Trigonometric functions in assembly?

[noodlesoup]

Okay, so maybe asking for it already in assembly is asking too much, but what I'm looking for is an equation(s) that approximates sin(x) and/or arcsin(x), preferably with controllable accuracy. Thanks.

[Fringe Pioneer]

I presume you can't just use existing assembly instructions to calculate the Taylor polynomial of sin(x) to some determined degree, centered at some arbitrary value a (i.e. using x=a as a reference point for calculation)?

f(x) = f(a)*(x + a)⁰/0! + f'(a)*(x + a)¹/1! + f''(a)*(x + a)²/2! + f'''(a)*(x + a)³/3! + f''''(a)*(x + a)⁴/4! + ...

Sine of x centered at some arbitrary a:
sin(x) = sin(a) + cos(a)*(x + a) - sin(a)*(x² + 2xa + a²)/2 - cos(a)*(x³ + 3x²a + 3xa² + a³)/6 + sin(a)*(x⁴ + 4x³a + 6x²a² + 4xa³ + a⁴)/24 + ...

Sine of x centered at 0:
sin(x) = 0 + x - 0 - x³/6 + 0 + x⁵/120 - 0 - x⁷/5040 + ...
sin(x) = sum((-1)ⁿx^{2n + 1}/(2n + 1)!, 0, infinity)

[noodlesoup]

Ah hah! So each prime mark on function f means the derivative, which explains why it goes sine-cosine-negative sine-negative cosine... Thanks for showing me how2Taylor, Veers.

so cos(x):

cos(x) = cos(0)*(x)^0/0! - sin(0)*(x)^1/1! - cos(0)*(x)^2/2! + sin(0)*(x)^3/3! + cos(0)*(x)^4/4! - sin(0)*(x)^5/5! - cos(0)*(x)^6/6! + sin(0)*(x)^7/7!...

cos(x) = 1 - 0 - x^2/2! + 0 +x^4/4! - 0 - x^6/6! + 0

cos(x) = 1 - x²/2! + x⁴/4! - x⁶/6!...

cos(x) = 1 - x²/2 + x⁴/24 - x⁶/720 ...



I would say accuracy is fine from -pi/2 <= x <= pi/2

---

tan(x):

tan(x) = [tan(0)*(x)^0/0!] + [sec^2(0)*(x)^1/1!] + [2tan(0)sec^2(0)*(x)^2/2!] - [2(cos(0)-2)sec^4(0)*(x)^3/3!] - [2(sin(0) - 11sin(0))sec^5(0)*(x)^4/4!] + [2(-26cos(0) + cos(0) +33)sec^6(0)*(x)^5/5!] + [2(302sin(0) - 57sin(0) + sin(0))sec^7(0)*(x)^6/6!] - [2(1191cos(0)-120cos(0)+cos(0)-1208)sec^8(0)*(x)^7/7!] - [2(-15619sin(0)+4293sin(0)-247sin(0)+sin(0))sec^9(0)*(x)^8/8!] + [2(-88234cos(0)+14608cos(0)-502cos(0)+cos(0)+78095)sec^10(0)*(x)^9/9!]...

tan(x) = 0 + x + 0 - 2x^3/3! - 0 + 16x^5/5! + 0 -274x^7/7! - 0 + 7936x^9/9!...

tan(x) = x - 2x³/3! + 16x⁵/5! - 274x⁷/7! + 7936x⁹/9!



Accuracy sucks past x=0.2, maybe using double-angle formula will help? Or dividing sine by cosine? They are relatively accurate... I will tackle the inverse functions next.

[Qwerty]

This math is making my brain hurt, but I think I understand it. Should be useful if we have to calculate gravitational trajectories.

[noodlesoup]

Or navigation, and possibly projectile/laser weapons in DCPU-16

Also Wolfram Alpha did a better job with approximating tan(x):

tan(x) = x + x³/3 + x⁵/5...

tan(x) = x + x³/3 + x⁵/5:



Using the function only for the range -pi/4 <= x <=pi/4 and then using the double-angle formula would probably be better than using the function for all values of x.

for x=1.2:

using the graph only:
tan(1.2) - [1.2 + 1.2³/3 + 1.2⁵/5] = 0.2984876221
percent error = .2984876221/tan(1.2) = .11604 = 11.604%

using the graph to find f(.6) then using the double-angle formula:
f(.6) = [0.6 + 0.6³/3 + 0.6⁵/5] = .687552
double-angle formula: tan(2x) = 2tan(x) / (1-tan²(x)): f(1.2) = 2(.687552) / (1-.687552²) = 2.607958236
tan(1.2) - f(1.2) = -0.03580661397
percent error = |(-0.03580661397)|/tan(1.2) = .01392 = 1.392%

---

Inverse Trigonometric Functions:

Arcsin(x) = x + 1²x³/3! + 3²x⁵/5! + 15²x⁷/7! + 105²x⁹/9! + 945²x¹¹/11!...



Arctan(x) = x - x³/3 + x⁵/5 - x⁷/7 + x⁹/9...

[~Memzak~]

Assembly I see... DCPU I see... Missiles I see...

Someone is preparing for 0x10c... >.> I understood some of the math behind your last couple posts, but didn't put much effort into reading/thinking about all of it. As for doing it in assembly... Yea... good luck with that.

[Fringe Pioneer]

Let me warn you about one thing, although you probably already know this: don't repeatedly alternate between taking the function of and the inverse function of some value, lest the error quickly build up and not return anything close to your starting value.

sin(sin^-1(sin(sin^-1(sin(sin^-1(sin(sin^-1(x))))))) should equal x if you have the exact functions for sin(x) and sin^-1(x) for any number of repetitions, but will diverge from the value of x if your error is sufficiently large enough...

[Hachi1]

wow I have no idea what you are all talking about... the most complex trig I know is simple identities

[Fringe Pioneer]

Well, the subject here, as you should have been able to get from M4's post nearly two months ago, was finding ways to approximate trigonometric functions so that he can find the sine or cosine of something in DCPU Assembly, an in-game programming language for a game by Mojang AB. I knew that non-polynomial functions could be approximated with Taylor polynomials, so I told him what a Taylor polynomial was and how to take the Taylor polynomial of any function. Everything past that point was either M4 trying to approximate many other trigonometric functions or others indicating that this is above their heads...

[Qwerty]

Veers, I don't think that'll help much...

M4 wanted to simulate a certain math function on a game that didn't support that function. Veers showed him a way to approximate said function using other functions.

[ganondorfchampin]

Jun 21, 2012 9:27:46 GMT Hachi1 said:

wow I have no idea what you are all talking about... the most complex trig I know is simple identities

The techniques being used here are calculus, not trig.

[Qwerty]

It's using calculus to simulate trigonometry.

[Hachi1]

Well that's perfect. It proves I have no idea what you're all talking about. Then I should say, 'the most complex calculus I know is basic integration to find areas under a curve'

[Fringe Pioneer]

Well, if you know of integrals, then surely you know of derivatives?

Generally, if you know one point on a curve and you know the derivative at that point on the curve, you can approximate where a nearby point will be if you don't know anything else about the curve.

As you know, the derivative at a point is approximately equal to the rise over the run of the tangent line from that point to a point some very small difference away.
δy/δx ~ y'(x)

Multiply both sides by the difference in x.
δy ~ y'(x)(δx)

Take the change in x to be the difference between some known x-value a and some x value whose y-value we wish to find. Substitute y(x) - y(a) for δy and substitute x - a for δx.
y(x) - y(a) ~ (x - a)y'(x)

Add both sides by y(x - a).
y(x) = y(a) + (x - a)y'(x)

Instead of merely tangent lines, you get closer approximations near your starting point if instead you use tangent parabolas, tangent cubics, tangent quartics, and so on. A pattern is revealed: this pattern is a Taylor polynomial.
y(x) = y(a) + (x - a)y'(x) + (x - a)²y''(x)/2 + (x - a)³y'''(x)/6 + ...

We have several knowns: x is a value we know, a is a value we know, y(a) is a value we know, and the n^th derivative of y(a) is a value we know, from n = 0 onward to infinity. With these values, we merely plug them into the Taylor polynomial, and we get the approximate the value y(x), which we don't know. For example, I want to find a general approximation for x. I know that sin(0) = 0, and I know all the derivatives of the sine function at a = 0. Let's plug the values in...
sin(x) = sin(0) + (x - 0)(1) + (x - 0)²(0)/2 + (x - 0)³(-1)/6 + (x - 0)⁴(0)/24 + (x - 0)⁵(1)/120 + (x - 0)⁶(0)/720 + (x - 0)⁷(-1)/5040 + ...
sin(x) = 0 + x + 0 - x³/6 + 0 + x⁵/120 + 0 - x⁷/5040 + ...
sin(x) ~ x - x³/6 + x⁵/120 - x⁷/5040

If I want to find sin(0.5), which should be close to sin(π/6) = 0.5, I just have to plug in 0.5 for x.
sin(0.5) ~ 0.5 - 0.5³/6 + 0.5⁵/120 - 0.5⁷/5040
sin(0.5) ~ 0.5 - 0.0208333 + 0.0002604 - 0.0000016
sin(0.5) ~ 0.4794255

We were able to approximate that sin(0.25) was very close to 0.4794. The approximation taken directly from the calculator (in radian mode, because derivatives with trigonometric functions assume radians) is 0.4794.

This all I did: help M4 approximate a function by using a polynomial that has derivatives in it...