help-10

[exotic oak]

ye I'm not really sure how to go from there

So not really

[balmy mortar]

'sen' D:

[raven spire]

Oh does the sen(x) mean sine of "x", considering your indeterminate form

[exotic oak]

oh

sorry I thought that was universal

sen = sin

[raven spire]

mhmm np

[exotic oak]

I thought the 2 ways were acceptable

[balmy mortar]

I think only in some languages it is written sen

[exotic oak]

Yes I'm Portuguese

Comes from seno

[raven spire]

$\lim_{x \to \frac{\pi}{2}} (\sin x - \cos x)^{\tan x}$

[warm shale]

Ansh

[balmy mortar]

tg is tan?

[exotic oak]

yes

[balmy mortar]

That is new to me lol

[raven spire]

Do you want to use this question as an example?

[exotic oak]

Yes that's alright

[raven spire]

kay-

[exotic oak]

ahahha from "tangente"

[raven spire]

so umm, first we start with this being a (+1)^infty form

[exotic oak]

exactly

that's where I got stuck

I tried L'Hopitale

But not the best way here for sure

[raven spire]

since we're more familiar with other indeterminate forms, preferably 0/0 or infinity/infinity forms... or 0times infinity

we want to convert this form into one of them

'kay?

[exotic oak]

k

[raven spire]

L'hopitale applies only in 0/0 or infinity/infinity forms. Unfortunately you don't have it here yet

[exotic oak]

Oh

[raven spire]

so, our first step is to use log properties

[exotic oak]

Some other guy here at discord told me with was applied with any indeterminations

[raven spire]

Remember $a^b = e^{b \ln a}$?

[warm shale]

Ansh

[exotic oak]

a should be above no?

[raven spire]

given.. the log of "a" is defined

[raven spire]

hmm?

[exotic oak]

so "e" and "ln" they dissapear and we would get b * a ?

[raven spire]

nope

Do you remember the log properties?

[exotic oak]

yes

[raven spire]

what does $a = \ln b$ imply?

[warm shale]

Ansh

[exotic oak]

e^a = b?

[raven spire]

Yep

now, my question is, what is: \
$e^{\ln a}$?

[warm shale]

Ansh

[exotic oak]

a

[raven spire]

Great

My next question is: what is- \
$e^{b \ln a}$

[warm shale]

Ansh

[raven spire]

you might have to use exponent laws here as well

once you answer this, we might very well be able to start with your actual question

[exotic oak]

Ik it's supposed to be a^b

[raven spire]

Yep

[exotic oak]

Just not clearly getting how

[raven spire]

okayyy

[exotic oak]

In my head

It should be

b * a

[raven spire]

So, you have: $e^{b \ln a}$ right?

[warm shale]

Ansh

[exotic oak]

y

[raven spire]

next, you use the exponent laws

$e^{b \ln a} = e^{b \times \ln a} = (e^{\ln a})^b$

[warm shale]

Ansh

[raven spire]

Got it!?

[exotic oak]

let's just another example

like

wait

[raven spire]

hmm

[exotic oak]

let me try to use the bot

[raven spire]

Let's take:

$2^{3\log_2 7}$

[warm shale]

Ansh

[exotic oak]

$4^{6 \times \x} = (4^{\x})^6$

[warm shale]

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

[raven spire]

$4^{6 \times x} = (4^{x})^6$

[warm shale]

Ansh

[raven spire]

Right ?

[exotic oak]

yes

[raven spire]

that's what the exponent laws say

[exotic oak]

that's what I wanted

Oh okay

No doubts then

[raven spire]

hence we could write

$e^{b \ln a} = e^{b \times \ln a} = (e^{\ln a})^b$

[exotic oak]

Got it

[warm shale]

Ansh

[raven spire]

$= a^b$

[warm shale]

Ansh

[raven spire]

Got it?

[exotic oak]

yes yes

[raven spire]

100% sure?

[exotic oak]

yes

I said before I wasn't understanding

[raven spire]

Now, back to the actual question

[exotic oak]

If I hadn't got it I would tell

[raven spire]

$\lim_{x \to \frac{\pi}{2}} (\sin x - \cos x)^{\tan x}$

[warm shale]

Ansh

[raven spire]

Right?

[exotic oak]

yes

[raven spire]

We use the same trick to send the stuff inside the parenthesis to the exponent

okay?

[exotic oak]

ok

don't say anything

let me try to work this out

[raven spire]

hmmm

[exotic oak]

then you say if I got it right

[raven spire]

okie

[exotic oak]

Just with this tip

okkkkk

tough

Cause it's so different

nope

@raven spire

Couldn't do it

[raven spire]

oop

So we do the same trick and get

$\lim_{x \to \frac{\pi}{2}} (\sin x - \cos x)^{\tan x}$

[warm shale]

Ansh

[exotic oak]

that's the same we had

[raven spire]

$=\lim_{x \to \frac{\pi}{2}} e^{\tan x \ln (\sin x - \cos x)}$

[exotic oak]

oh we need to change base

alright

makes so much sense

[warm shale]

Ansh

[raven spire]

Makes sense?

[exotic oak]

yes

[raven spire]

Yep

now, which indeterminate form do you see?

[exotic oak]

let me see

ln(1)

= 0

oo * 0?

[raven spire]

Yep

the exponent now has the indeterminate form: $\infty \cdot 0$

[warm shale]

Ansh

[raven spire]

Next, my question is: which indeterminate form do you want this to become... that will make it most comfy for you

[exotic oak]

That's a tricky question

[raven spire]

👀

[exotic oak]

I like 0/0 or oo/oo

For L'Hopitale

I guess

[raven spire]

Great

$=\lim_{x \to \frac{\pi}{2}} e^{\tan x \ln (\sin x - \cos x)}$

[warm shale]

Ansh

[raven spire]

the ln part is 0 already, what would you do to make the exponent 0/0 form

[exotic oak]

@raven spire

I don't seem to be getting anywhere

[raven spire]

$=\lim_{x \to \frac{\pi}{2}} e^{\frac{\ln(1 - \cot x) - \ln(\sin x)}{\cot x}}$

[warm shale]

Ansh

[exotic oak]

Thought of dividing by tan x

then multiplying again

[raven spire]

make any sense?

[exotic oak]

hm

not really

But that's my fault I think

[raven spire]

nope

so

First

$\tan x = \frac{1}{\cot x}$

[warm shale]

Ansh

[raven spire]

Yes???

[exotic oak]

Because cot I never used it before I'm being honest with you

But I guess it should be in my seventy

[raven spire]

okay never mind cot..

Tell me one thing

Have you dealt with substitution method during solving limits?

[exotic oak]

y -> 0

something like that?

[raven spire]

Yep

[exotic oak]

yes

[raven spire]

okay

so

Let $y = x - \pi/2$

[warm shale]

Ansh

[raven spire]

Then, $y \to 0$ right?

[warm shale]

Ansh

[exotic oak]

yes

[raven spire]

great

now, we substitute y = x - π/2 in the expression

$=\lim_{x \to \frac{\pi}{2}} e^{\tan x \ln (\sin x - \cos x)}$

[warm shale]

Ansh

[exotic oak]

but we don't have any x - pi/2 in the expression

[raven spire]

Okay you know

You could actually just-

break the tan into sine and cosine?

[exotic oak]

yes

[raven spire]

$=\lim_{x \to \frac{\pi}{2}} e^{\frac{\sin x}{\cos x} \ln (\sin x - \cos x)}$

[warm shale]

Ansh

[raven spire]

Now, it's 0/0 form because $\cos x \to 0$ and $\ln(\sin x - \cos x) \to 0$

[warm shale]

Ansh

[raven spire]

Use your favorite L'hopital (ヘ･_･)ヘ┳━┳

makes sense!?

[exotic oak]

I have one question

[raven spire]

mhmm?

[exotic oak]

sinx is the one multiplying ln(sinx-cosx) right?

[raven spire]

Yep

[exotic oak]

cosx is muliplying 1

[raven spire]

yep

[exotic oak]

so since sinx / cosx = 0

it's a 0 / 0

[raven spire]

$=\lim_{x \to \frac{\pi}{2}} e^{\frac{\sin x}{\cos x} \frac{\ln (\sin x - \cos x)}{1}}$

[warm shale]

Ansh

[raven spire]

No.. wait... what?

[exotic oak]

0 * 0

[raven spire]

how is sin x / cos x = 0 here?

"x" isn't going to 0

It's going to π/2

cos x -> 0

the denominator goes to zero

and the ln term in the numerator goes to zero

[exotic oak]

cos(pi/2) is zero right?

[raven spire]

yeah (ヘ･_･)ヘ┳━┳

[exotic oak]

so sin x * ln(sinx-cosx) / 0 * 1

[raven spire]

yep

[exotic oak]

alright

thank you

[raven spire]

wait

._. You ended that so vaguely I'm confused as heck

lemme write it up for you ✓

[exotic oak]

Lim is not really my strong, sorry if it gets annoying to you or so

[raven spire]

$=\lim_{x \to \frac{\pi}{2}} e^{\frac{\sin x}{\fbox{\cos x}} \frac{\fbox{\ln (\sin x - \cos x)}}{1}}$

hmm lemme see how I can fix it (ヘ･_･)ヘ┳━┳

[warm shale]

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

[raven spire]

Yeah

well anyways

@exotic oak the boxed terms make the indeterminate form 0/0

[exotic oak]

yes

I got that!

[raven spire]

Good luck !

And ping me with your answer lol

[exotic oak]

I will need too much luck

ahaha

Brb then

[obtuse pebble]

@exotic oak Has your question been resolved?

[exotic oak]

help @raven spire

[raven spire]

hehe

[exotic oak]

got hopefull that I was going the right way but solution is different

[raven spire]

Isn't it 1/e?

[exotic oak]

it is

I got "e"

[raven spire]

Well, looks like someone forgot how to differentiate as well 👀

[exotic oak]

no

[raven spire]

$=\lim_{x \to \frac{\pi}{2}} e^{\frac{\sin x}{\cos x} \frac{\ln (\sin x - \cos x)}{1}}$

[warm shale]

Ansh

[raven spire]

You had this?

[exotic oak]

Impossible 😂

yes I had

[raven spire]

$=\lim_{x \to \frac{\pi}{2}} e^{\frac{\sin x = 1}{\cos x} \frac{\ln (\sin x - \cos x)}{1}}$

[exotic oak]

then

[warm shale]

Ansh

[raven spire]

You put sin x limiting value 1?

[exotic oak]

d/dx of sinx / cosx

[raven spire]

whaaaaaaaaaat

nooooooo you dum dum

[exotic oak]

https://tenor.com/view/sad-baby-frown-cry-tantrums-gif-4649018

[raven spire]

$=\lim_{x \to \frac{\pi}{2}} e^{\frac{\sin x}{\fbox{\cos x}} \frac{\fbox{\ln (\sin x - \cos x)}}{1}}$

[warm shale]

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

[raven spire]

I told you only the boxed parts of the exponent are responsible for the indeterminate form

[exotic oak]

YES

but

[raven spire]

If you were gonna apply L'hopital, your numerator function should've been the log function

and the denominator must've been the cosine function

[exotic oak]

ohhhh

ffs

[raven spire]

the sine x limiting value is 1 so forget it

[exotic oak]

So

e^cos(x) + sin(x) / x / -sin(x) ?

[raven spire]

not sure how you got an x in there

$=\lim_{x \to \frac{\pi}{2}} e^{\frac{\sin x}{\cos x} \frac{\ln (\sin x - \cos x)}{1}}$

[warm shale]

Ansh

[exotic oak]

d/dx of ln

[raven spire]

$=\lim_{x \to \frac{\pi}{2}} e^{ \frac{\ln (\sin x - \cos x)}{\cos x}}$

[warm shale]

Ansh

[exotic oak]

yes

[raven spire]

$=\lim_{x \to \frac{\pi}{2}} e^{ \frac{\frac{\sin x + \cos x}{\sin x - \cos x}}{- \sin x}}$

[warm shale]

Ansh

[raven spire]

$= e^{ \frac{\frac{1+0}{1-0}}{-1}}$

$= e^{ -1}$

[warm shale]

Ansh

Ansh

[exotic oak]

why sinx + cosx / sinx - cosx?

[raven spire]

because differentiation chain rule

consider revising

[exotic oak]

dude

yes...

sorry

I'm just tired

I need to replace

1 / x with the expression

[raven spire]

:p

[exotic oak]

on (x)

thank you so much

finally it's done

[raven spire]

Have fun!

[exotic oak]

.close

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[near yoke]

on the topic of limits

[near yoke]

can you apply l'hospitals rule to only the numerator/denominator repeatedly until you get to an answer for the numerator/denominator only

[hazy marlin]

you can apply the rule repeatedly
so long as the conditions hold

[near yoke]

so if you

turn the denominator

into a fraction of itself

and that is 0/0 or inf/inf

you can use it

ah

[hazy marlin]

yea

[near yoke]

alr ty

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[upper widget]

how do i check how many roots a quartic function has?

[upper widget]

what is the method to do so

[sinful kraken]

use formula

[upper widget]

which formula

thats

[upper widget]

thats the quadratic formula

[sinful kraken]

thats what u asked

[upper widget]

how is that going to help

I said how do I check how many roots a quartic has

e.g. there is the discriminant for quadratics

what is there for quartics

[sinful kraken]

nvm then

[upper widget]

.close

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[sinful kraken]

what do you need help with ?

@timid silo

[timid silo]

Hi

[spiral maple]

<@&268886789983436800>

[timid silo]

I don’t get how to do the puzzle

[sinful kraken]

i think youre supposed to write the answer of each equation in the puzzle then i think you will see a pattern that will allow you to find the code

[timid silo]

Oh

Uh

Let me try that for a sec

I don’t see it

[sinful kraken]

what are your answers

[sinful kraken]

idk

your answers are correct tho

[timid silo]

Oh

Tanks

.close

[obtuse pebble]

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[crystal wind]

Part E, I am missing something.

[spice rune]

how did you get this answer?

[sour lantern]

should be a natural number

[crystal wind]

My first answer 4023 was from pluggin in 8 to the original equation. This current one was just me trying a difference/summation of the s(8) and s(11/4^1/3)

current answer has no real thought behind it. just plugging numbers in.

I wasnt sure if I had to only use numbers between (11/4^1/3, 8]

[spice rune]

have you tried s(8) - 2 * s((11/4)^(1/3)) ?

[crystal wind]

Why * 2?

But no, havent tried. just s(8) - s(the other)

i dont think that worked, unless I typed it wrong.

calculated

[spice rune]

Let $s(t_1, t_2)$ be the path travelled between $t1$ and $t2$, $T = \sqrt[3]{\frac{11}{4}}$

Then $s(t_1, t_2) = s(t_2) - s(t_1)$

Since the particle is moving in one direction for $t \in [0, T]$ and in the opposite for $t \in [T, 8]$, our answer should be:
[ \abs{s(T, 8) - s(0, T)} = \qty(s(8) - s(T)) - \qty(s(T) - s(0)) = s(8) - 2s(T) ]

[warm shale]

rept1d

[crystal wind]

Thank you! That makes sense, but where does the value of s(0) go in the final equality?

[spice rune]

oh, my bad

let's say s(t) = t^4 - 11 t

then the formula works

[crystal wind]

yeah, that makes sense!

The prior one still applies

and ill try that

Thank you! The opposite movement part really helped me conceptualize, and that answer worked!

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[meager belfry]

Asking this question again: Show that any distribution function F can be written as $F = aF_c+(1-a)F_d,$ where $F_c$ and $F_d$ are continuous and purely discontinuous distribution functions respectively, and $0\leq a \leq 1$

[warm shale]

MeowingLoudly

[meager belfry]

A function F is purely discontinuous if it equals the sum of its jumps, i.e. if F(x) = $\Sigma_{y\leqx} (F(y)-f(y^-))$ for all x

[warm shale]

MeowingLoudly
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[meager belfry]

that's the definition given for this

[last basalt]

This is the definition of convexity

Idk if that helps

[meager belfry]

maybe

any way to prove this using just tools from probability?

like baby tools

[last basalt]

I'm not sure

I believe convexity would indicate that the entire line segment between any two outputs of Fc and Fd would be contained in the overall distribution space

[obtuse pebble]

@meager belfry Has your question been resolved?

[last basalt]

So you can most likely prove this by showing that lim x-> infinity of aFc-(1-a)Fd = 1 and likewise for x-> 0 the expression is 0

Then you could show theyre increasing, bounded between 0 and 1 etc

[meager belfry]

Sure

I'll give that a shot, thanks

[warm shale]

Skid

[last basalt]

Like this

You should be able to do this for all conditions of a distribution

As the set is said to be convex by that definition

If you prove the necessary conditions for any arbitrary combination of distributions it means there would have to exist some combination to be that of the distribution F

@meager belfry

[obtuse pebble]

@meager belfry Has your question been resolved?

[obtuse pebble]

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[sage charm]

.

[sage charm]

Currently working on situation 1

These are my equations Standard form:
Equation 1: -(2x) + y - 2z = 5
Equation 2: x + 3y - z = 9
Equation 3: 4x + 4y – 4z = 36

Bit stuck on how to then get my x y and z

[green plank]

Use triple system of equations

So subtract both equations 2 times

Then sub

[obtuse pebble]

@sage charm Has your question been resolved?

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[limpid roost]

Is it possible to figure out if 3^9 is greater than or less than 4^7 without using a calculator?

[raven spire]

Yep

It's should be

[silk galleon]

it's useful to have $\log_{10} 2 = 0.3010$, $\log_{10} 3 = 0.4771$ memorized

[warm shale]

xdk1235

[silk galleon]

or tbh 4^7 = 2^14

and that's very doable

[limpid roost]

Ah, okay. Thank you.

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[dark charm]

How does Wolframalpha go from the summation to the alternate forms?

[dark charm]

More simplified:

I feel like I'm missing some rule about sums

[obtuse pebble]

@dark charm Has your question been resolved?

[hushed moat]

not really, my hint is to ||try writing out the terms (also ignore the 1-p factor since you can just take it out of the sum)||

[dark charm]

Yeah but I don't understand writing out the terms

[hushed moat]

like list them out

starting at i=0

then list the term for i=1

...

try to find things that cancel out or smth

[dark charm]

Oh ok

that makes sense I'm stupid lol

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[dark charm]

.reopen

[obtuse pebble]

✅

[dark charm]

Ok so I tried and now I'm here:

.close

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[fossil cipher]

Hi

[fossil cipher]

Can someone help with indices

Real quick

True?

Or

Help urgent

What’s the rule behind it

[balmy mortar]

look up exponent rules

Wrong

misread

why u got 2 different things

[fossil cipher]

Huh?

[balmy mortar]

You have 2 different things on the papers

I am super confused

anyways this is elementary exponent rules

look them up

[fossil cipher]

Ok

Are the all valid?

just confirming

Is this true?

Yes/No

Please 🙏

Hello?

I want to make g have no indices

[silent adder]

you'll want to square both sides

[balmy mortar]

no

[fossil cipher]

oh yes

[balmy mortar]

$$\left(g ^\frac{1}{2}\right)^\frac{-1}{2}=g^\frac{-1}{4}$$

[warm shale]

Shuri2060

[balmy mortar]

You apply the same exponents to both sides

so one of your first ones just doesnt work out cus of this

[fossil cipher]

Oh ic

Thanks

.close

[obtuse pebble]

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[ruby oyster]

is the inverse of y=f(x) x=f(y)

[spice rune]

no

[long sinew]

y = f(x) makes no sense, they serve similar purpose

[ruby oyster]

hmm

alright

wait

[long sinew]

f(x) means function

[timid silo]

@ruby oyster its x=g(y)

[ruby oyster]

yes thats why i deleted it cause ik that

[long sinew]

Ah ok

[ruby oyster]

srry i am learning about inverse functions

and to make an inverse function

[spice rune]

the inverse of f if f^(-1)
where f(x) = y and (f^(-1))(y) = x

[ruby oyster]

you rearange to find x

then you change x to y

[spice rune]

and f^(-1) is just a notation for another function (like g(y))

[ruby oyster]

ahh yes

why do we change all x to make y

like the inverse of y= 3/(x-1)

is x= (3+y)/y

but you chnage the y to x

[spice rune]

3/(x-1) = y means x = (y + 3)/y

and x = g(y) = (y+3)/y is the inverse function

[ruby oyster]

oh

but you can change y to x

[spice rune]

but since you usually use x as an argument and y as a function's value

[ruby oyster]

and say it is f-1

[spice rune]

you just rename it

so it is y = g(x) = (x+3)/x

[ruby oyster]

perfect

idky im getting so confused

this has cleared it up

thanks buddy

[obtuse pebble]

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[frigid oracle]

How valid is this proof? "Suppose there exists $N_0$ such that $s_n \le t_n \forall n > N_0$. Prove that if lim$t_n$ and lim$s_n$ exists, then lim$s_n$ $\le$ lim$t_n$."
Proof:
\\
Let lim$t_n$ = t and lim$s_n$ = s. Let $\epsilon > 0$, and observe that $|t_n - t| < \epsilon$ $\forall n > N_0$, $|s_n - s| < \epsilon$ $\forall n > N_0$
\\
$s_n \le t_n$ $\Rightarrow$ $s_n - s - t \le t_n - t - s$
\\
Thus: $|s_n - s| - t \le |t_n - t| - s$ $\Rightarrow$ $\epsilon - t \le \epsilon -s$ $\Rightarrow$ $-t \le -s$
\\
$\therefore s \le t$

[warm shale]

dulg

[balmy mortar]

\lim f(x)

looks a lot better in latex.

[frigid oracle]

i will keep that in mind, thanks

[balmy mortar]

How does this follow?

[frigid oracle]

you basically add - s -t to both sides

[balmy mortar]

oh nvm

i was reading -s+t in head

[balmy mortar]

https://i.imgur.com/k7S46YM.png

How does this follow from the above?

If s is greater than s_n or t greater than t_n

then ?

===
Something helpful for this proof might be to draw a rough diagram

And to mark epsilon

[obtuse pebble]

@frigid oracle Has your question been resolved?

[frigid oracle]

Good point. I kind of assume it since the textbook already went through it. Because of how absolute value is defined, where $|a| = a$ if a > 0, and $|a| = -a$ if a < 0, it follows that $-|a| \le a \le |a|$, since either $a = |a|$ or $a = -|a|$.

Thus: $-|s_n - s| \le s_n - s \le |s_n - s|$ $\Rightarrow$ $-|s_n - s| -t \le s_n - s - t \le |s_n - s| - t$
\\
Repeating the same process: $-|t_n - t| \le t_n - t \le |t_n - t|$

[warm shale]

dulg

[frigid oracle]

.
@balmy mortar ok wait

i am a bit stumped

realized something is a bit off halfway thru my last latex post

wait for me while i post this idea on how to prove $|s_n - s| - t \le |t_n - t| -s$

[warm shale]

dulg

[frigid oracle]

Consider case 1, where $s_n - s -t = |s_n - s| - t$. $s_n \le t_n$ implies $|s_n - s| - t \le t_n - t - s \le |t_n - t| -s$.
\\
Case 2, where I got stuck because I am not sure :(. Following the same process leads us to $s_n - s - t = -|s_n - s| - t \le t_n - t - s \le |t_n - t| - s$

[warm shale]

dulg

[frigid oracle]

The ambiguity comes from where $|s_n - s| - t$ would fit in Case 2

[warm shale]

dulg

[balmy mortar]

I would say to restart your proof

Because there really is no need to split cases

What you need to do

is draw a rough diagram

and indicate what you want to show with epsilon pictorially

[frigid oracle]

what would that look like?

I have never imagined real analysis using pictures

hmmm

[frigid oracle]

the entire proof? or this absolute value thing?

[balmy mortar]

Maybe all, cus there wasn't that much progress

I'm just saying restart in terms of ideas

Stuff will be reused

In particular, like you said

We will take this

Using this, we need to prove s <= t

idk if the picture will help you see what you need to do with these 2 inequalities

[frigid oracle]

not sure about the picture

i suppose they are converging to a set number, and one is def greater than the other

[balmy mortar]

Try to draw your own

and mark an epsilon, basically

@frigid oracle ah actually, I see an error in your inital proof

$$(\forall n > N_0)(\abs{s_n - s} < \varepsilon)$$
$$(\forall n > N_0)(\abs{t_n - t} < \varepsilon)$$

[warm shale]

Shuri2060

[balmy mortar]

You make these claims.

They are not necessarily true --- think carefully why.

This does not follow from the defn of limit.

[frigid oracle]

what is the difference between this and the defn of a limit?

where N is used, instead of N_0

[balmy mortar]

recall the defn of limit

Limit of sequence
$$(\forall\varepsilon\in\bR^+)(\exists N\in\bN){a_n:n\geq N}\subseteq(L-\varepsilon, L+\varepsilon)$$

[warm shale]

Shuri2060

[balmy mortar]

Or

[frigid oracle]

ah, N is a set of numbers in \bN, and N_0 is a one particular number?

[balmy mortar]

$$(\forall\varepsilon\in\bR^+)(\exists N\in\bN)(\forall n\in\bN) (n \ge N\implies \abs{L - a_n} < \varepsilon)$$

[warm shale]

Shuri2060

[balmy mortar]

No, you misunderstand

[balmy mortar]

You have probably seen this definition, correct

[frigid oracle]

yep

[balmy mortar]

Personally, I think this one is easier to visualize

The (L - e, L + e) represents an interval

In either case, you should realise the N_0 in the original question does not correspond to N

[frigid oracle]

how come?

[balmy mortar]

Let me write.

$$(\forall\varepsilon\in\bR^+)(\exists N_1\in\bN)(\forall n\in\bN) (n \ge N_1\implies \abs{s - s_n} < \varepsilon)$$
$$(\forall\varepsilon\in\bR^+)(\exists N_2\in\bN)(\forall n\in\bN) (n \ge N_2\implies \abs{t - t_n} < \varepsilon)$$

You should realise that the N's for different sequences are not necessarily the same

And moreover, then N0 in the original question has nothing to do with them

In fact --- the epsilons don't have to be the same either

But in this particular case, you can choose them to be the same if u like

[frigid oracle]

I see

[warm shale]

Shuri2060

[balmy mortar]

oops corrected

But what you CAN do

is construct a new N

that is helpful to you

[frigid oracle]

I just looked at the answer for this exercise

🤦

[balmy mortar]

The N's are differently located,not necessarily the same

[frigid oracle]

👍

[balmy mortar]

(N1 and N2 depend on epsilon)

So you the answer says you should take max(N0, N1, N2)

yes?

Then you can compare the inequalities

[frigid oracle]

you wouldnt mind if i shared u what my textbook says, right?

[frigid oracle]

its 9.9c, btw

sorry for all of this

[balmy mortar]

🤔

Your textbook doesn't even use epsilon-N to show

and uses some result instead

I mean it's honestly not hard to use epsilon-N

[frigid oracle]

Yeah, but atleast he proves the results in a more pedagogical way

[balmy mortar]

$$(\forall n\geq N_0) s_n \leq t_n$$
$$(\forall\varepsilon\in\bR^+)(\exists N_1\in\bN)(\forall n\ge N_1)(\abs{s - s_n} < \varepsilon)$$
$$(\forall\varepsilon\in\bR^+)(\exists N_2\in\bN)(\forall n\ge N_2) (\abs{t - t_n} < \varepsilon)$$

[warm shale]

Shuri2060

[balmy mortar]

You have these 3 things

You can combine them into one statement by setting a new N = max(N0, N1, N2)

$$(\forall\varepsilon\in\bR^+)(\exists N\in\bN)(\forall n\ge\bN)$$
$$(\abs{s - s_n} < \varepsilon \land \abs{t - t_n} < \varepsilon \land s_n \leq t_n)$$

[warm shale]

Shuri2060

[balmy mortar]

You work with just the inside

And you can use the triangle inequality

[frigid oracle]

what does the wedge mean?

[balmy mortar]

$$\abs{s - s_n} < \varepsilon$$
$$\abs{t - t_n} < \varepsilon$$
$$s_n \leq t_n$$

logical and

[warm shale]

Shuri2060

[balmy mortar]

So we have these 3 statements combined

$$s-t \leq t_n - s_n - t+s$$

[warm shale]

Shuri2060

[balmy mortar]

This idea

$$s-t \leq (t_n - t) - (s_n - s)$$

[warm shale]

Shuri2060

[balmy mortar]

$$(t_n - t) - (s_n - s) \leq \abs {(t_n - t) - (s_n - s)}$$

[warm shale]

Shuri2060

[balmy mortar]

I believe it's this? which would lead to the result

https://i.imgur.com/2UoEN1J.png

🤔

[frigid oracle]

thanks @balmy mortar

I truly appreciate helping me out

[balmy mortar]

no ive lost how to do it, but you should be able to show via this ^

👌

[frigid oracle]

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[obtuse pebble]

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[spring quest]

How do you solve this question?

[spring quest]

Trigonometric stuff aren't really my thing

[forest sinew]

which side will the largest angle be opposite

[spring quest]

I don't understand?

So you're saying the opposite will be the largest angle?

never mind this question

do you mind explaining what SOHCAHTOA this is?

is it sin or csc? <@&286206848099549185>

if i'm wrong please correct me.

[ruby finch]

first, draw the figure

[long sinew]

^^^^

[spring quest]

done

drew the figure gents

[ruby finch]

great

[long sinew]

Complementary angles

[long sinew]

Or that

[ruby finch]

alternate interior angles*

[long sinew]

Both work

🙂

[spring quest]

this is csc right?

if i'm correct

[ruby finch]

tan

[spring quest]

oh

[long sinew]

^^^^

Tangent

[ruby finch]

there's nothing to do with hypotenuse here

[spring quest]

thanks man :)

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[ruby finch]

you're welcome

[obtuse pebble]

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[strong vale]

What "otherwise" approach could be taken to express the quartic as a product of quadratic factors

[true rain]

You would do something like

$(ax^2+bx+c)(dx^2+ex+f)$

[warm shale]

KurtDee

[true rain]

But it's going to be uhhh pretty hard

[strong vale]

yh

[long sinew]

Sheesh thats a nightmare

[true rain]

So I mean like

if you know you have whole numbers

that makes it a lot easier

for example, a,d is pretty simple

[strong vale]

yh it would be easier

but no whole numbers

this question

[true rain]

and c,f would also be trivial

I'm not sure how unique this equation is

I dont think it is

I would start with those two assumptions

a,d is a pretty easy one to address

you can just factor by a,d

so you would get

$ad(x^2+\frac ba x+\frac ca )(x^2+\frac ed x+\frac fd)$

[warm shale]

KurtDee

[true rain]

which is effectively the same thing as just removing it

because a*d = 1

[obtuse pebble]

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[harsh phoenix]

I need some help whit this

[obtuse pebble]

@harsh phoenix Has your question been resolved?

[tardy epoch]

You'll more likely get help if you translate it to English

[obtuse pebble]

@harsh phoenix Has your question been resolved?

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[karmic scarab]

help pls

[obtuse pebble]

@karmic scarab Has your question been resolved?

[karmic scarab]

<@&286206848099549185>

I really dont know how to solve this

[sage geode]

A product is 0 whenever one of its term is 0, which is the case here

Can you see why?

[karmic scarab]

how did you get 0

i dont see why

can you show it to me visually

[sage geode]

I gave you a hint that one of the terms in this product is 0

(Try expressing nth term using n and then looking for its zeros)

[karmic scarab]

what is nth term

[long sinew]

n is a value

variable

to be exact

[sage geode]

For example here the first term would be considered to be (1 - 9/2)

The 2nd term then is (1 - 9/3)

And so on

[karmic scarab]

yeah

i understand tht

[sage geode]

Clearly the nth term in this case is (1 - 9/n)

[karmic scarab]

ok

[sage geode]

Now, what happens when n = 9?

[karmic scarab]

that the one tht is changing

then it is 1

then the value is 0

1-9/9 = 0

[sage geode]

9th term is 0, yes

Meaning the entire product is thus 0 too

[long sinew]

geometric series problem?

[sage geode]

0*anything = 0

[long sinew]

fun

[karmic scarab]

thx so much'

tht was fun

[sage geode]

This doesn't seem to be a geometric series tho

[karmic scarab]

thx

[sage geode]

Np

[karmic scarab]

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[obtuse pebble]

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[timid silo]

Hello

[timid silo]

Here, I do not understand third step:

You use the result from (1) and put it into (2)

@timid silo

Okay. Thank you 🙂

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[wraith gazelle]

I'm trying to work on 8

[slate zephyr]

Have you tried defining a homomorphim from <a> x <b> to G using the obvious generators of the former ans showing that it is bijective?

[wraith gazelle]

We haven't learned homomorphism yet

Oops

[slate zephyr]

How do you define isomorphic then?

[balmy mortar]

Do you get the Q?

The idea behind it

[wraith gazelle]

A slight idea about it but I'm not too sure

[balmy mortar]

I usually find it helpful to use small examples

Say C_3 and C_5

Then what does the direct product look like

Just to get you thinking in the right direction of what the question is asking

I think the hard bit is seeing what the group <a> x <b> looks like if you're unfamiliar with direct products

@wraith gazelle Sorry that was a bit badly written by me 😂

[wraith gazelle]

👀

C_5 and C_3 as in the cyclic groups?

[balmy mortar]

yh well...

Basically yes

My example is

m = 15

a has order 3
b has order 5

so <a> is the same as C_3

<b> is the same as C_5

if that makes sense? same structure/isomorphic

[wraith gazelle]

I'm a bit confused about cyclic groups

I have a slight idea but not well explained

[balmy mortar]

Alright let me see

${e, a, a^2}$

[warm shale]

Shuri2060

[balmy mortar]

This is what C_3 looks like

in general

The structure will always be the same

$C_n = {e, a, a^2, \cdots, a^{n-1}}$

[warm shale]

Shuri2060

[balmy mortar]

You get this part?

a is just any 'object', it doesn't really matter...

The group operation tells you how to handle multiplication between elements (making it work like 'normal multiplication')

[wraith gazelle]

Oh I see

[balmy mortar]

What's different is a group might not be abelian

But all cyclic groups are abelian

Since you've just got powers of the same element

So <a> represents the cyclic group generated by a

if a has order m (from the question) and b has order n

$$\langle a\rangle={e, a, a^2, \cdots, a^{m-1}}$$
$$\langle b\rangle={e, b, b^2, \cdots, b^{n-1}}$$

[wraith gazelle]

If a has an order m, does it also mean a^m=e?

[balmy mortar]

right.

You can see that must be the case in these 2 groups.

Note the group operation is 'borrowed' from G

If we're going by the question

We have the same operation, but just different domain

[wraith gazelle]

I see

[warm shale]

Shuri2060

[balmy mortar]

looks better

Alright, so now you want to think what the direct product of those 2 look like

You are familiar with the definition of direct product?

[wraith gazelle]

Not really

[balmy mortar]

you can think of it as like cartesian product

$\bR^2 = {(a, b) : a, b \in\bR}$

[warm shale]

Shuri2060

[balmy mortar]

Similar idea, but for groups

$$A\times B = {(a, b):a\in A, b\in B}$$

[warm shale]

Shuri2060

[wraith gazelle]

But in this case, we're doing this with order m and n?

[balmy mortar]

yh... but first of all you need to be familiar with how the group operation works

it turns out exactly how you expect in some ways

$(a, b)(c, d) = (ac, bd)$

[warm shale]

Shuri2060

[balmy mortar]

Note there are 3 different group operations going on here (but all 3 are referred to as 'multiplication)

$(a, b)\left(_{A\cross B}\right)(c, d) = (a\left({A}\right)c, b\left(*{B}\right)d)$

hmm that's ugly 😂

If that makes sense?

[warm shale]

Shuri2060

Shuri2060

[balmy mortar]

These are 3 different binary operations in general.

[wraith gazelle]

I'm a bit confused, which are from which?

[balmy mortar]

I am using usual function notation to show

Or I could use different symbols if that makes it clearer

[wraith gazelle]

Yeah

[balmy mortar]

$_1:A\to A$ \
$_2:B\to B$ \
$*_3:A\times B\to A \times B$

Just treat them as any symbols

[warm shale]

Shuri2060

[balmy mortar]

$(a, b)_3(c, d) = (a_1c, b*_2d)$

[warm shale]

Shuri2060

[balmy mortar]

This is an important thing to note

So $a *_1 c$ makes sense because a and c come from A

[warm shale]

Shuri2060

[wraith gazelle]

And these are the operations of the three groups?

[balmy mortar]

Yes

Because of how we defined A x B

a, c come from A

b, d come from B

And by closure, ac is in A

bd is in B

so (ac, bd) is still in A x B

[wraith gazelle]

Ah yes

I understand

[balmy mortar]

So I'm going to use a small example

$$C_2\times C_3 = {(a, b):a\in C_2, b\in C_3}$$

[warm shale]

Shuri2060

[balmy mortar]

$${(e, e), (a, e), (e, b), (a, b), (e, b^2), (a, b^2)}$$

[warm shale]

Shuri2060

[balmy mortar]

Can you see how this is the entire group?

[wraith gazelle]

How come there's no a^2?

[balmy mortar]

C2

is just

{e, a}

All good?

[wraith gazelle]

Ah yes

[balmy mortar]

So let's stick with this example

So in the original question let's say m = 6

So I now have <a> x < b>

a = 2, b = 3

and they are relatively prime.

[balmy mortar]

You now need to show this group must be isomorphic to the original group

[wraith gazelle]

I'm still confused how you get the whole thing

[balmy mortar]

ok

You understand I got this from the original definition, yes or no?

[wraith gazelle]

Well I understand {e,a} in C2 and {e,a,b} in C3?

[balmy mortar]

$$C_2 = {e, a}$$
$$C_3 = {e, b, b^2}$$

[wraith gazelle]

Or is it e b, and b2

[warm shale]

Shuri2060

[wraith gazelle]

Oh

[balmy mortar]

C_n is the cyclic group of order n

There is only 1 generator

for cyclic groups

You are familiar with that right?

A finite group of size n is cyclic if one element has order n. If that is true, then all the elements must be some power of that element

Then everything must be generated by it.

[wraith gazelle]

Oh okay

[balmy mortar]

So now this comes from me just writing out all the possible elements

[wraith gazelle]

Mhm

[balmy mortar]

So to familiarize yourself with this

What is (a, b)^2

[wraith gazelle]

Written in this way? Or?

[balmy mortar]

You can go step by step

So first

(a, b)^2 = (a, b)(a, b)

[wraith gazelle]

(a+a,b+b)?

[balmy mortar]

That all depends on the group operation

but we usually prefer to write like multiplication

$(a, b)^2 = (a, b)(a, b) = (a^2, b^2)$

So like this.

[warm shale]

Shuri2060

[balmy mortar]

For the general case it will be written like this (the multiplication sign is omitted, but they are different)

[wraith gazelle]

Oh okay

[balmy mortar]

But then, this must correspond to one of these

which?

[wraith gazelle]

(e,b^2)

[balmy mortar]

right

because a^2 = e

[wraith gazelle]

Mhn

[balmy mortar]

For clarity, sometimes it might be written like this

$${(e_A, e_B), (a, e_B), (e_A, b), (a, b), (e_A, b^2), (a, b^2)}$$

[warm shale]

Shuri2060

[balmy mortar]

Because e_A and e_B are not necessarily the same

They are just the identity elements from A and B

This is used when it is clear from context or doesn't matter.

So which of these is the identity element of A x B ?

[wraith gazelle]

(e,e)?

[balmy mortar]

yep

I think you get how this group works

[wraith gazelle]

I do now

[balmy mortar]

So the question...

You need to show this thing is isomorphic to G

Do you remember how an isomorphism works, roughly?

Just the main concept

[wraith gazelle]

Show it's bijective and morphic

[balmy mortar]

yah but the main concept is

If we have 2 groups

X and Y

and they are isomorphic

I can 'relabel' the things in X with some things in Y

And the group will still behave exactly the same

Just different labels (and group operation)

[wraith gazelle]

f(xy)=f(x)f(y)

[balmy mortar]

Yes, that is homomorphism. And then you also need bijection

So this isn't an easy question, there are several steps to it

You could focus on group G first.

It tells you $\gcd(m, n) = 1$

[warm shale]

Shuri2060

[balmy mortar]

a has order m
b has order n

Hmm

Thinking 😂

Do you have any idea what this could tell you about G?

[wraith gazelle]

I'm no sure

[balmy mortar]

So we have G is abelian also, forgot to mention

But think about the element $ab$

[warm shale]

Shuri2060

[wraith gazelle]

If this helps, these are the problems that the question hints about

[balmy mortar]

Right exactly

$(ab)^2 = (ab)(ab) = a^2b^2$

[warm shale]

Shuri2060

[balmy mortar]

First of all, the group is abelian

So when we take powers of ab

We can rearrange like this