Analyzing the Setup
The problem presents a piecewise function defined by alternating segments of sine and cosine curves. To ensure the bridge is perfectly continuous, we must enforce the condition that the left-hand limit equals the right-hand limit at every transition point.
The function changes its definition at every integer x. We must examine the continuity at two distinct types of points: even integers (x=2n) and odd integers (x=2n−1).
The Even Boundary
Let us stand at an even integer, x=2n. To the left, the bridge is defined by the red segment f(x)=bn+cos(πx), and to the right, it is defined by the blue segment f(x)=an+sin(πx).
For continuity, we set the limits equal:
x→2n−lim(bn+cos(πx))=x→2n+lim(an+sin(πx))
Substituting x=2n, we obtain:
bn+cos(2nπ)=an+sin(2nπ)
Since cos(2nπ)=1 and sin(2nπ)=0 for any integer n, this simplifies to:
This equation serves as our first pillar of stability.
The Odd Boundary
Now, we move to an odd integer, x=2n−1. At this point, the bridge transitions from a blue segment to a red segment.
To the left, we have the blue segment f(x)=an−1+sin(πx), and to the right, we have the red segment f(x)=bn+cos(πx). Equating the limits gives:
x→(2n−1)−lim(an−1+sin(πx))=x→(2n−1)+lim(bn+cos(πx))
Substituting x=2n−1, we get:
an−1+sin((2n−1)π)=bn+cos((2n−1)π)
Given that sin((2n−1)π)=0 and cos((2n−1)π)=−1 for any integer n, the expression becomes:
an−1+0=bn−1⇒an−1−bn=−1
The Grand Conclusion
We have successfully derived two fundamental relationships that govern the continuity of our function:
These constraints ensure the piecewise function remains continuous across the entire real line. By systematically analyzing the boundaries, we have reduced a complex trigonometric problem into a clear, elegant set of algebraic truths.
