6
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Irrational exponents are usually defined via logarithms/exponential functions. The general definition of $a^b$ for arbitrary $b$ (and $a\gt 0$) is $a^b = e^{b\ln(a)}$. $\endgroup$ – Arturo Magidin Jun 16 at 16:25
  • 2
    $\begingroup$ 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. $\endgroup$ – Toffomat Jun 17 at 10:44
  • 1
    $\begingroup$ 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. $\endgroup$ – Bladewood Jun 17 at 13:45
  • $\begingroup$ fixed both mistakes $\endgroup$ – Miguel Bartelsman Jun 17 at 19:18
14
$\begingroup$

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 $p$.

enter image description here


Also consider the inverse of this function,

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

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ This is illuminating! (+1) $\endgroup$ – mrtaurho Jun 17 at 12:03
5
$\begingroup$

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.....

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Something has gone wrong in your $g'(x) = (k-10)x^{k-1}$. The $(k-10)$ should have been simply $k$. $\endgroup$ – Ruslan Jun 17 at 9:14
  • $\begingroup$ 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.... $\endgroup$ – fleablood Jun 17 at 16:54
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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 ...

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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., $a\to\infty$).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.