Porisma

Worked solution

Observe an alternating convergent recurrence

A sequence of real numbers satisfies xn+1=−exnx_{n+1}=-e^{x_n} for n≥1n\ge1. Use a calculator to determine its behaviour when x1=−2x_1=-2.

CJC · Discussion · Question 2

Answer

Neither increasing nor decreasing; converges numerically to −0.567-0.567 (3 s.f.).

Generate the calculator sequence

Use sequence mode with x1=−2x_1=-2 and xn+1=−exp⁡(xn)x_{n+1}=-\exp(x_n). The source calculator table is reproduced to three decimal places:

nxₙ (3 d.p.)
1-2.000
2-0.135
3-0.873
4-0.418
5-0.659
6-0.518
7-0.596
8-0.551
9-0.576
10-0.562
11-0.570
12-0.565
13-0.568
14-0.567
15-0.567
16-0.567
17-0.567
18-0.567
19-0.567
20-0.567
21-0.567
How did this part go?

Describe the behaviour

The terms alternate on either side of a limiting value, so the full sequence is neither increasing nor decreasing. The calculator values approach −0.567-0.567 (3 significant figures). A fixed point satisfies ℓ=−eℓ\ell=-e^\ell; numerical solution gives ℓ≈−0.56714329\ell\approx-0.56714329. The table is numerical evidence of this behaviour, not by itself a proof of convergence.

How did this part go?

Common mistakes

Do not mistake alternating sides of a limit for divergence.

Read the hint