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i would also not like to ask, but if not ask, what should i do?

MaplePrimes Activity

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looking for Invariant matrix

this wiki is correct.


there is a case that we may not find

if power of 2 term has coefficient

such as 5*2^3+7*2^5 etc

is it possible to find?

if further assume these coefficients are prime only, can this be find?


besides above plot,

is there possible to convert above system to

state space form?

i intend to make well use of dynamic library and all functions in it such as checking controllability and observability, frequency plot , step plot etc

and is it possible to control the next output?


yes, all trials have errors.

i use example in simulate function dynamic system library in help file

in_t := Sine(1, 1, 0, 0);

i do not know how to plot, so, i use example directly

is there any other better in_t function to plot system?


in fact the system I posted at the bottom of above post

so you can use the system without knowing the variables

diff(y1(t), t) = 1.052936200*10^5*y1(t)+70106.19000*y2(t)+35169.00000*y3(t)
diff(y2(t), t) = 70106.19000*y1(t)+71031.61000*y2(t)+35511.00000*y3(t)
diff(y3(t), t) = 35169.00000*y1(t)+35511.00000*y2(t)+36100.00000*y3(t)




my mistake missing to subs it

however no graph 

how to know which range can show the graph?


thanks , it's my mistake


i guess swirl graph can be found from it

because it is a chaos system

how to plot this swirl graph from it except straight line?

@Carl Love 

i begin to guess some level of invariants that can not get trancendental functions

because they all have denominator which will be 0 and result in infinity these kind of singularity.

@Carl Love 

FunctionAdvisor(branch_points, g3);
Error, (in FunctionAdvisor) invalid input: generic_Topic["branch_points", "branch points"] expects its 1st argument, SF, to be of type {string, symbol, function}, but received 2*(1+exp(4*x))/(exp(4*x)-1)

i try to start with 1, but still calculating for a long time

g3 := 2*(1+exp(4*x))/(exp(4*x)-1);
a:=eval(diff(g3,x$n)/n!, x=1) assuming n>=0:
tanhx := sum(a*x^n, n=1..infinity):
tanhx2 := subs(x^n=subs(_C1=0, subs(t=n!, g2))*x^n, tanhx):
simplify( series(value(tanhx2),x,10) ); print(evalf(%));



Hi All,

g3 := 2*(1+exp(4*x))/(exp(4*x)-1);
a:=eval(diff(g3,x$n)/n!, x=0) assuming n>=0:
tanhx := Sum(a*x^n, n=0..infinity):
tanhx2 := subs(x^n=subs(_C1=0, subs(t=n!, g2))*x^n, tanhx):
simplify( series(value(tanhx2),x,10) ); print(evalf(%));

i want to show coefficient to confirm whether it can be calculated and plot graph

the reason i calculate it is that it is the i called fifth level of invariant

i use it to find trancendental function

i am automating to genreate large amount of transcendental functions

if you have interest to do this with me, please email me,

actually i have difficulty in guessing differential operator and integration operator for these new transcendental function

@Carl Love 



i find past question

to use numpoints

it can plot faster 

thanks again



a:=eval(diff(tanh(x+1),x$n)/n!, x=0) assuming n>=0;
tanhx := sum(a*x^n, n=0..infinity);
series(tanh(x+1), x=0, 10); evalf(%);
simplify( series(value(tanhx),x,10) ); evalf(%);


i look the expression different so have not checked decimal

after use evalf(%)

the result are correct

but more decimal places , the precision is incorrect

i uploaded here





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