05/16/2023, 11:13 PM

A typical simple limit for exp(x) is given by

(1 + x/n)^n

for n going to +oo.

How do we improve this and what do we mean by improving ?

exp(x) = exp(x/n)^n = (1 + x/n + ...)^n

So the above limit comes from the taylor series of exp(x).

We can improve this by getting more taylor terms.

Or we can replace n with 2n.

(1 + x/(2n))^(2n)

and simplify.

This also gives a similar limit but with different coefficients.

Notice that getting more taylor terms is better.

In general getting more terms from a series expansion is better than replacing n, although it depends.

Another way to improve is to use other series expansions.

It is also possible to get a limit of the form

f(x,n)^n = 1 + x + x^2 + ...( terms with n and x )

which is in a sense better. It depends on what we want, getting the first terms correct and not caring about the cubic approximation term or getting fast convergeance for all terms but none exact ?

How to achieve this might seem mysterious but it is usually not that hard.

Just combining series expansions and replacements of n.

This is reached for instance like this :

We take a simple pade for exp(x/n)^n

so we get

((2n + x)/(2n - x))^n

= 1 + x + x^2 /2 + (1/6 + 1/(12 n^2) ) x^3 ... ( terms with n and x )

we replace n with 2n

((4n + x)/(4n - x))^(2n)

and then similarly

( ((4n + x) (8n + x)^2)/((4n - x)(8n - x)^2) )^n

= 1 + x + x^2/2 + (1/6 + 5/(384 n^2)) x^3 + ( terms with n and x )

which is also just a second degree but the cubic terms is better !

We can go further

c = solution of 1 + 2x - 2 x^3 = 0.

[( (n - cx) (n + x)^(c + 1/2) )/( (n + cx) (n - x)^(c + 1/2) ) ]^n = 1 + x + x^2/2 + x^3/6 + x^4/24 + (1/120 - c^2 / (5 n^4) ) x^5 + ...

Now most functions are more complicated than exp.

And c is irrational.

Also I want the approximation to satisfy functional equations like exp or almost.

so I want

f(-x) = 1/f(x)

notice g(x)/g(-x) satisfies that.

We also want

f(x) = f(x)/f(-x) such that f(-x) ( or f(x) ) has no zero on the real line.

Therefore i choose the following approximation (based on pade of equal degree in denom and nom)

exp(x) === (1 + x/2 + x^2/9)/(1 - x/2 + x^2/9)

notice we have f(-x) = 1/f(x)

and g(x) = 1 + x/2 + x^2/9 has no zeros on the real line.

(min = 7/16 at x = - 9/4 )

So finally ( x ->> x/n , exp(x) = exp(x/n)^n )

exp(x) = [(n^2 + x/2 n + x^2/9)/(n^2 - x/2 n + x^2/9)]^n

Now since you guys have certainly thought about this , I wonder if you have other ideas ?

I know there is an iteration too, and ofcourse series and such but I am looking for simple iterations.

Also this method does not use too much data.

Notice this works for other bases too.

regards

tommy1729

(1 + x/n)^n

for n going to +oo.

How do we improve this and what do we mean by improving ?

exp(x) = exp(x/n)^n = (1 + x/n + ...)^n

So the above limit comes from the taylor series of exp(x).

We can improve this by getting more taylor terms.

Or we can replace n with 2n.

(1 + x/(2n))^(2n)

and simplify.

This also gives a similar limit but with different coefficients.

Notice that getting more taylor terms is better.

In general getting more terms from a series expansion is better than replacing n, although it depends.

Another way to improve is to use other series expansions.

It is also possible to get a limit of the form

f(x,n)^n = 1 + x + x^2 + ...( terms with n and x )

which is in a sense better. It depends on what we want, getting the first terms correct and not caring about the cubic approximation term or getting fast convergeance for all terms but none exact ?

How to achieve this might seem mysterious but it is usually not that hard.

Just combining series expansions and replacements of n.

This is reached for instance like this :

We take a simple pade for exp(x/n)^n

so we get

((2n + x)/(2n - x))^n

= 1 + x + x^2 /2 + (1/6 + 1/(12 n^2) ) x^3 ... ( terms with n and x )

we replace n with 2n

((4n + x)/(4n - x))^(2n)

and then similarly

( ((4n + x) (8n + x)^2)/((4n - x)(8n - x)^2) )^n

= 1 + x + x^2/2 + (1/6 + 5/(384 n^2)) x^3 + ( terms with n and x )

which is also just a second degree but the cubic terms is better !

We can go further

c = solution of 1 + 2x - 2 x^3 = 0.

[( (n - cx) (n + x)^(c + 1/2) )/( (n + cx) (n - x)^(c + 1/2) ) ]^n = 1 + x + x^2/2 + x^3/6 + x^4/24 + (1/120 - c^2 / (5 n^4) ) x^5 + ...

Now most functions are more complicated than exp.

And c is irrational.

Also I want the approximation to satisfy functional equations like exp or almost.

so I want

f(-x) = 1/f(x)

notice g(x)/g(-x) satisfies that.

We also want

f(x) = f(x)/f(-x) such that f(-x) ( or f(x) ) has no zero on the real line.

Therefore i choose the following approximation (based on pade of equal degree in denom and nom)

exp(x) === (1 + x/2 + x^2/9)/(1 - x/2 + x^2/9)

notice we have f(-x) = 1/f(x)

and g(x) = 1 + x/2 + x^2/9 has no zeros on the real line.

(min = 7/16 at x = - 9/4 )

So finally ( x ->> x/n , exp(x) = exp(x/n)^n )

exp(x) = [(n^2 + x/2 n + x^2/9)/(n^2 - x/2 n + x^2/9)]^n

Now since you guys have certainly thought about this , I wonder if you have other ideas ?

I know there is an iteration too, and ofcourse series and such but I am looking for simple iterations.

Also this method does not use too much data.

Notice this works for other bases too.

regards

tommy1729