Convergent Real Sequence/Examples/x (n+1) = x n^2 + k/Lemma 2

Example of Convergent Real Sequence

Let $a$ and $b$ be the roots of the quadratic equation:

$(1): \quad x^2 - x + k = 0$

Let:

$0 < k < \dfrac 1 4$

Then $a$ and $b$ are both strictly positive real numbers.

Proof

First we investigate the consequences of the condition $k < \dfrac 1 4$.

By Solution to Quadratic Equation with Real Coefficients:

In order for the quadratic equation $a x^2 + b x + c$ to have real roots, its discriminant $b^2 - 4 a c$ needs to be strictly positive.

The discriminant $D$ of $(1)$ is:

\(\ds D\)

\(=\)

\(\ds \paren {-1}^2 - 4 \times 1 \times k\)

\(\ds \)

\(=\)

\(\ds 1 - 4 k\)

Thus:

\(\ds k\)

\(<\)

\(\ds \dfrac 1 4\)

\(\ds \leadsto \ \ \)

\(\ds 1 - 4 k\)

\(>\)

\(\ds 0\)

and so when $k < \dfrac 1 4$, $(1)$ has real roots.

$\Box$

Next we investigate the consequences of the condition $0 < k$.

By Solution to Quadratic Equation:

\(\ds x\)

\(=\)

\(\ds \dfrac {-b \pm \sqrt {b^2 - 4 a c} } {2 a}\)

where $b = -1$, $a = 1$, $c = k$ in $(1)$

\(\ds \)

\(=\)

\(\ds \dfrac {-\paren {-1} \pm \sqrt {\paren {-1}^2 - 4 \times 1 \times k} } {2 \times 1}\)

\(\ds \)

\(=\)

\(\ds \dfrac {1 \pm \sqrt {1 - 4 k} } 2\)

We have that:

\(\ds 0\)

\(<\)

\(\ds k\)

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(>\)

\(\ds 1 - 4 k\)

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(>\)

\(\ds +\sqrt {1 - 4 k}\)

\(\ds -1\)

\(<\)

\(\ds -\sqrt {1 - 4 k}\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac {1 \pm \sqrt {1 - 4 k} } 2\)

\(>\)

\(\ds 0\)

That is, when $0 < k$ both roots of $(1)$ are strictly positive.

$\Box$

Hence when $0 < k < \dfrac 1 4$, both roots of $(1)$ are strictly positive real numbers.

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 5$: Subsequences: Exercise $\S 5.7 \ (2)$