# Does Taylor series around point zero (maclaurin series) always exist for differentiable function?

For every differentiable function $f(x)$,

is $f(x) = \sum_{n=0} ^ {\infty} \frac {f^{(n)}(0)}{n!} \, (x)^{n}$ always true and can be written?

The only thing we have to be concerned is just whether this taylor series (maclaurin series) converge or not? Therefore, we have to examinine some properties of the derivatives. But all in all $f(x) = \sum_{n=0} ^ {\infty} \frac {f^{(n)}(0)}{n!} \, (x)^{n}$ can always be written?

For example: Does taylor series of $y(t) = t^{3/2}$ exist at $t = 0$?

• If the function is infinitely continuously differentiable at $0$, then the series exists, but you are not guaranteed that it approximates the function very well. For instance, $f(x)=e^{-1/x^2}$ has Taylor series $0$. – Arthur Jul 26 '16 at 9:43
• The taylor series converges to the function within an interval, if all derivates are bounded on that interval (This condition is only sufficient). The second derivate of $t^{3/2}$ does not exist at $t_0=0$. – Peter Jul 26 '16 at 9:58
• @Peter does taylor series exist only for $C^{\infty}$ function? – Ka-Wa Yip Jul 26 '16 at 10:00
• Yes, the function must have arbitary high derivates. Otherwise, we must truncate the series at some point which can however give a reasonable approximation. – Peter Jul 26 '16 at 10:01
• – paf Jul 26 '16 at 10:04