Correct
What do you mean by "linear algorithmic complexity"? I can't find the term.
finallyCorrect
I meant the linear complexity of the TM algorithm
Linear Complexity - O(n) - Mastering Data Structures and Sorting Algorithms in JavaScript
An algorithm has linear complexity if the time taken increases linearly with the increase in the number of inputs. (Reading time: under 1 minute)
www.educative.io
Time complexity - Wikipedia
Ah okay.I meant the linear complexity of the TM algorithm
Linear Complexity - O(n) - Mastering Data Structures and Sorting Algorithms in JavaScript
An algorithm has linear complexity if the time taken increases linearly with the increase in the number of inputs. (Reading time: under 1 minute)Time complexity - Wikipedia
en.wikipedia.org
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
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):
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?
):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.I'm not mathcel but here's a statistics problem (i think it counts as math
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?
sorry bud, but I've seen the Monty Hall problem one too many times already
Math
Wdym?Math
Math is soy
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)
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.
You need to find it as an integer
I didn't see the "/" between cos(A/2 - C/2) and cos(A/2 + C/2). Hold on, I'mma solve it proper now.
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)
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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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.
Its Homer Simpson colored?my ballsack is racist
No, Your not Homer Simpson.Nigga are you seriously asking me about my ballsack?
the answer is "no" btw
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,y) = F(y,x).
(x + y)f(x, y) = yf(x, x + y).
Find F(14, 52)
Yeah, I realized it was a silly question, hence why I tried to delete it in time.
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
How close is it?
Very far away still, fortunately for me.How close is it?
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
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
. 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.
Can you still solve it?
S(n) = Σ (i = 1 to n-1) ( Σ (j= i+1 to n) i*j)
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@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
Wdym my summation does give me 11.
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