/sci/ - Science & Math

True or false? A subtractable number is less tractable than a tractable number.

>>12804990
I can't build muscle!!!!!!!!!

>>12779035
I am not your "fren" newfag

Is it time to give up newton's mechanics?

Nevermind, I figured it out.

>>12713731
>When was science political?
Aristotle fucked things up for a very long time. His lingering influence fucked the science culture for centuries. It was hard to break the mold, you were ostracized or buried. These days you have an internet, so the politics are weak.

>communities used to have a culture of sorts defined by: Sharing, communism, independent verification and free speech and progress
This is a myth you've bought into. The attitude was more or less invented by Pauli, Einstein, and friends in the early 20th century

>>12710196
>one, randomly generated version of reality is absolutely the same thing as a universe where all possible versions actually exist
I have no fucking clue what you are talking about. Is that you OP? Are you mentally incapacitated? Seems you were either unwilling or unable to read that post you responded to, maybe because you want to be a braindamaged popsci faggot who reads nothing, knows nothing, and gets off on being a drooling brainlet.

>>11444465
>>11444675
>>11445885
>>11446049
>>11446081
>>11447675
>>11447957
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>>11448215
>>11448220
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>>11448320
Consider this problem from a typical IQ test: [eqn] \text{2 5 10 17 26} ? [/eqn] What's the next number you expect in the sequence (this is not hard, you should do it). The n-th term in the sequence is given by: [eqn] n^2 + 1 [/eqn] as you can see by substituting [math] n=1,2,3,4,5, [/math] so the next term is [math] 37 [/math]. But if you did the problem, you probably noticed first that the differences are: [eqn] \text{5-2 = 3 10-5 = 5 17-10 = 7 26-17 = 9} [/eqn] and then filled in [math] 37 [/math] by adding [math] 11 [/math] to [math] 26 [/math]. This thing you did above, of finding the difference between successive terms, is called "taking the first difference", and given any sequence of numbers [math] A_n [/math], the derived sequence [eqn] \Delta A_n = A_{n+1} - A_{n} [/eqn] From the definition, you can check[eqn] \Delta 1 = 0 [/eqn]
[eqn] \Delta n = 1[/eqn]
[eqn] \Delta n^2 = 2n+1[/eqn]
[eqn] \Delta n^3 = 3n^2+3n+1 [/eqn]
[eqn] \Delta n^4 = 4n^3 + 6n^2 + 4n + 1 [/eqn]
[eqn] \Delta 2^n = 2^n [/eqn]
[eqn] \Delta {1\over n} = - {1\over n(n+1)} [/eqn]
and you can prove the general properties[eqn] A_n + \Delta A_n = A_{n+1} [/eqn]
[eqn] \Delta (A + B) = \Delta A + \Delta B [/eqn]
[eqn] \Delta cA = c \Delta A [/eqn]
This says that [math] \Delta [/math] is a linear operator. Further, you have a product rule
[eqn] \Delta (AB) = A \Delta B + B \Delta A + \Delta A\Delta B [/eqn]
[eqn] \Delta (AB)_n = A_{n+1} \Delta B_n + B_n \Delta A_n [/eqn]
So now you can see that
[eqn] \Delta (n^2 - n) = (2n+1 - 1) = 2n [/eqn]
[eqn] \Delta (n^2 2^n) = (2n+1) 2^{n+1} + n^2 2^n = (n^2 + 4n + 2) 2^n [/eqn]
And so on. It is good practice toward calculus to find the derived sequence of all common functions. This was done by early modern mathematicians, and this calculus of finite differences directly inspired calculus.

Consider this problem from a typical IQ test: [eqn] \text{2 5 10 17 26} ? [/eqn] What's the next number you expect in the sequence (this is not hard, you should do it). The n-th term in the sequence is given by: [eqn] n^2 + 1 [/eqn] as you can see by substituting [math] n=1,2,3,4,5, [/math] so the next term is [math] 37 [/math]. But if you did the problem, you probably noticed first that the differences are: [eqn] \text{5-2 = 3 10-5 = 5 17-10 = 7 26-17 = 9} [/eqn] and then filled in [math] 37 [/math] by adding [math] 11 [/math] to [math] 26 [/math]. This thing you did above, of finding the difference between successive terms, is called "taking the first difference", and given any sequence of numbers [math] A_n [/math], the derived sequence [eqn] \Delta A_n = A_{n+1} - A_{n} [/eqn] From the definition, you can check
[eqn] \Delta 1 = 0 [/eqn]
[eqn] \Delta n = 1 [/eqn]
[eqn] \Delta n^2 = 2n+1 [/eqn]
[eqn] \Delta n^3 = 3n^2+3n+1 [/eqn]
[eqn] \Delta n^4 = 4n^3 + 6n^2 + 4n + 1 [/eqn]
[eqn] \Delta 2^n = 2^n [/eqn]
[eqn] \Delta {1\over n} = - {1\over n(n+1)} [/eqn]
and you can prove the general properties
[eqn] A_n + \Delta A_n = A_{n+1} [/eqn]
[eqn] \Delta (A + B) = \Delta A + \Delta B [/eqn]
[eqn] \Delta cA = c \Delta A [/eqn]
This says that [math] \Delta [/math] is a linear operator. Further, you have a product rule
[eqn] \Delta (AB) = A \Delta B + B \Delta A + \Delta A\Delta B [/eqn]
[eqn] \Delta (AB)_n = A_{n+1} \Delta B_n + B_n \Delta A_n [/eqn]
So now you can see that
[eqn] \Delta (n^2 - n) = (2n+1 - 1) = 2n [/eqn]
[eqn] \Delta (n^2 2^n) = (2n+1) 2^{n+1} + n^2 2^n = (n^2 + 4n + 2) 2^n [/eqn]
And so on. It is good practice toward calculus to find the derived sequence of all common functions. This was done by early modern mathematicians, and this calculus of finite differences directly inspired calculus.