Inverse Hyperbolic Tangent Calculator

Inverse hyperbolic tangent is the function that returns the number y whose hyperbolic tangent equals x, for x strictly between −1 and 1. It is written tanh⁻¹(x), artanh(x), arctanh(x) or atanh(x): tanh(0.5493) = 0.5, so tanh⁻¹(0.5) = 0.5493.

The hyperbolic tangent is tanh(y) = (eʸ − e⁻ʸ)/(eʸ + e⁻ʸ). Its values fill the open interval from −1 to 1, and the inverse is half a natural log:

• Input: a number strictly between −1 and 1.
• Output: any real number, with the same sign as x.
• Special values: tanh⁻¹(0) = 0, tanh⁻¹(0.5) = ½ ln 3 = 0.5493, tanh⁻¹(3/5) = ln 2 = 0.6931, tanh⁻¹(0.8) = ln 3 = 1.0986.

As x approaches 1, tanh⁻¹(x) grows without bound: tanh⁻¹(0.99) = 2.6467 and tanh⁻¹(0.9999) = 4.9517. At x = ±1 the value is infinite.

On the unit hyperbola, tanh t is the slope of the line from the origin to the point (cosh t, sinh t). tanh⁻¹ turns that slope into t, twice the shaded area. 1/tanh(x) is coth(x), not tanh⁻¹(x).

To calculate inverse hyperbolic tangent, divide 1 + x by 1 − x, take the natural log, then halve it. Follow these 4 steps for x = 0.5:

1. Check that −1 < x < 1. The values ±1 and anything beyond have no real tanh⁻¹.
2. Divide: (1 + 0.5)/(1 − 0.5) = 1.5/0.5 = 3.
3. Take the natural log: ln 3 = 1.098612.
4. Halve it: 1.098612/2 = 0.549306.

Check the answer with tanh: tanh(0.549306) = (e^1.098612 − 1)/(e^1.098612 + 1) = (3 − 1)/(3 + 1) = 0.5.

Exact cases: when (1 + x)/(1 − x) is a perfect square, the ½ cancels. tanh⁻¹(3/5) = ½ ln 4 = ln 2, tanh⁻¹(0.8) = ½ ln 9 = ln 3 and tanh⁻¹(12/13) = ½ ln 25 = ln 5.

Type a value between −1 and 1 and the calculator returns tanh⁻¹(x) as you type, with the log formula, the exact value and the steps. The steps:

1. Type x in the field. It accepts decimals (0.5), fractions (3/5), roots (1/√2) and expressions.
2. Choose the rounding, from whole numbers to 15 decimal places.
3. Read the result, the exact log form, the 6 hyperbolic functions of the result and eʸ.
4. Switch the diagram between the unit hyperbola, the curve tanh(y) = x and the graph of tanh⁻¹.
5. Open the steps to see the log formula worked out for your value.

The calculator rejects 1, −1 and values outside them and states the reason. Results can be copied, shared as a link, printed or kept in the history list.

Fisher z mode is the statistics use of artanh. Enter a sample correlation r and the sample size n. The calculator returns z = artanh(r), its standard error 1/√(n − 3) and a 95% interval for r: r = 0.6 with n = 30 gives z = 0.6931 and r between 0.31 and 0.79.

The inverse hyperbolic tangent formula is tanh⁻¹(x) = ½ ln((1 + x)/(1 − x)) for −1 < x < 1.

Where the formula comes from: set x = tanh y and let u = e²ʸ. Then x = (u − 1)/(u + 1), so u = (1 + x)/(1 − x). Taking the log gives 2y = ln((1 + x)/(1 − x)).

Related forms:

• Identity: tanh(tanh⁻¹ x) = x for −1 < x < 1, and tanh⁻¹(tanh y) = y for every real y.
• Split log: tanh⁻¹(x) = ½[ln(1 + x) − ln(1 − x)].
• Odd function: tanh⁻¹(−x) = −tanh⁻¹(x).
• Reciprocal link: tanh⁻¹(x) = coth⁻¹(1/x) for 0 < |x| < 1.
• Sum rule: tanh⁻¹ a + tanh⁻¹ b = tanh⁻¹((a + b)/(1 + ab)).
• Logit link: tanh⁻¹(x) = ½ logit((1 + x)/2), where logit(p) = ln(p/(1 − p)).

The domain of inverse hyperbolic tangent is −1 < x < 1, and the range is all real numbers.

tanh y stays strictly between −1 and 1 for every real y, approaching ±1 only as y goes to ±∞. So tanh⁻¹(1), tanh⁻¹(−1) and tanh⁻¹(2) have no real value.

• Domain: (−1, 1), an open interval
• Range: (−∞, ∞)
• Monotonic: increasing over the whole domain
• Vertical asymptotes: x = −1 and x = 1
• Symmetry: odd, tanh⁻¹(−x) = −tanh⁻¹(x)
• Zero: tanh⁻¹(0) = 0

Software differs at the edges: JavaScript Math.atanh(1) returns Infinity, while Excel returns #NUM! and Python raises ValueError.

The graph of y = tanh⁻¹(x) is an increasing S-shaped curve through the origin, trapped between the vertical asymptotes x = −1 and x = 1.

• Passes through (0, 0), (0.5, 0.5493) and (−0.5, −0.5493).
• Slope 1 at the origin, the flattest point; the slope 1/(1 − x²) grows without bound near x = ±1.
• Inflection point at the origin: the curve bends down for x < 0 and up for x > 0.
• Mirror image of y = tanh x across the line y = x. The horizontal asymptotes y = ±1 of tanh become the vertical asymptotes of tanh⁻¹.

Near 0, tanh⁻¹(x) ≈ x, so the graph hugs y = x between −0.3 and 0.3.

The derivative of tanh⁻¹(x) is 1/(1 − x²) for −1 < x < 1. The slope is 1 at x = 0 and grows without bound toward x = ±1.

Proof: let y = tanh⁻¹ x, so x = tanh y. Differentiate: 1 = sech² y · dy/dx = (1 − tanh² y)·dy/dx = (1 − x²)·dy/dx.

• Chain rule: d/dx tanh⁻¹(u) = u′/(1 − u²). For tanh⁻¹(x/2): 2/(4 − x²).
• Integral it solves: ∫ dx/(1 − x²) = tanh⁻¹ x + C for |x| < 1, and ∫ dx/(a² − x²) = (1/a)·tanh⁻¹(x/a) + C.
• Integral of tanh⁻¹: ∫ tanh⁻¹ x dx = x·tanh⁻¹ x + ½ ln(1 − x²) + C.

On a scientific calculator, press hyp, then SHIFT (or 2nd), then tan, type the value and press =. On a Casio fx-82 or fx-991MS, the keys for tanh⁻¹(0.5) are hyp, SHIFT, tan, 0.5, ), = and the screen shows 0.5493061443.

1. Press hyp. The tan key now means tanh.
2. Press SHIFT (Casio, Sharp) or 2nd (Texas Instruments) to reach tanh⁻¹.
3. Type the value, close the bracket and press =.

Math ERROR (Casio) or ERR:DOMAIN (TI) means |x| ≥ 1. On Casio fx-991EX models, hyperbolic functions sit in the OPTN menu; on a TI-84, choose atanh( from the CATALOG.

The Android (Google) Calculator has no tanh⁻¹ key, so type ½ ln((1 + x)/(1 − x)). For x = 0.5, (1 + 0.5)/(1 − 0.5) = 3, so type ln(3) ÷ 2 and tap =: the result is 0.5493061443.

1. Tap the expand arrow, or turn the phone to landscape, to show the scientific keys.
2. Tap ln and type 3, the value of (1 + x)/(1 − x).
3. Close the bracket, tap ÷ and type 2.
4. Tap =.

For another x, type the full form, such as ln((1 + 0.8)/(1 − 0.8)) ÷ 2 for x = 0.8.

On the iPhone Calculator, switch to Scientific, type the value, tap 2nd, then tap tanh⁻¹. Type 0.5, tap 2nd and tap tanh⁻¹: the display shows 0.5493061443.

1. Turn the iPhone sideways, or tap the calculator icon and choose Scientific (iOS 18 and later).
2. Type 0.5.
3. Tap 2nd. The tanh key changes to tanh⁻¹.
4. Tap tanh⁻¹. The result appears at once.

A value of 1 or more shows Error.

In Windows Calculator, choose Scientific, type the value, open Trigonometry, select 2nd and hyp, then select tanh⁻¹. For 0.5, the result is 0.5493061443.

1. Open the menu (☰) and choose Scientific.
2. Type 0.5.
3. Open the Trigonometry list and select 2nd, then hyp.
4. Select tanh⁻¹.

A value of 1 or more shows “Invalid input”.

Software calls inverse hyperbolic tangent ATANH or atanh, and the result is a plain number.

• Excel and Google Sheets: =ATANH(0.5) returns 0.549306144. Excel's FISHER(r) is the same function.
• Python: math.atanh(0.5) returns 0.5493061443340548. NumPy uses np.arctanh.
• JavaScript: Math.atanh(0.5) returns 0.5493061443340548.
• MATLAB and Octave: atanh(0.5).
• C and C++: atanh(0.5) in C99 and C++11. Java uses 0.5 * Math.log((1 + x)/(1 − x)).
• Wolfram Language: ArcTanh[1/2].

At x = ±1, Excel returns #NUM!, Python raises ValueError: math domain error, and JavaScript returns Infinity or −Infinity. Beyond ±1, JavaScript returns NaN.

Without a calculator, use the Taylor series tanh⁻¹ x = x + x³/3 + x⁵/5 + x⁷/7 + … for |x| < 1. Every term is positive for x > 0, unlike the arctangent series, which alternates.

1. Series: tanh⁻¹ x = Σ x^(2n+1)/(2n + 1), summed from n = 0.
2. Exact cases: when (1 + x)/(1 − x) is a perfect square, tanh⁻¹ x is a plain log, such as tanh⁻¹(3/5) = ln 2.
3. Logs from the series: ln N = 2·tanh⁻¹((N − 1)/(N + 1)). This is how many calculators compute logarithms: ln 2 = 2·tanh⁻¹(1/3).

Example: tanh⁻¹(0.5) ≈ 0.5 + 0.041667 + 0.00625 + 0.001116 = 0.549033. The true value is 0.549306.

Near x = ±1 the terms shrink slowly. For x = 0.99, 100 terms still miss the true value by about 0.024.

The inverse hyperbolic tangent table lists tanh⁻¹(x) for common values of x between −1 and 1, with exact log forms. Click a row to load it into the calculator.

Values to know: tanh⁻¹(1/3) = ½ ln 2 = 0.3466, tanh⁻¹(1/2) = ½ ln 3 = 0.5493, tanh⁻¹(3/5) = ln 2 = 0.6931 and tanh⁻¹(4/5) = ln 3 = 1.0986.

The inverse hyperbolic tangent of a negative number is negative: tanh⁻¹(−x) = −tanh⁻¹(x). tanh⁻¹(−0.5) = −0.5493 and tanh⁻¹(−0.8) = −ln 3 = −1.0986.

tanh⁻¹ is an odd function, so its graph has point symmetry about the origin. In the log formula, swapping x for −x turns (1 + x)/(1 − x) into its reciprocal, and ln(1/u) = −ln u.

What is tanh⁻¹(−0.5)?

tanh⁻¹(−0.5) = ½ ln(1/3) = −0.549306. (1 − 0.5)/(1 + 0.5) = 1/3, and ½ ln(1/3) = −½ ln 3.

artanh accepts only −1 < x < 1 and returns any real number; arctan accepts any real number and returns an angle between −90° and 90°. The 2 functions swap their domain and range shapes.

The link through complex numbers is artanh(x) = −i·arctan(ix). The sum rules match up to a sign: artanh a + artanh b = artanh((a + b)/(1 + ab)), while arctan a + arctan b = arctan((a + b)/(1 − ab)).

Inverse hyperbolic tangent turns correlation coefficients into normally distributed scores, adds speeds in relativity and links tanh layers in neural networks to their inputs.

• Statistics (Fisher z-transformation): z = tanh⁻¹(r) makes a sample correlation r roughly normal with standard error 1/√(n − 3). For r = 0.6 and n = 30, z = 0.6931, and the 95% interval for r runs from 0.306 to 0.790.
• Relativity: the rapidity of an object moving at speed v is φ = tanh⁻¹(v/c). Rapidities add: 0.6c and 0.7c combine to tanh(0.6931 + 0.8673) = 0.9155c, not 1.3c.
• Machine learning: a tanh activation squashes any input into (−1, 1); tanh⁻¹ recovers the input from the output.
• Logarithms: ln N = 2·tanh⁻¹((N − 1)/(N + 1)) gives fast-converging log routines.
• Integrals: ∫ dx/(a² − x²) = (1/a)·tanh⁻¹(x/a) + C appears in growth and cooling models.