Solving Polynomials of Increasing Degree

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SUMMARY

This discussion focuses on solving a sequence of polynomials of increasing degree, specifically analyzing the patterns in the coefficients of polynomials such as \(x^2 + 2\), \(\frac{2}{3} x^3 + \frac{13}{3} x\), and others up to degree six. The key insight is that the constant term alternates between 0 and 2, and the leading coefficient can be derived from the preceding polynomial by multiplying by \(\frac{2x}{n}\). The proposed method involves multiplying each polynomial by \(n!\) to create a new sequence \(P_n(x)\) and then forming \(Q_n(x) = P_n(x) - 2x \cdot P_{n-1}(x)\) to simplify the coefficients.

PREREQUISITES
  • Understanding polynomial functions and their degrees
  • Familiarity with factorial notation and operations
  • Knowledge of sequences and series in mathematics
  • Basic algebraic manipulation skills
NEXT STEPS
  • Explore the concept of polynomial sequences and their properties
  • Learn about factorials and their applications in combinatorial mathematics
  • Investigate the method of generating functions for polynomial sequences
  • Study the relationship between polynomial coefficients and their roots
USEFUL FOR

Mathematicians, students studying algebra, and anyone interested in advanced polynomial theory and sequences will benefit from this discussion.

Xitami
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x^2+2\\\\<br /> \frac{2}{3} x^3 + \frac{13}{3} x\\\\<br /> \frac{1}{3} x^4 + \frac{14}{3} x^2 + 2\\ \\<br /> \frac{2}{15} x^5 + \frac{10}{3} x^3 + \frac{83}{15} x\\ \\<br /> \frac{2}{45} x^6 + \frac{16}{9} x^4 + \frac{323}{45} x^2 + 2\\\\<br /> \dots
?
 
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What is your question? We can't just guess.
 
But please guess :-) how to continue?
 
How are these polynomials supposed to be related?
 
Apparently, the OP is not a native english speaker and can't describe the problem well.

I guess that he encountered these polynomials while working and he is asking if somebody recognizes them or sees an easy pattern in them.
 
Last edited:
Thank you Micromass
 
Clearly the constant term alternates between 0 and 2. At the other end, you can get the leading term from the preceding line by multiplying by 2x/n. So the leading coefficient is 2n-1/n! So a natural thing to try is:
- multiply each line by n! (starting with n=2 in the first line) to form the poly sequence Pn(x)
- form a new sequence from this according to Qn(x) = Pn(x) - 2x*Pn-1(x)
The coefficients that result look a little friendlier. The highest prime that occurs in this sample is 19, a lot better than 83.
 

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