help-28

[minor cliff]

Wait so S_N doesn't equal zero?

[gritty rose]

the partial sums are not zero, no

[minor cliff]

The Nth Partial sum would be
1 + 1/2 + 1/3 + 1/4 and so on right..?

Alright so let me get this straight

[gritty rose]

probably yes

it would have been defined in part a) or earlier

[minor cliff]

Part a had like an entire different thing going on, cause the series ended up being different

Yeah it was two example problems

But wait lemme get this straight

.
.
To find out that S_(2N) = S(N)
You'd group the terms by two and what I mean by that is you put the paranthesis around them like the way you said.
Next, you find the Nth partial sum, and it turns out to be a sequence that seems to be clearly approaching infinity

So....

How do you find S_(2N) ?

Do you just multiply that Nth Partial sum by two ?

[gritty rose]

S_(2N) is the sum of the first 2N terms

2 * S_N would be multiplying the Nth partial sum by 2

S_(2*2) = S_4 = 1 + 1/2 + 1/3 + 1/4

[minor cliff]

But the problem says thats equal to S_(N)

[gritty rose]

2N = 2 times N

.

.

continue the pattern

[minor cliff]

So (2/1 - 1/1) counts as one term ?

cause im doing the simple math and S_(2N)

does not equal it

Wait what???

Thats not it either, what do you mean?

They say that (a_n) approaches zero, as n --> Infinity
Next they say they can argue that S_(2N) is equal to the Nth Partial Sum

So this is saying that
2/1 - 1/1 + 2/2 - 1/2 + 2/3 - 1/3 + 2/4 - 1/4 + ....
is equal to
(2/1 - 1/1) + (2/2 - 1/2) + (2/3 - 1/3) + (2/4 - 1/4) + ....

????

Oh and they both equal 2.083 + ....

Is that what they mean?

[gritty rose]

you need to keep reading

[minor cliff]

Ohhhh

So it's saying that

(2/1 - 1/1) + (2/2 - 1/2) + (2/3 - 1/3) + (2/4 - 1/4) + ....
is equal to THE divergent harmonic series
like THE literal Nth Partial sum of the divergent harmonic series
because that divergent harmonic series is easily recognized as (1/n) correct.

Because (1/n) would be like saying
1 + 1/2 + 1/3 + 1/4 + 1/5 + .... is a divergent series, therefore this S_(2N) partial sum is a divergent sequence, and then we can conclude the series diverges.

So this pretty much is what it's saying right

[gritty rose]

i don't think you know what partial sum means

[minor cliff]

bro is that wrong

[gritty rose]

https://www.mathsisfun.com/algebra/partial-sums.html

[minor cliff]

I mean it makes sense from my pov, cause that sigma of (1/n) is literally a divergent harmonic series

that has to be what they're saying no ?

[gritty rose]

S_N is always a finite sum

[minor cliff]

Right

[minor cliff]

and Im saying that S_(2N) is in other words simplified to 1 + 1/2 + 1/3 + 1/4 + .....

[gritty rose]

The N in this diagram is 4

[minor cliff]

Right

Just like the n in our example here is 1/n is infinity my fault

I mean

infinity

[gritty rose]

1 + 1/2 + 1/3 + 1/4 + .....
the ellipses is implies the sum goes on forever

[minor cliff]

Right, so

what im saying is

[gritty rose]

no

the partial sums of the harmonic series, S_N, is equal to 1 + 1/2 + 1/3 + ... + 1/N

[minor cliff]

Is equal to this

right yes or no

[gritty rose]

.

finite means the sum does not go on to infinity

[minor cliff]

Should be yes

Alright

Then we can say that.

If we were to put parentheses around these consecutive terms

As the way you described

This would equal to..

This.

[minor cliff]

Which is also equal to the harmonic series of 1 + 1/2 + 1/3 + ... + 1/N

Therefore it is divergent right.

[gritty rose]

you're still confusing partial sums with infinite sums

[gritty rose]

.

[minor cliff]

how

[gritty rose]

here

[minor cliff]

I put the dots there tho

[gritty rose]

and here

[gritty rose]

.

[minor cliff]

well I mean Keep in mind I didn't put 1/N for (S_2N) because thats how the problem gave it to me, it gave it as a sequence of numbers

like it literally left it as + ...

wait

Right

[minor cliff]

And if it isn't finite then it's divergent correct. That's how that works right ?

[gritty rose]

what is "it"

[minor cliff]

S_N

If S_N isn't finite

then it's divergent correct

[gritty rose]

.

finite sums are finite

[minor cliff]

Right.

And if it isnt.

Then.....

It's divergent right.

[gritty rose]

i don't know what you're talking about anymore

[minor cliff]

That's a question btw

I think you helped me solve my problem we just got lost in what answer we were both looking for

Mine was to argue that this alternating series diverges by saying S_(2N) is equal to S__(N)
And thanks to you I figured out that, the text was basically saying that..

the partial sums of the harmonic series ( 1 + 1/2 + 1/3 + ... + 1/N ) is equal to to..
(2/1 - 1/1) + (2/2 - 1/2) + (2/3 - 1/3) + (2/4 - 1/4) + ...

So we can say (2/1 - 1/1) + (2/2 - 1/2) + (2/3 - 1/3) + (2/4 - 1/4) + ... + 1/N = S_(2N)

Therefore the argument S_(2N) equaling S__(N) can be used to show that this series is divergent.

[austere cove]

I don't think this argument holds water.

Consider (a_n) = (0, 1, 0, 1/2, 0, 1/4, 0, 1/8, ...)

Clearly, this sum over all members of the sequence S_N = 2. Additionally, the sum over all even numbers of this sequence is S_2N = 2.

So we cannot conclude merely on the basis that S_N = S_2N that the series is divergent

[minor cliff]

I mean I don't know how else to word it then, cause like the book says it can be 'argued' that the series is divergent with the nth partial sum of the divergent harmonic series S_N = S_2N

Perhaps it is flawed, and maybe they went along with it cause for that problem in particular it works..?

Or am I just not understanding it correctly in the way I'm suppose to from their perspective/method/fking wtv.

Might have something to do with the direct comparison test, is what I assumed.

[austere cove]

So, the standard way to prove that the harmonic series diverges is as follows:

1 + 1/2 + 1/3 + 1/4 + ...

Take 1, and 1/2. Set them aside.

Now consider 1/3 + 1/4

These are all ≥ 1/4 so we have 1/3 + 1/4 ≥ 2/4 = 1/2

We have guaranteed that the sum is larger than 2 at this point.

Now consider 1/5 + 1/6 + 1/7 + 1/8

These are all ≥ 1/8 so we have 1/5 + 1/6 + 1/7 + 1/8 ≥ 4/8 = 1/2

We have guaranteed that the sum is larger than 2.5 at this point.

Now consider 1/9 + 1/10 + ... + 1/16

[minor cliff]

so its an infinite number right

[austere cove]

What now from this point?

[minor cliff]

well if its not finite then its got to be divergent right?

[austere cove]

So for a series to be divergent, what we really establish is not that it becomes infinite at any specific value. Just that for any sum we specify it grows larger than that

So let's say I wanted to prove that this series diverges

I say to you, "Hey bro, give me a big number, as big as you want."

And you say, being a cheeky jerk, a googol

Or whatever

And I say, well, I know that for each power of 2 number of terms, I can make 1 additional copy of 1/2 at least

So 2^(2*googol) terms should be sufficient to make this number larger than a googol

And you can play this game with any number

So this series eventually becomes larger than any number, which is what it means to be divergent

(it just takes its damn sweet time getting there)

[minor cliff]

But like what does that have to do with it being an alternating series and we just need to prove two conditions to see if it diverges or converges.

I think what they're trying to say is that S_(2N) is not less than
S_(N) Which would prove that the condition is not met I suppose..? But S__(2N) is the exact same thing..?

This is genuinely hurting my brain

[austere cove]

Ah

Sorry, I mistook your problem

Ok so the problem with alternating serieses that are not absolutely convergent is that you can rearrange them to converge to whatever value you like

[minor cliff]

these are the conditions btw if u need me to be specific on what they teach

Ohh Absolutely convergent I see

[austere cove]

Including divergent

[minor cliff]

I haven't gotten there yet. And this problem is given before that lesson is taught.

Perhaps that is why there was some miscommunication

[austere cove]

Absolute convergence is the idea simply that the absolute value of the terms is convergent as well

If you have an alternating series where the terms are not absolutely convergent, then you can rearrange the sum to give you whatever value you please

[minor cliff]

So, then in that case would my explanation to their justification have some ground ?

cause it does seem to me like they did rearrange it with S_2N

Or maybe not

Either way, they ended up comparing it to the Divergent harmonic series, we know that for a fact as they literally state it so..

[austere cove]

I'm not sure what the definition of S_2N is

[minor cliff]

Me neither, but riemann

[austere cove]

Can you carefully explain this notation

[minor cliff]

did bring up a good point

here I'll reply to it

He then proceeded to say that S_(2N) is the sum of the first 2N terms

So I believe what he said was

[austere cove]

Ok

Makes sense

So S_2N = H_N

Where H is the harmonic series

[minor cliff]

thats divergent, so because its not absolute we can conclude that it can be divergent as well

right 🥹 ?

[austere cove]

We conclude that S_2N = H_N, H_N is divergent, so S_2N is divergent

You want to also prove that S_N is divergent, right?

[minor cliff]

No 💀 thats like 6 units from me im psure srry

cause its the 1/n bs

[austere cove]

I just explained it though?

[minor cliff]

or I don't know, anyway thanks for confirming this.

[austere cove]

No matter

[minor cliff]

Thank you bru

[austere cove]

yw

[minor cliff]

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[minor cliff]

did ts even work

💀

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[shy flame]

Someone help me i get the first part on why id get dcostheta but i dont get the double angle bit like where does the pi/2 come from

[shy flame]

quite mad for 2 marks cant lie

unless im missing something stupid

[robust slate]

Consider drawing this

[hoary ember]

working backwards the sin part simplifies to -cos(2 theta) maybe that can give a hint on how to continue it

[shy flame]

mhh

well I made the angle at QP and the corner of the new triangle to be called alpha

then I solved for it knowing angles in the triangle add up to 180 which is pi

and got pi-2theta = alpha

but im kinda stuck on that

[robust slate]

Can you mark the distance on the diagram that gives the x-coordinate of the pen?

[shy flame]

aaaaaaah

its almost as if i compleatly forgot the question I just read it again and i think I get it

well I got the big purple side to be QA-dsin(alpha)

do I just ignore the y values since im just looking at the displacement in x?

[robust slate]

Yeah you don't care about vertical displacement

only horizontal displacement

@shy flame Hello?

uh

well I gotta dip so

ig I'll let someone else take over

[shy flame]

I did the rest on paper the rearranging was pretty easy it took me a bit

np

i got it its chillzo

thanks boss

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[tepid merlin]

I can't quite seem to figure out how it is I can find the compacted form of this series

I know it's wrong, just not where

[fathom saddle]

The index. They force you to start with n = 1

Just take all your n and replace with n-1

[tepid merlin]

oooooh

I see

[tepid merlin]

thank you!

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[teal basalt]

is it possible to approximate a function that look like this?

where I only need X between (0, 0.9900990099]

And I only need 1 digit precision for the Y

[full forum]

@teal basalt Has your question been resolved?

[teal basalt]

i guess no hope

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[torn jolt]

help

[torn jolt]

how do you even calculate that

[torn jolt]

getting the total angles of the hexagon

but that isnt gonna do me no good

You see angle d is a right angle and angle would be 120° as it is a regular hexagon

[onyx glen]

just knowing each single angle of the hexagon is gonna be enough ish

[torn jolt]

Angle D = 90°
Angle O = 120°

Now as O is the centre

and also since diagonals of a square bisect each other then ∆BOC and ∆DOC are congruent

@torn jolt

[torn jolt]

yeah I did draw that line

oh

[torn jolt]

You know about congruency?

yeah

okay how do you find what O is now

So , can you see that ∆BOC and ∆DOC are congruent

okay

[torn jolt]

It's a regular hexagon, so all angles are equal and also the sum of angles of a hexagon is 720°

no I mean after you draw the line

[torn jolt]

You have to do it before that

Doesn't matter it's still angle o

O is 120 but after you draw the line its like 30

arent you doing O + D + unkown = 180?

[sacred yarrow]

[torn jolt]

Just listen we are finding angle OCD

ok thats 45

[torn jolt]

No, if you find all the angles, then you see that OCDE is a quadrilateral

And using angle sum property you can find angle alpha

E is the point of intersection of hexagon and AD

[torn jolt]

what does knowing its a quadrilateral do again?

[torn jolt]

You know that the sum of all angles of a quadrilateral is 360°

[sacred yarrow]

The sum of its inner angle should be 360 degree

[torn jolt]

If you know the three angles and one is not given you can find it, in your case it's alpha

[torn jolt]

OCDE has the same angles as a square?

I know that but it doesnt look like it has 360

[sacred yarrow]

It’s a fact

[torn jolt]

You looking doesn't matter. It's math, the sum of angles of quadrilateral is 360°

so 105

[torn jolt]

Which angle?

E

360 - 120 - 45 - 90

Correct 💯

[torn jolt]

Yes that's correct

ok thats easy

I just forgot that quadrilaterals equal 360

[torn jolt]

Yep everything is hard until you understand.

[torn jolt]

I can give you something

?

Remember this whole solving problems related to finding angles in polygons

[torn jolt]

It's a very helpful chart

[cold wren]

there is also a formula ig

for angle sum

[torn jolt]

oh yeah we took that yesterday in class

n-2 * 180

[torn jolt]

Yep it's given in the poster

[cold wren]

i didnnt see that

[torn jolt]

Yes, for finding sum of angles of a polygon use it, where n is the no. Of sides

one more question it says sides does it mean all 4 sides or just left and right

[torn jolt]

Opposite sides of a rectangle are equal

[cold wren]

20 = x + y

and find the values of x and y that are prime numbers

so (13, 7), (17, 3)

[torn jolt]

so it means all 4 sides right

[cold wren]

area = 13*7

[torn jolt]

Yes , but as two pair of opposite sides are equal so yea

[cold wren]

In a rect. opposite sides are equal

[full forum]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[torn jolt]

its not the right answer anyway

[cold wren]

so perimeter = a + b + a + b

a + b + a + b = 2(a + b)

2(a + b) = 40

so a + b = 20

[torn jolt]

This tbh

alright thanks everyone

[torn jolt]

If your problem is resolved, then use .close

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[night musk]

I am doing question 9. So first I proved that the irrationals are not bounded above because I already proved that an irrational number plus a rational number is irrational. Now I know that x < y so I can use arch property to show that ny>nx+1. Now I am going to make a set A. I will say that the set has an element z in the irrationals such that z > nx. Now since we know the irrationals are unbound we know that given an real x we can find a z greater than x. So now I replace z with nx/n. So z > nx/n. So this set does have a z. But that doesn't correspond to my condition since it is greater than nx. But I need it to be nx/n so that I can do that nz>nx then nz-1< nx. nz< nx+1 so nz < ny. But then I have to change the condition of the set to z > nx/n which seems weird. So what should I do?

[grave elm]

I already proved that an irrational number plus a rational number is irrational.

Question 6 also says that there must be infinitely many rationals inbetween x and y

Using these 2, you should be able to get an irrational between x and y

Instead of always returning to the roots (archimedean property in this case), try using what you already proved

[night musk]

Ok I'll try that but what about my issue with the set I feel like something is off with the conditional

[grave elm]

I'll check..

[night musk]

Ok thank you

[grave elm]

you can choose some irrational z s.t. nx < z < nx + 1

then x < z / n < (nx+1) / n < ny/n = y

so z/n will be the wanted irrational

[grave elm]

so you'd just have to do this

[night musk]

Ok so that's the new condition. But now when I use the arch I have nz > x which proves that z> nz how do we know about the nx+1

How can we use well ordering principle if we only proved part of the condition

[grave elm]

well, thats the problem, its not that easy

One way to do it would be this:

choose 2 rationals a, b such that:
nx < a < b < nx + 1

then you should be able to explicitly find irrational between a and b

funny thing is that you can do exactly this to the original x and y

you can choose rational a, b, such that x < a < b < y, and then the problem reduces to finding an irrational between 2 rationals, which can be done explicitly

[grave elm]

if you need to go with this, then it could be reduced to 0 < z - nx < 1, and you'll need to find an irrational z

the main issue you'd face here is that you dont know whether nx is rational or not

[night musk]

Wdym by done explicitly?

[grave elm]

I could give you an expression for an irrational number between 2 rationals a and b

without applying archimedean property

but you can also do it by applying it

its up to you

[grave elm]

(it would be weighted average of a and b, with one of the weights being sth irrational, e.g. sqrt(2))

[grave elm]

You can still do this though

try finding an irrational z between rational a and b using arch property

[night musk]

So your saying for example a is rational b is irrational and then a+b /2

But that is just irrational but know you have just found an irrational number between a rational and irrational

[grave elm]

Yeah, this would actually also work

and is probably a bit easier

[night musk]

Wait I'm confused weren't we trying to find an irrational between two rationals though?

[grave elm]

The original question was finding an irrational between two reals

you can find one rational between those 2 reals

say the reals are x and y, and so you can find a such that
x < a < y

then (a+y)/2 is an irrational between a and y

x < a < (a+y) / 2 < y

and notice that the very same irrational also sits between x and y

so you're done

[night musk]

But doesn't this assume that x and y are irrational when they are real

[grave elm]

oh, sorry

[grave elm]

it actually doesnt need to assume that

i just typoed that

6 already proved that between any real numbers there is a rational

so we can now just choose it

you can make an explicit reference to 6 in your proof

[night musk]

Choose what?

[grave elm]

the rational a betweeen x and y

[night musk]

Yes but then I need to find a irrational between two rationals. All we found is an irrational between a rational and an irrational

[grave elm]

oh, yeah, i see, i assumed y is irrational

well that shouldnt be too difficult though

so say we have 2 rationals a, b

we need to find an irrational (z) between them, so we need
a < z < b

this should be doable by arch property

the trick is, we can subtract a

0 < z - a < b - a

now we only need to find an irrational between 0 and b-a (so irrational between 0 and another rational)

and then if we add a to it, we're gonna end up with irrational between a and b

[grave elm]

actually we can just construct it explicitly

pick your favorite irrational between 0 and 1 and multiply it by (b-a)

e.g. sqrt(2) / 2 * (b-a)

then a + sqrt(2)/2 * (b-a) is gonna be between a and b

[night musk]

One sec let me read this all and try to process it

[night musk]

Isn't this just rewriting the problem?

[grave elm]

Yeah, thats exactly what it is

we are reducing the problems to simpler problem, until we find a simple enough problem that we can solve it

first we reduced the problem of finding an irrational between 2 reals to finding an irrational between 2 rationals. Then this problem was reduced to finding an irrational between 0 and a positive rational. And that's already quite doable

if you wanted to, you could go one step further, divide 0 < z - a < b - a by b-a and get 0 < (z-a) / (b-a) < 1, reducing the problem merely to finding an irrational number between 0 and 1

[night musk]

I'm not seeing why we multiply by b-a

[grave elm]

just to make it fall between 0 and b - a

we need that irrational to be between 0 and b - a

if its not clear, lets go one step further.
0 < z - a < b - a
Divide both sides by b - a
0 < (z - a) / (b-a) < 1
Now we just need an irrational between 0 and 1. Pick e.g. sqrt(2) / 2
so sqrt(2) / 2 = (z - a) / (b-a)
(sqrt2/2) * (b-a) = (z-a)
(sqrt2/2) * (b-a) + a = z

and there is our irrational between a and b

$a<a+\frac{\sqrt{2}}{2}\left(b-a\right)<b$

[glossy valve]

MathIsAlwaysRight

[night musk]

Oh ok thats clear

Thank you

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[gray mauve]

i have to find the Domain that makes the function continuous

i know for the first interval (-infinit,-1] the function is continuous

but

i also found that x=-1 if a point where the function is disconitnuous

for the type II

i am not sure if in english is the same

and so when i write the solution do i say that D=(-infinit,-1]... or (-infinit,-1)...

i put ... beacuse i know there is more but this part interests me

[tacit siren]

it's discontinuos, since the right limit as x -> -1 doesn't exist

[pure basin]

We need to find the domain in which it is continuous

[tacit siren]

At x = -1

[gray mauve]

ok but i also said that for x in (-infinit,-1] the function is continuous

isnt that a contradiction?

one time it is continuous but the other time is not

[pure basin]

But RHL neq LHL at x=-1

[gray mauve]

?

[tacit siren]

Right hand limit, left hand limit

[gray mauve]

i dont understand RHL and LHL

oh

ok so then in x=-1 the function is not continuous

[pure basin]

ys

just check the same for

1

[gray mauve]

it is also not continuous

[pure basin]

Ok

So there's ur answer?

[gray mauve]

just to make sure

when i write D(continuous)=

i write all the intervals with () not []

[pure basin]

Domain ok

uh

[gray mauve]

i mean

[pure basin]

To write the domain??

[gray mauve]

yes

let me take a pic

wait

[pure basin]

I an pretty sure

You can write R - {-1,1}

[gray mauve]

oh yes

thats better

thank you for your help

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[gray mauve]

I have to find a,b from R so that the function is continuous on whole R

i need only one ideea and then i will do it myself

trying now to rewrite f(x) when x>0 in a simpler form

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[silver pecan]

$a=372\ca+cd=\frac{1}{93}a=4\c=1=\frac{1}{372}a\d=3=\frac{1}{124}a\\frac{1}{372}<\frac{1}{124}$

[glossy valve]

UCYT5040

[silver pecan]

i found my mistake

i reversed the inequality by mistake

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[desert valley]

I’m currently working on a mathematical model for a 5-reel, 3-row slot machine. The game will feature 20 pay lines, with payouts for 5, 4, 3, and 2 matching symbols from the left, and around 8 different symbols in total.

It also includes Wilds (which substitute for any symbol), Super Wilds (which fill an entire row and substitute as well), Bonus symbols (that trigger free spins), and Scatter symbols.

One of the biggest challenges I’m facing is ensuring the game meets specific RTP targets—including the base game RTP, Free Spins RTP, and Bonus Round RTP. I'm looking for guidance on how to approach the mathematical modeling, particularly with calculating probabilities, designing paytables, and accounting for the interactions between base gameplay, free spins, and bonus rounds.

I initially tried to simulate all possible outcomes in Python—including all paylines and how wilds and super wilds affect them—but it quickly became overly complex due to the huge number of combinations. I’m wondering if there’s a simpler or more structured way to approach this.

Any advice or pointers would be incredibly appreciated. Thank you!

[full forum]

@desert valley Has your question been resolved?

[torn jolt]

Step-by-Step Approach to Modeling Your Slot Machine

---

1. Define the Symbol Distribution

Start by specifying:

How many symbols (including Wilds, Super Wilds, Scatters, Bonus) appear on each reel.

How many times each symbol appears on each reel strip (these will determine the probabilities).

Use weighted reels or uniform probabilities depending on your game design.

---

2. Calculate Symbol Probabilities

For each symbol (S):

P(S on reel R) = (Number of S symbols on R) / (Total number of symbols on R)

Repeat for all 5 reels.

Wilds and Super Wilds increase complexity, so handle them separately.

---

3. Calculate Line Hit Probabilities (Without Wilds)

Focus on a single payline. For matching symbols (e.g. 5-of-a-kind):

Multiply the probabilities of the symbol appearing in sequence on each reel.

For example:

P(5 Aces on line) = P(Ace on Reel1) × P(Ace on Reel2) × ... × P(Ace on Reel5)

Repeat this for 4, 3, and 2 matches.

Then multiply by the number of paylines (20) and how many ways each match can occur.

---

4. Integrate Wilds and Super Wilds

Wilds: When calculating payouts, include Wilds as valid substitutes.

Super Wilds: Since they fill a whole row, they affect all paylines on that row.

Use conditional probability trees or simplify by estimating:

How often does a Wild appear on each reel?

What’s the probability that a Wild completes a 5-of-a-kind?

For Super Wilds, you can treat them like events:

E.g., "Probability of Super Wild on Row 1 = X%" → then calculate affected paylines.

5. Model Bonus and Scatter Mechanics Separately

Define the chance of landing enough bonus symbols to trigger a bonus round.

Estimate the average payout from bonus rounds.

Same for Scatters if they pay independently of lines.

Use:

P(bonus round) × Expected value of bonus round = Bonus RTP

---

6. Break Down the Total RTP

Split the RTP into:

Base Game RTP: From line wins, including Wilds.

Bonus Round RTP: From bonus rounds.

Free Spins RTP: If they’re separate, calculate them too.

Then:

\text{Total RTP} = \text{Base RTP} + \text{Bonus RTP} + \text{Free Spins RTP}

Adjust paytable values and probabilities to meet your target RTP (e.g., 96%).

---

7. Simplify with Monte Carlo Simulations (Optional)

Instead of brute-forcing all combinations:

Create a randomized reel spin simulator.

Run it 1–5 million times.

Track symbol positions, line hits, bonuses triggered, and payouts.

Use this to validate your theoretical math and fine-tune the balance.

---

Tools You Might Use

Excel (for rough calculations and payoff matrices)

Python (NumPy or Pandas to simulate and analyze results)

State machines or Markov chains (to model flow of base game → bonus rounds → return to base)

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[sharp osprey]

all roots for z^3 = -8i

[sharp osprey]

I have tried this:

[sudden condor]

shouldn't n just be 3 ?

oh wait my brain was off

[sharp osprey]

no you are all good

[sudden condor]

but that's not how fractions work

[sharp osprey]

im the one with the brain off

[sudden condor]

i guess we both are then

[sudden condor]

.

[sharp osprey]

im trying to work it out again

shoulf it be 4pik?

should*

[sudden condor]

nah should be 2pik it's just when you subbed for theta
you still have addition first so it should still be
[(3pi/2)+2pik]/n

[sharp osprey]

if i extend 2pik with 2

and write it over same denominator

then divide by n

[sudden condor]

then (3pi+pik)/6 yea ?

[sharp osprey]

no wouldnt it be 4pik?

[sudden condor]

man our brains are really off lol

[sharp osprey]

i promise it is usually mine

let me send photo

[sudden condor]

ok

[sudden condor]

i'm with you bro

[sudden condor]

yea but like isn't adding 4pi a little too much imo

[sharp osprey]

but if i added pik that would change the meaning?

pik instead of 4pik

[sharp osprey]

what you put here

[sudden condor]

pik was me calculating with an off brain

[sharp osprey]

ok but u r right that 4pik might be unnecessary

do i change to just 2pik

[sudden condor]

now that i look at it it's the actually the easier way lol

[sturdy valve]

can i give you the procedure for this with an example and can u figure it out from there?

or did u get it

[sharp osprey]

i got one earlier but i barely understood it

i had two people explain it to me

in two different ways

[sturdy valve]

oh lol

[sharp osprey]

im not sure if i was confused just cuz i was confused or

[sudden condor]

and i guess you are getting a third one here lol

[sharp osprey]

if i was confused because i mixed the explanations

[sturdy valve]

yeah so me explaining would be 4th one 💀

[sharp osprey]

^ yes but

is your 4rth good because right now

none of the others really stick

and i dont like any of them

i liked one where

there was this person named Ann

they made me understand the absolute value part

like how to get r

[sturdy valve]

yea lol she explains well with the bot text and stuff

if u want u can ping her

think shes on

[sharp osprey]

no i dont want to disturb them

i have been sitting with this for hours now

it's embarrassing

[sturdy valve]

no its not lol

was a tough concept for me as well

[lunar grove]

Still need help?

[sharp osprey]

i genuinely hate complex numbers

i just hate them so much

[sturdy valve]

💀

[sudden condor]

count me in

[lunar grove]

Tbh i can see why its weird but are you strong with modulus argument form

Ie i terms of modulus and tje angle cos(theta) and Isin(theta)

[sharp osprey]

im not strong with it

[lunar grove]

Hmmm do you know how to gst the r(cos(theta) +isin(theta)) form

Or do you know hoe to do a argand diagram

First of all

[sharp osprey]

i dont know how to show where it comes from, and i dont know what a argand diagram is- like hold on

so

r = |z|

[lunar grove]

Ab yes so no wonder ur confused lmao

[sharp osprey]

and it is the distance between a point and another

[lunar grove]

Yep

[sharp osprey]

for -8i

[lunar grove]

No its distance from origin to the point

[sharp osprey]

it would be the distance from 0 to -8i

so |z| = 8

[lunar grove]

Yep

But say if its 2+3i

[sharp osprey]

okay yeah then i need to do sqrt(4+9)

right?

[lunar grove]

You can represent that on a argand diagram (similar to x y axis)

[sharp osprey]

|z| = sqrt13

[lunar grove]

Two to the right and 3 up from origin

Then you can find distance using pythagoras

[sharp osprey]

is sqrt13 wrong?

[lunar grove]

Its correct but do you understand

[sharp osprey]

for 2 + 3i

[lunar grove]

It first of all

Feel like you just know how to compute it

If thats the case it won't gdt you far

[sharp osprey]

no i think i understand

i have the real and imaginary axis

[lunar grove]

And whats the argument

[lunar grove]

Yh thats what a argand diagram is

[sharp osprey]

the angle between the x axis and r?

is that how you say it

going in the

counter clockwise

direction

[lunar grove]

Well yeah sure you can do that

Usually its tan-1(y/x) but ur. Way would work

So you can write Z as r(costheta +isin(theta))

U familiar with demoivres theorem?

[sharp osprey]

yes familiar but i suck at it big time

i know you can do

[lunar grove]

Icl i cba typing ur should definitely watch a YouTube video

Its too much to explain

Via text

I reccomend tlmaths

[sharp osprey]

z^n = r^n(cos nTheta + isin nTheta)

but i dont quite understand what happens to the 2pi k part

with relation to n

because when i got z^3

i just wanna use n = 3

and plug everything in there

but i get really big numbers and again

i dont understand what happens to the 2pi k part

[clever turret]

Hi

[deep python]

Hello

[sharp osprey]

hi again Task Bot

[clever turret]

What is the problem?

[deep python]

Don't you have a problem to think about already ?

[sharp osprey]

z^3 = -8i, i think im about to solve it

[deep python]

2i and...other roots

[sharp osprey]

yes i got 2i and i - sqrt3

i need one more

[clever turret]

Can I see the sheet?

PLS

[deep python]

just do the division euclid algorithm thing

You know...to factor the polynomial z³ + 8i

[sharp osprey]

nah im stuck again on last step i will send sheet

it is too much for me

with the angles and division of angles

there are plenty of things wrong here

i can point out at least 4, probably 5

i just cant be asked to correct more

im just gonna start over on a new page

[clever turret]

Where does 5π/6 come from?

[sharp osprey]

that's one of the things wrong

previously

this part

it said 2pik

instead of 4pik

i got it from there

[clever turret]

Its 2πk

[sharp osprey]

for k = 1

i dont get why it is is 2pik

is it because any multiple of 2 of 2pik is the same as 2pik?

wont i end up in the wrong quadrant

if i dont call it 4pik

[clever turret]

?

[sharp osprey]

why is it 2pik

i got

4pik

from this

why is it 2pik

[clever turret]

because the cosine and the sine repeat after 2πk

[sharp osprey]

that's true if it only said 2pik

but it is divided

is it important that i don't write them together?

over the same division sign?

<@&286206848099549185>

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[frank lagoon]

Can someone help I’m not sure where to start for the next couple of ones

[tulip marlin]

you can simplify 3 algebraically for starters

by substituting y = x + 1

which is easier to understand

[frank lagoon]

so f(y) ?

[tulip marlin]

yes

whats the limit

y -> ?

[frank lagoon]

0?

[tulip marlin]

no

x -> 1 (from below)

[frank lagoon]

right

[tulip marlin]

x + 1 -> 1 + 1 (from below)

[frank lagoon]

wait hold on

i dont understand

so the lim x->1 from the left f(x) = 0

right

[tulip marlin]

x approaches 1 from the left
so
x + 1 approaches 1 + 1 = 2 from the left?

[tulip marlin]

I don't know where you get that = 0 from

[frank lagoon]

from the graph

[frank lagoon]

you can see that it approaches 0

[tulip marlin]

im looking at this question

[frank lagoon]

ya ur supposed to use the graph to answer it

[tulip marlin]

i was suggesting you first do a substitution

before looking at the graph

[frank lagoon]

sry i just dont understand how doing f(y) makes it easier

[tulip marlin]

well, forget about y then

as x approaches 1 from the left

[frank lagoon]

mhm

[tulip marlin]

you need to know what x+1 is doing

otherwise how do you take f of it

[frank lagoon]

ya

[tulip marlin]

so what is x + 1 doing

[frank lagoon]

oh

so 1+1?

2?

[tulip marlin]

from below

so its the same as limit of f(y) as y approaches 2 from the left

thats what i meant by substitution

then you can check the graph

[frank lagoon]

i see

so the answer is 2 right?

[tulip marlin]

yes

[frank lagoon]

ok got it, thank you

Can I do the other ones and could you check if I did them right?

[tulip marlin]

you can apply the same idea for the next questions

just post the answers and someone can check them yeah

[frank lagoon]

ok sounds good thank u!

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[misty flare]

Hey Guys
I need some help with this:
A pyramid with a square base of 100x100 and a height of 78 is placed in a rectangular coordinate system such that the base lies in the x-y plane and the top lies on the z-axis. The sides of the base are parallel to the x- and y-axes.
Hand-draw a sketch of the pyramid in a coordinate system.

I have kind of tried myself but it just doesn't make sense, I don't know how to draw.

[royal charm]

Show what you tried

[misty flare]

this is not really 3d, but I thought i kind of tried something

[royal charm]

Oh, that's pretty good

But yeah, you should draw the axes in perspective

[misty flare]

then I was told I should make it like this

[misty flare]

But now i don't know where to put the coordinates and how to draw the pyramid and such...

[royal charm]

Label the axes, then start at the origin and move in reach direction according to each axis

Like, for (3,5), you label the 3 on the x and the 5 on the y. Then you start at 0,0 and move 3 right and 5 up

It's the same thing here

[misty flare]

but I have 4 coordinates and 3 axes,

I don't understand where I should place them?

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[mossy loom]

HINTS ONLY for the next step. Part 2. How does the fact that M is locally finite come into play? I'm stuck here: well we gotta only check for points in the boundary. So, we need to show that every p in the boundary of the union (let's call it set 1) is in the union of boundaries (set 2). So, let's assume M is locally finite, yet there a point p in 1 but not 2. Where do I even proceed from there. How would I even use the fact that X is locally finite? Nah cuz why does it sound so irrelevant 😭? I presume we must also use results from (i).

[mossy loom]

I think I need to assume by contradiction that a p exists in boundary of 1, but not 2. And vice versa. I suppose proof by contradiction is one way

[devout valley]

(You can also ask this in #point-set-topology if you haven’t asked there )

[mossy loom]

But I need to show that both ways lead to breaking local finite ness

[mossy loom]

Oh, didn't know I could use multiple servers. I assumed it could have fallen under spamming. I'll be doing that

[devout valley]

You can use both, as long as you don’t take multiple help channels you’re good

[mossy loom]

Sure, it sounds that people there are somehow smarter, which is great. I've posted it there too.

Actually I think imma close it here just for some person that might need it.

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[night musk]

Can someone help me understand the phrasing of question 10?

[gritty rose]

Which part exactly confuses you

[night musk]

So he says n is even if it equals 2m then says n is odd if n+1 is even

[gritty rose]

for some integer m yes

[night musk]

Why is it phrased like this why not n is odd if it does not equal 2m

[mossy loom]

Well it could

But they're the same thing, you'll find many ways to define one concept

Both are right

[night musk]

How. So if we say n+1 is even then n+1 = 2m. n = 2m-1

[umbral dome]

that's another way of defining odd

[mossy loom]

But I suppose the way they phrased it because that way it is easier to solve the problem.

🤔

[mossy loom]

Tru

[night musk]

So let say I want to do part a. Let say I want to do proof by contradiction. I assume that an integer can be both even or odd. Using his definition n is even if n= 2m and n+1 is even. But then if n+1 is even. We have n+1 =2m and n= 2m. But how can that be?

[mossy loom]

Wait a minute

Nah, can you even assume something like that? I mean we know set of even and odd are disjoint. But let's pretend we didn't find an integer that's both even and odd. 🤔

@umbral dome what do you think?

I think that's the contradiction

Yeah, that's the contradiction

[night musk]

So your that's the whole prove but I feel like we didn't prove anything. Isnt that what we're trying to prove?

[mossy loom]

Honestly, yes.

Well if no such n exists that is both equal to 2m as well as 2m+1, then the sets are already disjoint

[night musk]

Is this fundamental enough though?

[mossy loom]

Yeah for sure. This works

[night musk]

Ok so either or works then

[mossy loom]

Wdym?

Oh, for the next question?

[night musk]

Nvm disjoint is the answer

I was referring to the first one

[mossy loom]

I mean you can do the same contradiction again for 2. Assume that you have an integer that's neither even, nor odd. Try to show this cannot happen. For the second question.

[night musk]

Wait doesn't disjoint work for both though

[mossy loom]

It does. But it's not fundamental enough

Given the question ask

You gotta show why they're disjoint

[mossy loom]

Which is what youve showed here.

[mossy loom]

You need to show why they're disjoint. That's what you have done

[night musk]

Haven't I just show that the definition of even is not the same?

[mossy loom]

Where have you shown that?

[night musk]

n = 2m and n+1 =2m

[mossy loom]

No

Ok fine

Let's make it more basic.

Since n = 2m, and n+1= 2m, then we can equate them.

So, n=n+1

But that means 0 = 1

Is that clear enough that it's a contradiction?

You haven't 'redefined anything '

Like you haven't redefined here that 0 equals 1.

This is the contradiction in your proof by contradiction

[night musk]

Isn't his definition the contradiction though that's what I was confused about

[mossy loom]

Yes it is

I understand it's very basic

Ok, let's make your proof a lot clearer.

Honestly, I can't even make it more basic.

If this is your first time writing proofs, I understand

[mossy loom]

Yes it is. But you have to prove it is a contradiction

For example, there could be such an integer that's both even or odd

And mathematicians have yet to discover that

But you have proved that no such integer exists, because one thing we know is that 0 isn't 1.

[night musk]

Is this what you meant by that the definition he gave helps with the proof

Also one final question why give a faulty definition

[mossy loom]

Which is the faulty definition you're talking about?

[mossy loom]

Short answer: Yes. Long answer: a different definition would have resulted in at most one more step.

[night musk]

His two definitions of even

[mossy loom]

No, ok fine

Let me try to make it more clearer

[mossy loom]

What are the two definitions?

[night musk]

That n=2m or that n is odd if n+1 is even. If something is even it is equal to 2m. Meaning n+1 = 2m

[mossy loom]

It says n is odd, if n+1 is even. Now, n + 1 is certainly 2m, but not n.

Oh wait a minute

I understand your question perhaps

Here, n and m are different 'variables'

Or rather m is the variable

Ok, let me put it in simple English

[night musk]

So case specific two different cases of even

[mossy loom]

An integer is even if it is twice any other integer. (This is what is represented by saying an integer n is even if n=2m). Now, another definition is 'an integer is odd if the next integer is even'. (This is what is represented by saying saying n + 1 = 2m.) It doesn't mean that n = 2m. It means the next integer, the even one is equal to 2m

Ok, yeah so they never gave two different definitions of even

One definition for even or one definition for odd.

[full forum]

@night musk Has your question been resolved?

[night musk]

So it is fine that when isolating for n you have a different equation since they are different n

[mossy loom]

Exactly

[night musk]

Ok thx so much

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[opaque terrace]

@devout valley @wide sundial
Hey there! I just want to say, with your help on the Gram-Schmidt orthogonalization from earlier, I was able to successfully complete the Schur trianglization problem. I promised to come back to share the results, and here they are!

[opaque terrace]

So as you can see, the Schur trianglization of a matrix allows us to rewrite any matrix in C as U = P^{-1} A P, where U is upper triangular

Thank you both for your help 😄

❤️

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[devout valley]

.Awwww, it was a pleasure to help you hope to see you around

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[old spade]

cosA - cosB = 1/2 and sinA - sinB = -1/4, find sin(A+B)

[torn jolt]

Have you learnt partial angles trigonometric identities?

[old spade]

maybe I have but i dont call them partial angles

[torn jolt]

i don't know the original name either but yeah you can find sin(A+B)/2 and subsequently sin(A+B) right?

[old spade]

how

[torn jolt]

oh wait it's sin(a-b)/2 that's common hold up

[torn jolt]

ok so first square both equations

then add them up

[old spade]

i probably tried that already

[torn jolt]

You'll get an equation with Cos(A+B) and numbers

sqrt(1-cos^2(A+B)) = sin(A+B)

[old spade]

no you get cos(A-B)

right

[torn jolt]

omg i forgot trigonometry my bad

[old spade]

generative ai used sum-to-product

ill probably just copy it

I just need someone to confirm that the answer is 4/5

[torn jolt]

i cant see that but i have a way to guarantee the solution

use the partial angles formula

divide the equations

You get value of tan((A+B)/2)

Which is sec^2((A+B)/2)-1

And you can take reciprocal of secant to get cos

1-cos^2x = sin^2x

you have both sin and cos, use sin2x = 2sinxcosx

[old spade]

so is the answer 4/5?

[old spade]

i think u did the same thing as the ai until here

anyways i have another question

The lengths of the sides of a triangle are successive terms of a geometric progression. Let A and C be the smallest and largest interior angles of the triangle respectively. If the shortest side has length 16cm and 9(sinA - 2sinB + 3sinC) = 19(sinC - 2sinB + 3sinA), find the perimeter of the triangle in centimeters. Is the perimeter 304/9 cm?

[torn jolt]

ah i didnt know that formula but fair
I think it's the same thing either way since 1+tan^2x is sec^2x

[old spade]

using sine rule to find sinA and sinB in terms of sinC then it's a quadratic for x (the common ratio in the geometric progression)

[torn jolt]

I got 184 as the perimeter

[old spade]

D:

[torn jolt]

if that's the correct answer I can show my procedure

[old spade]

i dont know the correct answer but show your procedure

[torn jolt]

I wrote it roughly ):

[old spade]

can someone else confirm the answer?

[torn jolt]

let me write it down once

took me like 4 minutes to solve

nvm i made a mistake in between

[old spade]

so my answer might be right

[torn jolt]

nvm i didnt make a mistake

[old spade]

show workingg

[torn jolt]

@old spade

[torn jolt]

ignore the sqrt(c) in the last equation on 96*sqrt(c)

[old spade]

want to see my working?

[torn jolt]

mine is correct

[old spade]

geometric progression means the sides have a common ratio?

[torn jolt]

If you want me to tell where yours is wrong I can tell

[torn jolt]

Yeah

c is largest angle, a is smallest angle

[old spade]

🤨

[torn jolt]

so a,b,c are in geometric progression

b/a = c/b

[torn jolt]

trust fr fr

or just see all my steps and check if im wrong

[torn jolt]

C is the largest angle right?

[old spade]

uh shit ur right

[torn jolt]

ye

this would also work tbh since x would just be the reciprocal value

[torn jolt]

how did you get c = -1/4 or 2/3

[old spade]

then that gives me 76

[old spade]

no i got x = -1/4 or 2/3

[torn jolt]

x sorry

how

[old spade]

12sinC x^2 - 5sinC x - 2sinC = 0

sinC(12x^2 - 5x - 2) = 0

12x^2 - 5x - 2 = 0

[torn jolt]

nvm

[old spade]

maybe where i put the lengths are wrong?

like are the 16, 16x, 16x^2 be at teh correct places

[torn jolt]

Shouldn't matter, since x is just a multiplier

I'm wondering if I did something wrong now

[torn jolt]

oh yeah wait it matters

Since you have 16 as a constant

[old spade]

x will differ

[torn jolt]

No, you're saying the length of side opposite to C is 16 when in fact it is opposite to A

[old spade]

the length of side opposite to C is 16x^2 and the length of side opposite to A is 16

?

[old spade]

ignore this

[old spade]

i redid everything with this and got 76

[torn jolt]

yes

[torn jolt]

What did you get for x

[old spade]

<@&286206848099549185> somebody check

[old spade]

1.5

[torn jolt]

Oh shit yea

I calculated c wrong

you get c = 36

a = 16

b = 24

my b

[old spade]

so am i correct

[torn jolt]

yeah

[old spade]

yay

[torn jolt]

🎊

[old spade]

i have uh 3 more questions that i need help with if ur down

[torn jolt]

Sure

ill just stop if I have anything else to do

[old spade]

gimme a min i have to write down the working

Find the value of 1/sin^2(0.5) - tan^2(0.5) + 1/sin^2(1.5) - tan^2(1.5) + ... + 1/sin^2(179.5) - tan^2(179.5)

(1/sinx + tanx)(1/sinx - tanx)

wait i think i see soemthing

[torn jolt]

symmetry?

[old spade]

no i think we have to rearrange the terms

i think im a genius

the answer is just 180?

[torn jolt]

i think its actually sin^x = cos^2(pi^2 + x)

[torn jolt]

how

[old spade]

nvm

😭

wait im a genius though

[old spade]

theres 1/sin^2(0.5), 1/sin^2(1.5), ..., 1/sin^2(179.5)

so you can replace sin with cos right

[torn jolt]

https://tenor.com/view/woody-woody-toy-story-sora-kh3-kh-gif-27223516

[torn jolt]

yea

[old spade]

1/cos^2(0.5), 1/cos^2(1.5), ..., 1/cos^2(179.5)

and 1/cos^2(x) - tan^2(x) = 1

[torn jolt]

oh makes sense

[old spade]

im not too sure about this part

[torn jolt]

you can

not

wait

nah you can

think of it as

sinx = cos(pi/2 - x)

After you transform 0-89.5 that way you get the same in sin

[old spade]

yeah yeah i was just worried about it might become -cos(x)

[torn jolt]

when you do it for 90.5-179.5 you get sinx = cos(pi/2-x) => sinx = cos(x-pi/2)

Then you get a repeating term for every 0.5-89.5 term

[old spade]

but basically it goes from cos 89.5 ... cos 0.5 ... cos 89.5 right?

[torn jolt]

do cosx = cos(pi-x) for every duplicate term

You don't need to worry about sign due to the squares

[torn jolt]

then boom 0.5-179.5

[old spade]

sin 0.5 ... sin 89.5 ... sin 179.5
cos 89.5..cos 0.5 ... cos 89.5?

or am I tripping

[torn jolt]

you have a corresponding term for every tan

[torn jolt]

Yes

read what I just said

[old spade]

aight cool

wait then it wont be so simple

cos you can't match tan^2(90.5), tan^2(91.5) ... tan^2(179.5)

right

[torn jolt]

read what I just said

[old spade]

i mean

prove it

how would you match the terms beyond 89.5

im stupid 😅

[torn jolt]

#help-28 message

[old spade]

nono i get it

[torn jolt]

cos(90.5) = cos(180-89.5)

[old spade]

its not that part its the

i get it it's fine

[torn jolt]

?

[old spade]

like i get what ur saying

so it's 180?

next question

find the value of ((cos10 + cos50 + cos70 + cos110)/cos20)^8

[torn jolt]

yea

[old spade]

calculator says its 81 but we gotta prove it sigh

[torn jolt]

has to involve cos(30-20), cos(30+20), cos(90-20), cos(90+20)

[old spade]

that means the cos10 + cos50 + cos70 + cos110)/cos20 = sqrt(3)

[torn jolt]

but gtg now if you still struggling with this when I come back will help

[old spade]

okay thanks for your help

Let A and B be acute angles. Suppose sin^2 A + sin^2 B = sin(A+B) = k. Show that k = 1.
Hint: let x = sin A and y = sin B. Prove by contradiction what happens if x^2 + y^2 > 1 or < 1

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@old spade Has your question been resolved?

[old spade]

<@&286206848099549185>

[gritty rose]

Did you try using the hint

[old spade]

i have no idea how to use it

[gritty rose]

,tex .sum diff trig/sin

[old spade]

wait

[glossy valve]

riemann

[old spade]

nvm

x^2 + y^2 = xcosB + ycosA

For x^2 + y^2 = xcosB + ycosA, cosB = x = sinA and cosA = y = sinB

?

[gritty rose]

Yea something like that. Do the rest of the hint

I'm not quite sure if you can safely conclude cosB=x actually

Similarly y=cosA can't be concluded

[old spade]

How T-T I don't do contradictions

[gritty rose]

I don't know what that means

https://www.mathsisfun.com/algebra/proof-by-contradiction.html
Read the example for sqrt 2

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@old spade Has your question been resolved?

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• 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.