x) + 2/x + C

let's see if it works

[vernal zephyr]

It isn't in the book answer

[glass carbon]

,w d/dx (x^2/2 + x + 2ln(x-1) - 2ln(x) + 2/x)

[flat frigate]

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[vernal zephyr]

,w derivative of
x^2/2 + x + 2ln(x-1) - 2ln(x) + 2/x + C

[flat frigate]

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[vernal zephyr]

,w derivative of
x^2/2 + x + 2ln(x-1) - 2ln(x) + C

[flat frigate]

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[vernal zephyr]

I see

Book and is wrong

[glass carbon]

that's something different

[vernal zephyr]

Ficking stupid book

[glass carbon]

maybe you rewrote the example wrong?

[vernal zephyr]

No I just checked

There's no mistaje

[glass carbon]

Okey, then yeah it might be wrong, since they're different, generally doing integrals answers can differ marginally but ONLY due to constant

[vernal zephyr]

You mean that there will only be the issues of constants

[glass carbon]

since derivative of a constant is 0

[vernal zephyr]

Like if there's a 2 in my anzwey

Answer

[glass carbon]

yes

[vernal zephyr]

And when I differentiate it

I don't get the 2

Is that it

[glass carbon]

yes, that's what I meant

[vernal zephyr]

Ok

[glass carbon]

but not only, it can be "hidden" e.g. in ln(...)

[vernal zephyr]

I see

[glass carbon]

but here, from what I've checked that's wrong

[vernal zephyr]

Is this correct

Or. Do I have to write Ax+B

@glass carbon

[glass carbon]

well technically it should be Ax+B and then Cx+D

[vernal zephyr]

How do you tell

Here it is just B

Over X squared

The degree is 2

[glass carbon]

because in the denom you have a trinomial with no roots

(real roots)

[vernal zephyr]

A trinomial

Or a binomial

[glass carbon]

doesn't matter, something of form ax^2 + bx + c, that's special case when b = 0

[vernal zephyr]

Can I do a quick check like this, whenever I am doing partial fractions I'll try to find the roots of the denominator to check if the numerators gotta be just A or Ax+B

[glass carbon]

yes, basically it's enough to factorise it

[vernal zephyr]

Ok so here I'm gonna get four unknowns A b. C and d

[glass carbon]

yeah but I guess in this case unknows next to "x" will be zero

since there's no "x" in the numerator

so it won't be that hard

[vernal zephyr]

Look

[vernal zephyr]

Is it okay to get zero value for A B or a variable in partial fracs?

[glass carbon]

B shouldn't be zero here, I guess

rather 1

[vernal zephyr]

I'm getting b 1

And d 1

[glass carbon]

yes, good

[vernal zephyr]

,w partial fractions (x^2 +2)/(x²+1)²

Hoorah ig

[flat frigate]

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[vernal zephyr]

,w Integrate (x^2 +2)/(x²+1)²

[flat frigate]

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[vernal zephyr]

,w differentiate (x^2 +2)/(x²+1)²

[flat frigate]

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[vernal zephyr]

Ok thanks a lot bud @glass carbon

Have a good one bye

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[icy skiff]

<@&286206848099549185>

[cinder lily]

ok

a intersection b is 20 right

[icy skiff]

yeah

[cinder lily]

its given in question

a intersection b dash is just
a - a intersection b

same story for b

b - b intersection a

a dash intersection b dash is just a union b the whole complement

using these find it urself ig

[icy skiff]

can you make it more simple

@cinder lily

so the 3 in middle are 20?

what does the apostrophy represent

@raven vessel

[cinder lily]

wat is 3 in the middle

[icy skiff]

forget that

can you tell me what the apostrophy represents

[cinder lily]

it means complement

for example if i say

A complement that means
it creates a set magically containing all the elements that are not in A

if A = {1,2,3,4,5}
and A is a subset of bigger set R
then A'
has R - {1,2,3,4,5}

[icy skiff]

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[grim bison]

$2x^{2} + 5x - 7 = 0$ Help I can't solve this

[flat frigate]

modified vessel

[south lotus]

!show

[safe radish]

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

[grim bison]

So I did factoring by grouping and I did 2x^2 - 2x + 7x - 7 = 0
and then 2x(x- 1) + 7(x -1) = 0
And then (2x + 7)(x-1) = 0
so x = -7/2 and x = 1

But I got it wrong and I don't know how

[south lotus]

2x-7x gives you?

[old chasm]

are you sure you got the sign correct?

yeah this is more like it

[south lotus]

Seems right

[grim bison]

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[lean otter]

Hi. Why can you compose g and h in this order given the codomain of h is different from the domain of g?

[lean otter]

Sure h is a bijection between A and B but there is no guarantee every element in B has an equal in A

[arctic raven]

yeah that looks goofy, the codomain of f should be the codomain of g, being B by context

is this notation consistent with the rest of the stuff in your source? like, maybe they take function composition in reverse order or something lol

[lean otter]

Just a minute, let me print the definition of composition

There

[arctic raven]

mmm maybe they mean the inverse of g in the def'n of f

[lean otter]

You see? I think the proof is incorrect

[arctic raven]

so f = g'h

that'd work

[lean otter]

The problem is that inverse functions are only defined in the next section

[brave wolf]

what does the theorem state?

[arctic raven]

I think that writing f = gh is certainly a typo, and since the codomain of f is explicitly given as the domain of g and g is restated to be a bijection onto its image, it must be that f = g^-1 h instead

[lean otter]

Yeah, that's the only way the proof makes any sense

Otherwise it just doesn't work

[arctic raven]

and for the sake of making the proof functional, you could show that bijections have inverses, if that's within the scope of the book/course and if it hasn't been done yet

[lean otter]

yeah, thanks for the help

.close

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[icy skiff]

can someone help me with this

[mossy fractal]

the intersection of these sets will be the set of elements that are in both sets

are there any elements that are in both sets?

[icy skiff]

6

[mossy fractal]

if no, then the answer is b

[mossy fractal]

yeah

any others?

[icy skiff]

no

[mossy fractal]

ok

so the intersection of those 2 sets is {6}

[icy skiff]

alright

[mossy fractal]

theres only 1 element in the intersection

[icy skiff]

is that what its asking for? the intersection?

[mossy fractal]

yeah

$\cap$ means the intersection of sets

[flat frigate]

eugene_krabs_has_cake

[icy skiff]

ohh

so the answer would simply be 6

[mossy fractal]

basically yeah, if i was being pedantic i would say that 6 is the element in the set, but the set is {6}

[icy skiff]

alright thanks

[mossy fractal]

np

[icy skiff]

this one confuses me as well

<@&286206848099549185>

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[lilac delta]

What steps should I take when integrating the reciprocal of a repeated irreducible quadratic

[lilac delta]

Example:

,tex $\int\frac{C}{(Ax^2+B)^2}\mathrm{d}x$

[flat frigate]

lILi

[lilac delta]

Where A, B, and C are constants

[royal kiln]

Idunno but I recommend partial fraction decomposition

[lilac delta]

My professor never covered this specific case

[royal kiln]

Get a fresh channel, ding-dong

[lilac delta]

The closest we got was linear over repeated irreducible quadratic

Never covered constant over such

[royal kiln]

Do you have a specific problem beyond this generalization?

[lilac delta]

I would rather do a generalization to understand the concept rather than just have someone do my homework for me

But I think I could just expand the denominator and decompose as a polynomial?

[royal kiln]

What makes this trickier is that you are using symbols we like to use when doing partial fraction decomposition, i.e. A, B, C, ...

[lilac delta]

Fair, I can rewrite

,tex $\int\frac{1}{(Ux^2+V)^2}\mathrm{d}x$

[flat frigate]

lILi

[royal kiln]

Also I would have expected, if this is the tail end of a partial fraction decomposition, for the numerator to be a linear expression

[zinc hornet]

I expect at the minimum you will want to consider different cases for the signs of A and B. If you don't the problem gets a lot more annoying

[lilac delta]

For this problem let's assume positive polarities for all constants

Hold on ill just get the full problem

[royal kiln]

Oh...fun-fact: you can Partial fraction decompose complex factors

[lilac delta]

,tex $\int\frac{1}{(\sqrt{4+x^2})^4}$

[flat frigate]

lILi

[lilac delta]

which I simplified to $\int\frac{1}{(4+x^2)^2}$

[royal kiln]

You would have to integrate those tho

[flat frigate]

lILi

[lilac delta]

oh i forgot the dx

[royal kiln]

It's not terribly complicated but probably hasn't come up for you yet

[lilac delta]

$\int\frac{1}{(\sqrt{4+x^2})^4}\mathrm{d}x=\int\frac{1}{(4+x^2)^2}\mathrm{d}x$

[flat frigate]

lILi

[lilac delta]

This is part of an exam review

[lime dust]

Trigonometric sub

[lilac delta]

RIGHT

i'm stupid

no, nevermind

that doesn't work here

[lime dust]

Are we with squared or to tje forth?

[lilac delta]

the first expression is on the question but i simplified it on the right

[lime dust]

In both trigonometric works

Tell me the exact one

[lilac delta]

but it would be trapped within the square, no?

[lime dust]

Tell me the exact one so i can check

Maybe i overlooked

[lilac delta]

$\int\frac{1}{(\sqrt{4+x^2})^4}\mathrm{d}x$

[flat frigate]

lILi

[lilac delta]

Using that one, even with trig sub, it would be trapped in the ()^4

oh

i see

[lime dust]

It works

I am right now walking in the street but i feel confident ive done this one with trigonometric before

You can use identities

[lilac delta]

Thank you, I can figure this out from here

.close

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[lean otter]

How can I get the second expression of b?

Given the polynomial...

If p(x) is a null polynomial, determine the numeric value of

[lean otter]

Ignore the obvious mistake

from here, how can I get the second expression?

[eternal carbon]

complete the square ||add 4 to both sides||

[lean otter]

is your plan finding b?

[eternal carbon]

no...?

[lean otter]

ok

[eternal carbon]

you asked how to get sqrt(b) + 2/sqrt(b)

[lean otter]

yess

thx

yes I got the anser now, thx

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[turbid bluff]

help

[turbid bluff]

i forgot basic math

[modest hazel]

x+(x+36)=180

no

wait

im wrong

[turbid bluff]

ok

[modest hazel]

I assume that because they are parallel, the angles that are x and x-36 are the same

[safe radish]

@turbid bluff Has your question been resolved?

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[gaunt lynx]

ive got no real ideas here, ive tried finding expressions for other line segments in terms of BC but i cant get enough to solve the problem

[jagged badger]

Hint: show triangle PBM is similar to triangle MCQ

[gaunt lynx]

ive thought of that but i dont see how to do it

i dont know the length of QC, QM, or angles QMC or CQM

[jagged badger]

suppose <QMC=alpha, what is <PMB then?

[gaunt lynx]

ah

that does do something yes

ok i think i got it thanks

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[jagged badger]

No problem

The right angle at a straight line like that is a telltale sign to look for similar right triangles because of the complementary angles

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[tepid walrus]

I want to derive the connection between the two, but I just can‘t see what to do next. I‘d like small a hint(s), as I‘d like to feel like I somewhat did it myself. I‘m pretty sure I‘m missing somethibg obvious as always

[cursive sand]

Hi :3

[tepid walrus]

Hello Clau

Can I ping helpers or are you here to save the day

<@&286206848099549185>

[white umbra]

Don't message here if you're not helping

[tepid walrus]

The purge

[white umbra]

connection between what two things? can you explain what I'm looking at lol

[fair dragon]

he's my friend, I'm sure it's fine

[tepid walrus]

Harmonic numbers and digamma function

[white umbra]

no, help or go away

[tepid walrus]

I‘m emotionally scarred

[fair dragon]

and I was gonna try

[tepid walrus]

Yeah I don‘t mind the memeing that much, but I need help so

[fair dragon]

what's phi?

[white umbra]

yeah, since the diagamma function and harmonic numbers satisfy the same recurrence relation, they differ only by a constant

[tepid walrus]

psi?

[white umbra]

namely the euler mascheroni constant

[fair dragon]

yea

[tepid walrus]

Ah well here‘s a fact I didn‘t know about reccurence

[white umbra]

well it applies to this specific recurrence

[tepid walrus]

So whether it is 1/(z+1) or 1/n it doesn‘t matter so long as the "function" parts are the same?

[white umbra]

oh, it'll also be shifted by one

so digamma(n) will relate to H_{n-1}

[tepid walrus]

Right but what‘s the intuition behind this, I‘m lost

[white umbra]

well think about it this way, we know digamma(1) = H_0 + some constant

let's call that constant C

digamma(1) = H_0 + C

[tepid walrus]

Wait wait why the first step

[white umbra]

it's just true about any two numbers

[tepid walrus]

I‘m not sure I got that "they have the same reccurence so they only differ by a constant" thing

[white umbra]

you can express one of them as the other one plus C

this is the base case

[tepid walrus]

Oh we‘re just saying for any two numbers you can represent one as C + the other?

[white umbra]

digamma(1) and H_0 are just two random numbers, so I can write digamma(1) = H_0 + C, where C = digamma(1) - H_0

yeah

[tepid walrus]

And in this case the numbers we pick are harmonic and digamma cause they seem to have a sinilar recurrence

[white umbra]

right

[tepid walrus]

Ok this does make sense

[white umbra]

so let's assume by induction that digamma(n) = H_{n-1} + C

then digamma(n+1) = digamma(n) + 1/n

and H_n = H_{n-1} + 1/n

[white umbra]

so adding 1/n to both sides of this equation, we get digamma(n+1) = H_n + C

that completes the induction

so that proves digamma(n) and H_{n-1} differ by the same constant for all n

[tepid walrus]

Then subtract log and take lim as n\to\intfy to get oily macaroni?

[white umbra]

it depends on what your definition of digamma is

but, you know when n=0 that digamma(1) = H_0 + C, and H_0 = 0, so digamma(1) = C

so all you really have to do is calculate digamma(1)

[tepid walrus]

I‘ll tinker around with this, but I was thinking of using

,w lim [n -> infty] H_n - ln(n)

[flat frigate]

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Requested by imtyp0

[tepid walrus]

Oh you said digamma, not gamma

Yeah I‘m only using the logarithmic derivative of the gamma function

[white umbra]

that works fine if you can prove digamma has the desired limit as n→∞

[tepid walrus]

I‘ll try some things and come back if need be. Thanks a bunch :)

[tepid walrus]

Well, i could use the fact that as n tends to inf, digamma behaves like ln(n) - 1/(2n) but that feels like cheating if I can‘t prove it

I‘m gonna try and find a way

[safe radish]

@tepid walrus Has your question been resolved?

[tepid walrus]

Eyy @white umbra idk if differentiating both sides of an asymptotic relationship is legal but I managed to find what digamma(n+1) is asymptotically equal to :D

[white umbra]

Lol

it doesn't sound like it makes sense

[tepid walrus]

Let me screenshot it

I abused Stirling‘s

[white umbra]

consider these two functions which are the same asymptotically but have very different derivatives

[tepid walrus]

Well idk man, it just worked 🫠

[white umbra]

plus that approximation only works for integer n afaik?

[tepid walrus]

Sitrling‘s?

[white umbra]

oh nvm

it works for any real number

[tepid walrus]

Right

Poor choice of variable on my end I assume

But yeah, at least I managed to do one thing :D

And from there it easily comes that digamma(z+1) + γ = H_n, so qed

My wording is weird bruh ignore it

So yeah, thanks for the help Eric :)

.solved

[safe radish]

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[hearty widget]

Hi. Can someone verify that the following is true?
Given 2 primes p and q of the form a1 * b1 + 1 = p and a2 * b2 + 1 = q where a1, a2 are small integers with a known prime factorization of a1 = c1 * d1, a2 = c2 * d2 and b1, b2 are prime:
n = p * q
phi(x) is the euler totient of x
e = phi(n) = (p - 1)(q - 1) = (a1 * b1)(a2 * b2)
f = phi(e) = (c1 - 1)(d1 - 1)(b1 - 1)(c2 - 1)(d2 - 1)(b2 - 1)

[lean otter]

I don't want to bother or anything but does anyone know math? and that he is between 17 and 20 years old? To help me with math please?

[empty gyro]

!occupied

[safe radish]

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).

[hearty widget]

try #help-4 ?

or #help-12 then lol

[lean otter]

good thanks but it would be a private tutor or something like that TT

[safe radish]

@hearty widget Has your question been resolved?

[safe radish]

@hearty widget Has your question been resolved?

[hearty widget]

Been a while

[hearty widget]

<@&286206848099549185>

[lunar sierra]

This?

[hearty widget]

yes

I think it should work given the definition for phi(x) and the way p and q are defined

[lunar sierra]

Verify the Euler quotient part?

[hearty widget]

yes

especially the last line there

[lunar sierra]

Yea for e but idk for f

Idk where b1,b2 come from for f

Oh nvm

[hearty widget]

b1 and b2 are used with a1 a2 to create p and q

p = a1 * b1 + 1, p = a2 * b2 + 1

[lunar sierra]

Yeah it works

[hearty widget]

Great!

[lunar sierra]

You just unrolled the defn

[hearty widget]

That means I can ditch using EED for modular multiplicative inverse then

.res

.close

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[lunar sierra]

What is eed

[hearty widget]

extended euclidean algorithm, sorry

trying to do RSA which involves computing e^-1 mod phi(n) where n = p*q

and that can be computed by e^(phi(phi(n))-1) mod phi(n) IIRC

[hearty widget]

so if I can just solve for phi(phi(n)) like here then a large part of the problem is solved(modular exponenitation is working)

[safe radish]

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[fervent edge]

good morning chat!

[fervent edge]

ive got a basic question about trigonometry using cosine law

AB=AC, i gotta find out what AB is, yet i've no clue

the tips mentioned using cos(180-α) = -cosα, but i dont really know what they mean? :D

[cinder lily]

so wat do u want me to find

wats even the question

[fervent edge]

find AB

[cinder lily]

hmm

is this an isoceles triangle

[fervent edge]

yes

[fervent hatch]

Find AC find AB

Use the law of cosines

[grim plover]

Find AC and AB in terms of alpha (alpha is angle ABC)

[cinder lily]

well use al kashi's theorem

@fervent edge

capital means Angle

small means side

use this for 2 sides

and solve simultaneous equations

[fervent edge]

ill uh, try it out

[cinder lily]

here in this example since this is an isoceles triangle
angle B and C are the same

so cos B = cos C

[fervent edge]

^_^

.close

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@worldly mango Has your question been resolved?

[worldly mango]

I still need help

.close

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[halcyon light]

1st stage to second stage

i dont understand how they factorised it

what happened to the 15 co efficient

[lean otter]

they factored out $3x(5x-3)^2$

[flat frigate]

pun pun

[halcyon light]

man my brain cant comprehend it

[vale oriole]

just go term by term and cancel out 3x(5x-3)^2

that leaves us with 15x+2(5x-3), thus this is whats left after 3x(5x-3)^2 has been pulled out

[halcyon light]

oh right

thanks

.close