Difference between revisions of "Harmonic numbers"
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Revision as of 04:19, 18 May 2014
The set $\mathbb{H}=\left\{\displaystyle\sum_{k=1}^n \dfrac{1}{k} \colon n \in \mathbb{Z}^+ \right\}$ of harmonic numbers is a time scale.
| Generic element $t\in \mathbb{T}$: | For some $n \in \mathbb{Z}^+, t =\displaystyle\sum_{k=0}^n \dfrac{1}{k}$ |
| Jump operator: | $\begin{array}{ll} \sigma(t) &= \sigma \left( \displaystyle\sum_{k=1}^{n} \dfrac{1}{k} \right) \\ &= \displaystyle\sum_{k=1}^{n+1} \dfrac{1}{k} \\ &= t + \dfrac{1}{n+1} \\ \end{array}$ |
| Graininess operator: | $\begin{array}{ll} \mu(t) &= \mu \left( \displaystyle\sum_{k=1}^{n} \dfrac{1}{k} \right) \\ &= t + \dfrac{1}{n+1} - t \\ &= \dfrac{1}{n+1} \\ \end{array}$ |
| $\Delta$-derivative: | $\begin{array}{ll} f^{\Delta}(t) &= f^{\Delta} \left( \displaystyle\sum_{k=1}^n \dfrac{1}{k} \right) \\ &= \dfrac{f \left( \displaystyle\sum_{k=1}^{n+1} \dfrac{1}{k} \right) - f \left( \displaystyle\sum_{k=1}^n \dfrac{1}{k} \right)}{\mu \left( \displaystyle\sum_{k=1}^n \dfrac{1}{k} \right)} \\ &= (n+1) \left[ f \left( \displaystyle\sum_{k=1}^{n+1} \dfrac{1}{k} \right) - f \left( \displaystyle\sum_{k=1}^n \dfrac{1}{k} \right) \right] \end{array}$ |
| $\Delta$-integral: | $\begin{array}{ll} \displaystyle\int_s^t f(\tau) \Delta \tau &= \displaystyle\int_{\sum_{k=1}^m \frac{1}{k}}^{\sum_{k=1}^n \frac{1}{k}} f(\tau) \Delta \tau \\ &= \displaystyle\sum_{k=m}^{n-1}\dfrac{1}{k+1} f \left( \displaystyle\sum_{k=1}^n \dfrac{1}{k} \right) \end{array}$ |
| Exponential function: | $\begin{array}{ll} e_p(t,s) &= e_p \left( \displaystyle\sum_{k=1}^n \dfrac{1}{k}, \displaystyle\sum_{k=1}^m \dfrac{1}{k} \right) \\ &= \exp \left( \displaystyle\int_{ \sum^m \frac{1}{k}}^{\sum^n \frac{1}{k}} \dfrac{1}{\mu(\tau)} \log(1 + \mu(\tau) p(\tau)) \Delta \tau \right) \\ &= \exp \left( \displaystyle\sum_{k=m}^{n-1} \log \left[1 + \mu \left( \displaystyle\sum_{j=1}^k \dfrac{1}{j} \right) p \left( \displaystyle\sum_{j=1}^k \dfrac{1}{j} \right) \right] \right) \\ &= \exp \left( \displaystyle\sum_{k=m}^{n-1} \log \left[1 + \dfrac{1}{k+1} p \left( \displaystyle\sum_{j=1}^k \dfrac{1}{j} \right) \right] \right) \\ &= \displaystyle\prod_{k=m}^{n-1} 1 + \dfrac{1}{k+1} p \left( \displaystyle\sum_{j=1}^k \dfrac{1}{j} \right) \\ \end{array}$ |
