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SuicideFuel Math thread problem (official)

I meant the linear complexity of the TM algorithm

Ah okay.
Calculate the linear algorithmic complexity of the following Turing machine
View attachment 738405
Does this Turing machine even halt when you input 1BBB...? I think it just propagates the 1 forward ad infinitum.
 
what is 2+2 nigga?
only niggers can try to answer this



it's over
 
I'm not mathcel but here's a statistics problem (i think it counts as math:feelsgah:):

A man is in a TV show and there are 3 doors, behind one of the 3 doors there is a car, he has to choose in which door the car is to get it. He chooses the door on the left. Once he has chosen, the host opens the door on the center and there is nothing on this door. Now the only doors that are unopened are left and right, and the host offers him the possibility of changing his choice before revealing where the car is. If the man changes the door will his chances of getting the car be the same as if he keeps his choice?
 
I'm not mathcel but here's a statistics problem (i think it counts as math:feelsgah:):

A man is in a TV show and there are 3 doors, behind one of the 3 doors there is a car, he has to choose in which door the car is to get it. He chooses the door on the left. Once he has chosen, the host opens the door on the center and there is nothing on this door. Now the only doors that are unopened are left and right, and the host offers him the possibility of changing his choice before revealing where the car is. If the man changes the door will his chances of getting the car be the same as if he keeps his choice?
Statistics certainly counts as math, but try something more original than the Monty Hall problem next time.
 
Statistics certainly counts as math, but try something more original than the Monty Hall problem next time.
baba sigh GIF by Puffin Rock
 
The sides a, b and c of a triangle are in AP in that order, being opposite the interior angles A, B and C, respectively. Find the value of the expression:


Cos(A/2 - C/2)/Cos(A/2 + C/2)
 
The sides a, b and c of a triangle are in AP in that order, being opposite the interior angles A, B and C, respectively. Find the value of the expression:


Cos(A/2 - C/2)/Cos(A/2 + C/2)
By prosthaphaeresis, cos((A - C)/2)cos((A + C)/2) = (cos A + cos C)/2. Since a,b,c are in arithmetic progression, let k be the common difference -- i.e., a = b - k and c = b + k. Note that the triangle inequality places restrictions on the size of k such as, e.g., k < b. Anyhoo, by the law of cosines, cos A = (b^2 + c^2 - a^2)/(2bc) and cos C = (a^2 + b^2 - c^2)/(2ab). Putting everything together and simplifying yields that cos((A - C)/2)cos((A + C)/2) = (b^2 - 4k^2)/(2b^2 - 2k^2).

Did I go wrong somewhere? I sort of expected b to cancel out.
 
By prosthaphaeresis, cos((A - C)/2)cos((A + C)/2) = (cos A + cos C)/2. Since a,b,c are in arithmetic progression, let k be the common difference -- i.e., a = b - k and c = b + k. Note that the triangle inequality places restrictions on the size of k such as, e.g., k < b. Anyhoo, by the law of cosines, cos A = (b^2 + c^2 - a^2)/(2bc) and cos C = (a^2 + b^2 - c^2)/(2ab). Putting everything together and simplifying yields that cos((A - C)/2)cos((A + C)/2) = (b^2 - 4k^2)/(2b^2 - 2k^2).

Did I go wrong somewhere? I sort of expected b to cancel out.
You need to find it as an integer
 
The sides a, b and c of a triangle are in AP in that order, being opposite the interior angles A, B and C, respectively. Find the value of the expression:


Cos(A/2 - C/2)/Cos(A/2 + C/2)
This is trivial by utilizing Mollweide's formula. Your ratio of cosines equals (a + c)/b = (b - k + b + k)/b = 2.
 
Find relation between m an n if Sum i from 1 to n of (i/i+n) is equal to sum i from 1 to m (1/i+m)
 
Find relation between m an n if Sum i from 1 to n of (i/i+n) is equal to sum i from 1 to m (1/i+m)
Let S(m) = sum from i=1 to m of 1/(i+m). By factoring out an m from the denominators, we see that S(m) is a lower Darboux sum of the integral from x=0 to 1 of 1/(1+x) which integrates to ln(2). As such, S(m) < ln(2) for all positive integers m. Turning our attention to n, it's not difficult to see that we can rewrite the sum from i=1 to n of i/(i+n) as n(1 - S(n)). Since S(n) < ln(2) it follows that n(1 - S(n)) > n(1 - ln(2)). Numerically we see that 3(1 - ln(2)) > ln(2) > S(m) so n cannot be greater than 2. Again, one can numerically check that the sum from i=1 to n of i/(i+n) > ln(2) if n=2. Ergo, n=1 and S(m) = ½. Since S(m) can be interpreted as the average of 1/(1+i/m) for i=1 through m, and every single one of those term except i=m is strictly greater than ½, it must follow that m=1 as well. So ultimately we find that m = n = 1.
 
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Let S(m) = sum from i=1 to m of 1/(i+m). By factoring out an m from the denominators, we see that S(m) is a lower Darboux sum of the integral from x=0 to 1 of 1/(1+x) which integrates to ln(2). As such, S(m) < ln(2) for all positive integers m. Turning our attention to n, it's not difficult to see that we can rewrite the sum from i=1 to n of i/(i+n) as n(1 - S(n)). Since S(n) < ln(2) it follows that n(1 - S(n)) > n(1 - ln(2)). Numerically we see that 3(1 - ln(2)) > ln(2) > S(m) so n cannot be greater than 2. Again, one can numerically check that the sum from i=1 to n of i/(i+n) > ln(2) if n=2. Ergo, n=1 and S(m) = ½. Since S(m) can be interpreted as the average of 1/(1+i/m) for i=1 through m, and every single one of those term except i=m is strictly greater than ½, it must follow that m=1 as well. So ultimately we find that m = n = 1.
I have been trying to figure out the answer for days even tried programming to solve it but didn't worked out i didn't even thought it would require this much analysis i tried to rewrite the summands by changing limits and got nowhere
High iq
 
I have been trying to figure out the answer for days even tried programming to solve it but didn't worked out i didn't even thought it would require this much analysis i tried to rewrite the summands by changing limits and got nowhere
High iq
Thanks. I had my computer tabulate the first hundred values for the m and n sums. By inspecting those tables it becomes readily apparent that S(m) increases toward some limit, which according to Wolfram Mathematica was ln(2). The n sums were also increasing by inspection and already from n=2 onward were they bigger than ln(2). From there it was just a matter of obtaining the right inequalities. Once you know that S(m) converges increasingly toward ln(2) guessing it's a lower Darboux sum comes pretty naturally as well, esp. if you've seen that trick used before (which I have). Hopefully this gives you some insight as to you might have been able to get at the answer yourself.
 
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.
 
F(x,x) = x.
F(x,y) = F(y,x).
(x + y)f(x, y) = yf(x, x + y).

Find F(14, 52)
 
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
 
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 :feelsokman:
 
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
 
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 n3
 
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
 
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
 
. 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 ?
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
 
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.
 

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