How to resolve the algorithm Roots of a quadratic function step by step in the Mathematica/Wolfram Language programming language
How to resolve the algorithm Roots of a quadratic function step by step in the Mathematica/Wolfram Language programming language
Table of Contents
Problem Statement
Write a program to find the roots of a quadratic equation, i.e., solve the equation
a
x
2
b x + c
0
{\displaystyle ax^{2}+bx+c=0}
. Your program must correctly handle non-real roots, but it need not check that
a ≠ 0
{\displaystyle a\neq 0}
. The problem of solving a quadratic equation is a good example of how dangerous it can be to ignore the peculiarities of floating-point arithmetic. The obvious way to implement the quadratic formula suffers catastrophic loss of accuracy when one of the roots to be found is much closer to 0 than the other. In their classic textbook on numeric methods Computer Methods for Mathematical Computations, George Forsythe, Michael Malcolm, and Cleve Moler suggest trying the naive algorithm with
a
1
{\displaystyle a=1}
,
b
−
10
5
{\displaystyle b=-10^{5}}
, and
c
1
{\displaystyle c=1}
. (For double-precision floats, set
b
−
10
9
{\displaystyle b=-10^{9}}
.) Consider the following implementation in Ada: As we can see, the second root has lost all significant figures. The right answer is that X2 is about
10
− 6
{\displaystyle 10^{-6}}
. The naive method is numerically unstable. Suggested by Middlebrook (D-OA), a better numerical method: to define two parameters
q
a c
/
b
{\displaystyle q={\sqrt {ac}}/b}
and
f
1
/
2 +
1 − 4
q
2
/
2
{\displaystyle f=1/2+{\sqrt {1-4q^{2}}}/2}
and the two roots of the quardratic are:
− b
a
f
{\displaystyle {\frac {-b}{a}}f}
and
− c
b f
{\displaystyle {\frac {-c}{bf}}}
Task: do it better. This means that given
a
1
{\displaystyle a=1}
,
b
−
10
9
{\displaystyle b=-10^{9}}
, and
c
1
{\displaystyle c=1}
, both of the roots your program returns should be greater than
10
− 11
{\displaystyle 10^{-11}}
. Or, if your language can't do floating-point arithmetic any more precisely than single precision, your program should be able to handle
b
−
10
6
{\displaystyle b=-10^{6}}
. Either way, show what your program gives as the roots of the quadratic in question. See page 9 of "What Every Scientist Should Know About Floating-Point Arithmetic" for a possible algorithm.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Roots of a quadratic function step by step in the Mathematica/Wolfram Language programming language
The provided Wolfram programming language code demonstrates various ways to solve polynomial equations and find roots. Here's a detailed explanation:
1. Solve[a x^2 + b x + c == 0, x]
This line solves a general quadratic equation a x^2 + b x + c == 0 for x. It finds the two possible values of x that satisfy the equation.
2. Solve[x^2 - 10^5 x + 1 == 0, x]
This line specifically solves the quadratic equation x^2 - 10^5 x + 1 == 0 for x. It again returns the two possible solutions.
3. Root[#1^2 - 10^5 #1 + 1 &, 1]
This line uses the Root function to find the first root of the polynomial #1^2 - 10^5 #1 + 1 (which is the same polynomial as above). The & symbol in Wolfram is used to create an anonymous function that takes one argument and returns #1^2 - 10^5 #1 + 1. The 1 argument tells Root to find the first root.
4. Root[#1^2 - 10^5 #1 + 1 &, 2]
This line follows the same principle as the previous one, but it tries to find the second root of the same polynomial.
5. Reduce[a x^2 + b x + c == 0, x]
This line simplifies a general quadratic equation a x^2 + b x + c == 0 by factoring out common terms and reducing it to its irreducible form. It does not explicitly solve for x.
6. Reduce[x^2 - 10^5 x + 1 == 0, x]
This line simplifies the specific quadratic equation x^2 - 10^5 x + 1 == 0 using the same rules as Reduce.
7. FindInstance[x^2 - 10^5 x + 1 == 0, x, Reals, 2]
This line tries to find two numerical approximations for the roots of x^2 - 10^5 x + 1 == 0 within the set of real numbers. The FindInstance function returns a list of approximate solutions.
8. FindRoot[x^2 - 10^5 x + 1 == 0, {x, 0}]
This line attempts to find a root of x^2 - 10^5 x + 1 == 0 starting with an initial guess of x = 0. FindRoot uses numerical methods to find a solution.
9. FindRoot[x^2 - 10^5 x + 1 == 0, {x, 10^6}]
Similar to the previous line, this one tries to find a root of the same equation but starts with a different initial guess of x = 10^6.
Source code in the wolfram programming language
Solve[a x^2 + b x + c == 0, x]
Solve[x^2 - 10^5 x + 1 == 0, x]
Root[#1^2 - 10^5 #1 + 1 &, 1]
Root[#1^2 - 10^5 #1 + 1 &, 2]
Reduce[a x^2 + b x + c == 0, x]
Reduce[x^2 - 10^5 x + 1 == 0, x]
FindInstance[x^2 - 10^5 x + 1 == 0, x, Reals, 2]
FindRoot[x^2 - 10^5 x + 1 == 0, {x, 0}]
FindRoot[x^2 - 10^5 x + 1 == 0, {x, 10^6}]
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