08/02/2014, 11:48 PM

Another try

tommy-sheldon method 1.0

f(x) is the half-iterate from the sinh method.

Start from a0 , a1 then solve a2 , a3 , ... with

a_(n-1) x^(n-1) + a_n x^n = f(x)

a_n x^n ( 1 + a_(n-1)/a_n 1/x ) = f(x)

ln(a_n) + n ln(x) + ln( 1 + a_(n-1)/a_n 1/x ) = ln(f(x))

ln(a_n) + n x + ln( 1 + a_(n-1) / a_n exp(-x) ) = f(x)

a_(n-1)/a_n = exp(C(n) a_n)

ln(a_n) + n x + ln( 1 + exp(-x + C(n) a_n) ) = f(x)

And now there are a few ways to continue.

Not sure which is the best at the moment.

But a good numerical method must clearly exist.

I considered how to test how good our equations are.

seems quite simple actually , we test a function from which we know the Taylor series.

For instance :

solve a_n x^n = exp(x)

vs

solve a_(n-1) x^(n-1) + a_n x^n = exp(x)

and compare both to 1/n!.

Btw this idea of estimating a Taylor series without much knowledge of taking derivatives , hence without using taylor's theorem seems very appealing to me.

In number theory this is also considered a bit , and maybe it should be considered more.

For instance when we have g(z) = g0 + g1 x + g2 x^2 + ...

and g(z)^7 = g0(7) + g1(7) x + ...

and g_n is either 1 or 0 then g(z)^7 might have

g_(n-1)(7) >= g_n(7) >= g(n+1)(7)

for all n.

This shows a connection to additive number theory.

HOWEVER g(z) is not entire.

THEREFORE usually we need some tricks to make the ideas work.

AND to make g*_(n-1)(7) >= g*_n(7) >= g*(n+1)(7) for the appropriate associated g*.

I know this is all a bit vague.

But it seems also to be correct.

By having an a_n depending on an a_(n-1) it seems every terms adapts depending on wheither the previous was estimated to high or too low.

Some optimism (OR ALOT) suggests then that these a_n are not Always an overestimate or not Always an underestimate but make a wave around the correct values.

I assume then that by tommy-sheldon method 1.0

we get TS1(x) = a_0 + a1 x + ...

TS1(x)/exp^[0.5](x) = O(ln(x)^2)

But It seems optimistic.

Maybe if it also depends on a_(n-2) and a_(n-3).

Oh btw ln(1+exp(z)) has an intresting Taylor series expansion.

Maybe that should be used.

regards

tommy1729

tommy-sheldon method 1.0

f(x) is the half-iterate from the sinh method.

Start from a0 , a1 then solve a2 , a3 , ... with

a_(n-1) x^(n-1) + a_n x^n = f(x)

a_n x^n ( 1 + a_(n-1)/a_n 1/x ) = f(x)

ln(a_n) + n ln(x) + ln( 1 + a_(n-1)/a_n 1/x ) = ln(f(x))

ln(a_n) + n x + ln( 1 + a_(n-1) / a_n exp(-x) ) = f(x)

a_(n-1)/a_n = exp(C(n) a_n)

ln(a_n) + n x + ln( 1 + exp(-x + C(n) a_n) ) = f(x)

And now there are a few ways to continue.

Not sure which is the best at the moment.

But a good numerical method must clearly exist.

I considered how to test how good our equations are.

seems quite simple actually , we test a function from which we know the Taylor series.

For instance :

solve a_n x^n = exp(x)

vs

solve a_(n-1) x^(n-1) + a_n x^n = exp(x)

and compare both to 1/n!.

Btw this idea of estimating a Taylor series without much knowledge of taking derivatives , hence without using taylor's theorem seems very appealing to me.

In number theory this is also considered a bit , and maybe it should be considered more.

For instance when we have g(z) = g0 + g1 x + g2 x^2 + ...

and g(z)^7 = g0(7) + g1(7) x + ...

and g_n is either 1 or 0 then g(z)^7 might have

g_(n-1)(7) >= g_n(7) >= g(n+1)(7)

for all n.

This shows a connection to additive number theory.

HOWEVER g(z) is not entire.

THEREFORE usually we need some tricks to make the ideas work.

AND to make g*_(n-1)(7) >= g*_n(7) >= g*(n+1)(7) for the appropriate associated g*.

I know this is all a bit vague.

But it seems also to be correct.

By having an a_n depending on an a_(n-1) it seems every terms adapts depending on wheither the previous was estimated to high or too low.

Some optimism (OR ALOT) suggests then that these a_n are not Always an overestimate or not Always an underestimate but make a wave around the correct values.

I assume then that by tommy-sheldon method 1.0

we get TS1(x) = a_0 + a1 x + ...

TS1(x)/exp^[0.5](x) = O(ln(x)^2)

But It seems optimistic.

Maybe if it also depends on a_(n-2) and a_(n-3).

Oh btw ln(1+exp(z)) has an intresting Taylor series expansion.

Maybe that should be used.

regards

tommy1729