No, it isn't. For one thing, sin(x) is an odd function while your series includes only even power of x. The Taylor's series for sin(x) about x= 0 is [tex]\sum_{n=o}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}[/tex]. What you have appears to be the Taylor's series, about x= 0, for [itex]cos(x^2)[/itex], except that the denominator should be (2n)! rather than 2n!.Expanding the series to the [tex]n^{th}[/tex] derivative isn't so hard, however I'm having trouble with the summation. Any tips for the summation?
e.g. taylor series for [tex]sinx[/tex] around x=0 in summation notation is [tex]\sum^\infty_{n=0} \frac{x^{4n}}{2n!}[/tex]
Thanks.
No, it isn't. For one thing, sin(x) is an odd function while your series includes only even power of x. The Taylor's series for sin(x) about x= 0 is [tex]\sum_{n=o}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}[/tex]. What you have appears to be the Taylor's series, about x= 0, for [itex]cos(x^2)[/itex], except that the denominator should be (2n)! rather than 2n!.
In any case, what do you mean "having trouble with the summation". What are you trying to do?
Yeah sorry turns out it was mistook for another expression.(edit: didn't notice HallsOfIvy had already answered)
No, the Taylor series sum around x=0 (i.e. the Maclaurin series sum) for ## \sin x ## is $$ \sum_{k=0}^{\infty} \frac{(-1)^k x^{(1+2 k)}}{(1+2 k)!} $$. How did you get to the expression you wrote?