Initial program 0.5
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}\]
Initial simplification0.5
\[\leadsto \frac{{\left(\pi \cdot \left(n \cdot 2\right)\right)}^{\left(\frac{1}{2} - \frac{k}{2}\right)}}{\sqrt{k}}\]
- Using strategy
rm Applied *-un-lft-identity0.5
\[\leadsto \frac{{\left(\pi \cdot \left(n \cdot 2\right)\right)}^{\left(\frac{1}{2} - \frac{k}{2}\right)}}{\color{blue}{1 \cdot \sqrt{k}}}\]
Applied unpow-prod-down0.6
\[\leadsto \frac{\color{blue}{{\pi}^{\left(\frac{1}{2} - \frac{k}{2}\right)} \cdot {\left(n \cdot 2\right)}^{\left(\frac{1}{2} - \frac{k}{2}\right)}}}{1 \cdot \sqrt{k}}\]
Applied times-frac0.6
\[\leadsto \color{blue}{\frac{{\pi}^{\left(\frac{1}{2} - \frac{k}{2}\right)}}{1} \cdot \frac{{\left(n \cdot 2\right)}^{\left(\frac{1}{2} - \frac{k}{2}\right)}}{\sqrt{k}}}\]
Simplified0.6
\[\leadsto \color{blue}{{\pi}^{\left(\frac{1}{2} - \frac{k}{2}\right)}} \cdot \frac{{\left(n \cdot 2\right)}^{\left(\frac{1}{2} - \frac{k}{2}\right)}}{\sqrt{k}}\]
- Using strategy
rm Applied sub-neg0.6
\[\leadsto {\pi}^{\left(\frac{1}{2} - \frac{k}{2}\right)} \cdot \frac{{\left(n \cdot 2\right)}^{\color{blue}{\left(\frac{1}{2} + \left(-\frac{k}{2}\right)\right)}}}{\sqrt{k}}\]
Applied unpow-prod-up0.5
\[\leadsto {\pi}^{\left(\frac{1}{2} - \frac{k}{2}\right)} \cdot \frac{\color{blue}{{\left(n \cdot 2\right)}^{\frac{1}{2}} \cdot {\left(n \cdot 2\right)}^{\left(-\frac{k}{2}\right)}}}{\sqrt{k}}\]
Applied associate-/l*0.5
\[\leadsto {\pi}^{\left(\frac{1}{2} - \frac{k}{2}\right)} \cdot \color{blue}{\frac{{\left(n \cdot 2\right)}^{\frac{1}{2}}}{\frac{\sqrt{k}}{{\left(n \cdot 2\right)}^{\left(-\frac{k}{2}\right)}}}}\]
Simplified0.5
\[\leadsto {\pi}^{\left(\frac{1}{2} - \frac{k}{2}\right)} \cdot \frac{\color{blue}{\sqrt{n \cdot 2}}}{\frac{\sqrt{k}}{{\left(n \cdot 2\right)}^{\left(-\frac{k}{2}\right)}}}\]
Final simplification0.5
\[\leadsto \frac{\sqrt{2 \cdot n}}{\frac{\sqrt{k}}{{\left(2 \cdot n\right)}^{\left(-\frac{k}{2}\right)}}} \cdot {\pi}^{\left(\frac{1}{2} - \frac{k}{2}\right)}\]