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(˙x(t),y˙(t)) be the closed regular plane curve formed by the velocity vectors Prove that the integral 1 2π I xd˙ y˙ −yd˙ x˙ x˙2 ˙y2 is an integer Point out geometric interpretations of this integer in terms of the velocity curve and of the original curve Solution Taking P = −y/˙ (˙x2 ˙y2) and Q = ˙x(˙x2 ˙y2) (where. Minimizing over y gives g(x) = infy f(x,y) = xT(A−BC−1BT)x g is convex, hence Schur complement A−BC−1BT 0 • distance to a set dist(x,S) = infy∈S kx−yk is convex if S is convex Convex functions 3–19 Perspective the perspective of a function f Rn → R is the function g Rn ×R → R,. 4 c g ï ò u 0 8 a 0 í í ñ î ò r h 8 ?.

When I think of y=f(x), i Think of y = f(x)= 1, x = 1, x =2, then y =f(x) =2, x =3, then y= f(x)=3, and so on Hi John, I find it helps sometimes to think of a function as a machine, one where you give a number as input to the machine and receive a number as the output. R q ܂߂ 05 ށE h N J e S ށE h N( я F i( ) 3 ^4 y W) 6,000 _ ȏ 𒴂 L x ȏ i Ŏ 葵 Ă ܂ I o ׂ S A V b v Ȃ ł͂ 100 ވȏ ̎ R q o ד Ă ͂ Ă ܂. C 3 O Ks/ r í õ.

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Y v Þ ¨ F Ì f h K b m G Ê Z ÷ Ä Ë É ø Þ í v Î Æ ¼ Î Â Î _ É d M É , $ A v d < É $ ¼ ñ I X ñ í Ì ° ´ æ æ ù I ® í Y v ´ æ ® Ã Æ ã å d < É & í Ë å ´ § $ Î Â $ ¼ Å Ç í Ü æ H ß í 0 ª º Ü É Ö O ¬ Ã Å ä î ® Ã í ¸ & ² Á æ > > Æ ó Æ Ê. ص ةروس C B A N M L K J I H G F E D C BA Z Y X W V U T S R Q P O ٢ \ Z Y XW V U T S R Q e d c b a ` _ ^ q p o n m l k j i h g f. Divide y, the coefficient of the x term, by 2 to get \frac{y}{2} Then add the square of \frac{y}{2} to both sides of the equation This step makes the left hand side of the equation a perfect square.

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X 5 X E X U r r y x { { r w { v x v r {. I H 5 M A ;. Electromagnetic energy flow the rate at which energy flows through a surface S is GG given by P = ∫ S da ⋅ S Useful Math Cartesian Gradient ∇t = ∂ t xˆ ∂ t yˆ ∂ t zˆ ∂ x ∂ y ∂ z Divergence ∇ ⋅ v z G = ∂ vx ∂ vy ∂ v ∂ x ∂ y ∂ z G ∂ vy ∂ v Curl ∇ × v = (x x ∂ vz ∂ vy ) xˆ ( ∂ v ∂ vz ) yˆ ( − )zˆ ∂ y ∂ z ∂ z ∂ x ∂ x.

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Thus 3 π y tan = ⇒ y = 3 x , where x > 0, y > 0 3 x Hence, locus of z is the part of y = 3 x in the first quadrant except origin 4 n –3 = (−i )4 n −3 = (−i )4 n (−i )−3 = (iv. Thus 3 π y tan = ⇒ y = 3 x , where x > 0, y > 0 3 x Hence, locus of z is the part of y = 3 x in the first quadrant except origin 4 n –3 = (−i )4 n −3 = (−i )4 n (−i )−3 = (iv. And since the P^(j) n all have the same distribution, taking expectations yields, E D(P^ n1kQ) E D(P^ nkQ) as claimed 5 Estimating the entropy The fact that EH(P^ n) H(P) follows from the concavity of the entropy (using Jensen’s inequality), upon noting that E(P^.

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