As a part of an excercise I am supposed to find the Taylor series expansion for $(1-t)^{\frac{1}{2}}$ on $[0,1]$.
According to the remainder theorem:
http://en.wikipedia.org/wiki/Taylor%27s_theorem#Explicit_formulae_for_the_remainder
The derivative has to be continuous on $[0,1]$ if we are to use that theorem to show uniform convergence.
But using induction I found that the derivative is:
$f^k(t)=\dfrac{(-1)^k*1*3*5*...*(2k-1)*(1-t)^{\frac{-(2k+1)}{2}}}{2^k}$.
But if we put t=1 we divide on 0, so the derivatives are not continuous in the interval? How do I solve this?
The issue is only at the endpoint $t=1$. For every $b<1$, Taylor's estimate shows convergence on $[0,b]$.
To deal with the endpoint, write down the binomial series explicitly: $$(1-t)^{1/2} = \sum_{k=0}^\infty \binom{1/2}{k} (-t)^k \tag1$$ The estimate $$\left|\binom{1/2}{k}\right|\le \frac{M}{k^{3/2}}$$ proved in the same wiki, implies that (1) converges uniformly on $[-1,1]$ by the Weierstrass test. As a consequence, the sum is continuous on this interval. By the first paragraph, the sum agrees with $(1-t)^{1/2}$ on $[0,1)$. By continuity, it agrees also at $1$.