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H u cxg. ¨\µ\Ô r > z\ r > Ñ\Õ Ì\Ã\õ é »\â\Ø r > « x\Ø ÿ \Ï\¹\ô t 6 à { ÷\î ½ ~ µ\ Ü Ñ k µ º Ì »\Ô\Ó\ Î Ì A\Ø ò\Ò Î ª ª\Ø v ­\Õ\ò\ô º ­ ý K £ ÿ\Ñ v ­\ü\°\»\õ Í\â\Ø ý K Ó M 0\Ø ² "\ D\ è 9 ÷\Ø / $ t × U2 S!. ' ( $ % # /, 0 * 1) 2 3 " 4 5 6 7 8 9;. ∫ (µ(t) y′ µ(t)p(t) y) dt = ∫ µ(t)g(t) dt → µ(t) y = ∫ µ(t)g(t) dt (**) Therefore, the general solution is found after we divide the last equation through by the integrating factor µ(t) But before we can solve for the general solution, we must take a step back and find this (almost magical!) integrating factor µ(t) We have.

¤ n Q @ â ¨ ½ t § Á ¨ ô ¨ ?. ìBŒÐ šÁñ}•Åþ”‘ZDe L ¨ôÁÿ 䬼=ªÇ{ ½” VE$äp3õ K ÍoE·ÖmÂKòJŸêä •ÿ @ ÿ†µ;. ¬ h v o O q ¤ ¥ ¹ è Ö ° P ± y ( ¤ v o O q { { { µ c 8 ^ Õ ë ¶ ¥ ­ Æ s V & y O Þ v o O q ¤ ¥ Q ( è i Q y ¬ è v R Q  þ · ¬ h y # ?.

µ,µ− are absolutely continuous with respect to Lebesgue measure, then there is a unique solution to the Kantorovich problem, which turns out to be also the solution to the Monge problem The same result holds true when µ is nonatomic (ie contains no atomics µ({x}) = 0 for all x∈ X). ó e L w C ä ;. Ñ Ì 5 h Í 1 ñ q / $ j @ * Ñ ) Æ a Ó Ì â * Ý j r > ½ Ð q · @ ù * ¹ µ q · ' q ï.

V o O q ¤ ¥ µ c ¤ Â Ü · ½ Õ ç ñ h y ê & s s Q y ß ± v o O q { { { ¤ y y s ß v o O q { { {. Ø ½ µ º _(ò m 0 ^ 8 G \ í9× _ ^ S u (ò m 0 \!F / b > Q 6 ~ x E @ Ø ½ µ º b ë c S6Û * _ k8 M G \ í Ø ½ µ º Æ _ Z K S x í y v ~7V 8 Z ë K ^ 8 \ # C%Ú o í!F!O í!ÿ$Î æ&g / b > Q 6 ~ 54 c K ^ 8 G \ í. ú ¨ 2 ¨ ¬ !.

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Ð ¨ ½ æ Y o ¨ ?. 2 Lecture 5 24 Relativistic Quantum Field Theory II Fall 10 Now we observe that DA = DA , as gauge transformations correspond to unitary transformations plus shifts, and that S A = S A , because of the defining gauge symmetryHence, Z = ˆ ˆ dΛ DA e iS A δ(f(A )) det δf(A ) δΛ ˆ ˆ dΛ DAeiSAδ(f(A)) det δf(A) δΛ as A is a dummy integration variable. × ³8( e ³ ´!.

µ K e8( ü ´!. Ñ ¼ Ý Z VOL2 (2) O d ã ° Í t ^ C Ì Ý Å ì Æ Ê à ­ È ¢O (3) ë ^ C v Å è, « ê Ë ¯ Ê Ï ª å « ¢O (4) ½ Ê I É Í, ¯ ¶ p ^ Ì J ) Ô µ ª14 K Ü Å (5)1 K Ì « ê ª É { Ý Ì º ®(S ¢) ª é¡ È Ç ª ° ç ê, « ê Ì Ý u ú Ô ª ñ í É · ¢ í è É Í. Cos1 6= 0, the latter equality implies that µ2 6= 1 and cos µ 6= 0 Hence it is equivalent to tanµ = 2µ.

ð £ é ½ ß É à g p µ ½ D A Ø z Ø z Ø Í C X p C t v ð g p µ C n g Ý µ ½ à Ì ð N É Ä Ý è ã ° K X ¯ Ú ð s Á ½ D g É Í S Ø Ý è ê p t N v ð g p µ C S Ä ³ « ê Å { H µ ½ D. Q N t L V u { b g A Ì p ¨ Ï » ð l ¶ µ ½ g o X g T { § ä Robust Control of Two link Flexible Robot Arm that Considers Change of Posture. µ, and if you define C(µ0) = fX µ0 2 l(X);u(X)g then this is a critical region (see Hypothesis Testing) for testing H0 µ = µ0 with level of significance fi The most important use of this duality is to find a CI First find a test (which is often easier because we have a lot of methods for doing this) and.

Then B is a basis for ¿ if and only if 1 B2B B ¶ X 2 For all A 2 ¿ and for all x 2 A;. ¤ n c µ Ò n Ô ú o º ¢ ñ þ Ò n æ ú ú 2 q µ T Ð a & ` G d n Ð 2 Q í ë @ â ¨ § ú æ Y Â ' I t ¨ § Á ¨ ô ( µ ¦ æ Y m æ Y ¨ § ¨ U ô ¨ ?. ª « ¬ ­ ® ¯ ° ± ² ³ ´ µ ¶ · ¸ ¹ º ¢ ¤ £ ª ¬ · µ ¸ ¦ ¥ ¹ ¡ « ´ § » ¼ ® ¢ ´ ¦ ª X Y Z \ ^ _ ` a b c d e f g h i j k l m n o p.

$ ì õ Õ Ê Ï ö Æ È Ï ÷ Ð Ê Ë Ê Ì È Æ Ç ø Ê Å Å Ò ù è Æ Ê Ç Ñ Ó Ì ú û Ô ü Î Æ Ç. 9·i txFà ¢€3¾És»oÙåϦÃ@ Úgu ù Ë CÝä 8ö h ޽ụ½DÓm íÒ%PÀŽOûÐ Ÿƒ4ËÝ>k¦¼·hƒª. MISSION STATEMENT—St Mary’s is a caring and welcoming family committed to building and strengthening a community of God, reaching out as disciples of Jesus Christ to Freeport and beyond.

Aå Fâ K² Pƒ Uû 9 aw gÛ m t" {o ‚¤ ‰~ ¹ —À"žv$¤‡&« (±_*·«,¾tÄv0ˆ2Ò&4Ø46Þd8äbêËñÞ>ø @ý•B UD PF ÂH sJ éL $èN ™P 2 R 8ŒT ?9V EŸX L Z R \ W ^ \ ` bŒb híd oAf u¸h j ‚{l ˆÉn Dp •»r ›Ít ¢bv ¨²x ¯oz µs »ú~ Â÷€ É«‚ Є × † ÞZˆ äæŠ ë Œ òBŽ øá ÿª. · e8(4 º 4 75 &!. G* e"© ¸ ¬ Û 6 pj pj pj ç Ü e i º v 50 >&'¨ "'>' red\dvkl 7 lvklprwr 0 6zhdulqjhq & )lorsrxorv 07 2kdud < 7rnxgd < hw do 'liihuhqfhv lq folqlfdo pdqlihvwdwlrqv.

µ @ r \Õ ±\ § \Ø ¨ ^ Q\Ø ± °\ ± °\ ± ¬\ü\»\õ\Ø \¶ ± ^. Jensen’s inequality and of the fact that MMDF,p,q = 0 ⇒ µp = µq While this result establishes the mapping µ p is injective for universal kernels on compact domains, this result can be shown to hold in more general cases, provided that we are using. MW « ·¥µ DNS dzC2 ¸³ mIP´ ­m% 2 ¨ Jk¤µ ¦¸ j ¥ °· O(´§¸ ²~³ ªX m9 3Exo ®¸³´§¸ UIPlxz$(µ¸©· ·¢®¸ xo¹renº UIP4lxz$(Active Directory¯¸ ren ·¢®¸ 3lxz$(µ¸ ³/ µ¸ ³9 ´§¸ lxz5>m;.

’óƒ houted €ØGoó€òïrålse – @„ „ „ „ „ „ „ „ ScreamsÁælood. < = > $) /?. Æ µ Ä Í J J Smolicz, R Radzik(04 j ª ã ° ç ê é B Å V Ì ¤ Æ µ Ä Í A » Ý Ì x V Ì ¾ ê ó µ ð ³ ç â } X f B A A o Å Ì ó µ È Ç à Ü ß L ­ T Ï µ ½ M3 Ó ¾ OTO Ba,.

Aå Fâ K² Pƒ Uû 9 aw gÛ m t" {o ‚¤ ‰~ ¹ —À"žv$¤‡&« (±_*·«,¾tÄv0ˆ2Ò&4Ø46Þd8äbêËñÞ>ø @ý•B UD PF ÂH sJ éL $èN ™P 2 R 8ŒT ?9V EŸX L Z R \ W ^ \ ` bŒb híd oAf u¸h j ‚{l ˆÉn Dp •»r ›Ít ¢bv ¨²x ¯oz µs »ú~ Â÷€ É«‚ Є × † ÞZˆ äæŠ ë Œ òBŽ øá ÿª. Ôò¡ € ¯^Ï LL 'Â_ € fÛÀ¨ óÍ5*‹cFÚ go microsoft com € ¯^ê ««Â_ € fÚÀ¨ òØ5* T go microsoft com ¯^û ««Â_ 66Â_ Á P L ¶­Qj¸É †6±cø¯sÓ#ç R—;éBo œ,ªi¥XÀ0$ÿ # http/11 Y0‚ U0‚ = Âm—lß@’4 0 *†H†÷ 0 ‹1 0 U US1 0 U Washington1 0 U Redmond1 0 U Microsoft Corporation1 0 U Microsoft IT1 0 U Microsoft IT TLS CA 50. "Ó 2( A í º b ¹ µ º í ª × ½ ª 4E m ± l g a#ú4E m ± '¼ b ¥ æ/² b g* \ M S T K º.

57 149 ì q í V § µ J w Z 2 58 137 Û t V § L _ w Z 2 59 152 ± ¼ Ñ Y ¾ V § á { w Z 2 105 2. ² Æ Ì È · p Æ µ Ä ß ç ê é ¡ ç p x É Â ¢ Ä Í C · J ì ç (1993) ª C ~ V K ^ u ò ð p ¢ ½ ° Å Í C U ^ Ì u ò É ä × ¡ ç p x ª å « ­ ( ° à Å Î ß É ¡ ç µ ½ ó Ô É) È é Æ ñ µ Ä ¢ é. Where k is the dimension of µ1Therefore, we reject H0 if ´2 obs >´ 2 1¡fi,where´ 2 1¡fi is the (1¡fi)th percentile of ´2 k Score test The score test is based on the fact that the score U(µ;X) has the following asymptotic distribution.

µ v à l q ã µ õ r B ã µ É Ë q J J / !. µ @ r \Õ ±\ § \Ø ¨ ^ Q\Ø ± °\ ± °\ ± ¬\ü\»\õ\Ø \¶ ± ^. § = ¸ 4 æ!.

Q(µ) = inf x∈X˜ f(x) µ′g(x), (64) which by weak duality, is a lower bound to the optimal value of the restricted problem minx∈X,g˜ (x)≤0 f(x) Because X˜ is finite, qis a concave piecewise linear function, and solving the dual problem amounts to minimizing the polyhedral function −qover the nonnegative orthant. $ ì OPî ¬ ¿ ³ ´ · « ² µ ° ½ À ® ¼ ­ ¹ ½ ® ¸ ± ² ¯ ¶ ± · ½ ¬ ½ º ½ ³ ± µ ± ¼ Ä ± ° å L !. 2 l h p S g µ Ü µ { Û Å ç ³ ï « º G O 8 ·% DN Ó ø w µ x þ { M ¼ p x K d { S Ö o s Ä ¿ Ó P í { ¤ ï Ø µ C » ª ` h µ Â ï è µ ¢464 £ Ä ¿ Ó x z ä t § X S Ö å « { ~ ( b s ;.

M v c s j d b u j p o 1 The first oil change should be performed after 500 hrs of operation;. B *( p F M ##ä @!. µ c ¤ y V Â È S ?.

@ 4( " 8 A 6 0 B 4 (!) < =, > 7 1 i j k l m n o j p q r s t u v w x y z {} ~ v} v w x y z. Autodesk Revit Autodesk Revit Grouping Revit " " " " ". C Æ å ­ µ º'¼ 7V d M >>0 G% è7F b2 Ü>< p @« Ë$× _ V$ö ì K Z 8 ^ 8 ?.

– if everything is normal, received means should be µ 0 and –µ 0 • action – ask the system to transmit a few 1s and measure X – compute the ML estimate of the mean of X • result the estimate is different than µ 0 = ∑ i Xi n 1 µ 17. Introduction to Convex Optimization for Machine Learning John Duchi University of California, Berkeley Practical Machine Learning, Fall 09 Duchi (UC Berkeley) Convex Optimization for Machine Learning Fall 09 1 / 53. ** Ø0rqwdjhdqohlwxqj Ù 2 f i e a g s g e l c u > t u c t i s Ú,qvwdoodwlrq lqvwuxfwlrqv Û,qvwuxfflrqhv gh prqwdmh Þ1rwlfh gh prqwdjh á6huhopvl ~wpxwdwy â,vwuxlrql shu lo prqwdjjlr é,qvwdoodwlhyrruvfkuliw ë,qvwuxnfmd prqwd px ì,qvwuxo}hv gh prqwdjhp í,qvwuxf xlxql gh prqwdm î ¼ À Á ¿ Â ¹ Å · Î ¾ ½ » ½ ¼ Á ¯ µ Â.

µ } b Ð ì " b p Õ Ó ô Ì ³ » Û ¼ Ð æ Y < p õ X ± õ s õ q b ;. # * j ¬ ` ý k ù * ¹ µ o n ¬ ' 1 Ó ) â ù Ý ¬ q !. J ñ µ ´!.

± µ ½ ½ ² « ® ± ° º ¹ ° ± ¶ · ¸ ³ ± · ± ¸ ± µ ¶ µ ¬ ¼ ñ ¯ ½ ¬ L !. ¹ µ ï Ó ý k µ d 4 ¯ ­ « è n Ô ³ ì ý k ¶ 1 !. 1800 m ± ¸¡ m² Ad.

Title Microsoft Word å ä¹ å· å¸ ã ã ã å 㠳簡æ é ç© å®¹å ¨ã ®è²¸ä¸ å ã ³è­²æ¸¡ã «é ¢ã ã è¦ ç¶±ï¼ R2111æ ½è¡ ï¼ doc. µ } b ö!. W Ù £ ÿ\Ø $\ë Í w Ù £ ÿ\Ø Ô \Ç\Ø N B ,c ï Ó $ ö ½ À ò Ì î Ø Ã À ò Ù µc Ö o \Ø à Ò w ¯\Ø í Ò.

µ, and if you define C(µ0) = fX µ0 2 l(X);u(X)g then this is a critical region (see Hypothesis Testing) for testing H0 µ = µ0 with level of significance fi The most important use of this duality is to find a CI First find a test (which is often easier because we have a lot of methods for doing this) and. Lemma 112 If ¿ is a topology on X and B µ ¿;. R_latin_01R ß°R ß°BOOKMOBI) ø#’ P S /K 0O 0‡ 1{ 2“ 3{ 4K 4_ 5_ 6C 6G ìô = •"= Á$= å&= ( MOBIø ýé5 †q.

Q@ Q@Š( Š( Š( Š( Š( Š( Š( Š( Š( Š( æ 'gh' Ê^EæXK²gclñ–È' ¯ä©º8·rz ?ʹne·µµƒ9&=Ì=2rjXâs¬ w aÑ àÃý }Ä»å㸠¤¸ù·1þè'ù Ö¥ 1› ?@>´Â˜FúÑ ÈTôpjn«'µ'£ WBÙðS ± ’8À;AÜÇÐVlPìHÜt5ÐZEþ‹,ØûÁUOë\õ §L ô8 ã–~™úTI‘Ö¬A#Ã*ȽTóQ Iy ô„xØ¡F1 ¹xÆÓþ ±QÃ. µ } b Ò 4 ô * Ì ³ Ö L Æ ã Ú M M ì " Ð æ Y < p õ X ± õ s õ q b ;. CHAPTER 3 ST 745, Daowen Zhang Then under H0, ´2 obs =(µ^1 ¡µ10)TC¡1 11 (µ^1 ¡µ10) »a ´2 k;.

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∂µ2 = −µ−2 i=1 x i < 0 Thus there is a local maximum at µ = ¯x We then note that as µ → 0 or µ → ∞, the loglikelihood ‘(µ;x) approaches −∞ Thus µ = ¯x is a global maximum, and the maximum likelihood estimate of µ is ˆµ = ¯x The maximum likelihood estimator in this example is then ˆµ(X) = X¯ Since µ is the. { ú w ) Î » S b b. 252 ConvexOptimizationAlgorithms Chap6 The dual problem is maximize q(µ) subject to µ∈ ℜr, µ≥ 0, (62) where the dual function qis given by q(µ) = inf x∈X L(x,µ), µ≥ 0, (63) and Lis the Lagrangian function defined by.

> @ â ÷ Á. Oracle_Cloudstrators_GuideW¢lWW¢lYBOOKMOBI % à4 ;. # j ± & f Ó r > ' ) j Ú ¬ è c ð ù * q ð ± j ' ' a * ü Á ¬ * j ¹ µ m f $ ß @ ¹ ð Ý ¬ ù * c Æ?.

There exists B 2 B with x 2 B µ A (Equivalently, for all A 2 ¿;. A µ is neutral, whereas, in the nonAbelian case, carries the group index, and so is charged under itself, leading to selfinteraction 2 In terms of F µ = F a T a and J = J aT a, we have from (4) that DµF µ = J , (7) with DµF µ = µF µ − ig Aµ,F µ , and under a gauge transformation, Fµ −→ VFµ V †, DµF µ −→ VDµF. MISSION STATEMENT—St Mary’s is a caring and welcoming family committed to building and strengthening a community of God, reaching out as disciples of Jesus Christ to Freeport and beyond.

Treasure_IslandS NšS NšBOOKMOBI a± Ø) /¯ 7R ¼ Aí Hœ P X `Æ i¿ rn z‘ ‚À ‹ “ ›‹ ¤Y"¬Ñ$µC&¾(Æ•*Ï,×Æß‘0çó2ðj4øŽ6 G8 – Ê Ñ> !4@ )¨B 2QD ¢F H J÷J RÛL Z;N b P iæR rbT zèV ƒyX ŒXZ ”º\ ^ ¥§` ®Kb ¶âd ¾Ïf Ç h Ï–j ×ül àfn é>p ñêr úTt žv @x z Œ #Ï~ °€ 4 ‚ 9„ D¬† Lkˆ TzŠ >Œ e„Ž m› uì’ ~O” †Ò. N À p Í B æ!.

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