help-4

[fleet junco]

also i just realied this simplifies to P(min >= x) = 1/x

so thats really cealn

so the answer is

1 + 1/2 + 1/3 + 1/4 + 1/5

damn thats a cool problem

[night patio]

are you sure thats the answer?

[fleet junco]

im pretty sure

i dont see anythting wrong with the logic

[night patio]

this sums up to 2.28333333333

and what i was told that the answer was 2.45

[night patio]

adding a 1/6 to this gets us to 2.45

I might be wrong tho

[night patio]

but this is what i was told

[fleet junco]

yeah but minimum

says

its including

the 5 roll

whihc means the minimum cannot go above 5

so ur 2.45 might be

not including

the 5 roll

[night patio]

i really don't know. Can you explain me your solution

[fleet junco]

ok

so

[night patio]

I have not studied this expected value stuff

[fleet junco]

do you see why this would be true?

[night patio]

let me think

i think i almost got it

just 1 min sorry

[fleet junco]

acutaly theres an additional constraint whihc is that P(X > N) = 0 and P(X < 1) = 0 and X discrete

[night patio]

yupp just 1 min

i think i got it

(i took the starting value n = 1 here)

but yes i get what you are saying

[night patio]

now if we pair up terms on the right, we'll get this expression

[fleet junco]

yep exactly

[night patio]

totally makes sense

[fleet junco]

liek each >= is just hte sum of the equalities

[night patio]

yes and each term will have n copies of itself

[fleet junco]

and the counts of easch

amke it exaclyt x

yep

ok

[night patio]

got it

[fleet junco]

so knowing

that we only

ahve to find a way

to calcualte

P(min >= x)

and the probelm is solved

so for min >= x

u have to have some number of rollls

= x

and a final roll of 5

so for a length n sequence

u have n-1 rolls >= x and a final roll equal to 5

also those n-1 rolls do not equal 5

so that leaves u with 6-x possible numbers to roll

[night patio]

like each one of those rolls should show a number on the die that is >= to the minimum

[fleet junco]

yes

and the chance u roll something >= x and not euqal to 5 is just (6-x) / 6

so u do that n-1 times with a final 1/6 chance to roll a five

[night patio]

yes

[fleet junco]

gives you the probabliltiy for any length n sequence

[night patio]

yupp

[fleet junco]

u add taht up, giving u this forumula

[night patio]

okay makes sense

[fleet junco]

and qed

[night patio]

we add that up

[fleet junco]

then u geometric series to find its just 1/x

[night patio]

yes i will simplify that part

and i totally got

i got how we found this

but what i dont get is

P(min >=5) + P(min >= 4) + P(min >= 3) + P(min >= 2) + P(min >= 1)

this

[fleet junco]

well thats just an appliaoitn of what u proved

https://cdn.discordapp.com/attachments/409596215697866773/1394176214729625701/image.png?ex=6875db48&is=687489c8&hm=5ce97384e61bae4ac4978b31808b9265fb30a7bee43eaa729fef58b7135546cf&

[night patio]

how does this give us the expected minimum

can you explain it a bit intuitively if you have time?

[fleet junco]

here i will write it more explicityl

[night patio]

like i totally get this part

this is simple

but why are we adding these in the first place? how do these give us the expected value?

[fleet junco]

P(min >= 1) = P(min = 1) + P(min = 2) + ... + P(min = 5)
P(min >= 2) = P(min = 2) + ... + P(min = 5)
P(min >= 3) = P(min = 3) + ... + P(min = 5)
P(min >= 4) = P(min = 4) + P(min = 5)
P(min >= 5) = P(min = 5)

u get that?

[unreal warren]

simple...?

gulp

[night patio]

I mean it was simple to prove😭

[unreal warren]

oh lol

[night patio]

totally get that

[fleet junco]

if u add all the equations

up

[night patio]

thats how i was able to prove that thing

[fleet junco]

u get our sum

=

1 * P(min=1) + 2*P(min=2) + ... + 5*P(min=5)

whihc is exactly

the definition

of expected value

[night patio]

okay thats what i didn't know

[fleet junco]

sry i shouldve made it celarrer

that this was the definition of epxected value

[night patio]

and why is it defined that way?

[fleet junco]

well its like an average

its

value * probablity of getting that value

[night patio]

like we are weighing stuff up

and adding it

[fleet junco]

so its bsaiclay an average

yeah a weighted avberage

cuz think of it like this

lets say u ahve a coin flip with 80% heads

then your expected number of heads

is

0 * .2 + 1 * .8

because tails = 0 heads

and heads = 1 heads

ig thats not a very good example

but its just a weighted average

[night patio]

okay its like we want to know how many heads do i get (given that i know how many percent i can get or like the probability of a head)

now since i want to know the number of heads, i put tails = 0 and multiply probability of getting 0 tails with 0 and add that to probability of getting heads * 1

something like that

did i just write a lot of nonsense

[fleet junco]

no ur right

but heres a better example

liek lets say you pull a ball from a box

50/50 chance theres two balls

but one ball is 10kg and the otehr is 1kg

then whats the expected weight of the ball

would be

10 * .5 + 1*.5 = 5.5

[night patio]

10* 1/2 + 1 * 1/2

[fleet junco]

yeha

[night patio]

totally makes sense

so the general formula is weight * P(weight)

[fleet junco]

yep exaclty

and also the reason its litearlly a weightred average is because

sum of probabilities = 1

cuz normally in a weighted avg u have like

(ax + by + cz) / (a + b + c)

is the weighted average of x,y,z with weights a,b,c

but in probablilty, the probablilites act as weights, and since they sum to one the denominator is 1, which is why there is no denomintaor

[night patio]

in this case we actually referred to the value or measure of the object to be the weight

not the probability

so is it the probability that acts as the weight?

i guess that makes sense

like if we increase the number of 10kg balls to 2

then 10 * 2/3 + 1 * 1/3

the probabilities change but the objects remain the same (i.e. the 10kg and 1kg balls)

[fleet junco]

well the weight was just my example

so value would be a better word

[fleet junco]

so yes, the weights in the wegihted average are the probabilities

[night patio]

hmm makes sense

and so we generalised this definition of expected value just like that?

[fleet junco]

yeah

[night patio]

w * P(w)

are there any other definitions too?

[fleet junco]

i dont think so

[night patio]

I mean this one makes so much sense for this example

[fleet junco]

like EV is defined to be the weighted average with respect to probabilities

nothing more nothing less

[night patio]

but I still can't wrap my head around the die problem

[fleet junco]

its just what it is

[fleet junco]

well

its hte same idea

like take expected value of just a single dice roll

each value 1,2,3,4,5,6 has probabliltiy 1/6

so its just liek drawing weighted balls

aucaltly the idea of EV also

is that

if u repeat

the trial

an infinite number of times

then EV is the average value you will get

so in your original problem

there maybe liek a 20% chance u get a minimum of 3

so if u reapeat ur problem an infinite number of times and note the minimum each time

and average them up normally (not weighted) then u will get teh EV

[night patio]

got it

so a distribution

right

of minimums

[fleet junco]

yes, in that given probelm

[night patio]

and the value 2.2833 lies in the middle

its the average for infinite number of trials

lets assume a very simple model for this problem

[fleet junco]

yes

[night patio]

since they want an average or expected value

[fleet junco]

so in the 20% of 3 case, 20% of all ur trials

will return a 3

and so it makes sense to assign it a weight of .2

[night patio]

i could have just done (1 + 2 + 3 + 4 + 5)/6

but that won't be right

[night patio]

we are actually multiplying each number in this by some weight

[fleet junco]

yes

[night patio]

and those weights are the probabilities for getting that respective number

[fleet junco]

yes

because probalbitly is the frequency/proportion

of ur infinite

trials

that will give that value

also to clarify, in this case, a single tirla is one sequence of rolling til u get a 5

[night patio]

(1 * P(1) + 2 * P(2) + 3 * P(3) + 4 * P(4) + 5 * P(5) + 6 * 0)/6

[fleet junco]

no

not /6

becuase u divide by the sum of the weights

which is 1 due to how probability works

[night patio]

yes makes sense

[night patio]

hmmm

i need to think more

thankyouu ^^

[fleet junco]

yeah just sleep on it

[night patio]

okay your answer is right

2.45 was wrong

wrote this code and ran it for a million trials

[vale dock]

@night patio Has your question been resolved?

[tender field]

@night patio was aa2 able to explain this problem? If not I can try to solve it

You can probably get away with a tail sum

$$E[X] = \sum_{i = 1}^{5} \mathbb{P}(X \geq i)$$

[night patio]

he explained it to me

[rocky lotus]

utop!a

[tender field]

alright :)

[night patio]

the final answer came out to be 1 + 1/2 + 1/3 + 1/4 + 1/5

@tender field can you confirm it

[tender field]

yep

i got the same answer

very beautiful question ❤️

[night patio]

thankyou for confirming :D

[night patio]

yea learned a lot. @fleet junco explained well

[vale dock]

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[static vigil]

.

[viscid spade]

Yes ?

Need help to write the integral?

[quick mica]

hold up lemme word this right

So for the second question in using chain rule u integrate it right?

[viscid spade]

What is the integral you wrote

[quick mica]

But the answers say it’s 2/9 instead of 9/2

I’m just confused how it’s 2/9

[viscid spade]

Split it and use the chain rule yes

[quick mica]

and this is the answers

[viscid spade]

To compensate

[glass kelp]

1/3 * 2/3 =2/9

It’s 1/3 * (1/(3/2)) =1/3*2/3

[desert stone]

Did you computed the first integral setup?

[quick mica]

I thought we needed to multiply it by 3/2 by 3

[desert stone]

I am confused as why there are two integrals on your scratch paper

[quick mica]

Bad working sorry

[desert stone]

Becasue when you use the curve above subtract the curve below at point of intersection it should give you the correct result, there is no change order of curve

[glass kelp]

[ \int_2^6 (3x-2)^{1/2} \dd x]
Let ${u = 3x-2}$. Then ${\dd u = 3\dd x}$, so ${\dd x = \frac{1}{3}\dd u}$
[ \int_{u=4}^{16} u^{1/2} \left(\frac{1}{3}\dd u\right) = \frac{1}{3}\int_{4}^{16} u^{1/2} \dd u = \frac{1}{3} \left[\frac{u^{3/2}}{3/2}\right]^{16}_{4}]
Simplifying,
[ = \frac{1}{3}\frac{2}{3} \left[ u^{3/2}\right]^{16}4 = \frac{2}{9} [u^{3/2}]{4}^{16}]

[desert stone]

So you set up the integral or you can’t compute?

[desert stone]

Why are there 2 integrals

[quick mica]

cuz ones the curve others the line

[desert stone]

You don’t need a second integral

[quick mica]

Ohhhh ok thank you so much

[desert stone]

Your first integral include the correct and sufficient information

[quick mica]

I think I finally get it

[quick mica]

u need to use chain rule no?

[rocky lotus]

k

[glass kelp]

There we go

[quick mica]

thanks man

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[pallid bronze]

Yo anyone here ?

[normal hollow]

Question plz

[pallid bronze]

are u math expert?

[normal hollow]

No i m not

But there r others

[jovial edge]

No... But we will try our best to help

[hollow shuttle]

it really depends on the question

[pallid bronze]

I don't have questions lol 😭

Sry

[hollow shuttle]

,

[jovial edge]

Oh...

[glass kelp]

Bruh

[normal hollow]

...

[glass kelp]

!close

[desert stone]

Do you want a question?

[pallid bronze]

But I need ur helps when I will be in class 9th

[glass kelp]

So u have a question😭

What kind of paradox is this gng💔

[normal hollow]

Future prep

[jovial edge]

Yeah.... We will help you with your problem when u have one

[normal hollow]

Booking a channel

[noble anvil]

.close

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[noble anvil]

Come when you have a question

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[night peak]

Hi, i do not understand how to solve this nor this solution? Could someone explain how is the 2nd step of the solution equal to 1, 3rd to 5?

[normal hollow]

U see log7(1) =0

Do u agree?

7 is the base here btw

[night peak]

ohh

[normal hollow]

Do u understand

[night peak]

yess

that made me understand the next step too

[normal hollow]

Yes

[night peak]

the only way to get 1 would be if log2(x) was 5

Thanks!

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[polar jasper]

Let x,y,z>0. Prove that (x+y)/z^2 + (z+x)/y^2 + (y+z)/y^2 >= 2(1/x+1/y+1/z)

[earnest ingot]

What have you tried?

[polar jasper]

I tried Cebishev

and rearranging inequality

this one

[polar jasper]

i think this one might work

[slate folio]

you don't need any of that

you add 1/x + 1/y + 1/z to both sides

[polar jasper]

oh

It was at the inequality theorems chapter so i thought i should use one of those

[slate folio]

nah I found out you don't need those

Only Cauchy-Schwartz/Bunyakovsky is enough

[polar jasper]

ok

[slate folio]

the left hand side will be (x+y+z)(1/x^2 + 1/y^2 + 1/z^2)

[polar jasper]

Ive gotten down to

(x+y+z)(1/x+1/y+1/z)>= 9

[slate folio]

yep

[frozen ledge]

then AM-HM

[polar jasper]

yeah

[frozen ledge]

divide both sides by 3(1/x + 1/y + 1/z)

[polar jasper]

alright

yeah i think i got it

i just overcomplicated it lol

thanks!

[slate folio]

no problem

[polar jasper]

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[next vortex]

have you tried anything?

[desert idol]

nope I have no clue

[next vortex]

what does it mean for f and g to have an intersection point?

of course, their graphs intersect at an intersection point, but what does it mean algebraically?

[desert idol]

they equal to each other

f(x) = g(x)?

[next vortex]

yup!

try setting the two functions equal and seeing where you can go

[desert idol]

I got k>√(2)-3 and k<-√(2)-3

[tidal swift]

i have a feeling you will have to consider the fact that those two graphs can intersect twice, meaning you may need a quadratic if you didn't use that already

[vale dock]

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[trim belfry]

Hi can someone help me with this question? Im having trouble dealing with area under the curve for polar coordinates

[pure mirage]

At which points do they intersect?

[trim belfry]

pi/2

This is how it looks like on desmos

Im confused with the constant and bound for integration

[slim summit]

yes and ?

there are two intersections

[trim belfry]

yea if its in a normal cartesian plane i could integrate it normally but idk how to deal with polar plane. the 4 axis on this plane is 0, pi/2, pi, 3pi/2

[slim summit]

Let us prove the formula for the area between two polar curves ( r_1(\theta) ) and ( r_2(\theta) ), where ( r_2(\theta) \geq r_1(\theta) ) on the interval ( \theta \in [a, b] ).

In polar coordinates, the area swept by a curve ( r(\theta) ) from ( \theta = a ) to ( \theta = b ) is given by:
[
A = \frac{1}{2} \int_a^b r(\theta)^2 , d\theta.
]

This comes from approximating the area of a small sector as:
[
dA = \frac{1}{2} r^2 d\theta.
]

Now, the total area enclosed by the outer curve ( r_2(\theta) ) is:
[
A_{\text{outer}} = \frac{1}{2} \int_a^b r_2(\theta)^2 , d\theta,
]
and the area enclosed by the inner curve ( r_1(\theta) ) is:
[
A_{\text{inner}} = \frac{1}{2} \int_a^b r_1(\theta)^2 , d\theta.
]

Therefore, the area between the two curves is:
[
A = A_{\text{outer}} - A_{\text{inner}} = \frac{1}{2} \int_a^b \left[ r_2(\theta)^2 - r_1(\theta)^2 \right] , d\theta.
]

[rocky lotus]

Clint

[trim belfry]

okay thanks for helping

.close

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[dusk cosmos]

why solve the negation? something to do with the equal-to? i got the same answer either way

[stark wedge]

no it's rather that |A| < B is somewhat easier to handle than |A| > B

[dusk cosmos]

oh ok yeah i see now, i wrote -3>...>3 in my second step and didn't notice that that makes no sense, it just happened to work out

thanks!

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[torpid scaffold]

1+1=3

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[glass finch]

How do I solve this question? I don't know how to factor.

[west cloud]

Has your teacher gone over how to factor yet?

He/she should have, if you're getting these types of problems.

An alternative is using long division for polynomials, but I doubt you'd know how to do that if you don't know how to factor.

[glass finch]

No, I'm applying for college. I am very rusty. I haven't done math since high school.

[west cloud]

ah

then you should definitely learn how to factor

[cloud coral]

I think easier way is to consider zeros

[old belfry]

yeah there are a few ways to do this but focus on factoring

[cloud coral]

But polynomial long division should be a standard skill

[west cloud]

https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratics-multiplying-factoring/x2f8bb11595b61c86:factor-quadratics-intro/v/factoring-quadratic-expressions

[glass finch]

I don't know how to do any of that. I was attempting to solve it using PEMDAS but I ended up very confused.

[cloud coral]

I see this and im like... ||just use quadratic formula||

[glass finch]

I skipped math class like every day in high school too on top of that... 😭 💀

I regret it

[west cloud]

yeah this might be the easiest way

[old belfry]

yeah def

[west cloud]

but OP should still learn to factor and do polynomial division

[glass finch]

teach me

[west cloud]

you can use khan academy

[old belfry]

khan academy is good

[glass finch]

ok

how would you solve that problem though?

[west cloud]

^^^

[old belfry]

in this case its quite easy to factorise but if it isnt, use quadratic formula

for this particular case id immediately factorise

[glass finch]

so how do you factor?

[west cloud]

right now, if you want to avoid factorizing and polynomial division, just find the roots

[cloud coral]

For this case ill immediately use quadratic formula. If it doesnt immediately pattern match something I can factor, i just take out the foolproof method

[noble anvil]

Watch a YouTube video instead of trying to learn from us

[west cloud]

since every quadratic is in the form (x-r1)(x-r2) where r1 and r2 are its roots, you can just find the roots and get the factors

[noble anvil]

Much less confusion

[west cloud]

^^^ for factoring

[old belfry]

yeah

[glass finch]

3x^2 x -18? = -54....

27 x -2 = 54 the number we just got and also 27 + -2 = 25

3x^2 + 27x - 2x -18

[vale dock]

@glass finch Has your question been resolved?

[glass finch]

No but you can go ahead and close it anyways

[zinc spear]

click on the ❌ to keep your help channel open

wait you want to close it, bruh i’m tired, you can type .close then

[glass finch]

I guess I'll keep it open if any one decides to help me

[weary swan]

hello hi

whats the question uhhh

[zinc spear]

this

[weary swan]

not this?

[zinc spear]

tbh either one

[weary swan]

yeah what everyone else is saying tbh

you would do this with factor by grouping

[zinc spear]

if you are lost by all the suggestion, i would say prerequisite knowledge is multiplying (foil) and adding polynomials, evaluating them (x^2-1 at x = 2 is 2^2-1 = 4-1 = 3), then the fact that if the polynomial p(x)=0 at a certain value « a » called a root, then you can factor a linear factor (x-a) out of the polynomial perhaps using long division example x^2-1= (x+1)(x-1). Then realizing this fact imply that finding the roots help you factor. then learning about the quadratic formula giving you the roots directly.

[vale dock]

@glass finch Has your question been resolved?

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[night patio]

You roll a standard 6-sided die repeatedly until you get a 5. What is the expected value of the minimum value rolled before (and including) the 5?

[night patio]

@fleet junco Hi. I needed to discuss about this problem

Can you help please. I had a doubt.

[fleet junco]

Sure

[night patio]

this was the expected value right?

What if we calculate the P(1)

and P(2)

Like all these 5 terms

I basically thought of this

to get the probability of getting a minimum of 1 and an n length sequence

Now to calculate P(min = 1), I summed P_n (1) from (n = 1 to inf)

I get P(1) = 1/2

so is this correct?

[silk zenith]

Is that n or m length sequence?

[night patio]

it is n

[night patio]

Also we did solve this question yesterday using another method

[night patio]

.

[night patio]

simplifies to this

[vale dock]

@night patio Has your question been resolved?

[night patio]

nvm solved it

.close

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[fleet junco]

yes that is one way

to do it

but

its way harder

[night patio]

no actually

it was not that hard

obviously your method was easier

but this works too

had to think about it a lot haha

[fleet junco]

oh wow ok

[night patio]

also I found this really cool method on stackexchange

https://math.stackexchange.com/a/3490283/1140083

[fleet junco]

but liek yeah thats really good

but liek yeah, these sorts of problems are always about splitting

into simpler cases

[night patio]

yuppp

thankyou for your help ^^

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[sour dune]

I want to understand how vector rotation works and im trying to rotate <1,0> by any angle, i found out that there is multiplication matrix thing. Anyways im trying to figure out the math way to do that but i always get cornered by sin(theta) and cos(theta). I dont get it how to get the value of sin or cos in a math way

[quasi valve]

you mean how to get a numerical result for cos(theta) and sin(theta)?

[sour dune]

yes, not the calculator way

[rustic olive]

θ

[quasi valve]

well there are only a few special angles where you can get an exact number, and even those will require a calculator if you don't want to leave it in terms of sqrt(3) etc

[sour dune]

I know that there is for triangles

[brisk lynx]

Ok I already answered that one

[sour dune]

I want a math solution, like in algebra, plug the values and get back a single digit in that case x,y of an angle

is there no formula for that?
i only know cos, sin, tan for triangles where the adjacent, opposite and hypotenuse is used
Im so confused

[lyric bloom]

I mean there is taylor expansion

[ebon glade]

just because you need a calculator doesnt mean that its not "a math way"

the formulas for sin and cos are just more involved than all the other stuff you have seen

so you do not ever want to do them by hand

[brisk lynx]

Think about it, where can you find a right triangle? Inside a circle

[sour dune]

well i want to do in programing but if you say so i guess i just use like math library, i just like doing things from scratch

[brisk lynx]

Start by drawing a circle on a graph centered at the origin 0,0 and let its radius be r

Then just pick any one x,y on that circle

Then draw a line from that center which was 0,0 to your point, that's your hypotenuse, and length between them is your radius r

[ebon glade]

I mean, did you also implement things like sqrts and other roots yourself?

or did you let a math library handle those

same thing

[sour dune]

i was planing do all things

[ebon glade]

well then have fun

read up on taylor series

[brisk lynx]

What are you confused about now

[sour dune]

i only understand how i could use cos sin if i got 2 vectors on unit circle, than i could think about a triangle between <0,0> start and 2 vectors

anyways, my only issue is that i wanted to calculate cos and sin in a formula way to get that value between 0-1

[ebon glade]

https://i.imgur.com/Wz4UC9s.png

here are the formulas you need

the series are infinitely long but the later terms are tiny so you can cut them off

the first 15 or so terms will be enough by far

[brisk lynx]

First it's a value between -1 and 1. You're already ngmi if you think it's just 0-1.

Yeah as he said what you're looking for is the taylor series expansion

It's a way to calculate this stuff with just basic arithmetic, you just have to do it a bunch of times

[sour dune]

oh right the dark moon side is -1

[brisk lynx]

And x has to be in radians not degrees don't be a normie @sour dune

[sour dune]

ok buddy, im europian xD

[brisk lynx]

You do understand what the ! means correct?

[sour dune]

Factorial?

[brisk lynx]

Ye

And the ... you can basically infer

Just making sure

[sour dune]

Anyways thanks for help, ill try to implement tailor series and make a vector rotation from scratch.

[sour dune]

but the more i do the more precise it is?

[ebon glade]

yes

you can just do a while loop depending on the precision you want

tho of course you will run into issues with double/floats anyway

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[karmic hemlock]

can someone help me simplify 52!/5!47! without the use of a calculator, my answer is 2, 140, 320 while my calc says 2, 598, 960. i simplified the 47 factorial first leaving 48 x 49 x 50 x 51 x 52 in the numerator and 1 x2 x3 x4 x5 in the denominator, so i cancelled out 2 with 48 which leaves 24, 3 with 51 which leaves 14, 4 with 52 which leaves 13, and 5 with 50 which leaves 10. this leaves me with 24 x 49 x 10 x 14 x 13 in the numerator which equals 2,140,320??? why is this wrong

[karmic hemlock]

<@&286206848099549185>

[midnight pier]

What school is that

[karmic hemlock]

wdym?

[midnight pier]

A calculator would get hard time while doing that 😭
But wait lemme think

[karmic hemlock]

i have a ti-nspire so not really

it can handle some big numbers

[slate folio]

prob rewrite 49*51 as 50^2 - 1

[karmic hemlock]

woudlnt that complicate it more? i dont think its a good idea to add an exponent into the equation

[spring jackal]

51/3 is not 14

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@karmic hemlock Has your question been resolved?

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Channel closed

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as this action is irreversible, and this channel may abruptly lock.

[somber lintel]

.close

[stark wedge]

it'll close by itself, just leave it

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[sick kernel]

feynman's trick for integration

[sick kernel]

can someone explain it to me as a high schooler

[ruby sleet]

Say you have an integral with set bounds (finite or infinite) and want to compute its value

Say $\int_0^\infty \frac{\sin x}{x}dx$

[rocky lotus]

Raphaelisius Maximus MMIII

[ruby sleet]

It would be easier to compute this if, say, the 'x' stuff wasn't there or was replaced with some exponential or trig function

Feynman's trick is you introduce an outside parameter

So that when you differentiate with respect to it

The problem part of your integrand goes away

So what if we defined $I(t) = \int_0^\infty e^{-tx} \frac{\sin x}{x}dx$

[rocky lotus]

Raphaelisius Maximus MMIII

[ruby sleet]

As when we differentiate by t

Some x gets outside, and will cancel with the x on the denominator

So we find $I'(t) = -\int_0^\infty e^{-tx}\sin x dx$

[rocky lotus]

Raphaelisius Maximus MMIII

[ruby sleet]

So now remember we're looking for this integral

[ruby sleet]

Which is just a particular value of this family of integrals at t = 0

Now, the idea is FTC (fundamental theorem of calculus)

To find I(0)

Find I(some appropriate value of t)

And find I'(t) for all t

Since we find that I'(t) = -1/(t²+1), I(t) = -arctan(t) + some constant

Here the interesting t value is "t = infinity"

Then the computation you can do on your own, but that's the gist

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@sick kernel Has your question been resolved?

[sick kernel]

Wait

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@sick kernel Has your question been resolved?

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[wraith dew]

Let KAM be a right triangle with angle K being right angle, side k is 24 cm and side m is 7 cm. Determine the length of side a correct to one decimal. Enter numerical value only.

[restive aurora]

Which angle is K?

Is it between sides A and K, M and K?

[noble anvil]

What do we do when we are given two sides of a right angled triangle and asked for the third?

[wraith dew]

the question just says angle k being riight

wait here

[restive aurora]

If you have a picture that'd be great

[wraith dew]

this is all it says

ive been doing these questions for like 3 hours i cant get them

[noble anvil]

It'll be the right angle with side K being the hypotenuse

[wraith dew]

its always off by a decimal or something

[restive aurora]

Yeah that makes sense

[tidal swift]

this sounds like it's calling for a certain law

[restive aurora]

Oh, so you know how to calculate this?

!status

[vale dock]

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 dew]

uh 3

[noble anvil]

Show work

[wraith dew]

a
2
=k
2
−m
2

𝑎
2

24
2
−
7
2

576
−
49

527
a
2
=24
2
−7
2
=576−49=527
𝑎

527
≈
22.9
 
cm
a=
527
​
≈22.9cm

[restive aurora]

Yeah, can you tell us what your answer was and how you got there?

[wraith dew]

why did it do that

[tidal swift]

oh my goodness

[restive aurora]

Lmao

[wraith dew]

i tried to copy paste what i typed in

[noble anvil]

Send a screenshot

[restive aurora]

Well that's essentially what I got

[wraith dew]

oiaky wait

[restive aurora]

How are you supposed to submit the answer to this question?

[wraith dew]

it says

[restive aurora]

Is it possible it wants you to enter it exactly? Like sqrt(527)?

[wraith dew]

correct to one decimal

[restive aurora]

Ah

[wraith dew]

numerical value on;ly

[restive aurora]

Well

[wraith dew]

thats all

[restive aurora]

I think you're rounding wrong

[noble anvil]

,calc (527)^(1/2)

[rocky lotus]

Result:

22.956480566498

[wraith dew]

im suppsed to write that whole thing ?

[tidal swift]

no

[restive aurora]

Nonono

[tidal swift]

round to nearest tenth

[restive aurora]

5 rounds up

[tidal swift]

your final answer is xx.x cm

[wraith dew]

23 ?

23.0

[tidal swift]

23.0

[restive aurora]

Yeah!

Try that

[tidal swift]

must have the .0

otherwise you are not meeting question reqs

[wraith dew]

wait but what didi do wrong withrounding on the question

[tidal swift]

you rounded down

[restive aurora]

Oh well look at the 2 decimal places, 22.95

To round to 1 decimal place we round up, since the last decimal is 5

So 22.(9+1) -> 23.0

[wraith dew]

what is this picture tooken on

[restive aurora]

My local calculator app, Qalculate

[wraith dew]

taken

[wraith dew]

oh ok then

[wraith dew]

ooh ok thanks alot

how do you guys know so much about math

[restive aurora]

Lots and lots and lots of practice

[wraith dew]

how old are you guys

[tidal swift]

personally, practice, good out of school teachers, practice, friends, and more practice

[restive aurora]

28

[tidal swift]

and books

[tidal swift]

i'm their age, halved, plus 1

[noble anvil]

High school

[wraith dew]

oh damn im only 15

[tidal swift]

oh hey, a peer

[wraith dew]

i turned 15 on the 13th

[restive aurora]

Yeah don't worry about being behind, I was a terrible student until I was like 23 and now I'm doing fantastic

[noble anvil]

Welcome to mathcord btw royal

[wraith dew]

yeah thanks i appreciate iit

[tidal swift]

i'm still terrible dw

together we improve

also welcome to the server!

[wraith dew]

thank you

[wraith dew]

yeah lol

[noble anvil]

!done

[vale dock]

If you are done with this channel, please mark your problem as solved by typing .close

[noble anvil]

I'll close it for ya

.close

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[proper marten]

can you solve E5=M^A if A=E

[stark wedge]

what is E5 supposed to mean?

...

!xy

[vale dock]

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.

[proper marten]

i mean E

it kept the typos

im so sorry

[stark wedge]

post the original question exactly as it was given to you

[proper marten]

E5=M^A if A=E then E5=E1+E2+E3+E4 find M

that was the question that was given

[opal pendant]

so they are just variables

[proper marten]

yea

...

[opal pendant]

what do u have to find M in terms of

cause it wont be a constant value as we dont know the values of anything here

[proper marten]

mass

M=mass

[opal pendant]

..

why didnt u specify that before

[proper marten]

convert eneergy into mass

forgot to tell

[opal pendant]

convert energy into mass?

what?

[proper marten]

i meant M^A

so can you help me?

ive been stuck on this question for a couple 2 days

[ebon glade]

so far this is just a bunch of nonsense

there is a lot of context missing

from E5=M^E and E5=E1+E2+E3+E4 with no other relationship between all those variables you can do absolutely nothing

[proper marten]

but i can swear he gave me E5=M^A

and said M= mass

so what do i do?

[stark wedge]

who's "he"

do you have a picture or screenshot of the question? yes or no.

[proper marten]

No

And hes my tutor

[stark wedge]

can you contact him again to ask for a clearer statement of the question

[proper marten]

He then said M=4 billion kg

Well I dont know how he got that

I think Im missing some info

And he said 4 billion kg is a approximate that rounded up

So can we backtrack to see the missing info or no?

[river shale]

Hi, can you screenshot the message between your tutor and you? We’d like to check out his clarification.

[proper marten]

he wants his info disclosed

[river shale]

The information is kinda scattered rn, it’d be better for you to rephrase and stick all of the information together.

[proper marten]

so sadly i cant

yea i could do that though

[river shale]

Appreciate it, big thanks

[proper marten]

E5=M^A if A=E E5=E1+E2+E3+E4 then if we know M=4 billion kg in the equation

E5=M^A

is that enough info though?

[desert knot]

there's not even a question to answer

[proper marten]

and then what would the mass be then? if E5=M^A

[drifting hornet]

is this part of some physics question?

we need the full context

[desert knot]

well apparently the mass would be 4 billion kg

since that's what you said 3 messages ago

[proper marten]

typo

i think

nvm

[desert knot]

[proper marten]

but still dont know how he got the answer

[river shale]

yeah….better screenshot the messages. It’s alr for you to hide his identity.

It’s not gonna work with these information

[proper marten]

also then he said noww found the enegy

energy

[desert knot]

my guy this is just absolute nonsense

[drifting hornet]

We know 0 context dude, what energy? Mass of what? What even was the original question? All we have is 2 equations including letters that werent even defined

we cant help with this

either you have to provide more context, or ask your tutor

[proper marten]

then said E= energy so E1 is E

[lyric sundial]

...

Don't you think we know nothing about the context of your question??

[proper marten]

im so sorry

[lyric sundial]

We've been telling you this from the beginning

[river shale]

No, don’t apologize. Let’s just make things clear. Is this question assigned by your tutor?

By any chance, could you provide the original question?

[proper marten]

nah he just said its a challange question

[lyric sundial]

It's not because we don't want to answer, but rather because we can't

[river shale]

We are not prophets 😭
I assume neither are you

[lyric sundial]

And so??

[proper marten]

i mean he said it a simple question

[drifting hornet]

so did your tutor give you the question? Or did you send the question to your tutor asking for help? What's the source of the question?

[proper marten]

nah he sent it to me

[river shale]

in text or pic?

[drifting hornet]

Can you either ss the message (censoring the sender ofc) or write it word for word in here?

[proper marten]

i cant because of the private policy contract would be violated

[lyric sundial]

Then we CAN'T solve your doubt.

[river shale]

Well

[drifting hornet]

Hmm, if everything is supposed to be private, i dont think we can help

[river shale]

Unfortunately

[drifting hornet]

we cant help with a private question we cant even see

You'll have to ask your tutor about it

[lyric sundial]

You should understand by yourself that the "info" you've given about the context is too little to understand what you're doing

[proper marten]

Im sorry for the confusion its just always on my mind

Because my tutor said that if I get one of the questions of that equation right then I'll easily pass his teachings and wouldn't need anymore tutoring

[noble anvil]

The question is figuring out what the question is

[proper marten]

In the same boat as yall

But atleast we know that acceleration is apart of the equation of energy ...right?

[desert knot]

[proper marten]

I mean energy5

[desert knot]

No, nobody knows anything like that because you haven't explained anything about the problem whatsoever

[noble anvil]

What is energy 5

[river shale]

No….there is 0 progress

[proper marten]

Energy5=energy+energy2+energy3+energy4

[river shale]

Is this related to kinetic energy?

[proper marten]

He said yes

I mean he did say that when he sent "now find energy"

Do i have to use chat gpt?

[river shale]

Alr so, just for these information, what does “E” stand for?

[proper marten]

Energy

[river shale]

Are you looking for E? Or E5

[proper marten]

E

Because E5 is energy5

[river shale]

So (4 billion)^E = E1 + E2 + E3 + E4

Is this everything?

[proper marten]

Yea

I mean no actual

Because where would E5 be in the equation if we just swapped it out

[river shale]

yeah, I revised it

[proper marten]

Okay

[river shale]

Is there anything missing?

[proper marten]

E1 is E and thats why we needed E5

[river shale]

Damn….you didn’t mention this earlier

[proper marten]

You see where it gets confusing?

[river shale]

But sure, lemme update the information

[river shale]

Too many variables.

What’s your native language btw

So (4 billion)^E = E + E2 + E3 + E4

[proper marten]

Yea

[river shale]

What’s your mother tongue

[proper marten]

Idk my native language

Got confused between English and Vietnamese

[river shale]

I see

[proper marten]

I keep connecting them to each other

But can we answer the equation? Or do we have to ask for more info

[river shale]

We need more information

Better screenshotted originally from your tutor.

[noble anvil]

You gonna go back to class and the tutors gonna say it was a test to see if you could realise if the info is sufficient or not

[proper marten]

I've tried that method and it didn't work:'"

I go up to him and ask that question of is the info sufficient or not?

And if that was the question but he didnt answer

Also he then said that ^E=Mlog7

[ebon glade]

oh great a random 7 from nowhere

[proper marten]

It meant to be a small 7 but the text font put it back to a big 7

Sorry

[ebon glade]

that makes even less sense

[proper marten]

Why do you think im so confused

[noble anvil]

what is a big 7

[ebon glade]

presumably log_7 vs log(7)

[proper marten]

Log(7)

I cant with him sometimes you know?

He changes his mind so many times because it used to be log_7

[noble anvil]

get a different tutor then

[proper marten]

But I do get what hes trying to teach me

Also is that enough info?

I feel like it is

[ebon glade]

this is still nonsense

[noble anvil]

no

[proper marten]

Do we times the M with Mlog(7)?

Or what do we do

Because im out of ideas

Wait he got me more info and said that if ^5=Mlog(7) then E=Mlog(7) times the original equation

Are we getting somewhere?

[river shale]

Hire @noble anvil as your physic tutor

[river shale]

Closer and closer, but still not enough

[proper marten]

So then the original equation was where A wasn't apart of the power

the pain

Can yall put that in one equation for me please I got so sidetracked with the messages that I forgot what the original equation was

And the he said E5=28th of the number of Mlog(7)

Number

I meant 28th number

[ebon glade]

a random 28th. it gets better

on the off-chance that you arent trolling, just fire your tutor

nothing of what you said made any sense

[proper marten]

I mean and then he said E5=28th number in Mlog(7) I put the wrong info in

That's why

That mean we know what E5 equals so then we have to find out what E+E2+E3+E4 is doing in this equation

Wait one min while I find out what E2,E3,and E4 is doing

..were cooked

We are cooked

[stark wedge]

you've been cooked from the beginning

now you are utterly charred

you've become a lump of carbon in the oven

a very well done steak

etc

[open abyss]

just fire this tutor😭

[proper marten]

So basically what he said was that these are all similar equations that equal to a similar rule of the temperature of the sun is the power of all of them in different times zones

Of that supposed sun

[stark wedge]

this is nonsense or bullshit or both.

[proper marten]

Then he gave me time zones

[stark wedge]

or maybe you could finally give us some screenshot of the tutor's message instead of continuing to play broken telephone!!!

[ebon glade]

Ann just give up. its pointless

[noble anvil]

.close

[proper marten]

Lemme finish these please stop texting for a little so we can get all of this in one part of the messages

[vale dock]

Channel closed

Closed by @noble anvil

Use .reopen if this was a mistake.

[stark wedge]

uh

[proper marten]

Thanks

[noble anvil]

this has been happening for an hour

there is 0 progress

.reopen

[vale dock]

✅

[proper marten]

Okay so E2=24 and we have to use the time zone of 12:00 July 15

As for the rule

Then E3=52 and the time zone for the rule is June 7th 12:50

2024

Then E4=8 and the time zone fo the rule is August 7th 12:30

Okay

Then he said use our sun of our solar system for a eazy estimate

August 2024 for the E4=8

Then we calulate

Okay I think we can talk now if this gave us all the info for E5=E+E2+E3+E4

To slove

their all am

amd E2=24 has the year of 2020

thats all the info i think

i need to go calulate all of this

wish me lcuk

he said it should take a couple hours to a month at least for one of these equations

i accidentally did 12:01 am July 15 time zone

:"""""

and used 2019

[river shale]

What does it mean?

[proper marten]

i used the wrong nuumbers

for the first equation

[river shale]

Alr alr, can you put every current information together?

We have to accumulate every piece of them

[proper marten]

that was the only one but it was actually E2=24 time zone of 12:00 a.m. July 24 and they are all a.m not pm

and at the year of 2016

[river shale]

I assume it’s not a pure physic question, more like a riddle

Do you contact your tutor in English or Vietnamese?

[proper marten]

actually in my new language im learning that is french

yep

[river shale]

Hmm….

How about this, you copy your tutor’s message instead of screenshotting them

Messages regarding the question ofc

cuz we have helpers that speaks French

[proper marten]

you sure i cant just translate it? because his "French" is too old that i had to ask a special translator to change his wording into actual reading words in english

he's even using slang that is not in date with the new ones

[river shale]

Don’t tell me the special translator is AI

We would be upset but at least this explains why the question looks so scattered brain

[proper marten]

i pay extra for a translator in Canada to tell me what he's saying and sometimes she doesn't even know some of the time

of wwhat hes saying

so what do we do now?

[river shale]

yeah, we’re not buying this.
Unless we get full content of the question, we’re not going anywhere. Sorry for your inconvenience.

[proper marten]

what other context do i need to show?

or do you need to know

[knotty storm]

bro

if the question is supposed to be private

why r u telling us thi

why r u asking us

[proper marten]

shhh

i want to make this help me in secret

[knotty storm]

bruh this guy is obviously trolling

[proper marten]

but not have it were he could take leagal action

[knotty storm]

its a new account

[stark wedge]

since when is he gonna take legal action over a fucking math problem

[noble anvil]

First day on discord and this is what you decide to do

[knotty storm]

because he is a troll

[proper marten]

could i show a legal form on this discord?

or does it not let ,e

[noble anvil]

Do not show any identifying information

Especially when it's of someone else

[proper marten]

so i cant show his contract? but yall call me a liar

[knotty storm]

show it

[proper marten]

I've been stuck with this crazy guy for so long trying to get at least one question right so i can leave his stupid tutoring

[stark wedge]

wdym "leave his stupid tutoring"

can you not... just leave???

[proper marten]

i made it my goal to atleast answer one question

and this one wa the most simple of the bunch

[stark wedge]

dude.

just leave.

[proper marten]

should i answer it my own?

[stark wedge]

these goofy-ass questions can honestly go fuck themselves

[drifting hornet]

get a better tutor with better questions

[proper marten]

ima try to answer it bymyself

trust that ima be back in about 3 years with the answer

[haughty fiber]

lmao just read this thread

this is nonsensical

[drifting hornet]

just make sure the timezones of the energies dont change during that time

[haughty fiber]

@proper marten I think the complaint is just simply that you're not explaining the question clearly, and in fact, you actively refuse to

mathematics is more collaboration than competition

if you can't help us help you, it will never work out

[proper marten]

trust and close help please

close.

[drifting hornet]

.close

[vale dock]

Channel closed

Closed by @drifting hornet

Use .reopen if this was a mistake.

[river shale]

.close

bruh, got sniped

[haughty fiber]

lmao

[proper marten]

,close

.close

[drifting hornet]

It's closed already

just wait

or keep writing .close. whatever you wish

[vale dock]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[open niche]

The integral on the right has a_k complex numbers so what would its integral be some solutions use just ln and not complex log

[open niche]

,w int(1/(x^24 +1)) dx

[rocky lotus]

Results provided by WolframAlpha

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Requested by taebek.8804

[open niche]

Here for example it uses half csc half of the logs (assuming complex ones mixed so thst its real inside )

[open niche]

But from what i see here there isnt no i arg

[open niche]

Oh conjucatives might cancel Im parts

[storm perch]

I guess you could say they are similar, but $\int\frac1{x-a_k}dx\neq\ln|x-a_k|$ apparently

[rocky lotus]

;(

[storm perch]

Yeah

[open niche]

It is

But ln isnt real ln its complex ln

When lnz = ln|z| + i arg(z)

[open niche]

Are these logs supposed to be ln?

Cause it doesn't mention base

[storm perch]

Yes, that's how wolfram treats it

[storm perch]

Of course mse lies to me

[storm perch]

I asked the exact same question and they said this, so I gave up

[vale dock]

@open niche Has your question been resolved?

[open niche]

If its refered to complex z then it is

[storm perch]

To answer the question, it depends on the notation that you're using (e.g. ln and log can be mixed up), and I would say that if it uses ln in the real sense, there would not be a proper integral

[open niche]

What form of roots should i use

[open niche]

When expanding this

e^iθ or csc

[open niche]

Whats best form you can use for representing all roots ak

e^iθ or csc

For easier expansion

[worn sparrow]

i would say something like

$\eta_k$ with a note "where $\eta_k, k=0,1,...,23$ are the 24th roots of -1

[rocky lotus]

gfauxpas

[vale dock]

@open niche Has your question been resolved?

[open niche]

Thats what it says lol

Just instead of η i have a