ComputerUser

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@vv 

when try this , return error in maple 12

a:=eval(diff(tanh(x+1),x$n)/n!, x=0) assuming n>=1;
tanhx := sum(a*x^n, n=1..infinity);

Error, (in RealRange) too many levels of recursion

and

if assume n >= 0, will it be incorrect?

@tomleslie 

moreover, I would like to know the patch about

how to make one dimensional data

with parametrization to 3 dimensional ?

If from sum of angles whether >,<,= Pi

 

@tomleslie 

yes it lack of equations

do you know the solution directly

for sum of triangle = Pi

And I would like to know solution for < Pi too

any other way to design this patch?

or how to find this patch?

@JohnS 

 

restart:
with(VariationalCalculus):
KineticEnergy := 1/2*m*diff(x(t), t)^2;
PotentialEnergy := (1/R^2)^2;
Action := KineticEnergy - PotentialEnergy;
EulerLagrange(Action, t, [x(t)]);
m := Weierstrass(Action, t, x(t), p);

 

EulerLagrange output two result, el := {-1/R^4, -x(t)/R^4 = K[1]}

there is no descriptions about these in help file in maple 12

which one is for dsolve?

moreover, Weierstrass do not return tangent for sphere case no matter i set kinetic energy zero or not.

where is the problem when calculate slope for triangle sum of inner angles = 180?

restart:
with(VariationalCalculus):
KineticEnergy := 0;
PotentialEnergy := subs(x=x(t),int((1/R^2)^2, x));
Action := KineticEnergy - PotentialEnergy;
EulerLagrange(Action, t, [x(t)]);
m := Weierstrass(Action, t, x(t), p);
 

 

 

@tomleslie 

i do not know how to formulate in mathematically way

i find it in help file

with(VariationalCalculus)

dsolve(Weierstrass(f, t, y(t), p) = tan(p), f);

but this can not be solved, is it impossible to solve for function f?

@vv 

updated: it is defintion of Weierstrass E 

Weierstrass E  function = tan(p) 

how to dsolve?

For example

if   f := 1/2*diff(x,t)^2+x*diff(x,t)+x+diff(x,t);
using above formula will be
f - subs(diff(x,t)=p, f) - (diff(x,t) - p)*(diff(subs(diff(x,t)=p, f), p));

so back to question

f - subs(diff(x,t)=p, f) - (diff(x,t) - p)*(diff(subs(diff(x,t)=p, f), p)) = tan(t)

how to dsolve to find back f?

@tomleslie 

please read my new question

then you must understand

http://www.mapleprimes.com/questions/222420-How-To-Calculate-This-New-Integration?sq=222420

please help to continue answer this question

@tomleslie 

sorry for previous and previous reply,

i make a mistake, and fix now

i find this operator, but can not act on differential

f := diff(g(x),x);
limit(diff((subs(x=h*(-(ln(-1-ln(exp(x)))-ln(-ln(exp(x)))+Ei(1, -1-ln(exp(x)))-Ei(1, -ln(exp(x))))/(exp(1)-2)), f)-f),h), h=0);
Error, (in diff/diff) invalid input: diff received -h*(ln(-1-ln(exp(x)))-ln(-ln(exp(x)))+Ei(1, -1-ln(exp(x)))-Ei(1, -ln(exp(x))))/(exp(1)-2), which is not valid for its 2nd argument
 
 
restart:
f := exp(x);
limit(diff((subs(x=h*(-(ln(-1-ln(exp(x)))-ln(-ln(exp(x)))+Ei(1, -1-ln(exp(x)))-Ei(1, -ln(exp(x))))/(exp(1)-2)), f)-f),h), h=0);
f := exp(x)^2;
limit(diff((subs(x=h*(-(ln(-1-ln(exp(x)))-ln(-ln(exp(x)))+Ei(1, -1-ln(exp(x)))-Ei(1, -ln(exp(x))))/(exp(1)-2)), f)-f),h), h=0);
f := exp(x)^3;
limit(diff((subs(x=h*(-(ln(-1-ln(exp(x)))-ln(-ln(exp(x)))+Ei(1, -1-ln(exp(x)))-Ei(1, -ln(exp(x))))/(exp(1)-2)), f)-f),h), h=0);
f := exp(x)^x;
limit(diff((subs(x=h*(-(ln(-1-ln(exp(x)))-ln(-ln(exp(x)))+Ei(1, -1-ln(exp(x)))-Ei(1, -ln(exp(x))))/(exp(1)-2)), f)-f),h), h=0);
f := exp(x^2);
limit(diff((subs(x=h*(-(ln(-1-ln(exp(x)))-ln(-ln(exp(x)))+Ei(1, -1-ln(exp(x)))-Ei(1, -ln(exp(x))))/(exp(1)-2)), f)-f),h), h=0);
f := x;
limit(diff((subs(x=h*(-(ln(-1-ln(exp(x)))-ln(-ln(exp(x)))+Ei(1, -1-ln(exp(x)))-Ei(1, -ln(exp(x))))/(exp(1)-2)), f)-f),h), h=0);
f := x^2;
limit(diff((subs(x=h*(-(ln(-1-ln(exp(x)))-ln(-ln(exp(x)))+Ei(1, -1-ln(exp(x)))-Ei(1, -ln(exp(x))))/(exp(1)-2)), f)-f),h), h=0);
f := x^3;
limit(diff((subs(x=h*(-(ln(-1-ln(exp(x)))-ln(-ln(exp(x)))+Ei(1, -1-ln(exp(x)))-Ei(1, -ln(exp(x))))/(exp(1)-2)), f)-f),h), h=0);
 

@tomleslie 

i let my defined function f a long equation including Ei.... just like exp(x)

transcendental 

then I would like to find an operator 

like diff(f,x), but this operator should be like diff(exp(x),x) = exp(x)

means finding an operator act on f equal to f itself

@tomleslie 

moreover, when i try and error , i discover q can be found in another form

restart:
f := -ln(-1-ln(exp(x)))+ln(-ln(exp(x)))-Ei(1, -1-ln(exp(x)))+Ei(1, -ln(exp(x))):
solve(limit(diff((subs(x=h*q, f)-f),h), h=0)=f,q);
 
but when i do for diff(g(x),x), it can not be calculated

f := diff(g(x),x);
limit(diff((subs(x=h/(exp(1)-2), f)-f),h), h=0);
limit(diff(Diff(g(x), h/(exp(1)-2))-(diff(g(x), x)),h), h=0);
 
 
restart:
f := exp(x);
limit(diff((subs(x=h/(exp(1)-2), f)-f),h), h=0);
f := exp(x)^2;
limit(diff((subs(x=h/(exp(1)-2), f)-f),h), h=0);
f := exp(x)^3;
limit(diff((subs(x=h/(exp(1)-2), f)-f),h), h=0);
f := exp(x)^x;
limit(diff((subs(x=h/(exp(1)-2), f)-f),h), h=0);
f := exp(x^2);
limit(diff((subs(x=h/(exp(1)-2), f)-f),h), h=0);
f := x;
limit(diff((subs(x=h/(exp(1)-2), f)-f),h), h=0);
f := x^2;
limit(diff((subs(x=h/(exp(1)-2), f)-f),h), h=0);
f := x^3;
limit(diff((subs(x=h/(exp(1)-2), f)-f),h), h=0);

@tomleslie 

you are correct,

is there any better method to guess this operator?

and if ignore this,

can the final line of code output without error if only for trial and error

 

@vv 

i Re the eigenvector

and result in

sys1 := -.736349402144656384+(.8336281837*1.000000000^po1+0.6583036207e-4*.5896103837^po1+.8336281837*1.000000000^po2+0.6583036207e-4*.5896103837^po2+.735533633151605248*Resid);
sys2 := .326676717828940144-(-.1665034762*1.000000000^po1-(3.399029877*10^(-17)*I)*1.000000000^po1+0.6585861689e-4*.5896103837^po1-.1665034762*1.000000000^po2-(3.399029877*10^(-17)*I)*1.000000000^po2+0.6585861689e-4*.5896103837^po2+.325144093024965720*Resid);
sys3 := .590327283775080036-(-4.793186903*10^(-8)*1.000000000^po1-(1.417087567*10^(-20)*I)*1.000000000^po1+.5896103554*.5896103837^po1-4.793186903*10^(-8)*1.000000000^po2-(1.417087567*10^(-20)*I)*1.000000000^po2+.5896103554*.5896103837^po2+.589610307487437146*Resid);

sys1 := -.736349402144656384+(.8336281837+0.6583036207e-4*.5896103837^po1+.8336281837+0.6583036207e-4*.5896103837^po2+.735533633151605248*Resid);
sys2 := .326676717828940144-(-.1665034762-(3.399029877*10^(-17)*I)+0.6585861689e-4*.5896103837^po1-.1665034762-(3.399029877*10^(-17)*I)+0.6585861689e-4*.5896103837^po2+.325144093024965720*Resid);
sys3 := .590327283775080036-(-4.793186903*10^(-8)-(1.417087567*10^(-20)*I)+.5896103554*.5896103837^po1-4.793186903*10^(-8)-(1.417087567*10^(-20)*I)+.5896103554*.5896103837^po2+.589610307487437146*Resid);
Minimize(sys1,{sys2,sys3},assume = nonnegative);
NLPSolve(sys1,{sys2,sys3},assume = nonnegative);
 

but can not solve even if no complex

@vv 

can Re((-0.99)^po1) be solved by adding Re() ?

@Kitonum 

thanks, it is correct.

 

@Kitonum 

A := Matrix([[-2,1,1],[0,2,0],[-4,1,3]]);

A:=[<-2,1,1>, <0,2,0>, <-4,1,3>];
S1 := LinearAlgebra:-Basis(A);
 

A:=[<-2,0,-4>, <1,2,1>, <1,0,3>];
S1 := LinearAlgebra:-Basis(A);

and

A:=`<|>`(`<,>`(-2, 0, -4), `<,>`(1, 2, 1), `<,>`(1, 0, 3));
S1 := LinearAlgebra:-NullSpace(A);

what is wrong with maple 12?


 

but output are not [1,4,0],[1,0,4],[1,0,1]
 

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