E (0,1)(z)=1/(1-z) for abs(z) less than 1
From specialfunctionswiki
Theorem
The following formula holds for $|z|<1$: $$E_{0,1}(z)=\dfrac{1}{1-z},$$ where $E_{0,1}$ denotes the Mittag-Leffler function.
Proof
References
- {{ #if: |{{{2}}}|H.J. Haubold}}{{#if: A.M. Mathai|{{#if: R.K. Saxena|, {{ #if: |{{{2}}}|A.M. Mathai}}{{#if: |, {{ #if: |{{{2}}}|R.K. Saxena}}{{#if: |, [[Mathematician:{{{author4}}}|{{ #if: |{{{2}}}|{{{author4}}}}}]]{{#if: |, [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]] and [[Mathematician:{{{author6}}}|{{ #if: |{{{2}}}|{{{author6}}}}}]]| and [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]]}}| and [[Mathematician:{{{author4}}}|{{ #if: |{{{2}}}|{{{author4}}}}}]]}}| and {{ #if: |{{{2}}}|R.K. Saxena}}}}| and {{ #if: |{{{2}}}|A.M. Mathai}}}}|}}: [[Paper:H.J. Haubold/Mittag-Leffler Functions and Their Applications{{#if: |/Volume {{{volume}}}|}}{{#if: |/{{{edpage}}}}}|Mittag-Leffler Functions and Their Applications{{#if: |: Volume {{{volume}}}|}}{{#if: |: {{{eddisplay}}} (2011)| ({{#if: |{{{ed}}} ed., }}2011)}}]]{{#if: |, {{{publisher}}}|}}{{#if: |, ISBN {{{isbn}}}|}}{{#if: Mittag-Leffler | ... (previous)|}}{{#if: E(1,1)(z)=exp(z) | ... (next)|}}{{#if: |: Entry: {{#if: |[[{{{entryref}}}|{{{entry}}}]]|{{{entry}}}}}|}}: $(2.1)$ (uses notation $E_0$ instead of $E_{0,1}$)
