I worked on this for a day, and hope you guys can give a hint on this strange thing.
y0, y1, and y2 are independently generated by the same method.
They are each split into 20 groups by the same method.
Yet, manova says they are significantly different? Why?
The summary of the Manova test (stored in variable s) says:
The Pr (>F) value is less than 2.2e-16.
y0 <- runif(100, 0, 1)
y1 <- runif(100, 0, 1)
y2 <- runif(100, 0, 1)
y0 <- c(y0, runif(100, 0, 10) )
y1 <- c(y1, runif(100, 0, 10) )
y2 <- c(y2, runif(100, 0, 10) )
y0=as.numeric(unlist(y0))
y1=as.numeric(unlist(y1))
y2=as.numeric(unlist(y2))
b=10
a=length(y0)/b
g=rep(1:a,rep(b,a))
m1 <- manova(cbind(y0, y1, y2) ~ g)
s=summary(m1, test = "Wilks")
a = s$stats
a = a[11]
s
a
The summary is here:
Df Wilks approx F num Df den Df Pr(>F)
g 1 0.37069 110.91 3 196 < 2.2e-16 ***
Residuals 198
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
I'm really not sure why, but I started running your code and the first three lines seem to shed some light as to why you are getting different results. If you run that part of the code and then simply ask
y0
y1
y2
you will see that all three objects have different elements. In fact, all elements are different. As to why this is so, I'm not sure cause your defining them the same way, but they are different for sure. Plot them out and take a look.
Hope this helps
Related
Without specifying units, I can express area and volume and have Maxima show the relationship:
(%i1) areaNoUnits: area = width * length$
(%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.
F(x1) > a;
F(x2) < b;
∀t, F'(x) >= 0 (derivative) ;
F(x) = ∑ ci*x^i; (i∈[0,n] ; c is a constant)
Your question is quite ambiguous, and stack-overflow works the best if you show what you tried and what problems you ran into.
Nevertheless, here's how one can code your problem for a specific function F = 2x^3 + 3x + 4, using the Python interface to z3:
from z3 import *
# Represent F as a function. Here we have 2x^3 + 3x + 4
def F(x):
return 2*x*x*x + 3*x + 4
# Similarly, derivative of F: 6x^2 + 3
def dF(x):
return 6*x*x + 3
x1, x2, a, b = Ints('x1 x2 a b')
s = Solver()
s.add(F(x1) > a)
s.add(F(x2) < b)
t = Int('t')
s.add(ForAll([t], dF(t) >= 0))
r = s.check()
if r == sat:
print s.model()
else:
print ("Solver said: %s" % r)
Note that I translated your ∀t, F'(x) >= 0 condition as ∀t. F'(t) >= 0. I assume you had a typo there in the bound variable.
When I run this, I get:
[x1 = 0, x2 = 0, b = 5, a = 3]
This method can be generalized to arbitrary polynomials with constant coefficients in the obvious way, but that's mostly about programming and not z3. (Note that doing so in SMTLib is much harder. This is where the facilities of host languages like Python and others come into play.)
Note that this problem is essentially non-linear. (Variables are being multiplied with variables.) So, SMT solvers may not be the best choice here, as they don't deal all that well with non-linear operations. But you can deal with those problems as they arise later on. Hope this gets you started!
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
I have the logistic map function in Maxima like so:
F(x,r,n):= x[n]=r*x[n-1]*(1-x[n-1]);
And when I input the correct variables it gives me the answer to, for example, x[0]:
(%i15) n:0$
x[n-1]:[0.1]$
F(x, r:3, n);
(%o15) x[0]=[0.27]
However, this answer does not stay memorized and when I enter x[0] I get
x[0];
(%o5) x[0]
How do I write a function that will calculate x[n] for me and store it in memory, so I can use it later? I am trying to make a bifurcation diagram for the logistic map without using any black boxes, i.e., the orbits functions.
Thank you!
There are different ways to go about it. One straightforward way is to create a list and then iterate, computing its elements one by one. E.g.:
(%i4) x: makelist (0, 10);
(%o4) [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
(%i5) x[1]: 0.1;
(%o5) 0.1
(%i6) r: 3;
(%o6) 3
(%i7) for i:2 thru 10 do x[i]: r * x[i - 1] * (1 - x[i - 1]);
(%o7) done
(%i8) x;
(%o8) [0.1, 0.2700000000000001, 0.5913000000000002,
0.7249929299999999, 0.5981345443500454, 0.7211088336156269,
0.603332651091411, 0.7179670896552621, 0.6074710434816448,
0.7153499244388992]
Note that : is the assignment operator, not =.
When I try to do this:
(%i1) declare (z, complex);
(%o1) done
(%i2) eq1: z^3 + 3 * %i * conjugate(z) = 0;
3
(%o2) 3 %i conjugate(z) + z = 0
(%i3) solve(eq1, z);
1/6 5/6 1/3 1/3
(- 1) (3 %i - 3 ) conjugate(z)
(%o3) [z = - -----------------------------------------,
2
1/6 5/6 1/3 1/3
(- 1) (3 %i + 3 ) conjugate(z)
z = -----------------------------------------,
2
1/6 1/3 1/3
z = - (- 1) 3 conjugate(z) ]
conjugates are not simplified. And the solution for z in terms of z isn't very useful. Is there a way to simplify it?
Also, how can I simplify out the (-1)^(1/6) part?
Also, this equation clearly has 0 as its root, but it's not in the solution set, why?
I don't think solve knows anything about conjugate. Try this to solve it with the real and imaginary parts of z as two variables. Like this:
(%i2) declare ([zr, zi], real) $
(%i3) z : zr + %i*zi $
(%i4) eq1: z^3 + 3 * %i * conjugate(z) = 0;
(%o4) (zr+%i*zi)^3+3*%i*(zr-%i*zi) = 0
(%i5) solve (eq1, [zr, zi]);
(%o5) [[zr = %r1,
zi = (sqrt(9*%r1^2-%i)+3*%r1)^(1/3)-%i/(sqrt(9*%r1^2-%i)+3*%r1)^(1/3)
+%i*%r1],
[zr = %r2,
zi = ((sqrt(3)*%i)/2-1/2)*(sqrt(9*%r2^2-%i)+3*%r2)^(1/3)
-(%i*((-(sqrt(3)*%i)/2)-1/2))/(sqrt(9*%r2^2-%i)+3*%r2)^(1/3)
+%i*%r2],
[zr = %r3,
zi = ((-(sqrt(3)*%i)/2)-1/2)*(sqrt(9*%r3^2-%i)+3*%r3)^(1/3)
-(%i*((sqrt(3)*%i)/2-1/2))/(sqrt(9*%r3^2-%i)+3*%r3)^(1/3)+%i*%r3]]
Note the variables%r1, %r2, and %r3 in the solution. These represent arbitrary values.