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

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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:

\(\displaystyle D\) \(=\) \(\displaystyle \paren {-1}^2 - 4 \times 1 \times k\)
\(\displaystyle \) \(=\) \(\displaystyle 1 - 4 k\)


Thus:

\(\displaystyle k\) \(<\) \(\displaystyle \dfrac 1 4\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 1 - 4 k\) \(>\) \(\displaystyle 0\)

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:

\(\displaystyle x\) \(=\) \(\displaystyle \dfrac {-b \pm \sqrt {b^2 - 4 a c} } {2 a}\) where $b = -1$, $a = 1$, $c = k$ in $(1)$
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {-\paren {-1} \pm \sqrt {\paren {-1}^2 - 4 \times 1 \times k} } {2 \times 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {1 \pm \sqrt {1 - 4 k} } 2\)


We have that:

\(\displaystyle 0\) \(<\) \(\displaystyle k\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 1\) \(>\) \(\displaystyle 1 - 4 k\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 1\) \(>\) \(\displaystyle +\sqrt {1 - 4 k}\)
\(\displaystyle -1\) \(<\) \(\displaystyle -\sqrt {1 - 4 k}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac {1 \pm \sqrt {1 - 4 k} } 2\) \(>\) \(\displaystyle 0\)

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.

$\blacksquare$


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