Question

Let \(f(x)=a_{1} \sin x+a_{2} \sin 2 x+\cdots+a_{n} \sin n x,\) where \(a_{1}, a_{2}, \ldots, a_{n}\) are real numbers and where \(n\) is a positive integer. Given that \(|f(x)| \leq|\sin x|\) for all real \(x\), prove that \(\left|a_{1}+2 a_{2}+\cdots+n a_{n}\right| \leq 1\)

Short answer

To show that \(\left|a_{1}+2a_{2}+\cdots+na_{n}\right| \leq 1\), we integrate the given inequality \(|f(x)| \leq|\sin x|\) over the interval [0, \(2\pi\)]. As \(sin(nx)\) integrates to 0 for \(n \neq 0\) over this interval, we obtain \((2\pi)\left|a_{1}+2a_{2}+\cdots+na_{n}\right| \leq 4\). Solving for \(\left|a_{1}+2a_{2}+\cdots+na_{n}\right|\), the result follows.

Step-by-step solution

Formulate the integral for the inequality

Start with the given inequality \(|f(x)| \leq|\sin x|\) for all x. Integrate both sides from 0 to \(2\pi\) on the interval \(0 \leq x \leq 2\pi\).

Evaluate the integral

Since \(f(x)\) is a sum of sin functions, when you integrate \(f(x)\), you get the sum of the coefficients times the integral of sin functions. And for \(|\sin x|\), its integral from 0 to \(2\pi\) is 4.

Utilize the result of the integral

On the left side, it is the sum of the coefficients times two, because the integral of \(sin(nx)\) for n≠0 from 0 to \(2\pi\) is 0. So you get \((2\pi)\left|a_{1}+2a_{2}+\cdots+na_{n}\right| \leq 4\).

Produce the final result

Solving the inequality for \(\left|a_{1}+2a_{2}+\cdots+na_{n}\right|\), we get \(\left|a_{1}+2a_{2}+\cdots+na_{n}\right| \leq \frac{2}{\pi}\). This is the final result and it completes the proof.

Key concepts

Inequality of Trigonometric Functions

Understanding the inequality of trigonometric functions is crucial in many areas of mathematics, including calculus. The essence of such inequalities is to compare the magnitudes of trigonometric functions over a specific interval. In the case of our exercise, we are dealing with an inequality where a composite trigonometric function, which is a linear combination of sine functions with different frequencies, is bounded by the absolute value of the sine function.

The initial given inequality is \( |f(x)| \leq |\sin x| \), where \( f(x) \) composes of multiple sine functions with coefficients \( a_1, a_2, \ldots, a_n \). The absolute value applied here ensures we are speaking about the magnitude or 'size' of the trigonometric values without being concerned about their sign. This means, regardless of whether the sine function is positive or negative in the interval of interest, its size is always less than or equal to that of \( |\sin x| \).

To work with such inequalities, we often proceed by integrating or differentiating, depending on what is required. Integration, in particular, allows us to analyze the behavior of the functions over an entire interval, rather than at just a single point.

Integration of Sine Functions

The process of integrating sine functions is a fundamental skill in calculus. When integrating a sine function, we look at the accumulated area under its curve over a certain interval. This integral can give us insights into the average value of the function or the total accumulated 'effect' over a period.

For the simple sine function \( \sin x \), its integral over one full period \( 0 \leq x \leq 2\pi \) gives us zero because the positive and negative areas cancel out. However, when considering the absolute value, as in \( |\sin x| \), the scenario changes – we only consider the positive area, which over the interval from \( 0 \leq x \leq 2\pi \) amounts to 4 for \( |\sin x| \).

In our exercise, we encounter a more complex function, comprising multiple sine functions with different frequencies. Yet, regardless of the frequency, the integral of \( \sin(nx) \), where \( n \) is a non-zero integer, over one full period from \( 0 \) to \( 2\pi \) is zero due to symmetrical cancellation. This property is central to our proof, as it helps us simplify the complex function into a manageable form after integration.

Mathematical Proof Techniques

Mathematical proof techniques encompass a variety of logical methods used to establish the truth of mathematical statements. In calculus and higher mathematics, proofs are essential for verifying the validity of conjectures, theorems, or inequalities. The proof in our exercise is a demonstration of how integrating both sides of an inequality can lead to new and useful insights.

The technique used in our step-by-step solution leverages the properties of integration applied to an inequality. By integrating both sides from \(0\) to \(2\pi\), we harness the fact that certain integrals of sine functions evaluate to zero over this interval, essentially 'filtering out' the coefficients of higher harmonics in the composite function \(f(x)\).

This method of proof is clean and elegant because it uses fundamental calculus concepts to boil down a complex inequality to a simple one. While the starting inequality involved a variable function \(f(x)\), the ending result is a dependable numerical inequality involving the coefficients of \(f(x)\). Understanding these proof techniques not only aids in solving calculus problems but also develops a student's logical and critical thinking skills.