The discussion revolves around finding the Taylor series representation for the function f(x) = cos(x) in powers of x - pi. Participants are exploring the implications of expanding the series around the point x = pi instead of the more common x = 0.
Some participants have provided clarifications regarding the expansion point and confirmed that the "a" value is indeed pi. Others have reiterated the formula for the Taylor series, indicating a productive exchange of ideas without reaching a final consensus.
There is an emphasis on understanding the specific point of expansion for the Taylor series, which is a shift from the typical expansion at x = 0. Participants are navigating the implications of this change in context.
Shyan said:That just means you should expand around [itex]x=\pi[/itex] rather than the usual [itex]x=0[/itex].
If you write [itex]f(a+\delta)=\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} \delta^n[/itex], then yes!TheRedDevil18 said:Does that mean the "a" value is pi ?
Yes, this is what the general term in your Taylor series will look like. Note that a Maclaurin series is a special case of a Taylor series, where a = 0.TheRedDevil18 said:I use this formula
f^(n)(a)*((x-a)^n)/n!
And sub in pi for a, thanks