Question #226019

Find McLaurin Series for Sin(x)


1
Expert's answer
2021-09-01T11:27:03-0400

f(x)=sin(x),  f(0)=0.f(x)=sin(x),\;f(0)=0.

f(x)=cos(x),  f(0)=1.f'(x)=cos(x),\;f'(0)=1.

f(x)=sin(x),  f(0)=0.f''(x)=-sin(x),\;f''(0)=0.

f(x)=cos(x),  f(0)=1.f'''(x)=-cos(x),\;f'''(0)=-1.

f(4)(x)=sin(x),  f(4)(0)=0.f^{(4)}(x)=-sin(x),\;f^{(4)}(0)=0.

Since f(4)=f(x)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=0f(n)(0)n!xn=Σn=0(1)nx2n+1(2n+1)!f(x)=\Sigma^\infin_{n=0}\frac{f^{(n)}(0)}{n!}x^n=\Sigma^\infin_{n=0}(-1)^n\frac{x^{2n+1}}{(2n+1)!}


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