Time for tommy's Q9 method.
The term Q refers to q-analogue as will be clear soon.
The 9 refers to post 9.
In post 9 we tried to find a good fake(f(x)) by using
a_n x^n < f(x).
We arrived at a solution that had the property
a_0 >= a_1 >= a_2 >= a_3 >= ... >= 0 (*)
We can use this property to find a better method.
a_0 + a_1 x + a_2 x^2 + ... + a_n x^n < f(x)
Using (*)
a_n + a_n x + a_n x^2 + ... a_n x^n < f(x)
Simplify
a_n (x^(n+1) - 1) / (x-1) < f(x)
a_n (x^(n+1))/(x-1) < f(x) + a_n/(x-1)
For x large we can ignore the a_n/(x-1) term on the RHS , that might give a worse fake for small x , but has almost no effect on the large x.
a_n (x^(n+1))/(x-1) < f(x)
a_n x^n x/(x-1) < f(x)
a_n x^n < f(x) (x-1)/x
ln(a_n) + n ln (x) < ln(f(x)) + ln(x-1) - ln(x)
ln(a_n) < ln(f(x)) + ln(x-1) - (n+1) ln(x)
ln(a_n) < Min( ln(f(x)) + ln(x-1) - (n+1) ln(x) )
This is an improvement.
Celebrate.
regards
tommy1729
The term Q refers to q-analogue as will be clear soon.
The 9 refers to post 9.
In post 9 we tried to find a good fake(f(x)) by using
a_n x^n < f(x).
We arrived at a solution that had the property
a_0 >= a_1 >= a_2 >= a_3 >= ... >= 0 (*)
We can use this property to find a better method.
a_0 + a_1 x + a_2 x^2 + ... + a_n x^n < f(x)
Using (*)
a_n + a_n x + a_n x^2 + ... a_n x^n < f(x)
Simplify
a_n (x^(n+1) - 1) / (x-1) < f(x)
a_n (x^(n+1))/(x-1) < f(x) + a_n/(x-1)
For x large we can ignore the a_n/(x-1) term on the RHS , that might give a worse fake for small x , but has almost no effect on the large x.
a_n (x^(n+1))/(x-1) < f(x)
a_n x^n x/(x-1) < f(x)
a_n x^n < f(x) (x-1)/x
ln(a_n) + n ln (x) < ln(f(x)) + ln(x-1) - ln(x)
ln(a_n) < ln(f(x)) + ln(x-1) - (n+1) ln(x)
ln(a_n) < Min( ln(f(x)) + ln(x-1) - (n+1) ln(x) )
This is an improvement.
Celebrate.
regards
tommy1729