# 3D Map Generator – Everest

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## What’s New In?

Q:

Newton’s method for finding an expansion point when not a power series

My problem is to use Newton’s method with a polynomial of the form
$x^n + ax^{n-1} + \ldots + ax + b$ with given real $a$ and non-negative real $b$.
Since Newton’s method typically assumes a power series expansion of the solution, it seems like the best way to approach this is to write $x = \frac{1}{n} + \frac{a}{n^2} + \frac{a^2}{n^3} + \ldots$, and use Newton’s method to find the exact solution.
However, since $a$ and $b$ are given, can we use the polynomial given to find all of the coefficients of this series? Or are there any tricks/observations to know that prevent this strategy from working?
I am having a hard time searching Google for any other discussion of this sort of problem.

A:

In the original Newton’s method, the idea was to use polynomials with known coefficients, so the example in your post is not the best example of Newton’s method. But one way to calculate the coefficients of polynomials by hand is to use Lagrange’s formula. With $p(x) = x^n + a_1x^{n-1} + \ldots + a_{n-1}x + a_n$, the following holds: $$p'(x) = nx^{n-1} + a_1x^{n-2} + \ldots + a_{n-1} = np(x) – a_1x^{n-2} – \ldots – a_{n-1}x$$
If we set $f(x) = np(x) – a_1x^{n-2} – \ldots – a_{n-1}x$ we have that $x= \frac{1}{n} + \frac{1}{n}f(x)$, and $x$ is the root of $x = \frac{1}{n} + \frac{1}{n} f(x)$ if and only if $f(x) = 0$ (if and only if you have \$nx^{n-1} + a_1x^{n

## System Requirements:

Supported OS:
Windows Vista, Windows XP SP2, Windows 2000
Mac OS X 10.4, 10.5, 10.6
Linux:
Red Hat Enterprise Linux 5.2
CUDA Version:
NVCC 3.0
CUDALibraries 3.0
x86 Compiler:
Visual C++ 2008 Express
Recommended System Specifications:
OS:
Windows Vista, Windows XP SP3, Windows 2000
Mac OS X 10.6
Linux: