1.- We begin by mapping out the chart of $latex e ^ x$ as well as $latex 2-x ^ 2$. The obstruct factors are the visual options to the issue.

Example 7 of Newton Raphson method solution 1

There are 2 x-intercepts, which represent the options of the provided formula. Among them has an x-coordinate worth near -1.5 as well as the various other, an x-coordinate near +0.5.

3.- To use the Newton-Rapson approach, we specify the feature:

$$ f( x) = e ^ {x} -2 + x ^ 2 $$

the worths where $latex f( x)= 0$ will certainly be the options to the provided formula.

4.- Discovering the by-product of f( x):

$$ f'( x) = e ^ {x} + 2x $$

5.- To locate the initial origin of f( x), we take as beginning factor of the approach the worth $latex x _ {0} =-1.5$.

6.- We use Newton Raphson’s formula together up until the preferred precision is gotten to:

$ latex x _ {i +1} = x_i– frac {f( x_i)} {f'( x_i)} $, beginning with the preliminary worth, $latex x _ {0} $ to obtain $latex x _ {1} $. After that with $latex x _ {1} $ to obtain $latex x _ {2} $. It is duplicated up until the distinction in between the last worth minus the previous one has an outright worth much less than or equivalent to 0.00001

The mathematical outcomes are revealed listed below:

Example 7 of Newton Raphson method solution 2

acquiring the worth of the initial remedy: x= -1.31597

8.- The 2nd remedy to the formula is close to +0.5, so this will certainly be made use of as a beginning worth to get the remedy with the asked for accuracy of at the very least 4 decimal areas.

9.- Considering that we began with a worth fairly near the specific origin, we got an outcome with an accuracy of 3 decimal areas in just 2 models:

Example 7 of Newton Raphson method solution 3

The 2nd remedy to the provided formula is: x= +0.53727

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