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
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.
The following integral does is not evaluated by Maxima:
integrate(charfun(x<1/2), x, 0, 1);
Is there a different trick to make it work, or is it simply un-implemented?
The share package abs_integrate can integrate some expressions containing signum, abs, and unit_step. In this case you can write charfun(x < 1/2) in terms of signum(1/2 - x) and then abs_integrate can handle it.
You'll need to load abs_integrate. Note that abs_integrate modifies the behavior of integrate; there isn't a separate abs_integrate function to call.
(%i2) load (abs_integrate) $
(%i3) integrate (signum (1/2 - x), x, 0, 1);
(%o3) 0
(%i4) integrate (signum (1/2 - x), x, -1, 1);
(%o4) 1
(%i5) foo (e) := (1 + signum(e))/2;
1 + signum(e)
(%o5) foo(e) := -------------
2
(%i6) integrate (foo (1/2 - x), x, 0, 1);
1
(%o6) -
2
(%i7) integrate (foo (1/2 - x), x, -1, 1);
3
(%o7) -
2
Note that foo corresponds to charfun here.
In Maxima I have an equation that is like:
eq : c0*a + d0*a + c1*b - c2*p - c4*q = c5*r
Is there a command that allows me to arrive at:
(c0 + d0)*a + c1*b = c2*p + c4*q + c5*r
In short, I want to choose which variables end on either the left or right hand side and I want to write it such that there is only one occurence of the variables I select (in this case a, b, p, q, r).
Perhaps coefmatrix is useful for this.
(%i1) display2d : false $
(%i2) eq : c0*a + d0*a + c1*b - c2*p - c4*q = c5*r $
(%i3) vars : [a, b, p, q, r] $
(%i4) coeffs : coefmatrix ([eq], vars);
(%o4) matrix([d0+c0,c1,-c2,-c4,-c5])
(%i5) coeffs . vars;
(%o5) (-c5*r)-c4*q-c2*p+a*(c0+d0)+b*c1
Note that both arguments of coefmatrix must be lists.
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.
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.