Sturm’s theorem. Its main objective is to discover the zeros of a polynomial function .
Summary
[ hide ]
- 1 Story
- 1 Notations
- 2 Sturm’s theorem
- 1 In practice
- 3 Example 1
- 1 Solution
- 4 Source
History
The method was developed by the French mathematician Jacques Charles François Sturm (September 29, 1803 – December 15, 1855 ) and has as its main objective: to determine intervals in which the real zeros of a polynomial function are found.
Notations
The divisions are made until f k + 1 (x) is a constant polynomial. We will consider the sequence f 0 , f 1 , …, f k + 1 which we will call the Sturm polynomial sequence :
- The sequence is evaluated at x = a and let W (a) be the number of sign changes. .
- The sequence is evaluated at x = b and let W (b) be the number of sign changes. .
- Null terms must be discarded to count sign changes. .
Sturm’s theorem
If the real numbers a and b (a <b) are not roots of the polynomial f (x), which lacks multiple roots, then:
- W (a) ≥ W (c).
- W (a) – W (c) gives the number of real roots of the equation f (x) = 0 on the interval (a, b). .
In practice
- Evaluate f in two numbers (one negative and the other positive) large enough in absolute value. This determines the amount of real roots.
- Evaluate at x = 0 and count the sign changes of the sequence.
- With 1 and 2 determine the number of real negative and non-negative roots.
- Evaluate in other numbers, negative and non-negative, the intervals in which the roots are found are refined.
Example 1
Determine the intervals where the zeros of the function defined by the equation f (x) = x 3 -9x 2 + 24x-36 are found.
Solution
Deriving and performing the divisions indicated by the method we have the polynomials: f 0 (x) = x 3 -9x 2 + 24x-36
f 1 (x) = 3x 2 -18x + 24
f 2 (x) = 18x + 108
f 3 (x) = -77760
Sign table is constructed with K> 0 “large enough”
| Signs | -K | 0 | K |
| f 0 (x) | – | – | + |
| f 1 (x) | + | + | + |
| f 2 (x) | – | + | + |
| f 3 (x) | – | – | – |
| Changes | 2 | 2 | one |
There is only one real root and this is positive. The remaining two are conjugated imaginary.
| Signs | 0 | 10 |
| f 0 (x) | – | + |
| f 1 (x) | + | + |
| f 2 (x) | + | + |
| f 3 (x) | – | – |
| Changes | 2 | one |
With this new evaluation we note that the root is in the interval (0.10). By carrying out other evaluations, the length of the interval where the root is found can be decreased.
