what is series of cos in maths
Answers
Answered by
1
Applying Maclaurin's theorem to the cosine and sine functions, we get
{\displaystyle \displaystyle \cos(x)=1-{x^{2} \over 2!}+{x^{4} \over 4!}-\cdots =\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n}}{(2n)!}}}{\displaystyle \sin(x)=x-{x^{3} \over 3!}+{x^{5} \over 5!}-\cdots =\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+1}}{(2n+1)!}}}
For both series, the ratio of the nth to the (n-1)th term tends to zero for all x. Thus both series are absolutely convergent for all x.
Many properties of the cosine and sine functions can easily be derived from these expansions, such as
{\displaystyle \displaystyle \sin(-x)=-\sin(x)}{\displaystyle \displaystyle \cos(-x)=\cos(x)}{\displaystyle \displaystyle {\frac {d}{dx}}\sin(x)=\cos(x)}
{\displaystyle \displaystyle \cos(x)=1-{x^{2} \over 2!}+{x^{4} \over 4!}-\cdots =\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n}}{(2n)!}}}{\displaystyle \sin(x)=x-{x^{3} \over 3!}+{x^{5} \over 5!}-\cdots =\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+1}}{(2n+1)!}}}
For both series, the ratio of the nth to the (n-1)th term tends to zero for all x. Thus both series are absolutely convergent for all x.
Many properties of the cosine and sine functions can easily be derived from these expansions, such as
{\displaystyle \displaystyle \sin(-x)=-\sin(x)}{\displaystyle \displaystyle \cos(-x)=\cos(x)}{\displaystyle \displaystyle {\frac {d}{dx}}\sin(x)=\cos(x)}
Similar questions