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Blericker74

Blericker74

Answered question

2022-07-09

ln ( x ) a x 1 / a a , which is good for large value of a. Where does it come from?

Answer & Explanation

gozaderaradiox5

gozaderaradiox5

Beginner2022-07-10Added 19 answers

For large a we have
x 1 a = e log  x a = 1 + log  x a + O ( ( log  x a ) 2 ) 
(According to Taylor's Theorem)
a x 1 a  a = log  x + O ( ( log  x ) 2 a ) 
so the approximation is accurate once a is large relative to ( log  x ) 2 . By the way, I'd put it exactly the other way around: that log  x is a good approximation to a x 1 a  a! It's not as if a x 1 a  a is particularly easy to compute for large a.

Raul Walker

Raul Walker

Beginner2022-07-11Added 7 answers

This is equivalent to saying that, if a > 0 and a is very small, then
log  ( x )  x a  1 a ,
which is true, since
lim a  0 + x a  1 a  = lim a  0 + e a log  x  1 a   = log  ( x ) ; 
Alternatively put, the derivative at 0 of a  x a is log  x.

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