f(x)=sin(x),f(0)=0.
f′(x)=cos(x),f′(0)=1.
f′′(x)=−sin(x),f′′(0)=0.
f′′′(x)=−cos(x),f′′′(0)=−1.
f(4)(x)=−sin(x),f(4)(0)=0.
Since f(4)=f(x) , the cycle of {0, 1, 0, -1} repeats itself, which means that every derivative of even degree gives 0 and that every derivative of odd degree alternates between 1 and -1. So, we have
f(x)=Σn=0∞n!f(n)(0)xn=Σn=0∞(−1)n(2n+1)!x2n+1
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