that's all. While there, Wikipedia: Sverhdorogoy computing, Moore's Law

asked October 8th 19 at 01:24

3 answers

answered on

Solution

No. If Moore's law continues to run, the number of operations performed by the machine per unit time will increase. But it remains finite. More precisely, the infinite it will be infinite time, but it is not for us.

ie little double productivity, you need more frequency doubling doubling - Birdie_Hahn commented on October 8th 19 at 01:29

answered on October 8th 19 at 01:28

I understand you came to this conclusion from the fact about convergent series on Wikipedia? Or rather, that capacity is doubled, and power * time = number of calculations => double capacity same number of calculations (unsuccessfully, but Oh well) is possible in half the time to do?

There are several problems. Well, firstly, computation is not infinitely divisible, and half of the "compute" is the same as zero "computing". And secondly, the logic is incorrect itself. After all, look today X, tomorrow capacity increased 2 times — 2X, then 4X. Yes, sum, get the infinity. But only if sum to infinity — this endless time. It is clear that if we take the continuous analog personal quality will not change. And your mistake is that you kakby trying to the underside of the to summarize a number of complications, but doing it wrong, because back to the series \sum_{n=0}^{\inf} \frac{2^n}, which is necessary to calculate not a number \sum_{n=0}^{\inf} \frac{1}{2^n}, they calculate you.

This is all if I correctly understood the gist of the error :)

There are several problems. Well, firstly, computation is not infinitely divisible, and half of the "compute" is the same as zero "computing". And secondly, the logic is incorrect itself. After all, look today X, tomorrow capacity increased 2 times — 2X, then 4X. Yes, sum, get the infinity. But only if sum to infinity — this endless time. It is clear that if we take the continuous analog personal quality will not change. And your mistake is that you kakby trying to the underside of the to summarize a number of complications, but doing it wrong, because back to the series \sum_{n=0}^{\inf} \frac{2^n}, which is necessary to calculate not a number \sum_{n=0}^{\inf} \frac{1}{2^n}, they calculate you.

This is all if I correctly understood the gist of the error :)

Well, firstly, computation is not infinitely divisible, and half of the "compute" is the same as zero "computing". In my opinion this is nothing, everywhere there is a view of computation speed, not an abstract number. So share you need not the actual calculation and the calculation time.

This is all if I correctly understood the nature of the error — Everything is easier, I didn't count the amount of rows, and at first glance, there were doubts. commented on October 8th 19 at 01:31

This is all if I correctly understood the nature of the error — Everything is easier, I didn't count the amount of rows, and at first glance, there were doubts. commented on October 8th 19 at 01:31

Yes, but I understand that you are intuitively considered instead of the sum of the series 1/(sum of inverse range), but the reverse was not true =) commented on October 8th 19 at 01:34

answered on October 8th 19 at 01:30

can perform an infinite number of steps in finite time

I don't think that the number of transistors will affect the fundamental foundations.

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