This root-finding algorithm is the Newton-Raphson method. It approximates the solution of f(x)=0 starting from an initial value x₁ and calculating xₙ₊₁=xₙ-f(xₙ)/f'(xₙ). In the example above, we have f(x)=17-x², f'(x)=-2x, and therefore xₙ₊₁=xₙ+(17-xₙ²)/2 as described.pic.twitter.com/sMvEp929jc
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By the way, I was eating Leibniz while tweeting this.
pic.twitter.com/DPtqg6kqDm
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This is just a consequence of Taylor’s theorem.pic.twitter.com/v6tjsMm0r2
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Is this an application of Newton-Raphson?
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Yes! f(x)=x²-y (f(√y)=0) f'(x)=2x, x_(n+1)= xn-f(xn)/f'(xn)=xn²+(y-xn²)/2xn
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√y~(y+x)/(2√x) X is the nearest perfect square √17~(17+16)/2.4=4.125 Just a simplified way to enunciate the same
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Simplified to the point where "." is used as both multiplication and a decimal point in the same line ;)
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Cont.) Square roots via Continued Fractions:* http://www.mathpath.org/Algor/squareroot/algor.square.root.contfrac.htm … ____ *https://en.wikipedia.org/wiki/Continued_fraction … https://archive.org/details/continuedfractio0000olds … (See p. 29)
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Ancient methods, my favourite :) People didn't wait for Newton or Taylor to calculate sqrt(2), and people have always been smart. I believe that sometimes, too much knowledge makes us blind.
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این یه جورایی همون تقریب مرتبه اول نیستش؟
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چرا دیگه، تقریب خطیه دقیقا به نظرم
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