Polynomials and Roots: Properties and Analysis

In summary, the conversation discusses the topic of polynomials and their roots. The speaker shares a method for generating a polynomial with roots decreased by one unit, using an example to illustrate the process. The resulting polynomial is shown to have the same sum and alternate coefficient values as the original polynomial. This method can be useful for analyzing roots and factoring polynomials.
  • #1
mente oscura
168
0
Hello.

I open this 'thread', in number theory, but he also wears "calculation".

I've done a little research, I share with you.

[tex]Let \ r_1, r_2, \cdots, r_n[/tex], roots of the polynomial.

[tex]P(x)=p_0x^n+p_1x^{n-1}+ \cdots+p_{n-1}x+p_n[/tex]

[tex]Let \ Q(x)=q_0 x^n+q_1x^{n-1}+ \cdots +q_n[/tex], such that its roots are:

[tex]r_1-1, r_2-1, \cdots, r_n-1[/tex]

[tex]Let \ T(x)=t_0^n+t_1x^{n-1}+ \cdots +t_n[/tex], such that its roots are:

[tex]r_1+1, r_2+1, \cdots, r_n+1[/tex]

I will assume:

[tex]p_0=q_0=t_0=1[/tex]Therefore:

[tex]\displaystyle\sum_{i=0}^n(p_i)=q_n[/tex]

and

[tex]\displaystyle\sum_{i=0}^n(p_i)(-1)^{i+1}=t_n[/tex], if “n” it's even.

[tex]\displaystyle\sum_{i=0}^n(p_i)(-1)^i=t_n[/tex], if “n” it's odd.

Also I have found how to calculate "complete", recurrently cited polynomials.

Example:

[tex]P(x)=x^9-33x^8+149x^7+4431x^6-45669x^5+9081x^4+1506119x^3-7038363x^2+12556936x-7987980[/tex]

Sum of coefficients=-995328, addition and subtraction of alternate coefficients=-29030400

Roots:-11, -7, 2, 2, 3, 5, 7, 13, 19.[tex]Q(x)=x^9-24x^8-79x^7+4634x^6-17676x^5-149768x^4+1177824x^3-2853504x^2+2833920x-995328[/tex]

Sum of coefficients=0, addition and subtraction of alternate coefficients=-7987980

Roots:-12, -8, 1, 1, 2, 4, 6, 12, 18.[tex]T(x)=x^9-42x^8+449x^7+2380x^6-67152x^5+296240x^4+931632x^3-10983168x^2+30862080x-29030400[/tex]

Sum of coefficients=-7987980, addition and subtraction of alternate coefficients=-71442000

Roots:-10, -6, 3, 3, 4, 6, 8, 14, 20.

This procedure can be useful for the analysis of the possible roots of the polynomial, and its factorization.

Regards.
 
Last edited:
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  • #2
Hello.

Continuing with the topic.

I'm going to show, with an example, as it would generate a polynomial, with the roots of the original polynomial, decreased in 1 unit:

[tex]P(x)=x^9-33x^8+149x^7+4431x^6-45669x^5+9081x^4+1506119x^3-7038363x^2+12556936x-7987980[/tex]

Sum of coefficients=-995328

Roots:-11, -7, 2, 2, 3, 5, 7, 13, 19.[tex]Q(x)=x^9-24x^8-79x^7+4634x^6-17676x^5-149768x^4+1177824x^3-2853504x^2+2833920x-995328[/tex]

Roots:-12, -8, 1, 1, 2, 4, 6, 12, 18.
1º) Separate term=Sum of coefficients of P(x)=-995328

2º) Coefficient of [tex]x[/tex]:

[tex]\dfrac{P'(x)}{1}=9x^8-264x^7+1043x^6+26586x^5-228345x^4+36324x^3+4518357x^2-14076726x+12556936[/tex]

Coefficient of [tex]x[/tex], the new polynomialQ(x)=Sum of coefficients [tex]\dfrac{P'(x)}{1}[/tex]=2833920.

3º) Coefficient of [tex]x^2[/tex]:

[tex]\dfrac{P''(x)}{2}=36x^7-924x^6+3129x^5+66465x^4-456690x^3+54486x^2+4518357x-7038363[/tex].

Sum of coefficients=-2853504.

4º) Coefficient of [tex]x^3[/tex]:

[tex]\dfrac{P'''(x)}{3!}=84x^6-1848x^5+5215x^4+88620x^3-456690x^2+36324x+1506119[/tex].

Sum of coefficients=1177824.

5º) Coefficient of [tex]x^4[/tex]:

[tex]\dfrac{P''''(x)}{4!}=126x^5-2310x^4+5215x^3+66465x^2-228345x+9081[/tex].

Sum of coefficients=-149768.

6º) Coefficient of [tex]x^5[/tex]:

[tex]\dfrac{P'''''(x)}{5!}=126x^4-1848x^3+3129x^2+26586x-45669[/tex].

Sum of coefficients=-17676.

7º) Coefficient of [tex]x^6[/tex]:

[tex]\dfrac{P''''''(x)}{6!}=84x^3-924x^2+1043x+4431[/tex].

Sum of coefficients=4634.

8º) Coefficient of [tex]x^7[/tex]:

[tex]\dfrac{P'''''''(x)}{7!}=36x^2-264x+149[/tex]

Sum of coefficients=-79.

9º) Coefficient of [tex]x^8[/tex]:

[tex]\dfrac{P''''''''(x)}{8!}=9x-33[/tex]

Sum of coefficients=-24.

10º) Coefficient of [tex]x^9[/tex]:

[tex]\dfrac{P'''''''''(x)}{9!}=1[/tex].

Therefore, the resulting polynomial is:

[tex]Q(x)=x^9-24x^8-79x^7+4634x^6-17676x^5-149768x^4+1177824x^3-2853504x^2+2833920x-995328[/tex]

Regards.
 

Related to Polynomials and Roots: Properties and Analysis

1. What are polynomials and how are they different from other types of functions?

Polynomials are mathematical expressions that consist of variables and coefficients, combined using addition, subtraction, and multiplication. They are different from other types of functions because they do not involve division, radicals, or other more complex operations.

2. What are the properties of polynomials?

Some key properties of polynomials include: they have a finite number of terms, the degree of a polynomial is determined by its highest exponent, they can be added, subtracted, and multiplied using standard rules, and they can be factored into simpler expressions.

3. How do you find the roots of a polynomial?

The roots of a polynomial are the values of the variable that make the polynomial equal to zero. One way to find the roots is by factoring the polynomial and setting each factor equal to zero. Another method is to use the quadratic formula for polynomials of degree 2 or higher.

4. What is the difference between real and complex roots of a polynomial?

Real roots of a polynomial are values of the variable that are real numbers. Complex roots, on the other hand, are values of the variable that are complex numbers, meaning they involve the imaginary unit, i. Complex roots always come in pairs, and they can be found using the quadratic formula.

5. How are polynomials used in real-world applications?

Polynomials are used in many real-world applications, including finance, physics, engineering, and computer science. They can be used to model and solve problems involving distance, time, cost, and many other quantities. In finance, polynomials are used to calculate interest and depreciation. In physics and engineering, they are used to model and analyze motion, forces, and electrical circuits. In computer science, polynomials are used to represent and manipulate data and perform calculations.

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