Nested Radicals and a Cosine Formula

Problem Statement

The numbers $a_n$, for integers $n \geq 1$, are defined as:

These numbers satisfy the relation $a_{n+1}^2 = 2 + a_n$, for $n \geq 1$. (Do NOT prove this.)

Use mathematical induction to prove that $a_n = 2\cos\left(\frac{\pi}{2^{n+1}}\right)$ for all integers $n \geq 1$.

Hints

Attempt the proof independently first. Focus on the key theorem, algebraic transformation, or contradiction setup that links the hypothesis to the target conclusion.

Solutions

Proof by Mathematical Induction

Base Case ($n=1$):

LHS: $a_1 = \sqrt{2}$ (given)

RHS: $2\cos\left(\frac{\pi}{2^{1+1}}\right) = 2\cos\left(\frac{\pi}{4}\right) = 2 \cdot \frac{\sqrt{2}}{2} = \sqrt{2}$

Since LHS = RHS, the statement holds for $n=1$. \checkmark

Inductive Hypothesis:

Assume the statement is true for $n=k$, where $k$ is a positive integer:

Inductive Step:

We must prove the statement for $n=k+1$:

Using the recurrence relation $a_{k+1}^2 = 2 + a_k$ and the inductive hypothesis:

Apply the double angle identity: $1 + \cos(2\theta) = 2\cos^2(\theta)$.

Let $2\theta = \frac{\pi}{2^{k+1}}$, so $\theta = \frac{\pi}{2^{k+2}}$:

Taking square roots:

Since $n \geq 1$, we have $0 < \frac{\pi}{2^{k+2}} < \frac{\pi}{2}$, so $\cos\left(\frac{\pi}{2^{k+2}}\right) > 0$.

Also, $a_n$ is defined as nested square roots of positive numbers, so $a_{k+1} > 0$.

Therefore:

This completes the inductive step. \checkmark

Conclusion:

By mathematical induction, $a_n = 2\cos\left(\frac{\pi}{2^{n+1}}\right)$ for all integers $n \geq 1$.

\hfill $\square$

Takeaways

• Induction with Recurrence: Use given recurrence relation in inductive step to relate $a_{k+1}$ to $a_k$
• Double Angle Formula: Identity $1 + \cos(2\theta) = 2\cos^2(\theta)$ is key to converting sum to square
• Sign Consideration: Must justify taking positive square root using domain/range analysis
• Angle Halving: Pattern shows $a_n$ relates to $\cos$ of successively halved angles ($\pi/4, \pi/8, \pi/16, \ldots$)

Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills:

• HSC Proofs: https://vumaths.com/booklets/hsc-proofs/
• HSC Polynomials: https://vumaths.com/booklets/hsc-polynomials/
• HSC Probability: https://vumaths.com/booklets/hsc-probability/

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