help-38

[tidal forge]

Is there any more efficient way to do?

I'm thinking to do with line sketching, but how?

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✅

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<@&286206848099549185>

<@&286206848099549185>

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[scarlet cape]

What are the main topics in Calculus at the start of university? Like is trigonometry the main thing?

[zinc ginkgo]

This is better for #discussion . Help channels are for math problems

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[maiden parrot]

Ok, so for context i go to a islamic school and we have a mosque in the school. Ok so on friday two previous gradutes form our school came back for friday prayer, after that they came in our class and sat down and made up a story that their grandpa died or something so they ahve not attended school much, then the teacher took us out and somebody told the teacher that they were not part of our class, then the teacher called them and asked them are you from the class and they said yes and then ignored her and then ran off in their car, now the admins are saying that they are going to take 10% overall mark off form the girls and 15% mark off from the boys since somebody didint tell the teacher that they were not from our school when she asked. ( I had not talked to them nor did i go and associate with them) what should i do now since this seems like a problem on the admin that a random stranger came in class and sat down.

[maiden parrot]

pls hlep me although not mah help

[pastel hedge]

I don't think this is a maths question

Sounds like something you should take up with an advisor at your school or something

Or whatever government agency is responsible for education where you live

[maiden parrot]

kk

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[maiden parrot]

contine\ue

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[uncut needle]

Can someone double check my work please? I’m trying to find the derivative of the original equation 2x / 1+x^2

[austere cedar]

2x•2x ≠ 4x

[uncut needle]

Would it be like this instead?

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[foggy herald]

can someone help me?

[rugged latch]

What’s the probability one die lands on a 2

[foggy herald]

1/6

[rugged latch]

What’s the probability one die lands on a 4

[foggy herald]

1/6

[rugged latch]

What’s the probability one die lands on a 2 or a 4

[foggy herald]

1/3

[rugged latch]

What’s the probability 2 dice lands on a 2 or a 4

[foggy herald]

2/3?

is that right?

@rugged latch

[rugged latch]

You tell me haha, what’s your reasoning

[foggy herald]

well because if theres an or in the question then you add the probabilities

so

1/3 + 1/3?

[rugged latch]

So there’s 4 outcomes that we desire: a 2 on die 1, a 4 on die 1, a 2 on die 2, or a 4 on die 2

Do u agree?

[foggy herald]

yes

I agree

[rugged latch]

How many total outcomes are there?

[foggy herald]

6+6+6+6

so 24

[foggy herald]

so 24/36?

nvm i found out what i was doing weong lol

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[main olive]

how to derive $\int \sec{x}dx$

[solid kiln]

Jash

[dull temple]

ibp i think?

$\int 1 \cdot \sec x \dd{x}$

[solid kiln]

haylsGhost

[zinc ginkgo]

Usual way is to multiply by sec(x) + tan(x) on the top and bottom

[main olive]

so $\int \frac{\sec²{x}+\sec{x}\tan{x}}{\sec{x}+\tan{x}}dx$

[solid kiln]

Jash

[main olive]

then what

[dull temple]

oh that's cute

now make a u-substitution

[main olive]

u=secx i think

[zinc ginkgo]

No u

[main olive]

du=secxtanxdx

bruh typing latex on phone is so hard

[main olive]

would u use secx

for u

[dull temple]

i don't think that will work because what are you going to do with the tan x?

[main olive]

ye so what u do you use

[dull temple]

secx + tanx

[main olive]

then du=(secxtanx+sec²x)dx

oh so ∫du/u

ln|secx+tanx|+C

ty

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[inner kestrel]

so if I have 3 < a < b
which is greater: a^b or b^a ?

idrk where to even start with this, I tried ln on both sides but it didnt rlly give me an answer

[dull temple]

the magic of multiple choice

means that you can just try an example lol

[inner kestrel]

naw its not multiple choice i gotta prove it💀

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[bronze hound]

i dont no clue what is going on

could someone explain this

[heady sedge]

do you know what cis(x) means

[bronze hound]

yes

[heady sedge]

ok so we have this right

[bronze hound]

right

[heady sedge]

which is this

and we are looking for theta

we dont want to deal with plus or minus

so what trig will relate -i and 1

[bronze hound]

uhhh pi/4?

[heady sedge]

so we have tan(theta) = -1

when we just take the magnitudes

so whats arctan(-1)

[heady sedge]

tan(pi/4) is 1

[bronze hound]

oh so -pi/4

[heady sedge]

yeah

except cis(-pi/4) is sqrt(2)/2 - sqrt(2)/2*i

so how do we get

1-i

sqrt(2)/2*r = 1

actually this is a bad way to do it

so we have this triangle

and we want this

what do we multiply by

[bronze hound]

you multiple by sqrt 2?

[heady sedge]

yeah

so in general what do we multiply by

[bronze hound]

im not getting this

wdym multiply in general

[heady sedge]

here

this is the general case

[bronze hound]

ok right

[heady sedge]

so we multiply by sqrt(x^2+y^2)

so to convert any rectangular coordinates x and y*i

we do

sqrt(x^2+y^2)*cis(arctan(y/x))

do you understand?

[bronze hound]

ya that makes sense

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[drifting rose]

can someone explain to me the justification for doing this part of the step for mathematical induction

[drifting rose]

original question is (n+3)^2 <= 2^(n+3) where n is an interger n=>1

[low gazelle]

You want to get a polynomial in the form (k+4)^2. This is the same as k^2+8k+16. If you compare term by term its obvious this is smaller than 2k^2+12k+18.

[drifting rose]

but how does that justify allowing you to replace it

because I understand replacing stuff from a given like a= 35 so therefore a+b = 5 is 35 + b = 5

[low gazelle]

It is not replaced. It is saying the 2k^2 polynomial is greater than the k^2 one

[drifting rose]

[low gazelle]

If I just said 2k^2+12k+18>k^2+8k+16 you would agree right

[drifting rose]

yes

[low gazelle]

Well thats exactly what that is saying

Its just written vertically

[drifting rose]

oh were following the lines aren't we

OHHHH

wait why was it done there not like earlier

is it because they are similar or...

[low gazelle]

Yeah

In those two forms it’s trivial to find which one is bigger

[drifting rose]

but why there can we do that

because before we subbed 2^(k+3) in as (k+3)^2

and thats obviously smaller than 2^(k+4)

am I only allowed to compare them when they are similar enough to each other

man

@low gazelle

soz for ping

[low gazelle]

You can compare them if you can prove one is greater than the other. In the case of 2x^2+12k+18 and k^2+8k+16 though it is trivial

So you dont need to prove it

[drifting rose]

oh so you only do it when you know one is greater than the other

ahh

oh boy thats going to be hard to see

alright thanks for the help

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[quiet oasis]

help :((

[quiet oasis]

in synthetic division pls

i solved it and i checked it with a calcu, but my answer is diff

how do u do itt

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[quiet oasis]

this my solution

<@&286206848099549185>

hihi

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[dull temple]

you divided by x+1

[dull temple]

you wanted to divide by (x+1)²

[quiet oasis]

how do u write that as a divisor??

[dull temple]

with the synthetic division you're doing it's kind of a pain

but you can just divide by (x+1) twice

[quiet oasis]

yea, buts unfortunately its what our teacher want

[dull temple]

so just do it twice

[quiet oasis]

wdym

[dull temple]

well like

if you divide by 5 and then divide by 5 again, it's like you divided by 25 in the first place

[quiet oasis]

oh so, after i divide it by the divisor and get the quotient, i just divide the quotient with that divisor again??

[dull temple]

yeah since you're trying to divide by that divisor squared

[quiet oasis]

imma try it

[dull temple]

there's a way to do it all at once but it's just as much effort and harder to understand

[quiet oasis]

yea, our teacher kinda sucks how she doesn't explain certain stuff

WAIT NO WAY IT WORKED

[dull temple]

unfortunately common

[dull temple]

of course it worked, it's algebra!

[quiet oasis]

wwww thank youuuuu

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[wraith hinge]

Question: You have 1 Red candy, 2 blue candies and 3 green candies in a jar. You are taking them ouut one by one, what is the probability that there are at least 1 blue and 1 green candy lfet when you have taken out all the red candies

[wraith hinge]

I've seen the answer is 3/6 * 2/3 + 2/6 * 3/4 but am struggling to understand it

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[wraith hinge]

<@&286206848099549185>

[wild quail]

Have you tried drawing a tree diagram?

[wraith hinge]

Not yet, would it have 3 branches initilly?

[shrewd ridge]

i don't understand it either

looks so short

[wild quail]

Aight let’s not draw a tree diagram for this

I just had a better look

The probability that the last of the 6 candies taken out is green would be 3/6

[shrewd ridge]

oh

and then 1/3 of the time red is last out of blue +red

[wild quail]

Then the probability that the last of the remaining 3 candies taken out is blue is 2/3

[wild quail]

Yh so 2/3 would be blue

Now we can also do it vice versa

That being the probability of the last of the 6 candies being taken out being blue would be 2/6

Then the probability of the last of the 4 remaining candies being green would be 3/4

[wraith hinge]

sorry if it's dumb but why is it 3/6?

[wild quail]

Because there’s 3 green candies out of the total 6

[wild quail]

For you were to draw a tree diagram, there would be 2 branches because there’s 2 possibilities but idk I wouldn’t for this

[wraith hinge]

why not 1/6 * 2/5 * 1/4 (first red then take out one of the blues then the last bblue) and then the sum of the different combinations ofthat

[shrewd ridge]

the probability that last is green is the same as first is green, or 4th is green

it's not obvious but that's how it works

it's not an easy problem to solve or explain

[wild quail]

You could Ofc try that but then it would be much longer

[wraith hinge]

ah okok

[shrewd ridge]

,calc 1 - (1/6 + (2/6)(1/5)+(2/6)(1/5)(1/4) + (3/6)(2/5)(1/4)(1/3)+(3/6)(2/5)(1/4) + (3/6)(1/5))

[solid kiln]

Result:

0.58333333333333

[wild quail]

It should still lead to the same probability at the end as long as it’s done right

[shrewd ridge]

that's what i did to get the answer

[wraith hinge]

that's the right answer woah

[wild quail]

You could have many combinations and as long as it’s done right it should be right ^^

You could do a 3 branch tree diagram for this too

But I wouldn’t

[wraith hinge]

So is this 1 - the scenarios where you take out all the blue or green before taking out the red?

[wraith hinge]

ah ok

[shrewd ridge]

we fail if we draw red last
also we fail if we it goes ...rbb and if it goes ...rggg
but also ...rb and ...rg and ...rgg

so six things

still look backwards as in the short solution

[wraith hinge]

ah right ofc

[wild quail]

Did it make sense?

The way you were going to do it would’ve been right too as long as you did all the possibilities just wasn’t the fastest way

[wraith hinge]

Or do we fail if it's rggg? as the problem said what is the probability that there are at least 1 blue nd 1 green

afterr all the red candies are taken out

[wild quail]

Yh so you can’t do rggg

Because we need exactly 1g left at the end

[wraith hinge]

But after you take out the red (assuming it's the first draww) you have 3 green and 2 blue which are both at leaast 1 each?

[wild quail]

I meant exactly 1g

You need to keep going until you have exactly 1g and 1b left

If you wanted to do it 1 at a time

You could have rbgg

Or rgbg

[shrewd ridge]

it's not the first draw in any of the six cases

[wild quail]

Or rggb Or grgb and on and on

[wild quail]

^^

[shrewd ridge]

you're confusing rggg... and ...rggg

[wraith hinge]

hmm ok

I have to think about this problem some more but thank you both so much for your help

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[smoky kite]

Is this correct for Gaussian elimination

[smoky kite]

And if so, are there no solutions because of the last row being absurd?

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[mystic veldt]

ik its physics, but its an easy one, wanted to ask why friction is not acting on the opposite side of friction in this FBD

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[silver raven]

<@&286206848099549185>

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[mystic veldt]

!show

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Show your work, and if possible, explain where you are stuck.

[silver raven]

i dorn really know hwo to solve it

[silver raven]

help me work out in steps

[wraith hinge]

i think it might be the 2nd last and last one but iim not too sure myself @mystic veldt is that correct?

[mystic veldt]

idk

[mystic veldt]

ok so

what can you write 8^4 as?

[silver raven]

8x8x8x8

[mystic veldt]

also can be written as$(2^3)^4$?

[solid kiln]

!Yajat!

[silver raven]

huhh

[mystic veldt]

,, 2^3=8

[solid kiln]

!Yajat!

[silver raven]

oh okay yeah

2x2x2= 8

yea i understand

[silver raven]

but why the four

[mystic veldt]

cuz this was the question

[silver raven]

okay

[mystic veldt]

okay now from laws of exponents can it be written as 2^12

[silver raven]

yes

no

i donr know

[silver raven]

arent we supposed to mulityy 3 n 4 ?

= 12

[mystic veldt]

,, (m^a)^b=m^{a\times b}

[solid kiln]

!Yajat!

[mystic veldt]

thats what i did

[silver raven]

nvm i got ocnfused on smt else

so ist 2 with the exponent 12 yeah

i do this exact steps on the 6 one too

[mystic veldt]

what?

,, 6 is already simplified, 6=2 \times 3

[silver raven]

so ist 2^12 3^2

[mystic veldt]

no

[silver raven]

i dont understand i understood the 2^12 but not the other

[mystic veldt]

wait firsly, let us simplify this term only

then we will jump to the 6 part

okay so now its $\sqrt{2^{12}}$

[silver raven]

okay

[solid kiln]

!Yajat!

[mystic veldt]

how would you solve this?

[silver raven]

64?

[mystic veldt]

yea, better write it as 2^6 as the answers are somewhat given in this form

[silver raven]

okay

[mystic veldt]

now simplify $\sqrt{6^4}$

[solid kiln]

!Yajat!

[silver raven]

36?

[mystic veldt]

please write it in the way i told you to

6^2

[silver raven]

sorry i keep forgetting

[silver raven]

yes

[mystic veldt]

so it is now $(2^6 \cdot 6^2)^{\frac{3}{2}}$

[solid kiln]

!Yajat!

[mystic veldt]

right?

[silver raven]

yes

[silver raven]

then i add the exponents?

6+2

[mystic veldt]

no you cant add the exponents until the bases are same

here the bases are 2 and 6

[mystic veldt]

what about you use this now?

[silver raven]

okay

[silver raven]

9?

[mystic veldt]

yes

[silver raven]

aslo 3 on the second one

[mystic veldt]

$2^9\cdot 6^3$

[solid kiln]

!Yajat!

[mystic veldt]

this your answer now

[silver raven]

yes

[mystic veldt]

so isnt that the second last option

[silver raven]

yes

[silver raven]

yes

[mystic veldt]

no it wouldnt be

why do you think so?

why are you deleting it

if you have a doubt ask it

[silver raven]

i over confused myself n i relaised it so i deleted it dw abut that

[mystic veldt]

ok

so you got your answer now.

[silver raven]

theres 2 anwser trying to figure out the second one

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[strange inlet]

Sarah places k objects into n boxes at random (i.e. with equal probability). Given that
the first box is empty, what is the probability that the second box is also empty?

[strange inlet]

I found that P=(n-2)^k/(n-1)^k but I have a doubt

because when I calculate manually for 4 boxes and 3 objects I don't get the same value that from the formula

[shrewd ridge]

i don't understand that formula

why (n-1)^k

makes no sense

oh, you want to place more than one object

[strange inlet]

yep

i can have multiple objects in a box

so what do you think @shrewd ridge

[shrewd ridge]

it looks right

the manual check must have gone wrong

[strange inlet]

yeah that's what I was thinking

thanks

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[odd galleon]

calc// find where the function decreases and increases f(x) = x^(1/3) +1

[odd galleon]

the derivative is 1/(3x^(2/3))

is it no solution?

[wooden mural]

Hello there I did a proof for L'Hopital's rule for Derivatives of the 0/0 case using the definition of the derivative. One trouble I am having is that I know we have to continously imply L'Hopitals rule till we arrive at a limit that exists, but am trying to figure out the proof for it. Any ideas.

[uncut aspen]

please get your own help channel

[wooden mural]

So what we are going to do here is observe the concavity of the function.

[uncut aspen]

refer to #❓how-to-get-help

[wooden mural]

You need to start by observing the critical points.

[odd galleon]

okay

[wooden mural]

And then you need to test points within those intervals and determine the sign of the derivative which will thus indicate the behavior.

[odd galleon]

yea how to find critical point

set the thing to 0

but it then becomes 1=0

[wooden mural]

Set the derivative to 0 or DNE.

[odd galleon]

so no solution?

[wooden mural]

Well then its an extraneous solution. Here Ima take a look and work it out.

[odd galleon]

okay

the graph exists though

[wooden mural]

Of course the graph will exist.

[odd galleon]

no sol

i canr read the end

is a cp?

and idk what triple dot means

[wooden mural]

So when we set the derivative equal to 0 there is no solution. However when we set the derivative equal to DNE we observe that x cannot equal 0.

And that is because our derivative would be infinity.

Now we set up our intervals

[odd galleon]

okay

first time i’ve heard of setting it equal to DNE

can you walk me through how to do

[wooden mural]

Interval: x > 0 & x < 0
f' sign: positive and positive
Behavior: increasing and increasing

[odd galleon]

how’d you find that

[wooden mural]

Sure, so the reason we also set it equal to Does Not Exist is because it gives us important information of the derivative.

I'll give you an example of a simpler function.

[odd galleon]

okay

yea

[odd galleon]

yeah

[wooden mural]

If we observe this function and picturize a tangent line from both ends as x --> 0 the derivative would be infinity.

But we can't have that as the derivative has to be finite.

And it would therefore not exist.

[odd galleon]

yes

[wooden mural]

Now what our concern is the behavior of the function cause our end goal is to sketch the function.

Do you know about conacvity of functions?

[odd galleon]

no

i may know the concept

[wooden mural]

taking the second derivative of the function have you learnt that yet?

[odd galleon]

yea

[wooden mural]

Concave up or concave down. Have they familiarized you with that.

Alright.

[odd galleon]

no

[wooden mural]

Mk are you a beginner in application of derivatives?

[odd galleon]

pretty much

[wooden mural]

Like just on the first unit sketching and analyzing functions?

[odd galleon]

yes

[wooden mural]

mk, ur gonna learn abt concavity later.

[odd galleon]

i think so

[wooden mural]

anwyays for now.

[odd galleon]

yeah

[wooden mural]

You always want to set the derivative equal to 0 and DNE. Remember that because there lies the possibility that a functions behavior can change.

for instance.

[odd galleon]

okay

[wooden mural]

Here is the general formula you want to always apply when sketching and analyzing functions. Gimme a sec. Ima write it neatly for you.

[odd galleon]

alright thank you

[wooden mural]

np.

[odd galleon]

ping me when you finish

[wooden mural]

Follow this procedure if you want to determine the behavior of a function.

[odd galleon]

alright ill read in a bit, i just got home

ill ping if i need any more help

[wooden mural]

I'll probs be busy, but I'm sure someone else will help if thats the case.

Gl wit it. Its quite an interesting unit man.

[odd galleon]

alright thanks anyway

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[daring brook]

either b or c i believe

i had it right

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[true bridge]

How do I prove that hyperspaces are convex but not affine, when I follow the same format as the proof given here it doesn't work

[true bridge]

i.e. why must theta be limited to [0,1] in the case of hyperspaces

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[left oriole]

take a point on the boundary and another point not on the boundary

if you allowed arbitrary affine combinations then the entire line containing those points would have to be in the hyperspace

but that line is not parallel to the boundary, so part of the line is outside the hyperplane

whereas if you consider only convex combinations (theta in [0,1]) then you only consider the line segment between those two points, not the entire line

[true bridge]

I see, intuitively I see why however when I try to formally prove it where is the contradiction.

I've allowed theta to be any real number but it needs to tonly be from 0 to 1

[calm anchor]

I need help with this

[true bridge]

use chain rule and quotient rule lil homie

@left oriole any ideas?

[calm anchor]

Yeah I’m stuck on that part let me just fix up the work and I’ll show where I’m stuck

Here it is idk what to do from where I am at

Was I supposed to use the quotient rule first ?

[true bridge]

$5 \left( \frac{x}{x+1} \right)^4 \frac{d}{dx} \left( \frac{x}{x+1} \right)$ chain rule then quotient rule

[solid kiln]

Master Wanky

[true bridge]

@left oriole any updates big bro 😦

[calm anchor]

wait i'm confused

was my x^5/(x+1)^5 not right?

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[trim hawk]

is the top row of pascal’s triangle row 0 or row 1?

[trim hawk]

ping me if u respond pls

[wraith hinge]

0th row is just [1]

then 1st is [1,1] etc

[trim hawk]

ty

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[warm kernel]

https://johnkerl.org/doc/eix.pdf

[warm kernel]

Can someone explain to me the definition, foiling, and collecting terms part?

I don't understand how it became
(cos a + i sin a)(cos B + i sin B)

(the doc is proving the cos and sin addition and subtraction rule)

ohh okay....

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[wind mortar]

hello

[wind mortar]

how do i find the domain of Arccos(cos(x)) ?

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[wind mortar]

<@&286206848099549185>

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[ornate quiver]

what does the U = {...}

mean?

like what should I do with it in my venn diagram

[boreal dome]

A and B are included in U

[ornate quiver]

meaning?

I have to create a larger circle than the venn diagram that represents U?

[boreal dome]

A have elements of U

[forest egret]

think of the universal set as elements you are focusing on

[ornate quiver]

uhh, what is "elements"

nvm

[forest egret]

an element is the thing in sets

[ornate quiver]

ok

[boreal dome]

I cant explain it better than this

[vagrant marsh]

Diagram is a little off, A and B share an element

[boreal dome]

i know

i was just explaining the "included in" part

[ornate quiver]

8 is not in both a and b

[forest egret]

so where is 8 in the diagram?

[ornate quiver]

out a and b

but insdie u

let me draw

[vagrant marsh]

Ah ok

[ornate quiver]

here

correct?

[forest egret]

correct

you should label the sets A and B though

[ornate quiver]

okee

thank you!

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[royal remnant]

x + y = 3

y = 3 - x

[sturdy holly]

ohh

thx

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[wooden plover]

<@&268886789983436800> spam

[fair bison]

banned

.close

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[noble vector]

https://cdn.discordapp.com/attachments/238956921581862913/1145689303482109983/20230828_172921.jpg

[noble vector]

what kinda sorcery is happening here

what's wrong here? LOL

[quasi wasp]

isn't it actually -1/12?

[warm kernel]

Distributive law doesn't work over infinite sums.

[quasi wasp]

why not?

[noble vector]

^

[warm kernel]

https://youtu.be/oaZKaV-oZ-g

Just watch this, it'll be pretty hard to explain without any visuals.

[noble vector]

is there a stack exchange page with that property or whatever u call with distributive thingy not working over infinite sum?

[noble vector]

in addition to this ofc

[warm kernel]

I can't really prove it tbh, it's an assumption.

[noble vector]

no not you

do you know what it’s called

[warm kernel]

I don't really know, sorry.

[noble vector]

so i can read stackexchange responses?

oh

rip

[warm kernel]

But there are also multiple evidence that proves it's wrong

https://youtu.be/YuIIjLr6vUA

[quasi wasp]

here, he is using the distributive property

[warm kernel]

not the distributive law, i'm not really sure about that

[noble vector]

lol i just noticed too

[warm kernel]

to show that it's wrong.

[quasi wasp]

oh, yeah... i didn't watch the whole vid

[warm kernel]

there are multiple videos and articles that has debunked all of those "proofs"

just search it up.

[quasi wasp]

@noble vector

he goes on to do it your way

[noble vector]

right right so basically the takeaway is infinite distributive law doesn't hold if the series doesn't converge

didn't quite get the riemann zeta part though

[quasi wasp]

yeahh

[noble vector]

like okay i get that we're using it beyond the domain it's defined for but hmm

analytic continuation?

okay nice video

thanks @warm kernel @quasi wasp

[quasi wasp]

thanks @warm kernel

didn't realize till now that -1/12 was actually kinda incorrect

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[shy thorn]

Hey guys im dealing with languages and regular expressions as well as regularity and irregularity.. In the first screen shot we prove the language is not regular via pumping lemma. However in the second screenshot the language satisfies all these definitions thus resulting in a regular language?

[dull temple]

the second thing is talking about a regular expression, and I don't see one for that language

[shy thorn]

What if I were to use this method instead. Try to come up with a regular expression for my language

[dull temple]

try it!

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[errant pike]

Did I verify this right? Or Do i have mistakes?

[wraith hinge]

my name is john as well

i thought i created this channel

[errant pike]

Haha what a coincidence

[lost wharf]

..

[wraith hinge]

how did you get to the third line?

[errant pike]

I grouped the 1/sinx-cosx/sinx

Resulting to 1-cosx/sinx?

[wraith hinge]

fourth equality i mean

[errant pike]

That's where i went random Haha

Dunno what i'm doing

I tried trial and errors But i can't find a suitable one

Hmm?

So Like it would be 1+cosx/1-cosx?

[wraith hinge]

explain to me the fifth and sixth equality

[errant pike]

I just tried my best to come up with that.... Thats my explaination

Idk also why icame up with that

Haha

I'm dead at trigo

[wraith hinge]

try doing the common denominator right from the beginning

[errant pike]

With Sine?

[wraith hinge]

without replacing csc and cot

[wraith hinge]

here it is how i would've solve it

sorry for the handwriting , i know it isn't the best i've got alot of complaints

[errant pike]

No no... You've helped Me a lot

I'm thankful

Ohhh @wraith hinge you used phytagorean

I didn't know you can use it here

Hahahaha

Thanks

[wraith hinge]

you can derive many equalities from it

$cos^2+sin^2=1$

[solid kiln]

john.

[wraith hinge]

if you devise the whole thing by sin^2 you get the identity that i have used

if you devise it by cos^2 you get 1+ tan^2 =sec^2

and i have used (a-b)(a+b)=a^2-b^2

[errant pike]

Ohhhhhh

I see

Imma learn It tom

Now I sleep

Hahaha

Thanks For helping me Dear same name

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[languid fog]

can sombody help me about how to integrate this liner over quadratic?

[delicate belfry]

So this is your integral: $$\int \frac{1-x}{(1+x)^2};dx$$

[solid kiln]

Good

[delicate belfry]

Are there any thoughts that you immediatly came up with when looking at this integral?

[languid fog]

yes

no

the squar is only on the x

[languid fog]

yeah but nothing worked

[delicate belfry]

If you got the integral:$$\int \frac{x}{(1+x)^2};dx$$ How would you solve for it?

[solid kiln]

Good

[languid fog]

maybe by parts

[delicate belfry]

Alright, that might be a way. But what if I asked you to do substitution integration?

Will it be a way for you to solve it?

[languid fog]

I don't think so

this showed up in my differential equations question

[delicate belfry]

Have you learned about substitution integration before?

[languid fog]

yes

[delicate belfry]

How does substituion integration work?

[languid fog]

something like u= 1+x then du= 1 then you replace and such...

[delicate belfry]

Alright: $$\int \frac{x}{u^2};du$$

[solid kiln]

Good

[delicate belfry]

You're on the right track. But how about x? What could you do to make it u? If u = x + 1?

[languid fog]

x= u-1

I'm not bad at math i took B= in calclus 1 and 2 but it is been a long time since i solved a question

I forgot some of the rules

thats it

[delicate belfry]

This is your algebra intuition. But is that really true if we convert u to x again?

Our u is x + 1, when making u to x again, you get x + 1 - 1, which is just x.

But that doesn't seem as what u is that we defined before, which is u = x + 1.

[languid fog]

so....

what are you truing to say

truing*

trying***

[delicate belfry]

The point here is converting u and x. Our definition is u = x + 1, which means when we convert u to x again, we want x + 1.

[languid fog]

okay

[delicate belfry]

If u = x + 1, then the numerator is missing the 1.

[languid fog]

what does this has to do with the problem I gave you

just give me an example that would be easier as english is not my mother tounge

[delicate belfry]

The question you gave is related to substituion integration.

[languid fog]

the square is just on the x not the whole x+1

[delicate belfry]

Apologies, you we're right here.

$$\int \frac{u-1}{u^2};du$$

[solid kiln]

Good

[languid fog]

that is fine

then what should we do now

[delicate belfry]

Then it's part integration that you mentioned.

[delicate belfry]

With the concept of the example, do the same with your question.

[languid fog]

so my main problem is integration by parts?

without supstitution

becuase it already looks like this

I can't use substitution on it

[delicate belfry]

I'm not sure if you could just dive into part integration, but it seems expect you to use substitution integration initially.

[languid fog]

I tried and it didn't work

[delicate belfry]

What if I did this instead? $$\int \frac{1-x}{(x+1)^2};dx$$

[solid kiln]

Good

[delicate belfry]

Are there any thoughts now?

[languid fog]

why do you keep putting the square on x+1

it is only on x

here I could just use partial fraction after playing with x+1 squared but that is not my problem

[delicate belfry]

My bad, I saw you used parathesis, so I thought it was (1+x)^2 the whole time.

[languid fog]

I mentioned it before but that is fine

let us try again

so any thoughts?

[delicate belfry]

So this question involves solving two integrals, meaning you're going to split the integration to two.

[languid fog]

how can I do this

[delicate belfry]

Just like fractions $\frac{a + 1}{b} => \frac{a}{b} + \frac{1}{b}$

[solid kiln]

Good

[languid fog]

there is a + constant on both sides

[delicate belfry]

What do you mean by that?

[languid fog]

1-x over 1+x square

you cannot do what you did here

[delicate belfry]

$\frac{a - 1}{b} => \frac{a}{b} - \frac{1}{b}$

[solid kiln]

Good

[languid fog]

how would you do this if the denominator is b+1 ?

you can not

[delicate belfry]

$$\int \frac{1-x}{1+x^2};dx$$ $$=>\int (\frac{1}{1+x^2}-\frac{x}{1+x^2});dx$$ $$=>\int (\frac{1}{1+x^2}); dx - \int (\frac{x}{1+x^2});dx$$

[languid fog]

Man please read the question

[delicate belfry]

$\frac{b - 1}{b} => \frac{b}{b} - \frac{1}{b} => 1 - \frac{1}{b} $

[languid fog]

THE SQUARE IS ONLY ON THE X

[delicate belfry]

My bad, I was copying the latex i did before

Does this satisfy you now?

[languid fog]

what makes me satisfied?

[delicate belfry]

That I finally did it correctly.

[languid fog]

thanks for the help

forgive me

I'm al little bit stupid

sorry for speaking to you like this

[delicate belfry]

All fine, it was my fault that I didn't see the expression correctly.

[languid fog]

the first is tan invers and the second is substitution right?

Thank you for your time and sorry again for being rude

[solid kiln]

Good

[delicate belfry]

All fine.

[languid fog]

.closed

Here is the final solution for the differential equation

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[sweet ferry]

How do I find 1 ?

<@&286206848099549185>

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[desert lichen]

How does this work?

[crimson cedar]

How do I go about starting step 3

U got it my bad

[fiery goblet]

https://www.youtube.com/watch?app=desktop&v=daUlTsnCNRQ

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[wraith hinge]

The mean and variance of the hypergeometric distribution $\map h{x;N,n,k}$ are [
\mu = \f{nk}{N} \tss{and} \sigma^2 = \f{N-n}{N-1} \cdot n \cdot \f k N \parens{1-\f k N}
]

[wraith hinge]

i am trying to prove this but i really have no idea

[zinc ginkgo]

do you know the formula for the pmf?

[wraith hinge]

can u show if u can

[zinc ginkgo]

https://en.wikipedia.org/wiki/Hypergeometric_distribution

[wraith hinge]

oh probability mass function

yes i do

wait

[
\map E X = \sum_{x=0}^m x\f{\binom k x \binom{N-k}{n-x}}{\binom N n}
]

[wraith hinge]

so like

ugh i have to remember some combinatoric bullshit identities dont i

okay one sec

[wintry stag]

It's usually just a matter of absorbing the x into the n choose k terms and then splitting it into up or changing the indices in such a way that you have something multiplied by a known probability distribution. These kind of problems are not really that enlightening in my opinion

[white merlin]

help #help-7｜zen1thxyz idk how to

i just need a small push xd

[zinc ginkgo]

!noadvert

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[wraith hinge]

oh i see lol

set $z = x-1$, [
\f{nk}N \sum_{z=0}^{n-1} \f{\binom{k-1}z \binom{(N-1)-(k-1)}{n-1-z}}{\binom{N-1}{n-1}}
]

i think

so you just get nk/N

oh wait

[wraith hinge]

ok so thats neat lol

[wraith hinge]

thanks doe

.close

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[wraith hinge]

and thank you riemann

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[sudden lagoon]

In measure theory, I want to find the lebesgue measure of the interval [0,1] without the numbers in which 7 appears in the first second and third decimal place

[sudden lagoon]

So with regards to the first decimal place I have an interval [0,0.7]U[0.8,1]

For the second decimal place I have [0,0.07]U[0.08,0.17]….

And similarly for the third decimal place

Where I have unions of 2, 11, and 101 intervals respectively

How do I combine these to get the total lebesgue measure in this case

[zinc ginkgo]

might be easier to find the measure of the complement of each

for the second decimal place, group the intervals into equal sized lengths

[weak kelp]

!status

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3. I got an answer but I was told that it's wrong.
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5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[zinc ginkgo]

!occupied

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[weak kelp]

mb

[sudden lagoon]

So we’re looking at the interval we’re discarding, namely 10 intervals of length 0.01?

So in the case of the first decimal place, we’ve discarded an interval of length 0.1, for the second decimal place we’ve discarded 10 intervals of length 0.01, and for the third decimal place we’ve discarded 100 intervals of length 0.001

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[sudden lagoon]

<@&286206848099549185>

[sudden lagoon]

<@&286206848099549185>

[calm anchor]

Help me

[sudden lagoon]

!occupied

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[naive crest]

if your number has 7 in the first, second, and third decimal place, you can write it 0.777... with ... meaning whatever after

unless you meant or

[sudden lagoon]

I meant or, apologies

[naive crest]

then you can follow riemann advice
the number 0.7... are [0.7, 0.8[ so 0.1 of measure
then the number 0.x7 are [0.x7, 0.x8[ are of measure 0.01, and we count them for x = 0, 1, 2, 3, 4, 5, 6, 8, 9 (we already counted for 7)
so the total measure is for now 0.1+9*0.01

and you do the same for the third place

and you subtract that from [0, 1]

[sudden lagoon]

And the third place we’d have [0.xx7,0.xx8] of measure 0.001

Where neither position contains an x?

[naive crest]

where neither x is a 7 yeah

since we already counted for 7 in first and 2nd

the xx can be 00, or 01, or ... 06 or 08 etc

81 choices

[sudden lagoon]

Ahhhh okay, okay

So then you’d have 81*0.001 for the third decimal place

[naive crest]

yes

also, there is another way of counting which was combinatorics, but I assumed you wanted to calculate some measure bc it was measure theory

[sudden lagoon]

Yea no you’re right, and it completely makes sense

I think again for today you’ve been a great help, thanks a million!

.close

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[trim jolt]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[wraith hinge]

3. I have an answer and I want to know why it's wrong

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Show your work, and if possible, explain where you are stuck.

[wraith hinge]

if X=number of diesel cars, we are looking for P(x>1), wich is equal to 1-(P(x=0)+P(x=1)) since we have 5% chance to have a diesel car, Poisson's probability is 0.05*4=0.2. I think my Poisson's probability is wrong though since my answer is 0.001209

here is what ive done so far

[naive crest]

the number of Diesel's among 4 already filled spots car follows binomial distrib since it's a repeated independent event with a probability of 0.05

[wraith hinge]

uhm intersting

how can you describe the binomial probability? x\sim B(4,0.05) or x\sim B(4,0.2) ?

[naive crest]

P(X > 1) where X ~ B(4, 0.05)

and should be around 0.014

[wraith hinge]

yup i think i have it

i have 0.014018749999999

it's correct!

i'll go on with the 2 other problems

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@wraith hinge Has your question been resolved?

[wraith hinge]

For q2, does this seams ok?

[naive crest]

it was still not poisson for q2

[wraith hinge]

how can you tell if its poisson or not?

[naive crest]

you're looking for a probability of first success, it's geometric

[wraith hinge]

i think thats the issue

[naive crest]

the probability that an appearing car is a diesel car is 0.05 for each car that comes, the cars are independent
and they ask for the probability that to see at least 15 cars to see a first diesel car
so they're asking for P(X >= 15) where X follows a law of first success
the law that describes this situation is called geometric distribution, and is the discrete equivalent of Poisson

[wraith hinge]

ok i'll try again

[naive crest]

oops I think we may have been trolled by first question

[wraith hinge]

first question is right though

[naive crest]

oh yeah ?

[wraith hinge]

yes

[naive crest]

the question is weird

bc there is only one spot for diesel at the gas station

so it's impossible that 4 cars at the gas station needs diesel

we could have interpreted the probability to be 0

[wraith hinge]

yes, that's what i initially thought, but they could be at the gas station but not refuelinh

[naive crest]

yeah I guess, that was ambiguous

for a sec I thought the question was a troll after reading for the 4 spots

[wraith hinge]

i saw it comming tried it first and lost a couple points

oops

is the geometric law for probability : p(X>=x)=(1-p)^(x-1) ?

[naive crest]

yes

[wraith hinge]

my result is 0.53670876984025

is it good?

[naive crest]

I find 0.48765, around that

[wraith hinge]

oh yeah

i found it too missed my formula

it's correct!

Last question now

i have an average profit of 1369,50$

[naive crest]

that's way too much

[wraith hinge]

there are 66 cars per hour

[naive crest]

on average, there are 66 cars per hour

oh yeah mb

yeah I find the same

[wraith hinge]

good

ok thanks for the help!

.close

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[winter tendon]

I don’t know what I’m doing wrong

[winter tendon]

I don’t know what I’m doing wrong

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[lilac coral]

how could I prove that the set is closed under addition? It makes sense that a linear combination of orthogonal vectors would also be orthogonal but I am unsure how to convey that mathematically.

[dusty vault]

you know that, say, $u_1 \cdot v = u_2\cdot v = 0$. to show closure under addition, you need to show $(u_1 + u_2) \cdot v = 0$.

[solid kiln]

gergill

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[rare fox]

I am having trouble understanding this problem. Not quite sure how to find Q

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[warped pawn]

Find the slope of the secant line first

[rare fox]

The formula: Mpq = f(x)-f(a) / x-a ?

?

[potent smelt]

yes

or wait

whats a?

Mpq = gradient of the secant line

not sure where you got a from

[rare fox]

a is 2?

[potent smelt]

okay good, thats p

as for q, you substituted x = 2. that would just be p again

x values of q are given, they are not 2

you’ve basically tried finding Mpp which is undefined on a curve

[rare fox]

Oh

Also is it even possible to find Q since the point (0,2) was unintentional. I just was working on it, no point was given besides P

[potent smelt]

P is given as (2, 0)
Q is given as (x, f(x)) ( cuz if x coordinate is x, y coordinate is f(x) as y = f(x))
youre given x values in the table. substitute them and find the corresponding y coordinates for Q if that is your x coordinate. then find the gradient (Mpq) between P and Q

using your formula, which is correct. except a = 2, f(a) = 0, which is P. x = whatever the table says, and f(x) you can find

[rare fox]

f(x) = 2- 1/2x^2 ?

[potent smelt]

as given, yes

[rare fox]

So Q can be (1.5, 2-1/2x^2)?

[potent smelt]

yes but if x = 1.5, what is 2-1/2x^2?

[rare fox]

Ohhh, so 0.875?

[potent smelt]

yes

so what’s Q?

in this case at least

[rare fox]

(1.5,0.875)?

[potent smelt]

yup

now you can find Mpq