Alapadatok
Év, oldalszám:2012, 46 oldal
Nyelv:magyar
Letöltések száma:8
Feltöltve:2024. április 06.
Méret:1 MB
Intézmény:
[BCE] Budapesti Corvinus Egyetem
[ELTE] Eötvös Loránd Tudományegyetem
Megjegyzés:
Csatolmány:-
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¸ª«f¡» û 1µ ¬¨ 1 T +1 1 1 2 1 è è è −1 e−ρ E (wt + Ft (x, ) − Gt+1 (Ft (x, ), ))−γ Mt−1 | Ht = (wt−1 + x)−γ Mt−1 µ3«ô¡ F (x) Sr«ÀÓ[£»£¡l®r¶ªrª»ü3®)¯ t = 1þ¹ô¡1r«Àµ¡ª¡6rª¿À¡ª¦ýÂt®f3¬»¨[£«À¦¬N3µ SÂr¶/3¡»º3û3®7¬ÔR¸ªå1® û 3µ 1 1 (w1 + F1 (x))−1 E e 1 H1 M1 ä¹13r´£¡6¶ªr3Ó3¨¦û F (x)þ ¡®fÀ÷iø¸ª¨¦û)®7¬ÔR¸ªå3®)û 3µ 1 = (w0 + x)− 2 F1 1 . M0 1 ³ª¡ x 1r«ÀµÂr¶ª w0 + x 2 E F1 (x) = e2 x − G1 (x) þ¹¡¿À¶ª¦¬Ò M0 H1 M1 1 + x − G1 (x) 2 F1 (x − G1 (x)) = E e2 Ò ø Ä&wrµr1«¡õ¬«¬ÔR¸»£ ÷i«¿À¶$[¬¡i¸ªå3® G1 (x) = E G1 (x) þ ¡ M1 1 + x − G1 (x) 2 E
M0 e2 x¡»¶»3Ó3r¨ô¦±Ârõ¬¾SÂr¶/3¡»º»Âr®¡õ®f[¬ª¨r[£«À¦¬Ò − w1 . M0 H1 . M1 M0 H1 M1 G0 . ~ E G1 (x) = 2 M0 M1 H1 E M1 M0 e2 + E 2 E M0 H1 M1 M0 M1 G0 + E H1 E M0 M1 H E G 1 0 M1 M0 õ¬¡¤£¶»¨¨rå3® SrS¬ ®3®f¶»Â¡¦£¶$3¬¡6½¾Âr¶§¡6®r®¡f M1 x M0 G0 . 1 · 0.998458 + 1 · 000152878 = 0.998458 + 1 1 · 0.998458 + 1 · 000152878 + 0.0000132 + 1.2 1 · 0.00152878 + 1 · 000152878 + 0.998458 = 0.9998, 1.1 ßÖ H1 þ E M1 H1 M0 1 = 0.998458 + 1 · 0.998458 + 1 · 000152878 1.2 + 0.0000132 + 1 · 0.998458 + 1 · 000152878 1.1 = 1.000168 + 0.998458 1 · 0.00152878 + 1 · 000152878 ʨr®ô¡ýc£¶w¦Rª»¨r¬¾¡»å3Ó7¸ªå3®u3«Àµ¡»¡»rª¿À¡63%®¶6£ ×3×
%rµr3«À¡63® × û)r¨Âr¶§¡ G1 (x) = 0.9999862 · 1000168 + 0999986 · E ä¹13r£¶t÷ir«¿À¶$[¬7¡6½¾¬ ÷iü3¦ÂfµÀf 0.9999862 · 1000168 + e2 H1 (x) M1 x M0 G0 = 0.1192+01192·E M1 x G0 = H1 (x) = 0.1353 1 + x − 01192 + 01192 · E M0 M1 = 0.1353 08808 + x − 01192 · E x G0 M0 ³ª¡)[¬¡6£¶»¨¨rå3® S rF¬kÔ.Â3¨rü3µá¡»º® ÷i«Àµ¬c¬¡6¡r¯îù,å3Ó¸ªå3®)ûG3µ ¯ X = H (X , ) X0 = 0 M1 x G0 . M0 ûýÂr3µ è 1 1 0 X1 = H1 (X0 ) = 0.1353(08808 + 0 − 01192 · 0) = 01353 · 08808 = 01192 õ¬¡¤£¶»¨¨rå3® SrS¬¨QÔ¡6c£«ÀT÷iµK¬»¨¡6£»÷i«ÀµK¬¬¡»¡·¬ Ùà&x¬«P¬ÔR¸»£7 c0 = 1 + 0 − 0.1192 · 1000168 = 088 c1 = 0 + 0.1192 − 0 = 01192
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¨:«»ºÂ¦¦ÂÂr ¬S½RÓ3»¿¢¡6¡ª¡A¸»£¶6¬Ó3Âr®K÷i«¢µK¬c¬¡õc [¬¡6£¶»¨r£6¬¾rRþ 1Âr« × Kµ«å3«¢¡¤¬ ×3× ¯ Ò Ô3¶ª × «ÀÂrc£¡<¬¨Q¸ªr«ÀrK¡6Hûg3µ ¬¨ r«À»º Âô¦¦ÂrÂY1r¡6å3Ó7¸ªå1® × À¨¡»»¶»¬ ¬¨¡QS3Ó[¬1 û 1µ «Àrª¨ x S«À«½ ,¡á®fd[¨rô¡6Ârªü33®)9û 3ª¨rr¦ [¬ 3ï¡»¼¶ª¡»Âr3À®´» [¬«£«U»r × ¬«»¡rû)¬®f®¶·3À3Þr·®fl3¨r¡»Âr¯ ʨÂr¶ª¡õ¡ª¡¬ ¦£/¶ [¬¡»½S®fd[¨rô¡6ÂrÂr¶§¡6®¦r«U»¨r£c«å33®)g¯ "l¬µÀ :X i i i i WX w̃0 = 1 + Π0 , w̃1 = −(0.998458 · 0 + 00000132 · 5 + 000152878 · 3) = −000465 ß ³
ª¡ Ô.Ó3u®¨rÓ3ô¡0¸ªü3®¬¨Y3Ó1å3®K¡6¿À¦§¡6¶ªå3®K¡6æ3¶»¬u÷ir«ÀÂrÔ3¿¢¡6»¡ ¬ «Àr«¹¸ªÂr¶»º« û ¬uc½RÓ3ª¿À¡6¡»¡ ¸»£¶6¬Ó3®F÷i«¢µK¬c¬¡»¡Ø3¬»¨[£«À¦¬1¯ Ê«ªº«ÂÔ.”¼L¦ô¡6®¨rºFrµr3«À¡ F (x) µ¼®Â3r® Srþ ¡6¬«P£«P£»Ò¬ 1 ³À¦« F 1 1 (w̃1 + F1 (x))− 2 E e 1 H1 1 H1 M1 þ c¶$3ô¡6º3û[¬ ®¼¦¡»®r¨º ¬Ó3½RÓ3À®¡ E2 M0 M1 1 = (w̃0 + x)− 2 · 1 . M0 H1 (w̃0 + x) − w̃1 , e2 x + 1 + Π 0 2 M0 F1 (x) = E H1 + 0.00465 = 01353(x + 1 + Π0 ) + 000465 e2 M1 F1 (x) = 乪cÂô¡ x 3 «ÀµÂ¶» x − G1 (x) þ ¡{3«Àµô¡»¡»r»¿¢¡»¦h x + 1 + Π0 − G1 (x) 2 M0 F1 (x − G1 (x)) = E H1 +
0.00465 e2 M1 = 0.1353(x + 1 + Π0 − G1 (x)) + 000465 úb¸ª¶6¬ª¨>£S«Ø¸ªå3® G1 (x) þ¹ô¡·H 0.1353E G1 (x) = 1 + x + Π0 ) G0 + 0.00465E M G 0 M0 . M1 1 + 0.1353 · E M G 0 0 M1 (1 M0 tf3«Àµô¡»¡»r»¿¢¡»¦¬c£¶±®¶6£ ×3× ¬Ã®fÀ»¨>£c«¢¡á¦£¶$[¬7¡6½´Â¶ª¡»Âr®®ô¡>û G (x) µR»¨r¶»Á ×3× ¬«P¬®f¶»¬ 1¨f[¬7¡6½ h 1 [«¿¶N¸ªå3® X þ¹¡Àf G1 (x) = 0.1192E H1 (x) M1 (1 + x + Π0 ) G0 M0 + 0.0041 M1 H1 (x) = 0.1353 1 + x + Π0 − 01192E G0 + 0.0041 + 0.0041 = M0 M1 = 0.1353 x + Π0 − 01192E (1 + x + Π0 ) G0 + 0.1388 M0 ³r [¬7¡¤£¶ª¨r¨å3®¬±Ô.Â3¨rü3µ¡6º®q÷i«Àµ¬c¬¡6¡rûK÷ir«3¬»¨[£«À¦¬1û 3µ X1 Âr X0 = 0 X1 = H1 (X0 ) M1 = H1 (X0 ) =
0.1353 0 + Π0 − 01192E (1 + 0 + Π0 ) G0 + 0.1388 = M0 = 0.119Π0 + 01226 ³ª¡c£¶X»¨r£c¿¢¡%[¬7¡6½¾¬¨áÔ¡6c£«ÀT÷iµK¬»¨¡6£»%÷i«¢µK¬c¬¡wf c0 = 1 + Π0 − (0.1191Π0 + 01226) · 1000168 = 08808Π0 + 08774 c1 = Y1 + (0.1191Π0 + 01226) − 0 = 00091Π0 + 012725 ÕKà ³ r [¬7¡¤£¶ª¨r¨å3®¬¨:¶»rÓ1¡6%þ ¡»º®Ô3P¬ÞrmcRÓ3«« × rª¨>£S«À¡§þlÔ1¡»c£«Àw÷iµ¬»¨ô¡¤£ªU÷i«¢µK¬c¬¡ 3¬»¨3»»£K£¡u§ÄÙh°¬«¬ÔR¸»£Dû,c¬¸ªÓ r¨r® ¬«P¬ÔR¸»£æ3µâ3¬¡¤£¶ª¨r¨å3® c ¬ Ó1¿â¸»¬¡rû:3µ¬ Π ×
¨¡»»¿¢¡¤£ªfÔ3P¬Þ<ª¨>£S«À¡UÔ¡6c£«À÷iµ¬»¨ô¡¤£ª÷i«¢µK¬c¬9¡ [¬ª¨r3»6£K¬ýr¨¨rr«crµr¨¨r¯ 0 U (c) = 1 X 1 2 c e−t 1t 2 t=0 √ √ c1 0.1192 = 2 c0 + 2 = 2 0.88 + 2 = 2.141 e e √ √ Ò »¡ º®Ô3P¬Þ3SÓ3«« × »¨>£c«¢¡ÊÔ1¡»c£«ÀU÷iµK¬ª¨¡6£»3÷i«¢µK¬c¬¡Ar»ô¡6 ¬N[¬»¨3ª6£ 2.141 µK¬3r¨ô¡:¬3¬»¨3»»£¡°®r««U¡6H[£¡Qr«ÀÂr¶ªRü13®õ< q¬«P¬ÔR¸»£ æ3µ û73µ Ó3¿Ø¸»¬¡ýª¨rrÓ1ü33® × Â·®£¶»®fl[¨¡»Âr»3® «Àf 3ô¡63®)¯ | p 0.1191Π0 + 012725 2 0.8808Π0 + 08774 + = 2.141 e
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¸ª«f¡» û 1µ ¬¨ 1 T +1 1 1 2 1 è è è −1 e−ρ E (wt + Ft (x, ) − Gt+1 (Ft (x, ), ))−γ Mt−1 | Ht = (wt−1 + x)−γ Mt−1 µ3«ô¡ F (x) Sr«ÀÓ[£»£¡l®r¶ªrª»ü3®)¯ t = 1þ¹ô¡1r«Àµ¡ª¡6rª¿À¡ª¦ýÂt®f3¬»¨[£«À¦¬N3µ SÂr¶/3¡»º3û3®7¬ÔR¸ªå1® û 3µ 1 1 (w1 + F1 (x))−1 E e 1 H1 M1 ä¹13r´£¡6¶ªr3Ó3¨¦û F (x)þ ¡®fÀ÷iø¸ª¨¦û)®7¬ÔR¸ªå3®)û 3µ 1 = (w0 + x)− 2 F1 1 . M0 1 ³ª¡ x 1r«ÀµÂr¶ª w0 + x 2 E F1 (x) = e2 x − G1 (x) þ¹¡¿À¶ª¦¬Ò M0 H1 M1 1 + x − G1 (x) 2 F1 (x − G1 (x)) = E e2 Ò ø Ä&wrµr1«¡õ¬«¬ÔR¸»£ ÷i«¿À¶$[¬¡i¸ªå3® G1 (x) = E G1 (x) þ ¡ M1 1 + x − G1 (x) 2 E
M0 e2 x¡»¶»3Ó3r¨ô¦±Ârõ¬¾SÂr¶/3¡»º»Âr®¡õ®f[¬ª¨r[£«À¦¬Ò − w1 . M0 H1 . M1 M0 H1 M1 G0 . ~ E G1 (x) = 2 M0 M1 H1 E M1 M0 e2 + E 2 E M0 H1 M1 M0 M1 G0 + E H1 E M0 M1 H E G 1 0 M1 M0 õ¬¡¤£¶»¨¨rå3® SrS¬ ®3®f¶»Â¡¦£¶$3¬¡6½¾Âr¶§¡6®r®¡f M1 x M0 G0 . 1 · 0.998458 + 1 · 000152878 = 0.998458 + 1 1 · 0.998458 + 1 · 000152878 + 0.0000132 + 1.2 1 · 0.00152878 + 1 · 000152878 + 0.998458 = 0.9998, 1.1 ßÖ H1 þ E M1 H1 M0 1 = 0.998458 + 1 · 0.998458 + 1 · 000152878 1.2 + 0.0000132 + 1 · 0.998458 + 1 · 000152878 1.1 = 1.000168 + 0.998458 1 · 0.00152878 + 1 · 000152878 ʨr®ô¡ýc£¶w¦Rª»¨r¬¾¡»å3Ó7¸ªå3®u3«Àµ¡»¡»rª¿À¡63%®¶6£ ×3×
%rµr3«À¡63® × û)r¨Âr¶§¡ G1 (x) = 0.9999862 · 1000168 + 0999986 · E ä¹13r£¶t÷ir«¿À¶$[¬7¡6½¾¬ ÷iü3¦ÂfµÀf 0.9999862 · 1000168 + e2 H1 (x) M1 x M0 G0 = 0.1192+01192·E M1 x G0 = H1 (x) = 0.1353 1 + x − 01192 + 01192 · E M0 M1 = 0.1353 08808 + x − 01192 · E x G0 M0 ³ª¡)[¬¡6£¶»¨¨rå3® S rF¬kÔ.Â3¨rü3µá¡»º® ÷i«Àµ¬c¬¡6¡r¯îù,å3Ó¸ªå3®)ûG3µ ¯ X = H (X , ) X0 = 0 M1 x G0 . M0 ûýÂr3µ è 1 1 0 X1 = H1 (X0 ) = 0.1353(08808 + 0 − 01192 · 0) = 01353 · 08808 = 01192 õ¬¡¤£¶»¨¨rå3® SrS¬¨QÔ¡6c£«ÀT÷iµK¬»¨¡6£»÷i«ÀµK¬¬¡»¡·¬ Ùà&x¬«P¬ÔR¸»£7 c0 = 1 + 0 − 0.1192 · 1000168 = 088 c1 = 0 + 0.1192 − 0 = 01192
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¨:«»ºÂ¦¦ÂÂr ¬S½RÓ3»¿¢¡6¡ª¡A¸»£¶6¬Ó3Âr®K÷i«¢µK¬c¬¡õc [¬¡6£¶»¨r£6¬¾rRþ 1Âr« × Kµ«å3«¢¡¤¬ ×3× ¯ Ò Ô3¶ª × «ÀÂrc£¡<¬¨Q¸ªr«ÀrK¡6Hûg3µ ¬¨ r«À»º Âô¦¦ÂrÂY1r¡6å3Ó7¸ªå1® × À¨¡»»¶»¬ ¬¨¡QS3Ó[¬1 û 1µ «Àrª¨ x S«À«½ ,¡á®fd[¨rô¡6Ârªü33®)9û 3ª¨rr¦ [¬ 3ï¡»¼¶ª¡»Âr3À®´» [¬«£«U»r × ¬«»¡rû)¬®f®¶·3À3Þr·®fl3¨r¡»Âr¯ ʨÂr¶ª¡õ¡ª¡¬ ¦£/¶ [¬¡»½S®fd[¨rô¡6ÂrÂr¶§¡6®¦r«U»¨r£c«å33®)g¯ "l¬µÀ :X i i i i WX w̃0 = 1 + Π0 , w̃1 = −(0.998458 · 0 + 00000132 · 5 + 000152878 · 3) = −000465 ß ³
ª¡ Ô.Ó3u®¨rÓ3ô¡0¸ªü3®¬¨Y3Ó1å3®K¡6¿À¦§¡6¶ªå3®K¡6æ3¶»¬u÷ir«ÀÂrÔ3¿¢¡6»¡ ¬ «Àr«¹¸ªÂr¶»º« û ¬uc½RÓ3ª¿À¡6¡»¡ ¸»£¶6¬Ó3®F÷i«¢µK¬c¬¡»¡Ø3¬»¨[£«À¦¬1¯ Ê«ªº«ÂÔ.”¼L¦ô¡6®¨rºFrµr3«À¡ F (x) µ¼®Â3r® Srþ ¡6¬«P£«P£»Ò¬ 1 ³À¦« F 1 1 (w̃1 + F1 (x))− 2 E e 1 H1 1 H1 M1 þ c¶$3ô¡6º3û[¬ ®¼¦¡»®r¨º ¬Ó3½RÓ3À®¡ E2 M0 M1 1 = (w̃0 + x)− 2 · 1 . M0 H1 (w̃0 + x) − w̃1 , e2 x + 1 + Π 0 2 M0 F1 (x) = E H1 + 0.00465 = 01353(x + 1 + Π0 ) + 000465 e2 M1 F1 (x) = 乪cÂô¡ x 3 «ÀµÂ¶» x − G1 (x) þ ¡{3«Àµô¡»¡»r»¿¢¡»¦h x + 1 + Π0 − G1 (x) 2 M0 F1 (x − G1 (x)) = E H1 +
0.00465 e2 M1 = 0.1353(x + 1 + Π0 − G1 (x)) + 000465 úb¸ª¶6¬ª¨>£S«Ø¸ªå3® G1 (x) þ¹ô¡·H 0.1353E G1 (x) = 1 + x + Π0 ) G0 + 0.00465E M G 0 M0 . M1 1 + 0.1353 · E M G 0 0 M1 (1 M0 tf3«Àµô¡»¡»r»¿¢¡»¦¬c£¶±®¶6£ ×3× ¬Ã®fÀ»¨>£c«¢¡á¦£¶$[¬7¡6½´Â¶ª¡»Âr®®ô¡>û G (x) µR»¨r¶»Á ×3× ¬«P¬®f¶»¬ 1¨f[¬7¡6½ h 1 [«¿¶N¸ªå3® X þ¹¡Àf G1 (x) = 0.1192E H1 (x) M1 (1 + x + Π0 ) G0 M0 + 0.0041 M1 H1 (x) = 0.1353 1 + x + Π0 − 01192E G0 + 0.0041 + 0.0041 = M0 M1 = 0.1353 x + Π0 − 01192E (1 + x + Π0 ) G0 + 0.1388 M0 ³r [¬7¡¤£¶ª¨r¨å3®¬±Ô.Â3¨rü3µ¡6º®q÷i«Àµ¬c¬¡6¡rûK÷ir«3¬»¨[£«À¦¬1û 3µ X1 Âr X0 = 0 X1 = H1 (X0 ) M1 = H1 (X0 ) =
0.1353 0 + Π0 − 01192E (1 + 0 + Π0 ) G0 + 0.1388 = M0 = 0.119Π0 + 01226 ³ª¡c£¶X»¨r£c¿¢¡%[¬7¡6½¾¬¨áÔ¡6c£«ÀT÷iµK¬»¨¡6£»%÷i«¢µK¬c¬¡wf c0 = 1 + Π0 − (0.1191Π0 + 01226) · 1000168 = 08808Π0 + 08774 c1 = Y1 + (0.1191Π0 + 01226) − 0 = 00091Π0 + 012725 ÕKà ³ r [¬7¡¤£¶ª¨r¨å3®¬¨:¶»rÓ1¡6%þ ¡»º®Ô3P¬ÞrmcRÓ3«« × rª¨>£S«À¡§þlÔ1¡»c£«Àw÷iµ¬»¨ô¡¤£ªU÷i«¢µK¬c¬¡ 3¬»¨3»»£K£¡u§ÄÙh°¬«¬ÔR¸»£Dû,c¬¸ªÓ r¨r® ¬«P¬ÔR¸»£æ3µâ3¬¡¤£¶ª¨r¨å3® c ¬ Ó1¿â¸»¬¡rû:3µ¬ Π ×
¨¡»»¿¢¡¤£ªfÔ3P¬Þ<ª¨>£S«À¡UÔ¡6c£«À÷iµ¬»¨ô¡¤£ª÷i«¢µK¬c¬9¡ [¬ª¨r3»6£K¬ýr¨¨rr«crµr¨¨r¯ 0 U (c) = 1 X 1 2 c e−t 1t 2 t=0 √ √ c1 0.1192 = 2 c0 + 2 = 2 0.88 + 2 = 2.141 e e √ √ Ò »¡ º®Ô3P¬Þ3SÓ3«« × »¨>£c«¢¡ÊÔ1¡»c£«ÀU÷iµK¬ª¨¡6£»3÷i«¢µK¬c¬¡Ar»ô¡6 ¬N[¬»¨3ª6£ 2.141 µK¬3r¨ô¡:¬3¬»¨3»»£¡°®r««U¡6H[£¡Qr«ÀÂr¶ªRü13®õ< q¬«P¬ÔR¸»£ æ3µ û73µ Ó3¿Ø¸»¬¡ýª¨rrÓ1ü33® × Â·®£¶»®fl[¨¡»Âr»3® «Àf 3ô¡63®)¯ | p 0.1191Π0 + 012725 2 0.8808Π0 + 08774 + = 2.141 e
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Módszertani útmutatónkból megtudod, hogyan lehet profi szakdolgozatot készíteni. Foglalkozunk a diplomamunka céljaival, a témaválasztás nehézségeivel, illetve a forrásanyagok kutatásával, szakszerű felhasználásával is. Szót ejtünk a szakdolgozat ideális nyelvezetéről és struktúrájáról és a gyakran elkövetett hibákra is kitérünk.
