ln ⁡ ( x ) ≈ a x 1 / a − a , which...

Blericker74

Blericker74

Answered

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

Expert

2022-07-10Added 19 answers

For large a we have
x 1 a = e log x a = 1 + log x a + O ( ( log x a ) 2 )
(by Taylor's theorem) which gives
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 would describe this in exactly the opposite way: 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

Expert

2022-07-11Added 7 answers

This is the same thing as asserting 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 ) ;
in other words, the derivative at 0 of a x a is log x.

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