Problem 11 - A History of Mathematics: An Introduction

Question

Given that tan \(\frac{1}{2} \Pi(x)=e^{-x}\), where \(\Pi(x)\) is Lobachevsky's angle of parallelism, derive the formulas $$ \sin \Pi(x)=\frac{1}{\cosh x} \quad \text { and } \quad \cos \Pi(x)=\tanh x $$ and show that their power series expansions up to degree 2 are \(\sin \Pi(x)=1-\frac{1}{2} x^{2}\) and \(\cos \Pi(x)=x\), respectively.

Step-by-Step Solution

Answer

Question: Find the power series expansions of \(\sin\Pi(x)\) and \(\cos\Pi(x)\) up to degree 2, given that \(\tan(\frac{1}{2}\Pi(x))=e^{-x}\). Answer: The power series expansions up to degree 2 are as follows: \(\sin\Pi(x) = 1-\frac{1}{2}x^2\) and \(\cos\Pi(x) = x\).

Step 1: Use the given equation

We are given the equation \(\tan (\frac{1}{2} \Pi(x)) = e^{-x}\).

Step 2: Derive the equations for sin and cos

Using the double angle formula for tangent, we have $$ \tan \Pi(x) = \frac{2 \tan (\frac{1}{2} \Pi(x))}{1 - \tan^2 (\frac{1}{2} \Pi(x))} = \frac{2e^{-x}}{1-e^{-2x}}. $$ Now, we will use the relation \(\tan^2 \alpha + 1 = \sec^2 \alpha\) to obtain $$ \sec^2 \Pi(x) = \frac{1}{\cos^2 \Pi(x)} = 1 + \tan^2 \Pi(x) = 1 + \frac{4e^{-2x}}{(1-e^{-2x})^2}. $$ Solving for \(\cos \Pi(x)\), we get $$ \cos \Pi(x) = \frac{1}{\sqrt{1 + \frac{4e^{-2x}}{(1-e^{-2x})^2}}} = \tanh x. $$ Now, using the Pythagorean identity \(\sin^2 \Pi(x) + \cos^2 \Pi(x) = 1\), we obtain $$ \sin^2 \Pi(x) = 1 - \cos^2 \Pi(x) = 1 - \tanh^2 x = \frac{1}{\cosh^2 x}. $$ Taking the square root, we find $$ \sin \Pi(x) = \frac{1}{\cosh x}. $$

Step 3: Power series expansions up to degree 2

For computing the power series expansions of \(\sin \Pi(x)\) and \(\cos \Pi(x)\), we can use the Taylor series expansions of the respective hyperbolic functions \(\cosh x\) and \(\tanh x\): $$ \cosh x = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots $$ $$ \tanh x = x - \frac{x^3}{3!} + \frac{2 x^5}{15} + \cdots $$ Now, let's find the power series expansions of \(\sin \Pi(x)\) and \(\cos \Pi(x)\) up to degree 2: For \(\sin \Pi(x) = \frac{1}{\cosh x}\), we have: $$ \sin \Pi(x) = 1 - \frac{1}{2} x^2 + O(x^4) $$ For \(\cos \Pi(x) = \tanh x\), we have: $$ \cos \Pi(x) = x + O(x^3) $$ Thus, the power series expansions up to degree 2 are \(\sin \Pi(x)=1-\frac{1}{2} x^{2}\) and \(\cos \Pi(x)=x\), respectively.

Key Concepts

Exploring Hyperbolic Functions

Hyperbolic functions are similar to trigonometric functions but are based on hyperbolas. Unlike the circular nature of sine and cosine, hyperbolic sine (\( \sinh x \) ) and hyperbolic cosine (\( \cosh x \) ) relate to hyperbolic geometry. These functions help describe geometric concepts in Lobachevskian and hyperbolic spaces.

In the hyperbolic world, the angle of parallelism, \( \Pi(x) \), parallels to angles in Euclidean geometry. We aim to relate \( \sin \Pi(x) \) and \( \cos \Pi(x) \) to hyperbolic functions.

• **\( \sinh x \):** Calculates as \( \frac{e^x - e^{-x}}{2} \).
• **\( \cosh x \):** Defined by \( \frac{e^x + e^{-x}}{2} \).

These functions, often encountered in physics and engineering, bridge calculus and geometry in space-time models.

Understanding Power Series Expansion

Power series expansions allow us to represent functions as infinite sums. It is a useful method for approximating functions, particularly when addressing limits and derivatives. In our exercise, we expanded the hyperbolic functions to derive Lobachevsky's formulas.

**Key Formulas:**

• **\( \cosh x \):** Expanded as \( 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots \)
• **\( \tanh x \):** Expanded as \( x - \frac{x^3}{3!} + \frac{2x^5}{15} + \cdots \)

Using these expansions, we approximated the functions\( \sin \Pi(x) \) and \( \cos \Pi(x) \) to the second degree. This step brings a complex concept into a manageable format. The results offer simpler expressions: \( \sin \Pi(x) = 1 - \frac{1}{2}x^2 \) and \( \cos \Pi(x) = x \).

The power series is vital for understanding and predicting behavior in physics, engineering, and mathematics.

Learning Trigonometric Identities

Trigonometric identities are formulas involving trigonometric functions, helping simplify complex equations. In this problem, identities support manipulation and integration of hyperbolic functions to derive key outcomes.

One identity used was the double angle formula for tangent, vital in step 2 of deriving the equations. This led us to.

• **\( \tan \Pi(x) = \frac{2 \tan (\frac{1}{2} \Pi(x))}{1 - \tan^2 (\frac{1}{2} \Pi(x))} \)**
• **\( \tan^2 \alpha + 1 = \sec^2 \alpha \)** was another key formula used in the solution.

These identities let us link hyperbolic functions and ordinary trigonometric concepts.

By applying these identities, we uncovered that \( \cos \Pi(x) = \tanh x \) and \( \sin \Pi(x) = \frac{1}{\cosh x} \).

Understanding these identities helps untangle trigonometric problems and provides deeper insights into geometric and algebraic relationships.