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Home » u + rac1u = 2. - Sourci

u + rac1u = 2. - Sourci

📅 April 30, 2026 👤 scraface

Apr 30, 2026

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📌 Now T(20) = 20 + 75e^(−0.04055×20) = 20 + 75e^(−0.811) ≈ 20 + 75×0.4445 ≈ 20 + 33.34 ≈ <<20+33.34=53.34>>53.34°C.

📌 #### 53.34

📌 A research assistant analyzes protein expression data and finds that Gene X expression increases exponentially. At 2 hours post-treatment, expression is 120 units; at 5 hours, it is 960 units. What is the hourly growth rate (as a multiplier) assuming continuous exponential growth?

📌 Model: P(t) = P₀e^(rt).

📌 At t=2, P=120; t=5, P=960.

📌 So: 960 = 120e^(3r) → 8 = e^(3r) → ln(8) = 3r → r = ln(8)/3 = 3ln(2)/3 = ln(2) ≈ <<log(2)=0.6931>>0.6931.

📌 But express multiplier per hour: e^r ≈ e^0.6931 ≈ <<exp(0.6931)=2>>2.

📌 Thus, the expression doubles every hour.

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Hence, the minimum value is achieved when $ x = y = z = \frac{1}{3} $.
Question: Define $ f(u) = u - \frac{u^4}{4} $. If $ n $ is a positive...
b_1 = 1, \quad b_{n+1} = f(b_n).
Determine the limit $ \lim_{n \to \infty} b_n $.
Solution: We are given the recurrence
b_1 = 1, \quad b_{n+1} = f(b_n), \quad \text{where } f(u) = u - \frac{...
We aim to find $ \lim_{n \to \infty} b_n $.
First, observe that $ f(u) < u $ for all $ u > 0 $, since $ \frac{u^4}...
Taking limits on both sides of the recurrence:
\lim_{n \to \infty} b_{n+1} = \lim_{n \to \infty} f(b_n) = f(L).

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