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A t x e e t x a e t x g x e(e t x b s) r1 (b t x e s t) q = 0 The stabilizing solution X puts all the eigenvalues L in the left halfplane Algebraic Riccati equations play a key role in LQR/LQG control, H2 and Hinfinity control, Kalman filtering, and spectral or coprime factorizations.

Uae tx p cxg. P, acting on C∞(Tx 0X,Ex0), which is obtained from the operator Hp by freezing coefficients at x0 This operator was introduced by Demailly 5, 6 Consider the trivial Hermitian line bundle L0 over Tx0Xand the trivial Hermitian vector bundle E0 over Tx0X with the fiber Ex0 We introduce the connection (14) ∇L p 0 v = ∇v −ipαv. Z p g × f ^ R r b v ð p µ ½ u ª « ¼ µ v b v R T C W ^ á ê å ÷ ¬ X c q ê Ì æ è g Ý ¬ ì W ¶1) E ¡ R l g c G ¥ 2) E Í ³ l § 1) \ _ Æ ü Ç y Z ^ \ ì x å ÷ ¬. V J S E Z y W w Z ރg RCOM x ł̓V J S ̐ Z Љ Ă ܂ B ǂ ́H Ǝv ͑ Ă ܂ B.

B Ô Þ ã K C b b j } Ô Þ ã K v p X ç F Ö É r Ô Þ ã K W ð è ` ) r d } ½ û ¼ 2 ¾ Ö $3 Ó ñ Ç ä ® ê È h ¹ ñ ¿ ã è y W « ã ³ ¢ ¸ Ô Þ Ö ä ® ê È z ç ¼ ê O v « ã ³ ¢ Ê C ¢ s ?. G P( ) (06) (04) 4 (06) (04) (this is a binomial expansion) (04 06) x x x x X x x t X xt t x t x t − − = = = = = ∑ ∑ ∑ b i Using the chain rule, G′ X (t)=4×06(04 063 24(04 3 So G′ X (1)=24(04 06)3 =24 Hence, mean=E(X)= ′G X (1)=24 (ii Again using the chain rule, G′′ X. And xc(t) is the general solution to the associate homogeneous system, x(t) = B(t)x(t) then x(t) = xc(t)xp(t) is the general solution Example 12 Let x0(t) = 4 ¡3 6 ¡7 x(t) ¡4t2 5t ¡6t2 7t1 x(t), x1(t) = 3e2t 2e2t and x2(t) = e¡5t.

Simple and best practice solution for g=(xc)/x equation Check how easy it is, and learn it for the future Our solution is simple, and easy to understand, so don`t hesitate to use it as a solution of your homework. P É v 9 a " , ã â U B q þ Ø v Á I Ü ;. 4 1 SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS If xp(t) is a particular solution of the nonhomogeneous system, x(t) = B(t)x(t)b(t);.

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And xc(t) is the general solution to the associate homogeneous system, x(t) = B(t)x(t) then x(t) = xc(t)xp(t) is the general solution Example 12 Let x0(t) = 4 ¡3 6 ¡7 x(t) ¡4t2 5t ¡6t2 7t1 x(t), x1(t) = 3e2t 2e2t and x2(t) = e¡5t. P X A E G C ` b v é ¢ Í t ¯ p ` b v ð p ¢, J b v u Î É æ é ½ Ê ¤ í ¨ æ Ñ ² t u Î É æ é ¤ Ê ¤ í ð Ý ½ B \3 É ã \ I È ¤ í Ê ð ¦ · B. YWCA ł́A O ˎ A c A w A w A Z Ώۂ̃v O s Ă ܂ B N L v i O j A p Ń g ~ b N A e q v O A q Ē ̐e Ώۂ̍u B i ꕔ ۈ t j ŋ ʂ Ă ̂ p ɐG Ă݂悤 I ŋ ̃e } ”The Giant Turnip” i u Ȃ ԁv j B V v ̂ X g łǂ̊w N ̎q y Q ł ܂ I @ t x ݂ 1 A l C e B u u t Ɖp ŗV ڂ.

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