Vai šādas funkcijas izliekts?

[thisnot]

Kā varētu i noteikt, vai šādu funkciju ir izliekts?

f (x) =- [det (A0 x1A1 ... xnAn)] ^ (1 / m) uz (x | A0 x1A1 ... xnAn> 0)~ Paldies

 

[me2please]

Es domāju, ka jums ir

<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$A_{n} (m \times m)' title="3 $ A_ (m \ reizes m)" alt='3$A_{n} (m \times m)' align=absmiddle>Izmantot
1.definīcija izliekta funkcija

<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 - \ alfa) y) \ leq \ alfa 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.Rādīt 1.izmantojot Minkowski's nevienlīdzības

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

un Pozitīvi 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$[[det(A B)]]^{\frac{1}{m}}\geq (detA)^{\frac{1}{m}} (detB)^{\frac{1}{m}}' align=absmiddle>Tad jums būs<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\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>un pati

<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$(1-\alpha ) f' title="$ 3 (1 - \ alpha) f " alt='3$(1-\alpha ) f' align=absmiddle>un<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$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 - \ alfa) y) = - [Det \ ((\ alpha (1 - \ alpha)) A_ (0) (\ alpha x_ (1) (1 - \ alfa) 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]

Kā jums šie vienādojumi uz pastu.tas ir pārsteidzošs!me2please rakstīja:

Es domāju, ka jums ir
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$A_{n} (m \times m)' title="3 $ A_ (m \ reizes m)" alt='3$A_{n} (m \times m)' align=absmiddle>

Izmantot

1.
definīcija izliekta funkcija
<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 - \ alfa) y) \ leq \ alfa 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.
Rādīt 1.
izmantojot Minkowski's nevienlīdzības
<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$A\neq 0, B\neq 0 (m\times m)' title="3 $ \ neq 0, B \ neq 0 (m \ reizes m)" alt='3$A\neq 0, B\neq 0 (m\times m)' align=absmiddle> un Pozitīvi 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>

Tad jums būs

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

un

<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 - \ alfa) y) = - [Det \ ((\ alpha (1 - \ alpha)) A_ (0) (\ alpha x_ (1) (1 - \ alfa) 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,

rakstot post tikai klick uz lateksu Poga pārskatu.Tekstā klick uz TeX, ievadiet formulu, kas aprakstīta pārskatu, vēlreiz nospiediet Tex * - tas arī viss.<img src='http://www.elektroda.pl/cgi-bin/mimetex/mimetex.cgi?3$\alpha \sum \int \sqrt[n]{abc}' title="3 $ \ alpha \ summa \ int \ SQRTn [n] (abc)" alt='3$\alpha \sum \int \sqrt[n]{abc}' align=absmiddle>
Mik