help-28

[west sequoia]

start by looking at each individual term

for $\sqrt[3]{x}$ , are there any real numbers that would make it not real?

oop

[glossy valve]

barometer

[fathom void]

oh so ur meant to solve it like that

[west sequoia]

well

you need to do that for each part and then see where the domains overlap

so we know that with an even root, like $\sqrt[4]{x-4}$, the argument $x-4$ needs to be positive, otherwise you get a non-real value

[glossy valve]

barometer

[fathom void]

i get it now

[west sequoia]

okay cool

[fathom void]

thank you 👍

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[sacred herald]

https://trinhngocthuyen.github.io/posts/tech/a-misuse-of-expectation/

not necessarily a problem, but I require some explanation. I understand math behind it, but as it happens with probability theory a lot, I don’t get the principle.

What is in fact E(Y/X)? What information does it give? Why it is different for E(X/Y)?

If it is essentially “average” of the X/Y why it is not just opposite of Y/X?

I understand the math, but I need some handwayvy explanation if that makes sense.

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@sacred herald Has your question been resolved?

[sacred herald]

<@&286206848099549185>

I’m sorry if this question doesn’t belong here. If anything, just delete it. I don’t know where else to ask.

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[muted flicker]

Hi, can someone explain to me what happened in the second step here?

[restive kelp]

they multiplied all terms with exp(-1/x), and a^-1 = 1/a

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@muted flicker Has your question been resolved?

[muted flicker]

Still not seeing it. Can you rephrase it perhaps? I get the second idea about the applying a negative power

[restive kelp]

do the multiplications, what u got?

[muted flicker]

I'm slapping an exponent of -1/x onto 1 let's say...

Or am I interperting exp() wrong

Ohhhh ok

There's an e in exp()

OK I'm trying again

[restive kelp]

yeah, sry, exp(y) means e^y

[muted flicker]

Np

So I'm obv stuck somewhere here...

[west sequoia]

what is 1/x - 1/x

[muted flicker]

Its a good day for me today

[west sequoia]

it's okay lol

[muted flicker]

Alr ok I see the approach now. Thanks!

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[alpine chasm]

Solution is incorrect and I cant spot my error. Any help?

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@alpine chasm Has your question been resolved?

[alpine chasm]

<@&286206848099549185>

Anyone?

Hello

[unreal seal]

sorry i wish i could help but this is way above my level

[alpine chasm]

It’s okay 👍 I hope someone else can assist

[sand silo]

ilk

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[lyric cipher]

hi

[gritty rose]

Do you have a math question?

[full forum]

@lyric cipher Has your question been resolved?

[swift fulcrum]

hi

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[steep fog]

What is the LS and RS formulas for a Riemann sums using sigma?

[swift fulcrum]

You mean something like this?

[steep fog]

Kind of

Like the formulas that you can use for this

[swift fulcrum]

I don't think there is a formula for that

It's just adding rectangles

Same for trapezoidal sum

[steep fog]

Figured it out, thanks!

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[prime fable]

Help asap plz

[prime fable]

The ans is D

But can plz someone explain it

[lyric turtle]

Search Integral calculator

Or try wolfram i’d suggest

[prime fable]

I don't think it is able to solve it and give one of these answers

[nocturne creek]

Do 3x=sect

[prime fable]

Can you give another hint

[hot wadi]

[lyric turtle]

[prime fable]

Like i got that sec^2t/sec^2t-1 = csc^2t but what after that

[nocturne creek]

Its more than a hint, its the entire process💀

[lyric turtle]

Lmfaoooo

[nocturne creek]

Can you send a picture

[lyric turtle]

Gold answer

[nocturne creek]

it was not intentional

[nocturne creek]

Im confused

[prime fable]

Same

[nocturne creek]

I meant this like a substitution

[prime fable]

Ok

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[dim compass]

Is slope of curve and tangent of curve same thing?

[dim compass]

Its a question of literature on mathematics

[stable heart]

typically the tangent is an equation for a line but a slope is just a number/expression

so for example say we have y=x^2+1, we then have at the point (0,1) that the slope is 0 but we have the tangent line y=1

in case you want a visual

[dim compass]

Ohh

Graphically I thought they are both a straight line that connects the small change of function

[stable heart]

well they are closely related

[dim compass]

So slope is tan theta of that tangent

If that's what you meant

but my chemistry sir says that tangent or slope of curve concentration vs time gives the instantaneous rate of curve. He literally writes "slope or tangent of curve"

[stable heart]

yeah i guess. slope is just tendency to increase in y when we increase x. tangent is a linear approximation of a function at a point

[dim compass]

Mathematically that seems to be so, thank you for your teachings

[stable heart]

np 👍

[dim compass]

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[torn jolt]

I dont understand why the answer to this question has to be x - p because surely x + p works just the same or am i missing smt?

[hushed spear]

The right part says something like "accept any of y = (x ± p)^3, doesn't it?

But yeah, since they didn't specify if p > 0 or not, either could work.

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@torn jolt Has your question been resolved?

[torn jolt]

Okay thank you I was just confused because it said 2 marks and the second mark was the x - p. Thank you for clearing that up

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[full forum]

Please don't occupy multiple help channels.

[slate reef]

Whoops, closed the other one

Anyways

The existence of that sqrt(1-y^2) means that I made a mistake calculating the inner integral

Symbolab says I can find this integral using trigonometric substitution but that’s not something we learned and my teacher doesn’t like it when we use things we didn’t learn in class during exams

Thus this must mean that I made a mistake

[tardy pier]

U in high school or uni?

[golden furnace]

multivariable calculus without learning the trig substitutions?

[hushed spear]

my teacher doesn’t like it when we use things we didn’t learn in class during exams

[tardy pier]

Honestly this question is a lot easier if you use cylindrical coordinates

Like way way way easier

From the start

[slate reef]

Uni, multivariable calculus yes

[tardy pier]

Have u done cyclindrical coords

[slate reef]

Uh, no

cos^2(x)+sin^(x)

How do I involve the +4 there

[tardy pier]

Like the r^2, theta, r stuff

[slate reef]

Oh, that’s later in this subject

[golden furnace]

hmm

[tardy pier]

Okay, and def no trig subs

[slate reef]

Are you guys sure that I even did the first integral correctly?

I went over it and got the same answer

But maybe I’m making the same mistake over and over again

[tardy pier]

Haven’t tried it yet but ill go for it now, have an exam on this on Friday so I’m kinda fucked if I can’t do it

[slate reef]

Same here

._.

[golden furnace]

uh yea you made a mistake

[slate reef]

[golden furnace]

you forgot the /2 part when evaluating the first part I think

so that sqrt(1-y²) terms cancels out that way

[slate reef]

I did forget the 1/2 yea

Does that cancel out the sqrt tho?

Let me try again

Thanks btw .3.

[golden furnace]

yea see it for yourself maybe I'm mistaken too

[slate reef]

This is the result

And yes

The sqrts cancel out

._. I am a n g e r y

Thanks though, both of you 😗

[golden furnace]

🦦

[slate reef]

and the answer is 2/3

🥳

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[tardy pier]

Defo 2/3 right

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[spring hamlet]

given that $8^{2x+5} ÷ 8^3 = 4096$ , find the value x

[glossy valve]

Katze

[spring hamlet]

help?

[stable heart]

$8^{2x+5} ÷ 8^3 = 4096 \implies 8^{2x+5-3}=4096$

[glossy valve]

Duh Hello

[neon crypt]

Then take the ln

[prime fossil]

8^{2x+5-3}=4096, 8^{2x+5-3}=8^4, 2x+5-3=4, x=1

como los exponentes son iguales

[spring hamlet]

$8^{2x+2} = 4096$

[glossy valve]

Katze

[golden furnace]

4096 is also a power of 2

[stable heart]

then you either take log base 8 of each side or you can try and find it in a simpler way

by just knowing what $8^y=4096$

[glossy valve]

Duh Hello

[prime fossil]

8^(2x+2)=8^4, entonces 2x+2=4, entonces x=1

[stable heart]

why are you writing in spanish?

[prime fossil]

I am from ecuador

[spring hamlet]

is it 2x+2?

[stable heart]

yes 2x+5-3=2x+2

[prime fossil]

yes, 2x+2=4

so x=1

[spring hamlet]

oh

so

[neon crypt]

Right

[spring hamlet]

its 4(1)

right?

[prime fossil]

??

[spring hamlet]

the x is 1

[prime fossil]

yes

[spring hamlet]

so the 1 doesnt matter right?

[golden furnace]

uh

[stable heart]

well if you know that $8^4=4096$ then you would end up with the equation $2x+2=4$ which you can solve for $x$

[glossy valve]

Duh Hello

[stable heart]

when you solve that equation for $x$ you get $x=1$ as the answer

[glossy valve]

Duh Hello

[golden furnace]

disregarding the fact that it's the solution. it's the identity element of the multiplicative set/group yes, in that matter it doesn't

[spring hamlet]

oh yeah

i forgot about the question

[golden furnace]

all good :3

[spring hamlet]

yes all good

thank you everyone

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[scenic wren]

Hey

[scenic wren]

I have a series

The index starts at 1

The series sums elements of a sequence

The sequence goes
1,1,0.5,0.5,0.5,0.25,0.25,0.25,...

So we add 1+1+0.5+0.5+0.5+0.25+0.25+0.25+...

The ones are the only exception where we only have 2 such elements

Every other element comes up 3 times

I have to show that this series diverges

[fast peak]

how does the sequence continue after that?

always half?

[scenic wren]

Im gonna send a picture

[fast peak]

and can you give the exact problem statement?

[scenic wren]

I have to show that this series diverges

[fast peak]

are you sure?

[scenic wren]

Yes

But i showed that it converges

That is why im confused

[fast peak]

well same

cause it does converge

$2+\sum_{n=1}^\infty 3\cdot \frac{1}{2^n}$

[glossy valve]

Denascite

[scenic wren]

That is what i did

I get 8

Im gonna think about the question a bit myself. Thank you^^

[fast peak]

5

[scenic wren]

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[unique pecan]

HELP

[unique pecan]

Oh kk

wait do yall do middle school math🤔

[golden furnace]

sure

[unique pecan]

okok

im stuck on this

[golden furnace]

so think it progressively

start on monday what is your probability that you are late?

[unique pecan]

0.7

[golden furnace]

progressing one day further you are on tuesday

and the probability that you are late on tuesday now has become?

[unique pecan]

0.6?

[golden furnace]

yes

since arriving on time is the probability of not being late

what does that wind up to be

[unique pecan]

Uhhh

0.2

[golden furnace]

hmm nope

rember that an event to happen certainly it has to have an probability of 1 (100%)

and for any event A, the chances are it's either that A happened or that A did not

so P(A) + P(A') = 1

where the notation P(X) means the probability of X and, X' means not X

are you with me so far?

[unique pecan]

wait so when u mean p(x) that means u cant have like 1+1=1

[golden furnace]

yea

[unique pecan]

oh okok

[golden furnace]

so you can't have A certainly happening and simultenously have A certainly NOT happening

this channel is occupied please book your own channel

!help

[full forum]

Please read #❓how-to-get-help

[golden furnace]

so

let's call event A the event of being late

[unique pecan]

ok

[golden furnace]

so that makes A' arriving at time, yes?

[unique pecan]

Yes

bc thats what u want to happen

[golden furnace]

sure

[unique pecan]

kk

[golden furnace]

it's more like arriving at time is the event of being late not happening but yes

recall P(A) + P(A') = 1

we know the value of P(A) for tuesday right?

[unique pecan]

0.6

[golden furnace]

yes

so 0.6 + P(A') = 1

[unique pecan]

wait whats the p

[golden furnace]

we can find P(A') from this

P is just the notation of asking the question "what is the probability of this?"

by arithmetic

[unique pecan]

oh ok

[golden furnace]

we get P(A') = 0.4 right

[unique pecan]

y 0.4

[golden furnace]

you can ask questions if there is something you don't get

[golden furnace]

because

0.6 + 0.4 = 1

we substituted 0.6 for P(A)

[unique pecan]

ok

So we did 0.7-0.1

[golden furnace]

yep

[unique pecan]

Wich i s6

is*

and we added 0.5+0.1

[golden furnace]

no?

0.6 is the probability of arriving late on Tuesday

[unique pecan]

wait then whats the 0.5 for

[golden furnace]

we still need to find the probability for arriving on time for Tuesday

I didn't mention a 0.5 anywhere

[unique pecan]

ik its in the problem thoigh

[golden furnace]

0.05*

[unique pecan]

ye sry

[golden furnace]

so we add 0.05 to 0.6 for the probability of arriving late on Wednesday

[unique pecan]

Thats 65

0.65

[golden furnace]

yes

[unique pecan]

ok

[golden furnace]

so far we got the probabilities:
Being late on Monday: 0.7
Arriving on time on Tuesday: 0.4
Being late on Wednesday: 0.65

right?

[unique pecan]

yes

[golden furnace]

since these are independent events we multiply them all

[unique pecan]

So its

0.7x0.4x0.65

[golden furnace]

yep

[unique pecan]

oh ok thx

uhm i also have another problem

that i missed a few times

[golden furnace]

well the chance pulling any ball is 1/49

hm

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@unique pecan Has your question been resolved?

[unique pecan]

yes

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[scenic wren]

$\frac{exp(z)}{1-z}=\sum\limits_{k=0}^{\infty}b_{k}\cdot z^{k}$

[glossy valve]

~Martin

[scenic wren]

I have to find b_k

I found it to be 1/((1-z)k!)

[rotund birch]

i believe you can take limit

of both sides

left side is a well known limit

[scenic wren]

I also have to show why this is only true for |z|<1

That is what i am stuck on

[rotund birch]

oh power series

what a nice question btw

[cloud shore]

b_k can't depend on z

[rotund birch]

well use D'alambert ratio test

were you taught that lim x->0 (e^z/1-z) = e^0 = 1

[scenic wren]

That is how i did it

But yeah, makes sense that b should not depend on z

[fast peak]

you could also multiply both sides by (1-z) and then compare coefficients

[scenic wren]

How would tht help?

[fast peak]

split the RHS into two sums, re-index the second one and then put them together again

it may help to explicitly write out each side tho

[gritty rose]

just multiply the two power series

[rotund birch]

is this analysis 3 ?

[scenic wren]

Ana 1

🙂

[rotund birch]

yeah i thought so

why wouldnt you just take limit of both sides then

its screaming at you to do so

[fast peak]

limit of what

how is taking any limit helping here

[rotund birch]

e^z/(1-z) is the well known limit in analysis 1

if z goes to 0

[fast peak]

and?

z to 0 is obvious

[scenic wren]

But i have to show for |z|<1

Not just z=0

[rotund birch]

i see

oh yeahhh

power series work

bk cancels out

use d'alambert with absolute value

because we dont know if bk pos or not

[gritty rose]

you can usually do both find the coefficients and find the interval/disk of convergence with the ratio test.

[rotund birch]

then if <1 its conv

$|\frac{a(n+1)}{an}| = |z| | \frac{b(k+1)} {bk}|$ = \frac {-ez} {1-z}

[glossy valve]

Cyrenux

[scenic wren]

I tried this and got weird stuff

[fast peak]

well the series is weird

so you now know that b_0 = 1

and b_k - b_(k-1) = 1/k!

so from that you can calculate all b_k

(at least in theory)

[scenic wren]

Why would b_0 be 1?

[fast peak]

well the series on the left is 1+z+z^2/2+...

the series on the right is b_0 + (b_1-b_0)z + (b_2-b_1)z^2+...

[scenic wren]

Oh yeah true

But i dont think k can derive a formula for bk like that

Wolframalpha gives weird stuf for that

Sonething with a gamma function

[fast peak]

well yes it is weird

[scenic wren]

Cyrenux said to use ratio test, do you think that would help?

[fast peak]

or rather, trying to find a closed form is weird

[gritty rose]

ratio test is only helpful once you have the coefficients

[fast peak]

but if you write them as a sum then it's obvious what they are

or rather what they approach

[gritty rose]

,w laurent series e^z / (1-z)

[glossy valve]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by riemann#2529

[gritty rose]

denominator looks like factorials. numerator looks difficult

[fast peak]

b_k = b_(k-1) + 1/k!

[gritty rose]

Numerator of Sum_{k=0..n} 1/k!.
https://oeis.org/A061354

[scenic wren]

Wouldnt bk just be the sum from n=0 to k of 1/n!

[fast peak]

yes

again, closed form sucks

but this is good enough to work with

[scenic wren]

Ok since i found a formula for bk i can now to ratio test to get the disk of convergence?

[fast peak]

yes

[scenic wren]

Thank you all very much^^

I think i can continue from here

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[fluid kite]

hi, mind to help? im stuck on the second part ;-;

[fluid kite]

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[thorny prawn]

I need help with this, I'm getting 2.8 as the answer in decimal form but don' tknow if that's right, and if it is, I don't know how to convert to the fraction form my platform wants.

[worthy tree]

it is

[grim pond]

2.8=28/10=14/5

[glossy valve]

Mehdi_Moulati

[thorny prawn]

somehow that answer didn't work

but okay , thanks

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[gritty rose]

just open a new channel

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[tulip beacon]

i got a maths test tmr yall pray for me fr fr

[round gust]

Gl

Do yoy have a specific question?

[tulip beacon]

yea

no actually

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[hybrid gulch]

,rccw

[hybrid gulch]

How do i get the x

[shrewd hamlet]

Is the top angle a right angle

[hybrid gulch]

It's just a normal triangle

No

[shrewd hamlet]

Is it isosceles

[hybrid gulch]

This is the full exercise it's just been a little used

Ex.6

[shrewd hamlet]

But we don’t have multiple side lengths, right?

[robust ridge]

That's right

[hybrid gulch]

I know it, but i don't think i need to use it.

[round gust]

With that information it isnt possible to find x

[hybrid gulch]

No just one

Well let's use cosine then

[robust ridge]

Is this not a right angled triangle though

[round gust]

You cant use cosine

You only know 1 side

[hybrid gulch]

I mean I don't think so

The exercise doesn't say anything about it having a right angle

[misty cosmos]

it's isoceles

isn't it?

[robust ridge]

I believe that opposite the 8.30 m there is a right angle

[robust ridge]

Can't be I don't think

That would be one funky door

[hybrid gulch]

8.3dm

Myb the teach was smoking pot while making the exercises

[misty cosmos]

yeah i get what you are saying😂

[round gust]

Listen

You need more info to solve this

If you only know 1 angle and 1 side you cant find anything

[hybrid gulch]

Aight chief, then I'm going ahead, thx for helping guys.

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[round gust]

Bye

Np

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[open knot]

can i do this

[viral jasper]

yes

That's a great strat

[open knot]

thanks

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[calm steppe]

Hi, i've been given the centre of a circle as(3, 0) an radius5 so i got the equation

[calm steppe]

$(x-3)^2 + (y)^2 = 25$

[glossy valve]

NightHorn

[calm steppe]

the line y = 2x + k was also given and we were told it intersected at 2 points

i'm not sure where ive gone wrong as i know im trying to get it into a quadratic formula to solve with discriminant but i cant seem to do it, i've ended up with

$5x^2 - 6x -16 + k^2 +4kx = 0$

[glossy valve]

NightHorn

[calm steppe]

im trying to find K, sorry i forgot to mention

[full forum]

@calm steppe Has your question been resolved?

[west sequoia]

can you post the complete problem and the work you've attempted?

[calm steppe]

7. A Circle has centre c(3, 0) and radius 5. The liney = 2x + k intersects the circle in two points. Find the set of possible values of k, giving your answers in surd form.

gimme a sec to type my workings

$(x-3)^2 + (y)^2 = 25
(x-3)^2 + (2x+k)^2 = 25
(x-3)(x-3)+(2x+k)(2x+k) = 25
x^2 - 6x +9 +4x^2 +4xk +k^2 = 25
5x^2 - 6x +9 +k^2 +4kx = 25
5x^2 - 6x -16 + k^2 +4kx = 0$

[glossy valve]

NightHorn

[calm steppe]

ffs it didnt space out, but i think you can get what i did from above

[west sequoia]

working on it 🙏

your expansion and substitution is right though

okay

so you are right about the discriminant

[calm steppe]

cool

but its in 5 parts not 3, how do i fix this

[west sequoia]

2 solutions means that the discriminant we get from the quadratic $(x-3)^2 + (2x+k)^2 = 25$ must be greater than five

[glossy valve]

baro | awake

[west sequoia]

factoring

[calm steppe]

how did you get 5?

is that just standard

[west sequoia]

\begin{align}
5x^{2}-6x-16+k^{2}+4kx&=0 \
5x^{2}+4kx-6x+k^{2}-16&=0 \
5x^{2}+(4k-6)x+(k^{2}-16)&=0
\end{align}

[glossy valve]

baro | awake

[west sequoia]

now you have a polynomial where $a=5, b=4k-6, \text{and } c=k^{2}-16$

[glossy valve]

baro | awake

[calm steppe]

thanks!

[west sequoia]

yup

do you know where to go from here?

[calm steppe]

yeah b^2 - 4ac

Thanks so much!

[west sequoia]

for sure!

this is an interesting problem so thanks to you too lol

[calm steppe]

haha have a good night

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[dense edge]

is it safe to say everything inside of an argument of a function is "reversed" from what you think it should be?

or is that a blanket statement and bad idea to go by for every single function argument?

for example with shift left with (x+12) instead of shift right

[wide sundial]

i see that but how about *12 and /12

how do you intuitively see that as stretch and compression?

also the idea of "opposite" gets really weird when you go ^-1

1 -> inf gets squished into 1 -> 0

i doubt most would consider that "opposite"

[dense edge]

well, I see *12 and /12 as the y value intuitively. so when I see f(x) = 12x^2 I think to myself "vertical stretch". this is a tall and narrow polynomial

[wide sundial]

honestly i see 12x^2 as stretching it upwards

which is why i pointed it out

i can see how most would say + is right and - is left

so it's opposite there

also for the sign on the constant, in terms of moving the function up or down

[dense edge]

sorry, wrong word. meant to say vertical stretch ^

[wide sundial]

that's just as you would think

x+2 is just move it up by 2

no wonky opposites happening here

[dense edge]

alright so naturally I would see f(x) = (x^2)/12 as the opposite of f(x) 12x^2

vertical compression, squashed down and wider polynomial

[dense edge]

yeah for y it's easier

but for x.. I plug in those same values into the brackets and i see weird things happening

like + is -

division is STRETCH, not compression

[wide sundial]

for moving left and right?

[dense edge]

yes

[wide sundial]

i mean in my mind division is compress

[dense edge]

yes, mine too

[dense edge]

but not for this third example

[wide sundial]

it's because the notation is actually x-

it's always x-

it really says (x-(-12))

[dense edge]

below that one

fourth line

yeah, I can live with x-, i just need to remember it's the opposite for x

[wide sundial]

horizontal stretch and vertical compress is the same thing

[dense edge]

according to WebWork it is not 😛

[wide sundial]

what is webwork

[wide sundial]

it's not because of the bracket i think

[dense edge]

we call that "inside the argument"?

when it's in the bracket like that?

#3 is stretched horizontally by a factor of 12

[wide sundial]

i'd say that is compressed by 12

[dense edge]

no

[wide sundial]

[dense edge]

that would be incorrect lol

believe me, I am just as confused

[wide sundial]

actually

give me an example of vertical changes

[wide sundial]

lmao

i see why

[wide sundial]

this is disgusting

so you see how in 1.

it says y = 12x^2

and that's vertical

[dense edge]

lol, care to elaborate?

[wide sundial]

if you rearrange y = 12x^2 so the coefficient is with one of the variables directly

[dense edge]

yes

[wide sundial]

it'll be y/12 = x^2

that's cutting all the y values to a twelfth of its original value

so they call this changing the vertical scaling

either compression or stretching

so x^2/12 = y

this is actually x^2 = 12y

so all the y values got multiplied by 12

aka stretched upwards by a factor of 12

so that's why they call it "stretched vertically"

[dense edge]

but we both said it's compression, when we see (x/12)^2. we are not wrong? or we are, with the terminology of what's being stretched and what's being compressed? x and y

https://www.desmos.com/calculator/fccz8bdpi6 here are all 4 together, they do look slightly different

[wide sundial]

it's arguable to say that the way they are saying is kinda right

the fact of the matter is

if you stretch is horizontally by 12

it's not the same as compressing it vertically by 12

it's the same as compressing it

but not by 12

[dense edge]

hmmmm

lol

[wide sundial]

because look

compressing vertically acts on y right?

[dense edge]

yea

[wide sundial]

y = (x/12)^2

this is equal to y = x^2/144

144y = x^2

it's actually stretching vertically by a factor of 144

[wide sundial]

same as compressing it horizontally by a factor of 12

[dense edge]

oh interesting, ya, these are the exact same

[wide sundial]

so compressing vertically = stretching horizontally,
and stretching vertically = compressing horizontally

it's just how they are defining them

[wide sundial]

that's why this is wrong

it's compressed, but not by 12

[dense edge]

wait, so how do we get from top example to bottom example again?

as per this screenshot

[wide sundial]

we want to move the coefficient to the y value

[dense edge]

it looks different to me, like x^2 has been moved to the left side of the equation where f(x) was

[wide sundial]

so we can see how it's stretch/compress ing vertically

so from y = (x/12)^2

we want to isolate the x^2

y = (x/12)^2
y = x^2/12^2
y = x^2/144
144y = x^2

now x^2 hasn't been changed but y has been changed

[dense edge]

oh we multiply by 144 to remove the denominator

got it, thanks

[wide sundial]

this is very disgusting wordplay, very confusing and unnecessary

i hate it, i hope you do too

[dense edge]

and since f(x) is the same thing as y, we can do that

[wide sundial]

yea

[dense edge]

haha, most of the time I am at war with math.. but when I get it I do enjoy it

[wide sundial]

haha

[dense edge]

trying to avoid emotions with math .. it's hard to do tho, we are emotional creatures by habit
the cup is half full (positive) or the cup is half empty (negative) mentality, just that the cup is half capacity and I need more work to make the cup reach it's maximum capacity with knowledge and wisdom so to speak haha

I dunno, I'm trying to stuff philosophy into a subject that doesn't need it here haha

[wide sundial]

haha definitely

it's definitely amazing when it clicks

[dense edge]

oh for sure, it's like a superpower has been unlocked, and I also find it's one of those things "use it or lose it", it's like "I just learned this one week ago and suddenly can't remember it at all anymore" haha. where it's good to keep practicing to keep the skills sharp

I like professor leonard's approach, the spiral method, good to go back on previous notes each day before learning new material

just to review, just write papers on

[wide sundial]

oh yeah he's hella good

i watch him too :D

[dense edge]

yeah, he's great. I think his videos are the only reason I have any hope of succeeding in calculus 1 this term haha

[wide sundial]

that's how i learned calculus too

watching his videos

[dense edge]

the university professor has kinda thrown in the towel, it's a bummer when you get in a classroom like that

[wide sundial]

they set me up for success in highschool

oh that's rough

[dense edge]

but at least we have internet, and wolfram alpha to help. I think learning maths today can be a lot easier than in the 1800s. but maybe back then they didn't have as many distractions either lol

gotta stay focused, that's the key

alright, thanks for the help today! I will keep plugging away at this

https://tenor.com/view/thank-you-so-much-thanks-thanks-so-much-thanks-so-very-much-gif-art-gif-25711651

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[quaint pendant]

Why are they neglecting writing the function v(t) and a(t) in those definite integrals? https://i.imgur.com/81c9xZU.png

That is, it should be (\int_{v0}^{v(t)} v(t) dt) for the LHS, and for the acceleration function, it should be, (\int{0}^{t} a(t) dt). So what's going on? Its really confusing me.

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@quaint pendant Has your question been resolved?

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[harsh halo]

hello, how can I solve this problem?

[severe basin]

Do you know what i^(any multiple of 4) is?

[harsh halo]

I know the sequence: i -1 -i 1

[severe basin]

Write i to the powers of natural numbers and observe.

[harsh halo]

I don't understand how I can simplify that

[severe basin]

Answer this.

[harsh halo]

I don't understand so I don't know

[severe basin]

Have you did this?

[onyx glen]

what is i^4?

[harsh halo]

1

[onyx glen]

and what is i^8?

[harsh halo]

1

[severe basin]

Right

[onyx glen]

do you understand that i^4 = 1 means i^(4n) = 1 for any natural n?

(or any integer n, really)

[harsh halo]

no it sounds a bit confusing can you give an example of what you mean

[onyx glen]

i^(40) = i^(4 * 10) = (i^4)^10 = 1^10 = 1

do you understand this?

[harsh halo]

no

[onyx glen]

would you like me to explain these manipulations in more detail?

[harsh halo]

yes

[onyx glen]

$i^{40} \overset{{\color{red}(1)}}{=} i^{4 \cdot 10} \overset{{\color{red}(2)}}{=} (i^4)^{10} \overset{{\color{red}(3)}}{=} 1^{10} \overset{{\color{red}(4)}}{=} 1$

[glossy valve]

Ann

[onyx glen]

i've numbered all of my steps here

do you understand the step marked with (1)? Y/N

[harsh halo]

yes

[onyx glen]

do you understand the step marked with (2)? Y/N

[harsh halo]

I understand it's a multiple of 4

but I don't understand how you find that out

[onyx glen]

hold on.

[harsh halo]

so still on 1st step

[onyx glen]

so you do not understand how to recognize 40 as a multiple of 4?

[harsh halo]

I don't understand the method you use. That I can do in my head but not for larger numbers

[onyx glen]

... okay let me try asking this again

the manipulations i've written out, independently of any method or lack thereof

do you understand the manipulations themselves? yes or no

(A) "yes, i understand them"
(B) "no, i don't understand one or more of your steps"
(c) "no, the fact that you don't adhere to a method i could replicate makes my brain unable to conceptualize any of this"

[harsh halo]

B

I don't understand step 3

[onyx glen]

you yourself said i^4 = 1.

in step 3, i am applying exactly that

i^4 = 1, thus (i^4)^10 = 1^10

do you understand this?

[harsh halo]

yes it's 1 so where does the 10 come from

[onyx glen]

what do you mean, 'where does the 10 come from'?

step 2 left us with (i^4)^10.

the 10 stays as it is.

[harsh halo]

okay yes mb

but can you explain the last step aswell

[onyx glen]

1^10 = 1

this step?

[harsh halo]

I got it now

[onyx glen]

okay, so then do you understand the sequence of steps as a whole?

[harsh halo]

yes

[onyx glen]

do you understand that a similar sequence of steps can be done to show that i^(4n) = 1?

[harsh halo]

yes

but you split 40 into 4 * 10 I don't know how to do that for my problem

[onyx glen]

well, we did not get to your problem yet.

227 is in fact not a multiple of 4.

does this fact need explanation?

[harsh halo]

I'll take your word for it

[onyx glen]

why take my word for it? you can see for yourself that 227 is odd.

[harsh halo]

so any number that is a multiple of 4 must be even?

[onyx glen]

yes, of course it does.

4 itself is an even number, after all.

[harsh halo]

okay

[onyx glen]

anyway,

you can write i^227 as, say, i^200 * i^20 * i^7, among other things

[harsh halo]

yeah

[onyx glen]

do you understand how to proceed from here?

[harsh halo]

no I understand that

but I don't know how to solve my problem

[onyx glen]

well, now you know that i raised to the power of a multiple of 4 gives 1, yes?

[harsh halo]

yes

[onyx glen]

can you use this knowledge to simplify, at least partially, the expression i^200 * i^20 * i^7?

if there is a part that you don't know what to do with, then leave it be.

[harsh halo]

1 * -i

[onyx glen]

well, it looks like you have just almost solved the problem...

[harsh halo]

how did you do this step?

[onyx glen]

i wrote 227 as a sum of several things most of which were hopefully easy to recognize as multiples of 4

i did not want to risk confusing you by writing 227 = 4*56 + 3

[harsh halo]

so when solving similar problems I should try and split the number into numbers that are divisible by 4?

[onyx glen]

yeah, that will work. i'm not going to say you "should" do that though

you can also just take the remainder mod 4 of the exponent

[harsh halo]

I got the next answer wrong

[onyx glen]

what did you do?

[harsh halo]

how can I know if the number is divisible by 4

I did i^2014 * i^2

[onyx glen]

are you familiar with divisibility tests?

[harsh halo]

no

only 3

and 2

[onyx glen]

well, you can look up more divisibility tests on your own time later, but

a number is divisible by 4 if and only if its last two digits form a number divisible by 4

and also, a number is divisible by 4 if and only if it is even and half of it is also even.

[onyx glen]

and you took i^2014 as 1, because you deemed 2014 to be a multiple of 4. is that what happened?

[harsh halo]

yes

[onyx glen]

yeah, that's where your mistake was.

2016 is a multiple of 4: its last two digits are 16, and 16 is 4*4

[onyx glen]

do you need me to explain why this is true in more detail?

[harsh halo]

even then how would I know if a larger number is divisible by 4 if the last 2 digits form a large number

[onyx glen]

how large a number do you think can be formed by the last two digits?

[harsh halo]

96

[onyx glen]

up to 99, but yeah

[harsh halo]

yeah but that is not even

[onyx glen]

still, it's not so large.

if you are unsure, you can simply divide it by 4, whether mentally or on paper, and see what the remainder is.

you know how to do long division, right?

[harsh halo]

yeah I think so

[onyx glen]

right

[harsh halo]

I should know

[onyx glen]

i can explain to you in more detail why the rule i said works, if you'd like to know.

[harsh halo]

yeah

sure

[onyx glen]

the difference between any number and its own last two digits always ends in 00, and hence is a multiple of 100.

100 is a multiple of 4, therefore all multiples of 100 (which are precisely those numbers that end in 00) are also multiples of 4.

two numbers that differ by a multiple of 4 have the same remainder mod 4, therefore they are either both multiples of 4 or both not.

do you understand this?

[harsh halo]

I don't see how the first line can be true

nvm

but shouldn't that work for all numbers that 100 is a multiple of?

[onyx glen]

indeed, and it does!

[harsh halo]

interesting

thanks for all the help

much appreciated

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[mossy sigil]

That's the question

My answer is

It is not marked as correct, but I don't know why.

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@mossy sigil Has your question been resolved?

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@mossy sigil Has your question been resolved?

[mossy sigil]

<@&286206848099549185>

[worn maple]

I guess section is a 2D-plane?

[mossy sigil]

yes it is

but the way the game works

is that I just have to trace the intersections between the 2d plane with the cube faces

[worn maple]

Yeah, and there are way to many of them no?

[mossy sigil]

I thought there's only 1 plane that satisfy all the conditions?

[worn maple]

I'm not sure though cause there are both dashed and light lines that were added

[mossy sigil]

dashed count, light line doesn't

[worn maple]

Yes, there is only one plane satisfying these conditions

[worn maple]

Ok, then explain how your 2D-plane traces an X on the bottom face of your cube? Seems pretty unlikely to me

[mossy sigil]

my understanding is that the answer is the triangle over here

that triangle is part of a plane, the plane intersect the cube at that triangle, that traingle passed through the 2 corner points, the triangle is perpendicular with the original white diagonal

[worn maple]

The line on the bottom face is dashed and not full
Also, are you sure that the third point of your triangle lies in the middle of the edge of the cube? How would you justify it?

[mossy sigil]

yes. dashed line still count. I think all lines not on the facing surface is dashed

diagonal is perpendicular with diagonal right? if it started from the middle, then it will end at halfway through the edge

[worn maple]

If you think this is the answer, why is the bottom red line (across the bottom face of the cube) not plain?

[mossy sigil]

All lines on the bottom will be dashed, the only plain lines are the line on the faces of the cube that is facing us

the bottom face is hidden (we cannot see it if the cube is solid)

thus all line there are dashed (this is automatic)

[worn maple]

Oh okay I get it, but then you added too many lines, by that I mean you added more lines that simply the trace of the 2D-plane section we want

[mossy sigil]

Yes, the game allows it. I need to add more lines as constructor

[worn maple]

Ok, now think about this:
Are you sure that the third point of your triangle lies in the middle of the edge of the cube? How would you justify it?

[full forum]

@mossy sigil Has your question been resolved?

[mossy sigil]

okay so. The new diagonal is perpendicular with the original diagnal right?
And then that thin dashed line, is parallel with the new diagonal. That means it is also perpendicular with the original diagonal

Ah I see

so

I think

my 1st assumption is wrong

the new diagonal is not perpendicular to the original diagonal

am I on the right track now?

[full forum]

@mossy sigil Has your question been resolved?

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[tulip beacon]

Nearly but not quite

(10x+5)^2 -(X+10)^2 = BD^2

[worthy tree]

(x-4)² - BD² = DC²

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@torn jolt Has your question been resolved?

[worthy tree]

DC² = (x - 4)² - BD²
= (x - 4)² - (10x+ 5)² + (x + 10)²
= x² -4x +16 -100x² - 25 -100x + x² + 100 +20x
= 2x² - 100x² -104x +20x - 9 + 100
= -98x² -84x + 91

so :
DC = sqrt(-98x² -84x + 91)

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✅

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[worthy tree]

anytime

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[weary stirrup]

How?

[weary stirrup]

I don't get that when I put it in the calc

[devout valley]

,calc (5/4877)^(4/5)

[glossy valve]

Result:

0.0040611945901459

[weary stirrup]

Must be me then, but I have a fx-991MS for reference

[light sonnet]

Recall that $\sqrt[n]{a^m} = a^{\frac{m}{n}}$

[glossy valve]

dldh06

[devout valley]

Can you show me what you typed in as the input please?

[rocky vale]

I'm guessing you didn't use enough parentheses

[weary stirrup]

I type 5 shift ^ ((5/4877)^4)

Took your advice with this ^

Didn't get it

[devout valley]

Do you have a picture of the expression in the calc?

[weary stirrup]

I'll take a pic rn

[devout valley]

Oh that is the same thing but in scientific notation

[weary stirrup]

How do I switch that off

[glossy valve]

chartbit

[devout valley]

,calc 4.06119459* 10^(-3)

[glossy valve]

Result:

0.00406119459

[weary stirrup]

Can I change that in my Calc settinfs

[devout valley]

Don't think you can tbh, those calcs default to that for really small numbers

[weary stirrup]

Bruh

[devout valley]

There might be a way to do it but I can't remember, it's been time since I had a calc like those ones

[weary stirrup]

Thx

.close

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[worthy tree]

like what ?

[devout valley]

The one in OP's picture, I have the ones where it does the "natural" looking outputs

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[hollow lintel]

i’m suppose to be using polynomial identities to multiply the expressions

[hollow lintel]

and i put it into mathway and don’t how it it go that answer

and idky what i’m doing at all and don’t even know where to start

with (x+6)^2

[kind jay]

How can u write (x+6)^2 also

[slow gulch]

I think they want you to use a^2 + 2ab + b^2

but if that confuses you you can use rainbow method

[hollow lintel]

x^2+2x6+6^2

[gritty rose]

2 * 6 = ?

[hollow lintel]

12

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@hollow lintel Has your question been resolved?

[mossy sigil]

.reopen

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[thorny horizon]

Hello, can someone help me understand the Poisson Distribution, like how it may be derived from the binomial distribution?

[thorny horizon]

say some event occurs lambda times per hour

then how do we find the probability of this event occuring say once in a minute?

[thorny horizon]

if we know this probability and it is equal to p, then the probability of observing the event occur k times in 60 minutes would be (60 choose k) * p^k * (1-p)^(60-k)

how can we get to the poisson distribution from here

https://youtu.be/3z-M6sbGIZ0
im watching this video by khan academy, and they say that the probability is lambda/60

but i dont see how that is possible, if lambda was like 120, 120/60 is 2 and that is certainly not a probability

[topaz valley]

the rate parameter in a poisson distribution is a measure of frequency

say you have a fixed time interval of 1

then λ tells you how many times you can expect the event to occur within that time interval

[thorny horizon]

right..and using this information can we model this process approximately as a binomial distribution?

how would we use lambda in doing so

[thorny horizon]

it says here that the poisson distribution is a limit of the binomial distribution

[topaz valley]

you model it as infinitely many binomial events occurring within a fixed time interval

[thorny horizon]

and with what prbability do we consider them to occur

[topaz valley]

the limit is taken by increasing the number of binomial RVs to infinity

while smearing the probability across all of them evenly

[thorny horizon]

but we do not have a probability to start with

we only know lambda events occur in this time

not a probability of them occuring

if this is difficult to answer can u give me a different intuition for like why the poisson distribution is defined the way it is

how does an e turn up in it etc

[topaz valley]

im just trying to recall the correct construction

you take the probability of each binomial to be p_n where in the limit np_n -> λ

so you can think of it as p_n = λ/n

the rate is smeared across all the binomials

[thorny horizon]

oh...

[topaz valley]

so in total you might expect λ to turn out to be successes

but you're limiting n to be infinity

so the probability of success of any particular binomial goes to 0

[thorny horizon]

oh ig i kinda get it

[topaz valley]

so like

it becomes a "rare event"

[thorny horizon]

by "binomials" do u mean like individual trials?

[topaz valley]

yes

[thorny horizon]

oh

[topaz valley]

it's like a random walk

[thorny horizon]

what is y and what is x here

X = num of successes in the jth trial?

[thorny horizon]

oh

[topaz valley]

but you just think of it like smearing a probability across a very large number of binomials

bernoulli trials

[thorny horizon]

okay..ig i get it a little better now

[topaz valley]

so like think of the jth X occurring at time j/n or something

within the unit interval

ultimately you'd expect λ of the Xs to succeed altogether

but there will be infinitely many of them in the limit

[thorny horizon]

and is there like a different way of making sense of the formula for the poisson distribution, withut involvin the binoial distribution?

[topaz valley]

well this is how you can relate binomials to poisson

but poisson at its heart is counting events which occur at some frequency

and the events are happening in continuous time

[thorny horizon]

what i really want is to like understand where the poisson distribution comes from, like why the formula makes sense

[topaz valley]

which formula

[thorny horizon]

$\frac{\lambda^x\cdot e^{-\lambda}}{x!}$

[glossy valve]

SilverSoldier

[topaz valley]

well that arises from the limit

or if you want to construct it from gamma RVs you can do that too

but that also would be a limiting thing from geometric RVs

[thorny horizon]

oh, idrk about gamma distributions

[topaz valley]

exponentials?

the one modelling waiting times

[thorny horizon]

no 😦

[topaz valley]

[thorny horizon]

i do know about like radioactive decay things

maybe this is kinda similar to that?

[topaz valley]

it should model that

[gritty rose]

Those are modeled with decaying exponentials

[topaz valley]

counting process is poisson tho

waiting times are exponential

the formula kinda just comes from computing the limit

it's called the poisson limit theorem

[thorny horizon]

oh

[topaz valley]

i think there's also another argument that can be made looking at generators of ctmcs