## 136 Reputation

14 years, 25 days

## oooh yes i see, thanks a...

oooh yes i see, thanks a lot Alec you an ancyclopedies of maple,

it is possible to know how they are construt the mathematics tools ( taylor, convert ,numapprox,......)

thanks

## but i want to collect all...

but i want to collect all the coefficients of an expansion of the form ( for example)

T=x^3+4x^2+x+6+(2/x)+(1/x^2)

so how i can do this

## but i want to collect all...

but i want to collect all the coefficients of an expansion of the form ( for example)

T=x^3+4x^2+x+6+(2/x)+(1/x^2)

so how i can do this

## but i want to collect all...

but i want to collect all the coefficients of an expansion of the form ( for example)

T=x^3+4x^2+x+6+(2/x)+(1/x^2)

so how i can do this

## BUT IT IS AT THE BEGINIG A...

BUT IT IS AT THE BEGINIG A TATIONAL FUNCTION WITH DENOMENATOR 1. SO WE CAN T OBTAIN THE [4;3] PA for this i suppose it eturn an error

## i know that we can have the...

i know that we can have the expansion of f at infinity if it has an expansion at o when we change the variable x by 1/x ,and it is the case the function in the program has o as limit at infinity

## why is not too hard to see...

why is not too hard to see why?

## oh thanks a lot how do you...

oh thanks a lot how do you find this

but our prof asked us to program the metrhod of pade approximant as it is defined but ,y program is just for one exa,ple and i want to do it for any function and anny degree

i a, not verry well in english excuse me

thanks

## BUT I DONT WANT TO USE...

BUT I DONT WANT TO USE NUMAPPROX I WANT TO CONSTRUCT MY PROPRE PROCEDUR

## IS THERE ANY ONE ABLE TO...

IS THERE ANY ONE ABLE TO CONSTRUCT WITH ME THIS PROCEDURE

## THIS IS AN ASYMPTOTIQUE...

THIS IS AN ASYMPTOTIQUE DEVELLOPEMENT WE MUST HAVE A+B1/x+C1/X2+D1/X3+.........

## BECAUSE THE FIRST TERME IS...

BECAUSE THE FIRST TERME IS ln(x)/x and in the devellope,ent we must have 0+°°X+§§X2+...... DO YOU UNDERSTAND WHAT I MEAN

## is there any answers to  my...

is there any answers to  my question

> hello

this is a program to calculate an [4.2]pade approximant to the function exp(x), i need to do a procedure for this. for example pade:=proc(f,..), ( if i want to calculate an other pade approximant i put in pade my function and the point and the ordre of denominateur and numerator.how i can do this?

> f := exp(x);
>   M := 1 + add(q[k]*(x)^k, k=1..2);
>   T := convert(taylor(M*f - L,x, 8),polynom);
>  H:=evalf(%) ;
f := exp(x)
2         3         4
L := p[0] + p[1] x + p[2] x  + p[3] x  + p[4] x
2
M := 1 + q[1] x + q[2] x
/1                     \  2
T := 1 - p[0] + (1 + q[1] - p[1]) x + |- + q[1] + q[2] - p[2]| x
\2                     /

/               1   1     \  3   /1    1        1            \  4
+ |-p[3] + q[2] + - + - q[1]| x  + |-- + - q[1] + - q[2] - p[4]| x
\               6   2     /      \24   6        2            /

/1         1    1      \  5   /1          1     1      \  6
+ |- q[2] + --- + -- q[1]| x  + |-- q[2] + --- + --- q[1]| x
\6        120   24     /      \24        720   120     /

/ 1      1          1      \  7
+ |---- + --- q[1] + --- q[2]| x
\5040   720        120     /
H := 1. - 1. p[0] + (1. + q[1] - 1. p[1]) x

2
+ (0.5000000000 + q[1] + q[2] - 1. p[2]) x

3
+ (-1. p[3] + q[2] + 0.1666666667 + 0.5000000000 q[1]) x

4
+ (0.04166666667 + 0.1666666667 q[1] + 0.5000000000 q[2] - 1. p[4]) x

5
+ (0.1666666667 q[2] + 0.008333333333 + 0.04166666667 q[1]) x

6
+ (0.04166666667 q[2] + 0.001388888889 + 0.008333333333 q[1]) x

7
+ (0.0001984126984 + 0.001388888889 q[1] + 0.008333333333 q[2]) x
> eqs:= {seq(coeftayl(H,x=0,j),j=0..6)};
>   S:=fsolve(eqs);