Having trouble understanding why the r-th mean tends to the geometric mean as r tends to zero I am having trouble understanding the proof of Theorem 3 in "Inequalities" by Hardy, Littlewood and Pólya. This theorem states that the r-th mean approaches the geometric mean as r approaches zero. I have seen the following post which makes things a little clearer (albeit using o(r) instead of O(r^2)): Why is the 0th power mean defined to be the geometric mean? However, I still cannot determine why: (a): a^r=1+r.log(a)+o(r) as r tends to zero and (b): lim_(r->0)(1+rx+o(r))^(1/r)=e^x. I have a pretty solid grasp of limits, as well as the log and exp functions, but I have never really been taught anything substantial on big/little-O notation, in particular as the variable approaches zero. Could some

clovnerie0q 2022-10-03 Answered
Having trouble understanding why the r-th mean tends to the geometric mean as r tends to zero
I am having trouble understanding the proof of Theorem 3 in "Inequalities" by Hardy, Littlewood and Pólya. This theorem states that the r-th mean approaches the geometric mean as r approaches zero.
I have seen the following post which makes things a little clearer (albeit using o ( r ) instead of O ( r 2 ):
Why is the 0th power mean defined to be the geometric mean?
However, I still cannot determine why:
(a): a r = 1 + r . l o g ( a ) + o ( r ) as r tends to zero,
and
(b): lim r 0 ( 1 + r x + o ( r ) ) 1 / r = e x
I have a pretty solid grasp of limits, as well as the log and exp functions, but I have never really been taught anything substantial on big/little-O notation, in particular as the variable approaches zero. Could somebody point me towards a suitable proof of (a) and (b) above please.
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Answers (1)

Mckenna Friedman
Answered 2022-10-04 Author has 10 answers
a) Use the definition of a r and the well known Taylor series for e x
a r = ( e log a ) r = e r log a = 1 + r log a + r 2 ( log a ) 2
b) Hint: Do you remember this formula?
lim n ( 1 + x n ) n = e x
Did you like this example?
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