Let $r>0$, $\varepsilon>0$ and $\alpha>0$. Assume that $0<\varepsilon
Convergence in binomial series
2
$\begingroup$
real-analysis
sequences-and-series
-
0Why don't you just use Taylor formula? – 2012-07-04
-
0How, precisely, is $\binom\alpha k$ defined for non-integer $\alpha$? The only reasonable way I can think of is with the Gamma function. – 2012-07-04
-
0@CameronBuie: $\binom\alpha k=\alpha(\alpha-1)\cdots(\alpha-k+1)/k!$. – 2012-07-04
-
0Interesting, Harald. Does that work out for a general binomial expansion? – 2012-07-04
-
0@Norbert the last power series is about $0$. Calculating the derivative of $x^{\alpha}$ at $0$ you would obtain zero in the denominator. – 2012-07-04
1 Answers
4
Such a series would converge for $|x|
-
0right, the power series is convergent in a symmetric interval about $0$, so it will convergent for $|x|
$\alpha$ ? The binomial series absolutely and uniformly converges for $|x-1|<1-\delta$ ($\delta>0$ is "small", this is the reason of the condition $0<\varepsilon$\varepsilon$ condition is important. – 2012-07-04 -
0@vesszabo the answer is about the series $\sum_{n=0}^{\infty} c_n x^n$. – 2012-07-04