Continuity of Rational Functions and the Role of Complex Roots

In mathematical analysis, the continuity of a function is a fundamental concept. We explore the continuity of the function defined as your given expression:

Let a1a1-xfrac{a_{2}}{a_{2}-x} cdots frac{a_{n}}{a_{n}-x}. This function is of particular interest, as it defines a rational function with several potential points of discontinuity.

Definition and Continuity

The expression given is a product of rational functions, each of which is defined as aiai-x. Each of these rational functions is continuous except at the points where the denominator is zero, i.e., at the points xi.

Continuity and Discontinuity Points

Given a set {x1,x2,...,xn}, the function a1a1-xfrac{a_{2}}{a_{2}-x} cdots frac{a_{n}}{a_{n}-x} is continuous in each (ai-epsilon1,ai epsilon1).

Proof of Continuity Using Bolzano's Theorem

To rigorously prove the continuity, consider a small enough epsilon0. For any xin(ai-epsilonai epsilon1), the term aiai-x is very small compared to ai, implying that aiai-x0.

On the other hand, the term a{i 1}a{i 1}-x is very large compared to a{i 1}. More specifically, for an xin(a{i 1}-epsilon1,a{i 1} epsilon1), a{i 1}a{i 1}-x0, and more specifically, let a{i 1}a{i 1}-x2015 for an xin(a{i 1}-epsilon1,a{i 1} epsilon1).

Application of Bolzano's Theorem

Bolzano's Theorem states that every continuous function that changes sign over a closed interval must have a root in that interval. This theorem is key to our proof. By considering the intervals where the function changes sign, we can apply Bolzano's Theorem and locate the roots of the function.

Role of Complex Roots

A crucial aspect of the function is the presence of complex roots. Complex numbers often come in conjugate pairs, ensuring that the overall function remains real-valued. However, the discontinuities in the rational functions can still cause the function to change sign, leading to the application of Bolzano's Theorem.

Conclusion

By applying Bolzano's Theorem and considering the behavior of the rational functions around their discontinuities, we can rigorously prove the continuity of the given function. The specific value 2015 used in the example is not critical, as any non-zero number would serve the same purpose.

Note: If you need further explanation or more detailed steps, feel free to ask!