Prove the following equation by using the following methods:
$\tanh ^{-1}(x)=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right) \quad-1 < x < 1$
(a) Using the method of this example:
Let $v=\tanh ^{-1}(v)$. Then we have the following.
$\begin{array}{l}
x=\tanh (\square)=\frac{\sinh (y)}{\cosh (y)}=\frac{\square}{\frac{\left(e^y+e^{-y}\right)}{2}} \cdot \frac{e^y}{e^y}=\frac{e^{2 y}-1}{e^{2 y}+1} \\
\Rightarrow \quad x e^{2 y}+x=e^{2 y}-1 \\
\Rightarrow \quad 1+x=e^{2 y}(1-x) \\
\Rightarrow \quad e^{2 y}=\frac{1+x}{1-x} \quad \\
\Rightarrow \quad 2 y=\ln \left(\frac{1+x}{1-x}\right)
\end{array}$
(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 $x=$ $\square$ , so we have the following.
$\begin{array}{l}
e^{2 y}=\frac{1+\tanh (y)}{\sqrt{\frac{1+x}{1-x}}} \\
2 y=\ln \left(\frac{1+x}{\sqrt{\frac{1+x}{1-x}}}\right) \\
y=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)
\end{array}$
Answer (2)
Method (a) fills in the blanks to show tanh − 1 ( x ) = 2 1 ln ( 1 − x 1 + x ) by expressing x in terms of exponential functions.
Method (b) uses the identity 1 − t a n h ( y ) 1 + t a n h ( y ) = e 2 y to derive the same result.
Both methods successfully prove the given equation.
The final result is tanh − 1 ( x ) = 2 1 ln ( 1 − x 1 + x ) .
Explanation
Problem Setup We are asked to prove the equation tanh − 1 ( x ) = 2 1 ln ( 1 − x 1 + x ) for − 1 < x < 1 using two methods, filling in the blanks in the provided steps.
Completing Method 1 (a) Method 1:
Let y = tanh − 1 ( x ) . Then x = tanh ( y ) . We have x = tanh ( y ) = cosh ( y ) sinh ( y ) = 2 e y + e − y 2 e y − e − y = e y + e − y e y − e − y Multiplying the numerator and denominator by e y , we get x = e 2 y + 1 e 2 y − 1 So the first blank should be filled with y and the second blank should be filled with 2 e y − e − y .
x = e 2 y + 1 e 2 y − 1 ⇒ x ( e 2 y + 1 ) = e 2 y − 1 ⇒ x e 2 y + x = e 2 y − 1 ⇒ e 2 y ( 1 − x ) = 1 + x ⇒ e 2 y = 1 − x 1 + x The third blank should be filled with 1 − x .
Taking the natural logarithm of both sides, we get 2 y = ln ( 1 − x 1 + x ) So, y = 2 1 ln ( 1 − x 1 + x ) .
Since y = tanh − 1 ( x ) , we have tanh − 1 ( x ) = 2 1 ln ( 1 − x 1 + x ) .
Completing Method 2 (b) Method 2:
Let y = tanh − 1 ( x ) . Then x = tanh ( y ) .
So the blank should be filled with tanh ( y ) .
We are given that 1 − t a n h ( y ) 1 + t a n h ( y ) = e 2 y . Since x = tanh ( y ) , we have e 2 y = 1 − x 1 + x Taking the natural logarithm of both sides, we get 2 y = ln ( 1 − x 1 + x ) So, y = 2 1 ln ( 1 − x 1 + x ) .
Since y = tanh − 1 ( x ) , we have tanh − 1 ( x ) = 2 1 ln ( 1 − x 1 + x ) .
Conclusion Therefore, we have proven the equation using both methods.
Examples
The hyperbolic tangent inverse function is used in machine learning, particularly in neural networks, as an activation function. It helps to map the input values to a range between -1 and 1, which can be useful for certain types of data normalization and scaling. Understanding and manipulating this function is crucial for designing and optimizing neural network architectures.
The equation tanh − 1 ( x ) = 2 1 ln ( 1 − x 1 + x ) is proved using two methods. Method (a) uses hyperbolic functions to derive the equation, while method (b) employs a key identity involving hyperbolic tangent. Both methods arrive at the same conclusion, confirming the equation's validity.
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