{"version":9,"randomSeed":"edd27ce0d0cdc22767f89977fd8a3989","graph":{"viewport":{"xmin":-2.5246846760500903,"ymin":-39.88301127201878,"xmax":68.91470059474499,"ymax":10.54734214230091}},"expressions":{"list":[{"type":"expression","id":"13","color":"#2d70b3","latex":"\\ln\\left(\\left|\\ln2-\\sum_{n=1}^{x}\\left(-1\\right)^{\\left(n-1\\right)}\\cdot\\frac{1^{n}}{n}\\right|\\right)","labelSize":"medium"},{"type":"text","id":"29","text":"ln(2) = ln(3/2) + ln(4/3)"},{"type":"expression","id":"17","color":"#c74440","latex":"\\ln\\left(\\left|\\ln2-\\sum_{n=1}^{x}\\left(-1\\right)^{\\left(n-1\\right)}\\cdot\\frac{\\left(\\frac{1}{2}\\right)^{n}+\\left(\\frac{1}{3}\\right)^{n}}{n}\\right|\\right)"},{"type":"text","id":"31","text":"ln(2) = ln(4/3) + ln(5/4) + ln(6/5)"},{"type":"expression","id":"24","color":"#388c46","latex":"\\ln\\left(\\left|\\ln2-\\sum_{n=1}^{x}\\left(-1\\right)^{\\left(n-1\\right)}\\cdot\\frac{\\left(\\frac{1}{3}\\right)^{n}+\\left(\\frac{1}{4}\\right)^{n}+\\left(\\frac{1}{5}\\right)^{n}}{n}\\right|\\right)"},{"type":"text","id":"33","text":"ln(2) = ln(5/4) + ln(6/5) + ln(7/6) + ln(8/7)"},{"type":"expression","id":"27","color":"#c74440","latex":"\\ln\\left(\\left|\\ln2-\\sum_{n=1}^{x}\\left(-1\\right)^{\\left(n-1\\right)}\\cdot\\frac{\\left(\\frac{1}{4}\\right)^{n}+\\left(\\frac{1}{5}\\right)^{n}+\\left(\\frac{1}{6}\\right)^{n}+\\left(\\frac{1}{7}\\right)^{n}}{n}\\right|\\right)"},{"type":"text","id":"37","text":"and so on"},{"type":"expression","id":"38","color":"#388c46","latex":"\\ln\\left(\\left|\\ln2-\\sum_{n=1}^{x}\\left(-1\\right)^{\\left(n-1\\right)}\\cdot\\frac{\\left(\\frac{1}{5}\\right)^{n}+\\left(\\frac{1}{6}\\right)^{n}+\\left(\\frac{1}{7}\\right)^{n}+\\left(\\frac{1}{8}\\right)^{n}+\\left(\\frac{1}{9}\\right)^{n}}{n}\\right|\\right)"},{"type":"text","id":"47","text":"compare the alternative series i developed in a different way: converges at the same rate as the two-value split (one bit per iteration)"},{"type":"expression","id":"48","color":"#2d70b3","latex":"\\ln\\left(\\left|\\ln2-\\sum_{n=1}^{x}\\frac{1}{n\\cdot2^{n}}\\right|\\right)"},{"type":"text","id":"50","text":"an analogous acceleration on the alternative series"},{"type":"expression","id":"51","color":"#6042a6","latex":"\\ln\\left(\\left|\\ln2-\\sum_{n=1}^{x}\\frac{1}{n\\cdot3^{n}}-\\sum_{n=1}^{x}\\frac{1}{n\\cdot4^{n}}\\right|\\right)"},{"type":"text","id":"40","text":"we can repeat the acceleration so many times that it forms an approximating meta-sequence of its own"},{"type":"expression","id":"41","color":"#000000","latex":"\\ln\\left(\\left|\\ln2-\\sum_{n=\\operatorname{round}\\left(x\\right)}^{2\\operatorname{round}\\left(x\\right)-1}\\frac{1}{n}\\right|\\right)"},{"type":"expression","id":"42","color":"#c74440","latex":"\\ln\\left(\\left|\\ln2-\\sum_{n=\\operatorname{round}\\left(x\\right)}^{2\\operatorname{round}\\left(x\\right)-1}\\frac{1}{n}+\\frac{\\sum_{n=\\operatorname{round}\\left(x\\right)}^{2\\operatorname{round}\\left(x\\right)-1}\\left(\\frac{1}{n}\\right)^{2}}{2}\\right|\\right)"},{"type":"expression","id":"43","color":"#2d70b3","latex":"\\ln\\left(\\left|\\ln2-\\sum_{n=\\operatorname{round}\\left(x\\right)}^{2\\operatorname{round}\\left(x\\right)-1}\\frac{1}{n}+\\frac{\\sum_{n=\\operatorname{round}\\left(x\\right)}^{2\\operatorname{round}\\left(x\\right)-1}\\left(\\frac{1}{n}\\right)^{2}}{2}-\\frac{\\sum_{n=\\operatorname{round}\\left(x\\right)}^{2\\operatorname{round}\\left(x\\right)-1}\\left(\\frac{1}{n}\\right)^{3}}{3}\\right|\\right)"}]}}