Induction: prove x^n + 1/x^n = 2cos(n*theta)

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In summary, the given statement states that if a and theta are real numbers such that x + 1/x = 2cos(theta), then the equation x^n + 1/x^n = 2cos(n*theta) holds true. The attempt at solving the problem through induction was made, but it was found that the base case n=1 was not sufficient to prove the statement for n+1. An alternate approach using complex analysis was suggested, but it was pointed out that the problem may not require this method. It was then noticed that the values for x and theta must be specific in order for the statement to hold true, specifically x = 1 or x = -1.
  • #1
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Homework Statement



If a and theta are real numbers such that x + 1/x = 2cos(theta), then:

x^n +1/x^n = 2cos(n*theta)

Homework Equations



x + 1/x = 2cos(theta)

The Attempt at a Solution



So I was able to show that the statement was true for n=1, but I'm stuck on how to even start proving it for n+1. I tried x^(n+1) + 1/x^(n+1), thinking I could solve it for n+1, but to no avail. Help, please?
 
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  • #2
IF
[itex]x^2 + 1 = 2x\cos(\theta)[/itex] ...(1)

THEN
[itex]x^{2n}+1 = 2x^k\cos(n\theta)[/itex] ...(2)

by induction:

n=1: [itex]x^2 + 1 = 2x\cos(\theta)[/itex] ... this is just (1) so true.

from pythagoras and trig relations - this also means:
[itex]\sqrt{4x^2-(x^2+1)^2} = 2x\sin(\theta) = (x^2+1)\tan(\theta)[/itex]n=k: [itex]x^{2k}+1 = 2x^k\cos(k\theta)[/itex] ...(3) assumed true.

consequences from pythagoras and trig?n=k+1: [itex]x^{2k+2}+1 = 2x^{k+1}\cos(k\theta+\theta)[/itex]

This is as far as you've got ... could see what happens if you use the difference in angles identity on the cosine ([itex]\cos(A+B)=\cdots[/itex])

You are going to have to do a lot of messing about before you'll see what to do: builds character ;)
 
  • #3
The trick is that [tex]x=\cos \theta + i\cdot \sin\theta[/tex], so
[tex]x^n+x^{-n}=\cos n\theta+i\sin n\theta+cos n\theta-i\sin n\theta=2\cos n\theta[/tex]
 
  • #4
ptolema said:

Homework Statement



If a and theta are real numbers such that x + 1/x = 2cos(theta), then:

x^n +1/x^n = 2cos(n*theta)

Homework Equations



x + 1/x = 2cos(theta)

The Attempt at a Solution



So I was able to show that the statement was true for n=1, but I'm stuck on how to even start proving it for n+1. I tried x^(n+1) + 1/x^(n+1), thinking I could solve it for n+1, but to no avail. Help, please?

My two cents...
Part of the problem is related to the way in which you framed it. To make induction work, you should put it following this structure (assuming that P is whatever property):

[itex] \forall n \in N ( P(n) \rightarrow P(n+1) ) [/itex].

The key is that you assume P(n), and then move from that to prove P(n+1).
Basing on how you framed it, you cannot do it: instead, you have to prove P(n).
 
  • #5
csopi said:
The trick is that [tex]x=\cos \theta + i\cdot \sin\theta[/tex], so
[tex]x^n+x^{-n}=\cos n\theta+i\sin n\theta+cos n\theta-i\sin n\theta=2\cos n\theta[/tex]

I'm guessing the point might be to prove the statement without resorting to complex analysis, which is what DeMoivre originally had to do.

Anyways, OP, to solve this by induction, the first thing you have to do is show the base case. Since n =1 is what defines x, you probably have to take n = 2 as the base case.

Then, you assume that

$$x^n + x^{-n} = 2\cos(n\theta)$$

is true. Let me call this the base equation. You want to show that it implies

$$x^{n+1} + x^{-n-1} = 2\cos((n+1)\theta).$$

I didn't do this myself yet, but my first guess at an approach would be to try to get an ##x^{n+1}## into the base equation, perhaps by multiplying through by x. This obviously doesn't get you the equation you need to prove - there's no ##x^{-n-1}##. You could alternately multiply the base equation through by 1/x, and end up with an equation that's missing the ##x^{n+1}## term. I'll let you figure out where to go from here. Remember that you are taking the base equation to be true by assumption, for any positive power ##\leq n##.
 
  • #6
Has anyone noticed that in the base problem the values of [itex]x[/itex] and [itex]\theta[/itex] are not at all arbitrary real numbers?

Since [itex]| x + \frac{1}{x} | \ge 2 [/itex] then the only possible values are [itex]x = \pm 1[/itex], and the corresponding values of [itex]\theta = 2k \pi[/itex], [itex](2k+1) \pi[/itex].

It seems to make the whole proof a bit moot, seeing as you can just substitute in the known values to verify the general proposition directly.
 
  • #7
You mean since [itex]1 \leq \cos(\theta) \leq 1[/itex], the equation is only valid for [itex]-2 \leq x+\frac{1}{x} \leq 2[/itex] ... ?

Presumably the exercise is not to prove the statement but to practice the method. Admittedly there are probably better examples...
 
  • #8
Simon Bridge said:
You mean since [itex]1 \leq \cos(\theta) \leq 1[/itex], the equation is only valid for [itex]-2 \leq x+\frac{1}{x} \leq 2[/itex] ... ?

Presumably the exercise is not to prove the statement but to practice the method. Admittedly there are probably better examples...

No I mean that considering just the function "[itex]x + 1/x[/itex]" in isolation, we can only only have either [itex]x+\frac{1}{x} \leq -2[/itex] or [itex] x+\frac{1}{x} \geq 2[/itex]. So when we include the range restriction of [itex]\cos(\theta)[/itex], we end up with only two possible values of "x". That is, [itex]x = 1[/itex] or [itex]x = -1[/itex].
 
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  • #9
Yes - sorry: since you'd already said that, I left it to context. I just filled in the other half.

I should have been more exact: here, let me practice:

Since, in the RHS, the cosine must be in [-1,1], then the LHS (x+1/x) must be in [-2,2] ... but LHS is not in (-2,2), therefore, there are only two values for x which can possibly make the relation true: x+1/x = {-2,2} i.e. x = {-1,1} (being a bit loose with the equals sign here).

I had noticed that the relation could only be true for specific values but had not bothered to calculate them. It is not that uncommon to give students an exercise which could be completed more easily by another method... mountaineers may climb a peak by a route other than the easiest one too ;)
 

1. What is induction?

Induction is a mathematical proof technique used to show that a statement is true for all natural numbers. It involves proving that the statement is true for a base case, usually n=1, and then showing that if the statement is true for n=k, it is also true for n=k+1.

2. How is induction used to prove x^n + 1/x^n = 2cos(n*theta)?

In this case, induction is used to prove that the statement is true for all natural numbers n. The base case is n=1, where x^1 + 1/x^1 = x + 1/x = 2cos(theta). Then, assuming the statement is true for n=k, we can use the identity (x^n+1/x^n)(x+1/x) = x^(n+1) + 1/x^(n+1) + x^(n-1) + 1/x^(n-1) to show that the statement is also true for n=k+1.

3. What is the significance of x^n + 1/x^n = 2cos(n*theta)?

This statement is known as the De Moivre's formula and is used to simplify complex trigonometric expressions. It allows us to write trigonometric functions in terms of powers of cos(theta) and sin(theta), making calculations and proofs easier.

4. Can induction be used to prove other trigonometric identities?

Yes, induction can be used to prove many other trigonometric identities that involve powers of cos(theta) and sin(theta). However, it may not be the most efficient or elegant proof technique for every identity.

5. Is induction the only way to prove x^n + 1/x^n = 2cos(n*theta)?

No, there are other proof techniques that can be used to prove this statement, such as direct proof or proof by contradiction. However, induction is a common and effective method for proving statements involving powers of integers.

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