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May
18
comment Nuking the Mosquito — ridiculously complicated ways to achieve very simple results
Sorry, but now it is 5 equations with 6 unknowns, still not good enough, and there is a typo in the second. If you fix the second to AB*BC=0 and add in BC*DC=0 it should work.
May
18
comment Nuking the Mosquito — ridiculously complicated ways to achieve very simple results
The official solution as written here can't be correct, it has 6 unknowns but only 4 equations. Obviously we also need some equations with AD, and we need to include CD to make any useful equations with AD. So add in AB+BC+CD+DA=0, |BC|=|AD| and |AB|=|CD|. That will give a solveable equation set of 9 equations with 9 unknowns. And anyone who isn't a mathochist is crying in a corner by now.
Apr
21
answered Non-literal applications of “Shortest Path” algorithm?
Mar
1
answered Why is not the answer to all probability questions 1/2.
Jan
31
comment Coin flipping probability game ; 7 flips vs 8 flips
Note that you have a problem with a trivial finite number of equally distributed outcomes, 32768 of them to be exact. In such a case you can iterate over each one of them and count the outcomes to calculate the exact probabilities. Unlike a Monte Carlo simulation it is a valid proof for the specific conditions. Though there is no formal way to extend it to a class of problems, and it doesn't provide much understanding, so a traditional proof is usually preferable.
Jan
14
comment Why does this not seem to be random?
Yes, and you wrote the opposite in the second to last paragraph.
Jan
14
comment Why does this not seem to be random?
I think you are missing a "not": "then the expected number of distinct numbers [not] chosen is"
Dec
28
comment How is the value of $\pi$ ( Pi ) actually calculated?
The answer you have marked as accepted is not a method for deriving $\pi$ since the formula depend on the value of $\pi$. I urge you to unmark it.
Dec
28
comment How is the value of $\pi$ ( Pi ) actually calculated?
-1 Your method use the degree $\sin$ function, which works by multiplying the input by $\pi/180$ before using a $\sin$ algorithm. Basically it is just a complicated way of extracting the built-in $\pi$ constant in your calculator or programming language. Archimedes did not use trigonometric functions, he used basic geometry. Since you can't delete an accepted answer I'd urge you to blank it until it is un-accepted.
Dec
8
comment How to prove convergence of $\sum_{n=1}^\infty \left(\frac{n}{n+1}\right)^n $?
Here is an answer to a similar problem that gets a bit more to the core of the answer: math.stackexchange.com/a/597781/3584
Dec
8
awarded  Critic
Dec
7
awarded  Commentator
Dec
7
comment How to prove convergence of $\sum_{n=1}^\infty \left(\frac{n}{n+1}\right)^n $?
-1 For calling me a coward. And for worrying more about speed than correctness of answer.
Dec
2
comment proof: center of a tree lies on the longest path
@bece Just imagine your graph is a web of strings of various length, and the nodes are blue beads, then fill every string with as many red beads as there is room for, now it is approximately a non-weighted graph.
Dec
1
answered proof: center of a tree lies on the longest path
Dec
1
comment proof: center of a tree lies on the longest path
Oh, you mean tree as in loop-free graph, and center as in the node(s) that are closest to the furthest away node. Now it makes sense.
Dec
1
comment proof: center of a tree lies on the longest path
I think you mean a weight-balanced tree, otherwise constructing a counterexample is pretty easy. So first things first, make sure that your description of the tree is accurate.
Dec
1
comment Probability of eventual winner
Where do you get the 1/6 from? This is all a big bunch of 1/36 and 35/36 probabilities.
Nov
30
comment Converting to NAND only
@user2437672 First you get rid of all the OR parts of your equation using the first identity I gave you, then you use the second identity to convert NOTs to NANDs, and where your ANDs lack the NOT part you double-negate, and then use the second identity to make one of the NOTs into a NAND.
Nov
30
answered Converting to NAND only