Explanation Given: x 0 = 0.5 And c = 1.5 Consider the logistic difference equation x t + 1 = c x t ( 1 − x t ) with x 0 = 0.5 and c = 1.5 . Substitute t = 1 , 2 , ... , 10 to find the corresponding value of x t . x 1 = x 0 + 1 = c x 0 ( 1 − x 0 ) = ( 1.5 ) ( 0.5 ) ( 1 − 0.5 ) = 0.375 x 2 = x 1 + 1 = c x 1 ( 1 − x 1 ) = ( 1.5 ) ( 0.375 ) ( 1 − 0.375 ) ≈ 0.3516 x 3 = x 2 + 1 = c x 2 ( 1 − x 2 ) = ( 1.5 ) ( 0.3516 ) ( 1 − 0.3516 ) ≈ 0.3419 x 4 = x 3 + 1 = c x 3 ( 1 − x 3 ) = ( 1.5 ) ( 0.3419 ) ( 1 − 0.3419 ) ≈ 0.3375 x 5 = x 4 + 1 = c x 4 ( 1 − x 4 ) = ( 1.5 ) ( 0

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ISBN: 9781133109631

Author: Stewart, JAMES, Day, Troy

Publisher: Cengage Learning,

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Differential Equations

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Chapter 1.6, Problem 27E

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To find:the value of xt for t=1,2,...,10 , for the logistic difference equation xt+1=cxt(1−xt) and also graph xt

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Chapter 1.6, Problem 26E

Chapter 1.6, Problem 28E

Chapter 1 Solutions

Biocalculus

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the value of x t for t = 1 , 2 , ... , 10 , for the logistic difference equation x t + 1 = c x t ( 1 − x t ) and also graph x t

Solution Summary: The author explains how to find the value of x_t for t=1,2,mathrm...,10, for the logistic difference equation, and graphs.

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To find:the value of xt for t=1,2,...,10 , for the logistic difference equation xt+1=cxt(1−xt) and also graph xt

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Chapter 1 Solutions