To obtain the value of y in more precision, This problem has result for n<=922, but I want to get the result for n>922. looking for another way takes less time and more degree of precision similar to sym or vpa, this takes many days for my program

example:

x=0.2147
n=?;
y=vpa((x^(n/2)+x^(-n/2)/2)^(-2/n))

x=sym(0.2147,'r');
n=950;
y=vpa((x^(n/2)+x^(-n/2)/2)^(-2/n))
=~0.2150

[Walter Roberson]

I do not understand? When I try I get a symbolic result in a fraction of a second.

[Hussein]

The program is something else but I wanna be sure if this is true that I am doing during my program to get the accurate precision or there is some else commands that do better and faster rather than sym or vpa. Thank you

[Walter Roberson]

sym / vpa is likely to be as accurate as you might hope, especially if you set Digits to 16.

Remember to

format long g

before running the computation, in order to see the result in fuller precision.