help-19

[steady arch]

log |tanx | + tan^2x/2 + C is what i get now

[mystic saffron]

This is indefinite integration so I don’t have limits

[brave gale]

then maybe u just need to assume that arcsin(sinx)=x

[odd edge]

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[mystic saffron]

how to solve this?

[limber glade]

i suppose you're stuck in part 2?

[mystic saffron]

yes

i moved the y on the right side

then i divided -3 on both sides

to get x

idk what to do next

[limber glade]

substitute that value of x in the other given equation

OR

even simpler

in equation 1, divide both sides by -3

and in equation 2, divide both sides by 9

[mystic saffron]

i got 0 = 0

[limber glade]

so what do you mean by that?

the lines are coincidental

[mystic saffron]

i distrubuted the 9

and then i was able to cancel out the 9

9y

[limber glade]

you don't even need substitution lmao

[mystic saffron]

then i got -3=-3

added 3

[limber glade]

just simplify the equations individually

[mystic saffron]

does this mean any value for x and y work for the equation

[limber glade]

not really

[limber glade]

.

don't sub

[limber glade]

.

[mystic saffron]

what

[limber glade]

i'm telling ya don't substitute

it'll confuse ya more

just make the coefficients of x and y 1 and you'll realise that the equations are coincidental

[mystic saffron]

i think i solved it now

x is -y - (1/3)

y is -x - (1/3)

[limber glade]

no you'll get constant values

see

yiu take the first question

-3x-3y = 1

-3(x+y) = 1

x+y=-1/3

now take the 2nd equation

9x+9y = -3

9(x+y)=-3

x+y = -1/3

see?

both the equations are same

so the solution is the line itself

x+y = -1/3

now the new line given by the question is y = x + 1

let us represent them as x1 and y1

so y1 = x1 + 1

x1 - y1 = -1

[limber glade]

now take this

multiply by 3

3x+3y = -1

[limber glade]

compare this

[limber glade]

and this

you'll get the values of x1 and y1

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close the previous channel

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[ancient rampart]

integrating a rather

[ancient rampart]

cumbersome definite integral

i am left with

how to approach?

from o to pi/4 btw

dont know if reduction formula will be useful

or i should do by parts or something since cosec^2x is -cotx

i can write it as tan^nx + tan^(n-2)x

[quasi sparrow]

simplify the integrand first

[ancient rampart]

idk how to after this

[quasi sparrow]

neither

[ancient rampart]

do i apply redn formula

[quasi sparrow]

tan^2 = sin^2 / cos^2

tan^2 csc^2 = ?

[ancient rampart]

what

ik that

[quasi sparrow]

answer then

[quasi sparrow]

.

[ancient rampart]

sec^2x

sin terms cancel out

[quasi sparrow]

brilliant

what's the derivative of tan(x) ?

[ancient rampart]

why derivative here?

sec^2x

damn dude u are one passive agressive fella

good day.

.close

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[quasi sparrow]

very straight forward

[quasi sparrow]

very straightforward

[quasi sparrow]

not at all

[ancient rampart]

nah

all it took was redn formula i was right

[ancient rampart]

the other term cancels out

[quasi sparrow]

great? you discovered there's more than 1 way to do a problem

[ancient rampart]

yeah but u could have just told

what was the alternative method

but u didnt for some reason

[quasi sparrow]

because you should use your brain and not be spoonfed answers

[ancient rampart]

obviously i did maybe i am stuck because i was not spotting something trivial, hence i asked here otherwise why would I

[quasi sparrow]

the solution was trivial so it's not so obvious

no one knows what you know, only you

[ancient rampart]

never mind

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[restive atlas]

What is a good process of determining if the external direct product Z_n1 and Z_n2 are isomorphic to Z_n3 and Z_n4. My first step was to determine if the result is of the same order (in my case they are) but i’m not sure what to do next?

[restive atlas]

Looking at part B

[sharp oak]

The usual things, when trying to show two groups are not isomorphic. Any property that one group has, but the other does not

Note that Z5 × Z8 ≈ Z40

[restive atlas]

sorry im looking at A

there are three parts in this problem and i cropped the first

so i’m looking at the two that have the same order

my bad

[sharp oak]

Za × Zb ≈ Z(a×b)
If a and b are coprime

[restive atlas]

wasn’t taught that yet. is this only for cyclic groups?

[sharp oak]

Essentially yes

[restive atlas]

Okay sweet, i appreciate the help. so my guess is show what group the external direct product is isomorphic to, and if they are isomorphic to the same group, then they themselves are isomorphic

both of those two are isomorphic to Z_210 so yes they are isomorphic

.close

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[native marten]

Hey guys, my cousin is working on his bachelor's thesis and needs some help. The topic is to construct an inequality as seen in pic 1, it is from a book called Concentration inequalities by Gabor Lugosi et al.. the second pic has the ideas of my cousin and where he is stuck. His first question is how to find the constant kappa. the 2nd question is, if he is on the right track with his idea; it is to show that a point x_i exists in every A_j.

Let me know if it's the right channel for this. Thank you in advance!

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[keen jacinth]

I need help 😭

[keen jacinth]

the hw question is just sooooo hard

the question is: we have a infinite series

but the derivative of the kth term is (k+2)th term times (k+2)(k+3)

how do I approach it

any hints?

oh yes, btw, every term if a function of t

so by the derivative, I mean, derivative wrt to t

[odd edge]

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.

[keen jacinth]

basically I'm trying to find $a_k(t)$

[clever fjord]

lisph

[keen jacinth]

I don't believe it is an exponential type function because the exponentiated part won't change with by differentiating

I think I'm close to the solution...

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@keen jacinth Has your question been resolved?

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[sage slate]

What is a more efficient way to calculate erfc^-1(x)? I need ~11500 iterations of newtons method to get an accurate result for small values of x erfc^-1(2^-16126) == ~105.7. This takes awhile especially when erfc(x) takes x^2 / 2 iterations to calculate.

[sage slate]

This is what I am using for newtons method, a is the approximation of erfc^-1(x)

[meager juniper]

Looking for a way to do inverse normal cdf right?

Why do you need it for such small values though?

[sage slate]

To ensure the entire exponent range of long double is covered. It can represent values as small as 2^-16383, which I believe is around erfc(106.7)

[meager juniper]

long double is an 80 or 128 bit floating point value?

[sage slate]

80bit, ieee 128bit floats have the same exponent range

[meager juniper]

@sage slate have you investigated the literature on this. I think Newton's method might just be a bad fit for this function.

[odd edge]

@sage slate Has your question been resolved?

[meager juniper]

Here's a paper on the icdf that I saw an implementation for. Depending on your accuracy requirements, this might be a much better option.

[sage slate]

it was a bit tricky since the arbritrary precision library mpfr doesn't have an inverse cdf or erf function, and wikipedia didn't list anything with the precision I needed (~126bits of precision)

[meager juniper]

Well, you might be able to make this work as a first guess then

Hmmm

There's another arbitrary precision library that might have it, let me check the documentation.

If I can remember the name

https://arblib.org/arb_hypgeom.html#c.arb_hypgeom_erfinv

@sage slate ^

If you're not married to mpfr I would recommend giving this library a look.

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[stiff kettle]

this is soo easy omg but how do u do the inverse function.

BRO IM LIKE STUCK ON INVERSE FUNCTIONS LIKE IM JUUST SO LOST

so i got to the point where

x - 1 = -y^2

but NOW IM STUCK.

bc idk if i should square root everything first, or multiply it all by -1or like

ohhh hii charbit

[brittle beacon]

I accidentally started typing, I was waiting for you to finish typing

[brittle beacon]

But multiply by -1

(Do they also give you a domain for g anywhere?)

[stiff kettle]

nope nothing about a domain

[stiff kettle]

hmm ok one sec

wait but

why is multiply by -1 first?

because isnt according to BEDMASS first its exponents?

ohhh waitttt its because im going the opposite way of bedmass right.... ?

[brittle beacon]

If you multiply by -1 first, it makes it easier to take square roots (as -y^2 can be negative, and you don't want to take square roots of negative numbers)

[stiff kettle]

ok so if i multiply -1 by the left side how would it look like

-(x-1)

and should i like open it up and make it x + 1

[brittle beacon]

It doesn't become x + 1, but expand it, yep

[stiff kettle]

oops

-x + 1

[brittle beacon]

Yep

[stiff kettle]

and finally

square root both sides

so it would be

+- sqrt - x + 1

im a little confused about why its + and - tho

i know square root has both + and - , i get that

but how can u determine the inverse function with that

[brittle beacon]

Aha, well, that's why I asked this question here

[stiff kettle]

well no they dont 🤔

[brittle beacon]

Annoyingly, they didn't give you a domain, which means that it's harder to choose whether it's the + one, or the - one

[stiff kettle]

can i somehow plug this into demos to see the + version and the - version?

would both verisons look like this?

oh

OH WAIT

THATS THE ANSWER

so u dont pick if its positive or negative..

[brittle beacon]

Well, remember that functions need to give one and exactly one output for all allowable inputs

[stiff kettle]

right so the inverse one isnt a function

[brittle beacon]

Yep, unless you do the whole "restrict the domain" thing

[stiff kettle]

oh i see ya we didnt learn that thankfully HAHA

but i see what u mean

[brittle beacon]

Fair enough I shall leave it to them to explain it to you when they get there then

[stiff kettle]

Yess ofc, im studying for a math test thats gonna be on Thursday so you'll probably see me here a few more times haha

but ur superrr good at explaining stuff

thank you so much again!!!

[brittle beacon]

Awwwww see ya around

[stiff kettle]

🥹

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[tribal nova]

Suppose 𝑣, 𝑤 ∈ 𝑉. Explain why there exists a unique 𝑥 ∈ 𝑉 such that 𝑣 + 3𝑥 = 𝑤.

[tribal nova]

Is my solution correct? Are there any intermidaye steps I’m missing v

?*

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[tribal nova]

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[naive field]

1+1=

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[leaden wigeon]

Hi, I need help with basic proofs. should be basic enough for help but idk how to do this lol. im 9th grade geomoetry honors class. its angle proofs and

[leaden wigeon]

like line segment proofs

ir wtv

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[native marten]

hey can i get some help for this? or should I ask somewhere else?

[inland rover]

maybe try asking in #advanced-probability looks like it involves some probability

[native marten]

i will try. thank you

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[misty marsh]

how do i express this as the square root of a single number?

[misty marsh]

im not even sure what its asking

[clever fjord]

faiyrose

faiyrose

[misty marsh]

so sqr root 25 x sqr root 2?

so sqr root 50?

the

express this as a square root of a single number

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[trim olive]

can someone double check my answers, please?

[tacit wasp]

All 4 are correct

[trim olive]

ty

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[narrow crypt]

Find all positive integers $a$ and $b$ that satisfy the equation $a^2 + b^2 < 23$. Is there a fast way to compute this?

[clever fjord]

Average Calc Student

[cursive field]

it's not that hard

the largest square below 23 is 16

so you only need to check (a, b) until 4

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[paper moth]

need help

[paper moth]

i dunno what to do tbh

i can't tell which switch controls which globe

[onyx cloak]

see which light bulbs are in the circuit when you close a switch

[paper moth]

so i think L2 will stop working

L3 as well

and L4

am i right?

@onyx cloak

[onyx cloak]

if u close S1, which bulbs should light up

[paper moth]

L1????

idk😭

if it is open switch what does that mean?

[onyx cloak]

switch open -> its off

the electrical switches at ur home are in open position (switch off) until you close them (switch on)

[paper moth]

so since s1 is open L2 won't work?

[onyx cloak]

if its open nothing would work

since the cicuit isnt completed

current can only flow once the circuit is completed

[paper moth]

so then wouldn't L2 and L1 work if switch s1 is closed

ugh i don't get it

.close

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[onyx cloak]

yes

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[violet elk]

I think you take the integral of it

But like

I’m so confused what to do

I know if it decreases

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[winged bison]

algebrically, what are the steps for this rewrite

[onyx cloak]

multiply and divide by $\sqrt[300]{x^{300}}$

[sterile blaze]

sqrt300(x^300)=x

[clever fjord]

Astar777

[uneven osprey]

x^299*

[onyx cloak]

no

theres a $\sqrt[300]{x}$ in denominator

[clever fjord]

Astar777

[onyx cloak]

so $\frac{\sqrt[300]{x^{300}}}{\sqrt[300]{x}} \implies \sqrt[300]{x^{299}}$

[clever fjord]

Astar777

[onyx cloak]

i mean urs will work too

[uneven osprey]

how does this gonna work

[onyx cloak]

what is $\sqrt[300]{x^{300}}$

[clever fjord]

Astar777

[onyx cloak]

solve this

what do you get

[odd edge]

@winged bison Has your question been resolved?

[winged bison]

Ohhh, thanks!!

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[royal palm]

Hi! I was wondering if there was a better way to calculate the modulus (and also after this, the argument) of re^(i2θ)?

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[royal palm]

Is this the best way to solve this?

<@&286206848099549185>

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[sharp lotus]

Hello, not sure if this belongs here... I'm following the course CFD with high performance programming in python and on the step 13 (cavity flow with Navier-Stokes), the pressure partial derivative $\frac{\partial p}{\partial x}$ is discretized with centered finite difference $\frac{\partial p}{\partial x} \approx \frac{p^n_{i+1,j} - p^n_{i-1,j}}{2 \Delta x}$. I tried with forward or backward first order and it doesn't work but I don't really understand why we have to take the central finite difference. I guess that the central finite difference is more 'stable', but in that case, why don't we take it for the velocity derivative also. I've looked at upwind scheme and I don't understant why the velocity finite difference doesn't work with negative velocities. If you can explain or have some ressources explaining finite differences intuitively and where/why I should use different formulations, I'd really appreciate it.

[clever fjord]

weareinthematrix

[echo ginkgo]

try asking in #numerical-analysis instead @sharp lotus

maybe seperate your questions a bit more tho, it feels like you're asking 12 things at the same time

[odd edge]

@sharp lotus Has your question been resolved?

[sharp lotus]

Ok, thanks, I'll ask there

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[unborn tiger]

1. Can there be an equilateral triangle with integral side length whose vertices have integral coordinates?
2. Find two primes whose sum is 81. (I know the answer is 79 and 2 but I want a proper process to solve this.)

[red surge]

for 1, the answer is no - think trigonometry

and what we can generalise

[ivory raven]

work with a equilateral triangle with one point on the origin and other points (a,b), and (c,d)

assume these are integer, expand things out with pythagoras, and work towards a contradiction

[red surge]

in fact, you can further assume one of the points if (k,0)

[red surge]

for 2, one of the primes has to be 2 cus otherwise odd+odd = even

so if it works, it has to be n-2 and 2

if ur asking about for general (i.e. finding 2 primes that sum up to say 80)

there is no general method of doing it

in fact there's a pretty famous unsolved problem regarding even numbers being written as the sum of 2 primes

[unborn tiger]

Okay

Thanks everyone

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[true brook]

quick question

[true brook]

i have a question that requires grouping

15x^3-25x^2+75x-125

i common factored first

a 5

which i think may of messed up the question i’m not sure though

i’m left with a 5(3x^3-5x^2+15x-25)

i csnt group out on the first term tho so do i leave it

and then i factor out another 5

[steep ginkgo]

I wldnt know how to motivate the thought but have you done rational root theorem?

[true brook]

no no i’m on factoring

grouping for this question

[steep ginkgo]

Ok do yk Abt plugging in numbers to evaluate if it's 0 or not and that proves if it is a root?

That whole idea

[true brook]

uh no

[steep ginkgo]

Alright if u can factor the 3x cubed term into a factor x- alpha, that means if u plug alpha into the original equation you will get 0

So u can quickly check if something is a factor to the original equation

[true brook]

yea i haven’t learned that

[steep ginkgo]

That sucks idk how you'd group it otherwise

[true brook]

i know i can group out a 5

[steep ginkgo]

Yh that's the right idea definitely

[true brook]

but like idk

what abt the 15-25

again?

[steep ginkgo]

If u plug in x=5/3 using the whole thing I said before

[true brook]

group out a 5

[steep ginkgo]

U can figure out that u can take 3x-5 out as well

Of the cubic

[true brook]

i actully noticed it gives a fraction

i doubt i’ll have something like this on the test then

[steep ginkgo]

Yh it's a bit hard if u haven't learnt

[true brook]

so i doubt i need to worry abt it

thanks anyway

[steep ginkgo]

For completions sake it shld be 5(3x-5)(x squared+ 5)

Fingers crossed haven't made an algebraic error

[true brook]

yea well i should be fine

thank you

.closr

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[true brook]

have a good one

[odd edge]

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[sage helm]

Hi. I'm reading a paper that makes use of some symbols I've never seen. The paper doesn't explain them:

[sage helm]

What are the symbols underlined in green?

[low locust]

I mean the sentence says what they stand for

do you want to know what their latex command is?

[sage helm]

yes please

[low locust]

try detexify

the middle one is \mapsto. not sure about the others

[sage helm]

The definitions mention the entire expression. But I'm asking about the specific symbols.

Like $\subset$ means something for sets.

[clever fjord]

Vulkanoid

[low locust]

each sentence is of the form "thing (then some string explaining what its a subset of) is this and that"

tho I think the last one may have a typo cause it includes TxQxT instead of just cartesian products with two sets

which would make more sense for partial orders

[sage helm]

kk, thanks

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[grizzled basin]

Guys need to prove the equality, how

[quasi sparrow]

well i was gonna respond

!15m

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[grizzled basin]

Ok im sorry

[mystic saffron]

for first one just write tg as s/c and multiply num and denom by cos22. then you get cos(45)/sin(30) after using the identities for cos(x+y) in num and sin(x+y) in denom

@grizzled basin

[grizzled basin]

Thanks bro

[mystic saffron]

@grizzled basin sure. and same for second, except multiply by sin(2pi/9) in num and denom.

[grizzled basin]

Ok

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[nova yarrow]

this is what I tried but I figured out it does not get me the answer I am looking for. This was bayes method and I tried the other conditioning formula which also did not work. Completely stuck

[tender carbon]

is 0.4 the correct answer?

[nova yarrow]

not sure, I just know that I got 6.36 and 2.25

which is not right

[tender carbon]

i used this and got .4

we know B happend

here

[nova yarrow]

sorry wrong thing

I used that method and I got 0.3+0.2/0.2

[tender carbon]

but nominatot is not right i guess

it should be 0.08

P((AB)C) = P(ABC) right ?

[nova yarrow]

Do I multiply with the intersection operator instead of add probability?

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[graceful parrot]

Am I on the right path for problem 40

[lethal spoke]

Yup

[graceful parrot]

Do I just find where the problem is 0

I mean derivative

[lethal spoke]

Yup

[graceful parrot]

I can’t find where it’s 0, is it a unit circle value?

[lethal spoke]

Divide through by $e^x$

[clever fjord]

[lethal spoke]

@graceful parrot

Sorry for the late response 😅

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[rigid brook]

can someone tell me what the apostrophe means

[wanton bison]

1st derivative

[rigid brook]

oh right thanks

[wanton bison]

wow chain rule is applicable with complex numbers

didnt know that

[rigid brook]

[wanton bison]

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[stiff kettle]

Advanced functions gr 12 lol im kinda lost

[eternal skiff]

The function will be continuos if the limits are equal so lim -3x -3k = lim x - k and both limits have x approaches 2

[stiff kettle]

umm

ok but when u put both to equal then u have two variables

[eternal skiff]

You can evaluate the limits though to get for example

[stiff kettle]

uh what

we didnt learn that lol

is there another way of doing it

[eternal skiff]

Uhhh I mean the simple way of doing it is to say that -3x -3k = x - k for x = 2 so that -3*2-3k = 2 - k

Although you should really do it with limits

[stiff kettle]

😀

well we never learned that we have to make it equal each other so im sure theres another way

of simple explaining

bc im looking at what another girl did and she wrote this

[eternal skiff]

Yeah ok than just this

[stiff kettle]

but. how do u get 2

[eternal skiff]

Because the function is defined around 2

[stiff kettle]

oh

[eternal skiff]

-3x-3k for values before 2 and x-k for values after 2

[stiff kettle]

ok

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[sterile blaze]

H < G, |H| = k. no other subgroup of G has order k. show H is normal in G.

[sterile blaze]

we have to show gH = Hg for all g in G

or gHg^-1 = H

im not too sure how

[odd edge]

@sterile blaze Has your question been resolved?

[red surge]

hint: ||think about why gHg^-1 is a subgroup of G||

[sterile blaze]

OH

yea because it acts on e with geg^-1=e

so it is a subgroup of order k as well and as there is only one subgroup of order k, gHg^-1 = H

yay!

thank you!!!!

<3

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[tribal nest]

Am I doing this right? I just learned limits today but this feels too easy

[tribal nest]

*I think it should be 4 not 5 but all my other work is the same

[zinc glacier]

looks good

other than the 5 lol

[tribal nest]

Yeah, I used the wrong factor the first time

[zinc glacier]

you dont need to simplify btw

[tribal nest]

But if I got 4, I'm right?

[zinc glacier]

yes

you would only need to simplify for y -> -3 since that would otherwise return undefined

[tribal nest]

Ah

Gotcha

I think I missed that in the lesson

[zinc glacier]

its best to try direct sub first

[tribal nest]

Just putting the value that the limit is approaching in for the variable?

[zinc glacier]

yes

[tribal nest]

Ok, good to know

Thank you!

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[zinc glacier]

youre welcome

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[untold cypress]

How do I do C and D?

[tribal nest]

You can sometimes use long division but I would suggest synthetic division

[untold cypress]

I've never learnt synthetic division, so i'll try it using long division first

-2x and 6x^2 cant be subtracted

did i do something wrong??

[mint mirage]

No

I suggest putting placeholders for the terms that aren't there, like 0x^2

So you can line up the terms

[untold cypress]

then what should I do?

[mint mirage]

Read the statements after that

See, now that the terms line up, you can do the math

[untold cypress]

ohh

i'll try again then

[untold cypress]

Is it right?

[mint mirage]

You messed up your sign at the highlighted part

The circle part is 10x and you said x goes into 10x, negative 10 times

[untold cypress]

I just fixed it

Ty!!

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[wooden tide]

How did they find this using kinematic equations

[amber pendant]

more information/context sssssssssssssssssssssniperwolf

[wanton bison]

or what 😝

[wooden tide]

Or ill take your cat

[brittle beacon]

Do you have a list of the kinematic equations you use?

[wooden tide]

Here!

[brittle beacon]

why do they use different labels everywhere

Maybe consider splitting it up into horizontal and vertical motion

[amber pendant]

yep, theres no ready equation to use here

Newton's 2nd law, find x(t), find y(t)

find time at which the projectile lands

and substitute that time in x(t)

(the time at which the projectile lands is the time at which y(t)=0 btw)

[wooden tide]

Yes but I don’t see which X equations and which Y equations I csn use

At least in the table I have

[wooden tide]

Oh I’ll try that

[brittle beacon]

(Remember what “launched horizontally” implies about the vertical initial velocity too)

[wooden tide]

Wait the equations I gave

Can they all be in x and y

[forest sky]

you can replace s with either x or y and get 2 equations out of each of the ones listed

[wooden tide]

oh aoky ty

very practical id say

[pastel orbit]

[amber pendant]

i really suggest you derive them x(t) and y(t) using Newton's 2nd law

instead of using ready equations or instead of memorizing

its good for you

[brittle beacon]

(Mind you, they ask you to do it using the equations in her question, so there’s that )

[amber pendant]

yeah yeah

but in exam they wont be given equations :d

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[twin swift]

For this question, I reduced it down (and checked it with an online calculator). The RREF matrix is:

1 -3 0 -9
0 0 1 -2
0 0 0 0
0 0 0 0

I put my columns in the same order as the question put them in my Matrix.

Why is my answer wrong?

[twin swift]

Since column 1 and 3 have pivots, that means that vectors 2 and 4 are redundant and removing them will make the set linearly independent right?

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[north river]

At the Los Angeles national laboratory in New Mexico, protons are accelerated from a rest to a velocity of 2.5x10^8m/s in an accelerator that is 800 m long, what is the average velocity?

[north river]

This is serious

Do you put the numbers in v^2 = u^2 +2(a)(s)

After that I’m kind of loss

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[haughty chasm]

What equations would I use for this question

[summer cradle]

thephalluspulverizer

[haughty chasm]

yes ma'am

[summer cradle]

what a name

[onyx cloak]

fr

[haughty chasm]

u should meet the flatus relinquisher

[west hull]

(2/3)=(1/2)^(10/HL)

HL being half life

[haughty chasm]

ermm

[onyx cloak]

use the exponential decay equation (N_o = Ne^(-kt))

find the constant (k)

then use half life formula (ln2/k)

[amber pendant]

yeah, this

but the expression is $N(t) = N_0 , e^{-kt}$

[clever fjord]

Emily

[onyx cloak]

thats what i wrote?

oh

[amber pendant]

you switched

[onyx cloak]

switched the variables

[amber pendant]

😛

[onyx cloak]

my bad

this is first order reaction (as the first sentence says)

thats why the half life is ln2/k

(@haughty chasm)

[haughty chasm]

YES

but I dont have the initial to find k?

[onyx cloak]

you do

see whats given

it takes 10 years for 1/3 rd of the substance to decay

[haughty chasm]

yurrp

wait so...

hol up lemme do the math

I'm still kinda stumped 😭

[summer cradle]

you start N_0 amount of substance (you could also just make this 1 if you want). after 10 years, how much is there?

[onyx cloak]

N_0 is initial amount, N(t) is amount left after t years

so what is N(t) (in terms of N_0) for the given data?

[haughty chasm]

1/3 No

?

[onyx cloak]

1/3 is decay

you have to write the remaining amount of substance

[haughty chasm]

2/3

[onyx cloak]

yes

[haughty chasm]

https://tenor.com/view/think-about-it-use-your-brain-use-the-brain-think-brain-gif-7914082

[onyx cloak]

so (2/3)N_0 = (N_0)e^(-10k)

now find k

[haughty chasm]

its -ln(2/3) / 10

but I was taught to keep it in terms of e^k

I suppose it wouldnt work that way

[onyx cloak]

yeah

how would u plug the k in ln2/k in e^k form

[haughty chasm]

yurr

ok so...

[haughty chasm]

I do not believe that my work is correct

I got a big ol number

[onyx cloak]

why is everything in power of e

[haughty chasm]

oh is it not?

I thought it replaces k

[onyx cloak]

its t (half life) = ln2/k

[haughty chasm]

ahhh yes

its 17

yeah Im failing this exam

[west hull]

This is simpler and yields 17.1 all the same

10/(log_0.5(2/3))

[haughty chasm]

yeah sry I was js unfamilar with HL

[west hull]

HL is half life

[haughty chasm]

😅

Its been a lonjg dayu

[west hull]

N(t)=N_{0}(\frac{1}{2})^{\frac{t}{HL}}\\
\text{or } N(t)=N_{0}(\frac{1}{2})^{\frac{t}{t_{\frac{1}{2}}}

[haughty chasm]

ohhh shoot

[west hull]

Where N(t) is 2/3, N_0 is 1 and t is 10

[haughty chasm]

the math is mathing

I see now

Im more of a visual learner

[clever fjord]

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

[west hull]

t_(1/2) is HL

[haughty chasm]

.close

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[naive field]

1+1=?

[mystic saffron]

3

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[night bone]

Is this the right answer

or did I make a mistake somewhere

[wanton bison]

looks right to me

[night bone]

ty

.close

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[stuck kraken]

why is the answer 3 in the blank?

[stuck kraken]

i understand that for each k there is a realisation yk of random variable Y but why do i need to connect it to sample variance of the 4 IID RVs?

and why is answer found by finding the expectation of Y?

[odd edge]

@stuck kraken Has your question been resolved?

[odd edge]

@stuck kraken Has your question been resolved?

[stuck kraken]

<@&286206848099549185>

[odd edge]

@stuck kraken Has your question been resolved?

[weary pelican]

law of large numbers: if $(X_k){k\in \bN}$ are independent and identically distributed random variables, then $\frac 1n \sum{k=1}^nX_k$ converges almost surely to $\mathbb{E}[X]$ when $n\to \infty$.

[clever fjord]

rafilou2003

[weary pelican]

here (y1,y2,...) is a sequence of iid random variables

so apply law of large numbers

E[y] = 3*sigma^2

[stuck kraken]

so is it true to say that sigma^2 = s^2 = (1/3) E[y]?

[weary pelican]

yes

[stuck kraken]

since we are accounting for all k individuals, why do we use sample variance 1/3 E[y] instead of population variance 1/4 E[y]?

[weary pelican]

?

the way y is defined gives you E[y] = 3sigma^2

it's not that we USE sample variance 1/3 E[y]

[stuck kraken]

oh ok

thank you!

.close

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[south plume]

hola

do you have a math question to ask?

[uneven lodge]

Can you show a picture of the original question

[red surge]

i think they want a picture cus it's hard to read this non-latexifed stuff

[odd edge]

@tight rock Has your question been resolved?

[red surge]

is this the original question?

i mean isn't this just an algebra problem then?

plug in y1 = etc. and y2 = etc. to find f(y) = etc. in terms of y1 hat and y2 hat

then rewrite that in terms of y1 and y2

[red surge]

yeah but i think ur missing a bracket

yeah

sure

i think it's just an algebra rearranging problem

[odd edge]

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[formal fable]

What is this notation for permutation groups?

[echo ginkgo]

the order of the permutation sigma prolly

[weary pelican]

more context?

like bigger screenshot?

[echo ginkgo]

one day we'll have 1 px screenshots

@formal fable forgot to ping

[echo ginkgo]

yeah alright

[red surge]

yeah it's just the order of sigma

what they've written is the size of the subgroup generated by sigma

which is the order of sigma in this case

[formal fable]

how do you find the order of a permutation

[red surge]

what do u know about permutations so far?

[solemn rose]

any examples of standard form, vertex form, and factored form (just for reviewer)

[red surge]

!occupied

[formal fable]

i read somewhere that you need to find the number such that the permutation applied to itself gives identity

[odd edge]

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).

[red surge]

feel free to post ur problem here

[formal fable]

I missed the lecture on permutations

[red surge]

ah

[formal fable]

but i understand how to compute composition of permutations

[echo ginkgo]

yeah, and that's quite slow if you just use the definition

[red surge]

if i say the thm "any permutation can be written as a product of disjoint cycles" does that ring a bell?

[formal fable]

no i didnt see that in my friend's notes

[red surge]

if not the intention of the exercise might just be to compute these the long&old fashioned way to get some intuition

[echo ginkgo]

the order can be quite huge potentially, so just computing sigma, sigma^2, ... until you get the identity might be a pain in the ass

Imma whip out sage real quick

[formal fable]

nice

[red surge]

well it's bounded by 6 in this case lol

[formal fable]

where did you get 6 come from

ohhh

finite could mean 10^50 xd

[weary pelican]

sigma = (1 3 4 5 6 2)

so...

[echo ginkgo]

yeah I was just scared for no reason

[weary pelican]

lucky we're in S_6, the bound in S_7 is 12 xdd

[formal fable]

how do you calculate that

[weary pelican]

decomposition into disjoint cycles

[red surge]

you don't really need to know how to calculate that yet if u're not sure how to compute the order of a permutation

do you know what cycles are?

[formal fable]

not really

[weary pelican]

oh

so cycles is just a permutation that is a single loop

for example 1->2->5->1

[formal fable]

like Z_3

[weary pelican]

?

what do you mean by Z_3

the group of remainders of integers by 3?

[red surge]

i think if u haven't covered them in ur book/lecture then i think the point of the exercise is for you to compute sigma, sigma^2, sigma^3 etc. to get some intuition for what cycles are

once u've learnt more theory there are much better ways of finding the order of a permutation

[formal fable]

i think she said we are moving to ring theory next

im not sure if were doing more permutations

[red surge]

oh ok

in which case yeah u should probably know what cycles are

[weary pelican]

in any case, the order of an element x is the smallest positive n such that x^n = the neutral

so to find the order of sigma

you compute sigma

[formal fable]

oh kk

[weary pelican]

sigma^2

etc...

as soon as you find the identity, you got the order

[formal fable]

What is this

[weary pelican]

any permutation can be written as cycle1*cycle2*...

where each cycle acts on different numbers than the other

for example, if cycle1 moves 1 to 5, then cycle2 will neither act on 1 nor 5

[formal fable]

ohh

close!

close!

.close

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[formal fable]

How is this possible if (1 5) means that 1 and 5 are switched in a the permutation

How can 1 be in more than one cycle

.reopen

[odd edge]

✅

[weary pelican]

because this is not disjoint cycle form

literally the question is about rewriting in such a form

[odd edge]

@formal fable Has your question been resolved?

[formal fable]

1 maps to 5

and 5 maps to 1

how does 1 map to anything else

that is like straight up false to say that 1 maps to anything else

[formal fable]

that's what (1 5) means right?

[odd edge]

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[atomic epoch]

Suppose the quadratic equation x^2 -5x +m = 0 has integral roots, where m is a constant. Peter claims that m must be 6. Do you agree? Explain your answer.

[atomic epoch]

ok so i used the quadratic formula thing and worked out that sqrt(25-4m) determines whether it is an integer

but i dont know what is a proper way to proof against peter's claim

i know that m can be 4

[mystic saffron]

that's not the only thing that determines whether it'll be an integer

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

[clever fjord]

45

[amber schooner]

yup

[atomic epoch]

oh wait true

[mystic saffron]

the discriminant will just tell you if it's rational or not (if b^2 - 4ac) is a perfect square

[atomic epoch]

it only makes it rational i guess

[amber schooner]

damn you typed that fast bro wtf

[mystic saffron]

been LaTeXing for years

[amber schooner]

i was sat there typing it for years

😭

you did it in like 3 seconds

[atomic epoch]

ok seems that both of m=4 roots are integers. so i should say that m=4 still has rational roots to prove peter is wrong?

[low locust]

what are the roots

[atomic epoch]

uh -1 and 4

[low locust]

1 and 4

[atomic epoch]

whops

[low locust]

but yes if you state that then you are good

[atomic epoch]

yay

thanks everyone

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[atomic epoch]

if the roots of the quadratic equation x^2 + kx + 12 = 0 are two integers, where k is a constant, write down all possible values of k

[atomic epoch]

not so sure how to approach this

[limber glade]

maybe you can use discrimininant

[atomic epoch]

yeah but i dont know what to do next

[amber schooner]

it says two integers

[atomic epoch]

k^2 - 48 should be a perfect square but im not sure if this results in the whole equation being integer

[amber schooner]

not that the roots are real

[limber glade]

then maybe the quad formula helps

[amber schooner]

also consider the factors of 12

[atomic epoch]

hmm

brb reading class notes

oh i looked it up and it seems that i can find the sum of roots and product of them

cool

it wasnt previously mentioned thats why im stuck ig

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[atomic epoch]

.close

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[soft wharf]

Hey guys please help

It must be liek this

<@&286206848099549185>

Guysss

[soft wharf]

Pls check if my answers are correct

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[severe bough]

It is asking me to find the limit of this function as x approaches infinity. So far, i only simplified the function by making it ^(1/5) but idk what to do next

[south plume]

you can make several approximations

$\frac{2x\left(2x^{3}+1\right)^{2}}{\left(3x^{2}-2\right)^{2}\left(8x^{3}+5\right)} \approx \frac{2x\left(2x^{3}\right)^{2}}{\left(3x^{2}\right)^{2}\left(8x^{3}\right)}$

[clever fjord]

artemetra

[south plume]

the limit would be the same

[south plume]

and this will simplify very nicely with all the x's

[severe bough]

ok so from here we multiply everything. numerator becomes (8x^7) and denominator becomes (72x^7). from here x7 goes with x7 and the limit will be 1/9?

[south plume]

yep

except root 5

$\sqrt[5]{1/9}$

[clever fjord]

artemetra

[severe bough]

ohhh yeah that will be the actual limit. much thanks!

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[south plume]

what have you tried?

can't you substitute gamma into the equation

you start with $\gamma \mathbf{X}$

[clever fjord]

artemetra

[south plume]

and through a chain of equalities get to Z

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[rough badge]

Hello, I've been stuck on this physics problem for 2 days now I have no idea what I'm doing wrong. The only solution I can think of is to get the velocity of the falling rain (-9j) and the velocity of the car (19.4i) and use cotan(o/a) to get the angle but apparently that's wrong

[rough badge]

Ive also tried all sorts of stuff like subtracting my answer from 90, 180, 360 but nothing works i feel like i've hit a wall

[mystic saffron]

i have smthg

@rough badge

still need help ?

[rough badge]

yeah still havent figured it out

[mystic saffron]

ok lemme write it ill send a pic

[hollow quarry]

24.9° would be the angle between the relative horizontal velocity of the rain and the relative velocity of the rain right?

[mystic saffron]

since all the velocities are constant, u can take dt as u want

[mystic saffron]

on the pic its written that it is not this answer

[rough badge]

i have it written as the angle between the velocity of the rain relative to the car and the velocity of the car relative to the ground

because the velocity of the rain relative to the car forms the hypotenuse and the velocity of the car forms the adjacent side, which is why I used cotan

wait that would be secant

[mystic saffron]

try 57.3 pls

[rough badge]

i used cotan with the velocity of the rain (opposite) and the car speed (adjacent)

[rough badge]

how did you get that?

[hollow quarry]

Not saying that the answer. Just confirming what exactly that measure referred to

[mystic saffron]

arctan((Vcdt)/(Vddt))

[mystic saffron]

with this

[hollow quarry]

In that case, it would have to be obtuse, no?

[rough badge]

why are you using derivatives?

[mystic saffron]

dt is an intrrval of time

u can take dt=1

[rough badge]

oh gotcha

[hollow quarry]

The horizontal components of those velocity are in opposite directions

[mystic saffron]

or whatever

[mystic saffron]

?

[rough badge]

didnt work

[rough badge]

i understand what you mean

[mystic saffron]

auugghhh

[rough badge]

with the dt

[hollow quarry]

Have you tried 180 - 24.8... ?

[rough badge]

yeah didnt work

[hollow quarry]

And are you sure it's the number that's the issue

[rough badge]

yeah its the number

[hollow quarry]

So the format, {number} deg is correct?

[rough badge]

yeah and it has 3 sig figs so 24.9 deg would be right if the number was right

[hollow quarry]

Hmm

[mystic saffron]

i tried the other angle of the triangle and i get 28.8 too

[hollow quarry]

Pretty sure the angle is this anyway