How can I tell Maxima to solve the following problem? (The "solve" part):
I did:
load(distrib);
fpprec: 100;
bftorat:true;
solve(2*bfloat(cdf_normal(x,0,1))-1=0.99999999999999999968130594071b0, [x]);
%,numer
Got:
(%o1) "/usr/share/maxima/5.32.1/share/distrib/distrib.mac"
(%o2) 100
(%o3) true
`rat' replaced -1.99999999999999999968130594071B0 by -199999999999999999968130594071/100000000000000000000000000000 = -1.99999999999999999968130594071B0
`rat' replaced 5.0B-1 by 1/2 = 5.0B-1
`rat' replaced 5.0B-1 by 1/2 = 5.0B-1
`rat' replaced 7.071067811865475244008443621048490392848359376884740365883398689953662392310535194251937671638207864B-1 by 118807941462947422469655519336079782367473013592460/168019802134529020067676914738440478110633605571601 = 7.071067811865475244008443621048490392848359376884740365883398689953662392310535194251937671638207864B-1
`rat' replaced 7.071067811865475244008443621048490392848359376884740365883398689953662392310535194251937671638207864B-1 by 118807941462947422469655519336079782367473013592460/168019802134529020067676914738440478110633605571601 = 7.071067811865475244008443621048490392848359376884740365883398689953662392310535194251937671638207864B-1
(%o4) [x=
(168019802134529020067676914738440478110633605571601*inverse_erf(99999999999999999968130594071/100000000000000000000000000000))/118807941462947422469655519336079782367473013592460]
inverse_erf: inverse_erf(1.0) is undefined.
-- an error. To debug this try: debugmode(true);
Further tried (to see if the rational replacement affects inverse_erf):
inverse_erf(9.9999999999999999968130594071b−1);
gamma_incomplete: continued fractions failed for gamma_incomplete(5.0b-1, 1.675965338889773975600843228854238162008399514002414970690458801529039878850010279559673197036592113916729947593696740535214189646774061729913734402353264492788885098143556404059170138591463120333687838496039284224858192931635551067412157341539627014907074717352945374476804912353312948754404927014555821149803440423871160460111635311071245528519256957555845418916034380535359079516879576795825857468710891474077746896341697834315575814989209244705740463652472196503944998297956825510866851943203353716451062616549067258800559231646552924469724160521456041856694702333938138297284123098699530288993519920353577729741393726951293645734447176179175560311787410907660921783058989196733914852345086206761158383783714360001646880454976428470630991611033582649957934844195398077660932657131767136966243415075424193909691302431604307524134764326959890911322928113456129784004060990585972427799869290459688878023964671879763390452043333273689436077956597199441415992202082578463153853017929328667898523224007b0).
-- an error. To debug this try: debugmode(true);
My advice is to try to solve the equation with a symbolic value and replace it with a numerical value later on. Here's what I get:
(%i1) load (distrib) $
(%i2) fpprec : 100 $
(%i3) solve (2 * cdf_normal (x, 0, 1) - 1 = a, x);
(%o3) [x = sqrt(2) inverse_erf(a)]
(%i4) %, a = 0.99999999999999999968130594071b0;
(%o4) [x = 6.33712711592763726142078700288254243769449484599872720866948195829\
6543071614144180808554952052800789b0 sqrt(2)]
(%i5) bfloat (%);
(%o5) [x = 8.96205111382716157629734310948891265577140990195458570850348910814\
5354833288951128435123731428126303b0]
Actually, after a bit of further study, replacing:
%,numer
with
%,bfloat
at the end of the original problem does the trick. I thought numer works with bfloats, but it only uses rational approximations, and that is why I received errors. However, using bfloat gives the correct answer.
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.
I want to display the partial derivative df/dx of f(x,y) = ln(y-1-x^2)-xy^2.
A comparable example for what I want is:
(%i0) f(x) := x^8$
(%i1) diff(f(x),x);
(%o1) 8*x^7
I've tried:
(%i1) f(x,y):=ln(y-1-x^2)-xy^2$
(%i2) g(x,y):=(diff(f(x,y), x));
(%o2) g(x,y):='diff(f(x,y),x,1)
(%i3) g(x,y):=''(diff(f(x,y), x));
(%o3) g(x,y):='diff(ln(y-x^2-1),x,1)
But it doesn't work (the summand -xy^2 got deleted?).
I want the derivated function. Something like this:
(2*x)
g(x,y):= ——————————— - y^2
(1 + x^2 - y)
the problem with your funciotn is that xy^2 instead of x*y^2for maxima is a variable you should write it as follows:
(%i5) f(x,y):=ln(y-1-x^2)-x*y^2;
(\%o5) f\left(x , \linebreak[0]y\right):=\mathrm{ln}\left(y-1-x^2\right)-x\,y^2
(%i6) diff(f(x,y), x);
(\%o6) \ifracd{d}{d\,x}\,\mathrm{ln}\left(y-x^2-1\right)-y^2
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.
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
In the Maxima session below, how come f(1) is not 0?
(%i1) eq: 2 * x + 1 = 3;
(%o1) 2 x + 1 = 3
(%i2) f(x) := lhs(eq) - rhs(eq);
(%o2) f(x) := lhs(eq) - rhs(eq)
(%i3) f(1);
(%o3) 2 x - 2
the process of function calling in maxima here binds x to 1 in the function
definition, lhs(eq)-rhs(eq). That has no x in it, so that binding does nothing.
Next, lhs(eq) is evaluated to 2*x+1. rhs(eq) is evaluated to 3. etc.
Do you always want the same equation eq? perhaps you want to do
define(f(x),lhs(eq)-rhs(eq));
to check what the definition is, try
grind(f);
If you want to vary the equation maybe something like
g(val, eq) := subst(val,x, lhs(eq)-rhs(eq)) ;
would do.