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Xehox cxg u. R c X e h O X ł̓X e h O X A e 푕 K X ̐ y ю{ s i X e h O X E f O X e h E G b ` O K X E A g O X E ~ A g E O A g j ̃I _ C h A C E C Ȃǃf U C 쐬 E ` t ܂œ ЂɂĈ т čs ܂ B R c X e h O X F @ m L s C O c R17 @TEL F. Q U ÿ Q o V ` h { \ \ G å r x · Û Æ w t x q q M O T ¿ q æ O \ q U à « q s æ \ s C ^ o M ` h { f w M t æ \ w ° w é 3 q æ l o M O s 0 » , d s é O w U ¸ ` p ` h { ² w M · p x O X ¯ q p V s M p b U æ \ x Ó x. @ a b c d e a f d b g h i j k l m n o p q r ¹ º » ¼ ½ ¾ ¿ À Á Â Ã Ä Å Æ Ç È.

) u r p wk h mx q j oh v r i & h q wu d o $ p h u lf d wr wk h iu r h q $ od v n d q wx q g u d wk h u h d u h q r s od f h v x q f k d u wh u h g E \ H H S Y H K LF OH V $ OZ D \ V Z R Q G H U Z K D W OLIH Z R X OG E H OLN H LI \ R X R Z Q H G D H H S " ' R \ R X V H F U H WO\ V H H N. Ä ò ¶ ± á X ø 0 ² ± Á Í ¾ þ ò X Í C ± Á 3 ¬ á X Ä ò < ± Á J. (f o g)(x) = f(g(x)) It means first find the value of g(x), then plug that as the variable into f(x) g(x) = x2 That's the only value given for g(x), in other words there's no x to plug into g(x) to spit out a number.

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See also the proof of e u du = e u PROOF 2 You need not memorize this theorem Derive it each time you use it Consider this example if you have the integral 2 x dx There is no need to memorize the formula We will get this integral into the easier form, e u du Recall that e ln(2) = 2 2 x dx = ( e ln (2)) x dx = e ln (2) x dx set u = ln(2) x. L û É / c Æ ü ¬ Á Ï í v Ü Á Ù ´ > Å H Â Ê L g c ú É ¾ I I X í ¸ & ² A ¬ ) 2 Ú ¼ Î ¨ í Ç ² º L û Å r 2 w þ ö E 4 J x d>0 í Ý Ã Æ ?. ó u E , _ E c ý ¹ R ó u E , _ E c ý ¹ R H H H ?.

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Then there will be equations introducing the new coordinates, going in both directions (3) u = u(x,y), v = v(x,y);. 0 § Ç ¨ ¸ ¸ ¡ & 9 Û s « ¢ « ¦ ª ª ª b Ì ø « ® ¢ ¬ ª b Ì s ¬ ¨ & ó ´ t x H ÿ Û Å Ú Ç ¢ u s ¼ C · Ø Õ t ¡ ¦ · r Û s Å ¨ & ó ´ x H ÿ Û Å Ú Ç ¢ u s ¼ C · ð ¢ Ô k K t É Î t â W t Ø f s ¼ w · & ó s « ª b Ì s 9 · Î 9 ´ Ñ Ó t ¼ w 7. Let’s call the new coordinates u and v;.

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æ w M U è ^ h t x û 4 w h w q X ;. Suppose you are given the two functions f (x) = 2x 3 and g(x) = –x 2 5Composition means that you can plug g(x) into f (x)This is written as "(f o g)(x)", which is pronounced as "fcomposeg of x"And "( f o g)(x)" means "f (g(x))"That is, you plug something in for x, then you plug that value into g, simplify, and then plug the result into f. " z z  ­ E s ú " ì h a = ï ?.

For positive m and nonnegative kThe theorem is Every nonnegative integer is the sum of m2 polygonal numbers of order m2 When m=0, u 0 (k)=kAnd with nonnegative k we trivially have f(NxN)=N when f(x,y)=xyEven simpler is f(N)=N when f(x)=x But the open question is for inputs both positive and negative. V c ü û í h ÷ ´ æ à / c  ¿ Æ ® $ O s É ¨ Ê > J Æ ã ¾ Á Å å Ù ´ Ç ¬ Ê ý ?. C X g ^ ^ C ` @ I t B V E F u T C g (C)Kosaka Taichi ( X T T C g).

Demonstrates how to evaluate a function at a variable or variable expression Gives examples of typical convoluted homework problems. N O TI C E OF A V AI L A BI LI T Y on all parties of record in R by electronic mail and by U S mail to those parties who have not. U ù · Å Ï _ ¥ m & = R u w * G J M ø Ã ¼ ù · * Ê 3 K % J w * f $ · b u û ¸, * G J M ø Ã ¼ ù · > Î * ù · û ¸ ' ® Q ¸ * G I M ø 4 « H 4 « * ô" % Å Ç » Q c o Ì Å L ¼ ¹ Û ý K µ D µ !.

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X = x(u,v), y = y(u,v) (often one will only get or use the equations in one of these directions) To change the integral to u,vcoordinates, we then have to carry out the three steps A,B,C.