help-26

[odd pagoda]

well, as rigorous as a conversation on discord tends to be

[desert apex]

You're funny

🙂

[odd pagoda]

to get a more rigorous proof you need a much more rigorous statement to prove in the first place

[thorn violet]

Where would you like more rigor?

[desert apex]

Thanks anyways you too
Love you
Bye

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[old fox]

is this correct?

[finite storm]

re-check implicit differentiation

[old fox]

I am not sure what youre referencing specifically

[finite storm]

${2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0}$

[thorny flame]

:D

[finite storm]

this should be the correct way to implicitly differentiate, yes?

try subsituting this again

[finite storm]

its 2x

not just 2

[old fox]

does that change the answer?

[topaz sinew]

@old fox Has your question been resolved?

[finite storm]

yes

its${20\frac{3}{4} + 20\sqrt{3}\frac{dy}{dt} = 0}$

[thorny flame]

:D

[old fox]

what do i do next?

[finite storm]

${\frac{dy}{dt} = \frac{-\frac{3}{4}}{\sqrt{3}}}$

[thorny flame]

:D

[finite storm]

${\frac{dy}{dt} = -\frac{\sqrt{3}}{4}}$

[thorny flame]

:D

[finite storm]

@old fox

[old fox]

7.5?

[old fox]

@finite storm

[finite storm]

,w -sqrt(3)/4

[thorny flame]

Results provided by WolframAlpha

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Requested by nuclearinteraction

[old fox]

sorry i meant -7.499

@finite storm

[finite storm]

could u show how

[finite storm]

isnt this just

dy/dt

[old fox]

but its 10 feet away

[finite storm]

?

It’s asking for the rate of how much y is changing

[old fox]

oh ok

thanks

[old fox]

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[sick delta]

hello,
how can i find the probability than some event with probability p occurs more than 5 times in n tries ?

[sick delta]

I looked online, i found a "choose" formula with factorials but i dont think it works

https://i.imgur.com/6TxlgHD.png

[gilded gorge]

there is a way to calculate it manually but assuming you dont need to

[sick delta]

i did this times (p)^5 * (1-p)^5

[gilded gorge]

https://stattrek.com/online-calculator/binomial

you can use binomcdf

[sick delta]

im trying to implement the formula into a google spreadsheet tho

[gilded gorge]

ohho k

i think google sheets might have it let me check

[sick delta]

they have FACT()

for factorials

but i dont think im doing it right

[loud oasis]

https://support.google.com/docs/answer/3093987?hl=en

[sick delta]

oh

XD wtf thats easy

[gilded gorge]

ah yes

[sick delta]

ty hahaha

for "prob sucesses" it wants probability for 1 event right

okit does

what is the diff between binomial cumulative distribution and binomial distribution.

[gilded gorge]

if you use the cumulative one

it gives you the probability of x successes or less

if you dont use cumulative then its just the probability of exactly x

[sick delta]

ohhk

uhh

[gilded gorge]

whats up

[sick delta]

its saying theres a higher chance to hit 5 out of 6 than 5 out of 30

[gilded gorge]

this is obviously not true but i'm not sure where you are going wrong since i can't see what you input

[sick delta]

=1-if(J13>5,BINOMDIST(5,J13,A$22,FALSE),0)

j13 is num trials

a22 is probability

a22 in this case is 0.95

[gilded gorge]

ohh

well in this case

actually itrs true

you have a higher chance to hit exactly 5 when you are only doing it 6 times than exactly 5 when you are doing it 30 times

because you would expect more than 5successes if you are oding it 30 times

[sick delta]

how can i tell it to include 5 or more sucesses for this formula

maybe i should do 4 with cumulative ?

[gilded gorge]

yes exactly

[sick delta]

ok

[gilded gorge]

and one minusthat

[sick delta]

like this or diff?

[gilded gorge]

ohh yeah you already have the 1 minus

so yes 4 with cumulative

[sick delta]

uhh

still similar issue

https://i.imgur.com/M64g0i3.png

the one highlighted is 6

6 num trials

(its rounding to two decimals but i would have imagined that theres a better chance to hit >5 with more num trials)

i think its still solving for exactly 5 trials / exactly 4 or less

but i need it to solve for 5 or more

actually lemme try smth

nvm i think solving for exactly for or less but not doing 1- it is the same as what im lookign for

[topaz sinew]

@sick delta Has your question been resolved?

[sick delta]

ok i did it manually

binomdist doesnt work

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[indigo bolt]

Is it possible to have a linear transformation T and linearly independent vectors u1, u2, u3 so that T(u1), T(u2), and T(u3) is a linearly dependent set?

[odd pagoda]

sure

send them all to zero

if you have linearly independent vectors you can freely choose what you want their images to be

and a linear map fitting that will exist

[indigo bolt]

i said it was possible and gave an example but my instructor said my answer was entirely incorrect

so i'm just very confused

[odd pagoda]

well then your example was probably incorrect

or you misread the question or something

!original

[topaz sinew]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[odd pagoda]

and show your work

[odd pagoda]

I have no idea what your example is supposed to mean

[indigo bolt]

it's just a matrix multiplication

though now that i look at it with fresh eyes

[odd pagoda]

I mean sure

[indigo bolt]

it's not clear

[odd pagoda]

but how is that relevant

[indigo bolt]

LOL

[odd pagoda]

btw, dont use arrows instead of =

[indigo bolt]

messed up the order for the matrices when i originally wrote them but this was the idea

wait

i should have just sent them all to 0 yeah

[odd pagoda]

even with labelling stuff as x, A and b I still have no idea what you are going for

[indigo bolt]

a linear transformation is Tx = Ax

i just show an example that turns linearly independent vectors into independent vectors

DEPENDENT*

[odd pagoda]

well and what are your three vectors u1,u2,u3?

[indigo bolt]

i intended it to be this

[odd pagoda]

thats a matrix

thats not three vectors

[indigo bolt]

OH

i see

ok ok

so if i did Au1, Au2, Au3 separately and then showed these three VECTORS were dependent, would that have worked?

[odd pagoda]

yes

[indigo bolt]

One last thing, I received partial credit for this other question in the same format. How could I improve this explanation?

[odd pagoda]

partial credit is generous for that

if u1,u2,u3 are linearly dependent, then there exist coefficients c1,c2,c3 not all zero with c1u1+c2u2+c3u3=0

but then also c1Tu1+c2Tu2+c3Tu3=0

[indigo bolt]

could you rephrase? i'm having a hard time parsing it

[golden blade]

He used the definition of linear independence

[indigo bolt]

i understand the first half now because if the only solution for Ax = 0 is the trivial then it's independent

[indigo bolt]

i dont understand this part's relation

[indigo bolt]

is the ^ just saying that the linear transformation doesn't change anything?

since that's the definition of a linear transformation basically

[golden blade]

yes

If you apply T on both sides

and apply its properties

T(0) = 0
T(av1+bv2) = aT(v1) + bT(v2)

[indigo bolt]

i see

ok, thank you both so much!

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[stray harbor]

help

[stray harbor]

had to find taylor serie and interval

found the taylor but i dont know how to continue for interval

[shadow salmon]

Interval as in where the series converges?

[stray harbor]

yes

[shadow salmon]

So for clarity, it’s the one on the bottom left?

[stray harbor]

yeah that the taylor i found

[opal vault]

how do you find the radius first of all

[stray harbor]

how do i find the radius without the the interval ? can i ?

[opal vault]

well yeah, there's a formula for the radius of convergence

[shadow salmon]

So since it’s written as a power series a good way to start is to examine the coefficients using something like the ratio test etc

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[golden trail]

Iam a starting with multi variable calculus and i need to cover these topics -Functions of two variables-limits and continuity-partial derivatives –total differential-Jacobian
and its properties. any youtube channels for this topic ?

[raven sparrow]

I think Organic Chemistry Tutor has a few videos on Calc 3, which will cover limits of functions of 2 variables and partial derivatives.

Dr. Trefor Bazett also has a playlist for Calc 3 and I think it covers essentially all the topics you've requested, although you could skip through the curvature stuff for curves if you don't need it.

And Paul's Online Notes covers all those topics in written format which can help getting another point of view on the subject if you're primarily learning from a textbook.

[golden trail]

thank you 😇

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[neon iron]

help

[neon iron]

Trying to solve by showing the sequence of partial sums is uniformly convergent

wanted to know if my way of finding n is legal

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[neon iron]

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[heavy stone]

Help with strong induction problem

I’ve been working through this problem and i don’t really know what to do.
I understand the base b notation and everything, but i can’t seem to find a way to prove it for n + 1.
I already know the inductive hypothesis in which i assume all k numbers with k <= n can br written in base b.

I’ve tried some extensions of the proposition but nothing really proves the induction

Any help would be appreciated

[heavy stone]

My work is messy though i’m not sure it’ll help

[distant nymph]

Once you write,
$n = \sum_{i=0}^{l-1} k_{i} \cdot b^{i}$

Just add 1 to the both sides and then manipulate the RHS to take the expected form.

[thorny flame]

Anonymous

[distant nymph]

I have totally forgotten Latex. Excuse me.

[heavy stone]

Yeah i thought about that because 1 in base b is still 1

I’ll try it

Thanks

[distant nymph]

You will have two cases though.

that $k_{0}$ is b-1 or less.

[thorny flame]

Anonymous

[distant nymph]

Two separate cases.

[heavy stone]

Yeah thats in the statement as the l numbers are up to b - 1 and l is bigger than 1

So thats true for numbers in base b

[distant nymph]

No. No. You are misunderstanding.

[heavy stone]

Ah that has to be a condition to be proven true

[distant nymph]

You shall have two cases as in,
case 1 : $k_{0} = b -1$

case 2: $k_{0} < b-1 $

[heavy stone]

I havent demonstrated it yet

Yes youre right

[thorny flame]

Anonymous

[heavy stone]

I dont see how the two cases could be applied to the sum? Sure they are there but i cant really see the implications

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[neon iron]

hello, for this question, if it says « divisible by x » is it correct to assume that the eqn = 0 when x = 1?

[neon iron]

otherwise i am not sure how to proceed with the question

[cursive patrol]

no. divisible by x means expression = 0 when x = 0, not x = 1

[neon iron]

thanks mate

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[neon iron]

.reopen

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✅

[neon iron]

nevermind ive just run into another problem

if

x=0 then eqn = 0

so r = 0

hollon lemme check if i did something wrong with the working at the side

[deft holly]

thats a pos

[neon iron]

yea i just realised

im blind sorry

okay i think i should be able to do the qn without any problem now,,,

thanks mate

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[bleak wedge]

Hiii! Are my identifications correct?

For 1st pic. No 3.
How would I answer that

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[bleak wedge]

No

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[clever citrus]

third line to 4th line how did they transform the numerator

from this

to this

wont the second one have sqrt(S_xx) in the denom of the upper fraction

Im tripping idk

wait nvm

ignore

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[sharp echo]

How do i find the domain and range of y = tan^-1(tanx) ?

[near harbor]

the domain of your function should be when tanx is defined, while the range of it is from −π/2 to π/2 rad

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[clever citrus]

Why did they go with that interval

instead of .01 to .025

it doesnt even look like its gonna be between (.025, .05)

so why .02 to .05

wait wtf the exact valye is 0.0354????

ai now Im lost

oh wait wtf (forgot to ultiply the bounds by 2)

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[devout bridge]

Let x(t)=h(t-k) and y(t)=x(2*t) then what is y(t) in terms of function h whether h(2(t-k)) or h(2t-k) i am looking for proper explanation...

[fallow heart]

The second one

x(2t) means: wherever you see t in the expression of x(t), replace t with 2t

[opal vault]

so you mean the second one

x(...) = h(... -k)

put 2t in there

h(2t - k)

[fallow heart]

Yeah yeah sorry

The second

[fallow heart]

The second one
x(2t) means: wherever you see t in the expression of x(t), replace t with 2t

[devout bridge]

Wait i recently asked this question if y(t)=x(2t) then wht is y(t-k) the answer tht i received was x(2(t-k)) rn iam little bit confused

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@devout bridge Has your question been resolved?

[opal vault]

we're doing the opposite now

y(...) = x(2(...))

so if you plug in t-k

y(t-k) = x(2(t-k))

[devout bridge]

.ha ok I got it

I messed up my brain with over thinking

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[humble saddle]

Okay so I have a square, I need to find the perimeter, area, and side length, but I only have the diagonal, which is 26m. How do I find them?

[wet harbor]

try to relate the diagonal to side length

if it worked you got all of them

[humble saddle]

Well, d² = 2x²

And I got 338

But the perimeter and side length is in decimals

[wet harbor]

that's great this means d = sqrt(2)x
you can now calculate the perim as 4x and area as x^2.
you don't have to write decimals you can keep it in terms of sqrt(2)

don't use calculator

[humble saddle]

d² = 2x²
x² = (1/2) d²
x² = (1/2) • 26 • 26
x² = 13 • 26
x² = 338

x² is the area right

[wet harbor]

yes

[humble saddle]

Okay well that leaves me with one answer

Because it's multiple choice

Thanks

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[balmy juniper]

Wait on the 3rd line where does the 2nd sin^2(theta) go?

[cunning kayak]

LCM i.e. similar to
1/2 + 3/2 = 4/2 = 2

[balmy juniper]

ohhhh

thank you, it just tripped me up there

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[silk tendon]

is there a straight forward way to find the minimum value of this

like if you do some algebra based on the partial derivatives it works, but what if you can't think of the algebra tricks used/factorizations

(please ping me)

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[silk tendon]

<@&286206848099549185>

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[inland valve]

is there any tricks to solve identities?

[near harbor]

trig identities?

[inland valve]

yes

[vernal shard]

Yeah, practice

[near harbor]

agreed, the only way

there's just too many

[inland valve]

the grade 11 one is already hard when i learned it and advance function one is even harder

[vernal shard]

Give an example and explain your thought process

[inland valve]

Like this one

[tender bison]

I think you have to apply the formula of tan alpha and beta

You can search this formula on internet and then apply the formula on left side

[inland valve]

hm ok

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[narrow kite]

It’s not 0.004 :((

[narrow kite]

I have no idea where to start with this problem :(

[hazy palm]

me neither

[topaz sinew]

@narrow kite Has your question been resolved?

[narrow kite]

<@&286206848099549185>

[prime flame]

Do you know that every wave can be decomposed into the sum of sine and cosine waves?

[narrow kite]

Yea it says that below the middle c is the sum of the harmonics

they are all sine waves but I dont know where to start in figuring out the numbers

[prime flame]

a sin(Ax) means a is the amplitude and A is the frequency (It is called standard form if you are interested)

amplitude is easy to guess from the graph

It means the highest value of your sine function in this case

[narrow kite]

no yea, I just cant seem to understand how to figure out the preiods

because they arent nice numbers

periods*

like the amplitude for the graphs are 3 1 1 respectivly

for asinAx+bsinBx+csinCx

but I dont know waht the A B and C values are

and I dont know how to figure them out

[prime flame]

Well, we can guess by counting the number of oscillations over some time interval

[narrow kite]

ohhhhhh

that makes so much more sense

ok well we have 0.008

and yellow oscillates 2 and a tiny bit times

green 4 and a bit times

and blue 6 times

bit the oscillations finish slightly before 0.008 so how do we deal with that

is this what you mean btw @prime flame

[prime flame]

Yeah, but the number is quite ....

And I think in some cases using period is easier

[narrow kite]

how would you figure out the period then

[prime flame]

Then you can compute frequency by 1/period

[prime flame]

period is the number of time the wave need to complete one cycle

[narrow kite]

Yea I got that, but how do I figure out how much time it takes when the graph doesnt show the exact number? Do I guess with the number of ocsilations?

[prime flame]

Example, for the yellow wave, if you choose a point at x = 0 then the wave completes one cycle at a point a little bit smaller than 0.004

But it is not 0.004 as they have hinted

So I am choosing another point

Is this all of the information?

[narrow kite]

this is all of the info

[prime flame]

Ok then it is a guessing game then

[narrow kite]

ok so we know yellow oscillates 2 times in 0.008 seconds

this means that one ocillation takes 0.004 seconds

1/0.004

right?

[prime flame]

Sadly, it is not that

it is a little greater than 2

[narrow kite]

so would we just guess?

how do you convert the number of oscilaltions to the period of the graph

wait nvm i got it

[prime flame]

frequency = number of oscillations/time and period = 1/frequency

and we need frequency

[narrow kite]

f(x)=sin((2pi/T)x)

[prime flame]

And damn looks like the frequency here is not an integer

[narrow kite]

yeaaa

well, it looks like its kinda 0.758 is one period

sorry not one period

the end of the graph

like I got that

so the colours match up

[prime flame]

btw do we need to be exactly, tho?

[narrow kite]

surely there is a way to figure out exactly what the period is though instead of guessing

[narrow kite]

I have no idea :(

[narrow kite]

wait wait

could we use simultaenous equations?

[prime flame]

What do you mean?

[narrow kite]

one sec, im writing something out

im not sure if itll work but im just going to try anyway

wait nvm it wont work

im just trying to wrap my head around how we can figure the periods out exactly

[prime flame]

If it is a real number then we can't figure it out exactly by eyeballing the graph, tho

[narrow kite]

Yea, im worried that theres just some trick that im missing

[prime flame]

Try to read through your book, again 😂

[narrow kite]

trueee

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[near root]

hi

[near root]

can someone help w this one?

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@near root Has your question been resolved?

[near root]

<@&286206848099549185>

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[neon iron]

how do i find the domain & range of -2y+4=1-5x ? i rewrote it in slope intercept form y=5/2x+3/2 , does it matter what number i plug in for x

[merry creek]

You say that because it's linear function, it's domain is ] -\inf, \inf [ .

[neon iron]

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[hard kelp]

shouldn't the t be in the denominator position?

[chilly walrus]

it's equivalent

$\frac{a/b}{c} = \frac{a}{bc}$

[thorny flame]

artemetra

[chilly walrus]

but in this case they put it there to make Q/t = power, which you are given

[hard kelp]

aha

[chilly walrus]

you aren't given Q or t but you are given Q/t

[hard kelp]

makes sense

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[median moat]

So you have equation of a plane in form Ax+By+Cz=D with a defining point P with x, y, and z as 0,0 on that plane, how would it be determined an x and y value on plane relative to P? This is in 3d space and all numbers are real and finite. To start I define a plane by 3 points and get said plane formula. If it helps this is in a 3d space with a view point and a view plane defined. How would I get xyz back from this? X E.

[median moat]

Sorry I have many questions, feel free to ask if you want to know anything. X E.

This is for finding centroid of a 3d planar shape in case that helps. X E.

If anyone knows what this is called might help too. X E.

3 points in plane are defined with x, y, and z known. X E.

<@&286206848099549185> Sorry I do this. Only one call this time, sorry about before. X E.

Is there a way without sine and cosine constantly? Trying for as exact as possible. X E.

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@median moat Has your question been resolved?

[median moat]

https://mathworld.wolfram.com/PolygonCentroid.html https://mathworld.wolfram.com/PolygonArea.html X E.

Those are 2d formulas, just how to do 3d?

https://tenor.com/view/stan-twitter-monkey-sad-depressed-bored-gif-14632661802414689947

I guess ping me if you have an answer. X E.

I have this 3d shape projected to a 2d view plane, would centroid in that 2d exactly reflect 3d centroid?

https://mathworld.wolfram.com/GeometricCentroid.html Found this, does this help?

[median moat]

Should I ask a new question? I really want to know how to find centroid of a 3d planar shape, x, y, and z. X E.

By centroid I mean if you defined all points in a shape then their average as would be. X E.

http://mechanicsmap.psu.edu/websites/A2_moment_intergrals/A2-3_centroids_3D/centroids3D.html I don't understand integrals but if someone wants to explain this feel free. X E.

What do I do since it has been an hour and a half?

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@median moat Has your question been resolved?

[median moat]

Found like my solution. X E.

.close

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[harsh shell]

how the fuck do you do this tech free

[lyric veldt]

change the bigger division bar into multiplication

so $\frac{\pi}{3} * \frac{1}{\pi}$

[thorny flame]

AwesomeRat

[harsh shell]

oh ok thanks

.close

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[formal peak]

is this math correct?

[icy sky]

looks fine

[topaz sinew]

@formal peak Has your question been resolved?

[vernal matrix]

Where does the negative sign come from here?

[icy sky]

ooh nice catch

[vernal matrix]

(also the integral limits, not entirely happy with them personally )

[halcyon lantern]

Isn't that the usual Laplace form?

[vernal matrix]

(as in, from the picture taken, the integral limits would be affected and changed)

[formal peak]

oh shit yea thanks

[formal peak]

how so

im not like too experienced with laplace im doing this for a highschool assignment

[vernal matrix]

For any integral substitution, you wanna find out what happens to u when t = 0 and as t -> inf

[formal peak]

so the lower limit would be -a?

[formal peak]

wouldn't that change the laplace form

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[vernal matrix]

It would as is, but also worth checking how your function f is defined, as likely you have it restricted somehow

.reopen

[topaz sinew]

✅

[formal peak]

how could i define it to make it work

could i assume that f(u), will be zero if u is less than 0

[vernal matrix]

Well, usually you have your Laplace transform defined for functions from the positive numbers to infinity, so likely your f isn't necessarily defined for negative numbers, but usually you choose it to be zero yes

[formal peak]

alright

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[crimson cosmos]

Translation: ABCD is a square. ED = 3√2m, calculate NC*

[crimson cosmos]

i dont think doing this ends anywhere

Please, help

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[tall iris]

still at it

what combo would even work bro💀

[noble laurel]

I'll tell you if you can beat me in a 1v1 clash royale

[tall iris]

bro💀

ok

here lemme pull it up

[noble laurel]

tell me your ID

[tall iris]

havent played in a bit

[tall iris]

hol up lemme re download the game]

[noble laurel]

You're cooked

I thought you played based on your pfp

[tall iris]

i did

havent played in a month

tho

[noble laurel]

Click this link to add as friend in Clash Royale! https://link.clashroyale.com/invite/friend/en?tag=29QVLCP8C&token=w89wkf62&platform=iOS

[tall iris]

bro

aint no way

ill try but i aint winnin

[noble laurel]

oops

im mad I suicided that dart goblin still

[tall iris]

eh ur fine

well gg

[noble laurel]

I guess you gotta brush up on your log properties alone

[tall iris]

yeah

my spells arent great

placement i mean

[noble laurel]

I'll rematch you with a trash deck

same deal

[tall iris]

nah

you'd win

whatever

ill try

[fallow igloo]

LMAO

don’t forget to share the result

[tall iris]

yeah bro

he won first time

when u said trash i didnt think u meant that trash bro

[noble laurel]

😂

firecracker deck

I guess I'll help you out after all

[tall iris]

thx

[noble laurel]

so log(a^b) is the top left box

and by log rules it equals

blog(a)

1 sec

[tall iris]

k

[noble laurel]

and the top middle box is log(ab)

which by log rules =log(a)+log(b)

so you want

blog(a) < log(a)+log(b)

agree?

[tall iris]

yes

[noble laurel]

and the third in the top row is

log(a/b)=log(a)-log(b)

so what you really want is

blog(a) < log(a)+log(b) < log(a)-log(b)

okay actually now im confused because looking at your thing it

it says we're increasing from left to right

but also increasing from top to bottom

but both can't be true

[tall iris]

is what i thought

but i guess not

every combo i try on one side doesnt work on another

[noble laurel]

yeah cause it is made wrong

look at this

circle in green is the same thing

if the one to the right in top right is bigger

then how could the one in the middle be simeltaneousy bigger and smaller

Oh nevermind I see, we are allowed to use different numbers in each box

Okay let's restart

[tall iris]

alr

[noble laurel]

log(a^b)

is top left

=blog(a)

[tall iris]

yes

[noble laurel]

log(cd)

is top middle

=log(c)+log(d)

log(e/f) is top right

=log(e)-log(f)

all by log properties

and you want to pick a,b,c,d,e,f from 0 to 9 such that

blog(a) < log(c)+log(d) < log(e)-log(f)

Understand?

[tall iris]

kind of

[noble laurel]

Let's pick b=0

then

log(a^(b))=log(a^(0))=log(1)=0

So it doesn't matter what we pick for a, if we pick b=0, the top left box is 0

that's good because we want it to be smaller than the other boxes

right?

[tall iris]

wouldnt we want top left to be smaller tho?

[noble laurel]

I meant top left

typo

[tall iris]

ah ok

but yes

[noble laurel]

okay so we've used 0 now in a blue box

now let's do like

2*1

so 2 in the red box, 1 in the blue box

for the top middle

log(2*1)=log(2)+log(1)=log(2)

so far so good since 0<log(2)

and now we've used 0, and 1 in blue boxes, and "a" and 2 in red boxes

[tall iris]

yep

[noble laurel]

now for the third lets use 5 in the red box and 2 in the blue box

then log(5/2) = by log rules =log(5)-log(2)
but it is actually more obvious to just simplify it to
log(2.5) to see that it's bigger than log(2)

so then we have in the top row,
0< log(2) < log(2.5)

and we've used 0, 1, 2 in the blue boxes, and "a", 2, 5 in the red boxes

[tall iris]

yes

[noble laurel]

so far so good?

[tall iris]

yes

looks nothing like what i did

which is probably a good thing

[noble laurel]

I'm going to retroactively change my choice of 9 to 5

nothing else changes, but it just saves us the bigger number

and it works out the same

[tall iris]

ok that works

[noble laurel]

let's continue down the right side

so right middle box

we'll use 3 in the blue and 1 in the red

so

log(3*1)=log(3)

log(2.5) < log(3)

so far so good

and again noting what we've used:
0,1,2,3 in blue boxes and 1, 2, "a", 5 in the red boes

[tall iris]

to be clear we're leaving "a" as a variable because either way it'll end up the same right?

[noble laurel]

yes, our choice of "a" so far affects nothing so we might aswell leave it to the end and pick whatever works

let's go to the left side going down now

our top left box was
log(a^0)=log(1)=0

so we want something bigger than 0

let's do 3 in the red box and 4 in the blue box

log(3*4)=log(12) > 0

so now let's write again what we've used:

0, 1, 2, 3, 4 in blue boxes and 1, 2, 3, "a", 5 in the red boxes

we need the bottom left corner to be bigger than log(12) now

ah damn looks like we're stuck

[tall iris]

bottom left is what gets me

[noble laurel]

Let me write this on paper instead of just free balling it here

1sec

[tall iris]

k

[noble laurel]

ok

Ready?

@tall iris

So I noticed a few things in this when I stopped guessing

[tall iris]

yep we back

[noble laurel]

the 1 in the middle is kind of important

First of all, we can't have a 0 as any of the multiplication boxes

since log(0) is undefined

right?

[tall iris]

yes

[noble laurel]

Which means in those multiplication boxes, we need the top middle, and left middle ones, to be less than 1, and be multiplying two numbers larger than 0

so they have to be 1*2 and 2*1 in whatever order you like

since log(3) > 1

and so is anything greater than 3

oh wait this is log base 10

bruh

I've been treating it as natural log

whatever this still should work

[tall iris]

wait

i thought u knew that

[noble laurel]

it should still wrok

I see log and I think natural log

its whatever

[tall iris]

ah ok

[noble laurel]

so we have this

but notice now we've used 1 and 2 in blue boxes

[lean pebble]

austin my mental health sir

[noble laurel]

whatts up sir

[noble laurel]

so now when we're picking for the blue boxes in the denominator, we want these to ideally be small as possible

so we pick 3 and 4

and now in the red boxes for the numerators, we need x/4 > 2 and y/3 > 2

so x has to be 9 and we'll let y=8

[tall iris]

alr

[noble laurel]

so now it isn't too hard to fill in the rest

I ended up going with this

but I'm sure you could also fill it in yourself from this point

[tall iris]

yeah

seriously tho big thanks

good cr matchs too

[noble laurel]

🫡

[wooden osprey]

man got a clash royale duel AND a lesson on logarithms

this is an insane W

[tall iris]

yeah bro

math teachers gotta pick this up

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[neon iron]

F

Eh the answer

Is mine wrong

Someone explain

[wet harbor]

yours is correct

he calculated a weird expression for it

his expression gives each one of the outcome twice

to get each once you gotta substitute n = 2k

you would then get $\theta = (4k-1)\pi$ which is the same as your expression but delayed by one, if you put n+1 in this expression you will get $\theta = (4k+3)\pi$

[thorny flame]

Mohamed Mohsen

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[neon iron]

hello, im doing e-maths and im not too sure how to start with this

[neon iron]

when i check the answers, i dont know where 30cm² comes from

[junior fable]

maybe the rectangle they drew was 6×5 cm^2

[neon iron]

oh

oh yea i needed to measure the square they gave me

sorry for wasting your time

.close

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[junior fable]

you didn't waste time, it's all good

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[lyric blaze]

how does one find a non-bashing solution?

[fallow igloo]

that's all you can do

[wary tulip]

well i wouldn't say that

[lyric blaze]

hmm

[cursive patrol]

listing out everything is very effective way to do this problem tbh

[pearl patio]

https://mathworld.wolfram.com/PartitionFunctionP.html partitioning n into k elements is a particularly hard problem. However there is some nice recurrence on it

but yeah listing is probably the best

[pearl patio]

note: this isn't quite the same as the problem but similar to some extent

[wary tulip]

pretty much is if you say at most k or something

[lyric blaze]

interesting but ig listing is the easy way out

[pearl patio]

there is this longgggg formula you can use

💀

P(n,4) is the partition of n parts, where the largest number is 4

but this is equivalent to your definition i think

[wary tulip]

is it?

[pearl patio]

list the parts as rows, there are 4 columns (largest is 4)

so we are just counting the columns instead of rows

[wary tulip]

ok fine

[lyric blaze]

so you substitute n as 0?

[pearl patio]

n is the number you are partitioning, in your question you are trying to find how many ways to partition 14 into 4 parts: P(14,4)

but this is off topic, but partitioning is a interesting subject if u wanna look up

[cursive patrol]

bro this solution isn't even right

ans is 23. it misses the case (3, 3, 3, 1)

[wary tulip]

LOL i got 23 with a generating function solution and have just been getting gaslit for the last 10 minutes trying to figure out what i was doing wrong

[cursive patrol]

crazy...

[wary tulip]

that's my solution

[pearl patio]

should be 14 mb, edited

[wary tulip]

though i was getting 23 with a simpler one before and this was supposed to be an alternative

[lyric blaze]

oh yahh

[pearl patio]

crazy you spotted the exact missing case

[wary tulip]

@pseudo jetty multivariable generating function

[lyric blaze]

well ig listing is one easy possibility, thank, i'll look into partitions

.close

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[wary tulip]

this also tells you how many solutions if there are only 3, 2, or 1 variables in the problem

[cursive patrol]

what was the simpler one? prod 1/(1 - x^i)

[wary tulip]

not quite

maybe it wasn't that much simpler lol. i'm not sure how you would track at most 4 parts with prod 1/(1 - x^i) though

that was the purpose of the variable y in what i posted above

[cursive patrol]

oh interesting

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[robust jasper]

It says

Drawing one and not replacing and then drawing another is the same as drawing two at the same time

Isnt drawing 2 whites

6/9 x 5/8

[neon iron]

you assume the first button is white

so you don’t need to multiply by 6/9

[robust jasper]

ok but then for the second one?

[neon iron]

it’s 5/8 like usual

[robust jasper]

so the probability of drawing 2 whites

is 5/8

[neon iron]

not quite

the probability of drawing two whites when you already know the first one is white is 5/8

[robust jasper]

OH

right

[neon iron]

so i pick a button out of the bag, show you that it’s white, and then we calculate the chances of the second one

[robust jasper]

ahhh

how would i find the probability of drawing two whites in the first 2 picks

Is it

6/9 x 5/8

[neon iron]

yes

[robust jasper]

Rightt

Thank you so much

https://tenor.com/view/cat-meow-fat-augh-gif-24948731

❤️

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[robust jasper]

.reopen

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✅

[robust jasper]

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[south narwhal]

I don't really know if this is even the right discord for this because it's partially related to biology. But does anyone know how long it would take for a population of 250-500 people reach 1,000,000?

[neon iron]

depends on how fast they procreate

more info needed

[topaz sinew]

@south narwhal Has your question been resolved?

[south narwhal]

As normal humans would. Nothing is stopping them, such as environment.

[shadow salmon]

Because of the vagueness I’m gonna assume you’re looking for a very rough estimate. To do this you can model the behavior according to an exponential model, and so whatever “as normal humans would” means I would start by figuring out a number on that by googling or just guessing I suppose

[south narwhal]

I think human growth rate has never gone over 2% per year. But it's vague because I don't need a pinpoint answer, and also because it's based on a story where a village of around 500 people have to repopulate a planet. I want to get an idea of how long it'd take to reach the first million if there were no hardships such as survival placed on them.

[still sparrow]

are all 500 able of procreation

[shadow salmon]

A very simple model would be something like $P(t) = C r^t$

[thorny flame]

Aslan

[south narwhal]

Yes.

[shadow salmon]

Where t is say amount of years, and r is that percentage you mentioned, so like r = 1.02 for example

And C here is the starting population

Which was around 250-500 according to you

[south narwhal]

Yes.

[shadow salmon]

So what you wanna do with this simple model in order to figure out the number of years it takes for P(t) = 1 million, is just to set up it as an equation and solve for the unknown amount of years t

[south narwhal]

Unfortunately math left me a long time ago. Is this a forum more teaching how to find the answer rather than giving the answer itself? If so, I apologize.

[still sparrow]

𝐺=𝑟×𝑁×(1−𝑁𝐾)

alternatively you could use this formula for a logistical growth if your saying as population grows resources are scarcer the need to populate drops

by the time it reachers a million

[shadow salmon]

It’s recommended we try and guide you through the answer but in this specific scenario it’s really not that serious so if you’re just looking for an answer I suppose it’s fine

[shadow salmon]

So using this very simple model which sort of assumes the most ideal scenario with the initial average grow rate at 2% and starting population at 500, then it would take around 384 years

[south narwhal]

Thanks! I'm a fiction writer so math isn't my strong suit. I appreciate the help.

[still sparrow]

if you really thought about it the population from innitially grow linerly for the first 18 or so years

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[mortal heart]

Guys, how to do this? Pls help, thankss

[mortal heart]

Am i correct?

[modern imp]

is this for an exam

[mint crescent]

It looks like a past paper

[modern imp]

ya ok

[mint crescent]

Yes, but it’s not that helpful

[modern imp]

you need to figure out which number x can possibly be

[mint crescent]

You’re better off making it as a quadratic in sqrt(x) or isolating sqrt(x) and squaring both sides

[mortal heart]

Yes im studying for my exam tomorrow

[modern imp]

ok cool

[patent trail]

substitute x for t^k where k is an appropriate number

such that you end up with a polynomial

here clearly k = 2

then you get t² - 4t + 1 = 0 which you can solve

[mint crescent]

(Note that’s the same thing as making it a quadratic in sqrt x)

[patent trail]

yeah

[mortal heart]

How to get 2 tho

[patent trail]

you want all powers to be integers

you have t^k - 4t^(k/2) + 1 when substituting

the smallest k such that k/2 is integer is k = 2

if you have x^(1/3) + x^(1/2) + 1 = 0 for instance, you would choose k = 6, and so on

ok @mortal heart ?

it's the same as substituting x^(1/2) = t as @mint crescent suggested

[mortal heart]

Hmmm do we get 6 from the 5/6 and 2 from the 1/2?

[mint crescent]

lcm(2,3)=6

[mortal heart]

Ohh yes

[patent trail]

6 is the smallest number such that both k/3 and k/2 are integers

[mortal heart]

Ohhhhh

[patent trail]

indeed the least common multiple of 2 and 3

[mortal heart]

Okokk

Im gonna try

Guys is that an AS level or A2 question

I felt like i know nothing

[patent trail]

I don't know what that is

but with practice you will remember things

and just trying things out

rewriting x - 4x^(1/2) + 1 as (x^(1/2))² - 4(x^(1/2)) + 1 is a very common trick

or rewriting sin(2x)/x as 2sin(2x)/(2x), and so on

always the same technique

my teacher called that the street sweeper technique for some unknown reason

add what you want to make it an expression you know how to deal with, and then subtract/divide/whatever what you added to keep the expression the same as before

here we rewrite x as (x^(1/2))^2

we want x^(1/2) to appear so that we get a polynomial in the variable x^(1/2), but to keep the expression equal to itself we square it too

[mortal heart]

okk i’ll try understanding this

thank you so much everyone

[mortal heart]

thank you so much

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[mortal heart]

guys

[mortal heart]

about this

do i multiply the whole equation by 2?

[patent trail]

no, why?

set k = 2

you get a quadratic

t^2 - 4t + 1 = 0

[mortal heart]

ohhhh

[patent trail]

$$t = 2\pm\sqrt{3}$$

[thorny flame]

suut

[patent trail]

you get that

then x is the square of that

[mortal heart]

okkk

[patent trail]

so $x = 5 \pm 4\sqrt{3}$

[thorny flame]

suut

[mortal heart]

btw are we allowed to set k whatever number we want for every question?

[patent trail]

yes

if it’s a multiple of 2

it will always give you the same result in the end

but whatever you do you will end back with having to solve the k = 2 case anyway

so take k = 2

[mortal heart]

ummm

[patent trail]

only quadratics are easy to solve

cubics and quartics can be solved too

but above that you have to use tricks to reduce to a lower degree

so we want k as small as possible

and hope to end up with a quadratic, maybe a cubic if you’re in university

[mortal heart]

ooohh

[mortal heart]

after this, i should substitute tbis to t and find k right?

[patent trail]

you want to find x

we already know what k is, it’s 2

and we just found what t is

so you just compute x = t^2

[patent trail]

as I did here

[mortal heart]

ohhh

btww thanks againn

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[grand blade]

i rly need help with the third inequality

ik im suppose to use the cauchy-schwarz inequality but idk how it works here

[patent trail]

the first vector is x

the second vector is (sgn(x1), …, sgn(xd))

sgn is the sign function

in C-S

in R

[grand blade]

right ok but where do u get the sign function from???

[patent trail]

in C it’s different I guess

but you can adapt it

[pseudo sonnet]

$||x||_2 = \sqrt{(|x_1|)^2 + \dots + (|x_d|)^2}$, right?

[grand blade]

yeh

[patent trail]

no

this is in C

[pseudo sonnet]

oh

[grand blade]

wdym C?

[patent trail]

you need to use the absolute value of coordinates

not just the square

[patent trail]

the vector space is C^d not R^d

[thorny flame]

higher!

[pseudo sonnet]

is this better?

[patent trail]

yes

[pseudo sonnet]

sure

and $||x||_1 = |x_1| + \dots + |x_n|$ ?

[thorny flame]

higher!

[pseudo sonnet]

I do not remember my norms well ngl

[grand blade]

is this right????

[pseudo sonnet]

yeah, that's what I thought

hmm

is that your work so far?

[grand blade]

yep 😦

[surreal mural]

yeah but not all that helpful I think

[patent trail]

yes

[grand blade]

i mean these r my lecture notes so i think so

[patent trail]

but it’s not extremely useful so far

you need to indeed use C-S

you understand my proof for R?

[grand blade]

sorry what is C-S???

[pseudo sonnet]

Cauchy-Schwartz

[patent trail]

for instance if x = (2, -3, 5) then the second vector in C-S is (1, -1, 1)