help-27

[novel goblet]

Hi. May I know if two coins are flipped at the same time. What is the probability of consecutively getting 3 times of "Head & Tail" in the first 3 trials?

I know the probability of getting "Head & Head" = 0.25, "Tail & Tail" = 0.25, "Head & Tail" = 0.5.
But am I correct if I just use 0.5^3?

[wicked turtle]

yes, that is correct, because each time you do the two-coin flip you get an independent outcome

[novel goblet]

I tried to draw out the possible outcomes where HT = "Head & Tail", HH = "Head & Head", TT = "Tail & Tail"

Possibility 1: HT HT HT
Possibility 2: HT HT HH
Possibility 3: HT HT TT
Possibility 4: HT HH HT
Possibility 5: HT HH HH
Possibility 6: HT HH TT
Possibility 7: HT TT HT
Possibility 8: HT TT HH
Possibility 9: HT TT TT
Possibility 10: HH HT HT
Possibility 11: HH HT HH
Possibility 12: HH HT TT
Possibility 13: HH HH HT
Possibility 14: HH HH HH
Possibility 15: HH HH TT
Possibility 16: HH TT HT
Possibility 17: HH TT HH
Possibility 18: HH TT TT
Possibility 19: TT HT HT
Possibility 20: TT HT HH
Possibility 21: TT HT TT
Possibility 22: TT HH HT
Possibility 23: TT HH HH
Possibility 24: TT HH TT
Possibility 25: TT TT HT
Possibility 26: TT TT HH
Possibility 27: TT TT TT

Why the probability isnt 1/27?

[wicked turtle]

because those possibilities don't all have the same probability

[next granite]

Because these scenarios don’t have equal probabilities of happening

[wicked turtle]

possibility 1 is the most likely because each HT is more likely than HH or TT

if you want to do it that way you have to distinguish between HT and TH, then all four of HH, TT, HT, and TH will have the same probability (1/4) and there are 64 different possibilities for three trials, instead of 27

[ruby citrus]

this is definitely an overcomplication of the problem. there's no need to do this!

each of the three trials is independent, correct?

the original solution of 0.5^3 was right

[wicked turtle]

I assume he was hoping to cross-check that solution with a more brute-force approach and was confused why they didn't agree

[novel goblet]

That's right

[wicked turtle]

if you distinguish between HT and TH then there are 64 total possible outcomes which are equally likely, of which 8 consist only of HT and/or TH

so that gives a probability of 8/64 = 1/8, consistent with the other argument

[novel goblet]

I see

[novel goblet]

If all the probability of all the 27 possibilities here are added up together I will get 1.00 right?

[wicked turtle]

yes for sure

just that they aren't all equal to 1/27

[novel goblet]

Okay, I understand now, thanks everyone

.close

[devout snow]

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[restive river]

I need to find the area of the surface obtained by rotating the curve about the x-axis. However, I don't know how to solve the integral 😦
Sorry if I'm not clear, English is not my first language.

[devout snow]

@restive river Has your question been resolved?

[wicked turtle]

sen = sine?

[restive river]

yep

[wicked turtle]

what happens if you substitute $u = \pi \cos(\pi x)$?

[woven radish]

OurBelovedBungo

[restive river]

Mmm

du = cos(pix) + xsin(pi*x)*pi dx

?

No

Wait

xd

du = -sin(pi * x) * (pi)^2
Right?

[devout snow]

@restive river Has your question been resolved?

[narrow plank]

@restive river Try writing this out neatly on a sheet of paper and show us where you are

[restive river]

Okay

[restive river]

Is it clear?

[narrow plank]

Looks good to me - don't forget the bounds when you make the trig sub

[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[timber pebble]

so working on $\int _C z^{\alpha -1} \dd z$ along the unit circle

[woven radish]

jan Niku (Shuri for Honorable)

[timber pebble]

my instructor has told us to place a branch cut at $theta_0$ and integrate from $\theta _0 \leq \theta \leq \theta _0 + 2 \pi$

this doesnt make sense right

[woven radish]

jan Niku (Shuri for Honorable)

[timber pebble]

since were gonna land on the branch cut twice

[pseudo basin]

no, we start at the cut and end at the cut

but never impale ourselves on it

[timber pebble]

okay

so its just 0 then

unless some weirdness happens in the non integer case

ill have to check

does it

this should be fine right

i dont understand the point of this problem

[timber pebble]

any chance you can confirm there is nothing special about this problem the way it is stated im sorry to ping

im just anxious im missing the point

[placid rover]

its not 0

[timber pebble]

? why not

[placid rover]

You can imagine something like this

(this is only an analogy)

[timber pebble]

oh well

[placid rover]

We start at one floor

as we move along it

[timber pebble]

i mean i say its 0 because this is what our teacher gave us in class

[placid rover]

we will end up up a floor

eh?

oh wait you had alpha - 1

[timber pebble]

this problem has a note to refer to an example we did in class

which states its 0 if alpha is an integer

but then

[placid rover]

if alpha = 0

i dont think its 0.

[pseudo basin]

$\int_{\theta_0}^{\theta_0+2\pi} \exp(i(\alpha-1)t) ie^{it} \dd{t}$

[woven radish]

Ann

[timber pebble]

alpha cant be 0

sorry did i forget to write that

[placid rover]

yes that is quite very important

[timber pebble]

sure

[placid rover]

funny things happen when alpha = 0. But since it isn't then ^

[timber pebble]

so we should get to

$\int i e^{i \alpha \theta} \dd \theta$

[woven radish]

jan Niku (Shuri for Honorable)

[pseudo basin]

from theta_0 to theta_0+2pi

[timber pebble]

yea

and since i guess its okay to be on the branch cut as long as we dont cross it

which im still nervous about

but then this is zero

why is it special if alpha is an integer or not

[pseudo basin]

is it really zero

[timber pebble]

yea

[pseudo basin]

the antideriv is $\alpha^{-1}e^{i\alpha \theta}$

[woven radish]

Ann

[timber pebble]

i mean if by the whole we can be on the branch cut thing then $e^{i \alpha \theta } = e^{i \alpha (\theta + 2 \pi}$

[woven radish]

jan Niku (Shuri for Honorable)

[timber pebble]

it gets eaten

by integration

so we stack a bunch of curves

oh is the problem that

so wait if alpha isnt an integer

we are racking up values

[placid rover]

I'm confused what you're describing

[timber pebble]

since alpha isnt analytic

err

f isnt analytic

[placid rover]

I would rather talk about this in terms of the algebra

[timber pebble]

on the entire inner region

[placid rover]

Since it is relatively simple. . .

[timber pebble]

what algebra?

[placid rover]

You convert to polar form.

[timber pebble]

sure

[placid rover]

Then integrate as shown above.

[timber pebble]

right

you get to $\frac 1 \alpha e^{i \alpha \theta} \bigg \vert _{\theta _0} ^{\theta _0 + 2 \pi}$

[placid rover]

You apply the fundamental thm of calculus

[woven radish]

jan Niku (Shuri for Honorable)

[timber pebble]

right but

i think its different

[pseudo basin]

no

[timber pebble]

depending on values of alpha

[pseudo basin]

thats the point

at the end you will have $\alpha^{-1} e^{i\alpha\theta_0} (e^{2\pi i \alpha} - 1)$

[woven radish]

Ann

[placid rover]

(Now I'm confused --- is this thing not 0, as long as alpha neq 0 ?????????????????)

[timber pebble]

huh?

[placid rover]

Thisis the antid

And we evaluate it at theta = 2pi

and evaluate it at theta = 0

and subtract? or did I miss something

[timber pebble]

wwell not necessarily 0

i do get there are problem alphas

[pseudo basin]

it is 0 precisely when α ∈ Z \ {0}

[timber pebble]

i dont understand the formula ann posted though

wait but

[placid rover]

Yes. Jan clarified alpha not 0

halfway through

[timber pebble]

what if alpha is 3/2

[placid rover]

[pseudo basin]

we evaluate at $\theta = \theta_0$ and $\theta = \theta_0 + 2\pi$

[woven radish]

Ann

[timber pebble]

idk if this is standard notation but

$\int _{C_1 (0)} \sqrt z \dd z$

[woven radish]

jan Niku (Shuri for Honorable)

[timber pebble]

is uhh

this isnt 0 right

hold on

[placid rover]

should be, im pretty sure

[timber pebble]

man im confused

[pseudo basin]

$\oint_{|z|=1} \sqrt{z} \dd{z}$

[woven radish]

Ann

[pseudo basin]

but yes this depends on where you take your branch cut

[timber pebble]

say its convenient

well i guess we cant say anything

so this might not be 0

[placid rover]

Proceed as above with polar 🤔

[pseudo basin]

i am doing exactly this btw

[timber pebble]

oh so

it is 0

but then why do some places say its not

this stuff is confusing as heck

so the point of the problem is it is always 0

and theres nothing special about alpha being an integer or not

[placid rover]

we are interested in the integer case because it is relevant to series expansions

[pseudo basin]

hold up what

are they really saying its always 0

[timber pebble]

no one is saying anything

im trying to understand what the answer is

[placid rover]

That aside - branch cuts are certainly relevant

[timber pebble]

i mean you can find something like this

but this doesnt seem to align with me just doing it and feeling like its 0 but then

the branch cut should always be an an okay place right

it doesnt matter where theta _0 is

just that it connects branch points

which it will so fine

[timber pebble]

its not always 0?

im trying to understand this

[pseudo basin]

yes! it's not always 0!

[timber pebble]

but were at the same place we were before

?

[pseudo basin]

we came full circle

[placid rover]

For a visualisation, I would recommend of thinking of 2 separate argand diagrams
One represents z. The other represents f(z) (in this case,its z^(alpha-1)).

[pseudo basin]

around a point-where-some-fuck-shit-happens

[placid rover]

To start with, an easy alpha would be like 3, 4.

f(z) = z^2 means as z goes around the unit circle, z^2 goes around twice

Then to have an idea of what the integral is... you have to sum up f(z) dz

Once you get your head around that (I would draw a small dz to check what f(z) dz looks like), consider what happens when alpha is non-integer

Hopefully you see the cancellation occuring when alpha is integer (not 0), but not in the other case. There will be some 'extra' that doesn't cancel

[devout snow]

@timber pebble Has your question been resolved?

[timber pebble]

sorry driving will read

[placid rover]

np. And my explanation wasn't the best (I do not have a crystal clear picture in my head atm, myself), but hopefully it makes sense

I remember justifying what these integrals looked like geometrically before using this. (later, for the alpha = 0 case, I think a geometrical understanding of the answer is v. helpful)

[timber pebble]

i believe i think

[devout snow]

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[jade halo]

Can someone help me on number one

So far this is what I have

[winged hazel]

not really, the picture is insanely low quality, pretty much illegible lol

[jade halo]

Oops

The question is asking for the linearization L(x) of the function at a

And the equation is

$f(x) = (x^3)-(x^2)+3$

[woven radish]

c0nc3ptsz

[winged hazel]

What are you stuck with? You have what it is written on the third line at the left

just sub in your values of f(-2) and f'(-2) and simplify

[jade halo]

Yeah sorry I was lost on the (x-a) part

But I just solved it after watching a video 🙂

Probably should’ve started off with that before asking for help lol

Sorry about that

!!

.close

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[native pike]

G is the center of CE as well

[supple knot]

can you repost the question again

what are you solving for or trying to prove?

[native pike]

prove that G is the intersection of AF and DB

from what i understand, this isn't a geometry question

we're supposed to go full algebra with vector calculations

[supple knot]

You're just supposed to prove G intersects lines AF and DB? Or that their intersection bisects line CE?

[native pike]

we do know that it bisects CE

we need to proof that G is the intersection of AF and DB

[supple knot]

ooh that's harder

[native pike]

i got a hint

but still didnt know where to go from it

1 sec

[devout snow]

@native pike Has your question been resolved?

[native pike]

i WISH

[devout snow]

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[supple knot]

Use this command

[devout snow]

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[native pike]

.reopen

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[native pike]

.reopen

[mortal tusk]

Hey I needed help on how to Evaluate the Riemann sum for: (x) = x − 6x, for 0 ≤ x ≤ 3
This has 6 subintervals, and I need to take these sample points from the right endpoint

[mortal tusk]

I have the first few steps completed I believe, but I think I got my answer "incorrectly" if that makes sense

I got the right values, but I got to it with the wrong method

[graceful cosmos]

0.5[f(0.5) + f(1) + f(1.5) +...

[mortal tusk]

Yeah i got that part correct

Bc of 6 subintervals and 3 is maximum, 3/6 --> 0.5 intervals

up to 3

The areas were verified to be correct, but the method of getting them was wrong?

also the sum of the area needs to be 3 decimal places, but i got 26.25

[devout snow]

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[graceful cosmos]

Are you sure x - 6x is typed correct? Odd way to write that

[devout snow]

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[mortal tusk]

.reopen

[devout snow]

✅

[mortal tusk]

Also yes

That was the practice question wording, not mine

[devout snow]

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[devout snow]

@mortal tusk Has your question been resolved?

[devout snow]

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[restive river]

can someone help explain to me the continued ratio thing it makes no sense to me

[devout snow]

@restive river Has your question been resolved?

[dusty violet]

ok so

a:b = 3:5

b:c = 4:7

so now, if you notice

we may also think of ratios as division

a/b = 3/5
b/c = 4/7

in our first equation, we get 1/b = 3/5a => b = 5a/3

b/c = 4/7 => b = 4c/7

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[ancient fractal]

I’m having trouble with a proof, it says prove by contrapositive that if (x-2)^2 is odd then x is even

[ancient fractal]

So the contrapositive would be if x is odd then (x-2)^2 is even right

however when you sub in x = 2n+1 you get (x-2)^2= 4n^2 -4n+ 1 which is an odd integer?

[placid rover]

(3 - 2)^2 = 1

there seems to be an error indeed.

[ancient fractal]

Maybe the question is asking to prove the statement false? have you ever seen something like that in an exam

this is an exam from last year i’m not currently in an exam situation

[placid rover]

the question is just erroneous.

[ancient fractal]

Ok thanks

.close

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[shrewd raft]

Hey, my question is: Can the area of ​​a quadrilateral be calculated if only the four sides are given? If so how?

[copper mango]

Probably

[placid rover]

Likely not.

Imagine 4 sticks whose ends are joined together. In most cases you can deform the area.

[graceful cosmos]

No. Think parallelograms. Squash them to have near 0 area

[placid rover]

^

[copper mango]

I'm pretty sure it's implied non-degenerate quadrilateral.

[serene mesa]

Ruler and determination

[shrewd raft]

Under what conditions can I decide whether a quadrilateral exists if only four sides are given? (such as triangular inequality)

[restive river]

.

so what do you do with the values

b = 5/3a and b = 4/7c

[devout snow]

@shrewd raft Has your question been resolved?

[devout snow]

@shrewd raft Has your question been resolved?

[placid rover]

no but that's not the point

The area varies as you squash

and goes to 0

The area varying is the issue

[devout snow]

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[coarse current]

can someone pls write me a grade 7 GCSE maths question

Non calculator pls

[coarse current]

I need practice

[shadow quest]

What is your question

I could help you

[coarse current]

I mean can u write me one to solve and if I get it wrong can u help me through it please

[shadow quest]

Dont you have homework from your teacher?

[coarse current]

Nope

He's taught us less than half the topics on the GCSE test we sit in 2 months

[shadow quest]

What is a gcse exam ?

[coarse current]

So I'm doing all my maths at home

[coarse current]

uh the ones you do at 16

The end of high school ones in the uk

[shadow quest]

you want me to invent an exam for you ?

[coarse current]

Nope just one question tbf

So I can see if my general maths is decent at all

[shadow quest]

Do you have a manual ?

[coarse current]

nope?

[shadow quest]

ugh i really dont know how i can help you

sorry

[coarse current]

It's ok

I need practice

But idk what to practise

[shadow quest]

You should have a manual or something

Homework or something

[coarse current]

my teacher is terrible

End of high school exams in 2 months

[shadow quest]

Does he give you exercises ?

[coarse current]

Top set / highest class in the final year

Averaging D grades

[coarse current]

Rarely

[shadow quest]

Ok

Well

Find a manual online and solve their questions ig

[winter patrol]

you should be able to find past papers

[devout snow]

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[devout snow]

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[tawdry tiger]

x=-b/2a
what if b is already negative, would it switch to positive?

[copper mango]

yes.

[tawdry tiger]

tysm!

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[young birch]

how do you curve sketch this function using calculus

[glass mauve]

quotient rule

do you know variation table?

[leaden aurora]

Firstly

You figure out whether the curve is undefined

then look for any zeros

then any oblique asymptotes

then minimum/max

[glass mauve]

Isn't that what a variation table is?

[young birch]

ye i know how to find the derivative and stuff

[leaden aurora]

what's a variation table

[young birch]

i just forgot what a function with the numerator one power over the denominator looks lie

[leaden aurora]

never heard of that

[young birch]

what is the asymptote

is it called oblique asymptote?

[steel sage]

It’s an ellipse rotated

[young birch]

answer says theres an asymptote at y=x

but i forgot what kind of asymptote thats called

[leaden aurora]

y=x is oblique

[young birch]

oh

so all functions with the numerator one power above the denominator have an oblique asymptote?

[leaden aurora]

To find oblique asymptotes of functions of this sort

you carry out long division

you will get a quotient function

and a remainder

[steel sage]

Sorry, hyperbola

[young birch]

hUh

is that grade 11 math sorry

[steel sage]

Just Diagonalize
(1 -1/2
-1/2 0)

[leaden aurora]

bro

[steel sage]

Oh..

My bad

[leaden aurora]

why do u think he has learned matrices

in grade 11

[steel sage]

Then ignore it

[young birch]

i was just told to sketch the function using first and second derivative

just have a question about the symptote

asymptote

[leaden aurora]

To find asymptote, carry out the divison. You will get a function of the form $(x-a) + \frac{b}{x-1}$

[woven radish]

azeem321

[leaden aurora]

As x approaches positive and negative infinity

[young birch]

are you talking about this kind of division

[leaden aurora]

Whatever method you know to divide polynomials will work

[young birch]

ight

[leaden aurora]

You can also do $x^2-3x+6= (x-1)(x-a) + b$ and compare the coefficients

[woven radish]

azeem321

[young birch]

so what is the answer in this case

[glass mauve]

why are you doing calc ingr 11

[young birch]

cause im in the ib program

😔

[leaden aurora]

i think only in US they do calc at college

in eu we do calc in school

[glass mauve]

The easiest way is to take first derivitve and second derivitive and set up a variaton table

[young birch]

bro why is everyone saying variation table

is it like a sign chart?

are you american

[leaden aurora]

Jub, what are you even confused about at this point

[young birch]

why theres a diagonal asymptote

i learned it in grade 10 but i forgot

i just wanted to refresh my memory

[glass mauve]

yes

[young birch]

oh

[leaden aurora]

An asymptote occurs when the function approaches a line but never crosses it

well it can cross it

[young birch]

yes

[leaden aurora]

but dont worry about that for now

[young birch]

nvm i found it

tysm for your help :D

.close

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[rugged shuttle]

.reopen

[sturdy chasm]

How do I find the convergence of something with 2 alternating series

[deft wagon]

you know about power laws?

[sturdy chasm]

im stupid

thanks

.close

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[deft wagon]

np

[sturdy chasm]

wait

is -1/n a harmonic series still?

or not

[devout snow]

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[placid narwhal]

The dragon has 2010 heads. A knight can cut off sections 2, 17, 21 or 33
heads, but after that the dragon grows new 9, 10, 0 or 47 heads in a row. Can at some point the knight cuts off all the dragon's heads?

[placid narwhal]

import random

heads = 2010

while heads > 0:
	heads -= random.choice([2 , 17 , 21 , 33])
	heads += random.choice([9 , 10 , 0 , 47])
	
print("dragon dead")

Judging by code it is

But how do you prove it

Mathematically

[polar bolt]

that is not what it is asking for is it?

[placid narwhal]

I think it is

[polar bolt]

its asking if the dragon can have the exact amount of heads left that the knight can cut off

[placid narwhal]

No its just asking if at one point dragon can loose all heads

[grizzled lantern]

do you have the original question?

[placid narwhal]

This is it

[polar bolt]

doubt it, but proof is dragon can grow 0 with non zero prob

[placid narwhal]

Edited a little so it makes sense translated

This is un edited

The dragon has 2010 heads. A knight can cut off sections 2, 17, 21 or 33 with one blow of a cat
heads, but after that the dragon grows new 9, 10, 0 or 47 heads in a row. Can u
At some point the knight cuts off all the dragon's heads?

Google translate is really bad

[polar bolt]

the answer is trivially yes if you read it like that, but I doubt this is the intended question

[grizzled lantern]

alternatively the reasoning could be that the expected value of heads cut of is greater than 1

take the avg of heads cut off and heads grown back since the are equally likely

[polar bolt]

that doesn't matter

[placid narwhal]

I mean i can manually translate it .

Dragon has 2010 heads. Knight can with 1 hit cut 2 , 17 , 21 or 33 heads , but after that dragon grows 9 , 10 , 0 or 47 heads in a row. Can knight at any point cut all dragons heads.

Maybe the catch is he grows like 1st round 9 then second 10 then 0 then 47

[polar bolt]

why would the Knight even have options then?

[polar bolt]

.

[grizzled lantern]

I thought they were all equally likely random events

[polar bolt]

EV doesn't matter

[grizzled lantern]

why not

[polar bolt]

non zero-prob for event to happen where heads cut off+regained<0

so it will happen infinite amount of times

hence it will happen that dragon has less than 0 heads at some point

[placid narwhal]

I guess that could be the catch so the dragon must have left 2 , 17 , 21 or 33 heads

[polar bolt]

like I was saying, yes

[grizzled lantern]

isn't that basically EV

[polar bolt]

no

EV of cut+regain dragon could be +10 heads

if one option is 0 then it will happen that the dragon loses all heads

as turns approaches inf

EV doesn't matter

[placid narwhal]

So i must prove that by adding 2 , 17 , 21 and 33 and subtracting 9 , 10 , 47 i can somehow get 2 , 17 , 21 or 33

From 2010*

[polar bolt]

Swap subtract and add

And yes

[placid narwhal]

Oh ye

2010 , remove 2000 (2 * 1000) we get 10 add 47 remove 30 (2 * 15) and remove 17

Would that work?

I mean chances are low but that can theoretically happen

[steel sage]

If you are asking for a possible situation , then yes, cut 2 grow 0 cut 2 grow 0…
But if the dragon want to live then no matter how knight does he grows 47 heads

[devout snow]

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[gritty comet]

does rank need to equal number of rows for it to have a unique solution

[gritty comet]

or is it just rank = number of columns

for a matrix

[proud notch]

rank = number of linearly independent rows = number of linearly independent columns

[gritty comet]

for example does the matrix 1 0
0 1
0 0

not have a unique solution if it was a homogenous system

@proud notch

what does linearly independent mean

[shadow terrace]

https://en.m.wikipedia.org/wiki/Linear_independence

It explains it in definition

[gritty comet]

@shadow terrace could you take a look at the example matrix I sent

[shadow terrace]

Could you rephrase your question please

[gritty comet]

i'll make it better

one sec

[shadow terrace]

I dont know too much linear algebra although (1, 0, 0) and (0, 1, 0) are linearly independent

[gritty comet]

say I have this as my homogenous matrix the number of rows is greater than rank but number of columns is equal to the rank

so does this matrix only have a unique solution

or is it infinite solutions

[shadow terrace]

I believe it would need a non zero determinant an afaik the row count must match the column count for determinant to be calculable

Could you elaborate what you mean by unique solution?

[devout snow]

@gritty comet Has your question been resolved?

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@gritty comet Has your question been resolved?

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[warped plaza]

Can somebody help me with this problem please?

[supple knot]

Did you already use a u-sub?

[devout snow]

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[restive river]

A 10-digit number M is 1 more than a square of another integer. Is it possible for all digits of M to be distinct?
could i solve this with mod 4?
M is either 1 or 2 mod 4
dont know if it helps
what other approaches could i take?

[restive river]

anyone pls

[devout snow]

@restive river Has your question been resolved?

[restive river]

bruh

[restive river]

.close

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[rigid lodge]

What would be the equation for something like this?

I'm kinda neo detergent and forgot to write it down if it came up before

It's probably something obvious

Mean is 60, StanDev is 5

They give me some ones I can practice that I try and "reverse engineer" but idk about this one

I've been trying all kinds of funky combos

[rigid lodge]

Like this one example that lets me see the answer, but whatever I try I can't seem to replicate it

This one guide seems to be wrong?

With the fake example, mean=90, StnDev=5

85-90/5 is -1

And 1 with 95

Not sure where I even go to get something with three decimal places

<@&286206848099549185>

[devout snow]

@rigid lodge Has your question been resolved?

[rigid lodge]

Is there some kind of division trick I'm missing or some ^2?

Supposedly there's some kind of table I can check? But I can't find it

[warped plaza]

Not yet😩

[devout snow]

@rigid lodge Has your question been resolved?

[devout snow]

@rigid lodge Has your question been resolved?

[devout snow]

@rigid lodge Has your question been resolved?

[ruby citrus]

if you're still wondering what the answer is, (if you have the right type of calculator) you can use the normalcdf function with the arguments of mean, standard deviation, lower z-score and upper z-score to get the probability value you're looking for
Otherwise, your only feasible option is to use a lookup table or memorize ncdf values at certain standard deviations/z-scores

[copper fossil]

is some work

[devout snow]

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[supple knot]

.close

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[muted blaze]

I need help with this problem

[exotic stump]

what have you tried

[tall talon]

recall some properties of parallelogram (or just parallel lines in general)

[devout snow]

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[desert field]

I need help on a calculus problem ,

[desert field]

this is all i have so far

[near trout]

let the side length of the triangle be s

find the area of the triangle in terms of s

then find the perimeter of the square in terms of s

from the perimeter you can get the side length, and so the area of the square as well

then add the area of the triangle to the area of the square to get total area

and find s for which the total area is minimised

[desert field]

ok lemme try

[devout snow]

@desert field Has your question been resolved?

[desert field]

ok i found an equation what’s the next steps

<@&286206848099549185>

<@&286206848099549185>

.close

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[rigid lodge]

@copper fossil sorry for the tag just hoping for some clarification

[rigid lodge]

How did you end up with the circled numbers?

[copper fossil]

the total probability of anything is 1

and that table does right tail probabilities

[devout snow]

@rigid lodge Has your question been resolved?

[rigid lodge]

So a different example, same mean and StnDev, 45 and 75

-3 and 3 respectively

I just find "3.0" on that table, which is apparently .9987?

.9987 - .9987 - 1 - = -1

Which doesn't feel right

[near trout]

should be 1 - 0.9987 - 0.9987

for the area not in either of the tails

[rigid lodge]

-0.9974

[near trout]

no

[rigid lodge]

Not sure if negatives are allowed lol

[near trout]

0.9974

1 - 0.9987 - 0.9987 is positive

wait i'm a fool

ok look

ok for z = 3, the blue area here is 0.9987

so the white bit is 1 - 0.9987 = 0.0013

now, imagine there were two white bits

this is called a two-tailed case

like, one white bit on each side

mirror the white bit across the middle vertically

so then the remaining blue area would be 1 - 0.0013 - 0.0013

ignore what i said before, this is it

[rigid lodge]

Oh yeah on the page above there was something I missed too I guess

That sneaky -3.0 and .0013

Very good work @near trout

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[heady ibex]

the sample size is 35, mean is 7.5 and the variance is 1.75 right?

[ivory tapir]

Yeah your sample will have a SD

And your mean will vary by the SD

Er sample mean

[heady ibex]

i still use z score?

[ivory tapir]

Yes but you use the sample SD

[heady ibex]

how can i use that to answer the questions?

[ivory tapir]

You’re comparing your population to your sample mean

What is your sample SD

It’s a function based on n

[heady ibex]

is the square root of 1.75

?

[ivory tapir]

No

Google it real quick

[heady ibex]

ohhh this

right?

[ivory tapir]

Yeah

The top is the variance

Sd/sqrt(n-1)

[heady ibex]

so variance over number of trials -1

sample size*

[ivory tapir]

Yeah

[heady ibex]

and after that we use z score?

[ivory tapir]

Yep

[heady ibex]

got it 👍 thank you

i will solve it right now

[ivory tapir]

I’m still at work so I can’t verify your solution

[heady ibex]

np i think i have the solutions for this question

[devout snow]

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[agile swan]

Can anyone tell me is that true?

[agile swan]

what if $ m != p $

How can this property be even valid as per stated in the above?

[wicked turtle]

why do you need m = p?

AB is defined if A is m x k and B is j x p as long as k=j

i.e. the number of columns of A equals the number of rows of B

[graceful cosmos]

I think you're correct, that you can't have both AB and A'B' unless m = p here

[wicked turtle]

the result will be m x p

[wicked turtle]

where does it mention A'B' though?

[graceful cosmos]

Where ' = transpose because lazy

Oh whoops yeah that's B'A'

Yes then m ≠ p seems possible

[wicked turtle]

yeah it's still true if m ≠ p. then AB will be rectangular of size m x p, and B'A' will be rectangular of size p x m

[agile swan]

Oh wait wtf....

I just noticed that the formula is B^T * A^T...

not A^T * B^T

well, in that case, it will form: p x m

.close

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[drowsy field]

Hey!

[drowsy field]

I feel kinda dumb asking this because I am bad at math

But let's go

imagine the scenario:

You have a world with 15000 cells in it

At water level 50 the amount of land cells is 12214 (81% of the world)

At water level 150 the amount of land cells is 2374 (16% of the world)

How we can deduce from this data what would be the ideal water level for the world to have 30% of it as land?

You know, it is supposed to be a deduction having these 2 data as a parameter, it will probably not be an exact result

Please help 😭

<@&286206848099549185>

This is simple!!! I need halp

[devout snow]

@drowsy field Has your question been resolved?

[drowsy field]

😭

<@&286206848099549185>

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[cosmic junco]

<@&286206848099549185>

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[restive river]

for this question, my ratio test expression looks ridiculous, and I don't know how to evaluate the limit I have from my root test

[analog trellis]

Since n is positive couldn't you just say let epsilon = 1/n > 0.

[restive river]

what

[restive river]

but there is no epsilon where epsilon < 1/n for each n

hence I can't do a comparison test

[analog trellis]

I mean that the question says it is convergent for each fixed value of epsilon. For n = 1, 2, 3, 4, ..., the epsilon can be found in a sense and all are positive and all will therefore converge. I think the problem wants you to use this fact otherwise it is telling you it for no reason.

[restive river]

how can I rigorously present this?

because I can't just say it works for the first finite terms

[analog trellis]

Well 1/n = epsilon > 0 for all n but I'm not sure how to produce a textbook answer since I learnt stuff like this on Paul's online notes a while ago and those aren't very rigorous.

Maybe you can use the comparison test instead with the given series and just present an appropriate relation of n and epsilon to make it smaller than the convergent one.

[restive river]

then does the same sequence, but replace 1/n with k and let k -> inf, diverge?

[analog trellis]

Try and produce

[split shoal]

I'm pretty sure the series is supposed to diverge. You can prove it using the limit comparison against 1/n.

[analog trellis]

It's just my idea I can't do this problem in a way I'd be happy with.

[split shoal]

In taking the limit comparison, we will get $n^{(1/n)} = e^{\ln n/n}$. Since $\ln n/n \rightarrow 0$, the limit tends to 1.

[woven radish]

PhenomPlasma

[restive river]

my brain is dead

one momebt

[split shoal]

Are you allowed to use the limit comparison test?

[restive river]

yea

sure

go for it

I regret taking this class LOL

but wasn't much of a choice to start with

ohh

I get it now

that's nice!

[split shoal]

Aye, nice.

[restive river]

I'll write it out and send it here

ah

you wrot eit

[split shoal]

That's what I meant. Was a bit hard to type.

[restive river]

no I got it!

thanks!

[split shoal]

No prob!

[restive river]

this works well for me!

[analog trellis]

I wonder the error in my logic. :/

[restive river]

@analog trellis thank you for trying too

[restive river]

your logic was quite based on intuition

it's not rigorously presented

but thanks for trying

apprecuate it

[analog trellis]

I think it's workable but there's a problem with logic after the start. I'll have to explore it further.

[split shoal]

The error is that you can always make 1/n less than any epsilon by making n large enough.

[analog trellis]

This is what I was thinking something like this was the error.

[split shoal]

Yeah, because of that, you can never prove 1 + 1/n to be more than 1 + epsilon for any fixed epsilon.

[analog trellis]

It would require epsilon < 1/n. and this means as n goes to infinity that epsilon < 0, a contradiction of the starting hypothesis.

Probably a error in that. I'll think it through more comprehensibly.

[split shoal]

Something like that, I guess. An intuitive argument is that for the series to converge, we need 1/n > epsilon for some positive epsilon. But since 1/n converges to 0, we can make 1/n as close to 0 as we'd like, which means we can make it closer to 0 than epsilon.

Hence, such an epsilon cannot exist.

[analog trellis]

I guess it kind of shows it has to diverge then. Thanks! 🙂

[restive river]

.close

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[pallid schooner]

Could anyone help me out? How do I make a formula out of this (and other) quadratic relationships? (I have a math test tomorrow and don’t understand this)

[devout snow]

@pallid schooner Has your question been resolved?

[pallid schooner]

<@&286206848099549185>

[main gull]

https://youtube.com/watch?v=rXumPPmfb_8

That should help

[pallid schooner]

Thanks dude

Seems clear, I’ll close it

.close

[devout snow]

Channel closed

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[balmy aspen]

I need help solving this

[exotic stump]

what have you tried

[balmy aspen]

The first one

[exotic stump]

what did you try

[balmy aspen]

Dividing it?

[exotic stump]

are you attempting to simplify logarithmic expressions without knowing log rules

[balmy aspen]

Oh you mean log rules

[balmy aspen]

I’ve tried the power rule

[exotic stump]

you're gonna need more than that

[balmy aspen]

Any help?

[exotic stump]

google log rules

[balmy aspen]

Quotient rule?

?

Sigh

[rustic epoch]

hey u need help

?

[balmy aspen]

Nah I finished it

[rustic epoch]

alright

if u need further help lmk

[balmy aspen]

.close

[devout snow]

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[ember shard]

Need gep

[ember shard]

Help

Not sure what its asking me to do

<@&286206848099549185>

[rustic epoch]

just plot the points

on the graph

[ember shard]

So

Plit all the coordinates

All right

[rustic epoch]

yes

[ember shard]

This doesnt look quite right

@rustic epoch

[rustic epoch]

you could have drawn it better

try doing it again on another graph

[ember shard]

Nah but still

How is that symetrical

It wouldnt be a good curve either

[rustic epoch]

it doesn't have to be

not all curves u see are perfect curves

[ember shard]

.close

[devout snow]

Channel closed

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[devout snow]

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[restive river]

I have the formulas log_a(1/2) and log a(2) for the half time and doubling time of a number. How do I utilize these in a question?

[restive river]

.close

[devout snow]

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[hollow comet]

I need help

[hollow comet]

Sry idk what happened

[devout snow]

@hollow comet Has your question been resolved?

[hollow comet]

.close

[devout snow]

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[long hatch]

can i please have some help on this?

[long hatch]

i know,

[novel goblet]

because both terms have x

[long hatch]

can i have an explaination as to why?

[torn vessel]

factor out x

[novel goblet]

so you just need to factor out x

[long hatch]

oh i see

[torn vessel]

$(x+4)(x-4)=x^2-16$ but you forgot about the $x$ at the end

[woven radish]

🙛𝕍ѳrtєx🙙

[long hatch]

oh

how do i solve sometyhing like this in the future?

[torn vessel]

you look for common factors

for example both terms have an $x,$ which means your instinct should be to factor out $x$ and obtain $x(x-16)$

[long hatch]

oh wait i think i got it

[woven radish]

🙛𝕍ѳrtєx🙙

[long hatch]

ok

[torn vessel]

Of course sometimes the first instinct may not be the right method or approach but you should at least see it as an option immediately

[long hatch]

ok

[torn vessel]

👍

gj and gl

[long hatch]

then how would you solve something like this ?

[torn vessel]

Firstly, what's ab+3b factored?

[long hatch]

a4b?

or is that not how it works

[torn vessel]

not exactly

if you have A groups of b and then you add 3 more groups of b you have a+3 groups of b

[long hatch]

no

[torn vessel]

hm?

[long hatch]

4b bc you al;ready have a group

dont u

?

oh wait

yeah

no i read the thing wrong

yeah a+3 groups of b

[torn vessel]

Good, so now you have $a(b+3)-a-3,$ right?

[woven radish]

🙛𝕍ѳrtєx🙙

[long hatch]

yeah

[torn vessel]

Do you agree that's equal to $a(b+3)-(a+3)?$

[woven radish]

🙛𝕍ѳrtєx🙙

[long hatch]

yeah

[torn vessel]

alright, so now what?

[long hatch]

put a into the bracket?

[torn vessel]

wait a minute

[long hatch]

(ab + a3) - (a + 3)

[torn vessel]

i wrote it wrong sorry

[long hatch]

o

[torn vessel]

when you factor ab+3b you should get b(a+3) right?

[long hatch]

yeah that makes sense, but how did u do that?

[torn vessel]

because of the distributive property of multiplication

also if you have A groups of b and another 3 groups of b you have A+3 groups of b in total

[long hatch]

yeah

[torn vessel]

so you end up with b(a+3)-a-3 as your equation which is equal to b(a+3)-(a+3)

following me?

[long hatch]

yep

[torn vessel]

now what does that equal?

[long hatch]

(ab+3b) - (a+3)

does this work?

[torn vessel]

yeah

but it's not factored

[long hatch]

b(a+3)-(a+3)
with this there is are 2 (a+3)s

[torn vessel]

factoring means that you want common factors of terms to be factored out. For example, if you have "ab" and "3b," they share a common factor of b, and so you want to write it as b(a+3) for simplification

[long hatch]

can i cancel them?

[torn vessel]

no, you can't

[long hatch]

oh yeah because b comes first

[torn vessel]

because the Order of Operation states that you multiply

[long hatch]

and you need to times that

ok

[torn vessel]

yep

However, can you see anything else?

try to think in groups, i.e. "B groups of (A+3)" and "1 group of (A+3)"

[long hatch]

b+1 groups (A+3)

right?

[torn vessel]

well, is it + or - between them?

the expression is b(a+3)-(a+3)

[long hatch]

-\

[torn vessel]

so would it be b+1 or b-1?

[long hatch]

b-1

[torn vessel]

right

so you have (b-1)(a+3)

[long hatch]

ohh i get it now

[torn vessel]

for these problems you just want to find different potential groupings of different pairs or terms

[long hatch]

yeah i got it

thank you so much]

[torn vessel]

no problem!

[torn vessel]

oh yeah you can type .close if you don't have any more questions

[long hatch]

.close

[devout snow]

Channel closed

Closed by @long hatch

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[torn vessel]

👍

[long hatch]

tyty

[devout snow]

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[restive river]

Ok I need quick help

[restive river]

so the half time of a material is 55,6 seconds

they are asking when only 3g of it are left (it had 10g to start with) aka they're asking when 30% will be left

so I just do a^55,6=0,5

then I get a

then I simply do f(t)=c*a^t

so I find out that it takes 96,575 seconds for 30% of it to be left

is that correct? am I calculating it correctly?

[distant harbor]

Yeah that should work

So long as a is positive

[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[languid ocean]

how do we calculate the fundamental group of a topological space

[languid ocean]

mainly i just want the details for R^2 \ (0,0)

i know it's Z and the equivalence classes of paths are based on the number of "loops" around the origin (does that make any sense?)

but, well, i still can't prove why 2 paths with different numbers of "loops" are not homotopic

[devout snow]

@languid ocean Has your question been resolved?

[languid ocean]

<@&286206848099549185>

anyone home?

(this can probably just be reduced to proving that a loop where the origin is outside the loop and one where the origin is inside the loop are not homotopic)

[haughty torrent]

you may want to try #point-set-topology

might get a faster answer there

[devout snow]

@languid ocean Has your question been resolved?

[devout snow]

@languid ocean Has your question been resolved?

[steel sage]

First select a base point, WLOG,we let it be (1,0), then we construct a continuous map f from R^2 \ {0,0} to S^1={(x,y):x^2+y^2=1}, f((x,y))=(1/sqrt(x^2+y^2))(x,y)

This derives a homomorphism g from G=π_1(R^2 \ {0,0},(1,0)) to H=π_1(S^1,(1,0))

mapping an equivalence class of a path [λ] to an equivalence class of a path [f•λ]

You need to proof that g is both surjective and injective, it’s easy to prove

So g is an isomorphism.

After that H , the fundamental group of S^1, can be found in any textbook of algebraic topology. Read the details yourself, basically it’s calculated by constructing its covering space {(x,y,z)=(cos(t),sin(t),t/2π): t from R }

[devout snow]

@languid ocean Has your question been resolved?

[languid ocean]

oh okay thanks cogwheels

[devout snow]

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[steel comet]

How to read math textbooks... Correctly??

[icy depot]

Do you have an example?

[broken roost]

Read, reread, do exercises, go back reread, do exercises. Note reread doesn’t have to be the whole chapter. Ask question in parts your confused about.

For reading proof, if it’s difficult, I need to reread a couple times to digest it. Digest doesn’t always mean fully understanding it.

Most of the time, moving on and coming back to it later always works.

[steel comet]

Thx

.close

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[mystic oak]

Got a problem from my linalg course:

[mystic oak]

Given $W=Sp{u,v}$ where $u=\begin{bmatrix}2\-5\1\end{bmatrix}$ and $v=\begin{bmatrix}4\-4\2\end{bmatrix}$

a) Find orthogonal basis for $W$.
Found: $\begin{bmatrix}2\-5\1\end{bmatrix}\begin{bmatrix}2\1\1\end{bmatrix}$

[woven radish]

Joachim

Joachim

[mystic oak]

b) Find the standardmatrix $\begin{bmatrix}P_W\end{bmatrix}$ for the orthogonal projection $P_W : \mathbf{R}^3 \to \mathbf{R}^3$ onto $W$

[woven radish]

Joachim

[mystic oak]

Was totally asleep for the lectures about this, trying to make sense of it