Note that $exp(it) = 1 + it - t^2/2 + O(t^3)$ uniformly in $t \in \mathbb{R}$. Thus $n^{exp(it)-1} = exp(it \cdot \log n - \log n \cdot t^{2}/2 + O(t^3 \cdot \log n))$ and also by Taylor's theorem $1/\Gamma(exp(it)) = 1 + O(t)$ when $t$ is small (but in fact also for all real $t \in \mathbb{R}$ by periodicity). Thus $$n^{exp(it)-1}/\Gamma(exp(it)) = exp(it \cdot \log n - \log n \cdot t^2/2 +O(t^3 \cdot \log n))$$ Multiplying both sides by $exp(- it \cdot \log n)$ and substituting $t := t \cdot (\log n)^{-1/2}$ we obtain as $n \rightarrow \infty$ the desired limit, $exp(-t^2/2)$.
|
2 | Added spaces to bump Tex to the next line | ||
|
|
||||
|
|
1 |
|
||
|
Note that $exp(it) = 1 + it - t^2/2 + O(t^3)$ uniformly in $t \in \mathbb{R}$. Thus $n^{exp(it)-1} = exp(it \cdot \log n - \log n \cdot t^{2}/2 + O(t^3 \cdot \log n))$ and also by Taylor's theorem $1/\Gamma(exp(it)) = 1 + O(t)$ when $t$ is small (but in fact also for all real $t \in \mathbb{R}$ by periodicity). Thus $$n^{exp(it)-1}/\Gamma(exp(it)) = exp(it \cdot \log n - \log n \cdot t^2/2 +O(t^3 \cdot \log n))$$ Multiplying both sides by $exp(- it \cdot \log n)$ and substituting $t := t \cdot (\log n)^{-1/2}$ we obtain as $n \rightarrow \infty$ the desired limit, $exp(-t^2/2)$. |
||||
