#6**+3 **

This link has both the original Latin and English translation for __Part 2__ of *Ars Conjectandi:*** The Doctrine of Permutations and Combinations: Being an Essential and Fundamental Doctrine of Changes. (1795). **(Navigate to page 217 for the English translation of the page Heureka references.) Despite the archaic spelling, syntax, and mathematical notation, this text is very readable. This book includes several related essays and theories from notable mathematicians, contemporary to, and proceeding, Jakob Bernoulli.

Complete English translations of ** Ars Conjectandi **are rare. This is required reading for Lancelot Link’s school of Mathematics, Physics, Chemistry, and Trolling. I had to brush-up on my Latin.

For anyone interested, here’s __Part 4__ of ** Ars Conjectandi **(English) Link

JB, will you sign my printed copies?

GA

GingerAle
Feb 2, 2018

#2**+1 **

The following presentation is adapted from conversational dialogues with Lancelot Link for stochastic measurement errors in the physical sciences.

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To calculate the error range for uncorrelated and random errors, use Gauss’s equations for normal distribution of errors. Using the Gauss equations, gives a 68% confidence interval the measurement error is within this range.

Here are some basic derivatives from the Gauss equations:

\(\small \text{If Q is a combination of sums and/or differences, then the sigma (error) of Q is equal to } \\ \sigma Q = \sqrt{(\sigma a)^2 + (\sigma b)^2 . . . + (\sigma x)^2 . . . } \\ \text{If } Q = b^2 \text{ then } \sigma Q = 2(\sigma (b)) \text{ where } \sigma (b) \text{ is the error of (b)}\\ \text{If } Q = \sqrt{(b)} \text{ then }\sigma Q = \frac{1}{2} (\sigma (b)) \text{ where } \sigma (b) \text{ is the error of (b)}\\ \text{Note how the exponent becomes the multiplier of the error.}\\ \small \text {For this equation, the easiest method is to calculate each error as a }\\ \small \text {decimal and then use the sum of the decimals to calculate the error range. }\\ \text {Uncertainties (In decimal) for a, b, and c. }\\ a_1 = \frac{0.1}{1.3} = 0.0769\\ b_1 = (2)\frac{0.2}{2.8} = 0.1428\\ c_1 = (0.5)\frac{0.1}{0.8} = 0.0625\\ Q= \frac{1.3 \times (2.8)^2}{\sqrt {0.8}} = 11.4 \pm ((0.0769 + 0.1428+ 0.0625) * (11.4))\\ Q= 11.4 \pm 3.2 \\ \small \text { note: pay attention to significant figures when recording the final error range. }\\ \\\)

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Relating to this question and others you’ve posted on here , another important thing you need to do is **Stop being a lazy dolt!**

I would have answered your chemistry questions, except I was too lazy. I caught this laziness from you—it’s more contagious than dumbness.

GA

GingerAle
Jan 6, 2018