Consider the following proposed proof that the sequence {tan(n)}diverges.1. First observe that tan(n) is a well-defined real number when n is a natural number, becausethe points where the tangent function is undefined are irrational numbers (namely, oddinteger multiples of π/2).2. Seeking a contradiction, suppose that the sequence {tan(n)}₁ converges to a limit L.3. A trigonometric identity implies thatStep 4 is faulty.Step 3 is faulty.tan(n + 1)4. Since {tan(n+1)} is a tail of the sequence {tan(n)}_1, both sequences have limit L,so the preceding identity implies thatStep 1 is faulty.Step 5 is faulty.The proof is valid.Step 2 is faulty.=L =tan(n) + tan(1)1 tan(n) tan(1)5. The preceding equation simplifies algebraically to 0 = (1 + L²) tan(1). Since I is a realnumber, 1 + 1² € 0, so it follows that tan(1) = 0. The points where the tangent functionequals 0 are integer multiples of #, so 1 must be an integer multiply of , contradicting thatTis irrational. OWhich of the following statements best describes this proposed proof?L+tan (1)1 - L tan (1)

Answered: Consider the following proposed proof…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Related questions

Question

Consider the following proposed proof that the sequence {tan(n)}
diverges.
1. First observe that tan(n) is a well-defined real number when n is a natural number, because
the points where the tangent function is undefined are irrational numbers (namely, odd
integer multiples of π/2).
2. Seeking a contradiction, suppose that the sequence {tan(n)}₁ converges to a limit L.
3. A trigonometric identity implies that
Step 4 is faulty.
Step 3 is faulty.
tan(n + 1)
4. Since {tan(n+1)} is a tail of the sequence {tan(n)}_1, both sequences have limit L,
so the preceding identity implies that
Step 1 is faulty.
Step 5 is faulty.
The proof is valid.
Step 2 is faulty.
=
L =
tan(n) + tan(1)
1 tan(n) tan(1)
5. The preceding equation simplifies algebraically to 0 = (1 + L²) tan(1). Since I is a real
number, 1 + 1² € 0, so it follows that tan(1) = 0. The points where the tangent function
equals 0 are integer multiples of #, so 1 must be an integer multiply of , contradicting that
Tis irrational. O
Which of the following statements best describes this proposed proof?
L+tan (1)
1 - L tan (1)

Expert Solution

Step 1

The objective is to find the option which describes the following proof:

1. ${\tan (n)}^{\infty}_{n = 1}$ is a well defined real number where n is a natural number because the points where the tangent function is undefined are irrational numbers i.e, odd integer multiples of $\frac{π}{2}$ .

2. Suppose ${\tan (n)}^{\infty}_{n = 1}$ converges to the limit L . (seeking a contradiction)

3. Trigonometric identity implies that $\tan (n + 1) = \frac{\tan (n) + \tan (1)}{1 - \tan (n) \tan (1)}$ .

4. Since, ${\tan (n + 1)}^{\infty}_{n = 1} is a tail of the sequence {\tan (n)}^{\infty}_{n = 1}$ . ${\tan (n)}^{\infty}_{n = 1} and {\tan (n + 1)}^{\infty}_{n = 1}$ both sequences have the same limit L . From the preceding identity, $L = \frac{L + \tan (1)}{1 - Ltan (1)}$ .

5. Preceding equation gives , $0 = (1 + L^{2}) \tan (1) . since, L is a real number (L^{2} + 1) \neq 0 . So it follows \tan (1) = 0 the points where tangent function is zero are integer multiples of π, so 1 must be integer multiple of π, contradicting that π is irrational .$