Question

Let $a_0 = 1$ and for all $n \ge 1$, let $a_n$ be the smaller root of the equation $4^{-n}x^2 - x + a_{n-1} = 0$. Given that $a_n$ approaches a value $L$ as $n$ goes to infinity, then the value of $L$ is

• A. $\frac{\pi^2}{4}$
• B. $\frac{\pi^2}{3}$
• C. $\frac{\pi}{2}$
• D. $\frac{3\pi}{4}$

Answer 1

Answer: (A)

$\frac{\pi^2}{4}$

Answer 2

The problem defines a sequence $a_n$ recursively.

Given $a_0 = 1$, and for $n \ge 1$, $a_n$ is the smaller root of the equation $4^{-n}x^2 - x + a_{n-1} = 0$.

We need to find the limit $L = \lim_{n \to \infty} a_n$.

The quadratic equation is $4^{-n}x^2 - x + a_{n-1} = 0$.

The roots are given by the quadratic formula $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$:

$x = \frac{1 \pm \sqrt{(-1)^2 - 4(4^{-n})(a_{n-1})}}{2(4^{-n})}$

$x = \frac{1 \pm \sqrt{1 - 4^{1-n}a_{n-1}}}{2 \cdot 4^{-n}}$

Since $a_n$ is the smaller root, we have:

$a_n = \frac{1 - \sqrt{1 - 4^{1-n}a_{n-1}}}{2 \cdot 4^{-n}}$

Let's calculate the first few terms of the sequence:

For $n=1$: The equation is $4^{-1}x^2 - x + a_0 = 0$.

Given $a_0 = 1$, we have $\frac{1}{4}x^2 - x + 1 = 0$.

Multiplying by 4, we get $x^2 - 4x + 4 = 0$.

This is $(x-2)^2 = 0$, which has a single root $x=2$.

So, $a_1 = 2$.

For $n=2$: The equation is $4^{-2}x^2 - x + a_1 = 0$.

Given $a_1 = 2$, we have $\frac{1}{16}x^2 - x + 2 = 0$.

Multiplying by 16, we get $x^2 - 16x + 32 = 0$.

The roots are $x = \frac{16 \pm \sqrt{(-16)^2 - 4(1)(32)}}{2(1)} = \frac{16 \pm \sqrt{256 - 128}}{2} = \frac{16 \pm \sqrt{128}}{2}$.

Since $\sqrt{128} = \sqrt{64 \cdot 2} = 8\sqrt{2}$, the roots are $x = \frac{16 \pm 8\sqrt{2}}{2} = 8 \pm 4\sqrt{2}$.

$a_2$ is the smaller root, so $a_2 = 8 - 4\sqrt{2}$.

Let's try to find a pattern or a transformation.

Consider the expression for $a_n$:

$a_n = \frac{1 - \sqrt{1 - 4^{1-n}a_{n-1}}}{2 \cdot 4^{-n}}$

This form reminds us of the half-angle formula for sine: $2\sin^2(\theta/2) = 1 - \cos\theta$.

Let $4^{1-n}a_{n-1} = \sin^2\theta_n$ for some angle $\theta_n$.

Then $a_n = \frac{1 - \sqrt{1 - \sin^2\theta_n}}{2 \cdot 4^{-n}} = \frac{1 - |\cos\theta_n|}{2 \cdot 4^{-n}}$.

For the smaller root, we typically choose the positive square root for $\sqrt{1-x}$ when $x$ is small and positive, so $\sqrt{1-\sin^2\theta_n} = \cos\theta_n$ (assuming $\theta_n \in [0, \pi/2]$).

Then $a_n = \frac{1 - \cos\theta_n}{2 \cdot 4^{-n}} = \frac{2\sin^2(\theta_n/2)}{2 \cdot 4^{-n}} = 4^n \sin^2(\theta_n/2)$.

From $4^{1-n}a_{n-1} = \sin^2\theta_n$, we have $a_{n-1} = \frac{4^{n-1}}{4} \sin^2\theta_n = 4^{n-1} \sin^2\theta_n$.

So, we have the relations:

$a_n = 4^n \sin^2(\theta_n/2)$

$a_{n-1} = 4^{n-1} \sin^2\theta_n$

Let's try to make the argument of sine consistent. Let's define $a_n = 4^n \sin^2 \phi_n$.

Then $a_{n-1} = 4^{n-1} \sin^2 \phi_{n-1}$.

Substitute this into the relation $4^{1-n}a_{n-1} = \sin^2\theta_n$:

$4^{1-n} (4^{n-1} \sin^2 \phi_{n-1}) = \sin^2\theta_n$

$4^{1-n+n-1} \sin^2 \phi_{n-1} = \sin^2\theta_n$

$4^0 \sin^2 \phi_{n-1} = \sin^2\theta_n$

$\sin^2 \phi_{n-1} = \sin^2\theta_n$.

Since we are dealing with a sequence that seems to be increasing and positive, we can assume $\phi_n$ and $\theta_n$ are in $[0, \pi/2]$.

So, $\sin\phi_{n-1} = \sin\theta_n$, which implies $\phi_{n-1} = \theta_n$.

Now substitute $\theta_n = \phi_{n-1}$ into $a_n = 4^n \sin^2(\theta_n/2)$:

$a_n = 4^n \sin^2(\phi_{n-1}/2)$.

But we also defined $a_n = 4^n \sin^2 \phi_n$.

Therefore, $\sin^2 \phi_n = \sin^2(\phi_{n-1}/2)$.

This implies $\phi_n = \phi_{n-1}/2$.

This is a simple geometric progression for $\phi_n$.

$\phi_n = \phi_0 / 2^n$.

Now we need to find $\phi_0$.

We have $a_0 = 1$.

The definition $a_n = 4^n \sin^2 \phi_n$ is for $n \ge 1$.

Let's use $a_0 = 1$ and $a_1 = 2$.

From $a_1 = 4^1 \sin^2 \phi_1 = 4 \sin^2 \phi_1 = 2$.

So $\sin^2 \phi_1 = 1/2$.

$\sin \phi_1 = 1/\sqrt{2}$.

Since we can choose $\phi_1 \in [0, \pi/2]$, we have $\phi_1 = \pi/4$.

Now using $\phi_n = \phi_0 / 2^n$:

For $n=1$, $\phi_1 = \phi_0 / 2^1$.

$\pi/4 = \phi_0 / 2$.

So $\phi_0 = \pi/2$.

Now we can write $a_n$ in terms of $\phi_0$:

$a_n = 4^n \sin^2 \phi_n = 4^n \sin^2 (\phi_0 / 2^n) = 4^n \sin^2 (\frac{\pi/2}{2^n}) = 4^n \sin^2 (\frac{\pi}{2^{n+1}})$.

We need to find the limit $L = \lim_{n \to \infty} a_n$.

$L = \lim_{n \to \infty} 4^n \sin^2 (\frac{\pi}{2^{n+1}})$

$L = \lim_{n \to \infty} (2^n)^2 \sin^2 (\frac{\pi}{2^{n+1}})$

$L = \lim_{n \to \infty} \left( 2^n \sin (\frac{\pi}{2^{n+1}}) \right)^2$

Let $x = \frac{\pi}{2^{n+1}}$. As $n \to \infty$, $x \to 0$.

We know that $\lim_{x \to 0} \frac{\sin x}{x} = 1$.

So, $\sin x \approx x$ for small $x$.

$L = \lim_{n \to \infty} \left( 2^n \cdot \frac{\pi}{2^{n+1}} \right)^2$

$L = \lim_{n \to \infty} \left( \frac{2^n \cdot \pi}{2^n \cdot 2^1} \right)^2$

$L = \lim_{n \to \infty} \left( \frac{\pi}{2} \right)^2$

$L = \frac{\pi^2}{4}$.

Final check:

If $a_n = \frac{\pi^2}{4}$, then $a_n \to L = \frac{\pi^2}{4}$.

The condition for roots to be real is $1 - 4^{1-n}a_{n-1} \ge 0$.

As $n \to \infty$, $4^{1-n}a_{n-1} \to 0 \cdot L = 0$. So $1-0 \ge 0$, which is true.

The values are positive, so our choice of $\sin\phi_n$ and $\cos\theta_n$ being positive is consistent.

The value of $L$ is $\frac{\pi^2}{4}$.