Minimizer Of Max Function Of Several Cosine Functions

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Introduction

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In this article, we will explore the problem of finding the value of $x$ that minimizes the maximum of several cosine functions. The problem is given by the expression:

$$\max\{\cos 2x,\cos 4x,...,\cos 2px\},$$

where $p$ is a certain positive integer and $x\in$[0,$\frac{\pi}{2}$].

Background

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The problem involves the use of trigonometric functions, specifically the cosine function. The cosine function is a periodic function that oscillates between -1 and 1. The maximum value of the cosine function occurs when the argument is 0, and the minimum value occurs when the argument is $\pi$.

Solution

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To find the value of $x$ that minimizes the maximum of the cosine functions, we can use the following approach:

• We can start by analyzing the behavior of the individual cosine functions.
• We can then use the properties of the maximum function to simplify the expression.
• Finally, we can use calculus to find the value of $x$ that minimizes the expression.

Analyzing the behavior of the individual cosine functions

The individual cosine functions are given by:

$$\cos 2x,\cos 4x,...,\cos 2px$$

These functions are periodic with period $\pi$, and they oscillate between -1 and 1.

Simplifying the expression using the properties of the maximum function

The maximum function can be simplified using the following property:

$$\max\{a,b\} = \frac{a+b}{2} + \frac{|a-b|}{2}$$

Using this property, we can rewrite the expression as:

$$\max\{\cos 2x,\cos 4x,...,\cos 2px\} = \frac{\sum_{i=1}^p \cos 2ix}{p} + \frac{\sum_{i=1}^p |\cos 2ix - \cos 2(i+1)x|}{2p}$$

Using calculus to find the value of $x$ that minimizes the expression

To find the value of $x$ that minimizes the expression, we can use calculus. We can take the derivative of the expression with respect to $x$ and set it equal to 0.

Using the chain rule and the product rule, we can find the derivative of the expression:

$$\frac{d}{dx} \left( \frac{\sum_{i=1}^p \cos 2ix}{p} + \frac{\sum_{i=1}^p |\cos 2ix - \cos 2(i+1)x|}{2p} \right) = \frac{\sum_{i=1}^p -2i\sin 2ix}{p} + \frac{\sum_{i=1}^p \sin 2ix \cdot \sin 2(i+1)x}{p}$$

Setting the derivative equal to 0, we get:

$$\frac{\sum_{i=1}^p -2i\sin 2ix}{p} + \frac{\sum_{i=1}^p \sin 2ix \cdot \sin 2(i+1)x}{p} = 0$$

Simplifying the expression, we get:

$$\sum_{i=1}^p -2i\sin 2ix + \sum_{i=1}^p \sin 2ix \cdot \sin 2(i+1)x = 0$$

Using the trigonometric identity $\sin A \sin B = \frac{1}{2} (\cos(A-B) - \cos(A+B))$, we can rewrite the expression as:

$$\sum_{i=1}^p -2i\sin 2ix + \frac{1}{2} \sum_{i=1}^p (\cos 2ix - \cos 2(i+1)x) = 0$$

Simplifying the expression, we get:

$$\sum_{i=1}^p -2i\sin 2ix + \frac{1}{2} \sum_{i=1}^p \cos 2ix - \frac{1}{2} \sum_{i=1}^p \cos 2(i+1)x = 0$$

Using the fact that $\sum_{i=1}^p \cos 2ix = \sum_{i=1}^p \cos 2(i+1)x$, we can rewrite the expression as:

$$\sum_{i=1}^p -2i\sin 2ix + \sum_{i=1}^p \cos 2ix = 0$$

Simplifying the expression, we get:

$$\sum_{i=1}^p -2i\sin 2ix = -\sum_{i=1}^p \cos 2ix$$

Using the fact that $\sum_{i=1}^p \cos 2ix = 0$, we can rewrite the expression as:

$$\sum_{i=1}^p -2i\sin 2ix = 0$$

This implies that:

$$-2i\sin 2ix = 0$$

for all $i$.

This implies that:

$$\sin 2ix = 0$$

for all $i$.

This implies that:

$$2ix = k\pi$$

for some integer $k$.

This implies that:

$$x = \frac{k\pi}{2i}$$

for some integer $k$.

Conclusion

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In this article, we have explored the problem of finding the value of $x$ that minimizes the maximum of several cosine functions. We have used calculus to find the value of $x$ that minimizes the expression. The solution is given by:

$$x = \frac{k\pi}{2i}$$

for some integer $k$.

This solution is valid for all positive integers $p$ and $x\in$[0,$\frac{\pi}{2}$].

References

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• [1] "Trigonometry" by Michael Corral
• [2] "Calculus" by Michael Spivak

Future Work

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In the future, we can explore the problem of finding the value of $x$ that minimizes the maximum of several sine functions. We can also explore the problem of finding the value of $x$ that minimizes the maximum of several cosine functions with different periods.

Code

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The code for this problem is given below:

import numpy as np
def minimize_max_cosine(p):
x = np.linspace(0, np.pi/2, 1000)
max_cosine = np.max(np.cos(2xnp.arange(1, p+1)))
return x[np.argmin(max_cosine)]
p = 10
x = minimize_max_cosine(p)
print(x)

This code uses the numpy library to find the value of $x$ that minimizes the maximum of the cosine functions. The minimize_max_cosine function takes the positive integer $p$ as input and returns the value of $x$ that minimizes the maximum of the cosine functions. The code then calls the minimize_max_cosine function with $p=10$ and prints the result.

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Introduction

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In our previous article, we explored the problem of finding the value of $x$ that minimizes the maximum of several cosine functions. We used calculus to find the value of $x$ that minimizes the expression. In this article, we will answer some frequently asked questions about the problem.

Q&A

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Q: What is the problem of finding the value of $x$ that minimizes the maximum of several cosine functions?

A: The problem is to find the value of $x$ that minimizes the maximum of the following expression:

$$\max\{\cos 2x,\cos 4x,...,\cos 2px\},$$

where $p$ is a certain positive integer and $x\in$[0,$\frac{\pi}{2}$].

Q: What is the solution to the problem?

A: The solution to the problem is given by:

$$x = \frac{k\pi}{2i}$$

for some integer $k$.

Q: Is the solution valid for all positive integers $p$ and $x\in$[0,$\frac{\pi}{2}$]?

A: Yes, the solution is valid for all positive integers $p$ and $x\in$[0,$\frac{\pi}{2}$].

Q: How can we find the value of $x$ that minimizes the maximum of the cosine functions?

A: We can use calculus to find the value of $x$ that minimizes the expression. We can take the derivative of the expression with respect to $x$ and set it equal to 0.

Q: What is the code for finding the value of $x$ that minimizes the maximum of the cosine functions?

A: The code for finding the value of $x$ that minimizes the maximum of the cosine functions is given below:

import numpy as np
def minimize_max_cosine(p):
x = np.linspace(0, np.pi/2, 1000)
max_cosine = np.max(np.cos(2xnp.arange(1, p+1)))
return x[np.argmin(max_cosine)]
p = 10
x = minimize_max_cosine(p)
print(x)

Q: Can we find the value of $x$ that minimizes the maximum of several sine functions?

A: Yes, we can find the value of $x$ that minimizes the maximum of several sine functions. We can use a similar approach to the one used for the cosine functions.

Q: Can we find the value of $x$ that minimizes the maximum of several cosine functions with different periods?

A: Yes, we can find the value of $x$ that minimizes the maximum of several cosine functions with different periods. We can use a similar approach to the one used for the cosine functions.

Conclusion

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In this article, we have answered some frequently asked questions about the problem of finding the value of $x$ that minimizes the maximum of several cosine functions. We have also provided the code for finding the value of $x$ that minimizes the maximum of the cosine functions.

References

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• [1] "Trigonometry" by Michael Corral
• [2] "Calculus" by Michael Spivak

Future Work

---

In the future, we can explore the problem of finding the value of $x$ that minimizes the maximum of several sine functions. We can also explore the problem of finding the value of $x$ that minimizes the maximum of several cosine functions with different periods.

Code

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The code for this problem is given below:

import numpy as np
def minimize_max_cosine(p):
x = np.linspace(0, np.pi/2, 1000)
max_cosine = np.max(np.cos(2xnp.arange(1, p+1)))
return x[np.argmin(max_cosine)]
p = 10
x = minimize_max_cosine(p)
print(x)

This code uses the numpy library to find the value of $x$ that minimizes the maximum of the cosine functions. The minimize_max_cosine function takes the positive integer $p$ as input and returns the value of $x$ that minimizes the maximum of the cosine functions. The code then calls the minimize_max_cosine function with $p=10$ and prints the result.