Can someone find the function that is equal to this Taylor series?
C Cadwalla New member Joined Jan 22, 2021 Messages 10 Aug 22, 2021 #1 Can someone find the function that is equal to this Taylor series?
D Deleted member 4993 Guest Aug 22, 2021 #2 Cadwalla said: Can someone find the function that is equal to this Taylor series? View attachment 28627 Click to expand... Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem.
Cadwalla said: Can someone find the function that is equal to this Taylor series? View attachment 28627 Click to expand... Please show us what you have tried and exactly where you are stuck. Please follow the rules of posting in this forum, as enunciated at: READ BEFORE POSTING Please share your work/thoughts about this problem.
H HallsofIvy Elite Member Joined Jan 27, 2012 Messages 7,763 Aug 23, 2021 #3 [imath]\sum_{n=0}^\infty \frac{(-x)^{6n+1}}{(2n)!}= -x\sum_{n=0}^\infty\frac{(x)^{6n}}{(2n)!}= -x\sum_{n=0}^\infty\frac{((x)^3)^{2n}}{(2n)!}[/imath]. Now, what is the Taylor's series for cos(x)?
[imath]\sum_{n=0}^\infty \frac{(-x)^{6n+1}}{(2n)!}= -x\sum_{n=0}^\infty\frac{(x)^{6n}}{(2n)!}= -x\sum_{n=0}^\infty\frac{((x)^3)^{2n}}{(2n)!}[/imath]. Now, what is the Taylor's series for cos(x)?
