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Cauchy induction, sometimes called backwards induction, works as follows:

  • show that $p(1)$ is true
  • show that $p(n)$ implies $p(2n)$ (which inductively implies $p(2^n)$ is true)
  • show that $p(n)$ implies $p(n-1)$

Then $p$ is true for all $n\geq 1$. The AM-GM inequality can be proven using this technique, as shown here.

In step $2$ there is of course nothing special about the number $2$, the same logic would work if we showed $p(n)\Rightarrow p(3n)$, for instance.

My question is, are there examples where this would be more convenient? In other words, where $p(n)\Rightarrow p(kn)$ for some $k>2$ would be easier to prove than $p(n)\Rightarrow p(2n)\,?$

Lavender Lullaby
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    Of course even more generally $2n$ can be replaced with $m(n)$ with $m(n)>n$ – Hagen von Eitzen Sep 03 '15 at 21:27
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    @HagenvonEitzen As far as the bounty, I would be happy to award it to any answer which used such a method. Basically I would love to see an example of a problem applying Cauchy inductive methods where it makes sense to use something other than $p(n)\rightarrow p(2n)$ (and of course where regular induction doesn't suffice or is otherwise terrible to implement or something). – Peter Woolfitt Sep 03 '15 at 22:10

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To answer this question somewhat more generally than posed; there is a completely bizarre backwards induction proof for the Fundamental Theorem of Calculus, due to Laplace (see, for example, p. 120 in this book).

The proof shows that $P(\frac{n(n-1)}{2}) \Rightarrow P(n)$, where $P(n)$ is the statement "Any degree $n$ polynomial with real coefficients has a complex root." This requires proving $P(n)$ for all odd $n$ as the base case; fortunately, when $n$ is odd, we know that we can find a real root by the Intermediate Value Theorem.

To show that all $P(n)$ follow from these base cases and the inductive step $P(\frac{n(n-1)}{2}) \Rightarrow P(n)$, observe that, if $n$ is even and $2^k$ is the highest power of $2$ that divides $n$, then $2^{k-1}$ is the highest power of $2$ that divides $\frac{n(n-1)}{2}$.

(The actual proof of the inductive step is a somewhat unenlightening bit of algebra that can be read from the Google Books link, so I'll skip it. The induction is the cool part.)

Misha Lavrov
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