Question

If \(a_{0}, a_{1}, \ldots, a_{n}\) are real numbers satisfying \(\frac{a_{0}}{1}+\frac{a_{1}}{2}+\cdots+\frac{a_{n}}{n+1}=0\) show that the equation \(a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n}=0\) has at least one real zero.

Short answer

By transforming the initial equation into a form suitable for Rolle’s theorem, and due to the nature of the given statement, the theorem allows to conclude that there exists at least one real root \(c\) for the given equation \(a_{0}+a_{1} x+ \ldots +a_{n} x^{n}=0\) within the range (0, 1).

Step-by-step solution

Function Formulation

Formulate a function \(f(x)\) built as follows: \(f(x)=a_{0} x+a_{1}\frac{x^{2}}{2}+\cdots+a_{n}\frac{x^{n+1}}{n+1}\). This can be obtained through integration of the given equation \(a_{0}+a_{1} x+ \ldots+ a_{n} x^{n}=0\). The integral runs from 0 to 1.

Evaluate f(x) at the Bounds

Evaluate \(f(x)\) at 0 and 1, which are the limits of our integral definition. We get, \(f(0)=0\) and \(f(1)=0\) using the initial given condition, which states that \(\frac{a_{0}}{1}+\frac{a_{1}}{2}+\cdots+\frac{a_{n}}{n+1}=0\). This implies \(f(0)=f(1)\).

Apply Rolle's Theorem

Since \(f(0)=f(1)\) and the function \(f(x)\) is continuous in the closed interval [0, 1] and differentiable in the open interval (0, 1), we can apply Rolle's Theorem, this states that there exists at least one \(c\) in (0, 1), such that \(f'(c)=0\). Where \(f'(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots+a_{n} x^{n}\), is our original equation.