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Approximate
solution of one equation using a quadratic or cubic regression
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Print!
| Objectives | To show how certain statistical knowledge can intervene in analysis |
| Public |
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| Prérequis | Knowledge of the method of least squares |
| Idea | B.Egger |
Unfolding:
That is to say to solve the equation x-cos(x)=0.
A fast study shows that this equation shows that this equation does not admit that a solution ranging between 0 and 1.
From some points of the curve representative of the function f(x)=x-cos(x), we will create polynomials of degree 2 or 3 passer by by the cloud formed by these points and will replace the preceding equation by polynomial equations.
Commencons by 4 points.
| One re-enters the column c1 in the form of a continuation and the
column c2 like image of c1 by F: c1=seq...... c2=f(c1) |
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| Determination of a quadratic regression for the scatter plot
considered. One rage the result in y1. |
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| Even thing with a cubic regression. One arranges the result in y2. |
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| The " real " solution of the equation and approximate
solutions (obtained much more quickly). Only one disadvantage: other solutions parasitic, but easy to eliminate. |
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Let us increase the number of points of the cloud, and remake calculations.
**time-out** one have improve the precision in increase the number of point, but it be not always the case... as it be possible also to find some situation for which from same cloud, the regression quadratic be good than the regression cubic.
We know that it is normal from the point of view of the statistics, but that can not be obvious for the pupils.
Let us examine the case of the function F defined by f(x) = ln(x) + 10x
