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> Control 2 Mate III 2009
rock&mati
mensaje Jan 16 2009, 10:29 PM
Publicado: #1


Dios Matemático
Ícono de Grupo

Grupo: Usuario FMAT
Mensajes: 316
Registrado: 2-May 08
Desde: a la &$€¬~#@ del mundo
Miembro Nº: 21.900
Nacionalidad:
Colegio/Liceo: Liceo Jose Victorino Lastarria
Universidad: Universidad de Chile-FCFM
Sexo:



Bueno, creo que no entendere jamas funciones, aca dejo el control 2, aporten soluciones xd.

TEX: % MathType!MTEF!2!1!+-<br />% feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn<br />% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr<br />% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9<br />% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x<br />% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaqjEa<br />% qaaiaabcfacaqGXaGaaeOlaiaab2caaaaabaGaaeikaiaabMgacaqG<br />% PaGaaeiiaiaabseacaqGLbGaaeyBaiaabwhacaqGLbGaae4Caiaabs<br />% hacaqGYbGaaeyzaiaabccacaqG1bGaae4CaiaabggacaqGUbGaaeiz<br />% aiaab+gacaqGGaGaaeysaiaab6gacaqGKbGaaeyDaiaabogacaqGJb<br />% GaaeyAaiaab+gacaqGUbGaaeiiaiaabghacaqG1bGaaeyzaiaabcca<br />% caqGOaGaeyiaIiIaamOBaiabgwMiZkaaigdacaqGPaGaaeiiaiaaik<br />% dacqGHflY1caaI3aWaaWbaaSqabeaacaWGUbaaaOGaae4kaiaaboda<br />% cqGHflY1caqG1aWaaWbaaSqabeaacaqGUbaaaOGaaeylaiaabwdaca<br />% qGGaGaaeyzaiaabohacaqGGaGaaeizaiaabMgacaqG2bGaaeyAaiaa<br />% bohacaqGPbGaaeOyaiaabYgacaqGLbGaaeiiaiaabchacaqGVbGaae<br />% OCaiaabccacaaI2aGaaiOlaaqaaiaabIcacaqGPbGaaeyAaiaabMca<br />% caqGGaGaae4qaiaabggacaqGSbGaae4yaiaabwhacaqGSbGaaeyzai<br />% aabccacaqGLbGaaeOBaiaabccacaqGMbGaaeyDaiaab6gacaqGJbGa<br />% aeyAaiaab+gacaqGUbGaaeiiaiaabsgacaqGLbGaaeiiaiaad6gaca<br />% GGSaGaaeiiaiaabogacaqGVbGaaeOBaiaabohacaqGPbGaaeizaiaa<br />% bwgacaqGYbGaaeyyaiaab6gacaqGKbGaae4BaiaabccacaWGUbGaey<br />% yzImRaaGOmaiaacYcacaqGGaGaaeiBaiaabggacaqGGaGaae4Caiaa<br />% bwhacaqGTbGaaeyyaiaabccacaqGZbGaaeyAaiaabEgacaqG1bGaae<br />% yAaiaabwgacaqGUbGaaeiDaiaabwgacaqG6aaabaaabaGaaeiiaiaa<br />% bccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaae<br />% iiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqG<br />% GaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabc<br />% cacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeii<br />% aiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGa<br />% GaaeiiaiaabccacaqGGaGaaeiiaiaabccadaaeWbqaamaabmaabaWa<br />% a0baaSqaaiaadUgaaeaacaWGUbaaaaGccaGLOaGaayzkaaGaam4Aam<br />% aaCaaaleqabaGaaGOmaaaaaeaacaqGRbGaaeypaiaabgdaaeaacaWG<br />% UbaaniabggHiLdaakeaaaeaacaqGGaGaaeiiaiaabccacaqGGaGaae<br />% iiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqG<br />% GaGaaeiiaiaabMeacaqGUbGaaeizaiaabMgacaqGJbGaaeyyaiaabo<br />% gacaqGPbGaae4Baiaab6gacaqG6aGaaeiiaiaadUgadaahaaWcbeqa<br />% aiaaikdaaaGccqGH9aqpcaWGRbGaey4kaSIaam4AaiaacIcacaWGRb<br />% GaeyOeI0IaaGymaiaacMcaaaaa!FA58!<br />\[<br />\begin{gathered}<br />  \boxed{{\text{P1}}{\text{. - }}} \hfill \\<br />  {\text{(i) Demuestre usando Induccion que (}}\forall n \geqslant 1{\text{) }}2 \cdot 7^n {\text{ + 3}} \cdot {\text{5}}^{\text{n}} {\text{ - 5 es divisible por }}6. \hfill \\<br />  {\text{(ii) Calcule en funcion de }}n,{\text{ considerando }}n \geqslant 2,{\text{ la suma siguiente:}} \hfill \\<br />   \hfill \\<br />  {\text{                                               }}\sum\limits_{{\text{k = 1}}}^n {\left( {_k^n } \right)k^2 }  \hfill \\<br />   \hfill \\<br />  {\text{              Indicacion: }}k^2  = k + k(k - 1) \hfill \\ <br />\end{gathered} <br />\]



TEX: % MathType!MTEF!2!1!+-<br />% feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn<br />% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr<br />% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9<br />% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x<br />% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaqjEa<br />% qaaiaabcfacaqGYaGaaeOlaiaab2caaaaabaGaaeikaiaabggacaqG<br />% PaGaaeiiaiaabofacaqGLbGaaeyyaiaabccacaWGgbGaaeiiaiaabw<br />% gacaqGSbGaaeiiaiaabogacaqGVbGaaeOBaiaabQgacaqG1bGaaeOB<br />% aiaabshacaqGVbGaaeiiaiaabsgacaqGLbGaaeiiaiaabYgacaqGHb<br />% Gaae4CaiaabccacaqGMbGaaeyDaiaab6gacaqGJbGaaeyAaiaab+ga<br />% caqGUbGaaeyzaiaabohacaqGGaGaaeizaiaabwgacaqGGaGaeSyhHe<br />% QaaeiiaiaabwgacaqGUbGaaeiiaiabl2riHkaab6cacaqGGaGaae4u<br />% aiaabwgacaqGGaGaaeizaiaabwgacaqGMbGaaeyAaiaab6gacaqGLb<br />% GaaeiiaiaabYgacaqGHbGaaeiiaiaabAgacaqG1bGaaeOBaiaaboga<br />% caqGPbGaae4Baiaab6gacaqGGaGaeuOPdyKaaeOoaiaadAeacqGHsg<br />% IRcqWIDesOcaqGGaGaaeyCaiaabwhacaqGLbGaaeiiaiaabggaaeaa<br />% caqGJbGaaeyyaiaabsgacaqGHbGaaeiiaiaabccacaWGMbGaeyicI4<br />% SaamOraiaabccacaqGSbGaaeyzaiaabccacaqGHbGaae4Caiaab+ga<br />% caqGJbGaaeyAaiaabggacaqGGaGaeuOPdyKaaiikaiaadAgacaGGPa<br />% Gaeyypa0JaamOzaiaacIcacaaIWaGaaiykaiaac6cacaqGGaGaaeii<br />% aiaabseacaqGLbGaaeyBaiaabwhacaqGLbGaae4CaiaabshacaqGYb<br />% GaaeyzaiaabccacaqGXbGaaeyDaiaabwgacaqGGaGaeuOPdyKaaeii<br />% aiaabwgacaqGZbGaaeiiaiaabwhacaqGUbGaaeyyaiaabccacaqGMb<br />% GaaeyDaiaab6gacaqGJbGaaeyAaiaab+gacaqGUbGaaeiiaiaaboha<br />% caqGVbGaaeOyaiaabkhacaqGLbGaaeyEaiaabwgacaqGJbGaaeiDai<br />% aabMgacaqG2bGaaeyyaiaab6caaeaaaeaacaqGOaGaaeOyaiaabMca<br />% caqGGaGaae4uaiaabwgacaqGHbGaaeOBaiaabccacaqGfbGaeyiyIK<br />% RaeyybIySaaeiiaiaabccacaqG5bGaaeiiaiaadgeacqGHgksZcaWG<br />% fbGaaeiiaiaacIcacaqGMbGaaeyAaiaabQgacaqGVbGaaiykaiaac6<br />% cacaqGGaGaae4uaiaabwgacaqGGaGaaeizaiaabwgacaqGMbGaaeyA<br />% aiaab6gacaqGLbGaaeOBaiaabccacaqGSbGaaeyyaiaabohacaqGGa<br />% GaaeOzaiaabwhacaqGUbGaae4yaiaabMgacaqGVbGaaeOBaiaabwga<br />% caqGZbGaaeiiaiaabccacaWGMbGaaeiiaiaabMhacaqGGaGaam4zai<br />% aabccacaqGKbGaaeyzaiaabccacqaHbpGCcaGGOaGaamyraiaacMca<br />% caqGGaGaaeyzaiaab6gacaqGGaGaeqyWdiNaaiikaiaadweacaGGPa<br />% Gaaeiiaiaab2gacaqGLbGaaeizaiaabMgacaqGHbGaaeOBaiaabsha<br />% caqGLbaabaGaamOzaiaacIcacaWGybGaaiykaiabg2da9iaadgeacq<br />% GHQicYcaWGybGaaeiiaiaabMhacaqGGaGaam4zaiaacIcacaWGybGa<br />% aiykaiabg2da9iaadgeacqGHPiYXcaWGybGaaiOlaiaabccacaqGGa<br />% GaaeiuaiaabggacaqGYbGaaeyyaiaabccacaqG0bGaae4Baiaabsga<br />% caqGVbGaaeiiaiaadIfacqGHgksZcaWGfbGaaiOlaaqaaiaacIcaca<br />% qGPbGaaiykaiaabccacaqGebGaaeyAaiaabohacaqGJbGaaeyDaiaa<br />% bshacaqGHbGaaeiiaiaabwgacaqGUbGaaeiiaiaabAgacaqG1bGaae<br />% OBaiaabogacaqGPbGaae4Baiaab6gacaqGGaGaaeizaiaabwgacaqG<br />% GaGaamyqaiaabccacaqGSbGaaeyyaiaabccacaqGWbGaae4Baiaabo<br />% hacaqGPbGaaeOyaiaabMgacaqGSbGaaeyAaiaabsgacaqGHbGaaeiz<br />% aiaabccacaqGKbGaaeyzaiaabccacaqGXbGaaeyDaiaabwgacaqGGa<br />% GaamOzaiaabccacaqGVbGaaeiiaiaadEgacaqGGaGaae4Caiaabwga<br />% caqGHbGaaeOBaiaabccacaqGZbGaae4BaiaabkgacaqGYbGaaeyzai<br />% aabMhacaqGLbGaaeOBaiaabogacaqG0bGaaeyAaiaabAhacaqGHbGa<br />% ae4CaiaabccacaqGOaGaae4yaiaabggacaqGKbGaaeyyaiaabccaca<br />% qG1bGaaeOBaiaabggacaqGGaGaaeiCaiaab+gacaqGYbGaaeiiaaqa<br />% aiaabohacaqGLbGaaeiCaiaabggacaqGYbGaaeyyaiaabsgacaqGVb<br />% Gaaeykaiaab6cacaqGGaGaaeOsaiaabwhacaqGZbGaaeiDaiaabMga<br />% caqGMbGaaeyAaiaabghacaqG1bGaaeyzaaqaaiaacIcacaqGPbGaae<br />% yAaiaacMcacaqGebGaaeyzaiaabshacaqGLbGaaeOCaiaab2gacaqG<br />% PbGaaeOBaiaabggacaqGYbGaaeiiaiaabwhacaqGUbGaaeiiaiaabo<br />% gacaqGVbGaaeOBaiaabQgacaqG1bGaaeOBaiaabshacaqGVbGaaeii<br />% aiaadgeacaqGGaGaaeiCaiaabggacaqGYbGaaeyyaiaabccacaqGLb<br />% GaaeiBaiaabccacaqGJbGaaeyDaiaabggacaqGSbGaaeiiaiaabcca<br />% caWGMbGaaeiiaiaabwgacaqGZbGaaeiiaiaabkgacaqGPbGaaeiDai<br />% aabwgacaqGJbGaaeiDaiaabMgacaqG2bGaaeyyaiaabYcacaqGGaGa<br />% aeyEaiaabccacaqGVbGaaeiDaiaabkhacaqGVbGaaeiiaiaabchaca<br />% qGHbGaaeOCaiaabggacaqGGaGaaeyzaiaabYgacaqGGaGaae4yaiaa<br />% bwhacaqGHbGaaeiBaiaabccacaWGNbGaaeiiaiaabwgacaqGZbGaae<br />% iiaiaabkgacaqGPbGaaeyEaiaabwgacaqGJbGaaeiDaiaabMgacaqG<br />% 2bGaaeyyaiaab6caaeaacaqGGaGaaeOsaiaabwhacaqGZbGaaeiDai<br />% aabMgacaqGMbGaaeyAaiaabghacaqG1bGaaeyzaiaab6caaaaa!E8AC!<br />\[<br />\begin{gathered}<br />  \boxed{{\text{P2}}{\text{. - }}} \hfill \\<br />  {\text{(a) Sea }}F{\text{ el conjunto de las funciones de }}\mathbb{R}{\text{ en }}\mathbb{R}{\text{. Se define la funcion }}\Phi {\text{:}}F \to \mathbb{R}{\text{ que a}} \hfill \\<br />  {\text{cada  }}f \in F{\text{ le asocia }}\Phi (f) = f(0).{\text{  Demuestre que }}\Phi {\text{ es una funcion sobreyectiva}}{\text{.}} \hfill \\<br />   \hfill \\<br />  {\text{(b) Sean E}} \ne \emptyset {\text{  y }}A \subseteq E{\text{ }}({\text{fijo}}).{\text{ Se definen las funciones  }}f{\text{ y }}g{\text{ de }}\rho (E){\text{ en }}\rho (E){\text{ mediante}} \hfill \\<br />  f(X) = A \cup X{\text{ y }}g(X) = A \cap X.{\text{  Para todo }}X \subseteq E. \hfill \\<br />  ({\text{i}}){\text{ Discuta en funcion de }}A{\text{ la posibilidad de que }}f{\text{ o }}g{\text{ sean sobreyenctivas (cada una por }} \hfill \\<br />  {\text{separado)}}{\text{. Justifique}} \hfill \\<br />  ({\text{ii}}){\text{Determinar un conjunto }}A{\text{ para el cual  }}f{\text{ es bitectiva}}{\text{, y otro para el cual }}g{\text{ es biyectiva}}{\text{.}} \hfill \\<br />  {\text{ Justifique}}{\text{.}} \hfill \\ <br />\end{gathered} <br />\]



TEX: % MathType!MTEF!2!1!+-<br />% feaafiart1ev1aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn<br />% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr<br />% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9<br />% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x<br />% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaadaqjEa<br />% qaaiaabcfacaqGZaGaaeOlaiaab2caaaaabaGaaeikaiaabggacaqG<br />% PaGaaeiiaiaaboeacaqGVbGaaeOBaiaabohacaqGPbGaaeizaiaabw<br />% gacaqGYbGaaeyzaiaabccacaqGLbGaaeiBaiaabccacaqGJbGaae4B<br />% aiaab6gacaqGQbGaaeyDaiaab6gacaqG0bGaae4BaiaabccacaWGbb<br />% Gaeyypa0ZaaiWaaeaacaaIXaGaeyOeI0YaaSaaaeaacaaIXaaabaGa<br />% amOBaaaacaGG8bGaamOBaiabgIGiolablwriLcGaay5Eaiaaw2haai<br />% aab6cacaqGGaGaaeyraiaab6gacaqGJbGaaeyDaiaabwgacaqGUbGa<br />% aeiDaiaabkhacaqGLbGaaeiiaiaabwgacaqGSbGaaeiiaiaabohaca<br />% qG1bGaaeiCaiaabkhacaqGLbGaaeyBaiaab+gacaqGGaGaaeizaiaa<br />% bwgacaqGGaGaamyqaiaabccacaqG5bGaaeiiaiaabchacaqGYbGaae<br />% yDaiaabwgacaqGIbGaaeyzaiaabccacaqGXbGaaeyDaiaabwgacaqG<br />% GaaabaGaaeyzaiaabAgacaqGLbGaae4yaiaabshacaqGPbGaaeODai<br />% aabggacaqGTbGaaeyzaiaab6gacaqG0bGaaeyzaiaabccacaqGLbGa<br />% ae4CaiaabccacaqGLbGaaeiBaiaabccacaqGZbGaaeyDaiaabchaca<br />% qGYbGaaeyzaiaab2gacaqGVbGaaeOlaaqaaaqaaiaabIcacaqGIbGa<br />% aeykaiaabccacaqGtbGaaeyzaiaabggacaqGGaGaamOqaiabgAOinl<br />% abl2riHkaab6cacaqGGaGaamOqaiabgcMi5kabgwGiglaab6cacaqG<br />% GaGaamOqaiaabccacaqGHbGaae4yaiaab+gacaqG0bGaaeyyaiaabs<br />% gacaqGVbGaaeiiaiaabohacaqG1bGaaeiCaiaabwgacaqGYbGaaeyA<br />% aiaab+gacaqGYbGaaeyBaiaabwgacaqGUbGaaeiDaiaabwgacaqGUa<br />% GaaeiiaiaabofacaqGLbGaaeyyaiaabccacqaHXoqycqGHiiIZcqWI<br />% DesOcaqGGaGaaeizaiaabggacaqGKbGaae4Baiaab6cacaqGGaGaae<br />% 4uaiaabwgacaqGGaGaaeizaiaabwgacaqGMbGaaeyAaiaab6gacaqG<br />% LbGaaeiiaiaabwgacaqGSbGaaeiiaiaabogacaqGVbGaaeOBaiaabQ<br />% gacaqG1bGaaeOBaiaabshacaqGVbGaaeOoaaqaaiaabccacaqGGaGa<br />% aeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccaca<br />% qGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaa<br />% bccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaae<br />% iiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqG<br />% GaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabc<br />% cacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaadkeacqGHRaWkcqaH<br />% XoqycqGH9aqpdaGadaqaaiaadkgacqGHRaWkcqaHXoqycaGG8bGaam<br />% OyaiabgIGiolaadkeaaiaawUhacaGL9baaaeaaaeaacaqGOaGaaeyA<br />% aiaabMcacaqGGaGaaeiraiaabwgacaqGTbGaaeyDaiaabwgacaqGZb<br />% GaaeiDaiaabkhacaqGLbGaaeiiaiaabghacaqG1bGaaeyzaiaabcca<br />% caqG0bGaaeyyaiaab6gacaqG0bGaae4BaiaabccacaWGcbGaaeiiai<br />% aabogacaqGVbGaaeyBaiaab+gacaqGGaGaamOqaiabgUcaRiabeg7a<br />% HjaabccacaqGWbGaae4BaiaabohacaqGLbGaaeyzaiaab6gacaqGGa<br />% Gaae4CaiaabwhacaqGWbGaaeOCaiaabwgacaqGTbGaae4Baiaab6ca<br />% aeaacaqGOaGaaeyAaiaabMgacaqGPaGaaeiiaiaabseacaqGLbGaae<br />% yBaiaabwhacaqGLbGaae4CaiaabshacaqGYbGaaeyzaiaabccacaqG<br />% XbGaaeyDaiaabwgacaqGGaGaae4CaiaabwhacaqGWbGaaeikaiaadk<br />% eacqGHRaWkcqaHXoqycaqGPaGaaeypaiaabohacaqG1bGaaeiCaiaa<br />% bIcacaWGcbGaaeykaiaabUcacqaHXoqycaqGUaaaaaa!58AC!<br />\[<br />\begin{gathered}<br />  \boxed{{\text{P3}}{\text{. - }}} \hfill \\<br />  {\text{(a) Considere el conjunto }}A = \left\{ {1 - \frac{1}<br />{n}|n \in \mathbb{N}} \right\}{\text{. Encuentre el supremo de }}A{\text{ y pruebe que }} \hfill \\<br />  {\text{efectivamente es el supremo}}{\text{.}} \hfill \\<br />   \hfill \\<br />  {\text{(b) Sea }}B \subseteq \mathbb{R}{\text{. }}B \ne \emptyset {\text{. }}B{\text{ acotado superiormente}}{\text{. Sea }}\alpha  \in \mathbb{R}{\text{ dado}}{\text{. Se define el conjunto:}} \hfill \\<br />  {\text{                                                }}B + \alpha  = \left\{ {b + \alpha |b \in B} \right\} \hfill \\<br />   \hfill \\<br />  {\text{(i) Demuestre que tanto }}B{\text{ como }}B + \alpha {\text{ poseen supremo}}{\text{.}} \hfill \\<br />  {\text{(ii) Demuestre que sup(}}B + \alpha {\text{) = sup(}}B{\text{) + }}\alpha {\text{.}} \hfill \\ <br />\end{gathered} <br />\]


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+Quote Post
rock&mati
mensaje Jan 16 2009, 10:30 PM
Publicado: #2


Dios Matemático
Ícono de Grupo

Grupo: Usuario FMAT
Mensajes: 316
Registrado: 2-May 08
Desde: a la &$€¬~#@ del mundo
Miembro Nº: 21.900
Nacionalidad:
Colegio/Liceo: Liceo Jose Victorino Lastarria
Universidad: Universidad de Chile-FCFM
Sexo:



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WaaabCaeaadaqadaqaamaaDaaaleaacaWGRbaabaGaamOBaiabgkHi<br />% TiaaikdaaaaakiaawIcacaGLPaaacaGGOaGaaGymaiaacMcadaahaa<br />% Wcbeqaaiaad6gacqGHsislcaaIYaGaeyOeI0Iaam4AaaaakiaacIca<br />% caaIXaGaaiykamaaCaaaleqabaGaam4AaaaaaeaacaWGRbGaeyypa0<br />% JaaGimaaqaaiaad6gacqGHsislcaaIYaaaniabggHiLdaakeaacqGH<br />% uhY2caWGUbGaaiikaiaaigdacqGHRaWkcaaIXaGaaiykamaaCaaale<br />% qabaGaamOBaiabgkHiTiaaigdaaaGccqGHRaWkcaWGUbGaaiikaiaa<br />% d6gacqGHsislcaaIXaGaaiykaiaacIcacaaIXaGaey4kaSIaaGymai<br />% aacMcadaahaaWcbeqaaiaad6gacqGHsislcaaIYaaaaaGcbaGamajG<br />% gsDiBlacqc4GUbGamajGgwSixlacqciIYaWaiajGCaaaleqcqcyaia<br />% jGcGaKaoOBaiadqcOHsislcGaKaIymaaaakiadqcOHRaWkcGaKaoOB<br />% aiacqcOGOaGaiajGd6gacWaKaAOeI0IaiajGigdacGaKakykaiadqc<br />% OHflY1cGaKaIOmamacqcihaaWcbKaKagacqcOaiajGd6gacWaKaAOe<br />% I0IaiajGikdaaaaakeaacqGHuhY2caWGUbGaeyyXICTaaGOmamaaCa<br />% aaleqabaGaamOBaiabgkHiTiaaigdaaaGcdaqadaqaaiaaigdacqGH<br />% RaWkdaWcaaqaaiaacIcacaWGUbGaeyOeI0IaaGymaiaacMcaaeaaca<br />% aIYaaaaaGaayjkaiaawMcaaiabg2da9iacqc4GUbGamajGgwSixlac<br />% qciIYaWaiajGCaaaleqcqcyaiajGcGaKaoOBaiadqcOHsislcGaKaI<br />% ymaaaakmacqcyadaqaiajGdGaKaUaaaeacqcOaiajGikdacWaKaA4k<br />% aSIaiajGd6gacWaKaAOeI0IaiajGigdaaeacqcOaiajGikdaaaaacG<br />% aKaAjkaiacqcOLPaaacqGH9aqpcaWGUbGaeyyXICTaaGOmamaaCaaa<br />% leqabaGaamOBaaaakmaabmaabaWaaSaaaeaacaaIXaGaey4kaSIaam<br />% OBaaqaaiaaisdaaaaacaGLOaGaayzkaaaabaGaeyi1HS9aauIhaeaa<br />% daWcaaqaaiaad6gacqGHflY1caaIYaWaaWbaaSqabeaacaWGUbaaaO<br />% GaaiikaiaaigdacqGHRaWkcaWGUbGaaiykaaqaaiaaisdaaaaaaaaa<br />% aa!5F37!<br />\[<br />\begin{gathered}<br />  {\text{(ii)  }}\sum\limits_{{\text{k = 1}}}^n {\left( {_k^n } \right)k^2  = } \sum\limits_{{\text{k = 1}}}^n {\left( {_k^n } \right)(k + k(k - 1)) = \left[ {\sum\limits_{{\text{k = 1}}}^n {\left( {_k^n } \right)k + \sum\limits_{{\text{k = 1}}}^n {\left( {_k^n } \right)} k(k - 1)} } \right]}  =  \hfill \\<br />   \Leftrightarrow \left( {\sum\limits_{{\text{k = 1}}}^n {\frac{{n!}}<br />{{k!\left( {n - k} \right)!}} \cdot k} } \right) + \left( {\sum\limits_{{\text{k = 2}}}^n {\left( {_k^n } \right)k(k - 1)} } \right) \hfill \\<br />   \Leftrightarrow \left( {\sum\limits_{{\text{k = 1}}}^n {\frac{{n!}}<br />{{(k - 1)!\left( {n - k} \right)!}}} } \right) + \left( {\sum\limits_{{\text{k = 2}}}^n {\frac{{n!}}<br />{{k!\left( {n - k} \right)!}} \cdot k(k - 1)} } \right) \hfill \\<br />   \Leftrightarrow \left( {\sum\limits_{{\text{k = 1}}}^n {\frac{{(n - 1)!}}<br />{{(k - 1)!\left( {n - k} \right)!}}}  \cdot n} \right) + \left( {\sum\limits_{{\text{k = 2}}}^n {\frac{{n!}}<br />{{(k - 2)!\left( {n - k} \right)!}}} } \right) \hfill \\<br />   \Leftrightarrow \left( {\sum\limits_{{\text{k = 1}}}^n {\left( {_{k - 1}^{n - 1} } \right)}  \cdot n} \right) + \left( {\sum\limits_{{\text{k = 2}}}^n {\frac{{(n - 2)!}}<br />{{(k - 2)!\left( {n - k} \right)!}}}  \cdot n(n - 1)} \right) \hfill \\<br />   \Leftrightarrow \left( {n\sum\limits_{{\text{k = 1}}}^n {\left( {_{k - 1}^{n - 1} } \right)} } \right) + \left( {\sum\limits_{{\text{k = 2}}}^n {\left( {_{k - 2}^{n - 2} } \right)}  \cdot n(n - 1)} \right) \hfill \\<br />   \Leftrightarrow n\sum\limits_{{\text{k = 1}}}^n {\left( {_{k - 1}^{n - 1} } \right)}  + n(n - 1)\sum\limits_{{\text{k = 2}}}^n {\left( {_{k - 2}^{n - 2} } \right)}  \hfill \\<br />   \Leftrightarrow n\sum\limits_{k = 0}^{n - 1} {\left( {_k^{n - 1} } \right) + n(n - 1)} \sum\limits_{k = 0}^{n - 2} {\left( {_k^{n - 2} } \right)}  \hfill \\<br />   \Leftrightarrow n\sum\limits_{k = 0}^{n - 1} {\left( {_k^{n - 1} } \right)(1)^{n - 1 - k} (1)^k  + n(n - 1)} \sum\limits_{k = 0}^{n - 2} {\left( {_k^{n - 2} } \right)(1)^{n - 2 - k} (1)^k }  \hfill \\<br />   \Leftrightarrow n(1 + 1)^{n - 1}  + n(n - 1)(1 + 1)^{n - 2}  \hfill \\<br />   \Leftrightarrow n \cdot 2^{n - 1}  + n(n - 1) \cdot 2^{n - 2}  \hfill \\<br />   \Leftrightarrow n \cdot 2^{n - 1} \left( {1 + \frac{{(n - 1)}}<br />{2}} \right) = n \cdot 2^{n - 1} \left( {\frac{{2 + n - 1}}<br />{2}} \right) = n \cdot 2^n \left( {\frac{{1 + n}}<br />{4}} \right) \hfill \\<br />   \Leftrightarrow \boxed{\frac{{n \cdot 2^n (1 + n)}}<br />{4}} \hfill \\ <br />\end{gathered} <br />\]
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Juαn Arcøζ
mensaje Jan 16 2009, 10:44 PM
Publicado: #3


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pucha hiciste justo la que me sabia... lo otro no le hemos visto u.u...XD



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Juαn Arcøζ
mensaje Jan 16 2009, 10:45 PM
Publicado: #4


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---doble-post--

Mensaje modificado por Juan Arcos el Jan 16 2009, 10:49 PM


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kbzoon
mensaje Jan 16 2009, 11:14 PM
Publicado: #5


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La sumita la habia posteado hace un par de dias atrás en este enlace : http://www.fmat.cl/index.php?showtopic=38488


Lo anterior es lo mismo que hizo coquitao wink.gif


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Abu-Khalil
mensaje Jan 25 2010, 02:57 PM
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2a

TEX: \noindent Sea $\alpha\in\mathbb{R}$. Como la función $f(x)\equiv \alpha,\forall x\in\mathbb{R},$ existe y $\Phi(f)=\alpha$, se concluye lo pedido.

2bii

TEX: \noindent $A=\emptyset$ y $A=E$ transforman a $f$ y $g$ en la función identidad, respectivamente. Luego, $f,g$ son trivialemente sobreyectivas.

3a

TEX: \noindent Es claro que 1 es cota superior del conjunto. Probaremos que $\sup A=1$. Sea $\epsilon>0$, entonces existe $N\in\mathbb{N}$ tal que $\forall n\geq N$ vale que<br />\[\frac{1}{n}<\epsilon.\]<br />Así,<br />\[1-\left(1-\frac{1}{n}\right)=\frac{1}{n}<\epsilon.\]<br />

3b

TEX: \noindent En efecto, como $B$ es acotado superiormente, existe $M\in\mathbb{R}$ tal que $b\leq M,\forall b\in B$ y por axioma del supremo, existe una cota que es la menor. Análogamente, podemos ver que $b+\alpha<M+\alpha,\forall b\in B$, de lo cual sigue que $B+\alpha$ es acotado superiormente y sigue el mismo argumento.\\<br />\\<br />Sea $M=\sup B$ y $\epsilon>0$. Por como fue definido $M$, existe $b\in B$ tal que<br />\[M-b\leq \epsilon\Rightarrow \left(M+\alpha\right)-\left(b+\alpha\right)<\epsilon.\]<br />

Para la 2bi me parece que basta con notar que TEX: $A\subseteq A\cup X$ y TEX: $A\cap X\subseteq A$ o no?


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VA Jiménez
mensaje Jan 19 2012, 07:38 PM
Publicado: #7


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u.u ojalá nos hubieran tirado este control!
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gamby
mensaje Jan 19 2012, 08:56 PM
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CITA(VA Jiménez @ Jan 19 2012, 08:38 PM) *
u.u ojalá nos hubieran tirado este control!

súbanse el control pos
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VA Jiménez
mensaje Jan 20 2012, 01:08 AM
Publicado: #9


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CITA(gamby @ Jan 19 2012, 09:56 PM) *
súbanse el control pos

****, no sé Latex !
He intentado con este http://www.codecogs.com/latex/eqneditor.php pero no cacho cuál de todos los "links" disponibles tengo que pegar en el foro xD.
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