The well known **William Lowell Putnam Mathematical Competition** (76th edition) took place this month.

Here is a Maple approach for two of the problems.

**1.** *For each real number x, 0 <= x < 1, let f(x) be the sum of 1/2^n where n runs through all positive integers for which floor(n*x) is even.*

*Find the infimum of f.*

(Putnam 2015, A4 problem)

**f:=proc(x,N:=100)**

**local n, s:=0;**

**for n to N do**

** if type(floor(n*x),even) then s:=s+2^(-n) fi;**

** #if floor(n*x) mod 2 = 0 then s:=s+2^(-n) fi;**

**od;**

**evalf(s);**

**#s**

**end;**

**plot(f, 0..0.9999);**

**min([seq(f(t), t=0.. 0.998,0.0001)]);**

** ** 0.5714285714

**identify(%);**

So, the infimum is 4/7.

Of course, this is not a rigorous solution, even if the result is correct. But it is a valuable hint.

I am not sure if in the near future, a CAS will be able to provide acceptable solutions for such problems.

**2.** *If the function f is three times differentiable and has at least five distinct real zeros,*

*then f + 6f' + 12f'' + 8f''' has at least two distinct real zeros.*

(Putnam 2015, B1 problem)

**restart;**

**F := f + 6*D(f) + 12*(D@@2)(f) + 8*(D@@3)(f);**

**dsolve(F(x)=u(x),f(x));**

We are sugested to consider

**g:=f(x)*exp(x/2):**

g3:=diff(g, x$3);

**simplify(g3*8*exp(-x/2));**

So, F(x) = k(x) * g3 = k(x) * g'''

g has 5 distinct zeros implies g''' and hence F have 5-3=2 distinct zeros, **q.e.d.**