问题
I am trying to write an algorithm which will find a(0),..., a(n-1), given the values of n, x_1, ..., x_n, a(n), such that:
a(n)*p^n + a(n-1)*p^(n-1) + ... + a(1)*p + a(0) = a(n)(p-x_1)(p-x_2)...(p-x_n)
for all real p.
After multiplying a(n)(p-x_1)(p-x_2) I've thought of using Viete's formulas to find the coefficients.
But it turns out writing the code down isn't as obvious as I expected.
I want to use only the basics in my code - that is loops, if-s addition and multiplication - no ready/ complex functions.
Here are the formulas:
First, I would like to emphasise that I only need a pseudocode, and I do not care about defining arrays for the root and coefficients. That's why I will just write a(n), xn. Oh, and I hope it won't bother you very much if I start indexing from i=1 not i=0 in order to be in synch with the mathematical notation. In order to start with i=0 I would have to renumerate the roots and introduce more brackets.
And this is what I've come up with so far:
a(n-1)=0;
for(i=1; i <= n; i++){
a(n-1) = a(n-1) + x_i;
}
a(n-1) = -a(n)*a(n-1);
a(n-2)=0;
for(i=1; i <= n; i++){
for(j=i; j <= n; j++){
a(n-2) = a(n-2)+ x_i * x_j;
}
}
a(n-2) = -a(n)*a(n-2);
a(n-3)=0;
for(i=1; i <= n; i++){
for(j=i; j <= n; j++){
for(k=j; k <= n; k++){
a(n-3) = a(n-3)+ x_i * x_j * x_k;
}
}
}
a(n-3) = a(n)*a(n-3);
...
a(0)=1;
for(i=1; i<=n; i++){
a(0) = a(0) * x_i;
}
if(n%2 == 0) a(0) = a(n) * a(0);
else a(0) = -a(n) * a(0);
As you can see, it doesn't look good.
I would like to link all those loops into one loop, because without I cannot write the full code, I cannot fill the gap between a(0) and a(n-j) for a fixed j.
Could you help me out?
This is what I have, based on Nico Schertler's answer:
for(i=1; i<=n; i++)
{a(i)=1;
for(j=1; j <= n; j++)
{b(i)= clone( a(i) );
a(i) = a(i-1);
b(i) = x_j * b(i);
c(i) = a(i) - b(i);
}
}
Would it be the same if instead we wrote
for(i=1; i<=n; i++)
{a(i)=1; b(i)=1;
for(j=1; j <= n; j++)
{t = a(i) ;
a(i) = a(i-1);
b(i) = x_j * t;
c(i) = a(i) - b(i);
}
}
(this is how we for example swap two elements of an array, by keeping the value of a[i] in some variable t).
回答1:
You can create the polynomial incrementally.
Start with p = 1. I.e. a(0) = 1.
In order to add a root, you have to multiply the current polynomial by x - x_i. This is:
p * (x - x_i) = p * x - p * x_i
So you need to support three operations:
1. Multiplication by x
This is quite simple. Just shift all coefficients by one to the left. I.e.
a(i ) := a(i - 1)
a(i - 1) := a(i - 2)
...
a(1 ) := a(0)
a(0 ) := 0
2. Multiplication by a scalar
This is equally simple. Multiply each coefficient:
a(i ) *= s
a(i - 1) *= s
...
3. Subtraction
Just subtract the respective coefficients:
c(i ) = a(i ) - b(i )
c(i - 1) = a(i - 1) - b(i - 1)
...
Altogether
Add root by root. First, clone your current polynomial. Then, do the operations as described above:
p := 1
for each root r
p' = clone(p)
multiply p with x
multiply p' with r
p := p - p'
next
回答2:
A static function in c# for this purpose. The roots of x^4-11x^3+44x^2-76x+48 are {2,2,3,4} and given the argument
roots = new Complex[4] {2, 2, 3, 4}
this function returns [48,-76,44,-11,1]
public static double[] FromRoots(Complex[] roots)
{
int N = roots.Length;
Complex[] coefs = new Complex[N + 1];
coefs[0] = -roots[0];
coefs[1] = 1.0;
for (int k = 2; k <= N; k++)
{
coefs[k] = 1.0;
for (int i = k - 2; i >= 0; i--)
{
coefs[i + 1] = coefs[i] - roots[k - 1] * coefs[i + 1];
}
coefs[0] *= -roots[k - 1];
if (Math.IEEERemainder(k, 2) == 1)
coefs[k] = -coefs[k];
}
double[] realCoefs = new double[N + 1];
for (int i = 0; i < N + 1; i++)
realCoefs[i] = coefs[i].Real; // Not sure about this part!
return realCoefs;
}
来源：https://stackoverflow.com/questions/33594384/find-the-coefficients-of-the-polynomial-given-its-roots