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I have to calculate the roots of the equation $x^3 + x^2 -3x -3 = 0$ in the interval $[1,2]$ using linear interpolation to six decimal places, by hand.

Now I know this is trivial in excel, but when doing it by hand it becomes very tedious. Is there a quicker way to do this using a calculator?

Bearing in mind I have to show some working to get marks, so it can't just be a plug in and press enter type of method.

I can not use any Ti calculators, I only have access to scientific casio calculators.

Any help would be appreciated, thanks.

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    $\begingroup$ Many root finding variants use linear interpolation, differing in the choice of endpoints to interpolate. Compare the secant method with regula falsi, for example. So the details may depend on the choice of root finding algorithm. But the linear interpolation computation is basically taking a proportion, so no more difficult using pencil and paper than a division and two subtractions. You also then evaluate the function at the new argument, assuming you are trying to refine the root approximation to a given accuracy. $\endgroup$ – hardmath Jun 21 '14 at 16:15
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So there are two key insights here.

Given $x^3 + x^2 - 3x - 3$ it is natural to ask if there are any 'obvious' factors. The way to find this is to remove common factors from the first 2 terms and last two terms and see if something in common is found:

$$x^2(x + 1) - 3(x+1)$$

Sure enough there is something in common and therefore we immediately know that the polynomial can be factored into:

$$(x+1)(x^2-3)$$

The second factor is clearly a difference of squares and therefore can itself be factored into

$$(x+1)(x - \sqrt{3})(x + \sqrt{3})$$

Thus the three solutions are $x = -1, \sqrt{3}, -\sqrt{3}$. Now to get their numerical values successive guess and checks can be useful. (ex: 1^2 < 3, 2^2 > 3, 1.5^2 < 3, 1.75^2 > 3, slowly converges to the correct value for $\sqrt{3}$)

There is a trick known as Newton's Method that may be of interest to you (it finds zeroes)

Which basically states that the next closest approximation for a solution to

$$f(x) = 0$$

is given by:

$$x_{n+1} = x_n - \frac{f'(x_n)}{f(x_n)}$$

Where $f'(x)$ is the derivative of the function $f(x)$. So for our purposes we then get the formula

$$x_{n+1}= x_n - \frac{3x^2 + 2x - 3}{x^3 + x^2 - 3x - 3}$$

Which allows to take a random test point of choice $x_0$ and successively apply this formula until it we get something sufficiently close to one of our target zeroes.

(There do exist generalizations of this formula involving second, third, fourth etc... derivative of functions for faster convergence under the name 'Householder methods'

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    $\begingroup$ Small mistake : $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$. I let you editing (don't forget $x_n$'s everwhere in the rhs). Cheers :) $\endgroup$ – Claude Leibovici Jun 21 '14 at 16:22

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