- #1
Simple Identity Proof for Homework | Equations & Solution Attempt
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SUMMARY
The discussion centers on a homework problem involving the inequality \(x^n \le a_0 + a_1x + a_2x^2 + \cdots + a_{n-1}x^{n-1}\). The initial attempt at a solution suggests proving this inequality, but it is established that the statement is not universally true for arbitrary values of \(x\) and coefficients \(a_0, a_1, \ldots, a_{n-1}\). Participants emphasize the need for clarification on the specific conditions under which the inequality holds.
PREREQUISITES- Understanding of polynomial inequalities
- Familiarity with algebraic manipulation
- Knowledge of mathematical proof techniques
- Basic concepts of limits and continuity
- Research conditions for polynomial inequalities to hold
- Study the role of coefficients in polynomial expressions
- Learn about mathematical proof strategies, particularly for inequalities
- Explore examples of polynomial functions and their behaviors
Students tackling advanced algebra, educators teaching polynomial inequalities, and anyone interested in mathematical proofs and their applications in homework problems.
- #2
HallsofIvy
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I take it that what you have in "The attempt at a solution",
[tex]x^n\le a_0+ a_1x+ a_2x^2+ \cdot\cdot\cdot+ a_{n-1}x^{n-1}[/tex]
is actually what you want to prove. I cannot open the first attachment. But that is certainly NOT true for general x, [itex]a_0[/itex], [itex]a_1[/itex], ... So what are you really trying to do?
[tex]x^n\le a_0+ a_1x+ a_2x^2+ \cdot\cdot\cdot+ a_{n-1}x^{n-1}[/tex]
is actually what you want to prove. I cannot open the first attachment. But that is certainly NOT true for general x, [itex]a_0[/itex], [itex]a_1[/itex], ... So what are you really trying to do?
- #3
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