help-0

[grizzled mauve]

actually you know what you can do

@unique dune

use perhaps my favorite inequality

[unique dune]

cauchy?

[grizzled mauve]

$$1-\frac 1 x \le \log x \le x - 1$$ with equality only at x=1

[ocean seal]

gfauxpas

[grizzled mauve]

just using the upper bound and replacing "x" with "1-x" gives

$\log(x-1) < -x \text{ for } x>2$

[ocean seal]

gfauxpas

[grizzled mauve]

then

$\prod\left({1-\frac 1 {3k}}\right)=\sum \log\left({1-\frac 1 {3k}}\right)$

[ocean seal]

gfauxpas

[grizzled mauve]

$\le \sum \frac {-1}{3k} = -\infty$

[ocean seal]

gfauxpas

[grizzled mauve]

aah no this just shows the product converges

becaue you would say exp(-infty)=0 but the original product isn't 0, it's 1, infinity is tricky like that

[unique dune]

infinity is always tricky 😭

[grizzled mauve]

trying to figure out what went wrong

[unique dune]

i have upper bound with ln(1)

but im struggling with lower bound..

[grizzled mauve]

okay so accordin to AI (grain of salt! but I was stuck)

pi(3k-1)/(3k) ~ (constant)/(cube root n)

and it is NOT asymptotically 1

so I apologize for not noticing that, but, i checked its work, it's right

[unique dune]

but

we got the correct radius of convergence?

[toxic verge]

Once in a blue moon

[grizzled mauve]

going over my work now to see if we got the right roc

[grizzled mauve]

actually chatgpt o4 is not bad at math, and grok is okay too, in think mode

[unique dune]

i checked in the appendix

[toxic verge]

Interesting, yeah ig it does help for parts of the reasoning but if you depend on it for the answer as a whole you doomed

[unique dune]

why cant they just make math easier...

[toxic verge]

Cause then other subjects would get harder and we would struggle even more on those

[unique dune]

subject economics

[toxic verge]

Manmade subjects are dumb

[grizzled mauve]

okay

i think the best thing to do here

is take one more term in the log approximation

[unique dune]

${\log\left( 1 - \frac{1}{3k + 3}\right)}$

[ocean seal]

k

[unique dune]

?

[grizzled mauve]

$\log(1-x) = -x-\frac{x^2}{2} -O(x^3)$

[ocean seal]

gfauxpas

[unique dune]

the log is in a sum tho

[grizzled mauve]

that's okay

$\frac{(3n-1)!!!}{(2n)!!}=\prod\left({ \frac {3}{2} n - \frac {1}{2n}}\right)$$

[ocean seal]

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

[grizzled mauve]

$=\frac 1 2 \prod \left({3n-\frac 1 n}\right)$

looking for mistakes

[unique dune]

can u just factor 1/2 out

shouldnt we raise it to some power

[grizzled mauve]

youre right

as I said, i was looking for mistakes, you foujdn it first, so, ignore that

$\prod_{n=1}^k \left({\frac 3 2 - \frac {1}{2n}}\right)$

lol nooooo

[ocean seal]

gfauxpas

[grizzled mauve]

take logs

$\sum_{n=1}^k \log \left({\frac 3 2 - \frac 1 {2n}}\right)$

[ocean seal]

gfauxpas

[grizzled mauve]

anyway you're PROBABLZY just supposed to say sometrhing like

$(3n-1)!!! ~ \text{constant} \cdot 3^n n!$

[ocean seal]

gfauxpas

[unique dune]

,w 3 - 1/4

[ocean seal]

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[unique dune]

this is allowed, right?

[grizzled mauve]

for just 2 numbers? sure

want to keep going this way?

[unique dune]

,w 3/2 - 1/4

[ocean seal]

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[unique dune]

anyway is fine

[grizzled mauve]

great, so, how to make a substitution so this looks like log(1-y) for y= something?

[unique dune]

we have to factor 3/2 tho, rgiht?

[grizzled mauve]

yes!

[unique dune]

in which case we have y = 1/3n

[grizzled mauve]

$=\sum_{n=1}^k \left({\log(3/2)+\log(1-\frac {1}{3n})}\right)$

[ocean seal]

gfauxpas

[unique dune]

yup

[grizzled mauve]

$\sum_{n=1}^k \left({\log(3/2) +\frac {1}{3n}+\frac{2}{9n^2} - O((3n)^{-3})}\right)$

[ocean seal]

gfauxpas

[grizzled mauve]

back to product sum land?

$$\prod_{n=1}^k \frac 3 2 + e^{3^{-1}n^{-1}} + e^{2^19^{-1}n^{-2}} + O((3n)^{-3})$$
and we can simplify $$O((3n)^{-3} = O(n^{-3})$, and then put $x^n$ and bring it back to an infinite sum

[ocean seal]

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

[grizzled mauve]

$$\prod_{n=1}^{+\infty}\left({\frac{3}{2}x^n}\right) \cdot \left({ \exp\left({\frac{1}{3n}}\right) \exp\left({\frac {2}{9n^2} }\right) \cdot O(n^{-3})}\right)$$

whew!

[unique dune]

im sorry but

im kinda confused about how we transformed the sum back to product

[grizzled mauve]

sure, ill explainit!

we just ... took e^everything

the way we go from a product to a log sum is taking the logarithm of everything

to go from log sum back to product you take exp of everything

[unique dune]

shouldnt the e^ be multiplied together then

[grizzled mauve]

oh, yes, you're absolutely right

[ocean seal]

gfauxpas

[unique dune]

why is there x^n

[grizzled mauve]

i put it back in from the original problem statement

[unique dune]

oh right

[grizzled mauve]

lol chatgpt found a shortcut

but

this was good brain exercise, it just, found a shortcut that didnt require knowing how to simplify (3n-1)!!!

[unique dune]

could u

explain it plz

[grizzled mauve]

my way or the short way?

[unique dune]

my brain is numbed 😭

[unique dune]

either way

[grizzled mauve]

step 1. okay with this?

[unique dune]

yup

wait

3/2 - 1/2n

no n at the front

[grizzled mauve]

oh i meant that

that mistake is fixed in the next step

but good eye

convert the product to log sum
here I discarded the n, because i never meant it to be there anyway

[unique dune]

yup

[grizzled mauve]

factor out the 3/2 so I get something of the form log(1-y), because that has a known power series

here y = 1/(3n)

[unique dune]

ye

and we do the expansion next, right?

[grizzled mauve]

yeah

this is the expansion

except diid I switch the sign

I did switch the sign but it doesnt change the answer

$$\sum_{n=1}^k \left({ \log(3/2)- \frac 1 {3n} - \frac 2 {9n^2} - O((3n)^{-3}) }\right)$$

[ocean seal]

gfauxpas

[grizzled mauve]

should have been this

follow? @unique dune

[unique dune]

Yup

[grizzled mauve]

now i go from logsum back to product. here I simplify O((3n)^-3)=O(n^{-3}), and I put back the x^n, and I put the product limits as to infinity, to get it equivalent to the original problem

[unique dune]

So this?

[grizzled mauve]

$\prod_{n=1}^\infty \left({\frac 3 2 x^n }\right)e^{(3n)^{-1}}e^{\frac{-2}{9n^2} \cdot O(n^{-3})$

[ocean seal]

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

[unique dune]

Yup

Things on the right always converge

Wait

It’s product

Nvm

😭

[grizzled mauve]

yeah actually I forgot the root test works on the log series

so it should have been log(x^n 3/2) you want to analyize

and the rest of the terms vanish because they shrink so fast

[unique dune]

So

I basically say that they are trivially the same

[grizzled mauve]

now, want to see the smart way to do it

that was all the stupid way because i didnt see the smart way

[unique dune]

Neither did i

😭

I couldn’t even get the radius of convergence

[grizzled mauve]

$\sum_{n=0}^\infty \frac{(3n-1)!!!}{(2n)!!} x^n$

[ocean seal]

gfauxpas

[unique dune]

Yup

[grizzled mauve]

so if the coefficient is (3n-1)!!!/(2n)!!

then the ratio test tells us that the radius is equal to

$L=\lim \sup \left \vert {\frac {a_{n+1}}{a_n}} \right \vert$

rather, it's 1/L

[ocean seal]

gfauxpas

[grizzled mauve]

here a_n is a ratio

of two sequences, call them N_n and D_n for numerator and denominator

N_n = (3n-1)!!!
N[n+1]=(3(n+1)-1)!!!
D_n = (2n)!!
D[n+1]=(2(n+1))!!

[unique dune]

,w (3n-3)/(2n+2)

[ocean seal]

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[grizzled mauve]

ratio test, take n->infity

[unique dune]

Its the ratio test all along?

[grizzled mauve]

https://tenor.com/view/the-whole-time-mrs-doubtfire-mrs-doubtfire-the-whole-time-sally-field-sally-field-the-whole-time-gif-727091483642243467

[unique dune]

Omg

[grizzled mauve]

let me cheer you up with a story

[unique dune]

Jesus

[grizzled mauve]

I had an okay-ish Calc II professor

but she was an unfair grader

and I worked so hard and got every problem right on the take-home midterm

and she gave me like a 70 or 80 or something for it because I didn't "simplify" enough and didn't "show my work" enough

even though to most math professors it would have been fine

[unique dune]

‘Simplify enough’ 🥀

[grizzled mauve]

including my calc I professor, who was a professional, a researcher, he taught me what standards he liked; the calc ii professor was a comp sci teacher

so

I was a teacher's pet but I wanted revenge anyway

in a passive aggressive way, not aggresive

so for the final

every single problem I used a method that was long and complicated and for sure was not on her answer key

using obscure hyperbolic trig identities

"weird" convergence tests

doing things like what we did with logarithm series

and on my grade, which of course I got an A on that exam, you can tell she marked "X" then erased it and marked "right" then erased it back and forth before she decided right

I'm quite sure it took her way longer to grade my test than any other student's

it's okay she liked me

she said, when I get to university proper, i should take topology, I for sure will love it - she was wrong lol

[unique dune]

But ure still very creative in math

That’s the thing I’m struggling with

Anyway

Thank u for everything

Since the start

U’ve helped a lot and i appreciate it

Wish you luck

.close

[grizzled mauve]

it takes practice

[lone heart]

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[livid marsh]

Please help me to understand and solve this problem
$Let X_i, i=1,2,3 be standard normals, let Y be defined Y_1=X_1+X_2+x_3, Y_2= X_1-X_2,Y_3=X_1-X_3. Obtain the joint of Y_1, Y_2, Y_3$

[ocean seal]

Alex

[keen idol]

please help us understand your problem.

[frank python]

Let $X_{i, i}=1,2,3$ be standard normals, let Y be defined $Y_1=X_1+X_2+x_3, Y_2= X_1-X_2,Y_3=X_1-X_3.$ Obtain the joint of $Y_1, Y_2, Y_3$

[livid marsh]

Thanks, first time using that option

[ocean seal]

walking rat

[livid marsh]

I'm not entirely sure, but from what I understand it's looking for the joint density function of Y, and as far as I know, I should use the Jacobian

[lone heart]

@livid marsh Has your question been resolved?

[lone heart]

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[crystal summit]

where does one go after differentiating, i need to find the max value somehow but idk how

[gilded ore]

Find the zeros/critical points

And then plug into F(x)

Whichever is largest value

Thats the maximum

[crystal summit]

yes but dont i need to plug it into F'(x) somehow?

[gilded ore]

Solve F’(x) for zeros

Set equal to zero

And solve which X’s make it zero

As you notice you have a polynomial you can solve multiplied by an X so you know atleast one x is zero

Factor the inside or use quadratic formula

[crystal summit]

wdym?

i just found the solution? i dont understnad the bottom half

[tacit arch]

Which line don't you understand first

[crystal summit]

From 4x^2+3x+1= ... i understand how to do it but how do we transition to that to the"Since F'(x) changes sign.."

sorry if im not making sense, but why do we have to do this and why does it help with the boxes below

[lone heart]

@crystal summit Has your question been resolved?

[crystal summit]

.close

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[calm glen]

See it is like F'(x) <0 for maximum first we find the critical points by doing F'(x)=0 since there is only one real rool x=0 critical point is 0 now put it in the function

You will get 26

[crystal summit]

thanks that makes a lot more sense :))

[calm glen]

Since we have proved that 4x^2+3x+1 is positive term we can apply wavy curve for only -x<0

[crystal summit]

ohhh yes thank you

[calm glen]

Welcome

[lone heart]

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[radiant oasis]

why is this wrong?

[lethal belfry]

cos(37π/12)= cos(3π+ π/12)

[lethal belfry]

It this a test

[radiant oasis]

no this is a deltamath

[lethal belfry]

can you work with this

[radiant oasis]

yes

that would simplify to cos(pi+pi/12) right?

[lethal belfry]

yup

[radiant oasis]

because you subtract 2pi as its a full circle

yes brain neoron synapse power

[calm glen]

Cos(pi/3 + 11 pi/4)

[radiant oasis]

pie

hee hee

wait but i cant use pi/12

[compact lotus]

hi peeps

oops wrong channel

[radiant oasis]

shame! shame!

[calm glen]

It should be -root 2 -root 6/4

(-root 2 -root 6)/4

[radiant oasis]

OHHHHHH i see now what i did wrong

[calm glen]

Check it

[radiant oasis]

it was in my late math

yeah when i did (2root2) x root 2

i did 2 and not 4

[calm glen]

Yup and pi was a typo

[radiant oasis]

lol yeah i assumed i just thought it was silly

thanks guys!!!

.close

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[gray tulip]

Hey guys! I don't really understand this solution, particularly the application of pigenhole principle

[gray tulip]

Like I understand that if it's prime that either g(ri) or h(ri) is a unit, but i don't understand why by PHP that g(ri) must be a unit for at least nM distinct indices

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[obtuse berry]

can someone help me with this question

my initial thought was to find where L1 and L2 intersect then we can determine the direction vector and a vector equation for P

but L1 and L2 are skew lines

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[rugged ermine]

Hey, what is overcounting? How do I know if I'm doing overcounting when im doing combinatorics questions?
For example, let us consider this question: "How many ways are there to split a dozen people into 3 teams, where each team has 4 people? How many ways are there to split a dozen people into 3 teams, where one team has 2 people, and the other two teams have 5 people each?"
Is there any method that I can follow to understand when there is overcounting going on, and how do I adjust for it? Please help! Thank you.

[vale wigeon]

overcounting is when you count the same arrangement/selection/whatever two or more times

there isn't really a "method" to recognizing when it happens

you learn it organically

[rugged ermine]

can you solve the question I mentioned? so that I can see at least how to approach

[vale wigeon]

ok so let's look at a 4-4-4 split

let's pretend for a moment that the teams have distinct identities (ie Team A, Team B, Team C)

then the number of team allocations could be phrased as, for example, $$\binom{12}{4} \cdot \binom{8}{4},$$or a bit more elegantly as $\frac{12!}{4!^3}$

this is not the answer yet, but i want you to tell me whether it makes sense to you thus far

@rugged ermine

[ocean seal]

Ann

[rugged ermine]

Uhh I dont get it...

I do get the 12 choose 4 part

[vale wigeon]

$\binom{12}{4}$ ways to fill team A, then $\binom{8}{4}$ ways to fill team B once we've already put 4 people into team A --- and then team C gets whoever is left.

[ocean seal]

Ann

[rugged ermine]

then where is the overcounting part here?

[vale wigeon]

im gonna get to it soon

[rugged ermine]

this one looks fine

and understandable

first take 4 ppl

then from the remaining 8 take 4 more

[vale wigeon]

ok, so when making this count we assumed that the teams themselves are distinct (ie there's a Team A, a Team B and a Team C)

this made it possible for us to do the count but it also made us overcount everything

[rugged ermine]

i dont get wdym by made overcount everything

[vale wigeon]

because you could take a team allocation like that and, for example, put everyone from team A into team B, and vice versa -- and the team allocation wouldn't be essentially different

[rugged ermine]

it looks normal to me

hmm

you mean, first select team b, then team a?

[vale wigeon]

no

[rugged ermine]

but they would have different members

i really suck at this kind of maths so please bear w/ me

[vale wigeon]

compare the following rosters:

Team A: Albert, Beth, Carl, David
Team B: Eve, Fred, George, Heather
Team C: Ian, Joan, Kelly, Larry

vs.

Team A: Eve, Fred, George, Heather
Team B: Albert, Beth, Carl, David
Team C: Ian, Joan, Kelly, Larry

do you agree that these are essentially the same team partition

[rugged ermine]

they dont look the same on paper...

but if teams are indistinguishable

then yeah

[vale wigeon]

thats the point

[rugged ermine]

if all i can see is groups of ppl

[vale wigeon]

the original question had the teams be indistinguishable

i MADE THEM distinguishable in order for counting to be possible

but now we need to correct for that

[rugged ermine]

huh

i feel smth

but not here yet

[vale wigeon]

and the way we do that is that we note that the teams can be shuffled around in 3! ways

(without changing who is teammates with whom)

and because of this, we divide our count by 3!

[rugged ermine]

so we always divide by (number of industinguishable)!?

like in banana

we have 3 indistuinguishable As

so divide by 3!

?

[vale wigeon]

like in counting the rearrangements of the word BANANA?

yeah, it's the same idea here.

[rugged ermine]

so always divide by number of industinguishable stuff

[vale wigeon]

we first pretend the A's are distinguishable and then correct for that by dividing by 3!

[vale wigeon]

hesitant to pose that as a rule tbh

[rugged ermine]

why

?

How many ways are there to split a dozen people into 3 teams, where one team has 2 people, and the other two teams have 5 people each?
Wait lets try this

lets assume indisinguishable

[vale wigeon]

assume distinguishable, you mean.

[rugged ermine]

yep, my bad

[rugged ermine]

that would be 12C5

[quasi vector]

well the group with 2 is always distinguishable from the groups of 5

[rugged ermine]

then, we have 7C5

then 2 remaining will be team

[rugged ermine]

hmm

[vale wigeon]

yeah good point

[rugged ermine]

so we divide the full answer by 2!

?

[vale wigeon]

yes we do.

12C5 * 7C5 * 1/2!

[rugged ermine]

the main realization is that when we are doing the formulas, we are accounting as number of ways to choose two teams of 5, but we are not including the point that during the calculation, for some unknown reason, we are overcouting, which happens only when indistinguishable stuffs are present

so we divide by that factorial

amiright?

[vale wigeon]

this wording looks a little off the mark to me but i don't know how to fix it.

[rugged ermine]

i think the same

i cant get a proper feel for how we are overcounting in the process

like, how does the binomial coeff formula forget to consider the overcounting

just tell me how which applying the binomial coeff formula, we are unintentionally overcounting

and why we are dividing instead of subtracting

[undone ledge]

`lemme ask a diff question , how many ways you can arrange the word ARA to form new 3 letter words

[rugged ermine]

RAA
ARA
AAR
thats all

[undone ledge]

how would you have counted it

3 words 3 spaces

3! -> 6

oh but wait you just counted 3 , can you explain what happened

[rugged ermine]

if it was abc, and three spaces

then for 1st place: 3 options
2nd place: 2 options
3rd place: 1 option

so 3!

so 6

but here it is ara

[pulsar tree]

in the answer sheet answer is (b)symmetric but why isnt reflexive?

[undone ledge]

!occupied

[lone heart]

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).

[undone ledge]

and...

can you see what you are overcounting

what casses are being overcounted

[rugged ermine]

yes, abc != cab
but ara == ara

[undone ledge]

yeah exactly thats what mainly happens when you over count

RAA' | RA'A
ARA' | A'RA
AA'R | A'AR

hence we divide the number of same terms after arranging

i.e. 3!/2 here

can you try mississippi now

after that try to overlap that into what ann was saying

[rugged ermine]

i get it now... but why divide, instead of subtract?

[rugged ermine]

well I can think of the two teams of 5 as A and the single team of two as R

[undone ledge]

dividing and subtracting are two peas in a pod

what does it mean to multiply for example

3*2
what are you doing here, adding 3 two times?

[rugged ermine]

yep

[undone ledge]

similar case is for dividing ,
15/3
what does it mean to divide here?

number of 3 required to vanish 15?

[rugged ermine]

yes!

[undone ledge]

so when we talk about dividing the ghost A' from ARA'
we are essentially saying to remove the extra number of A
if we go on subtracting its gonna take a lot of time for big words such as mississippi

[rugged ermine]

got it!

it makes sense now

thanks

but

one last question

im selecting k ppl from n ppl

[undone ledge]

nCk?

[rugged ermine]

yes

[undone ledge]

and..

[rugged ermine]

by multiplication rule

n(n-1)(n-2)...(n-k+1)

but then

prof says

he divides by k!

because "you can choose k ppl in any order"

and then the formula becomes nCk

[rugged ermine]

order? where did the order come from?

[undone ledge]

yeah cause thats choosing

permutation and combination are different thingsa

choosing means to just choose k data points from n set
for example i have a basket of 5 candies , i wanna eat 3
there are different type of candies in the basket , there is a candy corn, a candy cane , chupa chups , bubble gum , dark chocolate
now choosing 3 candies would mean , i am gonna take any 3 from the 5

however permutation or arranging them would mean the order i eat it in

hence arranging/permutating means to arrange the chosen data points in order

hence
nCk *k! = nPk

[rugged ermine]

got it...

thanks!

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[alpine sable]

Hi

[alpine sable]

How do math people find out where they are going wrong in calculations

[placid zinc]

Write clearly. You'd be shocked at how helpful it can be to put some effort into making your calculations pretty

Eventually you get into a habit of making every part very clear, and it helps with catching mistakes

[mossy reef]

It can also be helpful to see if your answer makes sense within the context, if it’s a word problem.

[alpine sable]

I am doing calculations programmatically

[vale wigeon]

what kind of calculations

[alpine sable]

More like statistics

Mean, average and all those kinda stuff

[undone ledge]

wdym programmatically

[alpine sable]

?

Excel and other kinds

[undone ledge]

ah

[vale wigeon]

you're... being a bit vague in what you are actually doing and actually want to know

[alpine sable]

^ How is this vague ??

I am calculating averages, when I do manually I get different answer, when I use excel I get different and when I do using python I get different

How do i know which one is correct

The problem is the numbers are in float and very precise

There could be rounding stuff going on I have no idea

[covert fossil]

If you're getting different values in all of them you're probably typing the formulas wrong

[alpine sable]

I cross checked that many times

[covert fossil]

Then you should get the same values or very similar to a lot of decimals

[vale wigeon]

well you have to show us your work for all 3

AND the data

[alpine sable]

Do you know about RSI

[vale wigeon]

i do not know what that abbreviation stands for, no.

[alpine sable]

Can you dm me ?

I cannot send stuff here

nvm

.close

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[alpine sable]

hey guys, I really don't know what to do from here

[alpine sable]

As far as I am aware

T (being the transitional matrix) * X_0 should be X_1, which is February

Then T * X_2 would be March, and T * X_3 would be April

But that's too tedious, I am guessing I can somehow make it easier by

diagonalizing T with

T = PDP^-1

when I tried to find the eigen values though i couldn't find values for lambda where the determinant would be = 0

i ended up with this equation = 0

this tells me nothing though

this just says x = 0.95 and x = 0.998

while the solution here says this instead

so im really confused

nvm had one wrong value

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[green oasis]

Says here that x+dx = 100*1

But i dont get why it gets multiplied instead of adding.

Also since dx =1/10 and that x is 100, i think that x+dx should look like 10/100. Somebody tell me how this works

[mortal trellis]

thats notation for 100.1

some countries use that

or its because the book is old

[green oasis]

Still dont get it, please explain it to me further

[mortal trellis]

100+0.1=100.1

[green oasis]

Nice

Alright that makes sense

Thank you

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[quasi sonnet]

What did i do wrong

It’s supposed to be 9d

[modern sedge]

Nevím cos dělal zde

|log(x+1)| > 1 když je log(x+1) > 1 nebo log(x+1) < -1

[quasi sonnet]

Idk isiel som iba podla postupu co mi ukazal doucovatel

[modern sedge]

tam má log(x-2) < 0

ty tam máš -log(x+1) < 0

když odděláš to - tak by to mělo fungovat

ale je to trošku přebytečný

[modern sedge]

jo, takhle by to mělo jít

[quasi sonnet]

Ale ajtak to je cele zle 💀

[modern sedge]

Vypadá to dobře, akorát to teď musíš dobře zkombinovat

[quasi sonnet]

(-inf,-2/3)U(2,inf)

[modern sedge]

akorát jsi zapoměl na podmínku

x > -1

[quasi sonnet]

(-1,-2/3)U(2,inf)

[modern sedge]

Toto by mělo být dobře

,w |log_3(x+1)| > 1

[quasi sonnet]

Dobre ale tu je spravna odpoved (0,1/9)U(1,inf)

[modern sedge]

Aha, ono je to totiž napsaný jako $\
\log_3({x}) + 1$

[ocean seal]

MathIsAlwaysRight

[modern sedge]

Proto nemám rád když se vynechávaj závorky

[modern sedge]

Toto je trošku jiný než log(x+1)

No, každopádně, absolutní hodnota něčeho je > 1 právě když je to něco buď menší než -1, nebo větší než 1

takže stačí řešit
log(x) + 1 > 1
log(x) + 1 < -1

(log je myšlen se základnou 3 samozřejmě)

[modern sedge]

log(x) + 1 není to samý jako log(x+1)

[quasi sonnet]

Ja som menil len ten prvy riadok

[modern sedge]

ano, ale nemůžeš změnit log(x) + 1 na log(x+1)

[quasi sonnet]

To chapem

[modern sedge]

tak co se stalo tady?

[quasi sonnet]

Takto

Ale ajtak mi to dajak nevychadza

[modern sedge]

Proč je v tom druhým sloupci log3(x) - 1?

Proč ne log3(x) + 1?

[quasi sonnet]

Nema sa to znamienko menit v abs hodnote

?

[modern sedge]

Tu metodu tvýho doučovatele vůbec nechápu, ale nemyslím si že by se tam mělo něco měnit

možná bude lepší to prostě udělat normálně

[quasi sonnet]

A to jak

[modern sedge]

jenom tady tento sloupec

bohatě to stačí

ten druhej sloupec je úplně zbytečnej

|něčeho| > 1 právě tehdy když

něco > 1
nebo
-něco < -1

takže ti stačí vyřešit ty dvě nerovnice a nemusíš řešit nic navíc

[lone heart]

@quasi sonnet Has your question been resolved?

[modern sedge]

Tady je moje řešení

[lone heart]

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[bright fiber]

I have a homework in measure but I am completely stuck

Can anyone give me an idea

[lone heart]

@bright fiber Has your question been resolved?

[bright fiber]

<@&286206848099549185>

[lapis jungle]

I would but i uh

Cant rly read it

Gimme a sec

[lapis jungle]

Which idea is the final

[bright fiber]

what do you mean ?

like how to construct that set

[lapis jungle]

One sec

[iron crystal]

can you please type it out

[lapis jungle]

Ohh

So you intergrated

And took the value to 0,1

I mean your on the right path

[bright fiber]

ok let f in the space L^1 the space of lebesg integrable function from X to [0,1 ] such that the integral of f over x is the same as the integral of f^2 over x show that there exists a measurable set E such that f = X_E

yah the problem i am stuck now i dont now how to pass from the set E to x to compleat the contrapositive

[lapis jungle]

Which after should be integral X then “fdų= intX f^2 dų then

[lapis jungle]

Did you prove theres an measurable set

[bright fiber]

do you mean the exsistence of E

[lapis jungle]

Yea

[bright fiber]

in the direct method no i didnt now how to construct one

[lapis jungle]

Integral , X E ų

[lapis jungle]

You get it after this

[bright fiber]

this is the measure of E

[lapis jungle]

Ye

Hol on

My phone abt to die

So if i stop talking

Yk why

[bright fiber]

ok

[lapis jungle]

But

Did you get to the part where their equal

I might use a different method OH

[bright fiber]

ok

[lapis jungle]

You didnt put

Well you had kt

It

But you swapped it

[bright fiber]

when

[lapis jungle]

You got intergal f = X E{-a.e

?

If not

[bright fiber]

you mean this one ?

[lapis jungle]

Ye

[bright fiber]

aha

[lapis jungle]

That should be the last part

Of where you need to go

Idk if you went off after that

Then you

[bright fiber]

but the thing this is the integral over E but i need the integral over X

[lapis jungle]

I menat

Meant

This what your supposed to do after

Lemme write it on smth

[bright fiber]

the main problem is that in the contrapositive proof i don't know how to pass from this integral to say that the integral of f over X is not the same as of f^2

ok

[lapis jungle]

Im making a pdf

[bright fiber]

whait can you give imeges instud

i just cant trust pdf

sorry about that

[toxic verge]

!nopdf

[lone heart]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

[lapis jungle]

O

Thats fair

Ye i’ll take a ss of it

[lapis jungle]

Back

[bright fiber]

yo did your phone died ?

[lapis jungle]

Is abt too

[bright fiber]

well in any case welcom back

[lapis jungle]

It bugged my words are we deadass

[bright fiber]

like did your space bare got broke ?

[lapis jungle]

no in there

i wrote it

o wait

i can read it

can you?

[bright fiber]

can you translate it

[lapis jungle]

this is measurable because f is measurable and the preimagine e x

we have

[bright fiber]

oh ok

what about this one

[lapis jungle]

this one says "always holds. but since the intergral are equal

[bright fiber]

how does the equality of the integrals implys that f(x) almost evry where

[lapis jungle]

i explain

further

[lapis jungle]

i think i did atleast

mb

if i didnt mbmb

i cant help nomo my battery gonna die

if it isnt solved later ill help

[bright fiber]

whait i dont thik that the equality of the integrals is equvalent to the equality almost evry wher

[lapis jungle]

well thats what i had wrote i might be wrong tho

[bright fiber]

ok you give me some clearness tho ty very much go charge your phone

[bright fiber]

I think i did it, @lapis jungle I just want to ty you gived me the idia of how to construct E, I am sure i was not gonna find it with out you bro ty so mush

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[alpine sable]

find alpha (all angles in black are also alpha)

[frail dagger]

!xy

[lone heart]

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.

[alpine sable]

It's exactly as I said. A zigzag line starts at point A, which is endpoint of a diameter AB of a circumference. Every angle of the jump is equal to alpha, as show in the figure. After 4 "jumps", the zigzag line ends at point B. What is the value of alpha

The answer is apparently 72

I think this problem might be wrong

Yeah, honestly, probably is

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[zinc haven]

how's my proof here?

[mortal trellis]

fine

[zinc haven]

alright thanks

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[obtuse sky]

I need help doing the integration here.
I have drawn the graph (hopefully its correct)
But the result I find is not even close to the answer [432/15]

[obtuse sky]

Oh the f function is
f(x,y) = x + y + 1
Btw

I forgot to write that

[rigid tree]

wait i need to type slowly

[obtuse sky]

Otherwise its just simple area finding 😅

[rigid tree]

uhm like domnt you just have to dom double integral on that area then done?

[obtuse sky]

Dom?

[rigid tree]

do*

sorry

[obtuse sky]

Okay lemme do the calculations

Aaand nope...

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[warm quarry]

I must proof this

I have this so far:

or maybe f: X -> Y , f(C,D,E) = (?, N \ {D u E}, D u E)

i am not sure about the functions :/

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[lone heart]

@warm quarry Has your question been resolved?

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[tacit arch]

did you try drawing a picture

[pseudo aurora]

Send the picture

[tacit arch]

yea no one can tell unless you share your picture

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[pseudo aurora]

Is a)ii) x=2?

[naive valley]

what is your reasoning?

[pseudo aurora]

When the equation is equated to 0 the value is 2

[naive valley]

what equation?

[pseudo aurora]

so the values are x<2 and x<2

[naive valley]

what's the difference btw those two inequalities?

[vital locust]

23456789 why are you setting it equal to zero that is polynomial solving for x

[pseudo aurora]

0^-3 is impossible

[naive valley]

(ii) is asking for the range of x values for which your expression in (i) is valid

so start by telling us your answer from (i)

[naive valley]

then how can x=2 be valid?

[vital locust]

i thought for these you didnt need part i for range of validity

now im learning

[vital locust]

ur doing A level maths?

[pseudo aurora]

yes

[vital locust]

id check TLmaths website he does free full course of video teachings i used it to teach myself alone. he has a entire section of ranges of validity

it is very good

[pseudo aurora]

This is the answer for i)

[vital locust]

https://youtu.be/9jHo7X-_1As?si=j9OttNUd1jHstvHQ

im like 90% sure first part doesnt affect 2nd part

[pseudo aurora]

Is the answer -1<x>1

[vital locust]

you have to take the validity of your own question, that is the validity of his question

he is saying this is fundemental validity

of the basic expression where x is just x

so if x is 2x, it changes the validity of x

as it becomes modulus 2x < 1

[pseudo aurora]

so the answer is -1/2<x<1/2

[vital locust]

your is not 2x

it is x/2

if you think (1+x)^n = (1+x/2)^-3, you can see n=-3 and x=x/2

so using the rule modulus x<1, and you know x=x/2. the new validity is modulus x/2<1

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[serene vault]

Im trying to understand this..

i get that hadamard basis is 1/sqrt(2) times the "amplitudes" but i just dont know

should i do 1/sqrt(2) * 2/3... then 1/sqrt(2) * sqrt(5)/3, then add the two... or do i need to apply borns rule and square each first?

[leaden bobcat]

What's the Hadamard basis?

[leaden bobcat]

So it seems to me you just apply to apply a change of coordinates

Unless I'm not reading well into the physics

[serene vault]

whats the point of giving the + sign here? ... if when we apply hadamard we need to do both plus and minus?

like would the outcome be any different if there was a negative sign in there?

[crimson jetty]

It's kinda like rotating the system 45 degrees i suppose

[serene vault]

rotating what system 45 degrees?

like im so puzzled on this whole thing

[crimson jetty]

Sorry that was probably not a useful way of explaining it

[serene vault]

i guess you are referring to the qubit itself?

[crimson jetty]

Its like, describing a point with respect to two different coordinate systems?

[serene vault]

should i think of all of this like a sphere?

[leaden bobcat]

You just have a vector written in components wrt |0>, |1> and are asked to express it in terms of |+> and |->

The value +2/3 and +sqrt(5)/3 have no special meaning here, that's just what the text has given you

[serene vault]

a 2d vector?

so given this: .... no matter what just disregard that plug sign in the middle? as it really doesnt matter?

[leaden bobcat]

If you change the sign you change the exercise

[serene vault]

hmm maybe i should take this a step further back

is this saying theres 2/3 of a chance its outcome is 0

and sqrt(5)/3 chance it is 1?

[leaden bobcat]

I'm sorry I don't know the phyisical interpretation, I'm just reading it as any linear algebra problem where you're changing the coordinates of a vector from 1 basis to another

[crimson jetty]

Think back to you basic linear algebra

[serene vault]

i didnt take linear algebra

[crimson jetty]

oof

[leaden bobcat]

Well that's rough

[crimson jetty]

You might find more luck in a physics help server then ^^;

[serene vault]

i was told it wasnt needed when signing up for this class.. and the linear algebra lecture was the only lecture labeled as "optional"

[serene vault]

i will find my way there then

[crimson jetty]

Yeah it might not be necessary, but it'd definitely help

[leaden bobcat]

I honestly think some basic linear algebra is too fundamental to be skipped in any physics course

[crimson jetty]

Yeah

[leaden bobcat]

But physicists always manage somehow

[serene vault]

like i think of a vector as a list

like an array

with elements inside

that can be strings

im so cooked

[crimson jetty]

Thats definitely one interpretation, but there's a ton about writing vectors in terms of a basis that you're missing out on

[leaden bobcat]

You can think of your "state" as just the list (2/3, √5/3)

Same goes for |+>, |->
But understanding what a basis is and how to switch from one another needs some more learning

[serene vault]

how about this question

this is all on a study guide review

[leaden bobcat]

That's straight up linear algebra

[serene vault]

that im so clueless on

[leaden bobcat]

There's a formula for the inverse of a matrix

[serene vault]

is that basically asking for the inverse of that matrix

oggg

[leaden bobcat]

Yh

[serene vault]

so basically i just plug it into the fraction there?

[leaden bobcat]

And the entries

[serene vault]

-1/2(1) - 1(1)

i dont even know what the final answer would look lik

[leaden bobcat]

Where did the extra - come from?

[serene vault]

ad-bc

or the minus sign before the fraction

[leaden bobcat]

The one before

[serene vault]

which equates to -1

[leaden bobcat]

Why a minus before the fraction?

[serene vault]

thats whats in the formula?

[leaden bobcat]

That's an equal sign

[serene vault]

jesus

ur right

why is it so dim lol

[leaden bobcat]

So, what's the inverse matrix?

[serene vault]

-1?

[winter light]

Huh?

Matrix, not determinant (which is +1, btw)

[serene vault]

wait

so i diregard the fraction?

is it as simple as just plugging in the numbers here?

[leaden bobcat]

You don't disregard it, you have to compute it as well
In this specific case it's 1 so it doesn't affect anything

[serene vault]

so its -1 followed by this with the plugged in numbers?

[leaden bobcat]

If, say, you got 5/3 as a result, you would have to multiply each entry by 5/3

[leaden bobcat]

Why do you keep saying it's -1?

[serene vault]

this equates to -1

[leaden bobcat]

We agreed that's not a minus before the fraction
It was an equal sign

[serene vault]

sorry equates to 1

wait

let me send a screen shot

is this the final answer?

[leaden bobcat]

1 times that matrix leaves it unchanged, so yes, that matrix there is the inverse you were looking for

[serene vault]

ohhhhhhh

ok thats not to bad

after working through it

[leaden bobcat]

Would you know what to do if that number was, idk, -3?

[serene vault]

yes, let me give the anseer to that

sorry

that would be if its 3

[winter light]

Correct

[lone heart]

@serene vault Has your question been resolved?

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[steel galleon]

problem with many typesetters

you gotta ozom in

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[lunar saddle]

It would be best if you moved to another channel

[lunar saddle]

.

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[opal coral]

can someone explain recursive sequences pls i forgot how to do it

[still finch]

It's been a while since I've done this but iirc, you start be giving a starting term (2 and 1 in this case) and then let the following term be the computation of the previous term (in this case you have previous term and 2 terms before sum together).

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[craggy egret]

Can i ask for help with a electrical math question?

[lone heart]

@craggy egret Has your question been resolved?

[vale wigeon]

!da2a

[lone heart]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[craggy egret]

can somebody explain this to me i dont get it

[gusty gorge]

what part don't you uget

like why V- is zero?

or where the equations come from?

[lone heart]

@craggy egret Has your question been resolved?

[craggy egret]

basicaly both

i didnt go onthe day this was taught because i was lick

sick

[gusty gorge]

did you learn about negative feedback?

[craggy egret]

and lecturer didnt hae voice on recording

[gusty gorge]

also have you done nodal analysis?

[craggy egret]

kirchoffs law?

no

[gusty gorge]

you don't know KCL and KVL?

[craggy egret]

i know kcl and kvl

sum in =sum out

[gusty gorge]

that's basically what the equations state

that the sums of the currents in/out of a node are zero

[craggy egret]

yes

i think the lecturer used kcl

[gusty gorge]

so does that explain to you where the equations come from?

[craggy egret]

oh yeah

[gusty gorge]

as for why V- is 0, it's because of the negative feedback

wait do you know what an op amp does?

[craggy egret]

amplifies signals

[gusty gorge]

sure, but that's a very general notion

like do you know how it decides its output voltage?

[craggy egret]

no

is it vout = smthng/vin

i forgot that the smthng was

[gusty gorge]

basically in an op amp, Vo = 1000000(V+ - V-)

where the 1000000 is some very large gain

[craggy egret]

hmm

[gusty gorge]

so it takes the difference between V+ and V-, scales it up by a really large number, and that's the output voltage

[craggy egret]

sorry if dont respond in time im also doing boolean elgebra atm

[craggy egret]

oh ok

[gusty gorge]

do you see that the output voltage is fed back into the V- slot through some resistors?

[craggy egret]

there is 1 beetween v2 and v-

is that the feedback resistor

[gusty gorge]

yes

now imagine that the V- is some nontrivial positive voltage

you see that V+ is 0

so that means the Vo is going to be a really large negative voltage

and that'll bring V- to some negative voltage

likewise, you can see if that V- is some nontrivial negative voltage, it's going to make Vo really positive and bring V- positive

there's "negative feedback" here in the sense that if V- is anything but really close to zero, the output swings to bring it toward zero

[craggy egret]

hmm

[gusty gorge]

so V- ends up being a really small value close to zero

[craggy egret]

hmm

[gusty gorge]

and in general, because the gain of the op amp is so high, V+ and V- are really close to each other because of negative feedback

[craggy egret]

hmm

[gusty gorge]

note if you have positive feedback, the op amp just produces one of the voltage rails

[craggy egret]

at v+ =v- = 0?

well close to 0

[gusty gorge]

You can assume V+ = V- for op amps in negative feedback

and in this case, V+ is tied to 0V, so V- is also 0

[craggy egret]

hmm

[gusty gorge]

and that's basically it

[craggy egret]

oh ok

wait

how did he get v1 = -2 v2

[gusty gorge]

rearrange this

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[lone heart]

@dry zephyr Has your question been resolved?

[lone heart]

@dry zephyr Has your question been resolved?

[hushed scroll]

@dry zephyr do you know the answer?

[dry zephyr]

i do but i just dont know how to find it

[hushed scroll]

is the answer -x+5y-22z+114=0?

[dry zephyr]

yes it is

[hushed scroll]

so its not a standard approach but i just thought how that plane would be and i figured if you take any point on one line say (2,-4,3) and take a random point on the other line (5k, k-6, 5), then the line formed by midpoints of these two points lies on the required plane.

[dry zephyr]

ok how did you find the midpoint?

because i tried finding it

[jaunty olive]

I would suggest helping him and not just handing him the idea

[hushed scroll]

similarly if you take any point on the 2nd line, say (0,-6,5) and any random point from 1st line (2+3t, 5t-4, t+3), so the line formed by midpoints of these two points also lies on the required plane

[jaunty olive]

Let him do it on his own

[hushed scroll]

the same way you find midpoint btw two points (x1, y1, z1), (x2, y2, z2)

[hushed scroll]

i am telling him a way to solve it, not giving him the answer directly

[dry zephyr]

by taking the average right?

[hushed scroll]

you will get two vectors which both lie on the required plane

[hushed scroll]

yes

[dry zephyr]

ok i just have a question

why cant i just take the two points given and take the midpoint of that>

might be a dumb question

[hushed scroll]

you can do that, but that will give a coordinate of a single point on the required plane, how would you find the equation of the plane by knowing the coordinate of just 1 point?

[jaunty olive]

Your first message was literally telling him what you did

[hushed scroll]

moreover as I said, my way isn't standard, so there is a better way sure

[jaunty olive]

Sigh nvm just continue let's not waste his time as well

[hushed scroll]

ok

[dry zephyr]

I tried plugging the two points into the plane equation and find the mid point.

[hushed scroll]

and what you got?
you should get something like
(x, y, z) where each of x,y,z are functions of t or k

[dry zephyr]

well i got a number doing that

[hushed scroll]

oh that, you plugged the given points and found midpoint

[dry zephyr]

i plugged (2,-4,3) and (0,-6,5) into -x+5y+22z=C

[dry zephyr]

and took the average of them

[dry zephyr]

yea it didnt give me the right answer tho

i have also tried finding the distance

[hushed scroll]

yes idk how to solve it that way but, from my way, you will get 2 vectors that lie on the required plane. Then you can convert those two vectors in parametric form and then take their unit vectors and take its cross product to get the normal vector from plane.

Now you have - Normal Vector from Plane in form (a,b,c)
and you also know coordinate of a point on the plane in form - (x1, y1, z1)

you can use point normal form of plane to find equation of plane

do you understand?

[dry zephyr]

isnt that a bit extra though?

for the normal

i just crossed the two direction vectors

unless that is what you meant

[hushed scroll]

yes, that's what cross product mean

(a, b, c) × (d, e, f) and you will get the normal vector from plane in form (g, h, i)

[dry zephyr]

ok so what do i do after finding the average between the two points

[hushed scroll]

you will not get any fix coordinate after finding average, instead you will get something a vector, did you find the midpoint?

yea, you can ping helpers and see if someone gives a normal answer, if you don't want to do it this way

[dry zephyr]

yea i got (1+(5/2)k, -5+(1/2)k, 4)

[hushed scroll]

convert it into parameteric form

[dry zephyr]

i dont mind exploring other ways

[dry zephyr]

okay

[hushed scroll]

what you got?

[dry zephyr]

x= 1+(5/2)k
y= -5+(1/2)k
z=4

i took midpoint of (2,-4,3) and (5k,-6+k,5)

[hushed scroll]

oh ok

you got both vectors? thisis one vector, you want the other one too

[dry zephyr]

ohhhh

mb mb

x= 1+(3/2)t
y= -5 +(5/2)t
z= 4+(1/2)t

[hushed scroll]

correct

now these are the two vectors that lie on the plane, you can easily take t=0 or k=0 to get exact coordinates of 1 point that lies on this plane
(because we need atleast 1 point to use point-normal form of plane equation)

a(x-x0) + b(y-y0) + c(z-z0) = 0
you know this? point normal equation of plane

[dry zephyr]

what is the point-normal form actually

[dry zephyr]

yea no, i was never introduced that

[hushed scroll]

where x0, y0, z0 is any point that lies on the plane and (a,b,c) is the normal vector to the plane