Talk:Aufgaben:Problem 15

Let \(t>0\) fixed. Define \( f(x) := x^{t-1}e^{-x} \), non-negative function on (0,\(\infty\)).

Part a)

Problem

For \(k = 1,2,... \) let $$ f_k(x) = \left\{ \begin{array}{l l} x^{t-1}(1-\frac{x}{k})^{k} & \quad 0<x<k\\ 0 & \quad k\leq x<\infty\end{array} \right. $$ Show that \(f_k(x) \rightarrow f(x) \) and \(f_k(x) \leq f_{k+1}(x), \forall x>0.\)

Hint: use the arithmetic-geometric mean inequality. You may not use the definition of the exponential function as a limit.

Solution

\textbf{To show:}\\

$ f_{k}(x) \rightarrow f(x) $ for k $ \rightarrow \infty$\\

\textbf{Proof:}\\

Take any x $ \in $ (0 ; $\infty$) and $ k > x $ ; consider:

$ \ln((1-\frac{x}{k})^{k}) = k\ln(1-\frac{x}{k}) = \frac{x}{x}k(\ln(1-\frac{x}{k})-\ln(1)) = -x\dfrac{(\ln(1)-\ln(1-\frac{x}{k}))}{\frac{x}{k}} $\\

The fraction is the difference quotient of ln evaluated in 1, thus:\\

$\Rightarrow \lim_{k \to \infty} \ln((1-\frac{x}{k})^{k}) = -x\dfrac{d}{dx}\ln(x)\mid_{x=1} = -x $ \\

$\Rightarrow \lim_{k \to \infty} f_{k}(x) = x^{t-1} \lim_{k \to \infty} \exp ( \ln((1-\frac{x}{k})^{k}) $

$ = x^{t-1} \exp(\lim_{k \to \infty} ln((1-\frac{x}{k})^{k}) = x^{t-1} e^{-x} = f(x)$\\

\textbf{Remark:} since exp is continuous, we can "bring the limes inside" $ \blacksquare $ \\

\textbf{To show:}\\

$ f_{k}(x) \leq f_{k+1}(x)$ $\forall x > 0$ \\

\textbf{Proof:}\\

\textbf{Case 1:} $ \forall x \in $ $ ]0;k[ $\\ The arithmetic-geometric mean equality says:\\ $ \forall x_{1},...,x_{n} \geq 0 $ it holds : $ ( x_{1}...x_{n})^{\frac{1}{n}} \leq \dfrac{x_{1}+...+x_{n}}{n}$\\

Consider:\\ $ ((1-\dfrac{x}{k})^{k}\cdot1)^{\frac{1}{k+1}} \leq^{AM-GM} \dfrac{1}{k+1}(k(1-\dfrac{x}{k})+1) = \dfrac{1}{k+1}(k + 1 - x) = \\ = (1 - \dfrac{x}{k+1}) $\\

$\Rightarrow (1-\dfrac{x}{k})^{k} \leq (1 - \dfrac{x}{k+1})^{k+1} $\\

\textbf{Remark:} we don't have any problem of sign or multivalue with the k+1-root because $ (1-\dfrac{x}{k})^{k} $ takes only real positive value, since $ 0 < x < k $.\\

\textbf{Case 2:} $ x \in [k;k+1[$\\ The inequality holds since $ f_{k}(x) = 0 $ and $ f_{k+1}(x) = x^{t-1}(1 - \dfrac{x}{k+1})^{k+1} $ is a positve function $ \forall x \in [k;k+1[$.\\

\textbf{Case 3:} $ x \in [k+1;\infty[$\\ Both the functions are equal to zero so the inequality still holds.\\

Part b)

Problem

Use the monotone convergence theorem to show that $$ \Gamma(t) = \lim_{k \to \infty} \frac{k!k^{t}}{t(t+1)...(t+k)}, t>0 $$ where \( \Gamma(t) \) is the Euler Gamma function.

Hint: consider \( I_{k} = k^{t} \int_0^1 \! u^{t-1}(1-u)^{k}du \).