help-38

[stable crescent]

repost it

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[civic hamlet]

if i have a form of (2x^3)(x+1)/((x+1)^3)^2 what is the order of operations because cutting the bottom factor before and after is different

[lone basin]

$$\frac{2x^3 \qty(x + 1)}{\qty(x + 1)^3)^2}$$

[solid kiln]

King Leo

[civic hamlet]

yes

i assume it would need to be simplified

to (x+1)^6

then it would be (x+1)^5

[zinc ginkgo]

after cancellation yes

[civic hamlet]

thanks

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[wheat pier]

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[stark salmon]

Double integral of cos(x) in the domain determined by the area between the curves y=x, y=2x and x=1. Solve using both methods (dxdy/dydx) and check the solution is the same. Can't get the same answer tried many times, I don't know what I'm doing wrong
These are the integrals that I did and the answers that I got. I don't have the paperwork right now but it's just solving simple integrals so there are no mistakes there I checked that 100 times.

[ionic pendant]

can you show a picture of the region you are working with? this will make any issues with the bounds more clear

[stark salmon]

I think I just realized the mistake sorry after an hour

In the dxdy integral I have to split it no?

[ionic pendant]

yes

[stark salmon]

Sorry for wasting your time. Thanks.

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[lone basin]

Can someone check my work

[violet gust]

All good except you forgot to put -2.5, and instead wrote (0.5, 2.5) for 14b

[lone basin]

I see, thanks!

But in general, do I properly understand basic vector stuff?

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[lone basin]

.reopen

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✅

[violet gust]

well, lets just say you got the scalar multiplication down properly for the most part

[lone basin]

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[untold adder]

hm

[lone basin]

.reopen

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✅

[lone basin]

?

[untold adder]

How do I solve the translation?

[lone basin]

-_-

!occupied

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[lone basin]

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[untold adder]

:/

Was I supposed to do it from the original points or the rotation?

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[jagged fulcrum]

ik i am being stupid but can someone pls explain how to find theta 😭 😭 😭 i have lost sm braincells doing this i know its easy but i genuinely cant

[jagged fulcrum]

the question then asks for the exact value of 7sintheta + 3costheta

which i think i can do

i just dk how to find theta

[jagged fulcrum]

ye i see but how do i work out the lengths of the red sides

<@&286206848099549185>

[whole rover]

hey

you dont need to calculate theta directly

[jagged fulcrum]

oh

what do i do then

[jagged fulcrum]

same as theta

[whole rover]

yeah

[jagged fulcrum]

erm

im still not getting it

😭😭

[whole rover]

try calculate a length a and b

in theta

[jagged fulcrum]

in terms of theta

okay

[whole rover]

and using the two sides you know (1.5) (3.5)

[jagged fulcrum]

so a = 3.5sintheta

b = 1.5costheta

ohhh

i see

so

3.5 sin + 1.5 cos = 2.5

times by 2

i see

okay

thank you <3

[whole rover]

https://tenor.com/view/exactly-gif-3397598315635453816

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[pastel stone]

hi i needed help with this
it wants the area of the rectangles

[dense breach]

what are those black symbols?

one of them looks sort of like the number 1, the other I have no clue

[pastel stone]

2

[dense breach]

That's a 2??

[gleaming fox]

its arabic

[pastel stone]

yeah persian numbers but they say arabic i dunno

[dense breach]

Aren't Arabic numerals 0123456789 though

[gleaming fox]

their number is different

[pastel stone]

but could be im not sure

could you say how you solved it?

[dense breach]

huh

[royal patio]

Tamil hasn't dsicovered 0 yet

[kindred pier]

There's a neat book worth a casual read: The History of 0

[dense breach]

fake number tbh

[kindred pier]

Prove some relation about triangles ABE and AFO

[dense breach]

oh rip i messed up nvm (deleted my fake sol)

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[pastel stone]

i found it

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[analog pier]

solve for the real numbers x>0 and y >0 the system:
2x+3y-8/x=0
2y+3x-14/y=0

[analog pier]

no idea how I should approach this

[limpid dawn]

solve both for y and equate them

[analog pier]

mb i made a mistake

[limpid dawn]

you can still solve 2y+3x-14/y=0 for y

multiply by y and apply the quadratic formula

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[peak sun]

Hey, please can someone help me with this question

[foggy zinc]

After two hours, how far will Gordon have moved?

[peak sun]

No clue

[foggy zinc]

And how far will his amigo Brian have moved?

Well, you've been given the speed.

Suppose I'm driving my car at 100 miles an hour because I'm crazy. In an hour, how far do I travel?

[peak sun]

100

[foggy zinc]

Correct. So how far have Gordon and Brian moved in two hours, knowing their speed per one hour?

[peak sun]

Where's it say that that was there speed after an hour though...?

[foggy zinc]

'Per hour' means 'given an hour, how far can I go'?

[peak sun]

Yeah but I could be doing 70mph and have been driving for half an hour

So I only done 35 miles

[foggy zinc]

Correct.

[peak sun]

Oh....

[foggy zinc]

So how far do you go if you're driving for 2 hours?

[peak sun]

Nvm
Silly mr

Me*

[foggy zinc]

Recall that speed = distance/time, if you want to use that. You have the speed and the time

[peak sun]

After 2hrs gordan has done 8.8km hr

And Brian has done 9.6

[foggy zinc]

Correct.

Now draw what has happened and see what you can see.

[foggy zinc]

There are a few more things you know.

[foggy zinc]

You have the bearings.

Put those in too

[peak sun]

Do I use cosine rule?

To find missing side?

[foggy zinc]

Yep.

[foggy zinc]

I believe that's right

[foggy zinc]

I haven't computed it, but I trust you as long as you plugged it in right

Just make sure you haven't used radians instead of degrees which is what a lot of people accidentally do

[peak sun]

Do I need to square root it?

[foggy zinc]

Yep

You got $a^2$

[peak sun]

Ok

[solid kiln]

Bete Puttigieg 🐢

[foggy zinc]

oh, that's unexpectedly large

[peak sun]

A=12.7

[foggy zinc]

12.7 kilometres

Please do not forget the unit.

[peak sun]

Ok

And that's the answer isn't it?

[foggy zinc]

If you plugged it in right then yes

[peak sun]

Ok. Thank you! 🙂

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[twilit olive]

are those two different equations

[stoic oxide]

what do you mean

thats a vector with 3 elements, there is another comma

[twilit olive]

oh

i didnt see that

would this be the standard matrix of t

[stoic oxide]

yes

(if i understand what they mean by standard matrix)

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[twilit olive]

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[rigid lance]

are angle BFC and angle BFD a linear pair with vertex F

[lone basin]

✅

[rigid lance]

thanks

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[peak ridge]

How do I start?

[zinc ginkgo]

do you know how to find slope when given 2 points

[peak ridge]

y1-y2 / x1-x2

[zinc ginkgo]

you know one of your points P

(1) is asking you to plug in x=0.5 into Q to get your 2nd point

[peak ridge]

So I just keep plugging the numbers in? What about the curve equation

[zinc ginkgo]

observe that Q is a point on the curve

[peak ridge]

So is Q x?

[zinc ginkgo]

Q is a point on the curve and x is the first coordinate of Q

[peak ridge]

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[late rivet]

So how did the 1/2 disappear in the 4th line

[late rivet]

im ngl im so confused

[limpid dawn]

1/2 = 2 × x⁴/2 × 1/(2x⁴)

so it resembles

a²+2ab+b² = (a+b)²

[silent canyon]

What is the original question?

[late rivet]

yeah that flew over my head 💀

wouldve never figured that out without your help bahaha

thank you

[silent canyon]

Arc length formula is 1+f'?

[late rivet]

the integral of sqrt(1+(f'(x))²)

[silent canyon]

I forgot how it works, it's been a while

[late rivet]

so have i 💀💀

my algebra has been failing me today

and i have an exam tmrw

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[glad grotto]

yo can someone help me out with 3, 4 & 5? And can you check if my answer for 2d is right?

[tranquil grove]

The big U stands for union

Which means or

And the upside down u means and

[glad grotto]

Yea ik that

Is my answer for 2d right tho?

I'm a bit confused on 3 tho

Brb

I'm back

<@&286206848099549185>

[sacred sapphire]

no

[glad grotto]

What's the right answer?

A & B are right its just the c that's wrong I think

What goes in there?

[sacred sapphire]

there are some elements in C that should also be in B

[glad grotto]

Really?

Which ones?

[sacred sapphire]

B is all even numbers to 20

and C is all 2 digit numbers

so numbers like 14 should be in both B and C

same for A and C

[glad grotto]

Ohh ok

I see now

I thought that the numbers in c are supposed to be different from both a & b

[sacred sapphire]

2c is also wrong

12 should not be in both A and C

12 is not odd

[glad grotto]

💀

Oh yea my bad

Idk why I wrote that

What about 3?

Can u help me with that?

[sacred sapphire]

just list the sets out

and write about it

[glad grotto]

Write what about it

What's that little line on top of the a?

Cardinality?

[sacred sapphire]

thats the complement

[remote wave]

i think thats "not A"

[sacred sapphire]

so everything that is not inside A

[glad grotto]

Ok so A complement is everything not inside of a

All even numbers to 20

A and C

Overlap

A or B all of a & b?

A or C complement idk this

What's d and f?

[sacred sapphire]

list the elements out if you're confused

[glad grotto]

e is all odd and double digits

Pls help me with d and f

I don't understand

Also for question 4. The cardinality is n(A) right?

Is that what it means

?

Anyways what's d and f?

[sacred sapphire]

ok d is (A union C)'

[glad grotto]

A or C complement

[sacred sapphire]

A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
C = {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

[glad grotto]

So all odd digits and double digits

The complements are the ones that are not in A or C?

2, 4, 6, 8

I'm lost

[sacred sapphire]

ok you have this

what does this mean

[glad grotto]

Even numbers

Idk

[sacred sapphire]

even numbers that are?

[glad grotto]

B

[sacred sapphire]

no

C is about two digit numbers

if you take the complement of that what do you say

[glad grotto]

All odd and even 1 digit numbers

Idk

[sacred sapphire]

well almost

[glad grotto]

Or just 1 digit numbers

1-9

[sacred sapphire]

its all even numbers that are 1 digit

thats d

[glad grotto]

Huh

Even?

So 2, 4, 6, 8

[sacred sapphire]

yea

[glad grotto]

So I was right?

[sacred sapphire]

i mean you listed out the elements

[glad grotto]

Alr

[sacred sapphire]

but you gotta write it in words

[glad grotto]

True

f)?

A complement and C complement

A is odd, C is 2 digits

So..

Even 1 digit numbers

Same thing as d

[sacred sapphire]

yep

[glad grotto]

Really?

4 is the same thing right?

[sacred sapphire]

yeah

but list the elements out instead

[glad grotto]

Didn't I already list them

My teachers asking for too much

I understand this part

So let's skip it

[sacred sapphire]

3 is to describe it in words without listing

[glad grotto]

Let's do 5

[sacred sapphire]

4 is to actually list it

[glad grotto]

If u have time

[glad grotto]

yea that's light

[sacred sapphire]

do you know what disjoint means?

[glad grotto]

I forgot

I started this today

Nothing in common?

[sacred sapphire]

disjoint means two sets have no overlapping elements

yep

[glad grotto]

Oh

So..

Hmm

B and A?

[sacred sapphire]

yep

[glad grotto]

Perfect

Thanks for ur help!

I gtg now

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[flint osprey]

How do I solve these?

[frozen plover]

I think the first two are ordered pairs (x, f(x))

[tulip perch]

the inverse flips the coordinates

[frozen plover]

so f(-5) = 3

take the inverse on both sides

[flint osprey]

what does this mean?

[solid kiln]

neon

[flint osprey]

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[brave arch]

Is this right? im not aware of the logrithimitic rule

[zinc ginkgo]

,tex .log rules

[solid kiln]

riemann

[brave arch]

oh thx

[crisp bronze]

A quicker way to that outcome is using the log(m/n) = log(m) - log(n) property.
ln(1/4) = ln(1) - ln(4) = 0 - ln(4) = ln(4)

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[shrewd ridge]

there are two line segments 1 cm apart, 2 cm long

[shrewd ridge]

i pick a point randomly on the left one, and i draw a line at a random angle

so I get a segment in between, sometimes

what's the average length

i did this

[lone basin]

Find average value of |tan(x)| on [0, pi]

[shrewd ridge]

and i did a simulation, 1.17106

[lone basin]

Average length of all possible segments, or just the ones that will touch the surrounding lines

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@shrewd ridge Has your question been resolved?

[shrewd ridge]

i don't understand

i bet the calc solution is wrong

but how

[slim temple]

hes asking if we take into account segments that dont connect to the other line > obviously not cause if not the average would be infinite under this problems "rules"

so just the segments that do connect between the two

Just asking, could we base the average length on the average angle?

[zinc ginkgo]

What does random angle mean? Unif([0,2pi])?

[shrewd ridge]

ok i get it

[slim temple]

figured a bit, the problem basically reduces to, chose two points at random from a segment with 2 units of length, what is the average difference/distance

based on that you can figure the average angle, and with that, the average length of the segment

[shrewd ridge]

it doesn't average by itself just because it would be neat

[shrewd ridge]

yeah, or [0,pi] same thing

[slim temple]

ill work with that distance as this average i do not know as "x", cause from that point onwards the problem ends up being pretty easy+

[ionic pendant]

the angle bounds seem wrong

maybe i'm imagining a different angle, though

[slim temple]

solution unless im heavily dumb, with x being the average distance between two points on a segment chosen at random of a segment of length 2

[shrewd ridge]

starts from 60° and reaches 90° at middlepoint

[slim temple]

i dont know statistics, so chatGPT will be my saviour

[shrewd ridge]

so total rad·cm divided by total radians

[slim temple]

the average length of the segment is 5/3 cm, neat

[ionic pendant]

i think i must have been measuring from a different starting place

[shrewd ridge]

yeah that happens

thanks everyone

.solved

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[reef oyster]

question 6 and 8

im having trouble doing these becaue i missed classes for being sick just wondering if someone can copme and help me understand the prosses to solving them

also questions 11 and 12

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[trim stratus]

I need some help with functions

[trim stratus]

So in task 2

It says i need to find the theme(?) Of the function

The zero poinz

Opening of the parabole

Axis od symetry

Extremhm of the function

And the minimum value of the function

Please guys help

[kind token]

I'm assuming you need to identify whether it is a parabola or hyperbola that sort of thing

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Nmv guys i figured it ouz

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[storm yarrow]

Can someone explain why it's e?

[safe loom]

you can use euler's formula to see why

[storm yarrow]

ok so e^iz = cosz + i*sinz

conjugate would be cosz - i*sinz

= e^-iz

[storm yarrow]

?

where am I wrong here

[safe loom]

Denote * as conjugate
(cos(z))* = cos(z*)
(sin(z))* = sin(z*)

[storm yarrow]

Shouldn't it only affect the imaginary part?

[safe loom]

that is imaginary as well because z is a complex number

[storm yarrow]

ok so if z = a + ib

It's cos(a-ib) - sin(a-ib)

which is e^-ai-b

ooh ok I got it

thx

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[warm valley]

Was wondering if this is sound reasoning, would appreciate any feedback, or any alternative methods that may be faster. And yes i understand that a step like checking for a difference greater than 1/3 is arbitrary for this question, but it was moreso as a general step for questions in such a style, as it may eliminate multiple options faster than subbing in x values.

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[warm valley]

.reopen

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✅

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[surreal spindle]

hi

[surreal spindle]

how do i find the GCF of over 2 number

for example 50 80 and 130

[remote wave]

find the biggest number that they are all divisible by

in this case it's 10

[surreal spindle]

yes but with bigger numbers it becomes harder and harder so is there a way to do it not in your head?

[surreal spindle]

more complex numbers too

[remote wave]

uh take the gcd of any two of them and then take the gcd of that and the 3rd number

[surreal spindle]

didnt learn gcd yet

ill keep searching google i

ig

[remote wave]

nah gcd and gcf mean the same thing

greatest common divisor or greatest common factor it doesn't matter

same meaning

[surreal spindle]

ok so i can only find the gcf of any number once

how would i find it again the second time for that gcf?

[remote wave]

like

for 50, 80, 130

gcf of 50 and 80 is 10

and then gcf of 10 and 130 is 10

[surreal spindle]

thats just doing it in the head though im trying to figure out how to do it manually so i can do it with far greater number

so ill give a more complex question

43 172 246

i think i have a method down though

lemme try it and see if i can find the gcf of that

[remote wave]

43 is prime

246 is 6 * 41

the gcf is just 1

[surreal spindle]

how do you find it quickly

i just try to guess or do in my head

but the same problem also occures here in larger numbers

[remote wave]

just divided 246 by 6

[surreal spindle]

i need a method though not the straight answer, it doesent help me to get the answer because i still dont have the method to find the factors of other numbers.

[remote wave]

why do you even need this

just try dividing by numbers to find the factors

[surreal spindle]

for advanced factoring techniques

one of the methods is finding the GCF

i understood how to find the GCF of 2 numbers with a rather easy method

but 3 or more is problomatic

[surreal spindle]

so it is just a guessing game

[remote wave]

how do you do it for 2 numbers

[surreal spindle]

https://youtu.be/7XgjqkvKHRQ?si=TtzmlG3g-TxOp7qP

tested it and it worked

[remote wave]

ok so if you have three numbers a, b, c then just do that on a, b and the do it again with whatever u get and c

[surreal spindle]

ok let me pick 3 numbers quickly and try that out for a sec

308 505 702

lemme do 308 and 505 first

didnt work

i think i fucked something up

i dont have the time to redo this cuz i have to go to work

but ill come back to you on this

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[vocal cobalt]

how do i solve this? do i use euler's method or i have no idea what this problem wants me to do...

[lone basin]

First, what’s your tangent line

[vocal cobalt]

y=2x-1

i found this using method somewhat similar (same?) as euler

[lone basin]

So the question is basically asking:

For what biggest $x$ value is:
$$\left| \qty(x^2 - 2x + 3) - (2x - 1) \right|<0.5$$

[solid kiln]

King Leo

[lone basin]

Idk what euler’s method is

[vocal cobalt]

its just some approximation method

but how do i determine the error if i dont know the correct one...?

[vocal cobalt]

i dont get this 😭

[lone basin]

The error in the vertical distance between the quadratic and your tangent line

[vocal cobalt]

oh so the tangent line and the parbolas distance should be less than 0.5 and we're finding the maximum x that makes the thing <0.5?

and are we using absolute values cause the distance is just distance here no signs needed

[lone basin]

The tangent line con be above or below by 0.5

[vocal cobalt]

this ones fast on calculator

thanks so much for helping!

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[vocal cobalt]

how do i solve this? (its calculator allowed question)

[vocal cobalt]

i dont know how to approach it

[lone basin]

,w maclaurin series e^x

[solid kiln]

Results provided by WolframAlpha

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Requested by .king.leo.

[vocal cobalt]

sign doesn't match?

[lone basin]

Oh wait

[lone basin]

@vocal cobalt use this series

[vocal cobalt]

what about the sign

and how am i supposed to know that it will be e^x or whatever how do i identify that?

[lone basin]

Okay so the sign is the problem

[lone basin]

You’re supposed to know that $e^x = \sum_{n = 0}^\infty \frac{x^n}{n!}$

[solid kiln]

King Leo

[lone basin]

This series can be simplified to:
$$\sum_{n = 0}^\infty \frac{(-x)^n}{n!}$$

[solid kiln]

King Leo

[vocal cobalt]

i should have known what e^x looks like its one of the MUST KNOW series 😭

what about the sign though

[lone basin]

If the sign has been changed

Then just change the sign

[vocal cobalt]

huh?

which sign?

[lone basin]

$x$ turned to $-x$. $e^x$ turns into?

[solid kiln]

King Leo

[vocal cobalt]

e^-x?

but how do i set it to - + - + - + mode

[lone basin]

$$e^{-x} = \sum_{n = 0}^\infty \frac{(-x)^n}{n!}$$

[solid kiln]

King Leo

[lone basin]

You don’t need to do that anymore

You have now found that the series is just e^(-x)

[vocal cobalt]

ohh so cause for -x ^ n sometimes its gonna be - and sometimes +?

[lone basin]

$$(-1)^n x^n = (-x)^n$$

[solid kiln]

King Leo

[vocal cobalt]

yea it makes sense

[lone basin]

So can you solve the problem now?

[vocal cobalt]

yeah i can find the intersection for those two

thanks for helping!

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[rose lotus]

I need to find monotonic intervals

purple is original funtion
red is derivitive

I have no idea how do i do it on paper because I can't draw something like this

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[pure fiber]

Hi, I would need help understanding the proof of Weierstrass theorem

[stable crescent]

post it

[pure fiber]

The professor told us to not learn the proof in a rigorous way, just understand the idea of it and how it works

[stable crescent]

sure post it

[pure fiber]

ok one moment

this is what he wrote, is it readable?

[stable crescent]

,rccw

[pure fiber]

oh it's upside down, sorry

I understand what we are doing but I don't understand how it proves the theorem

we end up with ${x_M} = \cap I_n$ so, since $I_n \subseteq [a, b]$ we have that ${x_M} \subseteq [a, b] \implies x_M \in [a, b]$

[solid kiln]

nico.alesi

[pure fiber]

but for this to work don't we already need to know that there is a maximum in [a, b] before starting?

[stable crescent]

State the theorem

[pure fiber]

Let $f: [a, b] \subseteq \mathbb{R} \to \mathbb{R}$ be continuous, then $\exists x_M, x_m \in [a, b] / f(x_m) \le f(x) \le f(x_M) \forall x \in [a, b]$

[solid kiln]

nico.alesi

[pure fiber]

basically if the function is continuous in a closed interval then that interval includes a point of maximum and a point of minimum

the part I don't get is the part that says "after splitting the interval, take the half so that max $f$ in $I_0$ = max $f$ in $I_1$"

[solid kiln]

nico.alesi

[pure fiber]

so we are actually pretending that there already is a maximum in $I_0$ = [a, b]

[solid kiln]

nico.alesi

[stable crescent]

it has to

[pure fiber]

but is it legal? to prove it pretending there is a maximum (to prove that there is a maximum)

[stable crescent]

well prior to bolzano's theorem, there is an even stronger theorem well its a fact tbh

consider any function that is continuous over a closed & bounded interval [a,b], that function has a min & max in that interval

[pure fiber]

yes but I am trying to prove that and I am assuming that what I am proving is correct

isn't it a loophole?

(sorry I'm not so acquainted with proofs, I don't study mathematics)

[stable crescent]

no

well wait a sec

[pure fiber]

ok, thank you for your time

[nova spire]

So let me recap, you want a proof of this Weierstrass theorem

My advice is to use Bolzano Weierstrass theorem

[pure fiber]

no no, our professor told us the idea of the bisection proof

[pure fiber]

this one

and I understood everything, my only perplexity is, can we assume there is a maximum to prove that there is actually one?

isn't it a loophole?

[nova spire]

they probably meant sup/inf?

[stable crescent]

well they kinda assumed f is bounded

[nova spire]

yes but it doesn't give existence of max/min

only sup/inf

[stable crescent]

still

[pure fiber]

the picture I sent is all I have

[nova spire]

yes so take I1 such that sup(f) on I0 = sup(f) on I1

and not max as stated

[stable crescent]

how would they know sup would exist in that interval though? OwO

[nova spire]

because the function is bounded, as you said

[stable crescent]

well thats the problem

they dont know if its bounded

yes ik

[nova spire]

yes we require BW

[stable crescent]

conti => bound

[pure fiber]

the sup exists for sure

[stable crescent]

yes thats what they try to proof

[stable crescent]

proof?

[nova spire]

weierstrass is a consequence of BW

not the opposite

[pure fiber]

because if we have $f: (a, b) \to \mathbb{R}$ then $f[(a, b)] = (c, d) \subseteq \mathbb{R}$

[solid kiln]

nico.alesi

[pure fiber]

this is another theorem we studied

[stable crescent]

add continuity there and yeah ur good to go

[nova spire]

and why would c,d not be infinite

[pure fiber]

well well well

nevermind

[nova spire]

especially since on this version you're considering open intervals

no, boundedness comes from BW

[pure fiber]

we straight up jumped that

[nova spire]

and again if you have BW

the "proof of Weierstrass theorem by bisection" is useless

[pure fiber]

we did 1. Weierstrass, 2. Existence of zeros, 3. Intermediate values

that's it

[pure fiber]

yo bro Idk what to say, this is what my professor told us, I have no idea

[nova spire]

then the teacher who didn't talk about bolzano-weierstrass did a mistake not talking about it

[pure fiber]

the thing is I am studying computer science and artificial intelligence so the role of these topics in my course is marginal

probably they selected the bare minimum

[nova spire]

and thus let out some half baked fake proofs to try to appease the math-oriented people in the course

[stable crescent]

its useful too tbh https://en.wikipedia.org/wiki/Pareto_efficiency

[pure fiber]

you didn't get what I meant

[stable crescent]

i do i do

[pure fiber]

I can't do anything about it bro ahahaha

[stable crescent]

well, it is marginal you'd say but it is what it is

[nova spire]

if you want to, I can try to give you a better idea of the proof. As easy to understand as possible

[pure fiber]

I didn't say all of it is marginal, some things are less useful than others, that's why, perhaps, they let out WB theorem Idk

Proofs are useless for me, I'm not studying mathematics or physics, I just need to understand how the theorem works and it's utility, how to apply it, not why it's true

the scope of that it's beyond my course, I mean, I trust the clever person who found that out ahahaha

[pure fiber]

we could try

[nova spire]

well weierstrass theorem is just about the existence of a global min and max of continuous functions on a closed interval

[pure fiber]

Ik but this half proof he gave us is just making confusion

that's why I think it would've been better not knowing anything about it

[nova spire]

well I can try to rectify the confusion in the best way possible

[nova spire]

Bolzano-Weierstrass theorem's proof is almost as easy as the proof of your theorem I'm gonna present

[pure fiber]

just explain it in a simple way, no need to be rigorous

[nova spire]

Bolzano Weierstrass tells you that from any bounded sequence

you can extract a convergent subsequence

that's all you need to know if you don't want the proof

[pure fiber]

ok

[nova spire]

so now take a continuous function $f:[a,b]\to \bR$

[solid kiln]

rafilou is not not born in 2003

[nova spire]

we're first gonna show it's bounded, and then we'll show it has a min and a max

[pure fiber]

ok

[nova spire]

ok so first of all, if f isn't bounded

that means there exists inputs such that the output is as big in the positive or in the negatives as I want

right

[pure fiber]

yes

[nova spire]

ok

so |f(x)| can be as big as I want

[pure fiber]

yes

[nova spire]

If I find the right x

so I can find x_n such that |f(x_n)| >= n for all n

[pure fiber]

yes

[nova spire]

still makes sense?

[pure fiber]

yes that's theopposite of the definition of bounded

[nova spire]

welp we just created a sequence $(x_n)$ in $[a,b]$ which is bounded

[solid kiln]

rafilou is not not born in 2003

[nova spire]

so we can extract a converging subsequence

name the limit of that subsequence x

it's pretty easy to see that since all x_n are between a and b

[pure fiber]

didn't understand this

[nova spire]

f is defined on [a,b]

[pure fiber]

yes

[nova spire]

so if find some input x_n for all n

the input must be in the domain of f

[pure fiber]

yes

[nova spire]

so I found x_1, x_2, etc...

that are all in [a,b]

[nova spire]

such that we have this

I picked them that way

[pure fiber]

yes, why is it bounded tho

[nova spire]

the domain of f is [a,b]

[pure fiber]

a function is bounded when the Image of the domain is bounded

[nova spire]

we're not talking about the image of the domain here

the x are INPUTS

so they come from the domain

not its image

[pure fiber]

yes but that's the definition

[nova spire]

and the IMAGE of those x_n

[pure fiber]

I know

[nova spire]

are bigger than n

because we assumed f to be unbounded

[pure fiber]

yes but what I'm not getting is why is the sequence bounded, for it to be bounded the image must be bounded

[nova spire]

the sequence (x_n) is bounded

because it's a sequence of INPUTS

so it lies inside the DOMAIN

[a,b]

which is bounded

[pure fiber]

oh ok

${a_n} = [a, b]$

[solid kiln]

nico.alesi

[pure fiber]

this?

[nova spire]

you mean subset?

${x_n}\subset [a,b]$

[solid kiln]

rafilou is not not born in 2003

[pure fiber]

no, I thought we were getting all of the inputs

[nova spire]

not at all

we're only getting a countable amount of inputs

we just want to pick a single x_n for each n

such that |f(x_n)| > n

[pure fiber]

ok

[nova spire]

ok

[pure fiber]

ok so the squence is bounded

[nova spire]

the sequence {x_n} is bounded

and so by BW

it has a convergent subsequence

we can name the limit of that subsequence x, which is also in [a,b]

[pure fiber]

mh

[nova spire]

so far so good?

[pure fiber]

yes

[nova spire]

ok

well

f is continuous

[pure fiber]

yes

[nova spire]

so |f(subsequence)| converges to |f(x)|

but at the same time |f(x_n)| keeps growing up to infinity

so it's impossible

and thus f must be bounded

[pure fiber]

okk

[nova spire]

now that we finally have f is bounded

you can either follow the stupid bisection proof, which works now that we did all this

replacing all mentions of max by sup

or

proving it this way:

since f is bounded, call M = sup(f)

[pure fiber]

ok

[nova spire]

create a sequence of x_n such that f(x_n) -> sup(f)

which is possible due to the characterisations of the supremum

{x_n} is once again a bounded sequence, as it's in [a,b]

take a convergent subsequence by bolzano-weierstrass

f(subsequence) -> f(limit) by continuity

but f(subsequence) -> sup(f)

sup(f) = f(limit)

[nova spire]

our limit serves as the desired x_M

repeat for the minimum

end proof

[pure fiber]

why sup becomes a maximum?

[nova spire]

because it's reached

sup(f) is in range(f)

as f(limit) = sup(f)

[pure fiber]

mmmhh

[nova spire]

an upper bound that is also an element of that set

[pure fiber]

yeah I know that

[nova spire]

is by definition, the maximum

[pure fiber]

I'm trying to understaned why it's reached

could we try the other path?

we know that f is bounded and ok

so we know that there is a sup

and we do the bisection of intervals

[nova spire]

ok so now that you have those intervals

we're gonna try to create a sequence {x_n}

such that each x_n is in I_n

and f(x_n) gets closer and closer to sup(f)

it's pretty standard

just take x_n in I_n such that f(x_n) > sup(f) - 1/2^n for example

since a_n <= x_n <= b_n

and {a_n}, {b_n} converge to the same limit

{x_n} converges also to that limit by squeeze theorem

call it x

and by squeeze theorem again, f(x_n) -> sup(f)

but f(x_n) -> f(x) by continuity

so once again we have f(x) = sup(f)

[pure fiber]

I think it'll take some time to understand this, so I'll read it again

thank you for your time tho

[trim jolt]

@pure fiber Has your question been resolved?

[pure fiber]

thanks @nova spire I kind of got it, I wouldn't be able to redo it and reexplain it in this way but since we have to know just the idea I will reproduce the main steps in a more detailed way

basically he said max of $f$ to simplify, it was $sup(f)$ instead

[solid kiln]

nico.alesi

[pure fiber]

and I kind of understood everything else

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[hot pebble]

Can someone explain why a circle or any shape with a curve is not a polygon? Aren't circles just shapes with infinite sides.

[lone basin]

I dont think infinity is a real number

[hot pebble]

It isn't but as you add more sides to a polygon it looks more like a rectangle. So wouldn't infinite just describe a circle.

[high patio]

It approqches a circle but never becomes one

[marble wharf]

frankly its just by definition

a polygon has finitely many sides

[hot pebble]

But if that's the case then is it that integrals we can find the area the curve with exact accuracy using infinitly small rectangles

[zinc ginkgo]

yes

[hot pebble]

I meant how is it that we can do integrals with infinite rectangle

[zinc ginkgo]

calculus ✨

[hot pebble]

I understand that calculus uses infinite intervals with rectangles to find the area but how can that be exact but using infinite sides of a polygon not be a circle

[zinc ginkgo]

this

[marble wharf]

by definition of the word a polygon just has finitely many sides

its just what the word means

[hot pebble]

Ok. But Am i right that a circle has infinite sides?

[zinc ginkgo]

no

[hot pebble]

Why

[zinc ginkgo]

a circle has 2 sides. inside and outside

[marble wharf]

sides are defined as straight edges of actual length

[hot pebble]

What does actual length mean

[marble wharf]

I mean not zero length

like your "infinitely many sides for a circle"

[hot pebble]

Ok

Thank you for your help

[marble wharf]

in integrals you never have infinitely many rectangles. you always have finitely many and then approach a limit

thats distinct from ever having actually infinitely many rectangles

[hot pebble]

Ok ty

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[wraith hinge]

If you have a bowl with six apples and you take away four, how many do you have?

[wraith hinge]

?

[wraith hinge]

how many

[jaunty scarab]

!noans

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[wraith hinge]

Ok just help me solve it

[bright quarry]

let me think about this one for a minute

[wraith hinge]

aight

[lone basin]

You take away four, and leave two in the bowl

So you have four apples

!done

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[wraith hinge]

Cuz I checked the solutions and it says 6

[lone basin]

:)

[bright quarry]

huh

you have a bowl

[wraith hinge]

so idk why

[bright quarry]

lmao

[lone basin]

But you dont have the apples in the bowl

Oh i see

Lol

[wraith hinge]

yeah

ohh

cuz you still have the 4 apples you took out

got it

how do i close

[bright quarry]

!done

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[lone basin]

!done

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[wraith hinge]

.close

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[boreal shuttle]

given $x > 0 ,ln_x \leq x - 1$ and $(x_i){1\leq i \leq n} , M = \frac{1}{n} \sum{i=1}^n x_{i} $

[solid kiln]

<rajel />

[boreal shuttle]

prove that $\frac{1}{n} \sum_{i=1}^n ln_{x_i} \leq ln\left(\frac{1}{n}\sum_{i=1}^n x_i \right)$

[solid kiln]

<rajel />

[boreal shuttle]

i mean i know i can start with the interval given

and would probably end up with
$$\sum_{i=1}^n ln{xi} \leq\left(\sum_{i=1}^n ln_{x_i}-1 \right)$$

[stable crescent]

AMGM

[nova spire]

this inequality is called Jensen inequality

[solid kiln]

<rajel />

[nova spire]

it doesn't have to do with ln(x) <= x-1, or at least it's not directly related to it

[boreal shuttle]

oh really

[nova spire]

have you studied second derivatives?

[boreal shuttle]

yep

but i dont think this inequality is studied in my class

thats why they probably gave us the ln(x) <= x-1

[nova spire]

problem is this is bounding ln(x) by some function that doesn't have "ln" in it

well maybe there is a way

but I don't see it at all, and it's not the most obvious

as was suggested, either you know about concave up/concave down functions and you use jensen

or AM-GM

[boreal shuttle]

ok thx for ya both

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[stable crescent]

use induction @nova spire

@boreal shuttle

[boreal shuttle]

what gonna be the variable

[stable crescent]

u suppose its true for n

and prove it is true for n+1

[boreal shuttle]

well in the exercice they insist to use the inequation given

[stable crescent]

yes

u use it when u have the induction to prove

[boreal shuttle]

i'll try that thx

[stable crescent]

$\frac{1}{n+1} \cdot \sum_{i=1}^{n+1} \ln(x_i) = \frac{n}{n+1} \frac{\sum_{i=1}^n \ln(x_i)}{n} + \frac{\ln(x_{n+1})}{n+1}$

[solid kiln]

Goëtia

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[blissful marsh]

how do we know angle C is 39.3 degrees?

[zinc ginkgo]

sine law here

[blissful marsh]

yes but im failing to understand why angle C is 39.3 degrees

[zinc ginkgo]

,calc asin(sin(25 deg) * 12 / 8) * 180 / pi

[blissful marsh]

this isnt helping me understand why angle C is 39.3 degrees

[solid kiln]

Result:

39.340476504462

[zinc ginkgo]

do you know what inverse sine function is?

also called arcsine

[blissful marsh]

oh is it
sinC/8 = sin25/8?

and then thats 8*sin25/8 ?

then you arc sin that to get the angle ?

[zinc ginkgo]

12 isntead of 8 somewhere

[zinc ginkgo]

yes this follows from sine law, once you fix 12 instead of 8.

[blissful marsh]

i understand how the sine law works. i dont understand why C is 39.3 degrees.

you are just giving me numbers and formulas. what did this person do to get 39.3 degrees as angle C

[zinc ginkgo]

yes sine law is a formula

[zinc ginkgo]

you have to solve for sinC before being table to take arcsine of both sides

if $\sin(C) = z$ then $\arcsin(\sin(C)) = C = \arcsin(z) $

[solid kiln]

riemann

[blissful marsh]

I got it now. I think I was misunderstanding why we cant use 8 instead of 12, but i kind of get it now intuitively but i cant explain it

I think sin B being equal to both 39.3 and 140.7 caught me off guard as 39.3 is also angle C

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[wintry zealot]

may someone help

[bright seal]

Where do you need help?

[wintry zealot]

like all of them

i don’t understand the process

[bright seal]

So from A)?

[wintry zealot]

c please

[bright seal]

Its correct

[wintry zealot]

but with bedmas don’t you do division first then multiplying ??

[bright seal]

No

[wintry zealot]

how

[bright seal]

Depends in the order of how division and multiplication are written

[west sleet]

Mult and division take the same precedence

[bright seal]

From left to right

[wintry zealot]

im confused

how do you know what to do then

[bright seal]

So if I leave from the left there is first ×, then I do the multiplication first

$10*5/ 25 = 50 /25 = 2$

[wintry zealot]

Oh

Wait so

[solid kiln]

Task Bot

[wintry zealot]

okay so let me try F

I don’t even know how to do f 😭

what first

[bright seal]

In the expressions the powers are calculated first, then multiplications and divisions, then sums and subtractions; Operations with the same priority must be carried out from left to right.

[wintry zealot]

okay i got 31

[bright seal]

If there are relatives first, the rounds are resolved, then the squares and finally the staples; In each case the general rule applies.

$15-2^2+5*4=15-4+20=31$

[solid kiln]

Task Bot

[wintry zealot]

do you know how to do G

I’ve done 6 ÷ 2 so far

and got 3

[bright seal]

Okay

Then did you finish the calculations?

[wintry zealot]

no because idk what to do next

because i did 3+ 11 (3 from dividing 6 and 2) then i got 14 which you cannot subtract 14 from 5+7=12