(b) Using the method of the equation $\frac{1+\tanh (x)}{1-\tanh (x)}=e^{2 x}$, with $x$ replaced by $y$ Let $y=\tanh ^{-1}(x)$. Then $x=\tanh (y) \quad$, so we have the following.
$\begin{aligned}
e^{2 y} & =\frac{1+\tanh (y)}{1-\tanh (y)}=\frac{1+x}{1-x} \\
\Rightarrow \quad 2 y & =\ln \left(\frac{1+x}{\square}\right)
\end{aligned}
Answer (1)
Given e 2 y = 1 − t a n h ( y ) 1 + t a n h ( y ) = 1 − x 1 + x .
Take the natural logarithm of both sides: 2 y = ln ( 1 − x 1 + x ) .
Identify the missing expression by comparing with 2 y = ln ( □ 1 + x ) .
The missing expression is 1 − x .
Explanation
Problem Setup We are given the equation 1 − t a n h ( x ) 1 + t a n h ( x ) = e 2 x and the substitution y = tanh − 1 ( x ) , which implies x = tanh ( y ) . We want to find the missing expression in the equation 2 y = ln ( □ 1 + x ) .
Taking the Natural Logarithm We have e 2 y = 1 − t a n h ( y ) 1 + t a n h ( y ) = 1 − x 1 + x . Taking the natural logarithm of both sides, we get 2 y = ln ( 1 − x 1 + x ) .
Identifying the Missing Expression Comparing this with the given equation 2 y = ln ( □ 1 + x ) , we can see that the missing expression is 1 − x .
Final Answer Therefore, the missing expression is 1 − x .
Examples
Imagine you're designing a communication system where signals are transmitted using hyperbolic tangent functions. Understanding how to manipulate and simplify expressions involving tanh ( x ) and its inverse is crucial for optimizing signal transmission and minimizing distortion. This problem demonstrates a useful identity that can help simplify calculations in such a system, allowing for more efficient signal processing and analysis.
