Do you know how to calculate square roots by hand?
>>16931343I used to know how to do cube roots by hand, but it was retardedly convoluted.Square roots are easy.
>>16931343no, will you teach me?
>>16931343I've just sort of memorized the most common ones at this point. I still have a print out guide in case my ass forgets how to do it by hand though.
>>16931355NTA, just look up something along the lines of "calculating square roots by hand pdf" and there should be some decent results.
>>16931343What exactly am I looking at here? I don't recognize all the commas next to some of the numbers.
They say they have to keep teaching children long division even though it makes them cry because of polynomial long division, but completing the square is just finding the square root of a quadratic with remainder.
I use the NR method. It's more reliable than doing that>le magical convoluted long division that totally doesn't have completely arbitrary steps for any given calculation trust me bro, and it totally works brooo *takes a rip of a bong*horseshit.For reference I'm a group theorist.
>>16931355To find [math]\sqrt{x}[/math] where [math]10^{2n} \leq x < 10^{2n+2}[/math], you first find the largest [math]y \in \{1,2,3,4,5,6,7,8,9\} \cdot 10^n[/math] such that [math]y^2 \leq x[/math]. Next you calculate [math]x - y^2[/math] and try to find a new estimate [math]y + z[/math] such that [math]x - (y+z)^2 \geq 0[/math]. The key is that [math]x - (y+z)^2 = (x - y^2) - (2y + z)z[/math], so you can pick the largest [math]z \in \{0,1,2,3,4,5,6,7,8,9\} \cdot 10^{n-1}[/math] such that [math](2y + z)z \leq x - y^2[/math]. Then you subtract [math](2y + z)z[/math] from [math]x - y^2[/math] to get [math]x - (y+z)^2[/math], and so on.
>>16931343https://youtu.be/-J_xL4IGhJA?t=57m10shttps://youtu.be/csInNn6pfT4?t=6m30s
[math]\begin{align}\sqrt{8720000} &= \sqrt{9000000 - 280000} \\&= 3000\sqrt{1 - \frac{28}{900}} \\ &\approx 3000\times\left(1 - \frac{1}{2}\cdot\frac{28}{900}\right) \\&\approx 3000\times\left(1 - \frac{1}{2}\cdot\frac{30}{900}\right) \\&= 3000\times\left(1 - \frac{1}{60}\right) = 2950 + \mathcal{O}\left(\frac{7}{225}\right)^2\end{align}[/math]
>>16932635One may refine this approach by recognizing this approach gives a lower bound to within about 3% error. If we take an upper bound to similar error, and average the results, we will hit the true value.[math]\begin{align} \sqrt{8720000} &= \sqrt{9000000 - 280000} \\ &= 3000\sqrt{1 - \frac{28}{900}}\\&= 3000\sqrt{1 - \frac{7}{225}}\\&\approx 3000\sqrt{1 - \frac{7}{231}}\\&= 3000\sqrt{1 - \frac{1}{33}}&\approx 3000\times\left(1 - \frac{1}{66}\right) = 2953\end{align}[/math]average the two for 2952.5
>>16931492erm, hypothesis:maybe it's Indian formatting, despite there being no 3 digit groupshttps://en.wikipedia.org/wiki/Indian_numbering_system#Decimal_formatting
yes they taught me this in elementary school
>>16931343Yes...with a table
>>16931343ok. next how do I calculate sines and cosines by hand? If society collapses and calculators disappear, I won't be able to rebuild society because I can't calculate a cosine without a calculator
>>16936394I use the Taylor series for [math]e^{i\theta}[/math], where [math]\theta[/math] is the angle you're calculating the cosine of.
>>16936387sqrt(3) is the year george washington was born
For calculating pi by hand, Machin's formula is convenient:[eqn]\pi = 16 \tan^{-1} \left( \frac{1}{5} \right) - 4 \tan^{-1} \left( \frac{1}{239} \right)[/eqn]This follows readily from[eqn]\frac{(5 + i)^4}{1 + i} = 478 + 2i.[/eqn][math]\tan^{-1}(x)[/math] can be approximated with the usual power series.
>>16932618It's also worth mentioning that this can be easily extended to cube and higher roots.To find [math]\sqrt[3]{N}[/math]:Start with a guess G.Solve [math]G \cdot G \cdot X = N[/math].Your new better guess is [math]\frac{G+G+X}{3}[/math].Repeat.This is equivalent to using Newton's method but doesn't require calculus to explain.
>>16931343My father taught me fifty years ago. I forgot, but then I worked it out again.
>>16931343do you know how to make a good thread?
