Equation not fully solved for - maxima
I've been trying to solve an equation for a 2D vector P.
But after solve there are still some P on the rhs.
Does this mean Maxima can't do it or I've done something wrong?
Here is it:
load("vect");
declare(".", commutative);
declare(P, nonscalar);
declare([v1,V1,r1], nonscalar);
declare([v2,V2,r2], nonscalar);
declare([w1,W1,m1,I1,w2,W2,m2,I2], scalar);
/* Revolute Constraint */
constraint: v2 + (w2~r2) - (v1 + (w1~r1)) = 0$
/* Velocities after impulse P */
eq1: v1 = V1 - P/m1$
eq2: w1 = W1 - (r1~P) / I1$
eq3: v2 = V2 + P/m2$
eq4: w2 = W2 + (r2~P) / I2$
eq: subst([eq1,eq2,eq3,eq4], constraint)$
solve(eq, P);
(I'm trying to get an equation for an impulse that satisfies the constraint.
I'm following Dirk Gregorius' 2nd post here: https://gamedev.net/forums/topic/469531-revolute-joint-usingimpulses/4086845)
I think I've worked out the details. I had to do some stuff by hand, and Maxima was mostly just checking that I did it correctly. If the goal is just to get to a solution, I guess that's okay.
Here's my script. You can execute this via: maxima --batch=foo.mac where foo.mac is the name of the script.
/* adapted from: https://stackoverflow.com/questions/69700162/equation-not-fully-solved
*/
load("vect");
/* I don't want to reorder arguments of cross product. */
remrule ("~", ?\~rule4);
/* I do want to flatten noncommutative multiplication.
* (It appears that's disabled by vect; ordinarily it happens by default.)
*/
dotassoc: true $
declare(P, nonscalar);
declare([v1,w1,V1,W1,r1], nonscalar);
declare([v2,w2,V2,W2,r2], nonscalar);
declare([m1,I1,m2,I2], scalar);
/* Revolute Constraint */
constraint: v2 - (r2~w2) - (v1 - (r1~w1)) = 0$
/* Velocities after impulse P */
eq1: v1 = V1 - P/m1$
eq2: w1 = W1 - (r1~P) / I1$
eq3: v2 = V2 + P/m2$
eq4: w2 = W2 + (r2~P) / I2$
eq: subst([eq1,eq2,eq3,eq4], constraint);
A(a) := matrix([0, -a[3], a[2]], [a[3], 0, -a[1]], [-a[2], a[1], 0]);
eqa: ev (eq, (u~v) := 'A(u).v);
matchdeclare (pp, lambda ([e], not freeof(P, e)));
matchdeclare (qq, lambda ([e], freeof(P, e)));
defrule (rp, pp + qq = 0, pp = -qq);
eqa1: expand (eqa);
eqa2: apply1 (eqa1, rp);
matchdeclare (aa, lambda ([e], matrixp(e) or listp(e)));
tellsimpafter (I(aa), ident (length (aa)));
matchdeclare ([aa, bb], all);
tellsimpafter (I(aa) . bb, bb);
tellsimpafter (aa . I(bb), aa);
M: -(1/I2)*'A(r2).'A(r2) - (1/I1)*'A(r1).'A(r1) + (1/m2)*I(P) + (1/m1)*I(P);
N: 'A(r2).W2 - 'A(r1).W1 - V2 + V1;
eqa2_factored: M . P = N;
expand (eqa2_factored);
?resimplify (%);
if % # eqa2 then error ("tried to factor eqa2, but something went wrong.");
solution: P = M^^-1 . N;
/* EXAMPLE: */
I1: 20 $
I2: 3 $
m1: 100 $
m2: 12 $
V1: [17, 19, -23] $
V2: [-5, -3, 11] $
W1: [8, 4, 14] $
W2: [-6, -16, 24 ] $
r1: [1/2, 2/3, 3/4] $
r2: [5, 7, 3] $
/* note various subterfuges to ensure evaluation with stated values */
example_M: ev (subst (I(P) = ident(3), M), nouns);
example_N: ev (N, nouns);
example_P: example_M^^-1 . example_N;
subst (P = example_P, ev (eqa2, eval, nouns));
if lhs(%) = rhs(%)
then print ("TEST PASSED: lhs = rhs")
else error ("TEST FAILED: lhs # rhs");
If you need to evaluate P for different parameters r1, r2, etc., my advice is to evaluate matrix M and vector N with whatever values you want to plug in, and then solve the equation P = M^^-1 . N. An explicit solution is probably going to be pretty messy.
Following Robert Dodier's advice, I broke up all the vectors and solved for P[1] and P[2] individually.
I've got something that gives me an answer but now how can I get it into nice vector form?
Here it is:
load("vect");
declare(".", commutative);
declare(P, nonscalar);
declare([v1,V1,r1], nonscalar);
declare([v2,V2,r2], nonscalar);
declare([w1,W1,m1,I1], scalar);
declare([w2,W2,m2,I2], scalar);
cross_scalar_vector(s,v) := [-s*v[2], s*v[1]]$
/* Revolute Constraint on Linear Velocity */
constraint: v2 + cross_scalar_vector(w2,r2) - (v1 + cross_scalar_vector(w1,r1)) = [0,0]$
/* Sub in velocities after impulse P. */
post_velocities: [
v1 = V1 - P/m1,
w1 = W1 - (r1~P) / I1,
v2 = V2 + P/m2,
w2 = W2 + (r2~P) / I2
]$
constraint: subst(post_velocities, constraint)$
/* Break up the remaining vectors for solve. */
vectors: [
P = [P[1], P[2]],
V1 = [V1[1], V1[2]],
r1 = [r1[1], r1[2]],
V2 = [V2[1], V2[2]],
r2 = [r2[1], r2[2]]
]$
constraint: subst(vectors, constraint)$
/* Break up vector constraint into x and y constraint for solve. */
xconstraint: lhs(constraint)[1] = 0$
yconstraint: lhs(constraint)[2] = 0$
/* Not sure why we need to do this again? */
xconstraint: subst(vectors, xconstraint)$
yconstraint: subst(vectors, yconstraint)$
/* Expand cross products for solve. */
xconstraint: express(xconstraint)$
yconstraint: express(yconstraint)$
solve([xconstraint,yconstraint], [P[1],P[2]]);
Related
How to substitute expressions containing units when using the ezunits package?
Without specifying units, I can express area and volume and have Maxima show the relationship: (%i1) areaNoUnits: area = width * length$ (%i2) volumeNoUnits: volume = area * height$ (%i3) volumeNoUnits, areaNoUnits; (%o3) volume = height length width (%i4) subst(areaNoUnits, volumeNoUnits); (%o4) volume = height length width Now I want to specify units so I will use the ezunits package. The ` (backtick) operator is the building block of ezunits: An expression a ` b represents a dimensional quantity, with a indicating a nondimensional quantity and b indicating the dimensional units. When I add units to the area and volume expressions, evaluation and substitution do not work: (%i1) load ("ezunits")$ (%i2) areaWithUnits: area ` m^2 = (width ` m) * (length ` m); 2 2 (%o2) area ` m = length width ` m (%i3) volumeWithUnits: volume ` m^3 = (area ` m^2) * (height ` m); 3 3 (%o3) volume ` m = area height ` m (%i4) volumeWithUnits, areaWithUnits; 3 3 (%o4) volume ` m = area height ` m (%i5) subst(areaWithUnits, volumeWithUnits); 3 3 (%o5) volume ` m = area height ` m The expected output is: volumeWithUnits, areaWithUnits; 3 3 volume ` m = height length width ` m I do not see a function in the ezunits package to do evaluation or substitution. What is the right way to do this?
I would phrase it like this: (%i2) load (ezunits) $ (%i3) width: W ` m; (%o3) W ` m (%i4) length: L ` m; (%o4) L ` m (%i5) area: width * length; 2 (%o5) L W ` m (%i6) height: H ` m; (%o6) H ` m (%i7) volume: area * height; 3 (%o7) H L W ` m I wrote each part as conceptualname: symbolforquantity ` unit and then wrote just conceptualname in further calculations, instead of conceptualname ` unit. The substitution you tried in %i5 didn't work because subst is a purely formal substitution -- if there isn't a literal subexpression which is the same as the substituted-for expression, it doesn't match; subst doesn't look for rearrangements or factorizations which could help make a match. There are ways to work around that, so it might be possible to make your original formulation work, but I think it's better overall to sidestep the problem and work with conceptualname and symbolforquantity ` unit. To say a little about what more one could do with expressions like %o7 above. There are at least two ways to replace symbols H, L, and W with specific values. One is to call subst: (%i2) load (ezunits) $ (%i3) volume: H*L*W ` m^3; 3 (%o3) H L W ` m (%i4) subst ([L = 20, W = %pi], volume); 3 (%o4) 20 %pi H ` m Another is to make use of ev. (%i5) ev (volume, L = 20, W = %pi); 3 (%o5) 20 %pi H ` m Note that at the input prompt, something, someflags, somevalues is equivalent to ev(something, someflags, somevalues). (%i6) volume, L = 20, W = %pi; 3 (%o6) 20 %pi H ` m This is just a convenience. Within a function, one has to say ev(...); the shorter syntax isn't understood there. ev is often convenient, but it's generally simpler to predict what the result is going to be with subst instead.
How does Roblox's math.noise() deal with negative inputs?
While messing around with noise outside of Roblox, I realized Perlin/Simplex Noise does not like negative inputs. Remembering Roblox has a noise function, I tried there, and found out negative numbers do work nicely for Roblox's math.noise(). Does anybody know how they made this work, or how to get negative numbers to work for Perlin/Simplex noise in general? The Simplex Noise I am using (copied from here but changed to have the bitwise and operation): local function bit_and(a, b) --bitwise and operation local p, c = 1, 0 while a > 0 and b > 0 do local ra, rb = a%2, b%2 if (ra + rb) > 1 then c = c + p end a = (a - ra) / 2 b = (b - rb) / 2 p = p * 2 end return c end -- 2D simplex noise local grad3 = { {1,1,0},{-1,1,0},{1,-1,0},{-1,-1,0}, {1,0,1},{-1,0,1},{1,0,-1},{-1,0,-1}, {0,1,1},{0,-1,1},{0,1,-1},{0,-1,-1} } local p = {151,160,137,91,90,15, 131,13,201,95,96,53,194,233,7,225,140,36,103,30,69,142,8,99,37,240,21,10,23, 190, 6,148,247,120,234,75,0,26,197,62,94,252,219,203,117,35,11,32,57,177,33, 88,237,149,56,87,174,20,125,136,171,168, 68,175,74,165,71,134,139,48,27,166, 77,146,158,231,83,111,229,122,60,211,133,230,220,105,92,41,55,46,245,40,244, 102,143,54, 65,25,63,161, 1,216,80,73,209,76,132,187,208, 89,18,169,200,196, 135,130,116,188,159,86,164,100,109,198,173,186, 3,64,52,217,226,250,124,123, 5,202,38,147,118,126,255,82,85,212,207,206,59,227,47,16,58,17,182,189,28,42, 223,183,170,213,119,248,152, 2,44,154,163, 70,221,153,101,155,167, 43,172,9, 129,22,39,253, 19,98,108,110,79,113,224,232,178,185, 112,104,218,246,97,228, 251,34,242,193,238,210,144,12,191,179,162,241, 81,51,145,235,249,14,239,107, 49,192,214, 31,181,199,106,157,184, 84,204,176,115,121,50,45,127, 4,150,254, 138,236,205,93,222,114,67,29,24,72,243,141,128,195,78,66,215,61,156,180} local perm = {} for i=0,511 do perm[i+1] = p[bit_and(i, 255) + 1] end local function dot(g, ...) local v = {...} local sum = 0 for i=1,#v do sum = sum + v[i] * g[i] end return sum end local noise = {} function noise.produce(xin, yin) local n0, n1, n2 -- Noise contributions from the three corners -- Skew the input space to determine which simplex cell we're in local F2 = 0.5*(math.sqrt(3.0)-1.0) local s = (xin+yin)*F2; -- Hairy factor for 2D local i = math.floor(xin+s) local j = math.floor(yin+s) local G2 = (3.0-math.sqrt(3.0))/6.0 local t = (i+j)*G2 local X0 = i-t -- Unskew the cell origin back to (x,y) space local Y0 = j-t local x0 = xin-X0 -- The x,y distances from the cell origin local y0 = yin-Y0 -- For the 2D case, the simplex shape is an equilateral triangle. -- Determine which simplex we are in. local i1, j1 -- Offsets for second (middle) corner of simplex in (i,j) coords if x0 > y0 then i1 = 1 j1 = 0 -- lower triangle, XY order: (0,0)->(1,0)->(1,1) else i1 = 0 j1 = 1 end-- upper triangle, YX order: (0,0)->(0,1)->(1,1) -- A step of (1,0) in (i,j) means a step of (1-c,-c) in (x,y), and -- a step of (0,1) in (i,j) means a step of (-c,1-c) in (x,y), where -- c = (3-sqrt(3))/6 local x1 = x0 - i1 + G2 -- Offsets for middle corner in (x,y) unskewed coords local y1 = y0 - j1 + G2 local x2 = x0 - 1 + 2 * G2 -- Offsets for last corner in (x,y) unskewed coords local y2 = y0 - 1 + 2 * G2 -- Work out the hashed gradient indices of the three simplex corners local ii = bit_and(i, 255) local jj = bit_and(j, 255) local gi0 = perm[ii + perm[jj+1]+1] % 12 local gi1 = perm[ii + i1 + perm[jj + j1+1]+1] % 12 local gi2 = perm[ii + 1 + perm[jj + 1+1]+1] % 12 -- Calculate the contribution from the three corners local t0 = 0.5 - x0 * x0 - y0 * y0 if t0 < 0 then n0 = 0.0 else t0 = t0 * t0 n0 = t0 * t0 * dot(grad3[gi0+1], x0, y0) -- (x,y) of grad3 used for 2D gradient end local t1 = 0.5 - x1 * x1 - y1 * y1 if t1 < 0 then n1 = 0.0 else t1 = t1 * t1 n1 = t1 * t1 * dot(grad3[gi1+1], x1, y1) end local t2 = 0.5 - x2 * x2 - y2 * y2 if t2 < 0 then n2 = 0.0 else t2 = t2 * t2 n2 = t2 * t2 * dot(grad3[gi2+1], x2, y2) end -- Add contributions from each corner to get the final noise value. -- The result is scaled to return values in the interval [-1,1]. return 70.0 * (n0 + n1 + n2) end return noise
The Lua programming language version that Roblox uses, LuaU (or Luau), is actually open-source since November of 2021. You can find it here. The math library can be found in this file called lmathlib.cpp and it contains the math.noise function along with internal functions to calculate it, perlin (main function), grad, lerp, and fade. It's a quite complicated thing I can't explain myself, but I have converted it into Lua here.
Maxima: Round like in Excel
Is there a function which rounds numbers (even decimal numbers) like round() in Excel? Example Round 1,45 to one decimal: 1,5 Round 2,45 to one decimal: 2,5 There is a similar question but they use a different algorithm.
OK, here's an attempt to reimplement Excel =ROUND function in Maxima. Some notes. (1) Values are rounded to 15 significant digits before applying the user's rounding. This is an attempt to work around problems caused by inexact representation of decimals as floating point numbers. (2) I've implemented excel_round and integer_log10 as so-called simplifying functions. That means that the calculation isn't carried out until the arguments are something that can be evaluated (in this case, when the arguments are numbers). (3) I didn't check to see what Excel =ROUND does with negative numbers -- does it round 5 upward (i.e., towards zero in this case), or away from zero? I dunno. I've posted this solution as the little package excel_round.mac on Github. See: https://github.com/maxima-project-on-github/maxima-packages and navigate to robert-dodier/excel_round. In the interest of completeness, I've pasted the code here as well. Here are a few examples. (%i1) excel_round (1.15, 1); (%o1) 1.2 (%i2) excel_round (1.25, 1); (%o2) 1.3 (%i3) excel_round (12.455, 2); (%o3) 12.46 (%i4) excel_round (x, 2); (%o4) excel_round(x, 2) (%i5) ev (%, x = 9.865); (%o5) 9.87 Here is the code. This is the content of excel_round.mac. /* excel_round -- round to specified number of decimal places, * rounding termminal 5 upwards, as in MS Excel, apparently. * Inspired by: https://stackoverflow.com/q/62533742/871096 * * copyright 2020 by Robert Dodier * I release this work under terms of the GNU General Public License. */ matchdeclare (xx, numberp); matchdeclare (nn, integerp); tellsimpafter (excel_round (xx, nn), excel_round_numerical (xx, nn)); matchdeclare (xx, lambda ([e], block ([v: ev (e, numer)], numberp(v)))); tellsimpafter (excel_round (xx, nn), excel_round_numerical (ev (xx, numer), nn)); excel_round_numerical (x, n) := block ([r, r1, r2, l], /* rationalize returns exact rational equivalent of float */ r: rationalize (x), /* First round to 15 significant decimal places. * This is a heuristic to recover what a user "meant" * to type in, since many decimal numbers are not * exactly representable as floats. */ l: integer_log10 (abs (r)), r1: round (r*10^(15 - l)), /* Now begin rounding to n places. */ r2: r1/10^((15 - l) - n), /* If terminal digit is 5, then r2 is integer + 1/2. * If that's the case, round upwards and rescale, * otherwise, terminal digit is something other than 5, * round to nearest integer and rescale. */ if equal (r2 - floor(r2), 1/2) then ceiling(r2)/10.0^n else round(r2)/10.0^n); matchdeclare (xx, lambda ([e], numberp(e) and e > 0)); tellsimpafter (integer_log10 (xx), integer_log10_numerical (xx)); matchdeclare (xx, lambda ([e], block ([v: ev (e, numer)], numberp(v) and v > 0))); tellsimpafter (integer_log10 (xx), integer_log10_numerical (ev (xx, numer))); matchdeclare (xx, lambda ([e], not atom(e) and op(e) = "/" and numberp (denom (e)) and pow10p (denom (e)))); pow10p (e) := integerp(e) and e > 1 and (e = 10 or pow10p (e/10)); tellsimpafter (integer_log10 (xx), integer_log10 (num (xx)) - integer_log10_numerical (denom (xx))); integer_log10_numerical (x) := if x >= 10 then (for i from 0 do if x >= 10 then x:x/10 else return(i)) elseif x < 1 then (for i from 0 do if x < 1 then x:x*10 else return(-i)) else 0;
The problem of rounding numbers is actually pretty subtle, but here is a simple approach which I think gives workable results. Here I define a new function myround which has the behavior described for Excel =ROUND. [1] (%i4) myround (x, n) := round(x*10^n)/10.0^n; n 'round(x 10 ) (%o4) myround(x, n) := ------------- n 10.0 (%i5) myround (2.15, 1); (%o5) 2.2 (%i6) myround (2.149, 1); (%o6) 2.1 (%i7) myround (-1.475, 2); (%o7) - 1.48 (%i8) myround (21.5, -1); (%o8) 20.0 (%i9) myround (626.3, -3); (%o9) 1000.0 (%i10) myround (1.98, -1); (%o10) 0.0 (%i11) myround (-50.55, -2); (%o11) - 100.0 [1] https://support.microsoft.com/en-us/office/round-function-c018c5d8-40fb-4053-90b1-b3e7f61a213c
cspline in Maxima giving me a result which indicates an error in Maxima
I have the following cubic polynomial f(x)=x³ - 3 x² + x -5 for which the cubic spline should provide the exact same polynomial assuming the following data: (-1, -10), (0,-5), (1, -6) with second derivative at the extremes f''(-1)=-12, f''(1)=0 (note that f''(x)=6x-6.) Here the piece of code that I tried on: /* polynomial to interpolate and data */ f(x) := x^3 - 3* x^2 + x - 5$ x0:-1$ x1:0$ x2:1$ y0:f(x0)$ y1:f(x1)$ y2:f(x2)$ p:[[x0,y0],[x1,y1],[x2,y2]]$ fpp(x) := diff(f(x),x,2); fpp0 : at( fpp(x), [x=x0]); fpp2 : at( fpp(x), [x=x2]); /* here I call cspline with d1=fpp0 and dn=fpp2 */ load(interpol)$ cspline(p, d1=fpp0, dn=fpp2); I expected the original polynomial (f(x)=x³ -3 x² + x -5) but I got the result: (%o40) (-16*x^3-15*x^2+6*x-5)*charfun2(x,-inf,0)+(8*x^3-15*x^2+6*x-5)*charfun2(x,0,inf) which does not agrees with the original polynomial. Evenmore. Here is a test on the results provided by Maxima. Code: /* verification */ h11(x) := -16*x^3 - 15* x^2 + 6* x - 5; h22(x) := 8* x^3 - 15*x^2 + 6* x - 5; h11pp(x) := diff(h11(x), x, 2); h11pp0: at( h11pp(x), [x=x0]); h22pp(x) := diff(h22(x), x, 2); h22pp2 : at(h22pp(x), [x=x2]); which throws 66 and 18 as the boundary conditions, which should be instead -12 and 0. Thanks.
It appears you've misinterpreted the arguments d1 and dn for cspline. As the description of cspline says, d1 and dn specify the first derivative for the spline at the endpoints, not the second derivative. When I use the first derivative of f to specify the values for d1 and dn, I get the expected result: (%i2) f(x) := x^3 - 3* x^2 + x - 5$ (%i3) [x0, x1, x2]: [-1, 0, 1] $ (%i4) [y0, y1, y2]: map (f, %); (%o4) [- 10, - 5, - 6] (%i5) p: [[x0, y0], [x1, y1], [x2, y2]]; (%o5) [[- 1, - 10], [0, - 5], [1, - 6]] (%i6) load (interpol) $ (%i7) cspline (p, d1 = at(diff(f(x), x), x=x0), dn = at(diff(f(x), x), x=x2)); 3 2 (%o7) (x - 3 x + x - 5) charfun2(x, minf, 0) 3 2 + (x - 3 x + x - 5) charfun2(x, 0, inf)
Fortran: matrix from symbolic functions
I need to transform a set of symbolic equations defining relations between \vec(a) = (a,b,c) and \vec(x) = (x,y), e.g. a = 1./2 * x b = -1./2 * x c = 1./2 * y into a matrix form so that I get the matrix A, when I write \vec(a) = A * \vec(x): / a \ / 1./2 0 \ / x \ | b | = | -1./2 0 | * \ y / \ c / \ 0 1./2 / Now the problem is, that the whole things needs to be in Fortran: reading the equations and transforming them to the matrix A. I have found the module fparser (https://www.sourceforge.net/projects/fparser/) to evaluate symbolic math expressions, but I could need some help figuring out how to most efficiently build these matrices without doing too much string parsing...
An approach (workaround?) in 100% pure Fortran might be... ! calc.f90 program main implicit none real avec( 3 ), xvec( 2 ), A( 3, 2 ) integer i do i = 1, size(xvec) xvec = 0 ; xvec(i) = 1.0 call calc() A(:,i) = avec enddo do i = 1, size(avec) print *, A(i,:) enddo contains subroutine calc() real a,b,c, x,y x = xvec(1) y = xvec(2) include 'eq.inc' avec = [a,b,c] end subroutine end eq.inc: a = 1./2 * x b = -1./2 * x c = 1./2 * y $ gfortran calc.f90 && ./a.out 0.500000000 0.00000000 -0.500000000 -0.00000000 0.00000000 0.500000000
Although it's long ago, I want to post what helped me solving the issue: I used fparser (http://fparser.sourceforge.net/).