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‎lib/node_modules/@stdlib/math/base/special/log1pf/README.md‎

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# log1pf
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> Evaluate the [natural logarithm][@stdlib/math/base/special/ln] of `1+x` in single‐precision floating-point number.
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> Evaluate the [natural logarithm][@stdlib/math/base/special/ln] of `1+x` as a single‐precision floating-point number.
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<section class="usage">
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#### log1pf( x )
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Evaluates the [natural logarithm][@stdlib/math/base/special/ln] of `1+x` in single‐precision floating-point number.
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Evaluates the [natural logarithm][@stdlib/math/base/special/ln] of `1+x` as a single‐precision floating-point number.
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```javascript
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var v = log1pf( 4.0 );
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#### stdlib_base_log1pf( x )
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Evaluates the [natural logarithm][@stdlib/math/base/special/ln] of `1+x` in single‐precision floating-point number.
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Evaluates the [natural logarithm][@stdlib/math/base/special/ln] of `1+x` as a single‐precision floating-point number.
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```c
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float out = stdlib_base_log1pf( 4.0f );

‎lib/node_modules/@stdlib/math/base/special/log1pf/docs/repl.txt‎

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{{alias}}( x )
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Evaluates the natural logarithm of `1+x`.
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Evaluates the natural logarithm of `1+x` as a single‐precision
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floating-point number.
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For `x < -1`, the function returns `NaN`, as the natural logarithm is not
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defined for negative numbers.

‎lib/node_modules/@stdlib/math/base/special/log1pf/docs/types/index.d.ts‎

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// TypeScript Version: 4.1
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/**
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* Evaluates the natural logarithm of `1+x`.
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* Evaluates the natural logarithm of `1+x` as a single‐precision floating-point number.
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*
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* ## Notes
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*

‎lib/node_modules/@stdlib/math/base/special/log1pf/include/stdlib/math/base/special/log1pf.h‎

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#endif
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/**
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Evaluates the natural logarithm of `1+x` in single‐precision floating-point number.
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Evaluates the natural logarithm of `1+x` as a single‐precision floating-point number.
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*/
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float stdlib_base_log1pf( const float x );
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‎lib/node_modules/@stdlib/math/base/special/log1pf/lib/index.js‎

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'use strict';
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/**
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* Evaluate the natural logarithm of \\(1+x\\) in single‐precision floating-point number.
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* Evaluate the natural logarithm of \\(1+x\\) as a single‐precision floating-point number.
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*
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* @module @stdlib/math/base/special/log1pf
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*

‎lib/node_modules/@stdlib/math/base/special/log1pf/lib/main.js‎

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/**
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* Evaluates the natural logarithm of \\(1+x\\) as a single‐precision floating-point number.
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*
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* ## Method
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*
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* 1. Argument Reduction: find \\(k\\) and \\(f\\) such that
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*
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* ```tex
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* 1+x = 2^k (1+f)
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* ```
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*
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* where
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*
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* ```tex
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* \frac{\sqrt{2}}{2} < 1+f < \sqrt{2}
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* ```
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*
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* <!-- <note> -->
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*
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* If \\(k=0\\), then \\(f=x\\) is exact. However, if \\(k \neq 0\\), then \\(f\\) may not be representable exactly. In that case, a correction term is needed. Let
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*
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* ```tex
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* u = \operatorname{round}(1+x)
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* ```
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*
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* and
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*
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* ```tex
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* c = (1+x) - u
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* ```
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*
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* then
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*
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* ```tex
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* \ln (1+x) - \ln u \approx \frac{c}{u}
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* ```
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*
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* We can thus proceed to compute \\(\ln(u)\\), and add back the correction term \\(c/u\\).
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*
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* <!-- </note> -->
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*
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* <!-- <note> -->
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*
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* When \\(x > 2^{24}\\), one can simply return \\(\ln(x)\\).
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*
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* <!-- </note> -->
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*
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* 2. Approximation of \\(\operatorname{log1pf}(f)\\). Let
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*
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* ```tex
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* s = \frac{f}{2+f}
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* ```
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*
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* based on
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*
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* ```tex
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* \begin{align*}
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* \ln 1+f &= \ln (1+s) - \lnf (1-s) \\
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* &= 2s + \frac{2}{3} s^3 + \frac{2}{5} s^5 + ... \\
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* &= 2s + sR \\
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* \end{align*}
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* ```
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*
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* We use a special Reme algorithm on \\(\[0,0.1716\]\\) to generate a polynomial of degree \\(14\\) to approximate \\(R\\). The maximum error of this polynomial approximation is bounded by \\(2^{-58.45}\\). In other words,
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*
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* ```tex
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* R(z) \approx \mathrm{Lp}_1 s^2 + \mathrm{Lp}_2 s^4 + \mathrm{Lp}_3 s^6 + \mathrm{Lp}_4 s^8 + \mathrm{Lp}_5 s^{10} + \mathrm{Lp}_6 s^{12} + \mathrm{Lp}_7 s^{14}
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* ```
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*
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* and
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*
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* ```tex
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* | \mathrm{Lp}_1 s^2 + \ldots + \mathrm{Lp}_7 s^14 - R(z) | \leq 2^{-58.45}
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* ```
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*
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* <!-- <note> -->
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*
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* The values of \\(Lp1\\) to \\(Lp7\\) may be found in the source.
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*
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* <!-- </note> -->
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*
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* Note that
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*
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* ```tex
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* \begin{align*}
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* 2s &= f - sf \\
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* &= f - \frac{f^2}{2} + s \frac{f^2}{2} \\
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* \end{align*}
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* ```
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*
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* In order to guarantee error in \\(\ln\\) below \\(1\ \mathrm{ulp}\\), we compute the log by
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*
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* ```tex
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* \operatorname{log1pf}(f) = f - \biggl(\frac{f^2}{2} - s\biggl(\frac{f^2}{2}+R\biggr)\biggr)
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* ```
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*
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* 3. Finally,
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*
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* ```tex
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* \begin{align*}
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* \operatorname{log1pf}(x) &= k \cdot \mathrm{ln2} + \operatorname{log1pf}(f) \\
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* &= k \cdot \mathrm{ln2}_{hi}+\biggl(f-\biggl(\frac{f^2}{2}-\biggl(s\biggl(\frac{f^2}{2}+R\biggr)+k \cdot \mathrm{ln2}_{lo}\biggr)\biggr)\biggr) \\
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* \end{align*}
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* ```
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*
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* Here \\(\mathrm{ln2}\\) is split into two floating point numbers:
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*
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* ```tex
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* \mathrm{ln2}_{hi} + \mathrm{ln2}_{lo}
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* ```
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*
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* where \\(n \cdot \mathrm{ln2}_{hi}\\) is always exact for \\(|n| < 2000\\).
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*
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* ## Special Cases
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*
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* - \\(\operatorname{log1pf}(x) = \mathrm{NaN}\\) with signal if \\(x < -1\\) (including \\(-\infty\\))
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* - \\(\operatorname{log1pf}(+\infty) = +\infty\\)
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* - \\(\operatorname{log1pf}(-1) = -\infty\\) with signal
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* - \\(\operatorname{log1pf}(\mathrm{NaN})= \mathrm{NaN}\\) with no signal
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*
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* ## Notes
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*
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* - According to an error analysis, the error is always less than \\(1\\) ulp (unit in the last place).
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*
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* - The hexadecimal values are the intended ones for the used constants. The decimal values may be used, provided that the compiler will convert from decimal to binary accurately enough to produce the hexadecimal values shown.
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*
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* - Assuming \\(\ln(x)\\) is accurate, the following algorithm can be used to evaluate \\(\operatorname{log1pf}(x)\\) to within a few ULP:
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*
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* ```javascript
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* var u = 1.0 + x;
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* if ( u === 1.0 ) {
204-
* return x;
205-
* } else {
206-
* return ln(u) * (x/(u-1.0));
207-
* }
208-
* ```
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*
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* See HP-15C Advanced Functions Handbook, p.193.
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*
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* @param {number} x - input value
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* @returns {number} the natural logarithm of `1+x`
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*

‎lib/node_modules/@stdlib/math/base/special/log1pf/lib/native.js‎

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// MAIN //
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/**
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* Evaluates the natural logarithm of \\(1+x\\) in single‐precision floating-point number.
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* Evaluates the natural logarithm of \\(1+x\\) as a single‐precision floating-point number.
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*
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* @private
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* @param {number} x - input value

‎lib/node_modules/@stdlib/math/base/special/log1pf/package.json‎

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{
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"name": "@stdlib/math/base/special/log1pf",
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"version": "0.0.0",
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"description": "Evaluate the natural logarithm of 1+x in single‐precision.",
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"description": "Evaluate the natural logarithm of 1+x as a single‐precision.",
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"license": "Apache-2.0",
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"author": {
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"name": "The Stdlib Authors",

‎lib/node_modules/@stdlib/math/base/special/log1pf/src/main.c‎

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/* End auto-generated functions. */
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/**
89-
* Evaluates the natural logarithm of \\(1+x\\) in single‐precision floating-point number.
90-
*
91-
* ## Method
92-
*
93-
* 1. Argument Reduction: find \\(k\\) and \\(f\\) such that
94-
*
95-
* ```tex
96-
* 1+x = 2^k (1+f)
97-
* ```
98-
*
99-
* where
100-
*
101-
* ```tex
102-
* \frac{\sqrt{2}}{2} < 1+f < \sqrt{2}
103-
* ```
104-
*
105-
* <!-- <note> -->
106-
*
107-
* If \\(k=0\\), then \\(f=x\\) is exact. However, if \\(k \neq 0\\), then \\(f\\) may not be representable exactly. In that case, a correction term is needed. Let
108-
*
109-
* ```tex
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* u = \operatorname{round}(1+x)
111-
* ```
112-
*
113-
* and
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*
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* ```tex
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* c = (1+x) - u
117-
* ```
118-
*
119-
* then
120-
*
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* ```tex
122-
* \ln (1+x) - \ln u \approx \frac{c}{u}
123-
* ```
124-
*
125-
* We can thus proceed to compute \\(\ln(u)\\), and add back the correction term \\(c/u\\).
126-
*
127-
* <!-- </note> -->
128-
*
129-
* <!-- <note> -->
130-
*
131-
* When \\(x > 2^{53}\\), one can simply return \\(\ln(x)\\).
132-
*
133-
* <!-- </note> -->
134-
*
135-
* 2. Approximation of \\(\operatorname{log1pf}(f)\\). Let
136-
*
137-
* ```tex
138-
* s = \frac{f}{2+f}
139-
* ```
140-
*
141-
* based on
142-
*
143-
* ```tex
144-
* \begin{align*}
145-
* \ln 1+f &= \ln (1+s) - \ln (1-s) \\
146-
* &= 2s + \frac{2}{3} s^3 + \frac{2}{5} s^5 + ... \\
147-
* &= 2s + sR \\
148-
* \end{align*}
149-
* ```
150-
*
151-
* We use a special Reme algorithm on \\(\[0,0.1716\]\\) to generate a polynomial of degree \\(14\\) to approximate \\(R\\). The maximum error of this polynomial approximation is bounded by \\(2^{-58.45}\\). In other words,
152-
*
153-
* ```tex
154-
* R(z) \approx \mathrm{Lp}_1 s^2 + \mathrm{Lp}_2 s^4 + \mathrm{Lp}_3 s^6 + \mathrm{Lp}_4 s^8 + \mathrm{Lp}_5 s^{10} + \mathrm{Lp}_6 s^{12} + \mathrm{Lp}_7 s^{14}
155-
* ```
156-
*
157-
* and
158-
*
159-
* ```tex
160-
* | \mathrm{Lp}_1 s^2 + \ldots + \mathrm{Lp}_7 s^14 - R(z) | \leq 2^{-58.45}
161-
* ```
162-
*
163-
* <!-- <note> -->
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*
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* The values of \\(Lp1\\) to \\(Lp7\\) may be found in the source.
166-
*
167-
* <!-- </note> -->
168-
*
169-
* Note that
170-
*
171-
* ```tex
172-
* \begin{align*}
173-
* 2s &= f - sf \\
174-
* &= f - \frac{f^2}{2} + s \frac{f^2}{2} \\
175-
* \end{align*}
176-
* ```
177-
*
178-
* In order to guarantee error in \\(\ln\\) below \\(1\ \mathrm{ulp}\\), we compute the log by
179-
*
180-
* ```tex
181-
* \operatorname{log1pf}(f) = f - \biggl(\frac{f^2}{2} - s\biggl(\frac{f^2}{2}+R\biggr)\biggr)
182-
* ```
183-
*
184-
* 3. Finally,
185-
*
186-
* ```tex
187-
* \begin{align*}
188-
* \operatorname{log1pf}(x) &= k \cdot \mathrm{ln2} + \operatorname{log1pf}(f) \\
189-
* &= k \cdot \mathrm{ln2}_{hi}+\biggl(f-\biggl(\frac{f^2}{2}-\biggl(s\biggl(\frac{f^2}{2}+R\biggr)+k \cdot \mathrm{ln2}_{lo}\biggr)\biggr)\biggr) \\
190-
* \end{align*}
191-
* ```
192-
*
193-
* Here \\(\mathrm{ln2}\\) is split into two floating point numbers:
194-
*
195-
* ```tex
196-
* \mathrm{ln2}_{hi} + \mathrm{ln2}_{lo}
197-
* ```
198-
*
199-
* where \\(n \cdot \mathrm{ln2}_{hi}\\) is always exact for \\(|n| < 2000\\).
200-
*
201-
*
202-
* ## Special Cases
203-
*
204-
* - \\(\operatorname{log1pf}(x) = \mathrm{NaN}\\) with signal if \\(x < -1\\) (including \\(-\infty\\))
205-
* - \\(\operatorname{log1pf}(+\infty) = +\infty\\)
206-
* - \\(\operatorname{log1pf}(-1) = -\infty\\) with signal
207-
* - \\(\operatorname{log1pf}(\mathrm{NaN})= \mathrm{NaN}\\) with no signal
208-
*
209-
*
210-
* ## Notes
211-
*
212-
* - According to an error analysis, the error is always less than \\(1\\) ulp (unit in the last place).
213-
*
214-
* - The hexadecimal values are the intended ones for the used constants. The decimal values may be used, provided that the compiler will convert from decimal to binary accurately enough to produce the hexadecimal values shown.
215-
*
216-
* - Assuming \\(\ln(x)\\) is accurate, the following algorithm can be used to evaluate \\(\operatorname{log1pf}(x)\\) to within a few ULP:
217-
*
218-
* ```c
219-
* double u = 1.0 + x;
220-
* if ( u == 1.0 ) {
221-
* return x;
222-
* } else {
223-
* return ln(u) * (x/(u-1.0));
224-
* }
225-
* ```
226-
*
227-
* See HP-15C Advanced Functions Handbook, p.193.
89+
* Evaluates the natural logarithm of \\(1+x\\) as a single‐precision floating-point number.
22890
*
22991
* @param x input value
23092
* @return output value

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