Recitation 17

Example 1: Find the exact area of the surface obtained by rotating the curve about the x-axis. y=\sqrt{1+e^x}, 0\leq x\leq 1.

Hint: Use the formula \int_{x_1}^{x_2}2\pi f(x)\sqrt{1+(f'(x))^2}dx for the area of the surface.

Example 2: Use Newton’s method with the specified initial approximation x_1 = -1 to find x_3, the 3rd approximation to the root of the equation x^7+4=0.

Hint: Use the iteration x_{n+1} = x_n - f(x_n)/f'(x_n), where f(x) = x^7 + 4.

Example 3: Investigate the equation \{a_n\} defined by a_1 = 2, a_{n+1} = \frac{1}{2}(a_n+6).

Solution: Suppose 2\leq a_n<6. Because a_{n+1}=\frac{1}{2}(a_n+6), \frac{1}{2}(2+6)\leq a_{n+1}<\frac{1}{2}(6+6), hence 2\leq a_n<6. The upshot is that if 2\leq a_n<6 then so is a_{n+1}. Because 2\leq a_1<6, so is a_n for all n. Therefore \{a_n\} is bounded. Because a_n < 6, we know that a_{n+1}=\frac{1}{2}(a_n+6)>\frac{1}{2}(a_n+a_n)=a_n, and that the sequence is increasing. By the monotonic sequence theorem, the sequence has a limit, say L. Take the limit on the recurrence relation a_{n+1}=\frac{1}{2}(a_n+6) and get L = \frac{1}{2}(L+6). Therefore L=6.

Remark: Note that 6 - a_{n+1} = 6 - \frac{1}{2}(a_n + 6) = \frac{1}{2}(6 - a_n). This means \{6 - a_n\} is a geometric progression with initial term 4 and common ratio 1/2. Therefore 6 - a_n = 8\cdot (1/2)^n and a_n = 6 - 8\cdot(1/2)^n. From this general formula of a_n, we can immediately see that the sequence is bounded and increasing, and that the limit is 6.

Leave a comment

Your email address will not be published. Required fields are marked *