# Is there a relation between fractional power and logarithm functions?

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.

• Irrational exponents are usually defined via logarithms/exponential functions. The general definition of $a^b$ for arbitrary $b$ (and $a\gt 0$) is $a^b = e^{b\ln(a)}$. – Arturo Magidin Jun 16 at 16:25
• Note that your first equation is wrong -- you want $\lim\left(\text{d}/\text{d}x a x^{\frac1a}\right)$ on the left-hand side, i.e. limit and derivative switched. – Toffomat Jun 17 at 10:44
• To be clear, "deriving" is not interchangeable with "differentiating." It's kind of unfortunate, but you want to use the second one here because we're talking about derivatives. – Bladewood Jun 17 at 13:45
• fixed both mistakes – Miguel Bartelsman Jun 17 at 19:18

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 $$p$$.

Also consider the inverse of this function,

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

• This is illuminating! (+1) – mrtaurho Jun 17 at 12:03

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.....

• Something has gone wrong in your $g'(x) = (k-10)x^{k-1}$. The $(k-10)$ should have been simply $k$. – Ruslan Jun 17 at 9:14
• I have no idea how I typed that. I was thinking $[x^k]' =$ "bring the $k$ as a coefficient to the front and have $x$ raised to the power of one less than $k$". ANd I thought "So bringing the $k$ to the front is $k -10$". Why on earth I thought that that moving $k$ from one area on the paper to another area on the paper would change it from $k$ to $k-10$ I have utterly no idea.... I guess by taking the $k$ in my hand and dragging it across the paper crumbs would rub off.... – fleablood Jun 17 at 16:54

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.

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 ...

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., $$a\to\infty$$).