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Searching for an asymptotic to exp[0.5]
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
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Messages In This Thread
RE: Searching for an asymptotic to exp[0.5] - by tommy1729 - 09/19/2014, 12:23 PM

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