Is there a relation between fractional power and logarithm functions?

Question

Something that has always bothered me is that there's no way to get $x^{-1}$ by differentiating $x^a$ for some $a$, even though all other negative powers of $x$ can be achieved by differentiating some power of $x$. So I began to mess around with equations trying to find some sort of connection.

I realized that $$\lim_{a\rightarrow\infty}\left(\frac{d}{dx}\left( ax^{\frac{1}{a}}\right)\right) = x^{-1}$$

And, upon further delving that $$\lim_{a\rightarrow\infty} ax^{\frac{1}{a}} - a = \ln(x)$$

Is there a some sort of relation here? I understand that logarithms are intrinsically linked to powers, but I don't understand why a very small power is equal to a scaled, offset logarithm.

Answer 1

Consider the function

$$f_p(x)=\frac{x^p-1}p.$$

It has the derivative

$$f'_p(x)=x^{p-1}$$

and is such that $f_p(1)=0$ and $f_p(0)=-\dfrac1p.$

Now if you let $p$ tend to $0$, you have that

$$\lim_{p\to0}f_p(x)=\ln(x)$$ and $$\lim_{p\to0}f'_p(x)=\frac1x.$$

Below, a pencil of curves for various positive and negative $p$.

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Also consider the inverse of this function,

$$g_p(x)=(1+px)^{1/p}$$ and see the connection with the exponential.

Answer 2

It might not be exactly what you're asking for, but as an observation it's too long for a comment. One way to see why there is no power of $x$ which gives $x^{-1}$ as derivative can be to think that, since the derivative maps $x^\alpha$ to $x^{\alpha-1}$ (omitting the constant factor), then the only possibility would be to have $x^{-1}$ as derivative of $x^0$. This is actually the case, but... There is a constant factor $\alpha$ in the derivative so that $0\times x^{-1} = 0$.

This can be seen as a justification for the fact that your limit works. What you are doing is to balance the way you're letting your exponent going to 0 with a very strong multiplicative factor, which aim at providing a non-vanishing scaling for the derivative. Indeed when you write $a x^{1/a} = \frac{1}{\alpha}x^\alpha$ (with $\alpha = 1/a$) you're just providing the right pre-factor so that the derivative is exactly $x^{\alpha-1}$, with unitary multiplicative coefficient. This being true for any $\alpha\neq 0$, it can make sense that it can be continuously extended to the case $\alpha = 0$ (i.e., $a\to\infty$).

Answer 3

A mechanical reason that $$\lim_{a\rightarrow\infty}ax^{\frac{1}{a}}-a=\ln(x)\text{:}$$

Take the binomial series $$\begin{align*} ax^{\frac{1}{a}}-a&=a(1-(1-x))^{\frac{1}{a}}-a\\ &=a\sum_{n=0}^{\infty}\binom{\frac{1}{a}}{n}(-1)^n(1-x)^n-a\\ &=a\left(1-\frac{1}{a}(1-x)+\frac{\frac{1}{a}\left(\frac{1}{a}-1\right)}{2!}(1-x)^2-\frac{\frac{1}{a}\left(\frac{1}{a}-1\right)\left(\frac{1}{a}-2\right)}{3!}(1-x)^3+\dotsc\right)-a\\ &=-(1-x)+\frac{\frac{1}{a}-1}{2!}(1-x)^2-\frac{\left(\frac{1}{a}-1\right)\left(\frac{1}{a}-2\right)}{3!}(1-x)^3+\dotsc \end{align*}$$ Now let $\epsilon=\frac{1}{a}$ $$\begin{align*} &=-(1-x)+\frac{\epsilon-1}{2!}(1-x)^2-\frac{(\epsilon-1)(\epsilon-2)}{3!}(1-x)^3+\dotsc \end{align*}$$ As $\epsilon\rightarrow0$, presuming it converges $$\begin{align*} &=-(1-x)+\frac{-1}{2!}(1-x)^2-\frac{(-1)(-2)}{3!}(1-x)^3+\dotsc\\ &=-\sum_{n=1}^{\infty}\frac{1}{n}(1-x)^n\\ &=\ln(1-(1-x))\\ &=\ln(x) \end{align*}$$ You've reconstructed the power series of $\ln(x)$, which is not an accident but I don't really have the deeper reasons why it should be so.

Answer 4

Yes. A basic and fundamental connection.

So much so that to define real (potentially irrational) power exponents most calculus books first define $\ln x$ as $\int_1^x \frac 1t dt$ and not assuming $\ln x$ has anything to do with powers at first.

Then define $e^x$ as the functional inverse of $\ln x$.

And finally the concept of power $b^x$ is defined to be $e^{x\ln b}$. And it's only as an outcome that $b^q; q\in \mathbb Q$ will have the property of "multiplying $b$ some number of times and taking a root" that we expect.

....

As for why this works....

Well, by the fundamental theorem of calculus if $f(x) = \ln x = \int_1^x \frac 1t dt$ then $f'(x) = \frac 1x$

but why should $h(x) = x^k= e^{k\ln x}$ have $h'(x) = kx^{k-1}$?

Welllll.....

$e^{\ln x} =x$ so if $f(x) = e^x$ and $g(x)=\ln x$ then $f(g(x)) = x$ and so

$f'(g(x))g'(x) = f'(g(x))\frac 1x = 1$ so $f'(g(x)) = x$. Let $y = g(x) = \ln x$ then $e^y = x$ and $f'(y) = e^y$

So if $h(x) = x^k = e^{k\ln x}=f(kg(x))$ the $h'(x) = e^{kg(x)}\cdot kg'(x)=kx^k\cdot \frac 1x = kx^{k-1}$.

.... oh wait... for integers $k$ why does $x^k = \underbrace{x\cdot....\cdot x}$?

Wellll.....

Answer 5

You can also see that in a "reverse way". Because of the fact that the derivative of $x^a$, which is $a x^{a-1}$ vanishes when $a=0$, $x^{-1}$ is the only power function not to have an antiderivative that is a power function.

On the other hand, the family of antiderivatives $$ f_a(x) = \int x^a \mathrm{d}x $$ is never $0$. They are all power functions multiplied by a constant except when $a=-1$, which is the logarithm. So the logarithm and the power functions are all parts of this family of antiderivatives.

Actually it is a good principle to remember, when you want to compare logarithm to power functions, that you can think of $\ln$ as "$x^0$". It appears a lot in analysis (critical Sobolev embeddings into $L^\infty$, space BMO ...), in physics, the effective Coulomb potential in dimension $2$ is sometimes $\ln(|x|)$ and it is $1/|x|^{d-2}$ in dimension $d$, and mathematically speaking, this is the solution of Laplace equation ...