SuicideFuel Math thread problem (official)

[Ahnfeltia]

View: https://www.reddit.com/r/facepalm/comments/12zrf4j/pov_you_missed_school_or_just_high/?utm_name=androidcss


2+2 is racist

my ballsack is racist

 

[Lonelyus]

my ballsack is racist

Its Homer Simpson colored?

 

[Ahnfeltia]

Its Homer Simpson colored?

Nigga are you seriously asking me about my ballsack?

Spoiler

the answer is "no" btw

 

[Lonelyus]

Nigga are you seriously asking me about my ballsack?

Spoiler

the answer is "no" btw

No, Your not Homer Simpson.

 

[Grim_Reaper]

A Turing machine is defined by the following scheme. Formulate an algorithm according to which this machine operates

View attachment 738404

Calculate the linear algorithmic complexity of the following Turing machine
View attachment 738405

Is this computational complexity?

 

[Ahnfeltia]

Is this computational complexity?

While computational complexity is more broad than Turing machines, the second question is indeed concerned with the computational complexity of a certain Turing machine.

 

[trying to ascend]

This is trivial by utilizing Mollweide's formula. Your ratio of cosines equals (a + c)/b = (b - k + b + k)/b = 2.

Correct .

Didn't know about this formula

 

[trying to ascend]

F(x,x) = x.
F(x,y) = F(y,x).
(x + y)f(x, y) = yf(x, x + y).

Find F(14, 52)

 

[trying to ascend]

What is the domain of F?

The problem didn't state it.

But you can solve it solely using N

 

[Ahnfeltia]

The problem didn't state it, but you can solve it solely using N

Yeah, I realized it was a silly question, hence why I tried to delete it in time.

 

[Ahnfeltia]

F(x,x) = x.
F(x,y) = F(y,x).
(x + y)f(x, y) = yf(x, x + y).

Find F(14, 52)

F(14,52) = F(14,38)*52/38 = F(14,24)*38/24*52/38 = F(14,10)*24/10*52/24 = F(10,14)*52/10 = F(10,4)*14/4*26/5 = F(4,10)*7*13/5 = F(4,6)*10/6*7*13/5 = F(4,2)*6/2*10/6*7*13/5 = F(2,4)*10/2*7*13/5 = F(2,2)*4/2*7*13 = 2*2*7*13 = 364

 

[trying to ascend]

F(14,52) = F(14,38)*52/38 = F(14,24)*38/24*52/38 = F(14,10)*24/10*52/24 = F(10,14)*52/10 = F(10,4)*14/4*26/5 = F(4,10)*7*13/5 = F(4,6)*10/6*7*13/5 = F(4,2)*6/2*10/6*7*13/5 = F(2,4)*10/2*7*13/5 = F(2,2)*4/2*7*13 = 2*2*7*13 = 364

Correct

 

[etbrute]

Once AI will be able to do math at the level of the best contemporary mathematicians, everything will be cope.

How close is it?

 

[Ahnfeltia]

How close is it?

Very far away still, fortunately for me.

 

[trying to ascend]

Find S(n), given that S(1) = 0. S(2) = 1.2, S(3) = 1.2 + 1.3 + 2.3. S(4) = 1.2 + 1.3 + 1.4 + 2.3 + 2.4 + 3.4

 

[Grim_Reaper]

Find S(n), given that S(1) = 0. S(2) = 1.2, S(3) = 1.2 + 1.3 + 2.3. S(4) = 1.2 + 1.3 + 1.4 + 2.3 + 2.4 + 3.4

S(n) = S(n-1) + Σ (i = 1 to n-1) i+(n/10) for n≥3

 

[trying to ascend]

S(n) = S(n-1) + Σ (i = 1 to n-1) i+(n/10) for n≥3

What is that summation adding up?

Also, it has to be as a function of n, not S(n).

 

[Grim_Reaper]

What is that summation adding up?

Also, it has to be as a function of n, not S(n).

Like for S(4) for example, the summation will be 1 + (4/10) + 2 + (4/10) + 3 + (4/10) = 1.4+2.4+3.4

 

[trying to ascend]

Like for S(4) for example, the summation will be 1 + (4/10) + 2 + (4/10) + 3 + (4/10) = 1.4+2.4+3.4

. as in a multiplication sign, it's 1 times 2, 1 times 3 and so on

 

[Grim_Reaper]

. as in a multiplication sign, it's 1 times 2, 1 times 3 and so on

Bruh I thought it was a decimal sign. You should've used the * sign for multplication.

 

[trying to ascend]

Bruh I thought it was a decimal sign. You should've used the * sign for multplication.

Can you still solve it?

 

[Grim_Reaper]

Can you still solve it?

S(n) = Σ (i = 1 to n-1) ( Σ (j= i+1 to n) i*j)

 

[Grim_Reaper]

S(n) = Σ (i = 1 to n-1) ( Σ (j= i+1 to n) i*j)

@trying to ascend ?

 

[trying to ascend]

@trying to ascend ?

Is it a doule summation? Assuming it is, then it's not correct, provided that S(3) = 11 and yours gives 14.

We will have sigma (upper equals 3 and lower 2) from (i)(i + 1) + Sigma (upper 3 and lower 3).

1.2 + 2.3 + 2.3.

You won't have pair of terms whose difference is higher than 1 in that summation.

Hint: The problem has a relation with arithmetic progressions

 

[Grim_Reaper]

Is it a doule summation? Assuming it is, then it's not correct, provided that S(3) = 11 and yours gives 14.

We will have sigma (upper equals 3 and lower 2) from (i)(i + 1) + Sigma (upper 3 and lower 3).

1.2 + 2.3 + 2.3.

You won't have pair of terms whose difference is higher than 1 in that summation.

Hint: The problem has a relation with arithmetic progressions

Wdym my summation does give me 11.

 

[trying to ascend]

Wdym my summation does give me 11.

Was this what you wrote:
View: https://www.youtube.com/watch?v=vGQMmpTInPU
?

 

[Grim_Reaper]

Was this what you wrote:
View: https://www.youtube.com/watch?v=vGQMmpTInPU
?

Yeah but j is dependent on i.

 

[Ahnfeltia]

Find S(n), given that S(1) = 0. S(2) = 1.2, S(3) = 1.2 + 1.3 + 2.3. S(4) = 1.2 + 1.3 + 1.4 + 2.3 + 2.4 + 3.4

(1 + 2 + ... + n)^2 = 1^2 + 2^2 + ... + n^2 + 2S(n). Since 1 + 2 + ... + n = n(n+1)/2 and 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6, one can easily solve for S(n) and find that S(n) = (n-1)n(n+1)(3n+2)/24.

S(n) = Σ (i = 1 to n-1) ( Σ (j= i+1 to n) i*j)

You have merely rewritten his function in terms of summation notation, which hardly counts as solving the problem.

 

[trying to ascend]

(1 + 2 + ... + n)^2 = 1^2 + 2^2 + ... + n^2 + 2S(n). Since 1 + 2 + ... + n = n(n+1)/2 and 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6, one can easily solve for S(n) and find that S(n) = (n-1)n(n+1)(3n+2)/24.

You have merely rewritten his function in terms of summation notation, which hardly counts as solving the problem.

Correct

 

[cripplecel]

f(x) = 2x^2 what is f’(x)

 

[Ahnfeltia]

Bruh I thought it was a decimal sign. You should've used the * sign for multplication.

@trying to ascend I concur. While using a "." for multiplication is fine when it comes to variables, using it this way between numbers is very confusing because virtually everyone will interpret it as a radix point. Moreover, I believe using a "." for multiplication is Indian notation, but Wikipedia suggests India uses the "." as the decimal separator as well, so I can't imagine using "." to indicate multiplication between numbers is common practice.

 

[Ahnfeltia]

f(x) = 2x^2 what is f’(x)

Hint: power rule.

 

[fukurou]

if the amount of sentences is x, and each param list is marked as a1...an with y elements per list.
how many senstence options are there?

    public PerChanceTest() {
        super();
        sentences.add("here is a  salad vegi1 #, vegi2 # and herb #.");
        sentences.add("how about this salad vegi1 #, vegi2 # and herb #. it goes well with tuna fish");
        UniqueItemSizeLimitedPriorityQueue temp = new UniqueItemSizeLimitedPriorityQueue();
        temp.setLimit(3);
        wordToList.put("vegi1", temp);
        temp = new UniqueItemSizeLimitedPriorityQueue();
        temp.setLimit(3);
        wordToList.put("vegi2", temp);
        temp = new UniqueItemSizeLimitedPriorityQueue();
        temp.setLimit(3);
        wordToList.put("herb", temp);
    }

output example:
output:
here is a salad cucmber, purple onion and cpriander.
here is a salad tomato, purple onion and irit.
here is a salad lettuce, pickle and cpriander.
how about this salad tomato, purple onion and cpriander. it goes well with tuna fish
how about this salad lettuce, pickle and irit. it goes well with tuna fish
how about this salad lettuce, pickle and parcely. it goes well with tuna fish
how about this salad tomato, onion and cpriander. it goes well with tuna fish
how about this salad cucmber, pickle and parcely. it goes well with tuna fish
here is a salad cucmber, purple onion and parcely.
here is a salad cucmber, purple onion and irit.

Process finished with exit code 0

 

[Ahnfeltia]

if the amount of sentences is x, and each param list is marked as a1...an with y elements per list.
how many senstence options are there?

If I understand you correctly, you have x different sentences, each with n variable slots and each slot permitting y options, correct? In that case, the total number of possible sentences is x*(y^n).

 

[trying to ascend]

f(x) = 2x^2 what is f’(x)

4x

@trying to ascend I concur. While using a "." for multiplication is fine when it comes to variables, using it this way between numbers is very confusing because virtually everyone will interpret it as a radix point. Moreover, I believe using a "." for multiplication is Indian notation, but Wikipedia suggests India uses the "." as the decimal separator as well, so I can't imagine using "." to indicate multiplication between numbers is common practice.

Yes, it indeed looked confusing, though I was just copying the original problem

 

[Ahnfeltia]

Yes, it indeed looked confusing, though I was just copying the original problem

fair enough in that case

 

[Grim_Reaper]

(1 + 2 + ... + n)^2 = 1^2 + 2^2 + ... + n^2 + 2S(n). Since 1 + 2 + ... + n = n(n+1)/2 and 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6, one can easily solve for S(n) and find that S(n) = (n-1)n(n+1)(3n+2)/24.

You have merely rewritten his function in terms of summation notation, which hardly counts as solving the problem.

I thought that was okay.

 

[Ahnfeltia]

I thought that was okay.

Generally not (although there technically are exceptions). Here's a rather extreme example: would you say that sum from j=2 to 4 of j is a valid solution to "find the value of 2 + 3 + 4"?

 

[Grim_Reaper]

Generally not (although there technically are exceptions). Here's a rather extreme example: would you say that sum from j=2 to 4 of j is a valid solution to "find the value of 2 + 3 + 4"?

No but the upper limit for your example is bounded.

 

[Ahnfeltia]

No but the upper limit for your example is bounded.

Does any of the limits being a variable change anything significant? In that case you could argue that summation notation is more formal as opposed to the ellipses, but that's about it, no? At the end of the day you have merely recast the given expression in a more formal form without really doing any kind of simplification.

 

[fukurou]

If I understand you correctly, you have x different sentences, each with n variable slots and each slot permitting y options, correct? In that case, the total number of possible sentences is x*(y^n).

that looks like a lot when it's put into a formula, maybe it could be used to write visual novel type of games, scripts

 

[Ahnfeltia]

that looks like a lot when it's put into a formula, maybe it could be used to write visual novel type of games, scripts

it is a lot, esp. for large n

 

[uranium235]

Consider 9 dots arranged as 3x3 what is the highest amount of closed figures that can be drawn by connecting the dots one by one in a single trial you can imagine this as drawing pattern lock in your phone i have achieved 10 closed figures so far

 

[Ahnfeltia]

Consider 9 dots arranged as 3x3 what is the highest amount of closed figures that can be drawn by connecting the dots one by one in a single trial you can imagine this as drawing pattern lock in your phone i have achieved 10 closed figures so far

I take it you want the number of essentially distinct ones? Mirror images counting as the same figure etc.

 

[uranium235]

I take it you want the number of essentially distinct ones? Mirror images counting as the same figure etc.

Wait i will upload an image to demonstrate

 

[uranium235]

There are 10 closed figures in this image

 

[uranium235]

I take it you want the number of essentially distinct ones? Mirror images counting as the same figure etc.

Notification

 

[Ahnfeltia]

There are 10 closed figures in this image

Aha. I see what you mean now. 11 is possible.

 

[uranium235]

Let P(x) be a polynomial with positive real coefficients. Prove that
Sqrt(P(a)P(b) )≥ Sqrt(P( ab)),for all positive real numbers a and b.

 

[Ahnfeltia]

Let P(x) be a polynomial with positive real coefficients. Prove that
Sqrt(P(a)P(b) )≥ Sqrt(P( ab)),for all positive real numbers a and b.

I'm not convinced this is true. Take P(x) = ¼ + x. In that case your desideratum amounts to ¼ + ab ≤ (¼ + a)(¼ + b) for positive a & b, but this is evidently false for a = b = ¼.