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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);
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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
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® 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 ?
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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
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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
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¤ ¥ 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
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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
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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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