Question

Find the Taylor series with a = 0 of sin(x) Find the Taylor series with a = 0 of sin(x) @Mathematics

Answer 1

x-(x^3/3!)+((x^5/5!)-(x^7)/7!+.......

Answer 2

express it as an infinite sum :-D

Answer 3

or an infinite series*

Answer 4

ok i got ur point

Answer 5

u mean in terms of summation

Answer 6

yes something like \[\sum_{n=0}^{\infty}\]

Answer 7

yes i am writing

Answer 8

Wow this crap is way over my head...

Answer 9

\[\sum_{0}^{\infty\[\frac{x ^{2n+1}}{(2n+1)!}\]}\]

Answer 10

\[\sum_{0}^{\infty}\]\[\frac{x^2n+1}{(2n+1)!}\]

Answer 11

this is right and as required

Answer 12

\[\sum_{n=0}^{\infty}{(-1)^n\frac{x^{2n+1}}{(2n+1)!}}\]

Answer 13

yes i forgot to put -1 for alternating sign