convexe functie

[thisnot]

Hoe zou ik bepalen of de volgende functie convex is?

f (x) =- [det (A0 x1A1 ... xnan)] ^ (1 / m) op (x | A0 x1A1 ... xnan> 0)~ thanks

 

[me2please]

Ik denk dat je

<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$A_{n} (m \times m)' title="3 $ A_ (m \ times m)" alt='3$A_{n} (m \times m)' align=absmiddle>Gebruiken
1.definitie van convexe functie

<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$f(\alpha x (1-\alpha)y)\leq \alpha f(x) (1-\alpha ) f' title="3 $ f (\ alpha x (1 - \ alpha) y) \ Leq \ alpha f (x) (1 - \ alpha) f " alt='3$f(\alpha x (1-\alpha)y)\leq \alpha f(x) (1-\alpha ) f' align=absmiddle>
2.

<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$det(cA)=c^m det(A)' title="3 $ det (CA) = c ^ m det (A)" alt='3$det(cA)=c^m det(A)' align=absmiddle>3.Toon 1.met behulp van ongelijkheid Minkowski's

<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$A\neq 0, B\neq 0 (m\times m)' title="3 $ A \ neq 0, B \ neq 0 (m \ times m)" alt='3$A\neq 0, B\neq 0 (m\times m)' align=absmiddle>

Positieve en semidefinite<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$[[det(A B)]]^{\frac{1}{m}}\geq (detA)^{\frac{1}{m}} (detB)^{\frac{1}{m}}' title="3 $ [[det (A B )]]^{ \ frac (1) (m)) \ GEQ (deta) ^ (\ frac (1) (m)) (detB) ^ (\ frac (1) (m))" alt='3$[[det(A B)]]^{\frac{1}{m}}\geq (detA)^{\frac{1}{m}} (detB)^{\frac{1}{m}}' align=absmiddle>Dan zult u<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\alpha f(x) = -[det(\alpha A_{0} \alpha x_{1}A_{1} \cdots \alpha x_{n}A_{n})]^{\frac{1}{m}}' title="3 $ \ alpha f (x) = - [det (\ alpha A_ (0) \ alpha x_ (1) A_ (1) \ cdots \ alpha x_ A_ (n })]^{ \ frac (1) (m))" alt='3$\alpha f(x) = -[det(\alpha A_{0} \alpha x_{1}A_{1} \cdots \alpha x_{n}A_{n})]^{\frac{1}{m}}' align=absmiddle>en hetzelfde voor

<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$(1-\alpha ) f' title="3 $ (1 - \ alpha) f " alt='3$(1-\alpha ) f' align=absmiddle>en<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$f(\alpha x (1-\alpha )y) = -[det\{ (\alpha (1-\alpha) )A_{0} (\alpha x_{1} (1-\alpha )y_{1} )A_{1} \cdots (\alpha x_{n} (1-\alpha )y_{n})A_{n} \}]^{\frac{1}{m}}' title="3 $ f (\ alpha x (1 - \ alpha) y) = - [det \ ((\ alpha (1 - \ alpha)) A_ (0) (\ alpha x_ (1) (1 - \ alpha) Y_ (1)) A_ (1) \ cdots (\ alpha x_ (1 - \ alpha) Y_ ) A_ \)] ^ (\ frac (1) (m ))" alt='3$f(\alpha x (1-\alpha )y) = -[det\{ (\alpha (1-\alpha) )A_{0} (\alpha x_{1} (1-\alpha )y_{1} )A_{1} \cdots (\alpha x_{n} (1-\alpha )y_{n})A_{n} \}]^{\frac{1}{m}}' align=absmiddle>

 

[firephenix405]

Hoe heb je deze vergelijkingen op de post.it's amazing!me2please wrote:

Ik denk dat je
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$A_{n} (m \times m)' title="3 $ A_ (m \ times m)" alt='3$A_{n} (m \times m)' align=absmiddle>

Gebruiken

1.
definitie van convexe functie
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$f(\alpha x (1-\alpha)y)\leq \alpha f(x) (1-\alpha ) f' title="3 $ f (\ alpha x (1 - \ alpha) y) \ Leq \ alpha f (x) (1 - \ alpha) f " alt='3$f(\alpha x (1-\alpha)y)\leq \alpha f(x) (1-\alpha ) f' align=absmiddle>2.
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$det(cA)=c^m det(A)' title="3 $ det (CA) = c ^ m det (A)" alt='3$det(cA)=c^m det(A)' align=absmiddle>

3.
Toon 1.
met behulp van ongelijkheid Minkowski's
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$A\neq 0, B\neq 0 (m\times m)' title="3 $ A \ neq 0, B \ neq 0 (m \ times m)" alt='3$A\neq 0, B\neq 0 (m\times m)' align=absmiddle> Positieve en semidefinite

<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$[[det(A B)]]^{\frac{1}{m}}\geq (detA)^{\frac{1}{m}} (detB)^{\frac{1}{m}}' title="3 $ [[det (A B )]]^{ \ frac (1) (m)) \ GEQ (deta) ^ (\ frac (1) (m)) (detB) ^ (\ frac (1) (m))" alt='3$[[det(A B)]]^{\frac{1}{m}}\geq (detA)^{\frac{1}{m}} (detB)^{\frac{1}{m}}' align=absmiddle>

Dan zult u

<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\alpha f(x) = -[det(\alpha A_{0} \alpha x_{1}A_{1} \cdots \alpha x_{n}A_{n})]^{\frac{1}{m}}' title="3 $ \ alpha f (x) = - [det (\ alpha A_ (0) \ alpha x_ (1) A_ (1) \ cdots \ alpha x_ A_ (n })]^{ \ frac (1) (m))" alt='3$\alpha f(x) = -[det(\alpha A_{0} \alpha x_{1}A_{1} \cdots \alpha x_{n}A_{n})]^{\frac{1}{m}}' align=absmiddle>
en hetzelfde voor
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$(1-\alpha ) f' title="3 $ (1 - \ alpha) f " alt='3$(1-\alpha ) f' align=absmiddle>

en

<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$f(\alpha x (1-\alpha )y) = -[det\{ (\alpha (1-\alpha) )A_{0} (\alpha x_{1} (1-\alpha )y_{1} )A_{1} \cdots (\alpha x_{n} (1-\alpha )y_{n})A_{n} \}]^{\frac{1}{m}}' title="3 $ f (\ alpha x (1 - \ alpha) y) = - [det \ ((\ alpha (1 - \ alpha)) A_ (0) (\ alpha x_ (1) (1 - \ alpha) Y_ (1)) A_ (1) \ cdots (\ alpha x_ (1 - \ alpha) Y_ ) A_ \)] ^ (\ frac (1) (m ))" alt='3$f(\alpha x (1-\alpha )y) = -[det\{ (\alpha (1-\alpha) )A_{0} (\alpha x_{1} (1-\alpha )y_{1} )A_{1} \cdots (\alpha x_{n} (1-\alpha )y_{n})A_{n} \}]^{\frac{1}{m}}' align=absmiddle>

 

[M! K]

Hi firephenix405,

bij het schrijven van een post net klik op de Latex-knop voor een overzicht.In de tekst klik op TeX, schrijf uw formule zoals beschreven in het overzicht, drukt u nogmaals Tex * - dat is het.<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\alpha \sum \int \sqrt[n]{abc}' title="3 $ \ alpha \ sum \ int \ sqrt [n] (abc)" alt='3$\alpha \sum \int \sqrt[n]{abc}' align=absmiddle>
Mik