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**Unformatted text preview: **Chapter 1
= âˆ’(1 + h)1/h = âˆ’(1 + h)1/h Hence 12 3 lim g (h) = lim âˆ’(1 + h)1/h âˆ’ + h âˆ’ h2 + Â· Â· Â· hâ†’0 hâ†’0 23 4 3 12 = âˆ’ lim (1 + h)1/h Â· lim âˆ’ + h âˆ’ h2 + Â· Â· Â· hâ†’0 hâ†’0 23 4 From calculus we know that e = lim (1 + h)1/h , which gives
hâ†’0 1 1 d 1 1 âˆ’ h + h2 âˆ’ h3 + Â· Â· Â· dh 2 3 4 12 3 âˆ’ + h âˆ’ h2 + Â· Â· Â· . 23 4 . hâ†’0 lim g (h) = âˆ’e âˆ’ 1 2 = e . 2 So we have e e âˆ’ yn =. nâ†’âˆž 1/n 2 lim 15. The independent variable in this problem is the time t and the dependent variable is the temperature T (t) of a body. Thus, we will use the recursive formulas (2) and (3) on page 25 with x replaced by t and y replaced by T . In the diï¬€erential equation describing the Newtonâ€™s Law of Cooling, f (t, T ) = K (M (t) âˆ’ T ). With the suggested values of K = 1 (min)âˆ’1 , M (t) â‰¡ 70â—¦ , h = 0.1, and the initial condition T (0) = 100â—¦ , the initial value problem becomes dT = 70 âˆ’ T, dt and so the recursive formulas are tn+1 = tn + 0.1 , Tn+1 = Tn + 0.1(70 âˆ’ Tn ). For n = 0, t1 = t0 + 0.1 = 0.1 , 24 T1 = T0 + 0.1(70 âˆ’ T0 ) = 100 + 0.1(70 âˆ’ 100) = 97 ; T (0) = 100, ...

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