help-23

[safe radish]

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[small epoch]

Does there exist a natural number n such that there exists a natural number x that satisfies the equation: x^2-x-1 = F_n where F_n is the nth Fibonacci number given that n>11 ?

[near sky]

this is from yesterday

do you have any progress

[small epoch]

Yeah, ik but I didn’t really get anywhere

[small epoch]

not really

[near sky]

ic

[wild girder]

you need help?

[safe radish]

@small epoch Has your question been resolved?

[small epoch]

I mean we could use the fact that a natural number z is a Fibonacci number if and only if 5z^2-4 or 5z^2+4 happens to be a perfect square. Then letting z = x^2-x-1, we have that 5(x^2-x-1)^2 ± 4 = y^2 where we have to solve for all (x,y). Now the thing is there exists some solutions to this equation, but I want to know whether there exists one such that z > F_11 = 89.

And I mean because of the discriminant of the quadratic equation in x, we have that 5 + 4 * F_n must be an odd perfect square

[desert belfry]

If you have some b^2 = 5+4*a, you can construct a sequence of a_i

That being a_i = i^2 + i - 1

And you need to check if any F_n fit that form

[small epoch]

You mean i^2-i-1

[small epoch]

And that wouldn’t really work because it is again a restatement of the problem. We need to find and prove all solutions

[desert belfry]

3^2 = 4*1 + 5
5^2 = 4*5 + 5
7^2 = 4*11 + 5
9^2 = 4*19+ 5
11^2 = 4*29 + 5
13^2 = 4*41 + 5

that's i^2 + i - 1

[small epoch]

Okay, I get that, but how would we find all Fibonacci numbers from the sequence: 1,5,11,19,29,41,….

[small epoch]

Your interpreting it as 2i+1, whereas I’m interpreting it as 2i-1. Either way you can see that it didn’t make a difference since they’re both odd

[safe radish]

@small epoch Has your question been resolved?

[safe radish]

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[twin prawn]

circumference + perimeter = 10 right

hmmm

if perimeter is 4x

can we say that

4x - 10 is the circumference of the circle already?

[brave wolf]

opposite

10 - 4x

[twin prawn]

oh crap right

so 10 - 4x + 4x = 10

[brave wolf]

okay so now you need to express the areas somehow

what would the area of the square be?

[twin prawn]

x^2

[brave wolf]

and now the circle, that one is slightly more difficult

you know that the circumfence is 10-4x

[twin prawn]

yeah we gotta do some algebra shit to relate circum to area of circle

10 - 4x = circumference of circle aka length of circle

OH I have an idea

pie * diameter = 10 - 4x

10 - 4x / pi = diameter

[brave wolf]

it's pi btw, not 🥧

[twin prawn]

yeah

LOL

[brave wolf]

radius might be a bit more useful for area calculations

[twin prawn]

yeah

radius = diameter /2

[brave wolf]

and its much better to use parenthesis, (10 - 4x) / pi = diameter

[twin prawn]

therefore,

[(10 - 4x) / pi] / 2 = diameter /2

[brave wolf]

= radius, yep

[twin prawn]

yeah

[brave wolf]

and the formula for area of circle is pi * radius^2

[twin prawn]

yes

we derive ts

we dont need to derive the area of a square no?

[brave wolf]

that one is just x^2

[brave wolf]

are you saying that we need to prove this?

or derive as in differentiate?

[twin prawn]

we need to derive to get critical value

differntiate yes

[brave wolf]

oh, we'll do that a bit later

we'll need to differentiate the total area

not the area of circle alone

and you still havent express the area of circle fully

[twin prawn]

woops

pi * ([(10 - 4x) / pi] / 2)^2

[brave wolf]

you forgot the ^2

[twin prawn]

yes yes

[brave wolf]

okay nice, now try to simplify it as much as you can

[twin prawn]

pi * ([(10 - 4x) / pi] / 2)^2 + x^2

hmm

[brave wolf]

yeah, this is the total area

[twin prawn]

pi * (2^-1 * [(pi)^-1 * (10 - 4x)])^2 + x^2

[brave wolf]

okay, now square it

(2^-1 * [(pi)^-1 * (10 - 4x)])^2
square this i mean

[twin prawn]

oh were expanding it bet

wait im fixing it a bit

can we cancel pi here?

[brave wolf]

(a / b) / c
can be rewritten as
a / (b * c)

so yeah, you can cancel one of the pi's

[brave wolf]

now you can cancel the 4

and then you can differentiate

[twin prawn]

right

btw, why do we differentiate the whole area instead of individually?

total*

[brave wolf]

because you are trying to minimize the combined area

not the square area or the circle area

you are minimizing the combined area - the total area

[twin prawn]

AHHHH

[brave wolf]

and so you have to differentiate the combined area - the total area

[twin prawn]

whats the difference between combined area here and total area...

[brave wolf]

its the same

i just used their original wording (which was combined) to make it clear for you

[twin prawn]

oh its just a dash, i thought subtraction lol

[brave wolf]

yeah, it indeed was just a dash lol

[twin prawn]

there ya go

now we equate to 0

[brave wolf]

perfect

you can cancel the 2

and then you can plug it in T(x) to get the minimal area

[twin prawn]

right

are my exponents correct lol

im not fully done but im cautious of my exponents in the denomin ator

[brave wolf]

looks correct, but i'd prolly sub it to the og form before that (2x - 5)^2 was expanded

$\frac{\left(2x-5\right)^{2}}{\pi}+x^{2}$

to this

[flat frigate]

MathIsAlwaysRight

[brave wolf]

that'd make it a bit less of a pain to simplify

[twin prawn]

fair

i think this is it for now, ill just use calculator for the exact number (i think our professor allowed)

so i think this is our minimum area

what do i do to find the maximum area?

[brave wolf]

okay sure

if you tried to simplify it further, you would get to 25 / (pi + 4)

[brave wolf]

well, strictly speaking, you actually dont know whether this minimum you found is the maximum or minimum area

its just one of the extrema

[twin prawn]

second deriv

yea

[brave wolf]

sure, you could do that

or you could reason about it geometrically

[twin prawn]

i think former is better cause thats what were restricted

were in Calc I for now

[brave wolf]

okay, sure. Use the 2nd derivative test then

spoiler alert: it'll be a minimum

and then where do you think the maximum will be?

because its still the same function

[twin prawn]

the function is a constant,,

[brave wolf]

it's not really a constnat

[brave wolf]

oh yeah, the second derivative is a constant

and its a positive constant

so that shows that the point you foudn is a minimum

[twin prawn]

right

[brave wolf]

but what about the original T(x) function? The one that gives total area

where will its maximum be?

[twin prawn]

I genuinely have no idea hahaha

[brave wolf]

well, using the derivative, we found only one extremum - and it was a minimum

[twin prawn]

yeah

[brave wolf]

also, notice that the function is just a parabola

so itll grow unbounded, without a maximum

[twin prawn]

so theres technically no maximum unless were considering infinity

[brave wolf]

but the problem gives some physical constraints

[twin prawn]

right

[brave wolf]

can u figure out the constraints here?

[twin prawn]

the wire is 10 ft long

cut into 2

one is a circle and one is a square

[brave wolf]

so under that constraint, what range of values could x take

[twin prawn]

local minimum upto 10

i assume

[brave wolf]

x isnt the perimeter

x is the side length of the square

[twin prawn]

oh right

1 and 2 only

4(1) = 1, 4(2) = 8

[brave wolf]

x can be any real number, not only an integer

[twin prawn]

[1, 2]

[brave wolf]

2.1 * 4 is 8.4, ig that would still be allowed

[twin prawn]

yup

OH

[brave wolf]

and 0.9 * 4 is 3.6, that should still be allowed

[twin prawn]

(0, 3)

[brave wolf]

3*4 is 12

that seems like a bit too much

[twin prawn]

3 is open its exclusive

[brave wolf]

2.9 * 4 is 11.6, still too much

well, you know that the perimeter 4x must be between 0 and 10, so x must be between..?

[twin prawn]

(0, 2.5]

[brave wolf]

[0, 2.5] I'd say

[twin prawn]

right

[brave wolf]

or well, that depends whether you allow to use the full wire to make only circle and only square

but lets suppose that we do allow that

anyway, now the maximum should occur on one of the endpoints

so either at x = 0 or at x = 2.5

you can just try them both and see which one is bigger

or you can use a bit of geometry and realize that making only circle out of the wire will be better, because circle makes the maximum area for a given perimeter

[twin prawn]

yeah 0 is bigger LOL

[brave wolf]

using all the wire on the circle reaches the maximum, yeah

and you can just plug x = 0 to T(X) to get that maximum

[twin prawn]

thanks guys

.close

[safe radish]

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[safe radish]

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[pulsar spade]

I have 6 shapes here. With a 1 inch border between each shape, what would be the smallest space I could fit them into?

[pulsar spade]

Pink box: 13"x3"x7.5"
Red box: 17"x5.5"x2.5"
Blue box: 24" tall, 6" diameter

I'm not actually sure how to go about this.

My initial thought was all the rectangles in a row, with the cylinder across the top

So something like this, but there's a lot of weird gaps on the sides.

[near sky]

wdym by 1 inch border

[pulsar spade]

1 inch away from each other

[near sky]

ok

[safe radish]

@pulsar spade Has your question been resolved?

[safe radish]

@pulsar spade Has your question been resolved?

[safe radish]

@pulsar spade Has your question been resolved?

[safe radish]

@pulsar spade Has your question been resolved?

[tardy mango]

I'd assume you can play around to look for numbers that repeat/add together to form the same total so that you can get nice edges. But I'm lazy, so some brute force yields this:

[safe radish]

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[brittle stump]

Hello. I am stuck on this problem. I tried to solve it, but got it incorrect. What’s written in red is correct answer.
In short, I used integral to find area of function above and below. Then I found a domain for part of the function 1+2cos2x, where curve goes down. And added it to final result. Yet, so how got wrong answer

[ionic blaze]

I can't read the solution it's kinda low res on my part, but the higher function is 2+cos2x, the lower part is 1+2cos2x, subratcting the higher part from the lower part then integrating from 0 to pi would give you the area, you started off right, although it seems like there's an arithmetic/part where you miscalculated on the last part(I can't read it)

wait why is there a second one

[small sandal]

is it not just integral [0, \pi] of 2 + cos(2x) - 1 + 2cos(2x) dx ?

[ionic blaze]

why is there a second thingy on the integral part, the 1+2cos2x from is that pi/3 and 2pi/3

[ionic blaze]

it is

[ionic blaze]

the subtraction of the two functions removes the need to make it positive, since positive - negative = positive

[brittle stump]

Her it states when area is under the curve and above the curve at the same time. I must calculate them separately

[random cobalt]

Watchu need

[small sandal]

does it want the total area like if negative is counted as positive? because you can just integrate from a to b regardless of wether its above or below the x axis and it will still give the correct answer (unless negative is counted as positive)

[ionic blaze]

well, when you subtract the two functions, it becomes positive

[random cobalt]

Big dawg js take the integral to A and subtract the integral A to B

[brittle stump]

Okay. If I will do your way. Answer will be pi. But in the book it states answer is 5pi/3-2*root of 3.
Problem 10

[ionic blaze]

it should be pi

[brittle stump]

It is very rare, but may be book is wrong.

[brittle stump]

Okay thank you. It was really confusing.

.close

[safe radish]

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[small sandal]

bit late but yea it should end up being \pi

[brittle stump]

Yeah. I also got same stuff before putting that thingy below.

[safe radish]

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[soft lava]

How do I add these?

[soft lava]

Also I am having trouble finding an angle that has angles in pi from known triangles I tried legit every combination and I don’t see anyway it could be subtraction either

I’m trying to make the angles pi/2 , pi/4 , pi/6 3pi/2 , pi , or 2pi

[safe radish]

@soft lava Has your question been resolved?

[safe radish]

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[safe radish]

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[pale bone]

i dont understand this

[pale bone]

generally dont understnd the concept

[burnt notch]

$$x^\frac{m}{n} = \sqrt[n]{x^m}$$

[flat frigate]

Alberto Z.

[burnt notch]

I guess you only need to use this for such exercises

[pale bone]

i know that formular but how does it work

[vague phoenix]

squareroot is a power of 1/2

[burnt notch]

Well, just plug in the numbers

[vague phoenix]

then 7(x^5)^(1/2)

[pale bone]

wait what do i do with this 7

[burnt notch]

Nothing, you just leave it there untouched

[round egret]

keep it

[pale bone]

what does the 7 mean

[burnt notch]

???

It's a number lol

What's the issue with it? 😅

[pale bone]

like does 7 mean 7 times square root of x to the power of 5

[round egret]

yeah

[burnt notch]

This is equivalent to $7 \cdot \sqrt{x^5}$

[flat frigate]

Alberto Z.

[pale bone]

OH OH

[burnt notch]

This should be taken for granted honestly

[pale bone]

then uhm

how the flip does this work

[round egret]

it just stays as it is

[pale bone]

but it said to change it to a positive

indices

[round egret]

oh

[pale bone]

x to the power of 1 half

[round egret]

I forgot how to do basic latex it seems

[round egret]

yeah that's right

[pale bone]

soo?

[round egret]

so apply it for sqrt(5)

[pale bone]

5 to the power of 1/2

[round egret]

yep

now you have a 5 timesing that as well

[pale bone]

so 25 to the powe of 1/2

[round egret]

not quite

look at this indice law

[pale bone]

ooooh

wait how do i sub that in

[round egret]

well 5 is just the same as 5^1

[pale bone]

ooo

i got it

thankkkks

🙂

wait why does that sound sarcastic

I ACTURALLY GET IT

byee

.close

[safe radish]

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[safe radish]

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[zealous grail]

Hello guys i would need help with this equations i have really no idea what to do there im sorry

[vague phoenix]

which one

[zealous grail]

all of those up

those three

[round egret]

do you know your log laws?

[zealous grail]

i mean i think i know what to do with the first one probably? after the "=" theres 2 • log5 ( x-2) so probably what i have to do is put the 2 before the log and raise the parentheses and square it?

[round egret]

yep

[zealous grail]

okay okay but what do i do woth the other ones

[round egret]

use rule 1

!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).

[round egret]

whoops

wrong one

!redirect

[safe radish]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[round egret]

<@&268886789983436800>

[zealous grail]

so i put the (9x -8) - (x-6) together like subtract that?

[round egret]

no, combine them into $\log_5(x+2)(x-4)$

[flat frigate]

MxRgD

[vague phoenix]

2log5(x-2) is actually log5(x-2)^2

[round egret]

is that not what they said?

[round egret]

oh i didn't read "put the 2 before part"

[zealous grail]

oh okay

so is this the correct way?

[vague phoenix]

,rcw

[vague phoenix]

correct

[zealous grail]

okay thank you and what do i do with the last one?

[vague phoenix]

e?

[zealous grail]

yes

[vague phoenix]

for the first term, its basically 3(log8x)^2

$3(log_8x)^2$

[flat frigate]

Minλ

[vague phoenix]

this

[zealous grail]

okay i have that

[vague phoenix]

now, you see $log_8x^{13}=13log_8x$

[flat frigate]

Minλ

[vague phoenix]

right? using log law

[zealous grail]

um where does the 4 go?

[vague phoenix]

let 4 be there

so now write down the equation after modification and ss bro?

[zealous grail]

wait so i do 3(log8x)2 - log8x13 + 4= 13log8x?

[vague phoenix]

correct

now, let $t=log_8x$

[flat frigate]

Minλ

[vague phoenix]

what would the equation be?

[zealous grail]

wait dont i change for example the log8x to some alphabetical letter so it would be like easier?

[vague phoenix]

this is what i meant

make it a quadratic equation

[zealous grail]

so it would be 3 • t2 - t13 + 4 =t?

[vague phoenix]

no no no

3 • t2 - t13 + 4 =0

[zealous grail]

oh okay

now i am so lost i need to do quadratic equation but i have no idea

theres like this ax2 +bx+c= 0

[vague phoenix]

Wdym

Find t

[safe radish]

@zealous grail Has your question been resolved?

[safe radish]

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[vague phoenix]

.reopen

[safe radish]

✅ Original question: #help-23 message

[vague phoenix]

Havent you studied quadratic equation yet?

[safe radish]

@zealous grail Has your question been resolved?

[zealous grail]

no

[vague phoenix]

Dude you studied log and havent got to quadratic equation?

[zealous grail]

yes istg i dont have it in my notebook anywhere and i didnt even hear her talking about it

[vague phoenix]

Damn how can you solve this without quadratic then

[safe radish]

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[hoary seal]

sooo let me send the working out

.close

[safe radish]

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[midnight steeple]

is this solvable?

[weary needle]

Just curious what math is this? (Not a helper sorry)

[midnight steeple]

$$E(X) = \int_0^1 ax^{-a} = \frac{ax^{1-a}}{1-a} |^0_1$$ which is divergent for a > 0

[midnight steeple]

probability theory

[weary needle]

Scary stuff bro I wish you all the best!

[midnight steeple]

it

it's fun 😄

[flat frigate]

astral

[weary needle]

I’m currently struggling with a math problem on webwork and it’s not fun 😭

[trail lotus]

it's only fun when you finally get the answer right lol

[dusky totem]

isnt the function ax^(-a-1)

[trail lotus]

$$E(X) = \int_0^1 ax^{-a-1} = \frac{ax^{1-a}}{1-a} |^0_1$$ which is divergent for a > 0 ?

[flat frigate]

garlicbread

[midnight steeple]

Yeah you multiply an x into the integral though, because you’re finding the expectation. So you’re integrating ax^-a

[safe radish]

@midnight steeple Has your question been resolved?

[dusky totem]

yeah i was saying this but obviously the integral changes, i meant the integrand to be what i said

[dusky totem]

but yes this makes sense

[dusky totem]

also it diverges for a>1 so youre good for a between 0 and 1

[safe radish]

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[trail lotus]

if it helps any in ur understanding, anything to the power of -1 would be an "a" divided by 1 based on the negative exponent rule

[safe radish]

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[dusky totem]

yh yh ik all that

[dusky totem]

im just saying the question is correct because x^(1-a) is well defined for a in 0 to 1 so the i ntegral is bounded there

.close

[safe radish]

Channel closed

Closed by @dusky totem

Use .reopen if this was a mistake.

[trail lotus]

oh so u solved the problem already? that was quick

[dusky totem]

no bruh i just used OP's result

also solving it isnt that hard its just one line right?

[trail lotus]

it's an integral with an expected value, it's a whole graph 💀

[dusky totem]

the expected value is gonna be a number

all you have to do is integrate x f(x) dx

[trail lotus]

whats the answer then

[safe radish]

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[dusky totem]

its literally this, just put the limits

[dusky totem]

@trail lotus im not tripping right

so its just a/(1-a)

[trail lotus]

that's just the equation for the expected value though, the OP didn't do the equation for the variance value yet

[dusky totem]

i still got the answer tho lol

variance is just $\sigma = \sqrt{E(X^2) - E(X)^2}$

[flat frigate]

robin.dabanc_

[dusky totem]

both of which are trivial to compute

$E(X^2) = \int_{0}^{1} x (ax^{2(1-a)}) dx$

[flat frigate]

robin.dabanc_

[trail lotus]

$\sigma is the symbol for standard deviation tho? I could be wrong lol

[dusky totem]

yh you know what i meant

var is just sigma squared

[trail lotus]

BRO I DON'T

💀

[dusky totem]

this sounded egoistic

thats why i assumed you know more than me

[near sky]

ts question?

[dusky totem]

yes

no no we solved it

theres no issue

.close

[safe radish]

Channel closed

Closed by @dusky totem

Use .reopen if this was a mistake.

[trail lotus]

$$variance is var [X] = \sqrt{E((X^-a)^2) a = answer$

[flat frigate]

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

[trail lotus]

I can't add filler words??

[dusky totem]

you can if it means what its supposed to mean

[flat frigate]

robin.dabanc_

[trail lotus]

y is it trivial, don't the OP want it to be solved

[dusky totem]

no, i meant for me, not for the OP

again, i assumed you knew probability theory so i thought it would be trivial for both of us

[trail lotus]

woww u only think about urself smh

we're meant to help the OP

[dusky totem]

no dude, im telling this because the help thread is already closed

its like we both are trying to solve the problem after the OP is already gone, so it has nothing to do with OP

[trail lotus]

the OP never closed the channel though

[dusky totem]

it got timed out and op never returned

WELL you got what im trying to say

forget it

[trail lotus]

so it's techically still going with you as the aficionado since your username is on the channel

[dusky totem]

not for long

[safe radish]

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[vestal musk]

Hi

[random eagle]

Excuse me my mathematicians, but can somebody tell me if to find the amount reduced or increased in a number via percentage, for ex. Out of 725 watermelons, 45% have been bought. Instead of Divding 725 by 100 to get the value of 1% and then multiply by 45% to get the answer, can you just multiply 725 by 45 and the divide the answer by 100?

[vestal musk]

Oh-

[simple gazelle]

peak

[obsidian oracle]

Unfortunately that channel just got occupied by someone else, try copypasting your message in another available one

[obsidian oracle]

Do you have a question?

[vestal musk]

Anyone familiar with normal standard deviation

[honest perch]

yes

[vestal musk]

Specifically finding the area?

[honest perch]

yes

[vestal musk]

Is there a specific formula?

[random eagle]

Ok, just to be sure

[vestal musk]

Or way of setting it up

[random eagle]

no i just thought of it while training for my percentage exam, im in the seventh grade in middle upper school,

so no major difficulty

can i use help-37 by any chance

[honest perch]

how old are you?

roymed

[honest perch]

are you allowed to use a z-table? calculator?

[vestal musk]

I believe so

But I am still unfamilar with my calculator

[honest perch]

is it a graphing calc

[vestal musk]

Yes but I... still havent opened it

[honest perch]

the formula is $\int_z^{\infty} \phi$ where $\phi$ is the standard normal pdf

[flat frigate]

mayervietors long exact sequence

[honest perch]

do you know integrals

[vestal musk]

In the previous question there is a table but i cant understand how you would get an area for something continuous

Hmmm not familar with it

[honest perch]

area for something continuous
sounds like integral

[vestal musk]

Ohhhhh

All I know is the mean is 0 and the sandard dev is 1

Ok I am simutaniously unsure if this subject before or after calculus

I'm doing introduction to statistcs and probablity

I apologize if confusing the way I am explaining

[tardy mango]

chances are there's a look up table of z-scores you can use

if you haven't been exposed to the integral defn

[vestal musk]

Hmm ok

Ok so if the z= -1.06 according to the chart it would be... 0.1446 correct?

[vestal musk]

I am still familizing myself with the z score chart

So would the area of the shaded region be 0.1446?

@tardy mango

[tardy mango]

Notice how the $z$-score chart gives probabilities that $z<Z$

[flat frigate]

Civil Service Pigeon

[vestal musk]

z < Z? Is it essentually negative < positive? Could you further explain please?

[tardy mango]

Probabilities of things like $z<0.75$, $z<-1.5$, $z<-1.06$, etc.

[flat frigate]

Civil Service Pigeon

[vestal musk]

Ohh I see

???

[tardy mango]

What part of the curve would that shade in

remember that z scores increase as you go to the right

[vestal musk]

Its to the right. Its positive. The standard dev is 1

I mean the z is negative

[tardy mango]

????

suppose that I asked you to shade the part of this curve where z<0.75

would you shade to the left or the right

[vestal musk]

If 0.75 is positive the shading would be to the left

[tardy mango]

It doesn't matter whether the z score is positive or negative

[vestal musk]

And the area would be...0.7734

[tardy mango]

you always shade to the left

because again, z scores increase as you go to the right

so if you draw the vertical line where z=Z

everything to the right is a z score of more than Z

[vestal musk]

But with negative its shaded to the right

[tardy mango]

and everything to the left is a z score of less than Z

notice how this is independent of the sign of Z

[tardy mango]

again, no

read what I just said

[tardy mango]

My point was for you to realise that the z score table giving you P(z<-1.06) gives you the area to the left of z=-1.06

you need to figure out how to use this to find the area to the right of z=-1.06

[vestal musk]

Im so sorry if Im am condusing

*confusing

But my question what is the difference between z and Z what do they stand for

I apologize

I know I know I am probably an idiot

I tried to use the example but its different

[hard crest]

this is a reasonable question, but explaining it might get you lost in the sauce a little bit. I'll do it anyway though.
Z is the location of the line you draw on the graph
z is the variable in question

i think you were mostly doing it right, saying that P(z < 0.75) was 0.7734. This means that P(z > 0.75) would be 1 - 0.7734, or 0.2266.

[vestal musk]

Ohhhhhhhh

[hard crest]

(the probability that z exactly equals 0.75 is 0, or you can think about this as being "infinitely small")

[vestal musk]

I see

Wait

Belay my dislexic last

you subtract 1 because it is the stanard dev correct? If the standard dev was say 2 or 3 you would subtract 2 or 3 instead?

[hard crest]

no, you subtract 1 because P(x < 0.75) + P(x > 0.75) = 1

because x is somewhere on the real axis, and it's either below 0.75 or it's above 0.75

so there's a 100% chance it's somewhere

[vestal musk]

So what would the mean be considered?

[hard crest]

the distribution we're working with is symmetrical around 0, with a standard deviation of 1

this is just for standardisation

[vestal musk]

I see

[hard crest]

if you imagine like.... the distribution of heights of humans, maybe that's centred around 165cm with a standard deviation of 10cm

working with z tables for this will involve transforming the distribution by shrinking and shifting it, so that it lines up with the "ideal" distribution

[vestal musk]

So then the problem would make z = -1.06 which then becomes 0.1446 have an area of...0.8554

[vestal musk]

I see

Ok hmm let me try this

[hard crest]

yep, the area to the left would be 0.1446, and since the entire area is 1, the area to the right is 0.8554

[hard crest]

for something like this, a question you might ask is "what is the chance of someone being over 180cm tall?"

[vestal musk]

The mean is 165 cm
Deviation is 10cm
And the formula is... let me check

[vestal musk]

Ohhhhhh I see

[vestal musk]

180 would be the x? Or the Z?

[hard crest]

oh right i shifted to x at some point instead of z. 180cm would be the Z, but you have to transform it.

Generally you'd think about it like "how many standard deviations away from the mean are we talking about?"

[vestal musk]

Hmmm

So first we need the z

[hard crest]

well, if the mean is 165cm and our line is drawn at 180cm, how many standard deviations away from the mean is that?

remembering that stdev is 10cm

[vestal musk]

So we would set it up as $z=x-mean/ó$

[flat frigate]

RainbowOcean

[daring thorn]

is area 0.85?

👀

[hard crest]

$\frac{x - \mu}{\sigma}$

[flat frigate]

professional attention seeker

[daring thorn]

this is Z score

😄

[vestal musk]

Yeah

[hard crest]

or you might use x-hat instead of mu for mean

[daring thorn]

so whats the problem

[vestal musk]

I am just trying to make sure I fully understand

[daring thorn]

oke

[vestal musk]

$\frac{165 -\mu}{10}$

[flat frigate]

RainbowOcean

[vestal musk]

Right?

[hard crest]

180, not 165

[vestal musk]

Ohhh

[hard crest]

remember, we said the mean (mu) was 165cm

[vestal musk]

165 is mean

Got it

So the z= 1.5

[hard crest]

yep!

[vestal musk]

: D

[hard crest]

so then you'd look that up in a table and get something like 0.80

[vestal musk]

So that makes 0.9332

[hard crest]

okay cool, so then what does that mean for the original problem?

remind yourself of what that problem was :D

[vestal musk]

Its to find the area

So then that makes

[hard crest]

well no, the original problem was "what is the chance of someone being over 180cm tall?"

[vestal musk]

Ohhhh

Its a probablity of -

[vestal musk]

Oh I'm dumb

Now I need to set up a different formula right?

Ok I dont understand where 0.80 came from...

Wait @hard crest is it supposed to look like this kinda visually?

[hard crest]

i made up that number

[vestal musk]

Oh.

[hard crest]

i knew it was somewhere between 0.68 and 0.95 and guessed, i was pretty wrong though lol

[vestal musk]

Ohhhh

[hard crest]

should be about 7%

[vestal musk]

Did you get to thag conclusion because it made area 0.0668? So rounded up it makes 0.07 which makes 7%?

[hard crest]

yep!

[vestal musk]

Yippee:D

[hard crest]

[vestal musk]

So area is another way of saying probablity? Or is it just something that happened to match for this example

[hard crest]

when we talk about probability density functions like the ones we have here, the area underneath a region* is the probability of a measurement being in that region, yes.

• this is called an integral which is a term you may or may not have heard before

probability density function is a function with the properties that the value is always ≥ 0, and that the total area underneath the entire thing is 1

[vestal musk]

Yep not familiar with integrals yet was thinking this was before calculus was planning to take precalc and trig this summer

[hard crest]

yeah that's fine don't worry about them

[vestal musk]

Ihonestly cant tell if this class was supposed to be taken after calculus

Hope not

[hard crest]

doesn't seem like it

[vestal musk]

I figured considering Im going into stem field and trying to get a bs in bio I might as well take this

Because data is a big part of the field

But I suck at deciphering word problems

Anyway moving on

So if there are 2 zs how would it be set up?

[hard crest]

well, what you have is the ability to calculate the area to the left of a certain line

[vestal musk]

The example shows something like

[hard crest]

if we consider the area to the left of 1.22 and the area to the left of -0.92, how can we use those to get the area between them?

[vestal musk]

Ok hmmm so then -0.92= 0.1814 or 0.8186

[hard crest]

let's be precise. P(x < -0.92) = 0.1814

[vestal musk]

Ooooohh

And then 1.22= 0.8869 or 0.1131
And more precise P(x < 1.22)=0.1131

[hard crest]

this isn't quite right

why did you choose 0.1131 here instead of 0.8869?

[vestal musk]

Well theres 2 of them

[hard crest]

yes. how did you decide which one to choose?

[vestal musk]

I thought we have to find both areas then subract

To fins the filled in space

[hard crest]

yeah but why is P(x < 1.22)=0.1131? why not P(x < 1.22) = 0.8869?

[vestal musk]

Ohhhhhhhh

Thats what you meant

[hard crest]

a helpful hint is that P(x < 0) = 0.5

so P(z < 3) would have to be bigger than that

[hard crest]

but then yes this is the idea

[vestal musk]

I see

[vestal musk]

How would this formula be set up as far as varables

Is x repreenting the two zs?

[hard crest]

i'm being naughty and interchanging between x and z freely

algebraically we have something like...
P(z < -0.92) + P(-0.92 < z < 0.122) = P(z < 0.122)

which makes sense because like. z is either below -0.92 or it's above -0.92

[vestal musk]

Ok iI see

I appriate the way you set it up

So it would be... uh

The area or I guess P= 0.7055

@hard crest is this correct?

[hard crest]

,calc 0.8869 - 0.1814

[flat frigate]

Result:

0.7055

[hard crest]

yeah

[vestal musk]

Yay:D

Uh

Its wrong

@hard crest

[hard crest]

this seems suspicious

[vestal musk]

Wrobg calculation?

[hard crest]

those aren't the numbers im seeing

have you misread your table by chance?

[vestal musk]

Probably

[vestal musk]

Oh I see my mistake

Wait

[hard crest]

counting

[vestal musk]

My chart is different

[hard crest]

looks the same to me

just with one fewer decimal place

[vestal musk]

Its not

[hard crest]

where is it different?

i'm spot checking numbers and they look the same to me

[vestal musk]

Im confused

[hard crest]

can you point out a spot where your chart is different than mine?

to cut to the chase, to me it looks like you were trying to read -0.92 and accidentally read -0.91

[vestal musk]

It starts off at with -0.9 for 0.0 being = 0.1841
0.01=0.1814
0.02 =0.1788

[hard crest]

yeah that's.... what mine says too

[vestal musk]

Ohhhh it just rounds up...

So 0.8869- 0.1788?

[hard crest]

yes

[vestal musk]

So the answer is: 0.7081

Incorrect again💀

It says the area between can be found by using the difference between area left of one z score and area to the left of other z score

@hard crest I cant figure out where I still messed up

[hard crest]

did you check your other z score?

[vestal musk]

I did

Oh.

I looked at wrong one.

Below it

Its 0.8888

So it would be 0.8888-0.1788

Welp last try before its just completely wrong and I have to do a new one

Ok the answer is

0.71????

That doesnt sound right.

[vestal musk]

Maybe Im mixing something again?

[hard crest]

why not?

[vestal musk]

Well its only 2 decimals but the others are 4 decimal points

Ok screw it lets see what the answer is

...

Ok nevermind

It was

I guess if you look at it it makes sense -0.92< (0.71)< 1.22

Its a true statement

Ok now with this one we dont have z scale recorded

We already have area

Would we set it up as

$z = \frac{0.9793 - 0}{1}$

[flat frigate]

RainbowOcean

[vestal musk]

@hard crest

Wait

Or could just use the scale

Which means z= 2.04

[safe radish]

@vestal musk Has your question been resolved?

[safe radish]

@vestal musk Has your question been resolved?

[vestal musk]

Sure. I guess

[safe radish]

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[balmy finch]

@fathom jewel yo if you're still on I had a quick question about the problem we solved earlier

[balmy finch]

I re-did it on my own

and got 0.083 again but I realized something this time

we didn't need to calculate the total probability of a drop occuring, or of an observer saying that a ball was dropped

we just needed to jump straight to the conditional total probability right?

my steps were basically:

1. P(Blow ^ Od)
2. P(Od | Blow) * P(Blow)
3. (P(Od | D, Blow) * P(D | Blow) + P(Od | ~D, Blow) * P(~D | Blow)) * P(Blow)
4. (P(Od | Blow, D) * P(D | Blow) + P(Od | Blow, ~D) * P(~D | Blow)) * P(Blow)
5. (P(Od | D) * P(D | Blow) + P(Od | ~D) * P(~D | Blow)) * P(Blow)
6. (0.9 * 0.9 + 0.2 * 0.1) * 0.1
7. 0.83 * 0.1
8. 0.083

[fathom jewel]

yeah i guess, also learn latex pls

the issue was more of i was confused what we were actually looking for vs. what we have and what we needed to derive

[safe radish]

@balmy finch Has your question been resolved?

[balmy finch]

latex ok I will try to learn it

might take a while though for me to write things out at first

[fathom jewel]

you will thanks yourself and others will thank you

[balmy finch]

yeah the problem is quite confusing, I have like 2 more to solve and I will try to write in latex for the next one

.close

[safe radish]

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[verbal briar]

i have tried part b like 5 times im still getting 4.43

[safe radish]

@verbal briar Has your question been resolved?

[blazing swallow]

can we see your work for part b?

[safe radish]

@verbal briar Has your question been resolved?

[safe radish]

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[proper sluice]

How do I know if something can use the alternating series test or not? I'm kind of bad with number logic, but I'm guessing A can use the alternating series test?

[proper sluice]

wait, no, 2/3 is larger than 1/2 right? Thus failing the first rule that it always needs to decrease

[fathom jewel]

yes

[fathom jewel]

another thing you can check is if the sequence converges to 0

that should help you with b)

[proper sluice]

I'm guessing B also fails the first rule, since sin(n) isn't always decreasing

[fathom jewel]

sure works

Though you'd need to find some examples to do that, which can become a bit of a hassle, so I'd rather prefer two other faster ways: The first is as mentioned the sequence must converge to 0, but the limit of sin(n) doesn't exist. Another thing: We know sin(x)<0 on [pi, 2pi] so for n=4 we would have sin(n)<0 but the sequence must be non-negative for all n.

[proper sluice]

hmm, I think C would be constantly decreasing just by looking at it, but I'm not sure about that 2. Wouldn't lim (2-1/n) converge to 2 and fail?

[fathom jewel]

yep exactly

[proper sluice]

b(n) of D would be 1/e^2 right? But there's no n! It's stuck in purgatory!

[fathom jewel]

yes haha

[proper sluice]

So,that means you can't take the limit of it, and it fails too?

[fathom jewel]

You can take the limit, it's just constant

[proper sluice]

so it only approaches itself, right?

[fathom jewel]

yes

[proper sluice]

and thus fails yet again

Alright, thanks! I think I'm doing better than I thought I would

.close

[safe radish]

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[safe radish]

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[hoary seal]

why did they change the inequality
when. they divided by 1-2x

[ionic blaze]

as x approaches infinity, the bottom part approaches negative infinity so it's dividing it by a negative number essentially which is why the inequality signs are swapped

[fathom jewel]

For x>=1 the 1-2x is negative, when you divide by negative numbers the inequality sign flips

[hoary seal]

also

why do they say this

they make an assumption

but why cant it be other form of numbers

[clever tinsel]

it's not an assumption.

[hoary seal]

its a fact

[fathom jewel]

assumptions can be

[clever tinsel]

there you go. it is a fact indeed.

[astral glacier]

The range of a cos is [-1,1] no matter what you plug into it

[clever tinsel]

to add to Xavier, if I may, this is true since you (or whoever wrote that statement) is explicitly considering real x only.

[safe radish]

@hoary seal Has your question been resolved?

[hoary seal]

well real x is evrything except complex numbers

[clever tinsel]

are you telling me that or are you looking for confirmation, or something else entirely?

[fathom jewel]

real numbers are complex numbers

but not every complex number is real

[hoary seal]

why did he remvoe the cos(x^2)

he said its at most 1

[fathom jewel]

to find an upper bound

[hoary seal]

but it doesnt mean its 1 every time

[fathom jewel]

also is this still related?

[hoary seal]

full ss

[fathom jewel]

ok

the idea is to find an upper bound that eventually goes to 0 because epsilon can be arbitrary

[clever tinsel]

compare the sizes of these two expressions.

[hoary seal]

fair enough

[hoary seal]

1/2x^2 is very small

and |cos(x^2)| is always going to be positive

and near 1

[clever tinsel]

ignoring 1/(2x^2) because both expressions have it, what is the relationship between the two expressions?

that will answer why they 'removed' |cos(x^2)| and changed the equal sign into a <= sign.

[hoary seal]

because both expressions have it?

wdym

[clever tinsel]

I mean what I said. I'm asking you to compare the sizes of 1/(2x^2) to that of [1/(2x^2)] |cos(x^2)|.
both of these expressions contain 1/(2x^2), and so its size is not really that important to us.

I would advise you to direct your attention to the possible sizes of |cos(x^2)| and what multiplying by such a number must do generally.

[hoary seal]

hm

because if u multiply by the value of cos(x^2) which ranges in value from 1 to -1 but since there is an abs, it will output a positive number

[clever tinsel]

so you agree that |cos(x^2)| ranges from 0 to 1?

[hoary seal]

so thats why the put a <= sign because it can be smaller or equal

[hoary seal]

yes

[clever tinsel]

correct then. so you've answered your own question.

[hoary seal]

also

why do they put a - and then bracket the numeratror

after removing the abs vlaue

ful ss

[clever tinsel]

because of this.

and if you don't already recognize it, -x - 4 = -(x + 4).

[hoary seal]

why havnt my prof wrote the full question

he just did |x+4|

he didnt divide by x+4

[clever tinsel]

because he wanted to first define the absolute value function in the numerator.

or, a better option: ask your lecturer!

but that's my best guess as to why he did it.

[hoary seal]

for this proof

in the black writing

how do we test

oh wait

i can sayy epslion = 3

we dont sub anything in here, this is just for display

we sub it in the new lines

and see if it holds the inequality

how do go about this, im already stuck

[astral glacier]

Well the square root of x³ isn't x

[hoary seal]

$x>\sqrt{3}{(\frac{1)(9)})$

[flat frigate]

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

[hoary seal]

wait what

[clever tinsel]

$x > \sqrt{3} \cdot \frac19$

[flat frigate]

Yukari

[hoary seal]

no uh

i meant

x > cube root of 1/9

[clever tinsel]

$x > \sqrt[3]{\frac19}$

[flat frigate]

Yukari

[hoary seal]

yes

then do we just keep pluggin the values until we reach

[astral glacier]

Rule of thumb, we almost never plug in values when doing analysis

Actually let's step back the question asks you to use the formal definition of a limit

Can you state that for me

[safe radish]

@hoary seal Has your question been resolved?

[hoary seal]

$|f(x)-L| < \epsilon$, where $\epsilon > 0 $and x>$M_\epsilon$

[clever tinsel]

the whole definition, I believe.

[hoary seal]

this?

[clever tinsel]

I don't see any mention of delta anywhere.

[flat frigate]

ø

[hoary seal]

im not sure why its doing that

[astral glacier]

The full definition please

It should be in your class notes

[astral glacier]

Do you understand this definition

[hoary seal]

yes

[astral glacier]

Good, then what do we do to prove what the question asks

What's the first step

[hoary seal]

should i just use this new question

I need to do that excercise anyways

first step is to

state the defintion of the formal limit

Correct?

[fathom jewel]

wrong order

For every eps there is delta

your order is unecessarily more restricting

[hoary seal]

ok ill use this

i just memorise it

?

[clever tinsel]

isn't his order weaker (sorry for the ping)?

[fathom jewel]

no because you say one delta for all epsilon like it must be very specific

it cannot depend on eps

[clever tinsel]

ah, alright, sorry for the misunderstanding then.

[hoary seal]

or what much is sufficient from this box

[fathom jewel]

In a sense, it's like having one key that opens all doors instead for evey door having a unique key

[fathom jewel]

Can u ask a proper question pls

[hoary seal]

I mean do we just need to write this

and prove it

[fathom jewel]

You have to understand and internalize

You dont get points for stating the def

[hoary seal]

I understand the definition

its just I dont have the practice in proving it

[fathom jewel]

The idea is you wanna find some upper bound that converges to 0 eventually

That way we can control eps

[hoary seal]

upper bound?

[fathom jewel]

Yes

[hoary seal]

what is that

[fathom jewel]

Generally speaking or what...?

[hoary seal]

relating the definition

[fathom jewel]

|f(x)-L|<g(x)

Thats all

[hoary seal]

epsilon isnt a function

[fathom jewel]

thats good news

what we are doing is trying to find something more managble

[hoary seal]

yeahh

[fathom jewel]

Like in this we found 1/(2x²) that is way better to work with than having a cosine term

[hoary seal]

yes I agree

[hoary seal]

< epsilon

[fathom jewel]

Obviously this is a bad start

so first step would be to combine the terms into one fraction

any ideas how

this is also just scratch work

@hoary seal

[hoary seal]

yes

i got up to here

[fathom jewel]

how can we combine both terms into one fraction

[fathom jewel]

the denom is 2x^2 not 2x

[hoary seal]

denom is 2x

[fathom jewel]

take a look at your f(x)

[hoary seal]

ohh

[fathom jewel]

this is way nicer

then from what you had in the beginning

[fathom jewel]

now similarly to this

how can you bound the expression

[hoary seal]

ok so we have |sinx/2x^2| < $\epsilon$

[fathom jewel]

not yet

[hoary seal]

hm

[fathom jewel]

ideally we want to rid sine

[hoary seal]

why

ik it would make it harder

but how

[fathom jewel]

what do you know about sin(x)

[hoary seal]

continuous function

starts from 0

[fathom jewel]

doesnt start from 0

e^x is also continuous

but what makes sin(x) stand out here

[hoary seal]

,plot sinx

[fathom jewel]

my blud

[burnt notch]

Recall the previous exercise with cosine @hoary seal

[fathom jewel]

for the 4th time

[burnt notch]

You can't have already forgotten!

[hoary seal]

its bounded

[fathom jewel]

ohhhhh

there we go

elaborate

[hoary seal]

from -1 to 1

[fathom jewel]

ohhhhh

[fathom jewel]

now use that

[hoary seal]

okk

but we have 2x^2 iin the denom

do i out it on the rhs of the inequality

put*

[fathom jewel]

it's not an issue

for example

1/2 < 2/2
i didnt touch the denominator

i just used that 1<2

the denominator can stay

if we wanna bound something from above we can ignore the denominator and simply focus on making the num greater that's all

[hoary seal]

ohh

okay so we just ignore it

[fathom jewel]

yup

[hoary seal]

but it doesnt follow up to it

[burnt notch]

The same thing as here, you know @hoary seal

[burnt notch]

...

[fathom jewel]

What doesn't follow up other than me rn with you

[hoary seal]

so this is correct?

[fathom jewel]

it's not what we want

we want to bound it up by something nicer and then bound that by eps

[hoary seal]

how is this not nice?

[fathom jewel]

fine, work with it

[hoary seal]

no i mean so I can know what to do

[fathom jewel]

I told you what to do

[hoary seal]

1/2x^2

thats the bound

[fathom jewel]

yeah