{"version":9,"randomSeed":"813dc2921985707565372a31e111b9a6","graph":{"viewport":{"xmin":-0.9995855972004141,"ymin":-5.068524554101321,"xmax":9.266660475969408,"ymax":7.371308784305999}},"expressions":{"list":[{"type":"text","id":"11","text":"https://en.wikipedia.org/wiki/Series_acceleration#Euler%27s_transform"},{"type":"expression","id":"12","color":"#000000","latex":"a\\left(n\\right)=\\frac{1}{\\operatorname{round}\\left(n\\right)+1}"},{"type":"expression","id":"13","color":"#c74440","latex":"F\\left(n\\right)=\\sum_{k=0}^{n}\\left(-1\\right)^{k}\\operatorname{nCr}\\left(n,k\\right)a\\left(n-k\\right)"},{"type":"text","id":"20","text":"thorough approximation by the raw series"},{"type":"expression","id":"15","color":"#388c46","latex":"\\sum_{n=0}^{100}\\left(-1\\right)^{n}a\\left(n\\right)"},{"type":"text","id":"22","text":"accelerated approximation (similar cost, since forward-difference implementation here is quadratic)"},{"type":"expression","id":"18","color":"#c74440","latex":"\\sum_{n=0}^{10}\\frac{\\left(-1\\right)^{n}F\\left(n\\right)}{2^{n+1}}"},{"type":"text","id":"24","text":"perfect to within numerical precision after 40 terms"},{"type":"expression","id":"16","color":"#6042a6","latex":"\\sum_{n=0}^{40}\\frac{\\left(-1\\right)^{n}F\\left(n\\right)}{2^{n+1}}"},{"type":"text","id":"27","text":"can't get that with even a million terms of the original!"},{"type":"expression","id":"25","color":"#000000","latex":"\\sum_{n=0}^{1000000}\\left(-1\\right)^{n}a\\left(n\\right)"},{"type":"text","id":"29","text":"target"},{"type":"expression","id":"17","color":"#000000","latex":"\\ln2"}]}}