@vv 

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?

@tomleslie 

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?

@tomleslie 

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)

@tomleslie 

o:=1

my mistake missing to subs it

however no graph 

how to know which range can show the graph?

@tomleslie 

thanks , it's my mistake

@tomleslie 

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):
print(tanhx2):
simplify( series(value(tanhx2),x,10) ); print(evalf(%));
 

@vv 

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):
print(tanhx2):
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 

thanks

@vv 

i find past question

to use numpoints

it can plot faster 

thanks again

@vv 

thanks.

precisionwhenmoredecimal.mw

@vv 

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

@vv 

i change to assumming n >= 0  and sum use n=0..infinity but where is wrong?

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);
simplify( series(value(tanhx),x,10) );
 

@vv 

in maple 12, it use over 1GB memory to evaluating, but return above error

it can return a result, which is

a := (eval(diff(tanh(x), `$`(x, n)), x = 0))/factorial(n)
 

i change to use maple 2015, it can compute result very fast and use only 38.7MB memory

it can compute

however, for tanh(x+1) , the result are not equal

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