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Seldom to ask question after retired math hobby Just waiting for beauty who born in 1994 And waited for her email to mavio@protonmail.com What is the difference in ownership among different universe?

## @vv when try this , return errora:=...

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...

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 equatio...

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(Vari...

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 f...

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 For exampleif   f := 1/2*d...

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?

then you must understand

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

## @tomleslie  sorry for previous and...

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 funct...

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 ...

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 t...

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 res...

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...

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

## @Kitonum Eigenvectors(A); return 1/...

thanks, it is correct.

## @Kitonum  A := Matrix([[-2,1,1],[0...

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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