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Prove that \(\tanh ^{-1} x=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right), \quad-1

Short Answer

Expert verified
\(\tanh ^{-1} x = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right), -1<x<1, has been proved by applying property of logarithms and the definition of \(\tanh ^{-1} x\).

Step by step solution

01

Rewrite the expression

The first step is to rewrite the given logarithmic equation \(\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)\) by using the properties of logarithm as: \(\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right) = \frac{1}{2} [\ln(1+x) - \ln(1-x)]\)
02

Applying definition of Arctanh

Arctanh is the inverse of hyperbolic tangent. In terms of logarithms it can be expressed as \(\mathrm{Artanh}(x) = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)\). So applying this definition to the expression obtained in Step 1 above, we get the Left Hand Side (LHS) of the given equation, i.e., \(\tanh ^{-1} x\).
03

Conclusion

Since both the sides of original equation have been proved to be equal, we can say with certainty that \(\tanh ^{-1} x = \frac{1}{2} \ln \left(\frac{1+x}{1-x}\right), -1<x<1.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Arctanh
The function known as arctanh is shorthand for the inverse hyperbolic tangent function. Hyperbolic functions are similar to trigonometric functions but are based on hyperbolas rather than circles.
They can often be useful in calculus and engineering calculations. The inverse of these functions helps us find values of the corresponding angles or areas in hyperbolic situations rather than circular ones.
The function \( anh x\) is defined as \( anh x = \frac{{e^x - e^{-x}}}{{e^x + e^{-x}}}\). Since arctanh, or \( anh^{-1}(x)\), is the inverse of \( anh(x)\), it helps to backtrack to find the original input given an output value of \( anh(x)\).
This inverse function has the range \(-1 < x < 1\) because that’s the only interval where it is defined and produces real number results. When graphed, arctanh has a shape that closely follows tanh but is flipped across the line \(y = x\).
  • Arctanh allows solving for angles in hyperbolic equations.
  • It is primarily used in advanced mathematics for solving integration and differential equations that are based on hyperbolic functions.
Exploring Logarithmic Identities
Logarithmic identities are tools that simplify the manipulation and comparison of logarithmic expressions. For instance, the exercise you encountered makes use of these identities to achieve its solution.
The identity \(\ln(a/b) = \ln(a) - \ln(b)\) allows the expression \(\frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)\) to be rewritten as \(\frac{1}{2} [\ln(1+x) - \ln(1-x)]\).
This ability to break down and separate complex logarithmic expressions is key in solving many mathematical proofs and equations.
Logarithmic properties such as power, product, and quotient rules provide a structured way to simplify the expressions themselves, often resulting in much easier arithmetic operations.
  • Quotient Rule: \(\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)\)
  • Product Rule: \(\ln(ab) = \ln(a) + \ln(b)\)
  • Power Rule: \(\ln(a^b) = b\ln(a)\)
These identities are especially important in calculus because they simplify integration and differentiation of logarithmic functions.
They also provide a framework for proving equations like the one in the exercise, connecting inverse hyperbolic functions to logarithmic expressions.
Applying Proof Techniques
Proof techniques are vital for verifying mathematical statements. They provide a way to demonstrate the truth of equations and inequalities logically and consistently.
When asked to prove that \(\tanh^{-1} x = \frac{1}{2} \ln\left(\frac{1+x}{1-x}\right)\), a logical step-by-step approach is necessary.
The approach often begins by transforming one side of the equation to make it match the other side. In this case, rewriting one side using logarithmic identities helped simplify the expression.
  • Begin by rewriting the expression using known identities or definitions.
  • Be consistent with symbol manipulation; step-by-step simplification helps ensure accuracy.
  • Check both sides of the equation thoroughly to ensure they are identical.
Eventually, both sides are compared to ensure they are equal in form and value. Once confirmed, the proof is complete.
Using these techniques, students can confidently solve similar proofs involving complex algebraic and calculus-based problems.

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Most popular questions from this chapter

Find \(\left(f^{-1}\right)^{\prime}(a)\) for the function \(f\) and the given real number \(a\). \(f(x)=\sin x, \quad-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}, \quad a=\frac{1}{2}\)

Evaluate the integral. $$ \int_{0}^{1} \cosh ^{2} x d x $$

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An object is projected upward from ground level with an initial velocity of 500 feet per second. In this exercise, the goal is to analyze the motion of the object during its upward flight. (a) If air resistance is neglected, find the velocity of the object as a function of time. Use a graphing utility to graph this function. (b) Use the result in part (a) to find the position function and determine the maximum height attained by the object. (c) If the air resistance is proportional to the square of the velocity, you obtain the equation $$ \frac{d v}{d t}=-\left(32+k v^{2}\right) $$ where \(-32\) feet per second per second is the acceleration due to gravity and \(k\) is a constant. Find the velocity as a function of time by solving the equation $$ \int \frac{d v}{32+k v^{2}}=-\int d t $$ (d) Use a graphing utility to graph the velocity function \(v(t)\) in part (c) if \(k=0.001\). Use the graph to approximate the time \(t_{0}\) at which the object reaches its maximum height. (e) Use the integration capabilities of a graphing utility to approximate the integral \(\int_{0}^{t_{0}} v(t) d t\) where \(v(t)\) and \(t_{0}\) are those found in part (d). This is the approximation of the maximum height of the object. (f) Explain the difference between the results in parts (b) and (e).

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