I am curious to know if there is a reason why $19^{\frac{1}{25}}$ is so close to $\frac{9}{8}$?

The reason is that $(9/8)^{25}\approx 19.0026$ is close to $19$.

This might sound like a boring answer, but odds are, there is no deeper reason. There are so many numbers and ways to combine them with common operations that it would be unthinkable that none of them were "close" to any others (and what defines how close is "close" anyway?) You happened to stumble across one of the many coincidences that are bound to happen, and that's (probably) it.

My personal favourite is $e^\pi - \pi \approx 20$, which I found through XKCD.

Of course, there are somewhat constructive ways to look for such coincidences, like using continued fractions and the like. And at least when comparing some expression to a rational number, there are objective tests for "closeness", like comparing their difference to the size of the denominator (the golden ratio is famously "far away" from any rational number). But that doesn't mean that the existence of any of them is part of some deeper pattern.

  • 1
    Sometimes there are reasons, though. For instance, $\exp\left(\pi\sqrt{163}\right) \approx 640320^3 + 744$. See this. – MathematicsStudent1122 Aug 30 at 8:38
  • 2
    I am going to go with coincidence. Other similar examples are $(11/10)^{37} \approx 34.0039$ and $(17/16)^{41} \approx 12.0084$. – gandalf61 Aug 30 at 8:43

I agree with Arthur that it is a coincidence. Here is one way to see why.

Take any fraction that is slightly bigger than $1$. You took $9/8$. Raise that to successive powers: $(9/8)^2,\ (9/8)^3,\ (9/8)^4,\ ...$
The first two digits after the decimal point are essentially random, so you expect $.00$ to occur once every hundred times. Occasionally you can be a bit lucky that it occurs a little early, say before you reach the $50$th power. Since the fraction you started with is close to $1$, the result is nearly integer that is not very large.

For example:
$(6/5)^{21} \approx 46$
$(11/10)^{37} \approx 34$
$(19/18)^{36} \approx 7$
and the best yet:
$(7/6)^9 \approx 4$
$(12/11)^8 \approx 2$

If you also look for cases where you have $.99$ after the decimal point, you get many more.

There is a connection between continued fractions and best rational approximations, e.g. see here.

You would need the continued fraction for $19^{1/25}$ of course, I am not sure if this is easy to derive. E.g. for roots $\sqrt[k]{\cdot}$ this would be easy.

Just to investigate from another point of view we have that

$$19^{\frac{1}{25}}\approx \frac98 \iff 19\approx \left(\frac98\right)^{25}$$

and by binomial expansion

$$\left(\frac98\right)^{25}= \left(1+\frac18\right)^{25}\approx 1+\frac{25}{8}+\frac{300}{64}+\frac{2300}{512}+\frac{12\,650}{4\,096}+\frac{53\,130}{32\,768}+\frac{177\,100}{262\,144}+\frac{480\,700}{2\,097\,152}+\frac{1\,081\,575}{16\,777\,216}\approx 18.98$$

which converges very slowly and confirm in some sense that the result obtained is really a "coincidence".

Note that if we take

$$\left(\frac{90}{89}\right)^{25}=\left(1+\frac1{89}\right)^{25}\approx 1+\frac{25}{89}+\frac{300}{7921}+\frac{2300}{704\,969}\approx \frac 43$$

and we obtain that

$$\frac 4 3\approx \left(\frac{90}{89}\right)^{25}$$

that is in some sense less a "coincidence" since the convergence is faster and the dominant term is

$$1+\frac{25}{89}\approx \frac43$$

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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