Each series is obtained by differentiating repeatedly, evaluating at x=0, and substituting into f(x)=∑n!f(n)(0)xn; part (vi) is easiest via the double-angle identity rather than raw differentiation.
Step 1 (i). f(x)=ex. Every derivative of ex is ex itself, so f(n)(0)=1 for all n.
ex=1+x+2!x2+3!x3+⋯=∑n=0∞n!xn.
Step 2 (ii). f(x)=sinx. f(0)=0,f′(0)=cos0=1,f′′(0)=−sin0=0,f′′′(0)=−cos0=−1,f(4)(0)=0,… (period-4 cycle 0,1,0,−1).
sinx=x−3!x3+5!x5−⋯=∑n=0∞(2n+1)!(−1)nx2n+1.
Step 3 (iii). f(x)=cosx. f(0)=1,f′(0)=−sin0=0,f′′(0)=−cos0=−1,f′′′(0)=0,… (cycle 1,0,−1,0).
cosx=1−2!x2+4!x4−⋯=∑n=0∞(2n)!(−1)nx2n.
Step 4 (iv). f(x)=log(1−x). f′(x)=1−x−1⇒f′(0)=−1; f′′(x)=(1−x)2−1⇒f′′(0)=−1; in general f(n)(0)=−(n−1)! for n≥1, giving n!f(n)(0)=−n1.
log(1−x)=−x−2x2−3x3−⋯,−1≤x<1.
Step 5 (v). f(x)=tan−1x. f′(x)=1+x21; expanding 1+x21=1−x2+x4−⋯ (a geometric series in −x2) and integrating term by term (equivalently, tabulating f(n)(0)) gives
tan−1x=x−3x3+5x5−⋯,−1≤x≤1.
Step 6 (vi). f(x)=cos2x. Use cos2x=21+cos2x and substitute the Step 3 series with x→2x: cos2x=1−2!(2x)2+4!(2x)4−⋯=1−2x2+32x4−⋯.
cos2x=21+(1−2x2+32x4−⋯)=1−x2+3x4−⋯.
✓Final answer
(i) ex=n=0∑∞n!xn. (ii) sinx=x−3!x3+5!x5−⋯. (iii) cosx=1−2!x2+4!x4−⋯. (iv) log(1−x)=−x−2x2−3x3−⋯, −1≤x<1. (v) tan−1x=x−3x3+5x5−⋯, −1≤x≤1. (vi) cos2x=1−x2+3x4−⋯.