Difference between revisions of "Talk:Aufgaben:Problem 15"
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''Hint'': use the arithmetic-geometric mean inequality. You may ''not'' use the definition of the exponential function as a limit. | ''Hint'': use the arithmetic-geometric mean inequality. You may ''not'' use the definition of the exponential function as a limit. | ||
| + | |||
===Solution=== | ===Solution=== | ||
| − | |||
| − | $ f_{k}(x) | + | '''To show:''' |
| + | |||
| + | $$ \lim_{k \to \infty} f_{k}(x) = f(x), \forall x>0 $$ | ||
| + | |||
| + | '''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 last fraction is the difference quotient of ln evaluated in 1, thus: | |
| + | $$ \lim_{k \to \infty} \ln((1-\frac{x}{k})^{k}) = -x\dfrac{d}{dx}\ln(x)\mid_{x=1} = -x $$ | ||
| − | + | Since exp is continuous, we can "bring the limes inside": | |
| − | $\Rightarrow \lim_{k \to \infty} \ln((1-\frac{x}{k})^{k}) = | + | $$ \Rightarrow \lim_{k \to \infty} f_{k}(x) = x^{t-1} \lim_{k \to \infty} \exp ( \ln((1-\frac{x}{k})^{k}) = x^{t-1} \exp(\lim_{k \to \infty} ln((1-\frac{x}{k})^{k}) = x^{t-1} e^{-x} = f(x) \: \blacksquare $$ |
| − | |||
| − | + | '''To show:''' | |
| − | \ | + | $$ f_{k}(x) \leq f_{k+1}(x), \forall x > 0 $$ |
| − | + | '''Proof:''' | |
| − | + | ''Case 1:'' \( x \in \: ]0;k[: \) | |
| − | \ | + | The arithmetic-geometric mean equality says that \( \forall x_{1},...,x_{n} \geq 0 \) it holds: $$ (x_{1}...x_{n})^{\frac{1}{n}} \leq \dfrac{x_{1}+...+x_{n}}{n} $$ |
| − | + | Consider: | |
| − | + | ||
| − | + | ||
| − | + | $$ ((1-\dfrac{x}{k})^{k}\cdot1)^{\frac{1}{k+1}} \leq^{AM-GM} \dfrac{1}{k+1}(k(1-\dfrac{x}{k})+1) = \dfrac{1}{k+1}(k + 1 - x) = (1 - \dfrac{x}{k+1}) $$ | |
| − | $ ((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} $ | + | $$\Rightarrow (1-\dfrac{x}{k})^{k} \leq (1 - \dfrac{x}{k+1})^{k+1} $$ |
\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{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 $.\\ | ||
Revision as of 15:59, 28 December 2014
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
To show:
$$ \lim_{k \to \infty} f_{k}(x) = f(x), \forall x>0 $$
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 last fraction is the difference quotient of ln evaluated in 1, thus:
$$ \lim_{k \to \infty} \ln((1-\frac{x}{k})^{k}) = -x\dfrac{d}{dx}\ln(x)\mid_{x=1} = -x $$
Since exp is continuous, we can "bring the limes inside":
$$ \Rightarrow \lim_{k \to \infty} f_{k}(x) = x^{t-1} \lim_{k \to \infty} \exp ( \ln((1-\frac{x}{k})^{k}) = x^{t-1} \exp(\lim_{k \to \infty} ln((1-\frac{x}{k})^{k}) = x^{t-1} e^{-x} = f(x) \: \blacksquare $$
To show:
$$ f_{k}(x) \leq f_{k+1}(x), \forall x > 0 $$
Proof:
Case 1: \( x \in \: ]0;k[: \)
The arithmetic-geometric mean equality says that \( \forall x_{1},...,x_{n} \geq 0 \) 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 \( I_{k} = k^{t} \int_0^1 \! u^{t-1}(1-u)^{k}du \).
