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--- quantum_integer [2014/12/09 22:25]
nikolaj
+++ quantum_integer [2014/12/10 12:02]
nikolaj
@@ Line 11: / Line 11: @@
 $[n]_{q} = q^{-f(n)/2}q^{-1}\sum_{k=1}^n q^k = n+\tfrac{n}{2}(n-1-f(n))\cdot(q-1)+\mathcal{O}\left((q-1)^2\right)$
 In fact this doesn't require $n$ to be an integer.
-The case $f=0$ is often considered, and also $[n]_{q^2}$ together with $f=n-1$, where the numbers are of the form $n+\mathcal{O}\left((q-1)^2\right)$. In the imaginary direction, $q\propto\mathrm{e}^{i\varphi}$, this corresponds to $\lim_{\varphi\to 0}\frac{\sin(n\varphi)}{\sin(\phi)}=n$.
+The case $f=0$ is often considered.
-Then, with $q=r\mathrm{e}^{i\varphi}$, along the positive real axis number $[n]_q$ is a valley with bottom at $q=1$, where $[n]_{1}=n$, and along $\varphi$ you have harmonic oscillations with period depending on $n$.
+Quantum aspect: $f=n-1$ gives $[n]_{q^2}$=n+\mathcal{O}\left((q-1)^2\right)$. (The $q^2$ isn't necessary.) In the imaginary direction, $q\propto\mathrm{e}^{i\varphi}$, this corresponds to $\lim_{\varphi\to 0}\frac{\sin(n\varphi)}{\sin(\phi)}=n$.
+With $q=r\mathrm{e}^{i\varphi}$, along the positive real axis number $[n]_q$ is a valley with bottom at $q=1$, where $[n]_{1}=n$, and along $\varphi$ you have harmonic oscillations with period depending on $n$.
 >I might change the exponent in $-f(n)/2$ to something else later