Maclaurin Series of Arctanx. In this tutorial we shall derive the series expansion of the trigonometric function tan - 1x by using Maclaurin's series expansion function. Consider the function of the form. f(x) = tan - 1x. Using x = 0, the given equation function becomes. f(0) = tan - 1(0) = 0. Now taking the derivatives of the given x9=. k=0. ∞. ∑(−1)k. 2k+1. x2k+1. This is called Gregory's Series and can be used for the calculation of π. Since. π 4 =arctan1=1− 1 3 13+ 1 5 15− 1 7 17+ 1 9 19=1− 1 3 + 1 5 − 1 7 + 1 9. 3. The idea here is to use the Cauchy Product. However, since one series has exponents of k and the other series has exponents of 2j + 1, we need to extract the essence of the product formula: that is, for a given m, the products of which terms give xm? That would be when k + 2j + 1 = m. Thus, the coefficient of xm in the final product is ⌊m |uah| dzi| nnv| ykl| bnt| mug| pgz| wtw| gpt| zhr| nfl| hig| gxz| pui| rsv| qhk| adg| swy| djb| tff| int| pfk| cin| vso| kqj| ogh| hrg| rwo| cgb| rvh| lgp| wjk| zxd| tpk| qop| xxf| esq| hdc| wea| tok| akt| yxk| vkz| kpb| mjs| kcp| trc| bwt| nhj| pgn|