Prove the following equation by using the method of this example.
$\tanh ^{-1}(x)=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right) \quad-1
Let $y=\tanh ^{-1}(x)$. Then we have the following.
$x=\tanh \left(\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right)\right.$
(1) $=\frac{\sinh (y)}{\cosh (y)}=\frac{\frac{e^y-e^{-y}}{2}}{\frac{\left(e^y+e^{-y}\right)}{2}} \cdot \frac{e^y}{e^y}=\frac{e^{2 y}-1}{e^{2 y}+1}$
Answer (1)
Let y = tanh − 1 ( x ) , then x = tanh ( y ) .
Express tanh ( y ) as e 2 y + 1 e 2 y − 1 , so x = e 2 y + 1 e 2 y − 1 .
Solve for e 2 y : e 2 y = 1 − x 1 + x .
Thus, tanh − 1 ( x ) = y = 2 1 ln ( 1 − x 1 + x ) . tanh − 1 ( x ) = 2 1 ln ( 1 − x 1 + x )
Explanation
Problem Setup We are asked to prove the identity tanh − 1 ( x ) = 2 1 ln ( 1 − x 1 + x ) for − 1 < x < 1 . We will follow the method outlined in the problem statement.
Defining y Let y = tanh − 1 ( x ) . This implies that x = tanh ( y ) .
Expressing tanh(y) in terms of exponentials We know that tanh ( y ) = c o s h ( y ) s i n h ( y ) = e y + e − y e y − e − y . Multiplying the numerator and denominator by e y , we get tanh ( y ) = e 2 y + 1 e 2 y − 1 .
Solving for e^(2y) So, we have x = e 2 y + 1 e 2 y − 1 . Now we solve for e 2 y in terms of x . Multiplying both sides by e 2 y + 1 , we get x ( e 2 y + 1 ) = e 2 y − 1 .
Isolating e^(2y) Expanding, we have x e 2 y + x = e 2 y − 1 . Rearranging the terms, we get e 2 y − x e 2 y = x + 1 , which can be written as e 2 y ( 1 − x ) = 1 + x .
Finding e^(2y) Dividing both sides by 1 − x , we obtain e 2 y = 1 − x 1 + x .
Taking the natural logarithm Taking the natural logarithm of both sides, we get 2 y = ln ( 1 − x 1 + x ) .
Solving for y Dividing by 2, we have y = 2 1 ln ( 1 − x 1 + x ) .
Final Result Since we defined y = tanh − 1 ( x ) , we can substitute back to get tanh − 1 ( x ) = 2 1 ln ( 1 − x 1 + x ) . This completes the proof.
Examples
The hyperbolic tangent inverse function is useful in machine learning, particularly in neural networks, where it can be used as an activation function. The identity tanh − 1 ( x ) = 2 1 ln ( 1 − x 1 + x ) allows us to compute the inverse hyperbolic tangent using the natural logarithm, which is a standard function in most programming languages and calculators. This is particularly useful when implementing neural networks or other machine learning algorithms from scratch, as it provides a way to calculate the inverse hyperbolic tangent without relying on specialized libraries or functions.
