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Extreme Value Problem


A B C D E F G H I J K L M N O P Q R S T U V W X Y Z overview

Example:

For which $ t \in \mathbb{R}$ is the distance between the point $ S_t(e^{t+1}+2e^{-t-1}/0)$ and the origin minimal?


For the objective function one calculates the distance between the point $ S_t$ and the origin, i.e. $ (e^{t+1}+2e^{-t-1})-0$

$\displaystyle d(t)$ $\displaystyle =e^{t+1}+2e^{-t-1} \qquad t \in \mathbb{R}$    
$\displaystyle d'(t)$ $\displaystyle =e^{t+1}-2e^{-t-1}$    
$\displaystyle d''(t)$ $\displaystyle =e^{t+1}+2e^{-t-1}$    

$ d'(t)=0$:

$\displaystyle e^{t+1}-2e^{-t-1}$ $\displaystyle =0$    
$\displaystyle e^{2t+2}$ $\displaystyle =2$    
$\displaystyle 2t+2$ $\displaystyle =\ln2$    
$\displaystyle t$ $\displaystyle =\frac{1}{2}\ln{2}-1$    

$ d''(\frac{1}{2}\ln{2}-1)=\frac{3}{2}\ln{2}>0$    minimum


Type of the minimum:

$\displaystyle \lim_{t \to -\infty} d(t)$ $\displaystyle =\lim_{t \to -\infty} e^{t+1}+2e^{-t-1} \rightarrow \infty$    
$\displaystyle \lim_{t \to \infty} d(t)$ $\displaystyle =\lim_{t \to \infty} e^{t+1}+2e^{-t-1} \rightarrow \infty$    
$\displaystyle d(\frac{1}{2}\ln{2}-1)$ $\displaystyle =2\sqrt{2}$    

Thus there exists an absolute minimum.


(Authors: Jahn/Knödler)

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  automatisch erstellt am 8.  7. 2004