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Homework Statement
"starting from the definition of tanhx in terms of exponentials, prove that...
arctanh(x)=1/2log((1+x)/(1-x))
This is such a simple equation but I'm having a hard time proving it, can anyone help?
Cheers
The formula for arctanh(x) is equivalent to the inverse hyperbolic tangent function, also denoted as tanh-1(x). It is defined as arctanh(x) = 0.5 * ln((1+x)/(1-x)), where x is the input value.
The formula for arctanh(x) is derived by taking the inverse function of the hyperbolic tangent formula, tanh(x) = (ex - e-x) / (ex + e-x). By solving for x in the tanh(x) formula and then taking the natural logarithm of both sides, we can derive the formula for arctanh(x).
The domain of arctanh(x) is all real numbers between -1 and 1, inclusive. The range of arctanh(x) is all real numbers between -∞ and ∞.
Arctanh(x) is a useful function in mathematics as it is the inverse of the hyperbolic tangent function and is used to solve equations involving hyperbolic functions. It is also used in statistics and physics to model certain relationships.
Yes, the formula for arctanh(x) has various applications in real life, particularly in fields such as engineering, physics, and statistics. It can be used to model and analyze data in various situations, such as in probability and risk assessment. It is also used in the design of certain structures and in calculating electrical properties of materials.