Using your Head is Permitted

March 2017 solution

Let us first find a complex function that satisfies the conditions.

A usual suspect for seeking solutions to this type of differential equations is any linear combination of functions of the form ezx.

Taking the tenth derivative leads to z10ezx, so we are looking for a z value that is a tenth root of unity: z10=1. This has 10 solutions, of which some are primitive roots, so satisfy the condition that lower order derivatives will not equal the original function.

Notably, the conjugate of a primitive root is itself a primitive root, so by adding the two together we still get a function that satisfies all the criteria listed so far, but is, additionally, also a real function.

Up to a factor of 2, this can simply be stated as Re(ezx), and z can be any primitive tenth root of unity, such as, for example, eiπ/5=cos(π/5)+isin(π/5).

So, now, it just becomes a question of wording the function in a way that makes no mention of the intermediate complex values:


Leading to one possible f being ecos(π/5)xcos(sin(π/5)x).

This is enough to solve the problem but not the bonus question, because π is not computable on the target pocket calculator.

Luckily, it is possible to calculate both cos(π/5) and sin(π/5): cos(π/5)=(1+sqrt(5))/4 and sin(π/5)=sqrt(1-cos2(π/5)).

Here are two ways how to prove these values. The first, purely algebraic, comes from Andreas Stiller:

For every x cos(2x)=2cos2(x)-1.


Substituting in the two connections we have between cos(π/5), which we will call c and cos(4π/5) we get:



8c4-8c2+c+1 = (c+1)(2c-1)(4c2-2c-1) = 0.

This equation has four solutions, but only one of them is in the search range between cos(π/4)=sqrt(2)/2 and cos(π/6)=sqrt(3)/2. It is c=(1+sqrt(5))/4.

A completely different take on this proof was sent in by JJ Rabeyrin. JJ's approach is geometric:

Consider triangle ABC, isosceles in A, where:

Let D be the intersection of AC and the bisector of angle ABC. We see that:

Moreover, we can observe that BCD is isosceles in B, so BD = BC = 1 = AD.

Looking at DAB, we have:

DB2 = AD2 + AB2 - 2AB AD cos(π/5)

and thus AB = 2cos(π/5).

Looking at DBC:

BC2 = BD2 + CD2 - 2BD CD cos(2π/5),

So, CD = 2cos(2π/5).

Because ABC is isosceles in A, we now have AB = AC = AD + DC, leading to

2cos(π/5) = 1 + 2cos(2π/5).

We can now integrate this geometric insight with

cos(2x) = 2cos2(x)-1
to conclude that c=cos(π/5) is a solution to 4c2-2c-1 = 0.

I will note that to actually calculate the required function with the designated pocket calculator does require x to be input twice.

Thank you, again, to all readers who have followed Using your Head is Permitted throughout these 10 years, and thank you to all readers who joined along the way.

And, of course, a big thank you to Ori Pomerantz for the anniversary riddle.

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