help-39

[dusk otter]

I believe they are!

Here is the second one!

[pearl pond]

@dusk otter Has your question been resolved?

[autumn trellis]

This looks good too

[dusk otter]

Great thanks!

[autumn trellis]

Except for the second line where you forgot to put the integral sign, but the integration is correct and the result too

[dusk otter]

oops yes sorry!

is the one above correct?

[autumn trellis]

Yep, looks good too

[dusk otter]

Okie 2 more to go!

Here is the next one

[autumn trellis]

For part a, you have to use the initial condition given and find the specific solution

[dusk otter]

Oh my I didn't even see that thanks!

[dusk otter]

How's this?

[autumn trellis]

Yeah

[dusk otter]

Okie the last one I am unsure about

I am not sure that I found the mass right

because in all my profs examples they have all been bounded by one curve not ever too

[autumn trellis]

You incorrectly substituted delta(x) here, also in this case the mass of the plate will be $\int_{0}^{1} \rho(x)(x^{1/4}-x^{2}) dx$

[jolly parrot]

smidgin

[dusk otter]

Ohh yess that makes sense

Since you have to do upper - lower

to find region in between

[autumn trellis]

The density stays constant for a given x coordinate, so the mass of a small slice with constant x coordinate will be (density function)(length of slice={x^1/4-x^2})(width of slice=dx)

[autumn trellis]

Yes

[dusk otter]

Is the moment part correct?

Wait wouldn't I have to write it in terms of why or no?

Because isn't dx for vertical slices?

[autumn trellis]

Yes, and the density for a vertical slice is constant

[dusk otter]

wait no either way

Yeah thank you!

Gotcha 🙂

[autumn trellis]

Yep the center of mass calculation looks right except for the mass of the plate part

You should also change the diagram you made for the finding mass part in case someone will be checking, since we took vertical slices to find the mass

[dusk otter]

mhm just did that!

It's just homework but no one will be checking but still important to fix!@

[dusk otter]

How's this?

[autumn trellis]

👍

[dusk otter]

Thanks for the help I appreciate it!

[pearl pond]

@dusk otter Has your question been resolved?

[pearl pond]

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[tired lance]

Help please

[pearl pond]

Please don't occupy multiple help channels.

[midnight haven]

could you .close? it appears you made a duplicate

[tired lance]

.close

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[pastel oriole]

Find the zeroes, and then find the difference between them; y=x^2+bx+c

[pearl pond]

@pastel oriole Has your question been resolved?

[midnight haven]

can you show your work?

[pastel oriole]

i have done the previous question from this where it shows to find the vertex

which was -b/2a

@midnight haven

[midnight haven]

Do you know the relation between the sum of the roots and product of roots?

[pastel oriole]

no

[midnight haven]

they have taken quadratic equation of the form
ax^2 + bx + c = 0

but your equation is x^2 + bx + c = 0

so a = 1 , in your case

so the sum of the roots x1 + x2 = -b

and the product is c

so

x1 + x2 = -b
x1*x2 = c

and you have to find the difference of the roots , which is x1 - x2

use the following identity and show your work

$$ (a-b)^2 = (a+b)^2 - 4ab $$

[jolly parrot]

just panic

[pastel oriole]

so its just 1

but what condition on the constants b and c must be satisfied for the difference to be exactly 1 if 1 is the answer?

[midnight haven]

noo

1 is not the answer

a , the coefficient is 1
in your case

the standard form is ax^2 + bx + c = 0

but your equation is x^2 + bx + c = 0

so a =1 , I said

[pastel oriole]

so the zeroes are 1

[midnight haven]

NOO

I have given you relation between the zeroes

let the zeroes be x1 and x2
so x1 + x2 = -b
x1*x2 = c

you have to find the difference

which is x1 - x2

and I have given you the identity to solve

how old are you , or which grade is this?

[pastel oriole]

im in yr 10 doing yr 11 work

[midnight haven]

then , u should be able to understand

[midnight haven]

read this again

and what i have sent before

[pastel oriole]

sorry im used to identifying zeroes in another forma

[pastel oriole]

do you want me to expand and simplify?

[midnight haven]

no no

use the formula to get the difference of zeroes

[midnight haven]

please read this again

[pastel oriole]

so am i going to subsitute

@midnight haven

[pearl pond]

@pastel oriole Has your question been resolved?

[pastel oriole]

<@&286206848099549185>

[pearl pond]

@pastel oriole Has your question been resolved?

[pastel oriole]

bruh

.close

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[velvet hound]

For which values of the real parameter λ, the vectors a_1 = (λ,-1,-1), a_2 = (-1,λ,-1) and a_3 = (-1,-1, λ) are not linearly dependant

[velvet hound]

ive got no clue how to solve this can someone pls help?

[sudden talon]

The three vectors are not linearly depenent.

Make sure that the combination of two vectors can't be the other one.

[velvet hound]

false. we need to find the parameters of lambda for which those 3 vecotrs are not linearly dependant

for the values -1 and 2 i found a solution online .

.close

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[thick copper]

Resolve this system

[thick copper]

Hello, I tried this

[cloud zephyr]

L2=L2-2L1
2 -1 1 -2 | 2
-) 1 1 -3 1 | 5 ×2
0 -3 7 -4| -8

[thick copper]

Hello, I don't really understand what is the step before that

[pearl pond]

@thick copper Has your question been resolved?

[feral leaf]

When reducing a matrix, you want to use one constant row to reduce the others. Meaning that in your first step, it's incorrect because you used row 1 to reduce row two and the original row 2 to reduce row 3. You forgot to take in consideration, at the fact that after you reduced row 2, it becomes something new, therefore, you can't use row 2 to reduce row 3

[cloud zephyr]

oh, it's your first step

you got it wrong

[thick copper]

Ohh makes sense thank you so much

[thick copper]

Oh ok, you did L2=L2-2L1, besides that it seems that you changed rows (?), and the last row I am really confused tbh

[cloud zephyr]

the last row is the resultant

i just write down the resultant of L2=L2-2L1 for checking

[thick copper]

Ohh lol okay thanks!

[cloud zephyr]

i think if you do it carefully again, you would get the result you want!

do ping if you have rewritten your work and wanna have it checked

[thick copper]

Thank you guys I did it ❤️

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[burnt raptor]

i found asnwer as C but it says it supposed to be A. where did i do wrong?

[light helm]

i can't read your work

[burnt raptor]

isn't this C?

[light helm]

yes

[burnt raptor]

same for these it says both supposed to be E

[light helm]

these two are indeed supposed to be E

finally deciphered your work
the end result you had is correct for the previous question, but your work is a bit flawed

[burnt raptor]

ok i did 73 again it's E but i didn't get 72

[burnt raptor]

how so

[light helm]

your g'(x) isn't applicable for all x

[burnt raptor]

i took derivative wrong?

[light helm]

you took an unnecessary loop, and ended up writing someting a bit invalid

in general, how would you differentiate
x^2 f(x)

forget about what's given in the question

[burnt raptor]

(x^2)' f(x) + x^2 f' (x)

[light helm]

yes

and f(1) can be read directly from the graph

[burnt raptor]

f' (x) is 1/squareroot 3 right?

[light helm]

f'(1) is 1/sqrt(3)

with the g(x) you had, you'd run into issues if you used another value than x=1
as that'd give you a point on the tangent line instead of the curve for the f(x) component

[burnt raptor]

isn't f(x) = x-1/sqrt(3) +4?

[light helm]

no

[burnt raptor]

why not

[light helm]

f(x) is the curve

ignoring your absence of parentheses
y = (x-1)/sqrt(3) + 4
is the equation of the tangent line

which is not the same as your curve

[burnt raptor]

but i can find slope of it

tilt angle

[light helm]

yes, but you already had the slope before you had your equation

from tan(30°)

[burnt raptor]

okay i get it now

you're right

but was my method really wrong?

[burnt raptor]

yes it was true

[light helm]

using what you wrote, you'd end up getting the red point instead of the blue point at a different value of x

for that f(x) component

[burnt raptor]

f'(x) is 1/squareroot 3 only for x=1

[light helm]

yes

well not necessarily only
just not all values as the slope could potentially be the same at another far off location

[burnt raptor]

that makes lot of sense

[light helm]

as for q72, consider chain rule or the use of a substitution

[burnt raptor]

why is 72 D here

okay give me a minute

yeah you're right

lol

thanks a lot you really helped me

.close

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[quaint crown]

Can we have negative triple factorial?

[quaint crown]

and how?

?

[kind lake]

sory broI shared it on the wrong channel

[quaint crown]

Okay

[quaint crown]

Like with double factorial (-1)!!=1

[pearl pond]

@quaint crown Has your question been resolved?

[quaint crown]

damn

[quaint crown]

.close

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[burnt raptor]

there should be some relation between numbers either in same column or line

[burnt raptor]

i can't find it

[light helm]

some of these numbers look familiar

64

125

512

do you recognise anything special about them

[burnt raptor]

oh for god's sake

lol

thank you

idk how i didn't see that

i was looking it wrong

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Channel closed due to the original message being deleted.
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as this action is irreversible, and this channel may abruptly lock.

[swift lodge]

Help

[light helm]

please read the bot message above

[swift lodge]

I did intend to delete it tho

[light helm]

regardless, the channel is set to close

so unless you want the potential convo to be cut off,
make a new channel

[swift lodge]

Ok

.close

[sharp smelt]

It's already closed

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[hard crystal]

1.Path functions are function which depend on the path followed b/w 2 states
2. Path function are function which does not depend upon initial and final states.

[hard crystal]

Which statement is true ?

[gilded temple]

1

[gilded temple]

r u indian

[hard crystal]

Why 2 is not true?

[gilded temple]

by definition

[hard crystal]

Yes

[gilded temple]

jee aspirant?

[hard crystal]

Yes

@gilded temple you too?

[gilded temple]

yea

lol

[tawdry orbit]

Jee aspirants here as well

[gilded temple]

11th?

[tawdry orbit]

No am 2024 aspirant

You guys?

[gilded temple]

2025

[gilded temple]

so u wrote jan attempt

[tawdry orbit]

Yup mains

[gilded temple]

nice

how much do u expect to get?

[tawdry orbit]

Around 200 31 jan s1

Did a lot of silly mistakes in physics and chemistry

[gilded temple]

thatss a rlly good mark actually

did u go to some coachiing

[tawdry orbit]

Ya and my shift was toughest so i hope to get good percentile

Although i'll work to improve in april

[tawdry orbit]

No

You enrolled in any?

[gilded temple]

damn 200 without coaching is unbelievable

[gilded temple]

yea

[tawdry orbit]

Which one

[gilded temple]

allen

[tawdry orbit]

Oh thats the best one as far as I know

[gilded temple]

but it still largely depends on the amount of work u put in ursellf

[tawdry orbit]

Right

[pearl pond]

@hard crystal Has your question been resolved?

[pearl pond]

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[primal sable]

whats your take

@waxen sequoia

[waxen sequoia]

really dont have one. The answer was given im trying to figure out how to solve it

[primal sable]

yeah but what did you try so far

[waxen sequoia]

im actually not sure where to start lol

[primal sable]

did you think how 51 is composed of 2 or more integers

[waxen sequoia]

yes we can have the lagerest number to be 50 and smallest to be 1

[primal sable]

what else

[waxen sequoia]

thats pretty much it

[primal sable]

did you think of sets containing more than 2 elements

subsets*

[nimble osprey]

hint: pigeon hole principle

[primal sable]

yeah i was just building the intuition

you did not had to spoil it

[waxen sequoia]

lol

[nimble osprey]

it is not already so trivial i think...

[primal sable]

huh

[waxen sequoia]

can we continue? @primal sable

[primal sable]

you may ask everg now

[waxen sequoia]

why?

[cyan vortex]

26 + 25 = 51

the size of the subset is equal to the highest value integer in it

[nimble osprey]

you can continue with your help...i only send a visual hint ...i don t think that my answer is full

[waxen sequoia]

how do i know if its 26 or 25 for the answer

[cyan vortex]

because if you consider 25, it would be 25 + 24 which is 49

[waxen sequoia]

and how did you arrive to 25 + 26 without doing this

did you do trial and error?

[cyan vortex]

you can do that but since its multiple choice its faster to just check the options

if that appears during a test

[waxen sequoia]

ahh so given the options you do n + (n-1)

[cyan vortex]

sorry, im not sure how to do this one

[waxen sequoia]

ill open another channel thanks

.close

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[waxen sequoia]

how would i start this?

[autumn narwhal]

Try to think how to construct a palindrome that is longer than an existing palindrome

Like, find a recurrence relation

[waxen sequoia]

how exactly?

[autumn narwhal]

If I give you a palindrome of length n,

What can you do to it, in order to create a longer palindrome?

[waxen sequoia]

adding same characters to the sides

why is it P_n-2?

[autumn narwhal]

This creates a new palindrome of length n+2, correct?

And how many "characters" do you have?

Like, how many new palindromes can you create from the one I gave you?

[waxen sequoia]

n-2?

[autumn narwhal]

No

[waxen sequoia]

then

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[fallen ibex]

i cannot comprehend how these 2 equal each other

[fallen ibex]

i understand that you could take the 2 in f(2x) and take it out

but i dont see how they change the

1 and 4

into 2 and 8

i dont get why thats allowed

[summer juniper]

Idk too but I see it this way
take 2x = t
2dx = dt
dx = dt/2
The limit should change to since we introduced a new variable t
We know 2x = t
Putting upper limit , x= 4.
2 *4 = t implies that t = 8
So upper limit is 8 for the new variable t
Now,Lower limit x = 1
Put in 2x= t implies that t = 2
So lower limit for it should be 2

[fallen ibex]

so when you have an integral and theres a constant inside of the integral you can take it out and divide the boundary points by the constant it seems

[summer juniper]

You can use it as a general property if you want but you might have to change limits too so be careful.

[feral leaf]

Do you know u sub?

Or understand the concept of it?

[fallen ibex]

yea

i understand it

i think

2xcos(x^2) u = x^2 then you can

say

cos(u) du

[feral leaf]

That's basically what it did, it did a sub

[summer juniper]

Fr

[fallen ibex]

it would be nice if they give me more explanation steps tho 😢

[feral leaf]

So u = 2x then du = 2 dx, therefore (1/2)du = dx

[feral leaf]

It doesn't help that the work used the same variable for the sub

[fallen ibex]

ohh

that does make it more confsuing

using the same

letter

yea

[feral leaf]

Anyways the $\int_2^8 f(x) dx$ is the result after doing the sub

[jolly parrot]

CaptainNova22

[feral leaf]

Because it's in terms of u, you have to change the bounds as well

[fallen ibex]

here is an example of me doing u subsitution

i think thast what i do wrong i just

dont change the boundaries xd

[feral leaf]

Yeah, you have to change the bounds

[fallen ibex]

i thought plugging the boundary points into the equation of u would work but i still get the wrong answer for this

like here the boundaries would be 9 and 6.25 i thought

but it coems out wrong

would become*

[feral leaf]

Because here, 1 to 12 is the bounds for when the expression is in terms of x

[fallen ibex]

bc 1/4 * 12 + 6 = 9

ohh

that actualy makes sense

im chaning the x axis into a u axis

in a way

and im using

x values for coordinates on the axis

[feral leaf]

6.25 to 9 are the bounds for terms of u

[fallen ibex]

so i did get that part right atleast

i think i understand it somewhat

i should be able to compute it now

[feral leaf]

So to answer your original question that's why it's 2 to 8 because that was for terms of u

Like mentioned, the solution using the same variable is confusing

[fallen ibex]

theyre basically saying u = 2x and then the

boundary points

become twice as big

bc you can plug them into

u = 2x

yes

that makes sense thank you

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[feral leaf]

Np

[pearl pond]

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[hushed atlas]

How? I can't understand what happened in the middle

[summer juniper]

Add +1 and subtract -1 on the LHS

1+(n/n+1)-1+1
2+(n/n+1) -1
2+ (n-n-1/n+1)
2+(-1/n+1)
2-(1/n+1)

[hushed atlas]

Thank you very much

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[muted night]

i know that for a subset A included in a subset B, it's 2^{n-1}-1 but idk how to do it for 3 subsets

[pearl pond]

@muted night Has your question been resolved?

[muted night]

<@&286206848099549185>

so this is my question

.

also the subsets have to be different than the set Z and the empty set

[pearl pond]

@muted night Has your question been resolved?

[muted night]

<@&286206848099549185>

[edgy stone]

every element must be in either A and B and C, or B and C but not A, or C but not A or B, or just Z

ie. every element must be in D or E or F or G

so how many ways are there to do that

and then you just have to restrict it a bit more because they aren't allowed to be empty or whatever

[pearl pond]

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[muted night]

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[paper sorrel]

Is it actually 41? sometimes when I do this it ends up being like 40 or 42 or sum

[brittle tinsel]

the formula for percentile is usually $k = 100 \cdot \frac{x + 0.5y}{n}$, where x is the number of data values below yours, and y is the number of data values equal to yours

[jolly parrot]

doaby

[paper sorrel]

so basically the answer is 40

[brittle tinsel]

well no, it says round to the nearest whole number

y in this case is 1, since there is 1 data value equal to 41

[paper sorrel]

I see so I should be good going 41 this time

Its always iffy for me with this one

[brittle tinsel]

is 41 what you got from the formula I gave?

[paper sorrel]

i must've been doing it wrong since i was gettign 129.68

damn it was 39

wow

i finally see what u meant by that formula

stupid of me mb

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[hearty breach]

Can someone explain this two problems to me

[west notch]

are you familiar with integration by parts?

[hearty breach]

Yes that what I was trying to do

[west notch]

ok what part are you confused on?

for 1

[hearty breach]

Im confused on the ∫ u dv = uv - ∫ v du on the (dv and du)

[plush bramble]

You'll have to do the IBP twice for the first problem

[west notch]

what did you make your u and dv

[west notch]

yeah it's a 2 cycle

[pearl pond]

@hearty breach Has your question been resolved?

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[grave elk]

Need help with 2.1, i asked previously how to do it and figured it out but im not sure why the same method isnt working now

[jolly parrot]

jan Niku

jan Niku

[snow sail]

can you show any work?

whats giving you trouble

[grave elk]

Ive just been inputting it into a calculator

[snow sail]

a calculator? what kind of calculator

like an integral calculator?

[grave elk]

A phone one since my ti 84 broke so maybe not reliable :/

someone told me before to calculate from 0 to 5 and then 5 to 9 but i think i’m just stuck

[snow sail]

well you can always use https://www.integral-calculator.com/

this one has better typesetting than most calculators so its easier to spot if youve made an error

[snow sail]

so, okay, there is one thing you should think of

[grave elk]

Yeah i did that and got 1422

[snow sail]

ah, yfoure integrating the wrong function

so, you wanna use the velocity function

not really even that directly but, thats the first piece

since the velocity tells you how position changes with time

and you are going to integrate with time

[grave elk]

Oh is it the 18t-90?

[snow sail]

it is, but you also want to be sure that you don't get cancellation

as in, if your velocity is 4 for 3 seconds

then -4 for 3 seconds

and you integrate this

youll get 0

can you think of how to check if were going to have cancellation?

where is the velocity negative?

i guess when is the velocity negative is really want im asking

[grave elk]

At 0 its negative?

[snow sail]

yea, thats definitely an issue

because the arm is moving backwards, right

but, its still moving, so thats positive distance

maybe we have a way to make the velocity always positive?

like the velocity function as it is, it tells you direction and speed

but we almost dont care about direction

5 feet forward is 5 feet backwards is 5 feet for this problem

[grave elk]

So we just do absolute value right?

[snow sail]

yea, thats one option!

[grave elk]

I did the calculator with the right equation of 18t-90 and i got -81 so i put 81 but its still incorrect 😞

[snow sail]

,w draw y=18x-90 from x=-0.5 to x = 9.5

[jolly parrot]

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[snow sail]

well i give up

[grave elk]

LMAO

dont worry

[snow sail]

but

in general

$\qty | \int f \dd t | \neq \int |f| \dd t$

[jolly parrot]

jan Niku

[snow sail]

i think this is the mistake you made

you took abs of the result

[grave elk]

I actually figured it out i just did 0 to 5 and then 5 to 9 with the right equation and added those absolute values

[snow sail]

but, we want the abs of the function

yea, this works!

glad you got it

[grave elk]

I was accidentally adding one without it being an absolute balue

So i kept getting 81

Thank you!!

[snow sail]

it happens

np

[grave elk]

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[final steeple]

Hello. I need a little help with, I’ve verified this with explicit numbers so i know it’s true but my first instinct would’ve been to do this.

[final steeple]

I know this is wrong but can anybody explain it to me

[pearl pond]

@final steeple Has your question been resolved?

[brittle tinsel]

okay this formula isn't quite correct, it's kinda hard to see what you're trying to do

oh wait nvm I see what you did

[brittle tinsel]

this is fine, but you would need to keep the 2 in the 1 • 2 • 3 • 4

[brittle tinsel]

if you factor out a 2 from here, you can only factor it out from one of the numbers being multiplied. you'd be diving 2 out twice if you did it like you said

[final steeple]

So basically, take 2 once from those multiplied. But if I had 1 * 2 * 3*... (n+1) +2 + 2 there. Taking 2 would yield =1 * 3 *...(n+1)+1+1 ???

[brittle tinsel]

yes that is true

[final steeple]

Thank you!! I hope your food is tasty and your life be filled with boundless happiness 🌟

[brittle tinsel]

lol thank you 🫡

[final steeple]

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[final steeple]

Noob question. Im going through the book How to Prove it by Velleman. He defines prime numbers as integers larger than 1 that cannot be written as a product of 2 smaller positive integers. My question is, isn't one a positive integer???

[misty stump]

yep

[final steeple]

Ohhh

I get lmao

So it can only be written by 1 smaller positive integer considering the other one is itself

Holy macaroni

[misty stump]

yep haha

[final steeple]

Thank u so much. Self studying makes me dumb

[misty stump]

i don't think i did anything but yw

[final steeple]

You are my lightning rod hahaha

Thanks

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[vestal terrace]

can someone tell me why the vectors span R^2?

[vestal terrace]

does span R^2 mean that any linear combination in the set should make any vector in R^2? so, how to make vector [1, 1] using the vectors in set?

or did i misunderstand something?

[pearl pond]

@vestal terrace Has your question been resolved?

[sharp vigil]

1. it does mean that you can construct any vector in R^2 using a linear combination of the vectors in the set (although you don't actually need all of them, any two will suffice)
2. to find a the coefficients needed to construct [1, 1] using a pair of vectors from the set, you can set the coefficients as variables, then expand out until you have a system of linear equations and solve as normal

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[vestal terrace]

thanks

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[shell bay]

how is this incorrect???

[light helm]

sin^2(a) - 1 isn't cos^2(a)

[summer juniper]

Yes

• sign

[shell bay]

should be using pythagorean identity tho no?

[light helm]

yes, but you need to use it properly

start with the common form
s^2 + c^2 = 1
from there what's
s^2 - 1?

[shell bay]

-c^2?

[summer juniper]

-c^2

exactly

[light helm]

yes

[shell bay]

so it would be -cot(a)?

[light helm]

yes

[shell bay]

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[shell bay]

Thank you!

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[boreal tapir]

hi

[boreal tapir]

how can i prove this

using a wonderful limit

[exotic roost]

can you write it out?

[boreal tapir]

lim x-> ∞ lnx / x =0

[exotic roost]

ok im awsnering

please wait

[waxen talon]

Use expansion of ln(x)

[boreal tapir]

kk

wdym ?

soo log 10 x /log 10 e

[exotic roost]

@waxen talon take over

[waxen talon]

Expansion series

[boreal tapir]

idk what is that

😦

in class we used the

but that limit tends to 0

[terse cedar]

do u know l'hopital rule ?

[boreal tapir]

no

we didnt learn it yet

[terse cedar]

are u familiar with big O and little o notations ?

[boreal tapir]

no

[midnight haven]

why l'hopital this is the definition of the derivative

[waxen talon]

I'm not sure

[terse cedar]

i was talking about this limit

[waxen talon]

If this method works

,w ln(1 + x) expansion

[jolly parrot]

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[midnight haven]

,, \m {f'}a = \lim_{x\to a} \4{\m fx - \m fa}{x-a}

just use this

its the definition of the derivative

[boreal tapir]

oh

ok

thx

[midnight haven]

a = 0 here

so just take the derivative of f(x) and plug in a

[boreal tapir]

k thx

bye

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[whole valley]

Alright. So we go to a casino with x money. Every bet gives out what you bet, so bet a dollar and you win a dollar or loose it. winning a bet has a 15% chance of happening. Im using the martingale system starting with a dollar and doubling down on every lose, and if i cant afford the lose i start over with a dollar bet again. If i win i start down at a dollar again, How much money i need for this system to yeild a 50% of winning 100k dollars profit while the other 50% chance is of going broke before i do it

[whole valley]

Theres sub questions but where im stuck is comparing xmoney and lose streak since losestreak is a function of xmoney, so i would like to know how you do it in short 😅 thanks

Statistics/probability question

[pearl pond]

@whole valley Has your question been resolved?

[whole valley]

<@&286206848099549185>

[pearl pond]

@whole valley Has your question been resolved?

[pearl pond]

@whole valley Has your question been resolved?

[whole valley]

<@&286206848099549185>

[midnight haven]

ya

tell

[pseudo oxide]

@midnight haven not helpful

the question is literally up there

[pearl pond]

@whole valley Has your question been resolved?

[midnight haven]

there are two

[whole valley]

What is the value of xmoney

<@&286206848099549185> question is what is the value of xmoney, number

[midnight haven]

multiply both nmr and dmr in with x then do Direct substituition

u will get 0

[whole valley]

This is a diffrent ticket now, he also got his answer

[vestal tapir]

it doesn't even sound like it's definitely possible

maybe you can't get an arbitrary chance

[whole valley]

If you put a number you will get the other number, you just dont use all of the info at the same time

From what i was told its all about functions, you play with the numbers and ratio @vestal tapir

I hope that makes any sense

[vestal tapir]

well ok 50% is not very arbitrary, i can see it being possible

[vestal tapir]

i give up

[whole valley]

Alright, you got any leads on the functions relationship before.you go?

[vestal tapir]

no sorry

[whole valley]

Alright, thanks for trying

<@&286206848099549185> has not yet been resolved

[midnight haven]

hello

i will see

[whole valley]

Thanks,

[midnight haven]

i dont think i can help

which topic is this

[whole valley]

It says above, probability/statistics

Also previous knowledge of martingales would help a lot

[midnight haven]

i j started as level stats

im sorry i cannot help

[whole valley]

Do you think i can ping a mod or someone? Its been 8 hours today with no answer

[midnight haven]

have u tried googling it

[whole valley]

Googling which part

[midnight haven]

the question

[whole valley]

I did but..

Didnt get a strightforward answer

i think i just need someone who studied martingales

[unborn abyss]

i don't think that's how martingales work

the amount of bankroll at the start doesn't change the odds of doubling your money before going broke

[whole valley]

It does not, it just effect the lose streak, and both have only 1 definite number that confirms all the previous data

[unborn abyss]

oh i misread 100k as 100%

[whole valley]

And its not about doubling your money, its about 100k

[karmic fern]

,w solve 0.85^n = 0.5

[jolly parrot]

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[karmic fern]

are you fcking kidding me

[whole valley]

Huh?

[karmic fern]

okay so, you have a ~56% chance of not going broke in 5 doublings

[whole valley]

What??

[karmic fern]

so at a starting bet of 100k, you need 3.2 million dollars

[vestal tapir]

the starting bet is 1

[karmic fern]

(might be off by an order of 2)

oh you can't start with 100k?

[whole valley]

Thats not how this works.. you cant just tree diagram your way out of this

[whole valley]

Must start at a dollar after every win

[vestal tapir]

this is about winning 100000 times, but when you lose you still have money, and that even without being able to use winnings

[karmic fern]

oh if you have to start at a dollar, then you cannot guarantee a certain probability of winning 100k

even if you have [an arbitrarily large amount of] money

[whole valley]

Its about profit so you actually can use the winnings, good point, i forgot to mention that

I guess i just didnt get to that part yet

[karmic fern]

like there's some cutoff

[whole valley]

You can. Its probability

[karmic fern]

basically there is a finite, nonzero probability that you go to ruin

[vestal tapir]

we want that

it's part of the task

50% ruin

[whole valley]

Yeah. Its a function of xmoney

[karmic fern]

no, you can't control the probability of ruin of the game because you don't control any of the betting

[whole valley]

Its a set game..

[karmic fern]

like there is a probability that you go down and never cross a certain threshold againn

it's not like the unbiased random walk where you continually re-cross every point in the space

in any case, we can calculate this

[whole valley]

Oooo

Im thrilled to hear how you fill all the variables

[vestal tapir]

it makes sense that you can't precisely control it, to arbitrary value

is thast what you're talking about?

we clearly can make it as high as we want

[whole valley]

Look at it as an equation, you put a number and its either wrong or true, are you saying there are no trues existing as an equation?

[karmic fern]

that's the point; you actually can't. There is a certain probability that you never cross the +100k mark, just as there is one where you never cross the +10 mark, no matter how much your initial balance is

[vestal tapir]

you;re confused

[whole valley]

0 mark?

[vestal tapir]

we have $1024, we have 52% chance to win 3 times immediately, and have $1027

[karmic fern]

you can cross +0 because your martingale increment is +1

[vestal tapir]

what are you talking about

i mean 2048 i guess

[whole valley]

This is a fixed scale therefore it has an answer, you could simply run it on code to find out

[karmic fern]

oh I see, you can reduce your probability of losing each martingale streak to arbitrarily close to zero

so you never see the negatives

[whole valley]

Thats martingales for ya

I dont know what your looking at but we.got 2 unknown numbers, xmoney and.losestreak, i would start from there

[karmic fern]

so you're basically asking for the median of the maximum losing streak if you have to repeat 100k martingales

[whole valley]

100k WINNING martingales

[vestal tapir]

so the problem is you're also allowed to lose the streak, you still have money, it's not necessary to always win, and worse, you can use your winnings

so i gave up

[karmic fern]

optimally, you don't want to lose the streaks because then you'd have to win back that streak and the next

at least this is the case where your p winning each round goes to close to 0, but not sure if 0.15 is close enough

[whole valley]

If you loose you just double down, every win adds a dollar to the profit

[vestal tapir]

yeah i mean completely lose

so you go to bet 1

[karmic fern]

i.e. you never "give up" on one of your streaks in that you take a loss and then go back to 1

[whole valley]

Maybe if we use the data right

[karmic fern]

the number of losses you take in one martingale round is geometrically distributed

[whole valley]

Maybe if we solve the first lose streak we could solve the final one

[karmic fern]

maximum of geometric distributions is kinda rough but probably solvable

[whole valley]

Yes.

[karmic fern]

do you care if this is approximated

I honestly don't want to be bothered by the discreteness

[whole valley]

So lets go from scratch, do you guys have any idea how to show what function lose streak is to xmoney? Because 1 unknown is solveable.

A lot more

[whole valley]

I do, everything matters in statistics

[karmic fern]

lol

tell that to every person who uses asymptotic approximations in their hypothesis testing

[whole valley]

Their task is not researching martingales

[karmic fern]

lmao

[whole valley]

Its to solve for a problem

[karmic fern]

in any case, it's just find the length such that P(max(G1, ... G100k) <= length) >= 0.5

[karmic fern]

if you knew what a martingale actually is

and P(max(G1, ..., G100k) <= length) can be factorized as P(G1 <= length)P(G2 <= length) ...

[whole valley]

Its a sequence

[vestal tapir]

,calc (1-0.85^74)^1e5

[jolly parrot]

Result:

0.54968269883732

[vestal tapir]

that's the dumb answer

[karmic fern]

hmm actually the length doesn't give you that much because you can use the previous winnings

but the previous winnings are pretty insignificant compared to the necessary bankroll

[whole valley]

Their not

Just as you can have 20 winnings you can have 80k winnings

[karmic fern]

yeah but your bankroll is at least in the trillions

your cumulative winnings are on the order of ~100k, no?

[whole valley]

Your winnings could also be a recover of what you lost.

[karmic fern]

in your scheme, you never lose until you go broke on one round

[whole valley]

Yes

[vestal tapir]

it's insignificant

[karmic fern]

like you will probably need 2^73 dollars or something to make it, no?

[vestal tapir]

the winnings don't help, so little help, so hopeless

[whole valley]

They could also include a lose. For a example you could recover.your entire first lose streak when you start again from a dollar, do you still count that as winnings?

Thats a pretty high precent

[karmic fern]

huh, no? you never lose in between the martingaling streaks

[whole valley]

What?

[karmic fern]

i.e. you basically just go back to a dollar 100k times

[vestal tapir]

if you lose the streak, you're done, what you have left plus what you won doesn't help

if we don;t account for that, we get about the same number

it's not satisfying, sure

[karmic fern]

yeah basically if you ever can't make the doubling bet, your remaining money is no way in hell going to get you all the way back up

[whole valley]

Are you sure? I havent calculated it but we should have between 1to40% of our original money after loosing an unaffordable streak

[karmic fern]

you can give a Markov bound on that

[karmic fern]

yeah but now you have to make back all of your original money instead of just the 100k

and your original money is in the order of the trillions or hundreds of trillions

[whole valley]

Are you sure were in the trillions?

[karmic fern]

absolutely

[whole valley]

Alright, so whats the lose streak we recieved?

[karmic fern]

like 74 or something according to frowny's calculation

[vestal tapir]

i can be off by one or even two

[whole valley]

We need the exact lose streak to find xmoney

Hmm

[karmic fern]

no you don't lol; 2^74 is more million-dollar bills than the world can print

[whole valley]

Bro that might aswell be 1/100000000 cents. That doesn't matter

I want to do it right

@vestal tapir How did you find xmoney?

[karmic fern]

it's still going to be 74; the discreteness gives you the benefit of discarding all of the insignificant effects lol

[whole valley]

.

[karmic fern]

what do you mean by do it right? you're asking for a whole number answer

and if you bound the probability P(winning with only enough bankroll for 73 losses) into some tight range that is less than 50%, then you need 74

do you have something against inequalities or something

the effect that you have some of your money leftover adds close to nothing to your probability of winning

if you could choose where you started in this martingale "strategy", maybe, but you can't

[whole valley]

Alright, so how did you calculate the lose streak?

[karmic fern]

how long each losing streak is is distributed geometrically with p = 0.15, i.e. your probability of winning

you can either look up the formula for the cdf of this distribution or figure it out yourself

it's something like (1-0.85^x), totally forgot

[vestal tapir]

you need 7 dollars to bet 3 times, so 74 is right i think

[karmic fern]

in any case, you know that this raised to the 100k has to be greater than 0.5

[whole valley]

Yeah, what did you do with that

[vestal tapir]

you're able to bet 74 times, so you lose with chance 0.85^74 so you win 1-0.85^74, so you never lose (1-0.85^74)^1e5

[whole valley]

Didnt get that last part

[vestal tapir]

winning 100000 streaks without losing one

1e5 means 100000

[whole valley]

You can loose,??

[karmic fern]

if you run out of money

this is the derivation of "what is the probability that I never lose one out of 100k streaks if I can double 74 times?"

[vestal tapir]

we've been through that, you said anyway

we know you;re not satisfied

[whole valley]

No its good thanks, im just a bit confused on this part.. your allowed to lose one out of the 100k wins

[vestal tapir]

if you lose once you're left with half your money sure, but you need to win sextillion dollars now, so you can't

[whole valley]

True

So leaving that out of the way

Did i misunderstood you or did you say a streak of 100k consecutive wins

[karmic fern]

yeah you need to win $1 100,000 times consecutively

[whole valley]

No 😭

[karmic fern]

in other words, you need to not go broke 100,000 times consecutively

[vestal tapir]

yes, it's what we're leaving out

[karmic fern]

split the game up into several subgames which encompass the whole cycle of doubling until you win: each game you have a probability of 0.85^74 of not being able to bet (and losing sextillions of dollars) and 1-0.85^74 of coming out $1 ahead

[whole valley]

If you loose you bet 2$ if you win there its +1 dollar, if not you bet again

[vestal tapir]

by lose i mean going to $1 bet

[whole valley]

Yes

Alright!

[karmic fern]

so you need to win all 100,000 of those subgames, or you won't win

[whole valley]

Yes

[karmic fern]

the probability that you win all of these is (1-0.85^74)^100000

and this comes out to slightly over 50%

[vestal tapir]

i'm not sure i trust the calculator to do power of 100k lol

[whole valley]

So how did you get that 74 number again

[karmic fern]

I think most calculators do exp of log multiply

[vestal tapir]

guess and check

[whole valley]

So there is no definite way to do this without trial and error?

[vestal tapir]

ok i will trust

[whole valley]

I thought i was doing something wrong.. no formula

[karmic fern]

you want (1-0.85^x)^100000 >= 0.5, so just take roots and use logarithms

you may need to solve it computationally in a smarter way than just taking the roots, however

[whole valley]

However?

[karmic fern]

because 0.85^74 is smallish and you might have computational inaccuracy, but just try it the naive way first

0.85^x = 1 - 0.5^(1/100000)

[vestal tapir]

i think we think the problem is technically impossible right?

there is no amount of money where it's 50%

[whole valley]

A near value is good enough

In that sense atleast

[karmic fern]

yeah I don't think so (there's no amount of money where it's 50%)

[whole valley]

What

[vestal tapir]

that's kind of the promise, what if there is, how surprising is that, well done on solving this

[karmic fern]

like you can do some sort of bounding of the probability that you recover from a loss

[whole valley]

Thanks it helped me a lot

[karmic fern]

probably even a weak bound like Chebyshev's inequality can show it

[whole valley]

From there i can probably use pie / and opposite of ^ to find a finite answer

[whole valley]

That part is cleared

[whole valley]

So thats how you found lose streak?

That doesnt seem right

[karmic fern]

the other related part is that this will never be feasible in practice, because if are forced to reset for reasons other than bankroll size, say at #bets = 50, then you can show that there is no initial bankroll size that will let you win with a probability greater than a certain amount

[whole valley]

Thats borders, not exactly true

Its just not exactly finite

You can find it using lose streak like we did here

[karmic fern]

like there are concentration inequalities from supermartingales that apply if your bet sizes cannot get arbitrarily large

[whole valley]

Hmm. Wont it just mean the number is not worth calculating for?

Because there is probably a number that answers this

[karmic fern]

no I mean that, for example, if the casino sets a limit at 50 doubles, and you want to win 1 million dollars, there may not necessarily be a bankroll amount to actually win this million with greater than 50% probability

[whole valley]

So about that..? There really is no way?

[whole valley]

50 double downs?

[karmic fern]

the equation I gave tells you how to calculate it for the original problem you gave

[karmic fern]

yes

[whole valley]

Oh i thought you said 50 max bet

Yeah that makes a lot of sense

How is it related to this problem tough

[karmic fern]

more generally, you need the ability to reduce your probability of actually taking a large loss to zero to get this to generalize arbitrarily

like if the casino decides to cut off your doubling if they flip a coin 20 heads in a row, you're also in trouble

[whole valley]

Yeah

You thought this was meant for personal use?

Im not sure i get where.you going with this

[karmic fern]

no I mean it's in a very strange regime of conditions where none of the concentration inequalities apply (rather, I should say they're not strong enough)

because things get unbounded

[whole valley]

Alright the answer is 73.1

[karmic fern]

so you need 74

[whole valley]

The answer is 73.0957

[karmic fern]

yeah but this is discrete

so you need 74

[whole valley]

I lost you,

Discrete to which part

[karmic fern]

74 represents the number of doublings you need to survive to win with >50% probability

if you can only survive 73 doublings, you get something less than 50% probability

this is just solving an inequality where the thing you're solving for can only be an integer

[whole valley]

73.1 is a number

[karmic fern]

is it an integer?

remember that the original task was to find the least integer such that (1-0.85^x)^10000 >= 0.5

[whole valley]

No

73.1 is acceptable tough,

[karmic fern]

no it isn't

it's a nonsense answer

you need at least 2^74 bankroll to do this

if you bring in 2^73.1, you will not win with 50% probability

[whole valley]

Isnt it accounting for what happens after the first lose streak?

[karmic fern]

no it's not

[whole valley]

Yeah your right..

[karmic fern]

the probability of recovery from a failure to double is infinitesimally small

like less than 1 in 1 sextillion probably

[whole valley]

Yeah

[karmic fern]

by the way, this whole game changes if you make it fair; then none of this analysis holds up

[whole valley]

But couldnt you theoretically run this program forever

[whole valley]

Ofcourse.

[vestal tapir]

your probability jumps over 50%

as you increase money

[karmic fern]

yeah but that 1 in 1 sextillion + (1-0.85^n)^100000 doesn't matter for the purposes of calculating n

[whole valley]

Yep..

[vestal tapir]

i'm explaining why we truly don't need to account for it

[whole valley]

Alright, are.you guys ready for sub questions? Strictly talk

[vestal tapir]

there's ~49.5% chance if you have between 2^73 and 2^74 and then it suddenly jumps to 55% and also stays there until the next jump

[whole valley]

Yeah i guess loose streak is all about jumps

I mistakenly thought the gap could be filled using the leftover money

Abs not true

Idk why i thought that

[vestal tapir]

yeah, every dollar raises the probability a bit but at some point you jump right over 50%

so accounting for it wouldn't give the required answer

[whole valley]

You could get the most true answer

[karmic fern]

no, you'd get exactly the same answer

[whole valley]

Which is the most true answer

[karmic fern]

the problem is that the dollars in between the powers of two only contribute to the recovery probability

[vestal tapir]

technically 49% is closer

[karmic fern]

so if you started with 2^74 - 4 dollars, you'd still have less than a 50% probability of winning

[vestal tapir]

so we want 2^74 − 1 −1

[karmic fern]

but as soon as you get above enough for another double, your probability jumps

[vestal tapir]

minus 4?

it's minus 1 from the start: 7 dollars to make 3 bets

[karmic fern]

-4 because I can't be bothered to figure out the exact bankroll for one of the doubling streaks lol

[vestal tapir]

oh i misunderstood

yeah, so 2^74 − 2

gives the closest number to 50

[whole valley]

What does the -2 accounts for

[vestal tapir]

minus one was already there: it's 7 dollars to make 3 bets

so 2^74−1 to make 74

and minus 1 to make the jump down

[whole valley]

Im confused, wouldnt it be 2^73+?

Since strictly theoretically if you put all the numbers .you get ^73.1

73 is closer then 74

[vestal tapir]

that comes not from accounting for recovery

it's just a fake number

[whole valley]

Yes, but it shows us 73 is closer to the fake number

[vestal tapir]

it's the formula not working right, it's super fake

[whole valley]

Really?

Calculator bug?

0.85^x = 1 - 0.5^(1/100000) thats what i used, its the way you told you me you found lose streak

[vestal tapir]

math bug?

[whole valley]

I dont know

But if its not then this equation is wrong and i still have no idea how to create a formula to find lose streak

So what equation is right?

[vestal tapir]

thank you Saccharine

[karmic fern]

yw

[whole valley]

Is that sarcastic? I cant tell its a computer 😔

[karmic fern]

it's just a consequence of the method you use to solve it

what you are solving is an inequality where only the integers are valid solutions

but you can force it into the real numbers, use methods from that, and then remember to convert it back to an integer

[whole valley]

And the integer i got is 73

[karmic fern]

no, the integer you get from solving the inequality is 74

it's like asking what's the lowest integer z such that 2z >= 8.2?

it ain't 4, even if you solve 2z = 8.2

[whole valley]

Were not looking for >=

Were looking for closest, even from down

[karmic fern]

did you not ask in your original question how much bankroll you needed to win at least 50% of the time?

[whole valley]

The closest to 50%

[karmic fern]

okay, this doesn't matter that much; just adjust your answer for that

try 74 and 73, see which is closer; if 73 is closer than 74, then bring the money down from 2^74 until you can no longer double the 74th time

you're quibbling over a 1/some sextillion difference at that point

[whole valley]

So from what i seen 73 is closer, so thats the correct way to build that equation?

[karmic fern]

it's the difference between solving the inequality for the largest integer x such that (1-0.85^x)^100000 < 0.5 and solving the smallest integer x such that (1-0.85^x)^100000 >= 0.5

one of those will get you closest to 0.5

then you need to add in the extra money to increase the infinitesimally small recovery probability

either way you do it, you get 2^74-2

[whole valley]

So 2^74 is closer?

[whole valley]

How do you add that money

Find it

[karmic fern]

the probability you win if you have 2^73-1 dollars is like 49.45%, and the probability that you win if you have like 2^74-1 dollars is like 54.97%

the probability that you win if you have 2^74-2 dollars is like 49.45% + [some really, really small number]

[whole valley]

So its the same?

[karmic fern]

since that [really, really small number] is still positive, you get closest to 50% with 2^74 -2 dollars

this is just some inequality stuff

[whole valley]

Alright i think i get it

Are you ready for sub questions?

[karmic fern]

don't know what they are

[whole valley]

I will tell them

Using the same money we got on the dollar equation which starting bet has the best chance of winning 100k profit? 10$ 11$ or 13$? Put them in a line from best to worst

[karmic fern]

$13

[whole valley]

You sure?

[karmic fern]

bigger starting bet = better chances

yes

subject to whatever discreteness nonsense, but there's none here

[whole valley]

But it could hit roofs like not being enough for 1 more lose streak

[karmic fern]

as an example, if you start with a starting bet of $100,000, you have a 15% chance of making it

and your starting bankroll is only $100k

[whole valley]

Wouldnt that effect

[karmic fern]

this isn't enough in the numbers you've given to counteract the effect that you don't need to win so many of subgames in a row

[whole valley]

So its stronger then 10^n+1 11^n?

11 stronger hmm

[karmic fern]

if you think about it, what you're changing is the exponent here: (1+0.85^x)^(100k / starting bet) while you're only reducing the number of times you can double by maybe 2 or 3 or 4 times

[whole valley]

So its greater?

How about 10 and 10.1

[karmic fern]

so if you have the same bankroll, you should increase the starting bet to increase your probability of winning 100k

[whole valley]

So its a golden rule even in small numbers? Like 0.0005 more

[karmic fern]

there's some bullshit where if you increase the starting bet a little, you could bring your bankroll right under the threshold for being able to double x number of times

so then you shouldn't

but the remedy there is to increase it even more

[whole valley]

Alright yeah lets keep it integers

Your right

[karmic fern]

the optimal strategy is to put it all in and try to double on a single round

[whole valley]

What is xmoney again? 2^?

[karmic fern]

the amount of money you need to survive x doublings is 2^x - 1 if I'm not mistaken

[whole valley]

Not if you got that much money and the goal is only a partial,

[karmic fern]

I mean if you want to make $150 profit, you bet $150 per round and do the doubling from that