I have the following Maxima code:
m:sum(x[i],i,1,N)/N;
and then I want to calculate $m^2$.
m2:m^2, sumexpand;
Then I get double summation:
sum(sum(x[i1]*x[i2],i1,1,N),i2,1,N)/N^2
What I want to achieve is to expand it into the two sums.
The first one is sum(x[i]^2,i,1,N) and the second is the rest over non-equal indices. How should I do that? How should I do that with arbitrary power of m?
sum is not declared linear by default; you can declare it linear and resimplify. Note that to get the expected effect, you have to declare the noun form of sum.
(%i1) m:sum(x[i],i,1,N)/N;
N
====
\
> x
/ i
====
i = 1
(%o1) --------
N
(%i2) m2:m^2, sumexpand;
N N
==== ====
\ \
> > x x
/ / i1 i2
==== ====
i1 = 1 i2 = 1
(%o2) ---------------------
2
N
(%i3) declare (nounify(sum), linear);
(%o3) done
(%i4) ''%o2;
N N
==== ====
\ \
( > x ) > x
/ i1 / i2
==== ====
i1 = 1 i2 = 1
(%o4) -----------------------
2
N
Related
Here's the initial premise: two sums s1 and s2 are added; the sum element expressions have a common factor a[n].
s1: sum(r1[m,q]*b[m,n]*a[n],n,0,N)$
s2: sum(r2[m,q]*c[m,n]*a[n],n,0,N)$
s1+s2;
I expect the sums to be combined and the common element expression a[n] factored out:
s12: sum(a[n]*(r1[m,q]*b[m,n]+r2[m,q]*c[m,n]),n,0,N);
However, I'm unable to make Maxima produce such contraction. The most simplification I was able to obtain was using sumcontract(s1+s2) and it results in two sums without the common element being factored out:
r1[m,q]*sum(b[m,n]*a[n], n,0,N) + r2[m,q]*sum(c[m,n]*a[n], n,0,N);
How to make Maxima produce the factored out expression from s1+s2 as in s12 above?
NOTE: If we remove the r1 and r2, then the factor(sumcontract(s1+s2)) indeed results in the expected s12 expression. However, with both present, it results in two sums and does not factor out the a[n] as mentioned.
How about this. I've applied sumcontract, intosum, and factor.
(%i1) s1: sum(r1[m,q]*b[m,n]*a[n],n,0,N)$
(%i2) s2: sum(r2[m,q]*c[m,n]*a[n],n,0,N)$
(%i3) s1 + s2;
N N
==== ====
\ \
(%o3) r2 > c a + r1 > b a
m, q / m, n n m, q / m, n n
==== ====
n = 0 n = 0
(%i4) intosum (%);
N N
==== ====
\ \
(%o4) > c r2 a + > b r1 a
/ m, n m, q n / m, n m, q n
==== ====
n = 0 n = 0
(%i5) sumcontract (%);
N
====
\
(%o5) > (c r2 a + b r1 a )
/ m, n m, q n m, n m, q n
====
n = 0
(%i6) factor (%);
N
====
\
(%o6) > (c r2 + b r1 ) a
/ m, n m, q m, n m, q n
====
n = 0
In this, intosum is pushing constant factors back into the sum.
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
Suppose you have a function defined by intervals, such as
f(x):=block(if x<0 then x^2 else x^3);
When we differentiate it with
diff(f(x),x);
we get
d/dx (if x<0 then x^2 else x^3)
whereas I'd like to get
(if x<0 then 2*x else 3*x^2)
Is there a way to obtain such result?
This may help in a simple case:
(%i1) f(x):= charfun(x<0)*x^2 + charfun(x>=0)*x^3$
(%i2) gradef(charfun(y), 0)$
(%i3) diff(f(x),x);
2
(%o3) 2 x charfun(x < 0) + 3 x charfun(x >= 0)
charfun, gradef
You can try also Pw.mac package from Richard Hennessy.
Here's a different approach using a simplification rule for "if" expressions. The unsolved part here is to detect discontinuities and generate delta functions for those locations. If you want to ignore those, you can define FOO to return 0. Note that I didn't attempt to implement the function discontinuities; that part is unsolved here. I can give it a try if there is interest.
(%i1) display2d : false $
(%i2) matchdeclare ([aa, bb, cc], all, xx, symbolp) $
(%i3) 'diff (if aa then bb else cc, xx) $
(%i4) tellsimpafter (''%, apply ("if", [aa, diff (bb, xx), true, diff (cc, xx)]) + FOO (aa, bb, cc, xx)) $
(%i5) FOO (a, b, c, x) := 'lsum ((ev (c, x = d) - ev (b, x = d)) * delta (d, x), d, discontinuities (a, x)) $
(%i6) diff (if x > 0 then x^2 else x^3, x);
(%o6) (if x > 0 then 2*x else 3*x^2)+'lsum((d^3-d^2)*delta(d,x),d,
discontinuities(x > 0,x))
Building on slitinov's answer I wrote this quite naive implementation for functions with more than two "pieces":
gradef(charfun(dummy),0)$
/* piecewise function definition */
itv: [[x<0],[x>=0,x<1], [x>=1]]; /* intervals */
fi: [ 1, x^2+1, 2*x ]; /* local functions */
/* creation of global function f and its derivative df */
f:0;
for i: 1 thru 3 do f:f+charfun(apply("and",itv[i]))*fi[i];
df:diff(f,x);
/* display of local functions and derivatives */
for i: 1 thru 3 do (
apply(assume,itv[i]),
newline(),
print(itv[i]),
print("f = ",ev(f)),
print("df = ",ev(df)),
apply(forget,itv[i])
);
plot2d([f,df],[x,-2,3],[y,-1,5],[style,[lines,4,3],[lines,2,2]]);