# Halving the Interval Method with Relative Accuracy

## Find the smallest positive root of \( x^3-2x + 1 = 0 \) with relative accuracy of \( 0.5 \% \) by finding good starting value from its graph.

According to the graph above, there are three negative and two positive roots.

Let’s give \( f(x) = x^3-2x+1 \), then \( f(1) = 1^3-2 \times 1 + 1 = 0 \).

Therefore \( x = 1 \) is the largest positive root as well as shown in the graph.

So the smallest positive root lies between \( x=0 \) and \( x=1 \).

Though this interval is good enough to start applying “halving the interval method”, it could be better to make the working more effective and efficient, say \( 0.5 < x < 0.9 \).

To find the rate of accuracy by taking the relative accuracy method, we need two values, the error and the “believed” true value.

In this case, the true value is \( \displaystyle \frac{0.5+0.9}{2} = 0.7 \) as we believe \( 0.7 \) is more accurate than both \( 0.5 \) and \( 0.9 \) and the error is \( \left| 0.5-0.7 \right| = 0.2 \) or \( \left| 0.7-0.9 \right| = 0.2 \).

Thus the relative accuracy is \( \displaystyle \frac{ \left| 0.5-0.7 \right| }{0.7} \times 100 \% =28.57 \% \).

## Applying Halving the Interval Method

Let’s start to find the smallest positive root!

\( f(0.5) = 0.125 > 0 \)

\( f(0.9) = -0.057 < 0 \)

The interval is \( 0.5 < x < 0.9 \) as the closing value of \( x \) lies between the opposite sign of its function values.

The midpoint is \( \displaystyle \frac{0.5+0.9}{2} = 0.7 \).

Its relative accuracy is \( \displaystyle \frac{|0.5-0.7|}{0.7} \times 100 \% = 28.57 \% \).

\( f(0.7) = -0.057 < 0 \)

The refined interval is \( 0.5 < x < 0.7 \).

The midpoint is \( \displaystyle \frac{0.5+0.7}{2} = 0.6 \).

Its relative accuracy is \( \displaystyle \frac{|0.5-0.6|}{0.6} \times 100 \% = 16.67 \% \).

\( f(0.6) = 0.016 > 0 \)

The refined interval is \( 0.6 < x < 0.7 \).

The midpoint is \( \displaystyle \frac{0.6+0.7}{2} = 0.65 \).

Its relative accuracy is \( \displaystyle \frac{|0.6-0.65|}{0.65} \times 100 \% = 7.69 \% \).

\( f(0.65) = -0.0253 \cdots < 0 \)

The refined interval is \( 0.6 < x < 0.65 \).

The midpoint is \( \displaystyle \frac{0.6+0.65}{2} = 0.625 \).

Its relative accuracy is \( \displaystyle \frac{|0.6-0.625|}{0.625} \times 100 \% = 4 \% \).

\( f(0.625) = -0.0058 \cdots < 0 \)

The refined interval is \( 0.6 < x < 0.625 \).

The midpoint is \( \displaystyle \frac{0.6+0.625}{2} = 0.6125 \).

Its relative accuracy is \( \displaystyle \frac{|0.6-0.6125|}{0.6125} \times 100 \% = 2.04 \% \).

\( f(0.6125) = 0.0047 \cdots > 0 \)

The refined interval is \( 0.6125 < x < 0.625 \).

The midpoint is \( \displaystyle \frac{0.6125+0.625}{2} = 0.61875 \).

Its relative accuracy is \( \displaystyle \frac{|0.6125-0.61875|}{0.61875} \times 100 \% = 1.01 \% \).

\( f(0.61875) = -0.0006 \cdots < 0 \)

The refined interval is \( 0.6125 < x < 0.61875 \).

The midpoint is \( \displaystyle \frac{0.6125+0.61875}{2} = 0.615625 \).

Its relative accuracy is \( \displaystyle \frac{|0.6125-0.615625|}{0.615625} \times 100 \% = 0.51 \% \ \)So close!

\( f(0.615625) = 0.002 \cdots > 0 \)

The refined interval is \( 0.615625< x < 0.61875 \).

The midpoint is \( \displaystyle \frac{0.615625+0.61875}{2} = 0.6171875 \).

Its relative accuracy is \( \require{AMSsymbols} \displaystyle \frac{|0.615625-0.6171875|}{0.6171875} \times 100 \% = \color{green}{0.25 \%} \).

As this \( \color{green}{0.25 \%} \) meets the minimum relative accuracy level of \( 0.5 \% \), the smallest positive root is \( \require{AMSsymbols}\color{green}{0.6171875} \)!

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