I have to find the inverse of a function which looks like:
T := ->x (x)^0.5/(x^0.5+(1-x)^0.5)^2.
As we can see from the polynomial, we have 4 solutions when solving y= f(x). In maple,I soled for the inverse of T(x)
V := x-> solve(t=T(x),x,useassumptions=true) assuming 0<=t<=1.
and I can evaluate V, i.e maple can do V(0)=0 V(1)=1 etc.
However, as discussed, there are four solutions to the inverse function, the output of V is an expressions sequence, which looks like (solution1, solution2, solution3, solution4).
In later part of the task, I have to find the derivative of V(x)and integrate it. When I apply diff(V(x),x), maple gives me an error, saying V(x) is not valid.As V(x) is an expression sequence. I tried to use the function D(V), but still no luck.
My questions is how would I be able to handle this V(x) as an expression sequence to finish the rest of the task. Is V(x) a piecewise function? If that's the case, how would I be able to convert this expression sequence to a piecewise function.
Regards,
restart:
T := proc (x) options operator, arrow; sqrt(x)/(sqrt(x)+sqrt(1-x))^2 end proc:
V := proc (x) options operator, arrow; solve(x = T(y), y) end proc:
sol := [allvalues(V(x))]:# Extract 4 solution, with command op(1, sol)->Only first solution is correct.
plot([x, T(x), op(1, sol)], x = 0 .. 2, legend = [typeset("Curve: ", "x"),
typeset("Curve: ", "T(x)"), typeset("Curve: ", "V(x)")]);
VV := proc (x) options operator, arrow; evalf(op(1, sol)) end proc;
eval(VV(x), x = 1/2); #Inverse function at point x=1/2
eval(diff(VV(x), x), x = 1/2);# Derivative of inverse function at point x=1/2
int(VV(x), x = 1/10 .. 1/2, numeric);# Integral of inverse function at range (1/10..1/2)
Mathematica 11.3 solution:
Related
I want to generalize some predicate written in swi-prolog to calculate the power of some function. My predicate so far is:
% calculates the +Power and the +Argument of some function +Function with value +Value.
calc_power(Value, Argument, Function, Power) :-
not(Power is 0),
Power is Power_m1 + 1,
Value =..[Function, Buffer],
calc_power(Buffer, Argument, Function, Power_m1), !.
calc_power(Argument, Argument, _, 0).
The call calc_power((g(a)),A,f,POW). gives so far:
A = g(a),
POW = 0.
My generalization should also solve calls like that:
calc_power(A1, a, f, 3).
the solution should be in that special calse A1 = f(f(f(a))). But for some reason it doesn't work. I get the error:
ERROR: Arguments are not sufficiently instantiated
in line
Power is Power_m1 + 1
it means probably in swi prolog it is not possible to take plus with two variables. How can I solve this problem?
Can delay the + 1 operation with:
int_succ(I0, I1) :-
( nonvar(I0) ->
integer(I0),
I0 >= 0,
I1 is I0 + 1
; nonvar(I1) ->
integer(I1),
I1 >= 1,
I0 is I1 - 1
; when((nonvar(I0) ; nonvar(I1)), int_succ(I0, I1))
).
Example in swi-prolog:
?- int_succ(I0, I1), I1 = 7.
I0 = 6,
I1 = 7.
This is more flexible than https://www.swi-prolog.org/pldoc/man?predicate=succ/2 , and can of course be modified to support negative numbers if desired.
Found some solution
:- use_module(library(clpfd)).
% calculates the +Power and the +Argument of some function +Function with value +Value.
calc_power(Argument, Argument, _, 0).
calc_power(Value, Argument, Function, Power) :-
Power #\= 0,
Power #= Power_m1 + 1,
Value =..[Function, Buffer],
calc_power(Buffer, Argument, Function, Power_m1).
I have a system of n equations and n unknown variables under symbol sum. I want to create a loop to solve this system of equations when inputting n.
y := s -> 1/6cos(3s);
A := (k, s) -> piecewise(k <> 0, 1/2exp(ksI)/abs(k), k = 0, ln(2)exp(s0I) - sin(s));
s := (j, n) -> 2jPi/(2*n + 1);
n := 1;
for j from -n to n do
eqn[j] := sum((A(k, s(j, n))) . (a[k]), k = -n .. n) = y(s(j, n));
end do;
eqs := seq(eqn[i], i = -n .. n);
solve({eqs}, {a[i]});
enter image description here
Please help me out!
I added some missing multiplication symbols to your plaintext code, to reproduce it.
restart;
y:=s->1/6*cos(3*s):
A:=(k,s)->piecewise(k<>0,1/2*exp(k*s*I)/abs(k),
k=0,ln(2)*exp(s*I*0)-sin(s)):
s:=(j,n)->2*j*Pi/(2*n+1):
n:=1:
for j from -n to n do
eqn[j]:=add((A(k,s(j,n)))*a[k],k=-n..n)=y(s(j,n));
end do:
eqs:=seq(eqn[i],i=-n..n);
(-1/4+1/4*I*3^(1/2))*a[-1]+(ln(2)+1/2*3^(1/2))*a[0]+(-1/4-1/4*I*3^(1/2))*a[1] = 1/6,
1/2*a[-1]+ln(2)*a[0]+1/2*a[1] = 1/6,
(-1/4-1/4*I*3^(1/2))*a[-1]+(ln(2)-1/2*3^(1/2))*a[0]+(-1/4+1/4*I*3^(1/2))*a[1] = 1/6
You can pass the set of names (for which to solve) as an optional argument. But that has to contain the actual names, and not just the abstract placeholder a[i] as you tried it.
solve({eqs},{seq(a[i],i=-n..n)});
{a[-1] = 1/6*I/ln(2),
a[0] = 1/6/ln(2),
a[1] = -1/6*I/ln(2)}
You could also omit the indeterminate names here, as optional argument to solve (since you wish to solve for all of them, and no other names are present).
solve({eqs});
{a[-1] = 1/6*I/ln(2),
a[0] = 1/6/ln(2),
a[1] = -1/6*I/ln(2)}
For n:=3 and n:=4 it helps solve to get a result quicker here if exp calls are turned into trig calls. Ie,
solve(evalc({eqs}),{seq(a[i],i=-n..n)});
If n is higher than 4 you might have to wait long for an exact (symbolic) result. But even at n:=10 a floating-point result was fast for me. That is, calling fsolve instead of solve.
fsolve({eqs},{seq(a[i],i=-n..n)});
But even that might be unnecessary, as it seems that the following is a solution for n>=3. Here all the variables are set to zero, except a[-3] and a[3] which are both set to 1/2.
cand:={seq(a[i]=0,i=-n..-4),seq(a[i]=0,i=-2..2),
seq(a[i]=0,i=4..n),seq(a[i]=1/2,i=[-3,3])}:
simplify(eval((rhs-lhs)~({eqs}),cand));
{0}
I have code similar to the one below, where a function with a parameter depending on the loop iteration is plotted after every iteration. I would like to save the plot with the name trigplot_i.ps where i is the iteration number, but don't know how.
I have tried trigplot_"i".ps but didn't work, and have not been able to find how to cast i to a string either.
I'm a beginner so any help is very welcome.
f(x) := sin(x);
g(x) := cos(x);
for i:1 thru 10 do
(plot2d([i*f(x), i*g(x)], [x,-5,5],[legend,"sin(x)","cos(x)"],
[xlabel,"x"],[ylabel,"y"],
[ps_file,"./trigplot_i.ps"],
[gnuplot_preamble,"set key box spacing 1.3 top right"])
);
code after edits gives an error:
f(x) := sin(x);
g(x) := cos(x);
for i:1 thru 10
do block([myfile],
myfile: sconcat("./trigplot_", i, ".ps"),
printf (true, "iteration ~d, myfile = ~a~%", myfile),
plot2d([i*f(x), i*g(x)], [x,-5,5],[legend,"sin(x)","cos(x)"],
[xlabel,"x"],[ylabel,"y"],
[ps_file, myfile],
[gnuplot_preamble,"set key box spacing 1.3 top right"])
);
error:
"declare: argument must be a symbol; found "./trigplot_1.ps
-- an error.
To debug this try: debugmode(true);"
Looks good. To construct a file name, try this: sconcat("./trigplot_", i, ".ps") or also you can try: printf(false, "./trigplot_~d.ps", i). My advice is to make that a separate step in the loop, and then you can use it in the call to plot2d, e.g.:
for i:1 thru 10
do block ([myfile],
myfile: sconcat("./trigplot_", i, ".ps"),
printf (true, "iteration ~d, myfile = ~a~%", i, myfile),
plot2d (<stuff goes here>, [ps_file, myfile], <more stuff>));
EDIT: Fixed a bug in printf (omitted argument i).
I want to define a function of a function ( maybe functional) in Maxima. For example, I want to define T in the code:
f(r,GN,M)::=(2*GN*M^2)*(2*GN*M/r-(2*GN*M/(2*r))^2);
X(r,GN,M) ::= 2*GN*M/(2*r)+2*GN*M/r*(1-2*GN*M/(4*r));
T(r,GN,M) ::= diff(f(r,GN,M),r)+X(r,GN,M);
But I don't know the code.
In maxima it is convenient to use expressions.
Define two expressions
(%i1) display2d: false $
(%i2) f: (2*GN*M^2)*(2*GN*M/r-(2*GN*M/(2*r))^2) $
(%i3) X: 2*GN*M/(2*r)+2*GN*M/r*(1-2*GN*M/(4*r)) $
Define a function which operates on two expressions
(%i4) T(f, X) := diff(f, r) + X $
Call it. Result is an expression
(%i5) dfpX: T(f, X);
(%o5) (2*GN*M*(1-(GN*M)/(2*r)))/r+(GN*M)/r+2*GN*M^2
*((2*GN^2*M^2)/r^3-(2*GN*M)/r^2)
You can create a function based on dfpX
(%i6) define(f(r, GN, M), dfpX);
(%o6) f(r,GN,M):=(2*GN*M*(1-(GN*M)/(2*r)))/r+(GN*M)/r
+2*GN*M^2*((2*GN^2*M^2)/r^3-(2*GN*M)/r^2)
and call it
(%i7) f(1, 2, 3);
(%o7) 2142
but you can stay with expressions
(%i8) subst([r=1, GN=2, M=3], dfpX);
(%o8) 2142
Define T as a function, rather than a macro, so that it evaluates its arguments.
(%i) T(r,GN,M) := diff(f(r,GN,M),r)+X(r,GN,M)$
Now define a function t on T which quotes ' its arguments (so that T knows which variable to differentiate with respect to) and then evaluates the result %% with ev.
(%i) t(r,GN,M) := (T('r,'GN,'M),ev(%%))$
(%i) t(1,2,3);
(%o) 2142
I have I simple problem but I cannot find a solution anywhere.
I have to integrate a function (for example using a Simpson's rule subroutine) but I am obliged to pass to my function more than one argument: one is the variable that I want to integrate later and another one is just a value coming from a different calculation which I cannot perform inside the function.
The problem is that the Simpson subroutine only accept f(x) to perform the integral and not f(x,y).
After Vladimir suggestions I modified the code.
Below the example:
Program main2
!------------------------------------------------------------------
! Integration of a function using Simpson rule
! with doubling number of intervals
!------------------------------------------------------------------
! to compile:
! gfortran main2.f90 -o simp2
implicit none
double precision r, rb, rmin, rmax, rstep, integral, eps
double precision F_int
integer nint, i, rbins
double precision t
rbins = 4
rmin = 0.0
rmax = 4.0
rstep = (rmax-rmin)/rbins
rb = rmin
eps = 1.0e-8
func = 0.0
t=2.0
do i=1,rbins
call func(rb,t,res)
write(*,*)'r, f(rb) (in main) = ', rb, res
!test = F_int(rb)
!write(*,*)'test F_int (in loop) = ', test
call simpson2(F_int(rb),rmin,rb,eps,integral,nint)
write(*,*)'r, integral = ', rb, integral
rb = rb+rstep
end do
end program main2
subroutine func(x,y,res)
!----------------------------------------
! Real Function
!----------------------------------------
implicit none
double precision res
double precision, intent(in) :: x
double precision y
res = 2.0*x + y
write(*,*)'f(x,y) (in func) = ',res
return
end subroutine func
function F_int(x)
!Function to integrate
implicit none
double precision F_int, res
double precision, intent(in) :: x
double precision y
call func(x,y,res)
F_int = res
end function F_int
Subroutine simpson2(f,a,b,eps,integral,nint)
!==========================================================
! Integration of f(x) on [a,b]
! Method: Simpson rule with doubling number of intervals
! till error = coeff*|I_n - I_2n| < eps
! written by: Alex Godunov (October 2009)
!----------------------------------------------------------
! IN:
! f - Function to integrate (supplied by a user)
! a - Lower limit of integration
! b - Upper limit of integration
! eps - tolerance
! OUT:
! integral - Result of integration
! nint - number of intervals to achieve accuracy
!==========================================================
implicit none
double precision f, a, b, eps, integral
double precision sn, s2n, h, x
integer nint
double precision, parameter :: coeff = 1.0/15.0 ! error estimate coeff
integer, parameter :: nmax=1048576 ! max number of intervals
integer n, i
! evaluate integral for 2 intervals (three points)
h = (b-a)/2.0
sn = (1.0/3.0)*h*(f(a)+4.0*f(a+h)+f(b))
write(*,*)'a, b, h, sn (in simp) = ', a, b, h, sn
! loop over number of intervals (starting from 4 intervals)
n=4
do while (n <= nmax)
s2n = 0.0
h = (b-a)/dfloat(n)
do i=2, n-2, 2
x = a+dfloat(i)*h
s2n = s2n + 2.0*f(x) + 4.0*f(x+h)
end do
s2n = (s2n + f(a) + f(b) + 4.0*f(a+h))*h/3.0
if(coeff*abs(s2n-sn) <= eps) then
integral = s2n + coeff*(s2n-sn)
nint = n
exit
end if
sn = s2n
n = n*2
end do
return
end subroutine simpson2
I think I'm pretty close to the solution but I cannot figure it out...
If I call simpson2(F_int, ..) without putting the argument in F_int I receive this message:
call simpson2(F_int,rmin,rb,eps,integral,nint)
1
Warning: Expected a procedure for argument 'f' at (1)
Any help?
Thanks in advance!
Now you have a code we can work with, good job!
You need to tell the compiler, that F_int is a function. That can be done by
external F_int
but it is much better to learn Fortran 90 and use modules or at least interface blocks.
module my_functions
implicit none
contains
subroutine func(x,y,res)
!----------------------------------------
! Real Function
!----------------------------------------
implicit none
double precision res
double precision, intent(in) :: x
double precision y
res = 2.0*x + y
write(*,*)'f(x,y) (in func) = ',res
return
end subroutine func
function F_int(x)
!Function to integrate
implicit none
double precision F_int, res
double precision, intent(in) :: x
double precision y
call func(x,y,res)
F_int = res
end function F_int
end module
Now you can easily use the module and integrate the function
use my_functions
call simpson2(F_int,rmin,rb,eps,integral,nint)
But you will find that F_int still does not know what y is! It has it's own y with undefined value! You should put y into the module instead so that everyone can see it.
module my_functions
implicit none
double precision :: y
contains
Don't forget to remove all other declarations of y! Both in function F_int and in the main program. Probably it is also better to call it differently.
Don't forget to set the value of y somewhere inside your main loop!