|
|
| Line 1: |
Line 1: |
| − | Let \(t>0\) fixed. Define \( f(x) := x^{t-1}e^{-x} \), non-negative function on (0,\(\infty\)).
| + | --[[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 11:38, 11 June 2015 (CEST) |
| − | ==Part a)==
| + | Just a remark. Wikipedia says \(T(\phi)\) continous \( \Leftrightarrow \exists k\in \mathbb{N}_0 \;|T(\phi)| \leq C \|\phi\|_{C_b^k(K)} := C \sum_{|\alpha| \leq k}^{} \sup_{x \in K}\left| \partial^\alpha \phi(x) \right|\) |
| − | ===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.
| + | That would probably relay more on part a) than your solution, but what you did, looks fine too to me. |
| − | | + | |
| − | ===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:
| + | |
| − | | + | |
| − | $$ \lim_{k \to \infty} \ln((1-\frac{x}{k})^{k}) = -x\dfrac{d}{dy}\ln(y)\mid_{y=1} = -x $$
| + | |
| − | | + | |
| − | | + | |
| − | ''An alternative version would be:'' \( \lim_{k \rightarrow \infty} \ln((1-\frac{x}{k})^{k}) = \lim_{k \rightarrow \infty} k\ln(1-\frac{x}{k}) = \lim_{k \rightarrow \infty} \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} \) and then use Bernoulli-L'Hopital to get: \( \lim_{k \rightarrow \infty} \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} = \lim_{k \rightarrow \infty} \frac{\frac{x}{k^2}}{(\frac{x}{k} - 1) \frac{1}{k^2}} = -x \)
| + | |
| − | | + | |
| − | | + | |
| − | 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: $$ (x_{1}...x_{n})^{\frac{1}{n}} \leq \dfrac{x_{1}+...+x_{n}}{n} \tag{$\dagger$}$$
| + | |
| − | | + | |
| − | Consider:
| + | |
| − | | + | |
| − | $$ ((1-\dfrac{x}{k})^{k}\cdot1)^{\frac{1}{k+1}} \overset{(\dagger)}{\leq} \dfrac{1}{k+1}(k(1-\dfrac{x}{k})+1) = \dfrac{1}{k+1}(k + 1 - x) = (1 - \dfrac{x}{k+1}) $$
| + | |
| − | | + | |
| − | $$\Rightarrow (1-\dfrac{x}{k})^{k} \leq (1 - \dfrac{x}{k+1})^{k+1} $$
| + | |
| − | | + | |
| − | Where we don't have any problem with the k+1-root because \( (1-\dfrac{x}{k})^{k} \) only takes real positive values, since \( 0 < x < k. \)
| + | |
| − | | + | |
| − | ''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 \( \forall x \in [k;k+1[. \)
| + | |
| − | | + | |
| − | ''Case 3:'' \( x \in [k+1;\infty[ \)
| + | |
| − | | + | |
| − | Both functions are equal zero so the inequality still holds.
| + | |
| − | | + | |
| − | \( \blacksquare \)
| + | |
| − | | + | |
| − | ==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 \( I_{k} = k^{t} \int_0^1 \! u^{t-1}(1-u)^{k}du \).
| + | |
| − | | + | |
| − | ===Solution===
| + | |
| − | | + | |
| − | '''Idea:'''
| + | |
| − | In a first step we show that \( I_{k} \) is equal to the fraction in the right hand side of the equality. Then we rewrite \( I_{k} \) in a more convenient way, looking for something containing the functions given in part a). Eventually we let \( k\rightarrow\infty \) and apply Lebesgue's monoton convergence theorem to conclude.
| + | |
| − | | + | |
| − | $$ I_{k} = k^{t} \int_0^1 \! u^{t-1}(1-u)^{k} \, \mathrm{d}u \overset{P.I.}{=} \\
| + | |
| − | =k^{t}\lbrace\dfrac{u^{t}}{t}(1-u)^{k}\vert_{0}^{1} - \int_0^1 \! \dfrac{u^{t}}{t}(-1)k(1-u)^{k-1} \, \mathrm{d}u \rbrace = \\
| + | |
| − | = k^{t}\lbrace 0 + \dfrac{k}{t}\int_0^1 \! u^{t}(1-u)^{k-1} \, \mathrm{d}u \rbrace =\\
| + | |
| − | = \dfrac{k^{t+1}}{t}\int_0^1 \! u^{t}(1-u)^{k-1} \, \mathrm{d}u \overset{P.I.}{=}\\
| + | |
| − | =\dfrac{k^{t+1}}{t}\lbrace\dfrac{u^{t+1}}{t+1}(1-u)^{k-1}\vert_{0}^{1} - \int_0^1 \! \dfrac{u^{t+1}}{t+1}(-1)(k-1)(1-u)^{k-2} \, \mathrm{d}u \rbrace = \\
| + | |
| − | = \dfrac{k^{t+1}}{t}\lbrace 0 + \dfrac{k-1}{t+1}\int_0^1 \! u^{t+1}(1-u)^{k-2} \, \mathrm{d}u \rbrace =\dfrac{k^{t+1}(k-1)}{t(t+1)} \int_0^1 \! u^{t+1}(1-u)^{k-2} \, \mathrm{d}u \overset{P.I.}{=}\\
| + | |
| − | = ... = \frac{(k-1)!k^{t+1}}{t(t+1)...(t+k-1)} \int_0^1 \! u^{t+k-1} \, \mathrm{d}u = \frac{(k-1)!k^{t+1}}{t(t+1)...(t+k-1)}\dfrac{u^{t+k}}{t+k}\vert^{1}_{0} = \frac{k!k^{t}}{t(t+1)...(t+k)} $$
| + | |
| − | | + | |
| − | Where we have used k times partial integration to reach the last but one step.
| + | |
| − | | + | |
| − | $$ I_{k} = k^{t} \int_0^1 \! u^{t-1}(1-u)^{k} \, \mathrm{d}u = k^{t-1} \int_0^k \! (\dfrac{x}{k})^{t-1}(1-\dfrac{x}{k})^{k} \, \mathrm{d}x = \\
| + | |
| − | = \int_0^k \! x^{t-1}(1-\dfrac{x}{k})^{k} \, \mathrm{d}x = \int_0^k \! x^{t-1}(1-\dfrac{x}{k})^{k} \, \mathrm{d}x + \int_k^\infty \! 0 \, \mathrm{d}x = \int_0^k \! f_{k}(x) \, \mathrm{d}x + \int_k^\infty \! f_{k}(x) \, \mathrm{d}x =\\
| + | |
| − | = \int_0^\infty \! f_{k}(x) \, \mathrm{d}x $$
| + | |
| − | | + | |
| − | Where the first step is a change of variable \( u = \dfrac{x}{k} \).
| + | |
| − | | + | |
| − | Letting \( k \rightarrow\infty \), using the monoton convergence theorem on \( f_{k}(x) \), we obtain:
| + | |
| − | | + | |
| − | $$ I_{\infty} = \int_0^\infty \! x^{t-1}e^{-x}\, \mathrm{d}x = \Gamma(t) $$
| + | |
| − | | + | |
| − | Where the conditions to apply the monoton convergence theorem on \( \Omega := ]0,\infty[ \) are satisfied (see part a), because: \( f_{k}(x) \) are non-negative, monotonically increasing functions \( \in L^1(\Omega) \) (because non-zero only on a bounded subset \( ]0,k[ \) and continuous).
| + | |
| − | | + | |
| − | '''Lebesgue monotone convergence theorem:'''
| + | |
| − | | + | |
| − | Let \( m \in \mathbb{N}, \Omega \subset \mathbb{R}^m, f_n \) be a sequence of functions \( \in L^1(\Omega) \) so that \( 0 \leq f_n(x) \leq f_{n+1}(x) \: \forall n \in \mathbb{N}, \: \forall x \in \Omega \) and \( \lim_{n \to \infty} f_n(x) =: f(x) \: \forall x \in \Omega \).
| + | |
| − | | + | |
| − | Then \( f \in L^1(\Omega) \) and \( \lim_{n \to \infty} \int_\Omega f_n(x)dx = \int_\Omega f(x)dx. \)
| + | |
| − | | + | |
| − | ''Note: some versions of the monotone convergence theorem require \( f \) to be integrable. For a proof that \( f \in L^1(\Omega) \) see the discussion.''
| + | |
| − | | + | |
| − | \( \blacksquare \)
| + | |
That would probably relay more on part a) than your solution, but what you did, looks fine too to me.
