Ask Uncle Colin: a fractional power

Dear Uncle Colin,

How would you work out ${\left(\frac{4}{5}\right)}^{10}$ in your head?

- Probability Estimation Needed, Relic Of Slide-rule Era

Hi, PENROSE, and thanks for your message!

That’s really two questions: how I would do it, and how the Mathematical Ninja would do it.

Me

Me, I’d probably use the fact that ${(1-\frac{1}{n})}^n$ goes to $\frac{1}{e}$ as $n$ gets large. What we have is ${(1-\frac{1}{5})}^{10}$, which should be approximately $e^{-2}$, which is about 0.135.

The Mathematical Ninja

“Ha! I don’t believe I have to tell you this, but five is not large.”

“Depends on what you compare it with, sensei.”

“You can do much better by saying $\ln \left(\frac{4}{5}\right)\approx -\frac{2}{9}$.”

“Why is that?”

A sigh. “$\ln (1+x)\approx x+\frac{1}{2}{x}^2$. $\ln (1-x)\approx -x+\frac{1}{2}{x}^2$. So $\ln \left(\frac{1-x}{1+x}\right)\approx -2x$, give or take a cubic term.”

“… and the ninth?”

“That’s not a ninth. That’s a ninth. Oh, er… yes, if you solve $\frac{1-x}{1+x}=\frac{4}{5}$, you find $x=\frac{1}{9}$ (shortcut: it’s the difference, 5-4 over the sum, 5+4).”

“OK, so I buy that $\ln \left(\frac{4}{5}\right)\approx -\frac{2}{9}$.”

“Then the logarithm of the thing we want is roughly $-\frac{20}{9}$, so we want $e^{-\frac{20}{9}}$.”

“I suppose you happen to know what that is?”

“Well, $e^{2.20}\approx 9$, so $e^{-2.222\dots }$ will be about 2% less than a ninth. A smidge less than 0.11 is about as close as I’d go.”

“0.1073, says the calculator I’m totally not looking at right now.”

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Hope that helps!

- Uncle Colin (and The Mathematical Ninja)