Explain the determine yellow 226Difference Equationsthen()kkak(7.92)eAgain, substitution of equation (7.92) into equation (7.80) gives- (k+1)(*)k +1Ck+1Ck.(7.93)= ekNote that for k → 0, this equation becomesCk+1 ~ Ck,(7.94)which implies that c ~ constant. Hence, in order to determine the dominantCk, higher-order terms must be included in the k dependent factorbehavior ofon the right side of equation (7.93).To proceed, define A(k) as follows:-(k+1)A(k) = e1+(7.95)Therefore, for k → 0, we havelog A = log(e) – (k +1) log ( 1+1= 1- (k + 1)k111(7.96)2k23k34k41112k6k212k3and111A(k)exp2k6k212k3(7.97)1112k6k212k3r=If only terms to order k-3 are retained, then1+2k11111A(k)1++..6k212k324k26k38k31712k3+...24k216k3(7.98)Hence, Ck obeys the difference equation:).1712k(7.99)Ck+1 =+...Ck•24k216k3APPLICATIONS227 l stc ksa2:37 PMC @ 1 65% 4equationak+1 == kak.(7.80)This follows directly from the results of Section 2.4. To determine the dom-inant or leading behavior of the solution to equation (7.80), we rewrite it inthe formCancelActual Size (410 KB)Choose

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Description: Explain the determine yellow 226Difference Equationsthen()kkak(7.92)eAgain, substitution of equation (7.92) into equation (7.80) gives- (k+1)(*)k +1Ck+1Ck.(7.93)= ekNote that for k → 0, this equation becomesCk+1 ~ Ck,(7.94)which implies that c ~ cons

When we take limk→∞ ek+1k-(k+1)ck =(1)ck=ck Hence, ck+1~ck Let us consider Λ(k)=ek+1k-(k+1) Taking…

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(7) k +1\-(k+1) Ck. Ck+1 = e (7.93) Note that for k → 0, this equation becomes Ck+1 ~ Ck, (7.94) which implies that Ch ~ constant. Hence, in order to determine the dominant behavior of c, higher-order terms must be included in the k dependent factor on the right side of equation (7.93). To proceed, define A(k) as follows: -(k+1) Aa) - (1+) = e (7.95) Therefore, for k o, we have log A = log(e) – (k +1) log ( 1+ %3D k 1 1 1 – (k +1) ( (7.96) = 1 - k 2k2 3k3 4k4 1 1 1 2k 6k2 12k3 and 1 1 1 A(k) = exp 2k 6k2 12k3 (7.97) 1 1 2k 6k2 12k3 If only terms to order k-3 are retained, then 1 1 1 1 1 1 1 A(k) = 1+ +.. 2k 6k2 12k3 4k2 6k3 8k3 1 3 +.. 1- + 2k 24k2 16k3 (7.98) Hence, Cr obeys the difference equation (1 1 7 3 +... (7.99) Ck+1 = Ck. 2k 24k2 16k3

(7) k +1\-(k+1) Ck. Ck+1 = e (7.93) Note that for k → 0, this equation becomes Ck+1 ~ Ck, (7.94) which implies that Ch ~ constant. Hence, in order to determine the dominant behavior of c, higher-order terms must be included in the k dependent factor on the right side of equation (7.93). To proceed, define A(k) as follows: -(k+1) Aa) - (1+) = e (7.95) Therefore, for k o, we have log A = log(e) – (k +1) log ( 1+ %3D k 1 1 1 – (k +1) ( (7.96) = 1 - k 2k2 3k3 4k4 1 1 1 2k 6k2 12k3 and 1 1 1 A(k) = exp 2k 6k2 12k3 (7.97) 1 1 2k 6k2 12k3 If only terms to order k-3 are retained, then 1 1 1 1 1 1 1 A(k) = 1+ +.. 2k 6k2 12k3 4k2 6k3 8k3 1 3 +.. 1- + 2k 24k2 16k3 (7.98) Hence, Cr obeys the difference equation (1 1 7 3 +... (7.99) Ck+1 = Ck. 2k 24k2 16k3

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section4.6: Variation

Problem 15E

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