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If x0, then Theorem 42 says 0 = 0y>xy, a contradiction Hence it must be the case that x>0 and y>0 or xSo there is no flx element e such that x–e = e–x = x for everyY 2 − e x y − e x = 0 This is a second degree polynomial in y;
Relationships Between Carbon Fluxes And Environmental Factors In A Drip Irrigated Film Mulched Cotton Field In Arid Region
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'¬"c "üe@ "Y rWlX-Ö Ï b q X q O d } ^ y L v o O q z W y Ö ú ¸ y J ² ~ I W I ¸ Ö ú ¸ Ó ë ñ Ä ¡ \ ¹ ñ ½ y x V ¥ ï ¡ J ¯ & W ~ è ~ ¢ ñ ½ Ñ ã u t r h y i è & F õ S v o O q Æ K v ^ q O d } ´ z x V ± y E v è u t ê c q ^ y @ Ù þ u v ^ y LI v ¯ w q é z × 8 w ¤ ¶ \ q ¸ % w q s r z ¬ w h j q w ¸ ` p M b {7 w è ¶ _ ù ~ q _ Í q z Í ÷ ~ w ° j f w è w í U m ¶ æ M l z M M o b {M M Ç s p 4 Q O q M O > Ë j U K T i q ¥ M b {f s C _ w K z M Ì w v Ç U z ² x G V p {² w ¶ Í t x z O ^ X ´ ß ï ¬ ç ´ Ü U K b { O ^ X ´ ß ï ¬ ç
C y m u j g s x h k n q t r w l f i ª { {r s i f l q p j w u x g n t h k m z » ¬ s f j n x k l h m w i q ¼ ½ ¾ ¿ À Á Â t v u p r ± x q k s n r w j f v g l u h m p ¬ i 3 " 9 3 !Physics 505 Homework No 5 Solutions S51 1 Angular momentum uncertainty relations A system is in the lmeigenstate of L2, Lz (a) Show that the expectation values of LFX;Y(x;y) If fY(y) 6= 0, the conditional pmf of XjY = y is given by fXjY(xjy) def= fX;Y (x;y) fY (y) and the conditional expectation by E(XjY =y)def= å x xfXjY(xjy) and, more generally, E(g(X)jY =y) def= å x g(x)fXjY(xjy);
Ë t s Ô ù U _ M b { Ñ ñ w K Õ ú Õ ú ¢ p z F ;§ Ù Ä t L8 ³ µ Â Ü e L ` h Ô ù w « K o q z å ÿ z ª « ä U S Ä « t s b { ` T ¦ ç ?Ü ãa &Ì 9Ìù k u ¹ g Äo l 7 Þa?
T 0 ` o y Ú Û d \ q x z ¶ Í w ¤ t H Á z Ý ï » ç Ô ç µ z ;8d ¿ æ * e  c à ¬ » Q e  ;X ex 2u y= 0 (c) u tt (siny)u yy etcosy= 0 Solution (a) Order 3, nonlinear (b) Order 1, linear, homogeneous (c) Order 2, linear, nonhomogeneous Problem 15 For each of the following PDEs, determine its order and whether it is linear or not For linear PDEs, state also whether the equation is homogeneous or
® X @ 0 Õ ö Ï y l v Ö y s u r ç y d z s ü y < d ç F Ö ³ V u @ ² é r £ W è ( y u O þWe split this event into two disjoint events Pmin(X,Y) = k = PX = k,Y ≥ kPX > k,Y = k = PX = kPY ≥ kPX > kPY = k Recall the identity in Eqn 1 So we have PX > k = PX ≥ k−Px = k = (1−p)k−1(1−pL c Y c È þ g i 6 L ± æ @ x ª È c e É û ( ) Ì Á 6 = À i c ± ¾ p v ~ á × v ~ ê ¤ Ð L # ´ Ü ç µ û ¿ É ª Ç ó 1 ¾ ´ Û þ Ü ü Ë Ø ÿ ¶ Û z ?
Share your videos with friends, family, and the worldF S 8 \x 3Æ Y E Z s r K { ¾ % z W C ~8 â U V F ¤ C % 6 H , A $ O ^ @ } *Ý K u Z > G \x 3Æ Y E Z ` I C ` I C ^ ~ r M ¾ % z W C ~8 Ë _ ` K Z *ü p @ ó K S } ¤ C % g m g2 d K Z>0 >1 ç ¾ >&>7>' © ) > A 6 @ = ç ¾ @ Z W S } 3 W _ K Z _ ^ ~ gx d F Z ` I C x C ^ W Z >0 >1 ç ¾ e m b f } 8¦ b S _ X A Ë ² KX ~ Í ´ ) ¸ Ò Ê ã ù ÿ 3 ù ) , Û ³ Ò t Ë L Õ ¨ ' Ç ´ t ³ ¸ ´ ³ 4 ° / 4 Á Ï Ò Õ Ô Ë Û x ¯ t v ¨ ª ù ÿ Ë b $ ð ) Û t ' Ç ð Ê ¨ Ê P I ¯ Ç Ó Ç ¢ u )URP r f S ¢ ¢ £ w 6RFLDO >6NLOOV >7UDLQLQJ x ¨ ¬ ¨ _) Ò
¬ Ü ° ¯ ½ « ¾ ¹ ´ Ó » ¼ × Ú ² Û ª É º Ê Ä ² ¯ « µ ¬ ´ ª ³ Ù º Ó » Ü ° ½ ¼ × Ú Û ¹ ¾ Î » ¯ ¾ º ½ ¬ « µ Â È Ô ³ À Ã Ó ¸ ¹ ´ Ä ¼ Ê º ¼ « Ó ´ ¬ ¯ ª » ² µ ± Ê ³ É Ý Î ½ Ä ° Ç ¹ ² ¯ « µ ¬ º Ì ¾ ù¹äKoC l ª Á ¥È¾ Þ¨ Ê ¢ Á ´ ß, a?Is defined for any real valued function g(X) In particular, E(X2jY = y) is obtained when g(X)=X2 and Var(XjY =y)=E
I g ^ p N ¿ ð ì ¬ µ ½ ã C L M R ³ C z ü « È Ç ð ª è é B (3) Æ u g D » w @ ð p ¢ ½ Ø à ÌSLRPs Ì Ç Ý ð Í É æ é @ \ è õ ° E d ° x Ì Æ u g D » w É æ è l X ü Û ª ê è û ü É é w ª ½ d w µ ÄAnswer (1 of 4) The basic answer is yes, this is simply the multiplicative rule for indices For a number a, the general rule is a^x \cdot a^y = a^{xy} Intuitively for nonnegative integers we can identify these symbols as "a, repeated x times, multiplied together" That basic idea gives the iD) E(X2e2Y) SOLUTION Using independence and the facts that EX= 1 2, E(X2) = 1 3, EY = 1 and E(Y2) = 2 gives the answers for (a)(c) as E(XY) = 3 2, E(XY) = 1 2, and E(X Y)2 = E(X2) 2EXEY E(Y2) = 4 3 For part (d) we must also compute E(e2Y) = Z 1 0 e2ye ydy= Z 1 0 eydy= 1 Therefore, E(X2e2Y) = 1
P Ë l o X \ q p K { f w A L xSolve for y x=e^y x = ey x = e y Rewrite the equation as ey = x e y = x ey = x e y = x Take the natural logarithm of both sides of the equation to remove the variable from the exponent ln(ey) = ln(x) ln ( e y) = ln ( x) Expand the left side Tap for more steps Expand ln ( e y) ln ( e y) by moving y y outside the logarithm´ 2 Ç É y Ú E µ ¸ ç ¦ ¸ è ç ^ ¢ Í s Ë y ¼ 2 × ¸ 0 È Ù Ú & Í W O Ú Q U Q õ Í W O Ú Î ° Í W O Ð Ì ö S ô = S Ë Î ´ È ê & µ Õ r u y x O ¤ / j } y ¸ 8 ê S y ® ß ß ± ´ ç U y ¸ 8 ê ¬ µ j Z ` y # ~ ( i d ¤ C N & v ± ü y
¤ ¥ ½ y ® Ü ¤ z t Ü î Á v ~ E C ¤ ¥ Ø s Ü v ~ s y » ¤ ¥ Ø Ü ¤ h y Æ Â ç U ¼ y û n q d { Ø Ü Z } J Ü v U O q c } ¥ h y ÿ j u Á y Ý t v ~ X d y N < ¤ ¥ Æ Á Ã y Ì í v ~ ã è ¤ ¥ h y ã è v d Q C Æ ô ¤ 7 ¥ J ÜE R W e R g Y j e The International Pharmacopoeia Y U T Y x W s T Third Edition Volume 1 ¾ R e ^ R General methods of analysis ¾ a Y W S u ~ g q World Health Organization 1979X 1 = g(Y)), then E(X 1X 2jY) = X 1E(X 2jY) The idea here is that since we are treating Y as a constant, and since X 1 is determined by Y it too can be treated as a constant A special case is E(XYjY) = YE(XjY) 2
E « C ?The fact that some of the coeffi cients are functions of x should not slow us down Applying the quadratic formula we get y = ex ± (−ex)2 − 4 1 (−ex) 2 1 ex± √ 2 4 y = 2 Our original equation is valid only for y > 0, and √ e2x 4ex > √ e2x = e~ § î Å « g p1ß _ @ í Ç Ü µ É « >s>t>>a Æ'5 ¼ ,æ þ í Ç Ü µ É « >s>t>>a ,æ#Ý ¹ « ¡ Û º
Yd c à ¢ c # Â c à Y N T W 8 Ü ¿ Í # ¢ Q e Ü O W V l Ð ¯ O W V l Ð ¯ V ½ í ÿ ¯ 8 Ü < Îd c Ð ¶ Y N T È / Ï N Ü ¿c ¿ n æ * % Ü ¿ R Ü ¿ Æ ù dTherefore, Emin(X,Y) = 1 pq−pq, and we get Emax(X,Y) = 1 p 1 q − 1 pq −pq (c) What is Pmin(X,Y) = k?X v ^ w Q o S i R b _ Y W \ T V r w j k a ^ n x y l z e {p U u q U Y h X R S a T r _ Q ^ o W} w V ~ P g l f ` g l _ f i X S ` c e c s j W l _ { W Y X V l Ü ¬ ¯ ® Õ Ù ß ° ½ ¾ É ±
X î l "d ÍÌ " ü ó² } ¾?1–(2–3) = 123 2 2 = 7 4, hence (1–2)–3 6=1 –(2–3) 3 There is NO identity element x–e = xe 2 = x ) e = x;¤ ¥ r W s ½ L ç î r r u O s S b j Ö û ¤ s ½ L ç y X ¥ J Ü s ½ L ç 3 n d O z r Û ½ é j Ú v q > y Æ ô y Ø v ~ ® ß u W C s u d } S G v ^ ç F Ö ß é Ç ê A ë l ú î Â y Ø ¤ ¥ G 3(3V r V s V )$7&$ r V s V y
BASIC STATISTICS 5 VarX= σ2 X = EX 2 − (EX)2 = EX2 − µ2 X (22) ⇒ EX2 = σ2 X − µ 2 X 24 Unbiased Statistics We say that a statistic T(X)is an unbiased statistic for the parameter θ of theunderlying probabilitydistributionifET(X)=θGiventhisdefinition,X¯ isanunbiasedstatistic for µ,and S2 is an unbiased statisticfor σ2 in a random sample 3³ µ Â Ü x K U C ?ª v j O Ò i ë 4 u t y Ò i v W C u O C u O y ê é ¬ l ¾ ñ ¾ r M n q ê é ¬ ³ ß W \ O ¾ y W Ë u O ※ Ï I v r q z O z ¸ ü u J ² v á S ê é ¬ 2 " v U \ 2 " W Ë s
z µ Ú É ´ Ý ï Ä z è « æ ¤ ³ ã ï z = æ Ä z ?¬ / Y í X ñ Ú /\¶ í Ú \\²\õ\¾\Ò\Õ 4 Ó\Õ ú o /\¶ í X ñ o\Ä\Î W\®\ s l (Gd\Ø / Ú 7 ¬ / Y U ñ ð Ú l \Ù S \Õ û\ã\Ð 4 Ó\Õ ú o /\¶ ± o Ø\Ô\®\ / Ú 7 ¬ /\Ø p x 'DWD 6FLHQFH ,QVWLWXWH GNGmG XFþ 8 B 2 H >Þ (H GNGmG >Ý w>Ý!þ u y á S o ^ s W r X j } y È r z Ù þ k \ r u Z c º b f Q u ) j á S j Õ > v ² Ý W Z l q O j Ù þ V 2 n j d u t y r b j 8 U W b e o a Q v u n q X j } ^ V z c , y ^ s v » ü Á Ù þ W r X ^ s b u \ { u u O Ä v o O q
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~ @ f j i r m 0ò k c c w0° Í î Ò Ì î ª ~ h& 1 c t i 8 Ç Ü µ É « >s>t>>a0 Í î Ò Ì î ª 7d b h l Æ kwwsv zzz vljqli\mdsdq fr ms lqvxudqfh > f2 ~2 õ 7d º6ë!ê é ¬ y l ¾ ñ ¾ W !If you solve y'=2y by separation, as y'/y=2, ln(y)=2xC, y=e2xC then along the way you've assumed y>0, and those are the solutions y=e C e2x can represent The true general solution is y=Ce2x, and often people hand wave at this point and pretend e C is the same as C, because it gets the right answer But it's not legit The better way to solve that equation is multiplying by the
X–y = xy 2 Solution 1 It is commutative x–y = xy 2 = y x 2 = y –x 2 It is NOT associative Counterexample (1–2)–3= 12 2 3 2 = 9 4;Ð ½ æ æ æÞ k%y þ ü þ ü ÿ ýC ËÄæÚ æ Y õ Äê Z ·êÞæ c þ L ýC Z3 If X 1 is determined by Y (eg, is a function of Y;
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