SPAM, SPAMMERS and their ASSOCIATES... - page 25
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https://forum.mql4.com/ru/29090 - inciting ethnic hatred
This thread isn't even spam, it's a VIRUS! And it needs to be deleted.
Удалено
That's good...
Эта тема даже не спам, а ВИРУС! И его надо удалить.
I didn't have time to finish reading it:(((( I'll have to google it
Ну вот, дочитать не успел:(((( придется гуглом
I haven't read it at all.
Удалено
Ukrainians and Tatars - Glory, Glory, Glory!!! Many national heroes committed terrible crimes during the struggle for independence, but were vindicated by the citizens of the new states and the generations that followed.
Ну вот, дочитать не успел:(((( придется гуглом
нЕчего здесь делать ни психиаторам ни социологам. ресурс - для программистов. лично мне не нравятся обсуждения вопросов межличностных и межнациональных отношений на ТЕМАТИЧЕСКОМ форуме. здесь это - ФЛУД и засорение ресурса. выяснять межличностные и межнациональные противоречия - в другом месте.
Believe it or not, Uncle Fyodor, for me the forum is not only a place of learning and technical communication, but also a kind of social environment. Programmers are human beings too, they need a smoking room, which not only brings down steam, but also naturally forms informal associations of people of the same cultural type, albeit with different political views. This will come in handy more than once in work and in life.
You, for example, also belong to one of these groups. And I know it mainly from your posts in the flood threads, which have no independent technical or cultural meaning. So there is some indirect usefulness of those topics.
Cube derivation, according to Mathemat
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I had to calculate the cube root and the calculator was an accounting calculator,
well, there's a lot of digits. I had to "break down the series". But they say there is a "secret" formula:)))
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Mathemat: Yes, there is one. I understand that it's off-topic, but I'm curious about it myself. I give it to you for nothing.
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Suppose we have a number 41. We need to extract a cube root from it with such precision that its cube differs from 41 by less than one millionth. How many terms of the binomial would we need? A lot, probably.
The algorithm is as follows. First we find a rough approximation, accurate to 1. It's either 3 or 4. Probably 3 is closer.
And then we start to refine this approximation using the formula:
x(n+1) = x(n) + (41 - x(n)^3) / (3 * x(n)^2)
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OK, we have x(0) = 3. Let's substitute 3 into the recurrence formula:
Then x(1) = 3 + (41-3^3)/(3*3^2) = 3 + 14/27 ~ 3.5185
(3.5185^3 ~ 43.56 - accuracy is still low. Moving on.)
3.5185 + (41-3.5185^3)/(3*3.5185^2) ~ 3.5185 - 2.56/37.1395 ~ 3.44957
(3.44957^3 ~ 41.048273 - that's not bad, let's iterate again).
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3.44957 + (41-3.44957^3)/(3*3.44957^2) ~ 3.44957 - 0.048273/35.696600 ~ 3.4482177
(3.4482177^3 ~ 41.000016395 - oh so great, let's do another iteration).
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3.4482177 + (41-3.4482177^3)/(3*3.4482177^2) ~ 3.4482177 - 0.000016395/35.67061592 ~ 3.44821724
(3.44821724^3 ~ 40.999999986 - fine, the cube error is just over one hundred millionth).
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It took us only 4 iterations after our first rough guess to get the accuracy to about 10^(-8). The main feature of this algorithm is that calculation error does not accumulate in it - it is automatically corrected in the right direction with each new iteration.
Danke sehr!
Ran to get a beer