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modified newton's method for finding roots
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src/num_methods/roots_newton.md

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@@ -18,7 +18,9 @@ We want to solve the equation. More precisely, we want to find one of its roots
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The input parameters of the algorithm consist of not only the function $f(x)$ but also the initial approximation - some $x_0$, with which the algorithm starts.
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![](roots_newton.png)
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<div style="text-align:center">
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![Image](roots_newton.png)
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</div>
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Suppose we have already calculated $x_i$, calculate $x_{i+1}$ as follows. Draw the tangent to the graph of the function $f(x)$ at the point $x = x_i$, and find the point of intersection of this tangent with the $x$-axis. $x_{i+1}$ is set equal to the $x$-coordinate of the point found, and we repeat the whole process from the beginning.
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src/num_methods/roots_newton.png

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