Proving The Decreasing Nature Of Trigonometric Summation

This article delves into the intricate problem of demonstrating that the function

$$v(x,5)=\sum_{k=0}^{4}\sqrt{1-c^2\sin^2\left(x+\frac{2k\pi}{5}\right)}$$

is decreasing over the interval $x\in[0,\frac{\pi}{10}]$. This problem elegantly intertwines concepts from calculus, trigonometry, and inequality, demanding a rigorous approach to establish the desired result. To showcase that a function is decreasing, we need to prove that its derivative is non-positive over the specified interval. This involves differentiating the given summation, analyzing the resulting expression, and leveraging trigonometric identities and inequalities to confirm its negativity.

Understanding the Function and Its Components

To effectively tackle this problem, let’s dissect the function $v(x,5)$ and its individual components. The function is a summation of five terms, each involving a square root of a trigonometric expression. The core of each term lies in the expression $1 - c^2 \sin^2\left(x + \frac{2k\pi}{5}\right)$, where:

• x represents the variable for which we want to prove the decreasing nature of the function.
• c is a constant, and its value plays a critical role in determining the behavior of the function. For the function to be real-valued, we must have $c^2 \sin^2\left(x + \frac{2k\pi}{5}\right) \leq 1$ for all $k$. This condition implies that $|c| \leq 1$, as the maximum value of $\sin^2$ is 1.
• k is an index that ranges from 0 to 4, generating five terms in the summation.
• The term $\frac{2k\pi}{5}$ introduces a phase shift within the sine function, effectively distributing the sine function's evaluations across different points within its period.

The Sine Function and its Properties

The sine function, $\sin(x)$, is periodic with a period of $2\pi$, meaning its values repeat every $2\pi$ units. The $\sin^2(x)$ function is also periodic, but with a period of $\pi$, and it is always non-negative, ranging from 0 to 1. Understanding these properties is crucial for analyzing the behavior of $v(x,5)$. The behavior of $\sin^2$ within our function significantly affects the overall trend. Specifically, $\sin^2(x)$ is an even function, meaning $\sin^2(x) = \sin^2(-x)$. It increases from 0 to 1 in the interval $[0, \frac{\pi}{2}]$ and then decreases from 1 to 0 in the interval $[\frac{\pi}{2}, \pi]$. These characteristics must be considered when we examine the derivative of $v(x, 5)$. In our summation, the phase shifts introduced by $\frac{2k\pi}{5}$ spread these increasing and decreasing behaviors across the five terms, creating a complex interplay that dictates the overall trend of the function. Understanding the sine function's behavior in different quadrants and its response to these shifts is vital for a comprehensive analysis.

Differentiating the Summation

To demonstrate that $v(x,5)$ is decreasing, we need to compute its derivative with respect to $x$ and show that it is non-positive over the interval $[0, \frac{\pi}{10}]$. Let’s first differentiate a single term in the summation:

$$\frac{d}{dx} \sqrt{1 - c^2 \sin^2\left(x + \frac{2k\pi}{5}\right)}$$

Using the chain rule, we get:

$$\frac{1}{2\sqrt{1 - c^2 \sin^2\left(x + \frac{2k\pi}{5}\right)}} \cdot (-c^2) \cdot 2\sin\left(x + \frac{2k\pi}{5}\right) \cdot \cos\left(x + \frac{2k\pi}{5}\right)$$

Simplifying, we have:

$$\frac{-c^2 \sin\left(x + \frac{2k\pi}{5}\right) \cos\left(x + \frac{2k\pi}{5}\right)}{\sqrt{1 - c^2 \sin^2\left(x + \frac{2k\pi}{5}\right)}}$$

Using the double angle identity $2\sin(\theta)\cos(\theta) = \sin(2\theta)$, we can further simplify this to:

$$\frac{-c^2 \sin\left(2x + \frac{4k\pi}{5}\right)}{2\sqrt{1 - c^2 \sin^2\left(x + \frac{2k\pi}{5}\right)}}$$

The Derivative of the Summation

Now, differentiating the entire summation $v(x,5)$, we have:

$$\frac{dv}{dx} = \sum_{k=0}^{4} \frac{-c^2 \sin\left(2x + \frac{4k\pi}{5}\right)}{2\sqrt{1 - c^2 \sin^2\left(x + \frac{2k\pi}{5}\right)}}$$

To prove that $v(x,5)$ is decreasing, we need to show that $\frac{dv}{dx} \leq 0$ for all $x \in [0, \frac{\pi}{10}]$. This involves analyzing the sign of the summation, which is not straightforward due to the presence of both sine functions in the numerator and the square root in the denominator.

Each term in the summation contributes to the overall sign of the derivative. The denominator is always positive, assuming $c^2 \sin^2\left(x + \frac{2k\pi}{5}\right) < 1$, which is a reasonable condition to impose. Thus, the sign of each term is determined by the sign of $-\sin\left(2x + \frac{4k\pi}{5}\right)$. The challenge lies in demonstrating that the sum of these signed terms is non-positive. Understanding the range of the argument $2x + \frac{4k\pi}{5}$ for each $k$ and $x \in [0, \frac{\pi}{10}]$ is crucial. The argument's range dictates the sign of the sine function, and we must show that the negative sine terms outweigh the positive ones within the summation.

Analyzing the Sign of the Derivative

Establishing that $\frac{dv}{dx} \leq 0$ for $x \in [0, \frac{\pi}{10}]$ requires a detailed analysis of the terms in the summation. We need to consider the signs and magnitudes of $\sin\left(2x + \frac{4k\pi}{5}\right)$ for $k = 0, 1, 2, 3, 4$. The interval for $x$ is $[0, \frac{\pi}{10}]$, so let’s analyze the arguments of the sine function for each $k$.

Range of Arguments

For $x \in [0, \frac{\pi}{10}]$, we have:

• $$k = 0: 2x + \frac{4(0)\pi}{5} = 2x \in [0, \frac{\pi}{5}]$$

• $$k = 1: 2x + \frac{4(1)\pi}{5} = 2x + \frac{4\pi}{5} \in [\frac{4\pi}{5}, \frac{4\pi}{5} + \frac{\pi}{5}] = [\frac{4\pi}{5}, \pi]$$

• $$k = 2: 2x + \frac{4(2)\pi}{5} = 2x + \frac{8\pi}{5} \in [\frac{8\pi}{5}, \frac{8\pi}{5} + \frac{\pi}{5}] = [\frac{8\pi}{5}, \frac{9\pi}{5}]$$

• $$k = 3: 2x + \frac{4(3)\pi}{5} = 2x + \frac{12\pi}{5} \in [\frac{12\pi}{5}, \frac{13\pi}{5}]$$

• $$k = 4: 2x + \frac{4(4)\pi}{5} = 2x + \frac{16\pi}{5} \in [\frac{16\pi}{5}, \frac{17\pi}{5}]$$

Analyzing the Signs

Now we analyze the signs of $\sin\left(2x + \frac{4k\pi}{5}\right)$ in these intervals:

• For $k = 0$, $\sin(2x)$ is positive in the interval $[0, \frac{\pi}{5}]$ because this interval lies in the first quadrant.
• For $k = 1$, $\sin\left(2x + \frac{4\pi}{5}\right)$ is positive in the interval $[\frac{4\pi}{5}, \pi]$ because it lies in the second quadrant (sine is positive in the second quadrant).
• For $k = 2$, $\sin\left(2x + \frac{8\pi}{5}\right)$ is negative. Since $\frac{8\pi}{5} = \frac{8\pi}{5} - 2\pi = -\frac{2\pi}{5}$ and $\frac{9\pi}{5} = \frac{9\pi}{5} - 2\pi = -\frac{\pi}{5}$, the interval is $[\frac{8\pi}{5}, \frac{9\pi}{5}]$ translates to $[-\frac{2\pi}{5}, -\frac{\pi}{5}]$ which falls in the fourth quadrant (sine is negative).
• For $k = 3$, $\sin\left(2x + \frac{12\pi}{5}\right)$ is negative. Since $\frac{12\pi}{5} = \frac{12\pi}{5} - 2\pi = \frac{2\pi}{5}$ and $\frac{13\pi}{5} = \frac{13\pi}{5} - 2\pi = \frac{3\pi}{5}$, the interval is $[\frac{12\pi}{5}, \frac{13\pi}{5}]$ translates to $[\frac{2\pi}{5}, \frac{3\pi}{5}]$. The sine is positive in the first and second quadrants, but we need to subtract $2\pi$ to bring it into a familiar range. Thus, the interval translates to $[\frac{2\pi}{5}, \frac{3\pi}{5}]$ which implies that $2x + \frac{12\pi}{5}$ is negative.
• For $k = 4$, $\sin\left(2x + \frac{16\pi}{5}\right)$ is negative. Since $\frac{16\pi}{5} = \frac{16\pi}{5} - 3\pi = \frac{\pi}{5}$ and $\frac{17\pi}{5} = \frac{17\pi}{5} - 3\pi = \frac{2\pi}{5}$, the interval is $[\frac{16\pi}{5}, \frac{17\pi}{5}]$ which translates to $[\frac{\pi}{5}, \frac{2\pi}{5}]$ implying that $2x + \frac{16\pi}{5}$ is also negative.

This analysis suggests that there are two positive sine terms (k = 0, 1) and three negative sine terms (k = 2, 3, 4). However, to conclusively show that $\frac{dv}{dx} \leq 0$, we need to demonstrate that the sum of the negative terms outweighs the sum of the positive terms. This is a non-trivial task and likely requires more advanced techniques such as bounding the magnitudes of the sine functions or using numerical methods to approximate the sum.

Proving the Decreasing Nature: Advanced Techniques and Numerical Methods

To rigorously establish that the negative sine terms outweigh the positive ones, we need to delve into more advanced techniques. One possible approach involves utilizing trigonometric inequalities to bound the magnitudes of the sine functions within the specified intervals. For instance, we can use the inequality $|\sin(x)| \leq |x|$ for all $x$, or explore specific inequalities that apply to the intervals we've identified.

Trigonometric Inequalities and Bounding

Applying trigonometric inequalities can provide upper and lower bounds for the sine terms, allowing us to compare their magnitudes more effectively. However, this approach often leads to complex expressions and might not yield a conclusive result without further simplification or the use of computational tools. Another approach involves considering the symmetry and periodicity of the sine function. The phase shifts $\frac{4k\pi}{5}$ are evenly spaced around the unit circle, which suggests there might be a symmetry argument that simplifies the summation. However, the $2x$ term breaks the symmetry, making this approach challenging.

Numerical Methods

Given the complexity of the analytical approach, numerical methods can provide valuable insights and potentially confirm the decreasing nature of the function. We can use computational software to plot the derivative $\frac{dv}{dx}$ over the interval $[0, \frac{\pi}{10}]$ and visually verify that it is non-positive. Additionally, we can numerically compute the summation for various values of $x$ within the interval and observe the trend.

Numerical verification is not a proof, but it can provide strong evidence and guide our analytical efforts. If numerical results consistently show that the function is decreasing, we can focus on specific aspects of the derivative's behavior to construct a rigorous proof. For example, we might identify critical points where the derivative is close to zero and analyze the function's behavior around those points.

Fourier Series Representation

Another sophisticated approach involves expressing the function $v(x,5)$ using a Fourier series. Fourier series represent periodic functions as an infinite sum of sines and cosines. If we can find the Fourier series representation of $v(x,5)$, we can analyze the coefficients to understand the function's behavior. If the coefficients of the sine terms are negative and decreasing, this would provide strong evidence that the function is decreasing. The challenge with this approach lies in the complexity of computing the Fourier coefficients for $v(x,5)$, which involves integrating the function multiplied by sine and cosine functions over a period. This integration might not have a closed-form solution and could require numerical integration techniques.

Conclusion

Proving that the function $v(x,5) = \sum_{k=0}^{4} \sqrt{1 - c^2 \sin^2\left(x + \frac{2k\pi}{5}\right)}$ is decreasing on the interval $[0, \frac{\pi}{10}]$ is a challenging problem that demands a blend of calculus, trigonometry, and inequality techniques. We have demonstrated the initial steps of computing the derivative and analyzing the signs of its terms. However, rigorously proving the decreasing nature likely requires advanced techniques such as trigonometric inequalities, numerical methods, or Fourier series representation. The interplay between positive and negative terms in the derivative makes this problem particularly intricate, highlighting the subtleties of trigonometric summations.

Further research could explore specific bounds for the sine functions, develop a numerical simulation to visualize the derivative, or attempt to derive the Fourier series representation of the function. These approaches may provide the necessary insights to complete the proof and fully understand the behavior of this intriguing trigonometric summation.