# Using Newton's Method to Approximate Solutions to Equations

## Key Questions

• The Newton-Raphson method approximates the roots of a function. So, we need a function whose root is the cube root we're trying to calculate.

Let's say we're trying to find the cube root of $3$. And let's say that $x$ is the cube root of $3$. Therefore,

${x}^{3} = 3$

For the Newton-Raphson method to be able to work its magic, we need to set this equation to zero.

${x}^{3} - 3 = 0$

Now we will recall the iterative equation for Newton-Raphson.

${x}_{n + 1} = {x}_{n} - \frac{f \left({x}_{n}\right)}{f ' \left({x}_{n}\right)}$

Substituting for $f \left(x\right) = {x}^{3} - 3$ gives us:

${x}_{n + 1} = {x}_{n} - \frac{{\left({x}_{n}\right)}^{3} - 3}{3 \cdot {\left({x}_{n}\right)}^{2}}$

Now, we pick an arbitrary number, (the closer it actually is to $\sqrt{3}$ the better) for ${x}_{0}$. Let's use ${x}_{0} = 0.5$. Then we substitute each previous number for ${x}_{n}$ back into the equation to get a closer and closer approximation to a solution of ${x}^{3} - 3 = 0$.

${x}_{1} = 0.5 - \frac{{\left(0.5\right)}^{3} - 3}{3 \cdot {\left(0.5\right)}^{2}} = 4.33333 \overline{3}$
${x}_{2} = {x}_{1} - \frac{{\left({x}_{1}\right)}^{3} - 3}{3 \cdot {\left({x}_{1}\right)}^{2}} \approx 2.94214333$
${x}_{3} = {x}_{2} - \frac{{\left({x}_{2}\right)}^{3} - 3}{3 \cdot {\left({x}_{2}\right)}^{2}} \approx 2.07695292$
${x}_{4} = {x}_{3} - \frac{{\left({x}_{3}\right)}^{3} - 3}{3 \cdot {\left({x}_{3}\right)}^{2}} \approx 1.61645303$
${x}_{5} = {x}_{4} - \frac{{\left({x}_{4}\right)}^{3} - 3}{3 \cdot {\left({x}_{4}\right)}^{2}} \approx 1.46034889$
${x}_{6} = {x}_{5} - \frac{{\left({x}_{5}\right)}^{3} - 3}{3 \cdot {\left({x}_{5}\right)}^{2}} \approx 1.44247296$
${x}_{7} = {x}_{6} - \frac{{\left({x}_{6}\right)}^{3} - 3}{3 \cdot {\left({x}_{6}\right)}^{2}} \approx 1.4422496$
${x}_{8} = {x}_{7} - \frac{{\left({x}_{7}\right)}^{3} - 3}{3 \cdot {\left({x}_{7}\right)}^{2}} \approx 1.44224957$

You can see that with only 8 iterations, we've obtained an approximation of $\sqrt{3}$ which is correct to 8 decimal places!

You can apply this same logic to whatever cube root you'd like to find, just use ${x}^{3} - a = 0$ as your equation instead, where $a$ is the number whose cube root you're looking for.

• Newton's Method is a mathematical tool often used in numerical analysis, which serves to approximate the zeroes or roots of a function (that is, all $x : f \left(x\right) = 0$).

The method is constructed as follows: given a function $f \left(x\right)$ defined over the domain of real numbers $x$, and the derivative of said function ($f ' \left(x\right)$), one begins with an estimate or "guess" as to where the function's root might lie. For example, suppose one is presented with the function $f \left(x\right) = {x}^{2} + x - 2.5$. This is similar to another function $g \left(x\right) = {x}^{2} + x - 2$, whose roots are $x = 1$ and $x = - 2$. Thus, thanks to this similarity, one might use $x = 1$ or $x = - 2$ as guesses to start Newton's Method with f(x).

(Alternately, if a graphical representation is available but the exact root is not listed, an acceptable approximation might be the nearest whole number to the root).

Whatever method used, we declare this initial guess to be ${x}_{0}$. We arrive at a better approximation, ${x}_{1}$, by employing the Method: ${x}_{1} = {x}_{0} - f \frac{{x}_{0}}{f ' \left({x}_{0}\right)}$. Essentially, by utilizing the derivative, one is able to increment closer to the actual value. In the above example, $f \left(x\right) = {x}^{2} + x - 2.5$, if we assume ${x}_{0} = 1$, then ${x}_{1} = 1 - f \frac{1}{f ' \left(1\right)} = 1 - \frac{- .5}{3} = \frac{7}{6} \mathmr{and} \approx 1.16667$.

Often, one may be able to find the root another way (by using a graphing calculator, for example), and an exam item or textbook problem may demand a certain degree of accuracy (such as within 1% of the actual value). In such a case, if ${x}_{1}$ is not an accurate enough approximation, one performs the iteration again, as often as needed for the desired degree of accuracy. The formula to find the general ${x}_{n}$, then, is ${x}_{n} = {x}_{n - 1} - f \frac{{x}_{n - 1}}{f ' \left({x}_{n - 1}\right)}$