I want to approximate the sine wave given by $\sin\left(\pi x\right)$ by applying a polynomial waveshaper to a simple triangle wave, generated by the function

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

where $\operatorname{mod}(x, 1)$ is the fractional part of $x$:

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

A Taylor series could be used as a waveshaper.

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

Given the above functions, $S_1(T(x))$ will get us a decent approximation of a sine wave. But we need to go up to the 7th power of the series to get a reasonable result, and the peaks are a little low and won't have a slope of exactly zero.

Instead of the Taylor series, we could use a polynomial waveshaper following a few rules.

  • Must pass through -1,-1 and +1,+1.
  • Slope at -1,-1 and +1,+1 must be zero.
  • Must be symmetrical.

A simple function meeting our requirements:


The graphs of $S_2(T(x))$ and $\sin\left(\pi x\right)$ are pretty close, but not as close as the Taylor series. In between the peaks and zero-crossings they visibly deviate a bit. A heavier and more accurate function meeting our requirements:


This is probably close enough for my purposes, but I'm left wondering if another function exists which approximates the sine wave more closely, and is computationally cheaper. I have a pretty good grasp on how to find functions meeting the three requirements above, but I'm not sure how go about finding functions which meet those requirements and also most closely match a sine wave.

What methods exist for finding polynomials which mimic a sine wave (when applied to a triangle wave)?

To clarify, I'm not necessarily looking only for odd-symmetrical polynomials, although those are the most straightforward choice.

Something like the following function could also fit my needs:


This meets the requirements in the negative range, and a piecewise solution could be used to apply it to the positive range as well; for example


where $P$ is the signed power function.

I'd also be interested in solutions using the signed power function to support fractional exponents, as this gives us another "knob to twist" without adding another coefficient.

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

Given the right constants, this could potentially get very good accuracy without the heaviness of fifth- or seventh-order polynomials. Here's an example meeting the requirements described here using some hand-picked constants: $a_0=1.\overline{666}, a_1=-0.\overline{666}, p_1=2.5$.

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

In fact, those constants are very close to $\frac \pi 2$, and $1 - \frac \pi 2$, and $e$. Plugging those in gives something that looks extremely close to a sine wave.

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

To put it another way, $x-\frac{x^e}{6}$ looks very close to $\sin(x)$ between 0,0 and π/2,1. Any thoughts on the significance of this? Maybe a tool like Octave can help discover the "best" constants for this approach.

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    $\begingroup$ so, what's your error term definition for "more close"? As far as I could tell, the Taylor series you quoted is the minimum L² error approximate for finite number of coefficients. (I think.) $\endgroup$ Commented Jan 23, 2018 at 11:30
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    $\begingroup$ What, by the way, is your goal? It might really help to tell us why you're looking for a polynomial wave shaper, on what technological basis, and what your main objectives are for the approximation. $\endgroup$ Commented Jan 23, 2018 at 11:32
  • $\begingroup$ @MarcusMüller I'm willing to sacrifice the accuracy of Taylor series for something significantly cheaper, if it's indistinguishable from a sine wave to the human ear. The peaks of the Taylor series approximation also bother me. I'm interested in finding something "more close" than the other two functions I listed. I suspect it won't get any cheaper than $S_2$. $\endgroup$
    – Guest
    Commented Jan 23, 2018 at 11:33
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    $\begingroup$ "To the human ear" is critical here :) Why do the peaks "bother" you? Again: give us an idea of why / for what purpose and under which restrictions you're doing this. Without enough background your question is simply too broad to be properly answered! $\endgroup$ Commented Jan 23, 2018 at 11:34
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    $\begingroup$ Why are you starting with a triangle wave? Sine-generators are simple and common, square waves are trivially filtered to the fundamental harmonic, etc. $\endgroup$ Commented Jan 23, 2018 at 19:20

5 Answers 5


about a decade ago i did this for an unnamed music synthesizer company who had R&D not too far from my condo in Waltham MA. (can't imagine who they are.) i don't have the coefficients.

but try this:

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

this guarantees that $f(-x) = -f(x)$.

To guarantee that $f'(x) \Big|_{x=\pm1} =0$ then

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

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

That's the first constraint. To guarantee that $|f(\pm 1)| = 1$, then

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

That's the second constraint. Eliminating $a_0$ and solving Eqs. (1) and (2) for $a_2$ in terms of $a_1$ (which is left to adjust):

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

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

Now you have only one coefficient, $a_1$, left to twiddle for best performance:

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

This is the way I would twiddle $a_1$ for best performance for a sine wave oscillator. I would adjust use the above and the symmetry of the sine wave about $x=1$ and place exactly one entire cycle in a buffer with a power of two number of points (say 128, i don't care) and run the FFT on that perfect cycle.

The FFT result bin 1 will be the strength of the sine and should be about $N/2$. Now you can adjust $a_1$ to bring your 3rd harmonic distortion up and down. I would start with $a_1 \approx \tfrac{5}{\pi} - 2$ so that $a_0 \approx 1$. That's in bin 3 of the FFT results But the 5th harmonic distortion (value in bin 5) will be consequential (it will go up as the 3rd harmonic goes down). I would adjust $a_1$ so that the strength of the 5th harmonic level is equal to the 3rd harmonic level. It will be around -70 dB from the 1st harmonic (as I recall). That will be the nicest-sounding sine wave from a cheap, 3-coefficient, 5th-order, odd-symmetrical polynomial.

Someone else can write the MATLAB code. How does that sound to you?

  • $\begingroup$ i will definitely not have time to do the MATLABing to hunt for the optimal $a_1$ so that the 3rd harmonic is equal to the 5th harmonic, about 70 dB below the fundamental (1st harmonic). someone else needs to do that. sorry. $\endgroup$ Commented Jan 23, 2018 at 16:07
  • $\begingroup$ Great answer, still digesting it. Actually starting to wonder if it needs to be a 3-coefficient, 5th-order, odd-symmetrical polynomial ... Could your f'(x) actually be f(x) and be a piecewise deal around 0? Rough sketch here. Maybe this is what Ced has in mind? Still catching up to you guys. $\endgroup$
    – Guest
    Commented Jan 23, 2018 at 16:39
  • $\begingroup$ This is a beautiful approach. I wonder if instead of taking the FFT and solving iteratively you could form the third- and fifth-order Chebyshev polynomials from your $f(x)$, then equate the two and solve for $a_1$? $\endgroup$
    – Speedy
    Commented Jan 23, 2018 at 16:44
  • $\begingroup$ Must have been half asleep when I posted that "sketch," I meant to do something like this, but corrected to run through ±1 and have zero slope (can just take the derivative, fiddle around with it, integrate it again). Not sure if there's any advantage over fifth-order, just something I hadn't considered yet. $\endgroup$
    – Guest
    Commented Jan 24, 2018 at 1:12
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    $\begingroup$ This really is a brilliant solution, just took a while to sink in. I hope marking it correct won't stop someone else from coming along and writing the code. $\endgroup$
    – Guest
    Commented Jan 25, 2018 at 4:11

What is usually done is an approximation minimizing some norm of the error, often the $L_{\infty}$-norm (where the maximum error is minimized), or the $L_2$-norm (where the mean squared error is minimized). $L_{\infty}$-approximation is done by using the Remez exchange algorithm. I'm sure you can find some open source code implementing that algorithm. However, in this case I think a very simple (discrete) $l_2$-optimization is sufficient. Let's look at some Matlab/Octave code and the results:

x = linspace(0,pi/2,300);    % grid on [0,pi/2]
x = x(:);
% overdetermined system of linear equations
% (using odd powers only)
A3 = [x,x.^3];
A5 = [x,x.^3,x.^5];
b = sin(x);
% solve in l2 sense
c3 = A3\b;
c5 = A5\b;
f3 = A3*c3;    % 3rd order approximation
f5 = A5*c5;    % 5th order approximation

The figure below shows the approximation errors for the $3^{rd}$-order and for the $5^{th}$-order approximations. The maximum approximation errors are 8.8869e-03 and 1.5519e-04, respectively.

enter image description here

The optimum coefficients are

c3 =


c5 =

So the third-order approximation is

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

and the fifth-order approximation is

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


I had a look into approximations with the signed power function, as suggested in the question, but the best approximation is hardly better than the third-order approximation shown above. The approximating function is

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

where the constants were chosen such that $f'(0)=1$ and $f'(\pi/2)=0$. The power $p$ was optimized to achieve the smallest maximum error in the range $[0,\pi/2]$. The optimal value for $p$ was found to be $p=2.774$. The figure below shows the approximation errors for the third-order approximation $(1)$ and for the new approximation $(3)$:

enter image description here

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.


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:

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

  • $\begingroup$ That's very helpful, thanks. This is the first time I've used Octave. I followed most of it, but how did you get the approximation error plots and maximum values? $\endgroup$
    – Guest
    Commented Jan 23, 2018 at 16:27
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    $\begingroup$ @Guest: The errors are just b-f3 and b-f5, respectively. Use the plot command to plot them. $\endgroup$
    – Matt L.
    Commented Jan 23, 2018 at 16:30
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    $\begingroup$ @Guest: And the maxima you get from max(abs(b-f3)) and max(abs(b-f5)). $\endgroup$
    – Matt L.
    Commented Jan 23, 2018 at 16:31
  • $\begingroup$ @Guest: I played around with the signed power function, but the result is not significantly better than the third-order approximation I had before. Check out my edited answer. As for complexity, would it make such a big difference? $\endgroup$
    – Matt L.
    Commented Jan 24, 2018 at 11:07
  • $\begingroup$ Thanks for looking into it. Complexity isn't a huge deal, just curious how accurate the approximation can get with relatively low complexity. I'm not quite sure how you came up with (3), but it works nicely. I'd need to use 2.752 instead for p, since anything above that will send the peaks over 1 (clipping). $\endgroup$
    – Guest
    Commented Jan 24, 2018 at 11:50

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):

Comparison of approximations

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 "Best Solution is:"
print bestSolution
print bestUnevaluatedRelativeAmplitude
print bestRelativeAmplitude
  • 1
    $\begingroup$ wow, solving it analytically. i woulda just used MATLAB and an FFT and sorta hunt around for the answer. $$ $$ you did very well. $\endgroup$ Commented Jan 29, 2018 at 4:38
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    $\begingroup$ YO! DSP AUDIO SYNTHESIS FOLKS!, unless the OP can uncheck my answer (i think it's too late) and check this one, EVERYONE should upvote this. $$ $$ this answer has very valuable information. $\endgroup$ Commented Jan 29, 2018 at 17:41
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    $\begingroup$ i did the Horner thing myself. thanks for doing this analytically and for taking it to the 7th-order. with less than -123 dB of harmonic distortion, i think this 7th-order odd-symmetry polynomial can be the last word on waveshaping a triangle into a beautifully clean sinusoid. $\endgroup$ Commented Jan 29, 2018 at 18:12
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    $\begingroup$ actually @OlliNiemitalo, i think -79 dB is good enough for the implementation of a digital synth sine wave oscillator. it can be driven by a triangle wave, which is generated easily from the abs value of a sawtooth, which is most easily generated with a fixed-point phase accumulator. $$ $$ no one will hear a difference between that 5th-order polynomial sine wave and a pure sine. $\endgroup$ Commented Jan 30, 2018 at 3:51
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    $\begingroup$ Polynomials in general as $f$ have the advantage that by increasing the order, the error can be made arbitrarily small. Rational functions have the same advantage, but a division is typically more costly to compute than multiplication. For example in Intel i7, a single thread can do 7-27 times as many multiplications and additions than divisions in the same time. Approximating some alternative $f$ means decomposing it to elementary ops, typically multiplications and additions which always amount to polynomials. Those could be optimized to approximate sine directly versus via $f$. $\endgroup$ Commented Feb 3, 2018 at 16:12

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.




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




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.

  • $\begingroup$ Good work! I fixed that bug ("176 of 120"). $\endgroup$ Commented Feb 3, 2018 at 20:12
  • $\begingroup$ Nice update, this makes more sense to me now. The $x-\frac{x^e}{6}$ probably doesn't need to be tested, I just threw it out there because I was trying to figure out the significance of $e$ which seemed to keep popping up while I was playing with this. A better rational expression to test might be something like this: $f_0\left(x\right)=\left|x\right|^a\operatorname{sign}\left(x\right)$ ; $b=f_0'\left(1\right)$ ; $f_1\left(x\right)=f_0\left(x\right)-bx$ ; $c=\frac{1}{f_1\left(1\right)}$ ; $f_2\left(x\right)=f_1\left(x\right)c$ ... now $a$ should be set to about $2\frac{2}{3}$... $\endgroup$
    – Guest
    Commented Feb 4, 2018 at 8:38
  • $\begingroup$ ...or $f_0(x)$ can be pretty much any other odd-symmetrical function; sigmoids seem to work well, like $\frac{a^x-1}{a^x+1}$ (but then the right value for $a$ needs to be found, of course). Here's a plot... as Olli mentions, this probably isn't practical for on-the-fly computation, but I guess it could be useful for building a lookup table. $\endgroup$
    – Guest
    Commented Feb 4, 2018 at 8:42
  • $\begingroup$ Or a more accurate 2-param version of that, $\frac{a_0^x-a_1^x}{a_0^x+a_1^x}$ looks pretty good with $a_0\approx\frac{1}{3}$ and $a_1\approx\frac{10}{9}$ $\endgroup$
    – Guest
    Commented Feb 4, 2018 at 10:23
// Recursive approximation,

// draw point (0, 0)
for (float x=0.0; x<10; x+=0.01) {
   int y = ((int)x) % 2;
   // draw line to (x, y) from previous point

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