MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 6 September 2014

Lecture 38

Formula 1

f(x)=f(a)+ (dfdx|x=a) (x-a)+12! (d2fdx2|x=a) (x-a)2+ (d3fdx3|x=a3!) (x-a)3+ If a=0 then f(x)=f(0)+ (dfdx|x=0)+ (d2fdx2|x=02!) x2+ (d3fdx3|x=03!) x3+ (d4fdx4|x=04!) x4+ So if we want to find the series expansion for ex: ex = e0+ (dexdx|x=0)x+ (d2exdx2|x=02!)x2+ (d3exdx3|x=03!)x3+ = e0+ e0x+ e02!x2+ e03!x3+ = 1+x+x22!+ x33!+ x44!+ If we want to find limx0 ex-1x = limx0 (1+x+x22!+x33!+x44!+)-1x = limx0 x+ x22!+ x33!+ x44!+ x55!+ x = limx0 1+x2!+ x23!+ x34!+ = 1+0+0+ = 1. If we want to find the series expansion for sinx, sinx = sin0+ (dsinxdx|x=0)x+ (d2sinxdx2|x=02!)x2+ (d3sinxdx3|x=03!)x3+ (d4sinxdx4|x=04!)x4+ = sin0+ cos0x- sin02!x2- cos03!x3+ sin04!x4+ = 0+x-0-x33!+0 +x55!-0- x77!+0+ = x-x33!+ x55!- x77!+ If we want to find limx0 sinxx = limx0 x-x33!+ x55!- x77!+ x = limx0 1-x23!+ x45!- x67!+ = 1-0+0-0+0- = 1. If we want to find the series expansion for 11-x: 11-x = 11-x|x=0+ (d11-xdx|x=0)+ (d211-xdx2|x=02!)x2+ (d311-xdx3|x=03!)x3+ (d411-xdx4|x=04!)x4+ = 1+ (1(1-x)2|x=0)x+ (2(1-x)3|x=02!)x2+ (3·2(1-x)4|x=03!)x3+ = 1+x+x2+x3+x4+x5+. If we want to find the series expansion for 11+x: 11+x = 11-(-x)=1+ (-x)(-x)2+ (-x)3+(-x)4+ = 1-x+x2-x3+x4-x5+ If we want to find the series expansion for ln(1+x): 11+xdx = ln(1+x) = (1-x+x2-x3+x4-x5+)dx = x-x22+ x33-x44 +x55-x66+ So, in particular, 1-12+13-14 +15-16+17- 18+=ln(1+1) =ln2. Also limx0 ln(1+x)x = limx0 x-x22+x33-x44+x55-x66+ x = limx01-x2+ x23-x34+ = 1-0+0-0+ = 1. If we want to find the series expansion for tan-1x: tan-1x = 11+x2dx = (1-x2+x4-x6+x8-)dx = x-x33+x55- x77+x99- x1111+ If we want to find π: π = 4·(π4)=4tan-1(1) = 4(1-13+15-17+19-111+) = 4-43+45-47 +49-411+413-

Notes and References

These are a typed copy of Lecture 38 from a series of handwritten lecture notes for the class MATH 221 given on December 11, 2000.

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