I am calculating the following function's Laplace transform, but I get some strange output with limit and realpart etc on (%t5) and (%t6). What I expect is something like a_n = 0 and b_n = -1/(n*%pi).
Why is this happening? Am I defining the function at (%i1) incorrectly? Or is this a limitation of Maxima?
(%i1) f(t) := mod(t, 1);
(%o1) f(t) := mod(t, 1)
(%i2) plot2d(f(t), [t, -2, 2]);
(%o2) [/tmp/maxout1266174.gnuplot_pipes]
(%i3) load(fourie);
(%o3) /usr/share/maxima/5.43.0/share/calculus/fourie.mac
(%i4) fourier(f(t), t, 1);
1
(%t4) a = -
0 2
/
[
(%t5) a = limit I cos(%pi n t) realpart(floor(t)) dt
n t -> - 1 ]
/
/
[
- limit I cos(%pi n t) realpart(floor(t)) dt
t -> 1 ]
/
/
[
(%t6) b = (- limit I sin(%pi n t) realpart(floor(t)) dt)
n t -> 1 ]
/
/
[ 2 sin(%pi n) 2 cos(%pi n)
+ limit I sin(%pi n t) realpart(floor(t)) dt + ------------ - ------------
t -> - 1 ] 2 2 %pi n
/ %pi n
(%o6) [%t4, %t5, %t6]
(%i7)
Related
Solving Equations with (wx)Maxima: Control stack exhausted
I'm trying to solve equations with (wx)Maxima: formulate the equation, then let it insert the variables and solve the equation for the missing variable. But I'm having a hard time. Somehow it's having problems in the last line:
Control stack exhausted (no more space for function call frames).
This is probably due to heavily nested or infinitely recursive function
calls, or a tail call that SBCL cannot or has not optimized away.
That's my code:
kill(all);
load(physical_constants);
load(unit);
setunits([kg,m,s,N]);
showtime: false;
α: 30*%pi/180;
/*α: 30*°;*/
masse: 1000*kg;
g: 9.80665*m/(s*s);
b: 0.3*m;
B: 0.5*m;
L: 0.1*m;
F_g: masse*g;
F_H: masse * g;
kill(S, x);
S: solve(0=F_H-2*x*sin(α), x);
S: assoc(x, S);
kill(H, x);
H: solve(0=-F_g+2*x, x);
H: assoc(x, H);
kill(Ly, x);
Ly: solve(tan(α)=x/(B/2), x);
Ly: assoc(x, Ly);
kill(FN, x);
FN: solve(0=H*B/2-x*(L+Ly)+S*sin(α)*B/2+S*cos(α)*Ly, x);
FN: assoc(x, FN);
If I calculate it "directly", it works though:
kill(all);
load(physical_constants);
load(unit);
setunits([kg,m,s,N]);
showtime: false;
kill(FN, x);
FN: solve([α=30*%pi/180, H=196133/40*N,
B=0.5*m, L=0.1*m,
Ly=sqrt(3)/12*m, S=196133/20*N,
0=H*B/2-x*(L+Ly)+S*sin(α)*B/2+S*cos(α)*Ly],
[x, α, H, B, L, Ly, S]);
FN: assoc(x, FN[1]);
FN: float(FN);
(FN) 1934473685/128529*N
Unfortunately the unit package has not been updated in some time. I'll suggest to use instead the package ezunits, in which dimensional quantities are represented with a back quote. To solve equations, try dimensionally which goes through some gyrations to help other functions with dimensional quantities, e.g. dimensionally (solve (...)). (Note that dimensionally isn't documented, I'm sorry for the shortcoming.)
I've modified your program a little to remove some unneeded stuff and also to use rational numbers instead of floats; Maxima is generally more comfortable with rationals and integers than with floats. Here is the program:
linel: 65 $
load(ezunits) $
α: 30*%pi/180;
masse: 1000`kg;
g: rationalize(9.80665)`m/(s*s);
b: 3/10`m;
B: 5/10`m;
L: 1/10`m;
F_g: masse*g;
F_H: masse * g;
S: dimensionally (solve(0=F_H-2*x*sin(α), x));
S: assoc(x, S);
Ly: dimensionally (solve(tan(α)=x/(B/2), x));
Ly: assoc(x, Ly);
FN: dimensionally (solve(0=H*B/2-x*(L+Ly)+S*sin(α)*B/2+S*cos(α)*Ly, x));
FN: assoc(x, FN);
subst (x = S, F_H-2*x*sin(α));
subst (x = Ly, tan(α)=x/(B/2));
subst (x = FN, H*B/2-x*(L+Ly)+S*sin(α)*B/2+S*cos(α)*Ly);
ratsimp (expand (%));
and here is the output I get. Note that I substituted the solutions back into the equations to verify them. It looks like it worked as expected.
(%i2) linel:65
(%i3) load(ezunits)
(%i4) α:(30*%pi)/180
%pi
(%o4) ---
6
(%i5) masse:1000 ` kg
(%o5) 1000 ` kg
(%i6) g:rationalize(9.80665) ` m/(s*s)
5520653160719109 m
(%o6) ---------------- ` --
562949953421312 2
s
(%i7) b:3/10 ` m
3
(%o7) -- ` m
10
(%i8) B:5/10 ` m
1
(%o8) - ` m
2
(%i9) L:1/10 ` m
1
(%o9) -- ` m
10
(%i10) F_g:masse*g
690081645089888625 kg m
(%o10) ------------------ ` ----
70368744177664 2
s
(%i11) F_H:masse*g
690081645089888625 kg m
(%o11) ------------------ ` ----
70368744177664 2
s
(%i12) S:dimensionally(solve(0 = F_H-2*x*sin(α),x))
690081645089888625 kg m
(%o12) [x = ------------------ ` ----]
70368744177664 2
s
(%i13) S:assoc(x,S)
690081645089888625 kg m
(%o13) ------------------ ` ----
70368744177664 2
s
(%i14) Ly:dimensionally(solve(tan(α) = x/(B/2),x))
1
(%o14) [x = --------- ` m]
4 sqrt(3)
(%i15) Ly:assoc(x,Ly)
1
(%o15) --------- ` m
4 sqrt(3)
(%i16) FN:dimensionally(solve(0 = (H*B)/2-x*(L+Ly)
+(S*sin(α)*B)/2
+S*cos(α)*Ly,x))
1 1
(%o16) [x = (----------------------------------------- ` --)
140737488355328 sqrt(3) + 351843720888320 2
s
2
(351843720888320 sqrt(3) H ` s
3/2
+ 1150136075149814375 3 ` kg m)]
(%i17) FN:assoc(x,FN)
1 1
(%o17) (----------------------------------------- ` --)
140737488355328 sqrt(3) + 351843720888320 2
s
2
(351843720888320 sqrt(3) H ` s
3/2
+ 1150136075149814375 3 ` kg m)
(%i18) subst(x = S,F_H-2*x*sin(α))
kg m
(%o18) 0 ` ----
2
s
(%i19) subst(x = Ly,tan(α) = x/(B/2))
1 1
(%o19) ------- = -------
sqrt(3) sqrt(3)
(%i20) subst(x = FN,(H*B)/2-x*(L+Ly)+(S*sin(α)*B)/2+S*cos(α)*Ly)
1 1
(- ---------) - --
4 sqrt(3) 10 1
(%o20) ((----------------------------------------- ` --)
140737488355328 sqrt(3) + 351843720888320 2
s
2
(351843720888320 sqrt(3) H ` s
3/2 H
+ 1150136075149814375 3 ` kg m) + -) ` m
4
2
690081645089888625 kg m
+ ------------------ ` -----
281474976710656 2
s
(%i21) ratsimp(expand(%))
2
kg m
(%o21) 0 ` -----
2
s
EDIT. About converting kg*m/s^2 to N, you can apply the double back quote operator. For example:
(%i25) F_g `` N
690081645089888625
(%o25) ------------------ ` N
70368744177664
By the way, to convert back to floats, you can apply float:
(%i26) float(%)
(%o26) 9806.649999999998 ` N
Converting FN to N is a little more involved, since it's a more complex expression, especially because of H which doesn't have units attached to it yet. Some inspection seems to show the units of H must be kg*m/s^2. I'll apply declare_units to say that's what are the units of H. Then I'll convert FN to N.
(%i27) declare_units(H,(kg*m)/s^2)
kg m
(%o27) ----
2
s
(%i28) FN `` N
351843720888320 sqrt(3) qty(H)
(%o28) (-----------------------------------------
140737488355328 sqrt(3) + 351843720888320
3/2
1150136075149814375 3
+ -----------------------------------------) ` N
140737488355328 sqrt(3) + 351843720888320
(%i29) float(%)
(%o29) (1.023174629940149 qty(H) + 10033.91548470256) ` N
The notation qty(H) represents the unspecified quantity of H. One could also just subst(H = 100 ` kg*m/s^2, FN) (or any quantity, not just 100) and go from there.
How to expand taylor series/polynomials about Q=0 , and then extract coefficients as a list
example :
taylor ( (sin(q)), q, 0, 9); //taylor expansion for first 9 terms gives the next line
(%o1)/T/ q\-q^3/6+q^5/120\-q^7/5040+q^9/362880+...
then using coeff ((%o1), q ^n); gives me the coefficient at n only, what i want is a list for all the coefficients of that expression
Try coeff plus makelist, e.g. something like: makelist(coeff(%o1, q, n), n, 0, 9);
Edit:
I see now that I misread your question and there is already an answer. Nevertheless I will keep it because it is related to your question.
Use powerseries instead of taylor:
(%i1) expr:powerseries(sin(x),x,0);
inf
==== i2 2 i2 + 1
\ (- 1) x
(%o1) > -----------------
/ (2 i2 + 1)!
====
i2 = 0
You can access the coefficient by the args or part function
(%i2) op(expr);
(%o2) sum
(%i3) args(expr);
i2 2 i2 + 1
(- 1) x
(%o3) [-----------------, i2, 0, inf]
(2 i2 + 1)!
(%i4) part(expr,1);
i2 2 i2 + 1
(- 1) x
(%o4) -----------------
(2 i2 + 1)!
(%i5) args(expr)[1];
i2 2 i2 + 1
(- 1) x
(%o5) -----------------
(2 i2 + 1)!
If you want to change the index variable:
(%i6) niceindices(expr),niceindicespref=[n];
inf
==== n 2 n + 1
\ (- 1) x
(%o6) > ---------------
/ (2 n + 1)!
====
n = 0
newbie Maxima question
I have a transfer function in Maxima
E1 : y = K_i*s/(s^2 + w^2);
I'd like to have the closed-form of the equation affter applying the bilinear transform
E2 : s = (2/Ts*(z-1)/(z+1));
I would like to get the transfer function for z, by substituing s by equation E2. How should I proceed?
Regards
Note that subst can apply one or more substitutions stated as equations. In this case, try subst(E2, E1).
That will probably create a messy result -- you can simplify it somewhat by applying ratsimp to the result.
Here's what I get from that.
(%i2) E1 : y = K_i*s/(s^2 + w^2);
K_i s
(%o2) y = -------
2 2
w + s
(%i3) E2 : s = (2/Ts*(z-1)/(z+1));
2 (z - 1)
(%o3) s = ----------
Ts (z + 1)
(%i4) subst (E2, E1);
2 K_i (z - 1)
(%o4) y = ------------------------------
2
4 (z - 1) 2
Ts (z + 1) (------------ + w )
2 2
Ts (z + 1)
(%i5) ratsimp (%);
2
2 K_i Ts z - 2 K_i Ts
(%o5) y = -----------------------------------------------
2 2 2 2 2 2 2
(Ts w + 4) z + (2 Ts w - 8) z + Ts w + 4
I have an matrix in Maxima, let´s say (for simplification of the problem):
A: matrix([2*(a^2+b^2+c^2)])
But I know that:
a^2+b^2+c^2 = 1
How do I simplify that matrix in Maxima in terms of that equation, in order to obtain A = [2]?
I found the solution:
A: matrix([2*(a^2+b^2+c^2)]);
eq: a^2+b^2+c^2 = 1;
scsimp(A, eq);
You can use tellrat.
(%i1) A:matrix([2*(a^2+b^2+c^2)])
(%o1) [ 2 2 2 ]
[ 2 (c + b + a ) ]
(%i2) a^2+b^2+c^2 = 1
2 2 2
(%o2) c + b + a = 1
(%i3) solve(%,a^2)
2 2 2
(%o3) [a = (- c ) - b + 1]
(%i4) tellrat(%[1])
2 2 2
(%o4) [c + b + a - 1]
(%i5) algebraic:true
(%o5) true
(%i6) rat(A)
(%o6)/R/ [ 2 ]
(%i7) untellrat(a)
(%o7) []
I'm trying to update a matrix of real valued numbers in a for loop using torch.Tensor.
Here is what I'd like to do:
-- W and P are of size NxN, r is of size N
delta_W = P * r:view(N, 1) * r:view(1, N) * P -- this is an NxN
denominator = 1 + r:view(1, N) * P * r:view(N, 1) -- this is a number
delta_W = delta_w / denominator -- ## THIS ONE RAISES ERROR ##
W = W + delta_W
Just to be clear:
denom -> [torch.DoubleTensor of size 1x1]
P, delta_W, W -> [torch.DoubleTensor of size 200x200]
The error when I do the division is:
bad argument #2 to '?' (number expected at /usr/local/torch/pkg/torch/generic/TensorOperator.c:145)
I'm a heavy numpy users so I thought "broadcasting" was the issue, therefore I tried simulating it through torch.repeatTensor(denom, N, N) with no luck. If denom is just a number (not a DoubleTensor) everything works fine anyway. Using the element doesn't work either, delta_P / denom[1] gives the same error.
What am I doing wrong?
Edit:
I tried using
denominator = (1 + r:view(1, N) * P * r:view(N, 1)):apply(function(x) return x^(-1) end)
delta_w = delta_w * torch.repeatTensor(denominator, N, N)
which doesn't throw an error but the results are wrong. To see this, try the following:
torch.linspace(0, 3, 4):view(2, 2) * torch.Tensor(2, 2):fill(0.5)
I managed by using apply, repeatTensor and finally cmul for element-wise multiplication
rPr = r:view(1, N) * P * r:view(N, 1)
denominator = (1 + rPr):apply(function(x) return x^(-1) end)
delta_w:cmul(torch.repeatTensor(denominator, N, N))
though I wonder if this can be transferred to GPU with cutorch.
If you do denominator[1][1], you will get a 'number' instead of a 'torch.Tensor'. Then you can just write the division statement normally.
-- W and P are of size NxN, r is of size N
delta_W = P * r:view(N, 1) * r:view(1, N) * P -- this is an NxN
denominator = 1 + r:view(1, N) * P * r:view(N, 1) -- this is a 1x1
delta_W = delta_w / denominator[1][1]
W = W + delta_W
By the way, in the 1st statement did you want one of the P matrices transposed (P:t())?