Encontrar aproximaciones polinómicas de una onda sinusoidal

Quiero aproximar la onda sinusoidal dada por aplicando una forma de onda polinómica a una onda triangular simple , generada por la función$\sin\left(\pi x\right)$

$$T\left(x\right)=1-4\left|\tfrac{1}{2}-\operatorname{mod}(\tfrac{1}{2}x+\tfrac{1}{4},\ 1)\right|$$

donde es la parte fraccionaria de :$\operatorname{mod}(x, 1)$$x$

$$\operatorname{mod}(x, y) \triangleq y \cdot \left( \left\lfloor \frac{x}{y}\right\rfloor - \frac{x}{y} \right)$$

Una serie de Taylor podría usarse como formador de ondas.

$$S_1\left(x\right)=\frac{\pi x}{2}-\frac{\frac{\pi x}{2}^3}{3!}+\frac{\frac{\pi x}{2}^5}{5!}-\frac{\frac{\pi x}{2}^7}{7!}$$

Dadas las funciones anteriores, $S_1(T(x))$ nos dará una aproximación decente de una onda sinusoidal. Pero necesitamos subir a la séptima potencia de la serie para obtener un resultado razonable, y los picos son un poco bajos y no tendrán una pendiente exactamente cero.

En lugar de la serie Taylor, podríamos usar un formador de onda polinomial siguiendo algunas reglas.

• Debe pasar por -1, -1 y + 1, + 1.
• La pendiente en -1, -1 y + 1, + 1 debe ser cero.
• Debe ser simétrico.

Una función simple que cumple nuestros requisitos:

$$S_2\left(x\right)=\frac{3x}{2}-\frac{x^3}{2}$$

Los gráficos de $S_2(T(x))$ y $\sin\left(\pi x\right)$ están bastante cerca, pero no tan cerca como la serie Taylor. Entre los picos y los cruces por cero, se desvían visiblemente un poco. Una función más pesada y precisa que cumple nuestros requisitos:

$$S_3\left(x\right)=\frac{x(x^2-5)^2}{16}$$

Probablemente sea lo suficientemente cercano para mis propósitos, pero me pregunto si existe otra función que se aproxime más a la onda sinusoidal y sea computacionalmente más barata. Tengo una buena comprensión de cómo encontrar funciones que cumplan los tres requisitos anteriores, pero no estoy seguro de cómo encontrar funciones que cumplan esos requisitos y que también coincidan más con una onda sinusoidal.

¿Qué métodos existen para encontrar polinomios que imitan una onda sinusoidal (cuando se aplica a una onda triangular)?

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Para aclarar, no necesariamente busco solo polinomios de simétrica impar, aunque esas son la opción más sencilla.

Algo así como la siguiente función también podría satisfacer mis necesidades:

$$S_4\left(x\right)=\frac{3x}{2}+\frac{x^2}{4}+\frac{x^4}{4}$$

Esto cumple con los requisitos en el rango negativo, y también se podría usar una solución por partes para aplicarlo al rango positivo; por ejemplo

$$\frac{3x}{2}-\frac{P\left(x,2\right)}{4}-\frac{P\left(x,4\right)}{4}$$

donde es la función de potencia firmada .$P$

También me interesarían las soluciones que utilizan la función de potencia con signo para admitir exponentes fraccionales, ya que esto nos da otro "mando para girar" sin agregar otro coeficiente.

$$a_0x\ +a_1P\left(x,\ p_1\right)$$

Dadas las constantes correctas, esto podría obtener una muy buena precisión sin la pesadez de los polinomios de quinto o séptimo orden. Aquí hay un ejemplo que cumple los requisitos descritos aquí usando algunas constantes seleccionadas a mano: .$a_0=1.\overline{666}, a_1=-0.\overline{666}, p_1=2.5$

$$\frac{5x-2P\left(x,\ \frac{5}{2}\right)}{3}$$

De hecho, esas constantes están muy cerca de , y , y . Conectarlos da algo que se ve extremadamente cerca de una onda sinusoidal. 1-$\frac \pi 2$$1 - \frac \pi 2$$e$

$$\frac{\pi}{2}x\ +\left(1-\frac{\pi}{2}\right)P\left(x,e\right)$$

Para decirlo de otra manera, ve muy cerca de entre 0,0 y π / 2,1. ¿Alguna idea sobre el significado de esto? Quizás una herramienta como Octave pueda ayudar a descubrir las "mejores" constantes para este enfoque. pecado(x$x-\frac{x^e}{6}$$\sin(x)$

Hace aproximadamente una década hice esto para una compañía de sintetizadores de música sin nombre que tenía I&D no muy lejos de mi condominio en Waltham MA. (No puedo imaginar quiénes son). No tengo los coeficientes.

pero prueba esto:

$$\begin{align} f(x) &\approx \sin\left(\tfrac{\pi}{2} x \right) \qquad \qquad \text{for }-1 \le x \le +1 \\ \\ &= \tfrac{\pi}{2} x (a_0 + a_1 x^2 + a_2 x^4) \\ \end{align}$$

esto garantiza que .$f(-x) = -f(x)$

Para garantizar que entonces$f'(x) \Big|_{x=\pm1} =0$

$$f'(x) = \tfrac{\pi}{2}( a_0 + 3 a_1 x^2 + 5 a_2 x^4)$$

$$a_0 + 3 a_1 + 5 a_2 = 0 \tag{1}$$

Esa es la primera restricción. Para garantizar que , entonces$|f(\pm 1)| = 1$

$$a_0 + a_1 + a_2 = \tfrac{2}{\pi} \tag{2}$$

Esa es la segunda restricción. Eliminando y resolviendo Eqs. (1) y (2) para un 2 en términos de un 1 (que se deja ajustar):$a_0$$a_2$$a_1$

$$a_0 = \tfrac{5}{2 \pi} - \tfrac{1}{2} a_1$$

$$a_2 = -\tfrac{1}{2 \pi} - \tfrac{1}{2} a_1$$

Ahora solo tiene un coeficiente, , restante para girar para obtener el mejor rendimiento:$a_1$

$$f(x) = \tfrac{\pi}{2} x \left((\tfrac{5}{2 \pi} - \tfrac{1}{2} a_1) + a_1 x^2 - (\tfrac{1}{2 \pi} + \tfrac{1}{2} a_1)x^4 \right)$$

Esta es la forma en que giraría para obtener el mejor rendimiento para un oscilador de onda sinusoidal. Ajustaría usar lo anterior y la simetría de la onda sinusoidal sobre x = 1 y colocar exactamente un ciclo completo en un búfer con una potencia de dos puntos (digamos 128, no me importa) y ejecutar el FFT en ese ciclo perfecto$a_1$$x=1$

El resultado de FFT bin 1 será la fuerza del seno y debe ser de aproximadamente . Ahora puede ajustar un 1 para subir y bajar su distorsión armónica tercera. Comenzaría con un 1 ≈ $N/2$$a_1$para quea0≈1. Eso está en el bin 3 de los resultados de FFT, pero la distorsión armónica 5 (valor en bin 5) será consecuente (aumentará a medida que baje el 3er armónico). Ajustaríaun1para que la fuerza del 5 ° nivel armónico sea igual al 3 ° nivel armónico. Será de alrededor de -70 dB desde el primer armónico (como recuerdo). Esa será la onda sinusoidal de mejor sonido de un polinomio barato, de 3 coeficientes, 5º orden, simétrico impar.$a_1 \approx \tfrac{5}{\pi} - 2$$a_0 \approx 1$$a_1$

Alguien más puede escribir el código MATLAB. ¿Cómo te suena eso?

Lo que generalmente se hace es una aproximación que minimiza alguna norma del error, a menudo la forma (donde se minimiza el error máximo) o la forma L 2 (donde se minimiza el error cuadrático medio). La aproximación L ∞ se realiza utilizando el algoritmo de intercambio Remez . Estoy seguro de que puedes encontrar algún código fuente abierto que implemente ese algoritmo. Sin embargo, en este caso creo que una optimización muy simple (discreta) l 2 es suficiente. Veamos un código de Matlab / Octave y los resultados:$L_{\infty}$$L_2$$L_{\infty}$$l_2$

x = espacio de lins (0, pi / 2,300); % cuadrícula en [0, pi / 2]
x = x (:);
% sistema sobredeterminado de ecuaciones lineales
% (usando solo poderes impares)
A3 = [x, x. ^ 3];
A5 = [x, x. ^ 3, x. ^ 5];
b = sin (x);
% resolver en sentido l2
c3 = A3 \ b;
c5 = A5 \ b;
f3 = A3 * c3; % Aproximación de tercer orden
f5 = A5 * c5; % Aproximación de quinto orden

La siguiente figura muestra los errores de aproximación para el orden de y para las aproximaciones de orden de 5 t h . Los errores de aproximación máxima son y , respectivamente.$3^{rd}$$5^{th}$8.8869e-031.5519e-04

Los coeficientes óptimos son

c3 =
   0.988720369237930
  -0.144993929056091

y

c5 =
   0.99976918199047515
  -0.16582163562776930
   0.00757183954143367

Entonces, la aproximación de tercer orden es

$$\sin(x)\approx 0.988720369237930x-0.144993929056091x^3,\quad x\in[-\pi/2,\pi/2]\tag{1}$$

y la aproximación de quinto orden es

$$\sin(x)\approx 0.99976918199047515x-0.16582163562776930x^3+\\0.00757183954143367x^5,\quad x\in[-\pi/2,\pi/2]\tag{2}$$

EDITAR:

Eché un vistazo a las aproximaciones con la función de potencia con signo, como se sugiere en la pregunta, pero la mejor aproximación es apenas mejor que la aproximación de tercer orden que se muestra arriba. La función aproximada es

$$f(x)=x - \frac{1}{p}\left(\frac{\pi}{2}\right)^{1-p}x^p,\qquad x\in[0,\pi/2]\tag{3}$$

donde las constantes fueron elegidas de manera tal que y f ′ ( π / 2 ) = 0 . La potencia p se optimizó para lograr el error máximo más pequeño en el rango [ 0 , π / 2 ] . Se encontró que el valor óptimo para p era p = 2.774 . La siguiente figura muestra los errores de aproximación para la aproximación de tercer orden ( 1 ) y para la nueva aproximación ( $f'(0)=1$$f'(\pi/2)=0$$p$$[0,\pi/2]$$p$$p=2.774$$(1)$ :$(3)$

The maximum approximation error of the approximation $(3)$ is 4.5e-3, but note that the third-order approximation only exceeds that error close to $\pi/2$ and that for the most part its approximation error is actually smaller than the one of the signed power function.

EDIT 2:

If you don't mind division you could also use Bhaskara I's sine approximation formula, which has a maximum approximation error of 1.6e-3:

$$\sin(x)\approx\frac{16x(\pi-x)}{5\pi^2-4x(\pi-x)},\qquad x\in [0,\pi/2]\tag{4}$$

Start with an otherwise general, odd-symmetry 5th-order parameterized polynomial:

$$\begin{align} f(x) &= a_0 x^1 + a_1 x^3 + a_2 x^5 \\ &= x\big( a_0 + a_1 x^2 + a_2 x^4 \big) \\ &= x\bigg( a_0 + x^2 \Big(a_1 + a_2 x^2\Big) \bigg) \\ \end{align}$$

Now we place some constraints on this function. Amplitude should be 1 at the peaks, in other words $f(1) = 1$. Substituting $1$ for $x$ gives:

$$a_0 + a_1 + a_2 = 1 \tag1$$

That's one constraint. The slope at the peaks should be zero, in other words $f'(1) = 0$. The derivative of $f(x)$ is

$$a_0 + 3 a_1 x^2 + 5 a_2 x^4$$

and substituting $1$ for $x$ gives our second constraint:

$$a_0 + 3 a_1 + 5 a_2 = 0 \tag2$$

Now we can use our two constraints to solve for $a_1$ and $a_2$ in terms of $a_0$.

$$a_1=\frac{5}{2}-2a_0 \\ a_2=a_0-\frac{3}{2}\tag{3}$$

All that's left is to tweak $a_0$ to get a nice fit. Incidentally, $a_0$ (and the slope at the origin) ends up being $\approx\frac{\pi}{2}$, as we can see from a plot of the function.

Parameter optimization

Below are a number of optimizations of the coefficients, which result in these relative amplitudes of the harmonics compared to the fundamental frequency (1st harmonic):

In the complex Fourier series:

$$\sum_{k=-\infty}^\infty c_k e^{i\tfrac{2\pi}{P} kx},$$

of a real $P\text{-}$periodic waveform with $P = 4$ and time symmetry about $x = 1$ and with half a period defined by odd function $f(x)$ over $-1 \le x \le 1,$ the coefficient of the $k\text{th}$ complex exponential harmonic is:

$$c_k = \frac{1}{P}\int_{-1}^{-1+P}\left(\cases{f(x)&if $x < 1$\\-f(x-2)&if $x \ge 1$}\right)e^{-i\tfrac{2\pi}{P}kx}dx.$$

Because of the relationship $2 \cos(x) = e^{ix} + e^{-ix}$ (see: Euler's formula), the amplitude of a real sinusoidal harmonic with $k > 0$ is $2\left|c_k\right|,$ which is twice that of the magnitude of the complex exponential of the same frequency. This can be massaged to a form which makes it easier for some symbolic mathematics software to simplify the integral:

$$2|c_k| = \frac{2}{4}\left|\int_{-1}^{3}\left(\cases{f(x)&if $x < 1$\\-f(x-2)&if $x \ge 1$}\right)e^{-i\tfrac{2\pi}{4}kx}dx\right|\\ = \frac{1}{2}\left|\int_{-1}^{1}f(x)e^{-i\tfrac{\pi}{2}kx}dx - \int_{1}^{3}f(x-2)e^{-i\tfrac{\pi}{2}kx}dx\right|\\ = \frac{1}{2}\left|\int_{-1}^{1}f(x)e^{-i\tfrac{\pi}{2}kx}dx - \int_{-1}^{1}f(x+2-2)e^{-i\tfrac{\pi}{2}k(x+2)}dx\right|\\ = \frac{1}{2}\left|\int_{-1}^{1}f(x)e^{-i\tfrac{\pi}{2}kx}dx - \int_{-1}^{1}f(x)e^{-i\tfrac{\pi}{2}k(x+2)}dx\right|\\ = \frac{1}{2}\left|\int_{-1}^{1}f(x)\left(e^{-i\tfrac{\pi}{2}kx} - e^{-i\tfrac{\pi}{2}k(x+2)}\right)dx\right|\\ = \frac{1}{2}\left|e^{i\tfrac{\pi}{2}x}\int_{-1}^{1}f(x)\left(e^{-i\tfrac{\pi}{2}kx} - e^{-i\tfrac{\pi}{2}k(x+2)}\right)dx\right|\\ = \frac{1}{2}\left|\int_{-1}^{1}f(x)\left(e^{-i\frac{\pi}{2}k(x-1)}-e^{-i\frac{\pi}{2}k(x+1)}\right)dx\right|$$

The above takes advantage of that $|e^{ix}|=1$ for real $x.$ It is easier for some computer algebra systems to simplify the integral by assuming $k$ is real, and to simplify to integer $k$ at the end. Wolfram Alpha can integrate individual terms of the final integral corresponding to the terms of the polynomial $f(x)$. For the coefficients given in Eq. 3 we get amplitude:

$$= \left|\frac{48\left((-1)^k - 1\right)\left(16\,a_0\left(\pi^2 k^2 - 10\right) - 5\times(5\pi^2 k^2 - 48)\right)}{\pi^6k^6}\right|$$

5th order, continuous derivative

We can solve for the value of $a_0$ that gives equal amplitude $2|c_k|$of the 3rd and the 5th harmonic. There will be two solutions corresponding to the 3rd and the 5th harmonic having equal or opposite phases. The best solution is the one that minimizes the maximum amplitude of the 3rd and above harmonics and equivalently the maximum relative amplitude of the 3rd and above harmonics compared to the fundamental frequency (1st harmonic):

$$a_0 = \frac{3\times(132375\pi^2 - 130832)}{16\times(15885\pi^2 - 16354)}\approx 1.569778813,\\ a_1 = \frac{5}{2} - 2a_0 = \frac{79425\pi^2 - 65416}{8\times(-15885\pi^2 + 16354)}\approx -0.6395576276,\\ a_2 = a_0 - \frac{3}{2} = \frac{15885\pi^2}{16\times(15885\pi^2 - 16354)}\approx 0.06977881382.$$

This gives the fundamental frequency at amplitude $\frac{13679616}{15885\pi^6 - 16354\pi^4} \approx 1.000071420$ and both the 3rd and the 5th harmonic at relative amplitude $\frac{1}{8906}$ or about $-78.99\text{ dB}$ compared to the fundamental frequency. A $k\text{th}$ harmonic has relative amplitude $\frac{\left(1 - (-1)^k\right)\left|8177k^2 - 79425\right|}{142496 k^6}.$

7th order, continuous derivative

Likewise, the optimal 7th order polynomial approximation with the same initial constraints and the 3rd, 5th, and 7th harmonic at the lowest possible equal level is:

$$\begin{align} f(x) &= a_0 x^1 + a_1 x^3 + a_2 x^5 + a_3 x^7\\ &= x\big( a_0 + a_1 x^2 + a_2 x^4 + a_3 x^7 \big) \\ &= x\bigg( a_0 + x^2 \Big(a_1 + x^2 \big(a_2 + a_3 x^2 \big) \Big) \bigg) \\ \end{align}$$

$$a_0 = \frac{2a_2 + 4a_3 + 3}{2} \approx 1.570781972,\\ a_1 = -\frac{4a_2 + 6a_3 + 1}{2} \approx -0.6458482979,\\ a_2 = \frac{347960025\pi^4 - 405395408\pi^2}{16\times(281681925 \pi^4 - 405395408 \pi^2 + 108019280)} \approx 0.07935067784,\\ a_3 = - \frac{16569525\pi^4}{16\times(281681925\pi^4 - 405395408\pi^2 + 108019280)} \approx -0.004284352588.$$

This is the best of four possible solutions corresponding to equal/opposite phase combinations of the 3rd, 5th, and 7th harmonic. The fundamental frequency has amplitude $\frac{2293523251200}{281681925\pi^8 - 405395408\pi^6 + 108019280\pi^4} \approx 0.9999983752,$ and the 3rd, 5th, and 7th harmonics have relative amplitude $\frac{1}{1555395} \approx -123.8368\text{ dB}$ compared to the fundamental. A $k\text{th}$ harmonic has relative amplitude $\frac{\left(1 - (-1)^k\right)\left|1350241k^4 - 50674426k^2 + 347960025\right|}{597271680k^8}$ compared to the fundamental.

5th order

If the requirement of a continuous derivative is dropped, the 5th order approximation will be more difficult to solve symbolically, because the amplitude of the 9th harmonic will rise above the amplitude of the 3rd, 5th, and the 7th harmonic if those are constrained to be equal and minimized. Testing 16 different solutions corresponding to different subsets of three harmonics from $\{3, 5, 7, 9\}$ being of equal amplitude and of equal or opposite phases, the best solution is:

$$f(x) = a_0 x^1 + a_1 x^3 + a_2 x^5\\ a_0 = 1 - a_1 - a_2 \approx 1.570034357\\ a_1 = \frac{3\times(2436304\pi^2 - 2172825\pi^4)}{8\times(1303695\pi^4 - 1827228\pi^2 + 537160)} \approx -0.6425216143\\ a_2 = \frac{1303695\pi^4}{16\times(1303695\pi^4 - 1827228\pi^2 + 537160)} \approx 0.07248725712$$

The fundamental frequency has amplitude $\frac{1080430592}{1303695\pi^6 - 1827228\pi^4 + 537160\pi^2} \approx 0.9997773320.$ The 3rd, 5th, and 9th harmonics have relative amplitude $\frac{7}{263777} \approx -91.52\text{ dB},$ and the 7th harmonic has relative amplitude $\frac{726083}{31033100273} \approx -92.6\text{ dB}$ compared to the fundamental. A $k\text{th}$ harmonic has relative amplitude $\frac{\left(1 - (-1)^k\right)\left|67145k^4 - 2740842k^2 + 19555425\right|}{33763456k^6}.$

This approximation has a slight corner at the half-cycle boundaries, because the polynomial has zero derivative not at $x = \pm 1$ but at $x \approx \pm 1.002039940.$ At $x = 1$ the value of the derivative is about $0.004905799828$. This results in slower asymptotic decay of the amplitudes of the harmonics at large $k,$ compared to the 5th order approximation that has a continuous derivative.

7th order

A 7th order approximation without continuous derivative can be found similarly. The approach requires testing 120 different solutions and was automated by the Python script at the end of this answer. The best solution is:

$$f(x) = a_0 x^1 + a_1 x^3 + a_2 x^5 + a_3 x^7\\ a_0 = 1 - a_1 - a_2 - a_3 \approx 1.5707953785726114835\\ a_1 = - \frac{5\times(4374085272375\pi^6 - 6856418226992\pi^4 + 2139059216768\pi^2)}{16\times(2124555703725\pi^6 - 3428209113496\pi^4 + 1336912010480\pi^2 - 155807094720)} \approx -0.64590724797262922190\\ a_2 = \frac{2624451163425\pi^6 - 3428209113496\pi^4}{16\times(2124555703725\pi^6 - 3428209113496\pi^4 + 1336912010480\pi^2 - 155807094720)} \approx 0.079473610232926783079\\ a_3 = -\frac{124973864925\pi^6}{16\times(2124555703725\pi^6 - 3428209113496\pi^4 + 1336912010480\pi^2 - 155807094720)} \approx -0.0043617408329090447344$$

The fundamental frequency has amplitude $\frac{16991801282396160}{2124555703725\pi^8 - 3428209113496\pi^6 + 1336912010480\pi^4 - 155807094720\pi^2} \approx 1.0000024810802368487.$ The largest relative amplitude of the harmonics above the fundamental is $\frac{50}{2400688077}\approx -133.627\text{ dB}.$ compared to the fundamental. A $k\text{th}$ harmonic has relative amplitude $\frac{\left(1 - (-1)^k\right)\left|-162299057k^6 + 16711400131k^4 - 428526139187*k^2 + 2624451163425\right|}{4424948250624k^8}.$

Python source

from sympy import symbols, pi, solve, factor, binomial

numEq = 3 # Number of equations
numHarmonics = 6 # Number of harmonics to evaluate

a1, a2, a3, k = symbols("a1, a2, a3, k")
coefficients = [a1, a2, a3]
harmonicRelativeAmplitude = (2*pi**4*a1*k**4*(pi**2*k**2-12)+4*pi**2*a2*k**2*(pi**4*k**4-60*pi**2*k**2+480)+6*a3*(pi**6*k**6-140*pi**4*k**4+6720*pi**2*k**2-53760)+pi**6*k**6)*(1-(-1)**k)/(2*k**8*(2*pi**4*a1*(pi**2-12)+4*pi**2*a2*(pi**4-60*pi**2+480)+6*a3*(pi**6-140*pi**4+6720*pi**2-53760)+pi**6))

harmonicRelativeAmplitudes = []
for i in range(0, numHarmonics) :
    harmonicRelativeAmplitudes.append(harmonicRelativeAmplitude.subs(k, 3 + 2*i))

numCandidateEqs = 2**numHarmonics
numSignCombinations = 2**numEq
useHarmonics = range(numEq + 1)

bestSolution = []
bestRelativeAmplitude = 1
bestUnevaluatedRelativeAmplitude = 1
numSolutions = binomial(numHarmonics, numEq + 1)*2**numEq
solutionIndex = 0

for i in range(0, numCandidateEqs) :
    temp = i
    candidateNumHarmonics = 0
    j = 0
    while (temp) :
        if (temp & 1) :
            if candidateNumHarmonics < numEq + 1 :
                useHarmonics[candidateNumHarmonics] = j
            candidateNumHarmonics += 1
        temp >>= 1
        j += 1
    if (candidateNumHarmonics == numEq + 1) :
        for j in range(0,  numSignCombinations) :
            eqs = []
            temp = j
            for n in range(0, numEq) :
                if temp & 1 :
                    eqs.append(harmonicRelativeAmplitudes[useHarmonics[0]] - harmonicRelativeAmplitudes[useHarmonics[1+n]])
                else :
                    eqs.append(harmonicRelativeAmplitudes[useHarmonics[0]] + harmonicRelativeAmplitudes[useHarmonics[1+n]])
                temp >>= 1
            solution = solve(eqs, coefficients, manual=True)
            solutionIndex += 1
            print "Candidate solution %d of %d" % (solutionIndex, numSolutions)
            print solution
            solutionRelativeAmplitude = harmonicRelativeAmplitude
            for n in range(0, numEq) :                
                solutionRelativeAmplitude = solutionRelativeAmplitude.subs(coefficients[n], solution[0][n])
            solutionRelativeAmplitude = factor(solutionRelativeAmplitude)
            print solutionRelativeAmplitude
            solutionWorstRelativeAmplitude = 0
            for n in range(0, numHarmonics) :
                solutionEvaluatedRelativeAmplitude = abs(factor(solutionRelativeAmplitude.subs(k, 3 + 2*n)))
                if (solutionEvaluatedRelativeAmplitude > solutionWorstRelativeAmplitude) :
                    solutionWorstRelativeAmplitude = solutionEvaluatedRelativeAmplitude
            print solutionWorstRelativeAmplitude
            if (solutionWorstRelativeAmplitude < bestRelativeAmplitude) :
                bestRelativeAmplitude = solutionWorstRelativeAmplitude
                bestUnevaluatedRelativeAmplitude = solutionRelativeAmplitude                
                bestSolution = solution
                print "That is a new best solution!"
            print

print "Best Solution is:"
print bestSolution
print bestUnevaluatedRelativeAmplitude
print bestRelativeAmplitude

Are you asking this for theoretical reasons or a practical application?

Usually, when you have an expensive to compute function over a finite range the best answer is a set of lookup tables.

One approach is to use best fit parabolas:

n = floor( x * N + .5 );

d = x * N - n;

i = n + N/2;

y = L_0 + L_1[i] * d + L_2[i] * d * d;

By finding the parabola at each point that meets the values for d being -1/2, 0, and 1/2, rather than using the derivatives at 0, you ensure a continuous approximation. You could also shift the x value, rather than the array index to deal with your negative x values.

Ced

=================================================

Followup:

The amount of effort, and the results, that have gone into finding good approximations is very impressive. I was curious as to how my boring and bland piecewise parabolic solution would compare. Not surprisingly, it does much better. Here are the results:

   Method    Minimum    Maximum     Mean       RMS
  --------   --------   --------   --------   --------
     Power   -8.48842    1.99861   -4.19436    5.27002
    OP S_3   -2.14675    0.00000   -1.20299    1.40854
     Bhask   -1.34370    1.63176   -0.14367    0.97353
     Ratio   -0.24337    0.22770   -0.00085    0.16244
     rbj 5   -0.06724    0.15519   -0.00672    0.04195
    Olli5C   -0.16367    0.20212    0.01003    0.12668
     Olli5   -0.26698    0.00000   -0.15177    0.16402
    Olli7C   -0.00213    0.00000   -0.00129    0.00143
     Olli7   -0.00005    0.00328    0.00149    0.00181
    Para16   -0.00921    0.00916   -0.00017    0.00467
    Para32   -0.00104    0.00104   -0.00001    0.00053
    Para64   -0.00012    0.00012   -0.00000    0.00006

The values represent 1000x the error between the approximation and the actual evaluated every .0001 from a scale of 0 to 1 (inclusive), so 10001 points in all. The scale is converted to evaluate the functions from 0 to $\pi/2$, except for Olli Niemitalo's equations which use the 0 to 1 scale. The columns values should be clear from the headers. The results don't change with a .001 spacing.

The "Power" line is the equation: $x-\frac{x^e}{6}$.

The rbj 5 line is the same as Matt L's c5 solution.

The 16, 32, and 64 are the number of intervals that have parabolic fits. Of course there are insignificant discontinuities in the first derivative at each interval boundary. The values of the function are continuous though. Increasing the number of intervals only increases the memory requirements (and initialization time), it does not increase the amount of calculation needed for the approximation, which is less than any of the other equations. I chose powers of two because a fixed point implementation could save a division by using an AND in such cases. Also, I didn't want the count to be commensurate with the test sampling.

I did run Olli Niemitalo's python program and got this as part of the printout: "Candidate solution 176 of 120" I thought that was odd, so I am mentioning it.

If anybody wants me to include any of the other equations, please let me know in the comments.

Here is the code for the piecewise parabolic approximations. The entire test program is too long to post.

#=============================================================================
def FillParab( argArray, argPieceCount ):

#  y = a d^2 + b d + c

#  ym = a .25 - b .5 + c
#  y  =                c
#  yp = a .25 + b .5 + c

#  c = y
#  b = yp - ym
#  a = ( yp + ym - 2y ) * 2

#---- Calculate Lookup Arrays

        theStep = pi * .5 / float( argPieceCount - 1 )
        theHalf = theStep * .5

        theL0 = zeros( argPieceCount )
        theL1 = zeros( argPieceCount )
        theL2 = zeros( argPieceCount )

        for k in range( 0, argPieceCount ):
         x  = float( k ) * theStep

         ym = sin( x - theHalf )
         y  = sin( x )
         yp = sin( x + theHalf )

         theL0[k] = y
         theL1[k] = yp - ym
         theL2[k] = ( yp + ym - 2.0 * y ) * 2

#---- Do the Fill

        theN = len( argArray )

        theFactor = pi * .5 / float( theN - 1 )

        for i in range( 0, theN ):
         x  = float( i ) * theFactor

         kx = x / theStep
         k  = int( kx + .5 )
         d  = kx - k

         argArray[i] = theL0[k] + ( theL1[k] + theL2[k] * d ) * d

#=============================================================================

=======================================

Appendum

I have included Guest's $S_3$ function from the original post as "OP S_3" and Guest's two parameter formula from the comments as "Ratio". Both are on the 0 to 1 scale. I don't think the Ratio one is suitable for either calculation at runtime or for building a lookup table. After all, it is significantly more computation for the CPU than just a plain sin() call. It is interesting mathematically though.