Why is the growth of exps & logs so fast & slow, respectively?

[mcastillo356]

TL;DR

I don't understand the theorem that proves the question of the thread's title.

Hi, PF, there goes the theorem, questions, and attempt
THEOREM 4 If $x>0$, then $\ln x\leq{x-1}$
PROOF Let $g(x)=\ln x-(x-1)$ for $x>0$. Then $g(1)=0$ and

$g'(x)=\displaystyle\frac{1}{x}-1\quad\begin{cases}>0&\mbox{if}\,0<x<1\\ <0&\mbox{if}\,x>1\end{cases}$ (...). Thus, $g(x)\leq{g(1)=0}$, $\forall {x>0}$ and $\ln x\leq{x-1}$ for all such $x$

Question: The proof, why implies the quick growth of $e^x$ and the steady growing of $\ln x$?

Attempt: $g$ grows in $x\in{(0,1)}$ and decreases in $x\in{(1,\infty)}$ ($g'(x)>0\,\forall{(0<x<1)}$ and $g'(x)<0\,\forall{(1<x<\infty)}$

Greetings!



I will post without preview, apologizes. If it makes no sense, I will repeat promptly

 

[jbriggs444]

Question: The proof, why implies the quick growth of $e^x$ and the steady growing of $\ln x$?

Let us take the latter question first: "How does the proof imply the steady growth of $\ln x$?

What do you even mean by the "steady growth" of $\ln x$?

Certainly the proof examines the difference between the first derivative of $\ln x$ and the first derivative of $x-1$. It concludes that the "growth rate" (first derivative) of $\ln x$ is less than the growth rate of $x-1$.

By itself, this does not assure us that $\ln x$ even has a growth rate. But we can easily see that indeed, $\ln x$ is strictly monotone increasing.

I suppose that "strictly monotone increasing" with a growth rate that is always less than 1 might count as "steady growth" to you.

Personally, I'd use the adjective "steady growth" for something with a growth rate that converged on a positive constant. $\ln x$ does not qualify.We are in a position now to attack the other question: "How does the proof imply the quick growth of $\ln x$?

That answer is shorter. It does not.

Roughly speaking, the proof of the first part puts a bound on the graph of $\ln x$. It bounds it below a 45 degree line, the graph of $f(x) = x-1$. The graph of $e^x$ will be the mirror image, obtained by swapping the x and y axes. The proof puts a lower bound on $e^x$ at the mirror image 45 degree line, the graph of $f(x) = x + 1$.

There are plenty of "steadily growing" functions with such a lower bound.

 

[scottdave]

It depends on where on the x-axis whether it is increasing rapidly or slowly. I've forgotten a lot of Spanish, but I think it is asking about when x is large.

 

[apostolosdt]

You might like to check Courant’s “Differential and Integral Calculus”, Vol. I, Rev. Ed., about the order of magnitude of functions, pp. 189.

 

[mcastillo356]

Hi, PF, @jbriggs444, @scottdave, @apostolosdt.

Let us take the latter question first: "How does the proof imply the steady growth of $\ln x$?

What do you even mean by the "steady growth" of $\ln x$?

It's been an own's and free translation, @jbriggs444. I agree with your statement:

I suppose that "strictly monotone increasing" with a growth rate that is always less than 1 might count as "steady growth" to you.

This is, for all $x$ and $y$ such that $x\leq y\rightarrow{f(x)\leq{f(y)}}$. One question: how could I get some info on growth rate?

Now, the question I do not solve is that $e^x$ increases more slowly than any positive power of $x$ (no matter how large the power), while $\ln x$ increases more slowly than any positive power of $x$ (no matter how small the power). The fact that $g(x)\leq{g(1)=0}$ for all $x>0$ and $\ln\leq{x-1}$ for all such $x$ is bounds, but not a proof of the singular growing properties of $\ln x$ and $e^x$.

Now I think that it is all about introducing exponential growth and decay models, a few paragraphs forward.

Greetings!

 

[jbriggs444]

This is, for all $x$ and $y$ such that $x\leq y\rightarrow{f(x)\leq{f(y)}}$. One question: how could I get some info on growth rate?

The terminology I learned (in the 70's) for that attribute is "monotone increasing".

If one strengthens the inequality to $x \lt y \rightarrow f(x) \lt f(y)$ then the term is "strictly monotone increasing".

So a constant function such as $f(x) = 1$ would be "monotone increasing" but not "strictly monotone increasing". A function such as $f(x) = x^3$ would be both monotone increasing and strictly monotone increasing despite the inflection at $x=0$.

Now, the question I do not solve is that $e^x$ increases more slowly than any positive power of $x$ (no matter how large the power), while $\ln x$ increases more slowly than any positive power of $x$ (no matter how small the power).

Here, you appear not to be worried about any constant multiplier or fixed offset. So, for instance, you would say that $x^2$ increases more rapidly than $10x + 15$.

That sounds like an application for "big O notation".