[sp]Let $Q_n(x) = (x-1)^2(x^{2n-2} + 2x^{2n-4} + 3x^{2n-6} + \ldots + n) + (n+1)$. Then $Q_n(x)>0$ for all $x$. (In fact, $Q_n(x)$ has minimum value $n+1$, when $x=1$, because each term in the expression for $Q_n(x)$ is an even power of either $x$ or $x-1$.) So $Q_n(x)$ has no real roots. It will therefore be sufficient to show that $P_n(x) = Q_n(x).$lfdahl said:
Opalg said:
. Please explain how on Earth you came up with the expression for $Q_n(x)$.[sp]That just came from looking at what happens for small values of $n$. You soon find that each $P_n(x)$ (for $n=1,2,3$) has its minimum value when $x=1$, and that this minimum value is $n+1$. That leads to the discovery that $P_n(x) - (n+1)$ has a factor $(x-1)^2$. The quotient of $P_n(x) - (n+1)$ by that factor is $Q_n(x).$[/sp]lfdahl said:I am curious
[sp]That should be \(\displaystyle (1+x)P_n(x) = x\frac{x^{2n+1}+1}{x+1}+2n+1\), which certainly shows that $P_n(x)>0$ for $x>0$. But it is not so clear from that expression what the minimum value of $P_n(x)$ should be (in fact, it is $n+1$, not $2n+1$).[/sp]lfdahl said:
A polynomial is an expression that contains variables and coefficients and is raised to positive integer powers. Real roots refer to the values of the variables that make the polynomial equal to zero. Therefore, a polynomial with no real roots means that there are no values of the variables that will satisfy the equation and make it equal to zero.
There are several methods to prove that a polynomial has no real roots. One way is to use the Descartes' Rule of Signs, which states that the number of positive real roots of a polynomial is equal to the number of sign changes in the coefficients, and the number of negative real roots is equal to the number of sign changes in the coefficients or less than that by an even number. If the number of positive and negative real roots is zero, then the polynomial has no real roots.
Yes, a polynomial can have no real roots but still have complex roots. Complex roots are solutions to the polynomial equation that involve the imaginary number i, which is the square root of -1. These complex roots can be found by factoring the polynomial and solving for the roots using the quadratic formula. However, if a polynomial has no real roots, it will also have no real solutions, whether they are complex or not.
Yes, there are several methods for determining if a polynomial has no real roots. Apart from Descartes' Rule of Signs, another method is to use the Intermediate Value Theorem, which states that if a continuous function takes on positive and negative values at two points, then it must take on the value of zero at some point in between. If a polynomial only takes on positive or negative values, it cannot equal zero, and therefore, it has no real roots.
Proving that a polynomial has no real roots is essential in various fields, such as engineering, physics, and economics. For example, in engineering, a polynomial with no real roots can represent a physical system that has no equilibrium point, meaning it is unstable. In economics, a polynomial with no real roots can represent a demand function that has no intersection with the supply curve, indicating that there is no market equilibrium. Understanding the behavior of these systems is crucial in making informed decisions and solving real-world problems.