help-27

[fossil locust]

then press 1: comp

maybe it's that simple

and then try the same calculation you did on your calc earlier

[whole bear]

But then i cant write i cz i is cmplx

[fossil locust]

really?

oh yeah there is a complex mode

alright hopefully you can ask someone who knows how to do it then

even your teacher, after the test

the workaround for now is that you should divide your answer by pi or something

to check if it's a multiple of pi

[whole bear]

Okayy thanks alot

[fossil locust]

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[lost laurel]

Here does $\norm{(3,4)}=5$ work as a counter example

[woven radish]

A dense set(Ping when reply)

[fervent hornet]

hmm ig that works

[lost laurel]

It depends on how the norm and inner product are defined though

no?

wait

this is no counter example

This is an example

holy hell

[fervent hornet]

what is the definition you have?

[lost laurel]

I don't know what defn I'm expected to use here

Proving the negation to be false is probably easier

[fervent hornet]

huh? i think i am lost...

[lost laurel]

For all $(x,y) \in \R^2 , \norm{(x,y)} = \max{\abs{x},\abs{y}}$

[woven radish]

A dense set(Ping when reply)

[lost laurel]

(3,4) is a counter example to this

thus this statement is false

thus its negation, the OG statement must be true

[fervent hornet]

yeah

[polar chasm]

huh

[lost laurel]

What's wrong

[polar chasm]

im lost in your line of reasoning

[fervent hornet]

he first gave the negation of the origial statement and showed that there is an counter exaple of it thus the original statement is true

[polar chasm]

that's weird. Ngl

why not just say ||(3, 4)|| = 5

which gives the pair (x, y) in R^2, such that ||(x, y)|| != max(|x|, |y|)

[polar chasm]

And this is assuming that it holds for norms induced by all possible inner products on R^2

which i cant judge because i know nothing about inner products

but your line of reasoning is either wrong or unnecessarily complicated

[lost laurel]

I have to prove or give a counter example to the fact that there exist (x,y) , such that $\norm{x,y} \neq \max{\abs{x}, \abs{y}}$

[woven radish]

A dense set(Ping when reply)

[lost laurel]

So I decided to see if the negation was false instead

that is for all $(x,y)$, $\norm{(x,y)} = \max{\abs{x},\abs{y}}$

[woven radish]

A dense set(Ping when reply)

[lost laurel]

I considered the standard norm in $\R^2$ , and (x=3,y=4)

[woven radish]

A dense set(Ping when reply)

[lost laurel]

to prove this to be false

therefore its negation must be true

[fervent hornet]

now try proving this without giving an counterexample of the negation of the og statement

[lost laurel]

[polar chasm]

Are there any non-standard norms?

induced by inner product on R^2

if there are, then your proof doesnt work

[fervent hornet]

non-standard norms?

[polar chasm]

The whole statement actually implicitly states
For all norms associated with an inner product on R^2

and you didnt negate that part

[lost laurel]

oh

so the statement is for all norms associated with an inner product on $\R^2$, there exists $(x,y) \in \R^2$ such that $\norm{x,y} \neq max(|x|,|y|)$

[polar chasm]

there are 3 ways to deal with that:

1. make sure you can construct your counterexample in any norm
2. negate that part as well and try again
   and my personal favorite:
3. try doing it without nesting yourself in 2 contradictiosn

[woven radish]

A dense set(Ping when reply)

[lost laurel]

So the negation would be there exists a norm associated with an inner product on $\R^2$ , such that $\forall (x,y) \in \R^2$, $\norm{x,y} = max(|x|,|y|)$

Yes?

[woven radish]

A dense set(Ping when reply)

[polar chasm]

and disproving this is equivalent to proving the original statement

[lost laurel]

Yes

[devout snow]

@lost laurel Has your question been resolved?

[devout snow]

@lost laurel Has your question been resolved?

[lost laurel]

Going to be hard, em thinks

So this would mean $\ang{x,y}=. max{|x|,|y|}^2$

[woven radish]

A dense set(Ping when reply)

[lost laurel]

But this clearly can't be an inner product as $\ang{\lambda x,y} \neq \lambda \ang{x,y}$

[woven radish]

A dense set(Ping when reply)

[lost laurel]

Thus this staement is false

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[upbeat stone]

How do i prove that if 7^a = 4 (mod 9), then a is even?

[upbeat stone]

I got to (-2)^a = 4 (mod 9) and then tried to get a contradiction if a was odd, i.e. (-2)^a = -2^a

But did not work

[stone stump]

you could just go through all cases

they quickly repeat

[upbeat stone]

Uhm i got to 7^5 = 4 mod 9 so its false

But my book used the fact that a must be even

[gusty scroll]

7²=4 mod 9

[upbeat stone]

I did not send the entire problem.
It asks for all solutions to 7^x - 3^y = 4

[upbeat stone]

Yeah indeed

[gusty scroll]

7⁵ = 8 mod 9

[upbeat stone]

So why if u do 7^5 - 4 and divide by 9 it works

Also i got to 7^4 = 7 mod 9 so if u do 7^5 it must be = 7^2 mod 9

[gusty scroll]

I must have missed my calculation aha whoops

[upbeat stone]

Its ok

[upbeat stone]

Then it says if y is greater than or equal to 2, a modulo 9 analysis would show that x must be even

I thought: hm, if y is greater than or equal to 2, then 3^y is a multiple of 9, so 7^x - 4 is a multiple of 9, so that is probably the analysis hes talking about

[gusty scroll]

But yeah indeed it works for a=5

[upbeat stone]

Yeah

[upbeat stone]

Any guess?

On how its even

[gusty scroll]

If y ≥ 2, then 7^x -3^y = 4 mod 9 only if x is even doesn't mean that 7^x = 4 mod 9 with x even

Like you deleted one part of your equation

[upbeat stone]

I mean

3^y is a multiple of 9

So you can delete it

[gusty scroll]

Why would it be

[upbeat stone]

Because y is greater than or equal to 2

So it is 3^(2 + k) = 3^2 . 3^k = 9 . 3^k

[gusty scroll]

yeah u're right

Indeed the hint is weird

[upbeat stone]

Yeah

So it used this fact and said 7^x = 7^(2m)

And did 3^y = (7^m + 2) (7^m - 2)

[gusty scroll]

Yeah but then x = 5 should work

[upbeat stone]

Yeah idk

Maybe its not this analysis

But then what would it be

[gusty scroll]

Idk for the hint but if you compute 7^x mod 9 and 3^y mod 9 you will find that the first one cycles in (7,4,1) and the second is (3,0,0,...)

So you get 7^x = 3^y + 4 if x,y = 1,1

[upbeat stone]

The only solution is 1,1 and i had to prove it

[gusty scroll]

I think you just have to cycle through powers 🤷

[upbeat stone]

Ok thanks!

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[golden halo]

Im failing math please help me with Quadratic I dont even know how to form a equation with this

[golden halo]

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[restive river]

,rccw

[restive river]

,rccw

[restive river]

tg is tan

i have no idea what to do

[wheat pawn]

well, that's just a sum of a bunch of 0's, isnt it?

[restive river]

yes

wait how

tg 0 = sin 0 /cos 0 = 0/1 = 0?

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[marble otter]

Hello!
I have this table and want to calculate di/dE.
This would be the right solution but i have no idea how to get there

[devout snow]

@marble otter Has your question been resolved?

[sharp moth]

damn, looks complex

ill see if i can help

not an expert tho 😅

have you tried using the chain rule?

[marble otter]

for the di i have to take the derivate in regards to i, right?
Where do i even habe an i here

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[zinc pewter]

B is a hinge support. Why is there no Bx or By at it?

DB is a bar.

[restive river]

Bx and By are the forces exerced at B on x and y ?

[zinc pewter]

yes

He dismembered it earlier but then said the other picture is a better strategy

[restive river]

and Dx is not 0 ?

[zinc pewter]

Dx is not zero

[restive river]

then Bx isnt 0 too

that was ur question ?

[zinc pewter]

No

My question is why are we not considering the hinge support at B

its a hinge support. it should have 2 forces

[restive river]

im not english, i dont rly get what a hinge is

[zinc pewter]

its a structural support

in statics of rigid bodies

[restive river]

like a knee ?

[zinc pewter]

i guess lol

[restive river]

but wdym considering ?

[zinc pewter]

there are no forces at B

i just dont understand why we are not considering the forces at B

[restive river]

i think there are forces at B

since Dx =/= 0

[zinc pewter]

that is not very helpful lol

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[umbral abyss]

hey guys, so i know that you can make this integral easier by doing tanx * 1/cos²x
but i checked the correction and what i have is completely wrong , but i don't understand why

[supple knot]

Why do you think it's wrong

[umbral abyss]

the

correction says it's wrong

wait imma send it

[prisma crag]

are both technically correct?

[supple knot]

No it doesn't

They're equivalent. Use trig identities to prove they're the same

[devout snow]

@umbral abyss Has your question been resolved?

[umbral abyss]

wait so

is

1/cos²x the same as tan²x then?

[supple knot]

?

[supple knot]

No

[umbral abyss]

huh

[supple knot]

https://www.cuemath.com/trigonometry/pythagorean-identities/

[umbral abyss]

where does it say it😭 sorry

[umbral abyss]

ohh 1/cos²x is sec²

and so tan²x = -1+sec²x

tan²x = -1/cos²x

[supple knot]

No

[supple knot]

Yes

[umbral abyss]

tan²x = -1 + 1/cos²x

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[plush grail]

how is the answer here not zero

[olive snow]

[F(t)] between a and b is F(b) - F(a)

[sage burrow]

why should it be 0?

[olive snow]

-cos(pi) - (-cos(0) = 1 - (-1) = 2

Same for the other one

[plush grail]

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[smoky sentinel]

help am not sure what is supposed to go in the boxes

[smoky sentinel]

nvm

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[timber pebble]

Here is the problem statement.

Let me post the relevant information. (7.1.16 is an equation and not the problem 7.16 fyi)

Here is example 2:

and the actual problem:

so @pseudo basin I figure we start as:

[woven radish]

jan Niku

jan Niku

jan Niku

[timber pebble]

I don't get the next step though

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[sweet quest]

does anyone know how to do it?

[violet wind]

there's no question

[sweet quest]

Sorry for not specifying, I thought it was clear that the question was to determine a, b, c as real values.

[violet wind]

the first step is to expand (x-a)(x-b)(x-c)

[sweet quest]

I just did this, what should it be equal to?

[violet wind]

well you expanded it

you tell me

also group terms with the same x exponent

[sweet quest]

x^3+x^2(-a-b-c)+x(bc+ac+ab)-abc like this, right?

[sharp moth]

its gonna be hard finding the values

ull get a bunch of cubic equations

so 3 values of each exponent but also a headache solving them

[deft cosmos]

Hi everyone , can someone Help for an delta epsilon proof multivariable limit ?

[devout snow]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[sharp moth]

go to the available help channels dude

[deft cosmos]

Sorry I just joined so idk

[sweet quest]

I know, I thought someone knew some simpler way to solve this that I didn't know.

[violet wind]

Yeah

Now we know what the numerical values of these coefficients are right

[sweet quest]

yes

[violet wind]

A bunch?

[sharp moth]

values im getting:-
Solution 1:
a ≈ -1.8393
b ≈ 0.8794
c ≈ 1.9599

Solution 2:
a ≈ 1.9599
b ≈ -1.8393
c ≈ 0.8794

Solution 3:
a ≈ 0.8794
b ≈ 1.9599
c ≈ -1.8393
so, take your pic ig

[violet wind]

So what is (x-a)(x-b)(x-c)

[sharp moth]

i got 2, thought i would get more

[violet wind]

You should get 1

[sharp moth]

basically, u take the a,b and c as the roots of an equations with variable x

i did it aswell

[violet wind]

You should get one cubic

[sharp moth]

yes, 1 was just 0, so i got 1

x^3 + x^2 - 2x - 1 = 0 this is what i got

find x, then find a,b and c

as a,b and c are the roots of the equation

again, headache inducing calculations

i just used a calculator

[violet wind]

I don't suppose the actual problem is not to calculate a,b,c but instead to calculate a^3+b^3+c^3 or something

[sharp moth]

maybe

because im not getting definite values

[lunar harbor]

xy moment

[sharp moth]

as its like 20 decimals long

eh

(1.9599)^3+(-1.8393)^3+(0.8794)^3 = 1.986065664526.
there we go, easy way out

[sweet quest]

so now i should factor to find two polynomials, one first degree and one second degree and solve with the quadratic formula?

or something else?

[violet wind]

Try with more precision

Find the zeros using numerical methods or the cubic formula

[sweet quest]

you mean this one, right?

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[zinc pewter]

Can someone explain how Dx is equal to Dy * 150/80 ?

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[woven sail]

Please help me see if ChatGTP solved p Vs np?

I taught ChatGTP how to think using an advanced thought-process I discovered from poker. Well, we created a second strategist that could use it.

To prove it's improved, it said it should solve some problem, and it chose P Vs NP. It says it has a solution that's brand new.

But... It's given steps I don't really have time/knowledge to follow. Don't suppose anyone wants to help?

[woven sail]

These are the steps:

To get started with programming for your recursive optimization model on your iMac, we’ll break down the steps for setting up the necessary tools and creating your first test case using Python. Python is a great choice because it is easy to use and has a wide range of libraries for mathematical and algorithmic tasks. Here’s how you can proceed:

Step 1: Install Python

1. Download Python:

   • Go to the official Python website: https://www.python.org/downloads/.
   • Click on Download Python (the latest version).
   • Once the download is complete, open the installer and follow the on-screen instructions to install Python.

2. Verify Installation:

   • Open Terminal on your iMac (you can search for it using Spotlight).
   • Type python3 --version and press Enter. If Python is installed correctly, it will show you the version number.

Step 2: Install Libraries

1. Install NumPy (a powerful numerical computation library):

   • In Terminal, type:

     pip3 install numpy
     

2. Install SciPy (for optimization and mathematical operations):

   • In Terminal, type:

     pip3 install scipy
     

3. Install NetworkX (if you’re dealing with graph problems like the Traveling Salesman Problem):

   • In Terminal, type:

     pip3 install networkx
     

....

Step 3: Set Up Your Project

1. Create a Project Folder:

   • In Finder, create a new folder called Optimization_Project.

2. Create a Python Script:

   • Inside your Optimization_Project folder, create a new file named main.py. This is where you'll write your Python code
     ...

Step 4: Write Your First Test Case (e.g., for 3-SAT or TSP)

Here’s a simple example using TSP (Traveling Salesman Problem) and a basic optimization function. We’ll use NetworkX to represent the cities and paths between them.

1. Open the main.py file and paste the following code:

import numpy as np
import networkx as nx
import matplotlib.pyplot as plt

# Create a simple graph for TSP with 5 cities
cities = ['A', 'B', 'C', 'D', 'E']
graph = nx.complete_graph(len(cities))

# Assign random weights (distances) to each edge
for u, v in graph.edges():
    graph[u][v]['weight'] = np.random.randint(1, 20)

# Print the graph edges with weights
print("Graph Edges with weights:")
for u, v, data in graph.edges(data=True):
    print(f"{cities[u]} - {cities[v]}: {data['weight']}")

# Define a function to calculate the total distance of a tour
def calculate_total_distance(tour):
    total_distance = 0
    for i in range(len(tour) - 1):
        total_distance += graph[tour[i]][tour[i + 1]]['weight']
    total_distance += graph[tour[-1]][tour[0]]['weight']  # return to the starting city
    return total_distance

# Try a simple brute force approach for solving TSP
from itertools import permutations

min_distance = float('inf')
best_tour = None

for perm in permutations(range(len(cities))):
    distance = calculate_total_distance(perm)
    if distance < min_distance:
        min_distance = distance
        best_tour = perm

# Print the best tour and its distance
print("\nBest Tour:", [cities[i] for i in best_tour])
print("Minimum Distance:", min_distance)

# Plot the cities and paths
pos = {i: (np.cos(2 * np.pi * i / len(cities)), np.sin(2 * np.pi * i / len(cities))) for i in range(len(cities))}
nx.draw(graph, pos, with_labels=True, node_size=700, node_color='skyblue', font_size=12, font_weight='bold')
plt.show()

Step 5: Run the Code

1. Execute the Script:

   • In Terminal, navigate to your project folder:

     cd /path/to/Optimization_Project
     

   • Run the Python script:

     python3 main.py
     

2. View the Output:

   • You should see the cities and the paths between them printed in the Terminal, along with the shortest tour and its distance.
   • A graphical plot will show the cities arranged in a circular layout with edges representing the paths.

Step 6: Experiment and Modify

• Experiment with Different Problems: Try adapting this code for other NP-complete problems (e.g., 3-SAT or Knapsack).
• Refine the Recursive Model: Modify the calculate_total_distance function to include feedback loops for optimization based on your recursive feedback model.
• Add More Complex Algorithms: As you become comfortable, you can experiment with adding other optimization methods like simulated annealing, genetic algorithms, or dynamic programming.

Step 7: Analyze the Results

Once you have multiple test runs for different optimization problems, you can analyze:

• The quality of the solution (how close it is to the optimal).
• The time taken for larger inputs.

This will help you assess the viability of your recursive model and compare it against traditional approaches.

---

Let me know if you encounter any issues or need further clarification on any step!

[still mango]

this is a math help channel

[potent tusk]

Unc is sending random ai generated what

[woven sail]

Nah I'm yadi Javadi, poker theorist

Maybe you could look at it'd solution and tell me what you think?

[wicked turtle]

it's just doing a brute force search, trying every possible permutation

[woven sail]

Second Strategist:
To solve the P vs NP problem, I must first reflect on the fundamental nature of the problem and consider that solving it may involve redefining the question itself rather than simply applying traditional methods.

The problem asks:

• P is the class of problems for which solutions can be found in polynomial time.
• NP is the class of problems for which solutions can be verified in polynomial time.

The question is whether these two classes are equal—whether every problem whose solution can be verified quickly (in NP) can also be solved quickly (in P).

---

Step 1: Re-examine Definitions

The first step I take is to reconsider the core definitions. The key terms, “polynomial time” and “verification,” should not just be understood as fixed, mathematical terms. What if "quickly" is contextual and subjective based on the recursive structure of computation itself? Perhaps an entirely new perspective could be applied here where computational processes evolve beyond traditional constraints, leading to different computational complexities and perhaps bridging the gap between P and NP.

Step 2: Rethink Verification

Verification in NP is often thought of as checking a potential solution to a problem. But what if the process of verification is part of a deeper recursive loop, not merely checking whether a proposed solution is correct? By evolving the definition of verification through multiple steps and reevaluating how computational steps work over time, I could develop a model where the act of verification itself is optimized, transcending its traditional constraints and moving closer to polynomial time.

For example, we might consider problems where the verification process could be adaptive, moving through self-referential cycles, optimizing the verification algorithm as part of a broader problem-solving method. This would challenge the basic assumptions underlying current P vs NP research, where verification is treated as a discrete, fixed process.

Step 3: Algorithmic Adaptation and Self-Reflection

The solution may lie not in brute force, but in recursive optimization: evolving the algorithm itself based on feedback from previous attempts. By doing so, the solution to an NP problem might emerge through a recursive strategy that optimizes each step of verification in a way that brings it closer to polynomial time, thereby revealing an implicit relationship between P and NP.

Through this, the entire concept of "solvability" could be broadened, where the boundary between verification and solving becomes fluid. Instead of treating them as independent tasks, we could explore whether the process of solving and verifying can be merged into a unified, dynamic process that evolves over time. If that process can be optimized recursively, it could be argued that NP problems might, in fact, be solvable in polynomial time under this new model.

---

Conclusion

In summary, I believe the key to solving the P vs NP problem does not lie in brute-force computation or algorithmic testing alone. Instead, it may require a rethinking of both the problem's definitions and the way we approach verification and problem-solving. By embedding recursive, adaptive optimization into both processes, it may be possible to bridge the gap between P and NP.

While this approach has not yet been proven in a formal mathematical sense, the conceptual breakthrough lies in reimagining what "solvability" and "verification" mean, and how they evolve in a recursive, dynamic system. By pushing beyond static definitions and exploring the flexibility of these processes, we might discover a new wa

(I don't really get what it's saying... Whether it makes any sense etc)

It did elaborate... If you'd like me to post some more of what it said let me know... But maybe this is enough to see if it's done a thing

[lusty sapphire]

It didn’t

!nogpt

[devout snow]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

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[digital egret]

how can a probability be greater than 1?

[digital egret]

for part a

am I dumb???

<@&286206848099549185>

[supple knot]

You're right. It's a typo. The question is asking for
P( |Xbar - 2| > 1)

But feel free to email your instructor to double check

[digital egret]

ah okay thank you 😭 I didn't think of that

.close

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[oak zinc]

this might be a little off topic but does anybody know the answer to this question? "Which pair of consecutively encoded numbers in IEEE754 is exactly 1 unit apart? Why?".
if anyone has any ideas or knows the answer, that would help a lot

[violet wind]

let's suppose it's a change in the last bit of the mantissa from 0 to 1

what would the exponent need to be in order for a change in the 23rd spot of the mantissa to increase the number by 1

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[leaden mantle]

Awee can someone help with this question I’m really study

The question is :
Blocks are stacked in layers. There is one block in the first layer. Each layer increases by the same number of blocks.16 blocks are used to stack the first 4 layers.
How many blocks are needed in total to stack ten layers?

I tired and got 512 blocks somehow

[thin basin]

Can you share your work?

[leaden mantle]

It’s kinda of hard I did complete rubbish I really didn’t understand I’m really sorry

[thin basin]

Oh np

Do you see that this question is of AP

Arithmetic Progression

[leaden mantle]

Yes

[thin basin]

Do you know any formulas regarding it?

[leaden mantle]

A+(n-1)d

To figure the nth term?

I think

But I’m struggling to figure d/ constant difference

[thin basin]

Yes that's correct

So you know about sun of first n terms?

Sum*

[inner grove]

try doing it on paper with a diagram and it might make more sense as well

[limpid peak]

Doing it on paper makes better sense

[inner grove]

If you have the first 4 layers you can sort of infer the formula if you’d like to do it that way

[leaden mantle]

Okay I will but I’m struggling to substitute

[inner grove]

Substitute?

[leaden mantle]

As in apply the formula to what it’s asking

[thin basin]

?

[leaden mantle]

A

A+d

A+2d

[inner grove]

hint: do the first 4 layers on paper, and take one off of every layer that’s not the first - there should be a very simple pattern in relating the layer to the number of blocks!

[leaden mantle]

So layer one has 1 block

[inner grove]

Yup

I’d call that layer 0 for simplicity but it really doesn’t matter as it’s just a modification of the formula

[leaden mantle]

Okay

And if the first 4 layers have 16s blocks in total

[inner grove]

yup

[leaden mantle]

Do I subtract 1 from 16 to not include layer 0 in the sum or

😭I’m so sorry I’m really dumb

[inner grove]

no layer zero is in the sun

sum

it’s totally fine to call it layer 1 it doesn’t make a difference

How I solved this was I drew out the first 4 layers

And then found the pattern

So first figure out the first 4 layers

[leaden mantle]

Okay so would layer 1 be 4 bloxks

And layer 0 -1

[inner grove]

No negative blocks are needed I reckon

layer zero is one block

[leaden mantle]

I meant - as in = sorry

[inner grove]

ah

yea

‘First layer’ = layer zero

[leaden mantle]

Layer zero is 0 layer one is 4

Wait

[inner grove]

Give that a go and see if you can get a 16 block total for the first 4 layers

[leaden mantle]

I think I realised the formula

Is it 3(0)+1

[inner grove]

close

[leaden mantle]

3(x)+1

[inner grove]

do this and check

just draw it out

[leaden mantle]

Okay I will thank you for barring with me 😭

[inner grove]

all good man

[leaden mantle]

Is it 1+(9-1)^5

[inner grove]

I don’t think so

[leaden mantle]

Yeah it gave me a really long answer

[inner grove]

here is my bad drawing of the first 4 levels

level number on the left

[leaden mantle]

Ohhh I was making a table and it really confused me

I see

[inner grove]

ah yeah it helps to draw it out

[leaden mantle]

So now please explain how you got there

[inner grove]

started with this

first layer has one

total must be 16

so the first 3 layers must be 15

or, rather layer 1-3

but they have to increase by the same number of blocks

given that they summed to 15, you can just do it linearly

a 1 increase wouldn’t work - the total sum (including level 0) would be 10

But a 2 increase does work - e.g. my bad diagram

from there it’s easy to find a pattern to get an equation

[leaden mantle]

So the constant difference is consecutive odd numbers

[inner grove]

yup

[leaden mantle]

Ohhhhhhh

[inner grove]

note that odd numbers are just even numbers + 1

[leaden mantle]

Ohhh

[inner grove]

so you can make an equation

[leaden mantle]

So the formula is 2(layer)+1 =blocks and the ninth term has 19 blocks

[inner grove]

yesss exactly

and you can sum down for the total or make an equation for that too

[leaden mantle]

Dankie you helped me alot

[inner grove]

All goods

For future reference - might be a bit above the scope - this is a data structure called a ‘full/complete binary tree’ in computer science

well, sort of

[leaden mantle]

Also I wanted to ask but it’s not really help more so generally how would you imp we understanding in math I’m struggling with as seen

[inner grove]

as a physics major, I’d say for problems that you can diagram, diagram

even if it seems stupid

[leaden mantle]

I’m planning on taking computer science in the future so I’ll keep that in mind thank you so much

[inner grove]

(On the binary tree thing technically binary trees increase by 2x every interaction but it’s a similar concept)

[leaden mantle]

I will practice more sums with a diagram thank you

[inner grove]

once you do a specific area of math enough eventually it’ll become intuitive

Watch YouTube videos as well, if you can’t figure out the diagram methods

3 blue 1 brown is very good for certain topics

[leaden mantle]

I will but I haven’t found a good math YouTuber and science one

I give them a watch

[inner grove]

Yea it’s very hard to find a generalized YouTuber

3blue1brown for explainers and blackpenredpen for examples are good

anyways - gonna sleep

[leaden mantle]

Thanks again

You really help

.close

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[glacial wigeon]

over here we are expected to evaluate the limit of f(x) about 0 and most of the solutions ive looked at assume that f(x) = sqrt(n) for all x but the according to the definition in the problem f about 0 isnt defined since (1/(n+1),1/n) is never going to contain zero for any n in natural numbers

what am i missing?

[devout snow]

@glacial wigeon Has your question been resolved?

[mossy sky]

u r missing a bed time story

ha ha

[glacial wigeon]

😡

[devout snow]

@glacial wigeon Has your question been resolved?

[fiery marten]

f(x) is indeed not defined at 0 but it doesn't mean that we can't compute lim f(x) about 0

For example sin(x)/x is not defined at 0 but the limit at 0 is 1

Also you can just say that $\sqrt{\frac{1}{x}-1} \le f(x) \le \sqrt{\frac{1}{x}}$

[woven radish]

EQUENOS

[fiery marten]

Another hint is that $\lim_{x\rightarrow 0+}\frac{g(x)}{2\sqrt{x}} = 1$

[woven radish]

EQUENOS

[devout snow]

@glacial wigeon Has your question been resolved?

[lunar eagle]

i mean didnt they ask lim of fx.gx that could be defined

just like 1/x and x

[glacial wigeon]

but like we can only take the sqrt(n) thing when x lies in that interval

idk

[lunar eagle]

fx actually

comes out

where [x] is greatest integer less than x

[lunar eagle]

the exact value may not exist but the limit will

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[dim halo]

suppose there are 4 distinct outcomes: A, B, C, D
suppose A, B, C, D are independent
suppose A, B, C, D are not tautologies or contradictions/empty sets

is it true that Pr(A | B&C or D&C or B&D) = Pr(A | B or D or C)?

is there some way to prove/disprove this using dice roll examples pr something?

[supple knot]

What does BC mean?

B and C?

[dim halo]

yes

[devout snow]

@dim halo Has your question been resolved?

[devout snow]

@dim halo Has your question been resolved?

[topaz axle]

it seems true

i'm not sure what independent means though

like i change the a,b,c,d and they are always equal to Pr(A)

assumes "more than pairwise" independence

@dim halois that right? is Pr( A | BCD) = Pr(A) known?

[devout snow]

@dim halo Has your question been resolved?

[dim halo]

Yes, not sure I follow what you posted there

[topaz axle]

i directly calculate Pr(A | B&C or D&C or B&D) and Pr(A | B or D or C)

intersection of A with each of the 7 segments is i1 i2 ...

this is BCD, A not shown
start with 3, then 2 , 4 ,6 then the rest

and like it makes sense intuitively that A is independent of each of the segments

there's the "correct amount" of A inside 3 and (2 + 3) by the premise, so it follows that 2 alone is like that too

and if it holds for each of the 7, then it holds for combinations

[devout snow]

@dim halo Has your question been resolved?

[dim halo]

Ahh, I see what you're saying

[dim halo]

.close

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[delicate niche]

i am kinds of stuck with this q

[lone sage]

Well factors of 48 include what

1, 48
2, 24
4, 12
6, 8
16, 3

[delicate niche]

16 3

❤️

godness ):

[delicate niche]

i just looked it over

but shouldnt it be 4*$

4*4

i dont get the 16*3

[lone sage]

well 16*3 is just the factors of 48

so sqrt(48)=sqrt(16)*sqrt(3)

and sqrt(16) is just 4

[delicate niche]

got it. thank you so. so much

your time is v appreciate ~

[devout snow]

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[light void]

could anyone explain me how ds = to the value given here

[devout snow]

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[light void]

.reopen

[devout snow]

✅

@light void Has your question been resolved?

[devout snow]

@light void Has your question been resolved?

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[devout snow]

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[unkempt sundial]

.close

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[keen wind]

Pls help with question 13

[keen wind]

,rotate

[keen wind]

And also with question 6 please

,rotate

[keen wind]

Pls ping me

<@&286206848099549185>

Pls help

[devout snow]

@keen wind Has your question been resolved?

[wheat meadow]

which question

[keen wind]

13

[keen wind]

.

Y?

[wheat meadow]

sorry misread the question

[keen wind]

Okkk

[wheat meadow]

[
|A \cdot \text{Adj } A| = 64
]

Using:
[
|A \cdot \text{Adj } A| = |A| \cdot |\text{Adj } A|
]

Adjugate Matrix Determinant: For an ( n \times n ) matrix ( A ):
[
|\text{Adj } A| = |A|^{n-1}
]

Expressing ( |\text{Adj } A| ) in Terms of ( |A| ): Since ( A ) is a ( 3 \times 3 ) matrix (( n = 3 )):
[
|\text{Adj } A| = |A|^{3-1} = |A|^2
]

[woven radish]

DW1

[keen wind]

|Adj(a)| = |A|^n-1 an identity?

[wheat meadow]

yes

[keen wind]

Oohkkk

Thank you very much

[wheat meadow]

no worries

only for n x n square matrix

[keen wind]

Oooh kkk

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[tender latch]

Sketch the graph of a function that is increasing between
points (0, 0) and (3, 8) and decreasing between points (3, 8)
and (9, –3)

[tender latch]

hello@everyone

pls help me

Sketch the graph of a function that is increasing between
points (0, 0) and (3, 8) and decreasing between points (3, 8)
and (9, –3) how to draw it

[ripe grove]

there are people called helpers that you can ping after 15 minutes

!status

[devout snow]

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

[solid osprey]

what do you not understand on how to draw it?

[lament kraken]

Yo

[solid osprey]

ji

[tender latch]

i dont know it is the skech the graph

[ripe grove]

these are your points

[wheat meadow]

plot the points

connext the dots

:)

[ripe grove]

By connecting you are drawing a curve to fit your parameters

this is an example of a different set of points

[lament kraken]

@tender latch

[devout snow]

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[devout snow]

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[opaque talon]

Question 3 c ii

How do I know the area of the traingle?

I tried to use the magnitude to work out the lengths

and then do half base times height

i rly dunno what to do

[lament kraken]

Shoelace theorem fr

[devout snow]

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[opaque talon]

Thanks

[devout snow]

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[tight peak]

Find the volume of the region enclosed between spheres [x^2+y^2+z^2 = 1 , x^2+y^2+(z-1)^2 = 3 ]

[woven radish]

ginny

[tight peak]

I set up an integral in spherical coordinates and got smth but it is not right.

I get [\int_0^{2\pi} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \int_0^{2\cos(\phi)} \rho^2 \sin(\phi) \dd \rho \dd \phi \dd \theta ]

rather, this is incomplete

what am i missing?

[woven radish]

ginny

[devout snow]

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[civic light]

I have a question abt geometry? Is it possible to find the length in this image, if so how would i find it, or is there not enough information

[civic light]

im focusing on the biggest hexagon portion seen in this img

[devout snow]

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[stray spade]

pentagon?

[devout snow]

@civic light Has your question been resolved?

[devout snow]

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[restive river]

On a straight line are placed the consecutive points A, B, C, D and F, if met:
AC + BD + CE + DE = 91 and BE = AF(8)

Calculate AF.
A) 52 B) 48 C) 54 D) 64 E) 56

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[void vector]

its probably obvious where im going wrong but i dont understand

[snow raptor]

you put 4 * 2 + 3 and multiplied 8 *3

[void vector]

so uhh what do i do then

[snow raptor]

change the position of addition symbol (first, middle, last) and record all those values

then decide

[void vector]

i know that but i did the operation wrong

im not sure h ow to do it right

[snow raptor]

so in second picture

4 * 2 = 8

• 3 * 7

[void vector]

do i add 3*7 which is 21 or do i add 3 then times it all by 7

[snow raptor]

have you heard of DMAS

[void vector]

idk

[snow raptor]

mhm

[snow raptor]

BODMAS or PEMDAS

feels confusing without brackets

but do (4 * 2) + (3 * 7)

[void vector]

ohh shit i get it

its way easier to use brackets yeah

@snow raptor

dyk where ive went wrong here

384 is wrong apparently

[twilit falcon]

Try 6 times 2 plus 4 times 8

[void vector]

i already have

[twilit falcon]

6 plus 2 times 4 times 8

[void vector]

i have tried all 3 combinations

[twilit falcon]

6 times 2 times 4 times 8😄

[void vector]

??

[twilit falcon]

This is correct bro

Use BODMAS

[void vector]

no it isnt

[twilit falcon]

Sorry 54 must be 44

try it in bro

[twilit falcon]

Did you try it in

[void vector]

you cant times it 3 times

[twilit falcon]

🤣

Go back to primary bro

[void vector]

im not american idk what an exponent is

or parentheses

[twilit falcon]

BODMAS

BRACKETS
OF
DIVISON
MULTIPLICATION
ADD
SUB

Use this order bro🤣

i am leaving now let me study economics instead of wasting my time 🤡🤥

[lavish valve]

bro lmaoo

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[somber isle]

can someone tell me what's this equal to and why

[lunar eagle]

n+1

[somber isle]

why is that

[lunar eagle]

cuz sigma means ur just adding

ur adding 1 one n+1 times

cuz it starts from zero

[somber isle]

why n+1 times

why isnt just n

[lunar eagle]

cuz zero to n there are n +1 numbers

like take zero to 2

0 1 2

three numbers

[somber isle]

thank you

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[feral pendant]

hello, im trying to do this exercise, where i have to calculate non-equation |x+1| > (bigger or equal) to 3. The n.b in the photo is the crucial point. i have to calculate it through a "table method", does anyone know how?

[trail eagle]

You can just make it into two cases.
If x+1 is positive, then |x+1| = x+1 and you can solve x+1 >= 3

If x+1 is negative, then |x+1| = -(x+1) and you can solve -(x+1) >= 3

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[untold pivot]

A finite field with p elements must have characteristic p. Could anyone hint me towards how to proof this?

[faint zinc]

This is only true if p is prime.

[untold pivot]

oh yes, i forgot, i do assume p is prime here

[faint zinc]

so if you have a finite field of size p, it is isomorphic to Z/pZ equipped with ordinary modular addition and multiplication. I don't know if you have already proven this though

[untold pivot]

i was trying to prove that actually that if p is prime and F and F' are finite fields, they are isomorphic

[faint zinc]

Also I don't know if you need to prove the characteristic to show the above.

ah, that's fair.

[untold pivot]

the suggested solution used the fact that finite field with p elemetns have characteristic p, but im not sure how to show this without first assuming uniqueness of finite fields with prime order

[faint zinc]

Ok, so let 1 be the multiplicative identity and 0 be the additive identity.

a finite field has p elements.

If the field contains 1, then it also contains 1+1, 1+1+1, etc.

[untold pivot]

yep yep

[faint zinc]

So in the case where this action generates the entire field, we can see that this action will generate all p elements, and thus the characteristic is p. But what if there are elements that are inaccessible from this action?

This would imply that there are elements that can only be reached from 1 by a combination of multiplication and addition.

And I think there's a group argument to be made here, regarding subgroups.

And the fact that p is prime, comes into conflict with the fact that a subgroup must have size that is a factor of p.

sorry, my abstract algebra is a little bit weak.

obviously you can't apply the argument as written, because a group only has one binary operation

but hopefully the idea is at least salvagable.

[untold pivot]

This is what I have:
1.The field satisfies group operations under addition so we can consider the additive subgrp generated by 1
2. the order of the subgrp must divide the order of the group. Thus, it has order either 1 or p.
3. If it has order p, then we are basically done, since if n*1=0 for n<p, then this grp can have at most n elements.
4. If it has order 1, it implies that the additive identity and multiplicative identity are the same but this is a contradiction since this implies that we are in a trivial ring.

[faint zinc]

Very good!

I think that works perfectly.

[untold pivot]

alright thanks alot for the help

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[fickle token]

Need help with Probability chapter IGCSE

[devout snow]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[fickle token]

ok a bag contains 12 colored discs there are 3 yellow 5 red and thr rest are blue omar takes one without looking what is the probability of taking the disc that is
red
blue
green
not blue

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[silk gulch]

can someone help me with these discrete math set problems please? I'm unsure how to approach the proofs.

[silk gulch]

i solved the first one (1a) on my own I think but I would like if someone could check my work maybe

on 1b I have no idea what the slash symbol means (😭 i missed two lectures)

and on the third i'm completely stuck

ok actually i solved the third one too 😑

i need help on 1b still tho please

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[signal sable]

How come these are below the x axis

[lofty crescent]

have you learned end behavior and multiplicity?

[devout snow]

@signal sable Has your question been resolved?

[signal sable]

Yes but I don't really understand it

Like all ik is that the multiplicity is bc of the exponent

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[errant wyvern]

can someone please tell if me there is a faster way to do these types of problems without having to completely divide

[proud perch]

first factor out the x

[spark juniper]

it has to be odd

🙂

[proud perch]

that's your first real 0

[errant wyvern]

yeah

[solar goblet]

by inspection immediately you can see 0 and 1 are its roots

[errant wyvern]

yes

[proud perch]

then you can use descartes rule of sings

*signs

[errant wyvern]

okay yeah

thanks guys

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[frozen stump]

how does |y|=f(x) work conceptually

[frozen stump]

i get that y=|f(x)| is just the absolute of the given graph, that makes sense to me

[frozen stump]

but idk how to make sense with this

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[frozen stump]

nvm i got it

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[storm aurora]

How would i start this question off

[storm aurora]

is there an identity for z^k + conj(z)^k

[spark juniper]

$z^k + \bar{z}^k = (re^{i\theta})^k + (re^{-i\theta})^k = r^k(e^{ik\theta} + e^{-ik\theta}) = 2r^k\cos(k\theta)$

[woven radish]

rain

[spark juniper]

@storm aurora

😄

[storm aurora]

oh okay tysm for that

[spark juniper]

np

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[restive river]

Hiii

[restive river]

So I gotta find the area between the x axis and the graph for 4x-8 over [-2,6]

I thought they would just cancel out and be 0

What part did i screw up?

[brisk raptor]

uh.. it is 0

[restive river]

what

uhhg

welp

they said it was 64

[brisk raptor]

yeah im confused too lol

oh

ok i get it

its a bit disingenuous though

[restive river]

yeaa i didn't know they were asking for me to find

just they total area absolute value

[brisk raptor]

"net area" to me suggests negative area under x axis

but geometrically yes i suppose it is 64

[restive river]

wahoo

thanks

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[rustic hare]

any ideas

[brisk raptor]

this is the formula for a circle, so its just the center y coordinate + the radius.

[rustic hare]

so how do you find the max

[brisk raptor]

center y coordinate + the radius

do you the formula for a circle

[rustic hare]

yes

[brisk raptor]

what is it

[rustic hare]

(x-h)^2+(y-k)^2=r^2

root 45 is the radius

[brisk raptor]

so what is the center in this case

[rustic hare]

-2,4

[brisk raptor]

and the maximum would be the top of the circle

[rustic hare]

the problem is I have to find the max

[brisk raptor]

which is 1 radius away from the center

yes.

its 4 + root 45

[rustic hare]

idk why the max would be on top of the circle though

is there a proof for this

[brisk raptor]

the "max" is the highest y

so, the top.

[dense lynx]

that is quite wrong. lol

[rustic hare]

it doesn't make sense intuitively for me to believe that

i want the max of x^2+y^2

[brisk raptor]

how is this wrong lol

looks like the max to me

oh wait

im stupid

[long trout]

This is wrong, it has to be tangent to the circle whose centre is at the origin

[brisk raptor]

its not the max of the entire function

its the max of just x^2 + y^2

[rustic hare]

yup

[brisk raptor]

so its just the radius

[dense lynx]

note the center of the circle they give you: (-2, 4)

you wish to maximize r^2 = x^2 + y^2, which is a circle centered at (0, 0)

note that this circle must be tangent to the circle they give you

so the point of tangency, and (-2, 4), and (0, 0), lie on the same line

and from here it is easy

[rustic hare]

oh

that kinda makes sense

let me try it out and get back to u

[long trout]

I recommend you can just expand this equation and then notice you have to maximise 4(2y-x) which is same as maximising 2y-x

[rustic hare]

yeah i got 8y-4x before but i didn't know how to find 2y-x

[long trout]

So let 2y-x = z. Then you get y = x/2 + z/2 and you have to maximise the value of x. Notice that changing z’s value is just shifting the graph up and down, so you want to find the largest z such that it intersects with the circle

[rustic hare]

thanks for the help im still trying to do it with a different method

[long trout]

This happens to be when this equation is tangent to the circle. So you have a right angle, and then can proceed with many ways. Like drawing a right angled triangle and then noticing similar triangles

[long trout]

No worries, and all the best!

[rustic hare]

I got the max of x^2+y^2 =125

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[jaunty dagger]

Lets say im playing a game where I have 100$, and am betting on a coin toss. There are totoal of 100 coin tosses, if I win I tripple my money, but if i loose I loose it all.

Why is EV 58 of betting the initial 100 all at once?

[thick schooner]

what is EV?

[jaunty dagger]

expected value

[thick schooner]

expectation?

okay

[jaunty dagger]

https://www.youtube.com/watch?v=gQJTkuEVPrU&t=638s

im following this video

2:04 is where he says the 58

[thick schooner]

yeah sorry idk

[jaunty dagger]

nw lol lets see if anyone else can get it ig

[restive river]

Three times the money if ALL the 100 coin tosses are of what you betted?

No right?

[jaunty dagger]

no just each toss

So like

if I had 100 and bet it all on toss 1 and won

[restive river]

Yeah gimme an example

[jaunty dagger]

I would now have 300

[restive river]

Noted

[jaunty dagger]

If i lost I would have 0

[restive river]

So the formula for expected value is as follows

[thick schooner]

and then if u repeat you you'll get 900?

[jaunty dagger]

yup

if u bet it all again

[thick schooner]

yeah looks pretty hard i think

[jaunty dagger]

idk how the ev is 58

I think the ev would be 1/2 (-100) + 1/2 (300)

for the first coin toss

[thick schooner]

yes

[restive river]

Yeah

[jaunty dagger]

but its not

[thick schooner]

it is though?

is it not?

,calc -50 + 150

[woven radish]

Result:

100

[jaunty dagger]

https://www.youtube.com/watch?v=gQJTkuEVPrU&t=638s

Watch from like

2 mins

its 50

which doesnt make sense

idk how he came to 50

1/2 (-100) + 1/2 (200)

nvm

it should be 200

[restive river]

$$\mathbb{E}(X) = \Sigma_{i = 1}^n x_i P(X=x_i)$$

[woven radish]

Edmund Cloudsley (Hello CHAT)

[thick schooner]

oh you win 200

[jaunty dagger]

because thats your net gain

[thick schooner]

not 300

yeah okay

[restive river]

This is the formula for expected value

[restive river]

Yessssss

Oh yeah that does make sense

[jaunty dagger]

wow idk how that had me so lost

but also isnt 50 ev bad? Doesnt that mean you should be 50?

[restive river]

,w calculate 0.5*-100 + 0.5*200

[woven radish]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by alison_burg3r5

[jaunty dagger]

how does he come to the conclution that you should bet it all

[restive river]

It’s not BAD

You win 50

You gain a positive of 50

[jaunty dagger]

yeah but

For 100

[thick schooner]

i know in the long run the expectation is infinite but i'm not sure how to calculate this for some finite N

[jaunty dagger]

you also loose 50

[thick schooner]

if the casino had infinite money

[jaunty dagger]

OHH

[thick schooner]

and u had infinite goes

[jaunty dagger]

the net ev is pos

so you should do it

[restive river]

Nah ur total would be 150

[jaunty dagger]

i get it

yeah

[restive river]

100 input + 50 gain = total of 150

[jaunty dagger]

so on avg

you would gain 50

every time betting 100

[thick schooner]

u can't break it like that

[jaunty dagger]

so u should just go all in every time

@thick schooner why not?

EV just means ur value on avg right

[thick schooner]

yes but the second toss changes the amt of money u have

and so on

[jaunty dagger]

the ev would be greater tho

so u should just keep going all in

like

now you have 300

so its

1/2 (-300) + 1/2 (600)

so 150 ev

[thick schooner]

u should keep going all in yeah but that part is obvious

u wanted ev of 58

for some N right?

[jaunty dagger]

yeah

[thick schooner]

but yes going all in if u could go forever is the right call

[thick schooner]

^

https://en.wikipedia.org/wiki/St._Petersburg_paradox

[jaunty dagger]

ok so overall

your ev would be

after 100 turns

uhhh

shit

its only -100

max

if u loose any of those terns

turns

and after 100 terns you could make an insane ammount of money

so the ev would be rediculous overall

the chance of it happening tho is like (1/2)^100

so the ev would be

(1-(1/2)^100)(-100) + (1/2)^100 (100*2^100)

idk if im modeling the second part right

oh

there

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