clovnerie0q

2022-10-03

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): $\underset{r\to 0}{lim}(1+rx+o(r){)}^{1/r}={e}^{x}$

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 somebody point me towards a suitable proof of (a) and (b) above please.

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): $\underset{r\to 0}{lim}(1+rx+o(r){)}^{1/r}={e}^{x}$

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 somebody point me towards a suitable proof of (a) and (b) above please.

Mckenna Friedman

Beginner2022-10-04Added 10 answers

a) Use the definition of ${a}^{r}$ and the well known Taylor series for ${e}^{x}$

${a}^{r}=({e}^{\mathrm{log}a}{)}^{r}={e}^{r\mathrm{log}a}=1+r\mathrm{log}a+{r}^{2}(\mathrm{log}a{)}^{2}\dots $

b) Hint: Do you remember this formula?

$\underset{n\to \mathrm{\infty}}{lim}(1+\frac{x}{n}{)}^{n}={e}^{x}$

${a}^{r}=({e}^{\mathrm{log}a}{)}^{r}={e}^{r\mathrm{log}a}=1+r\mathrm{log}a+{r}^{2}(\mathrm{log}a{)}^{2}\dots $

b) Hint: Do you remember this formula?

$\underset{n\to \mathrm{\infty}}{lim}(1+\frac{x}{n}{)}^{n}={e}^{x}$

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