# 物理代写|计算物理代写Computational Physics代考|PHY410

## 物理代写|计算物理代写Computational Physics代考|Bisection–Newton–Raphson

From the previous section, we note that the Bisection method is robust to initial guesses but slow to converge. Whereas the NewtonRaphson method will converge rapidly but only if the initial guess is relatively close to the root, and in some circumstances may fail to converge at all. Also, the Newton-Raphson method may fail to converge if the gradient of the function in the neighborhood of the root approaches zero. We would therefore like to combine the reliability of the Bisection method with the rapid convergence of the Newton-Raphson method so that for any general function we can find its roots with relative ease.

We can do this by making a hybrid method that decides whether to take a Newton-Raphson step or a Bisection step. As computers cannot think for themselves, we as programmers must provide some logical criteria to determine the step to take. Crucially, if an NR step takes the next approximation outside of our interval, then we should discard it and apply a Bisection step instead; else we accept the NR step. To do this, let us consider our Bisection interval $[a, b]$ with some best approximation to the root, $r$, contained within that interval. To accept the NR step the following inequality has to be satisfied
$$a<r-\frac{f(r)}{f^{\prime}(r)}<b .$$

Now while we could use this inequality as the conditional expression in an if statement the coding becomes lengthy and rather difficult to read. We can make our lives easier by rearranging the inequality into the form
$$y \geq 0 \geq z .$$
In other words, to satisfy the inequality (4.8) such that the NR step is accepted, the left-hand expression, $y$, must be positive or zero, and the right-hand expression, $z$, must be negative or zero. We can therefore apply the trick of comparing the product of $y$ and $z$ to zero to determine whether the inequality has been satisfied. To rearrange inequality (4.8) into the form of $(4.9)$ we subtract $r$, multiply through by $-f^{\prime}(r)$, and lastly, subtract $f(r)$ resulting in
$$(r-a) f^{\prime}(r)-f(r) \geq 0 \geq(r-b) f^{\prime}(r)-f(r)$$

## 物理代写|计算物理代写Computational Physics代考|Brute Force Search

In the previous sections, we have developed solid methods to accurately compute the roots of a function, so long as we know the rough locations of those roots in advance. The problem then is finding those rough locations. One straightforward technique is to graph the functions either by hand or using a plotting program and obtain those bounds by eye. This is recommended when finding the roots of a function is the problem to solve. But what if the root-finding is only one part of a bigger problem? It would be impractical to manually locate the rough location of roots for numerous functions in this case.

Typically, we use an exhaustive root search across a region of interest (ROI) for the function. That is, starting at the minimum value of the ROI we step the value of $x$ by some small amount and check to see if the function has changed significantly within that small step. If it has, we have found the bounds of at least one root, if not we continue the search. This continues until the whole ROI has been covered. How then do we decide on the step size? Too small and we make our rapidly converging root-inding algorithms redun(ant; too large and we run the risk of stepping over multiple roots (for an even number of roots this means missing them entirely, for step size is an educated guess and is very much dependent on the function under investigation.

The program rootSearch.cpp showcases our root searching classes on the Legendre polynomial:
$$P_{\mathrm{s}}=\frac{6435 x^5-12012 x^6+6930 x^4-1260 x^2+35}{128}$$

## 物理代写|计算物理代写Computational Physics代考| 等分牛顿拉弗森定律

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$$a<r-\frac{f(r)}{f^{\prime}(r)}<b .$$

$$y \geq 0 \geq z .$$

$$(r-a) f^{\prime}(r)-f(r) \geq 0 \geq(r-b) f^{\prime}(r)-f(r)$$

## 物理代写|计算物理代写Computational Physics代考|蛮力搜索

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$$P_{\mathrm{s}}=\frac{6435 x^5-12012 x^6+6930 x^4-1260 x^2+35}{128}$$

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