- Let
*a*be the initial thickness of the grass, let*b*the rate of growth of the grass, and let*c*be the rate at which the cows eat the grass (in the appropriate units). Let*n*denote the number of cows that will eat the third field bare in 18 weeks. Then we have10( *a*+ 4*b*)/3= 12*4 *c*10( *a*+ 9*b*)= 21*9 *c*24( *a*+ 18*b*)= *n*18*c*

If we multiply the first equation by -27/5 and the second equation by 14/5, we obtain 10(*a*+ 18*b*) = 270*c*, so (*a*+ 18*b*)/*c*= 27. We conclude that*n*= 36, so the answer is 36 happy cows. - The exact number
*N*of minutes to complete the puzzle is ∑_{x=0}^{999}3(1000 -*x*)/(1000 +*x*). Since 3(1000 -*x*)/(1000 +*x*) is a non-negative monotonic decreasing function for 0≤*x*__<__1000, we see that*N*- 3≤∫_{0}^{1000}-3 + 6000/(1000 +*x*)*dx*≤*N*.*N*/60≈50(2 ln 2 - 1). Using ln 2≈.69, we conclude that it takes approximately 19 hours to complete the puzzle. - One can quickly check that
*f*(2) = 2 and*f*(3) = 3, so it seems reasonable that*f*(*n*) =*n*, so let us try to prove this. Certainly if*f*(*n*) =*n*, then*f*(1) = 1, so we will prove the result by induction on*n*; we assume that the result is true for all integers ≤*n*. Then*f*(*n*+ 1) =*f*(*f*(*n*)) +*f*(*n*+ 1 -*f*(*n*)) =*n*+*f*(1) =*n*+ 1*f*(*n*) =*n*for*n*= 1, 2,.... - Write
*P*(*x*) =*ax*^{3}+*bx*^{2}+*cx*+*d*, where*a*,*b*,*c*,*d*∈**Z**. Let us suppose by way of contradiction that*a*,*b*,*c*,*d*≥ - 1. From*P*(2) = 0, we get 8*a*+ 4*b*+ 2*c*+*d*= 0, in particular*d*is even and hence*d*≥ 0. Since 4*b*+ 2*c*+*d*≥ - 7, we see that*a*≤ 0. Also*a*≠ 0 because*P*(*x*) has degree 3, so*a*= - 1. We now have 4*b*+ 2*c*+*d*= 8 and*b*+*c*+*d*= 1 from*P*(1) = 0. Thus -2*c*- 3*d*= 4, so -2*c*= 4 + 3*d*≥4 and we conclude that*c*≤ - 2. The result follows. - (a)
- For small positive
*x*, we have*x*/2 < sin*x*<*x*, so for positive integers*n*, we have 1/(2*n*) < sin(1/*n*) < 1/*n*. Since ∑_{n=1}^{∞}1/*n*^{p}is convergent if and only if*p*> 1, it follows from the basic comparison test that ∑_{n=1}^{∞}(sin 1/*n*)^{p}is convergent if and only if*p*> 1. - (b)
- It is not difficult to show that any real number
*x*, there exists an integer*n*>*x*such that | sin*n*| > 1/2. Thus whatever*p*is, lim_{n-> ∞}| sin*n*|^{p}≠ 0. Therefore ∑_{n=1}^{∞}| sin*n*|^{p}is divergent for all*p*.

- (a)
- If
*y*^{*}is a steady-state solution, then*y*^{*}=*y*^{*}(2 -*y*^{*}), so*y*^{*}= 0 or 1 = 2 -*y*^{*}. Therefore the steady-state solutions are*y*^{*}= 0 or 1. - (b)
- Suppose 0 <
*y*_{n}< 1. Then*y*_{n+1}/*y*_{n}= 2 -*y*_{n}> 1, so*y*_{n+1}>*y*_{n}. Also*y*_{n+1}= 1 - (1 -*y*_{n})^{2}, so*y*_{n+1}< 1. We deduce that*y*_{n}is a monotonic positive increasing function that is bounded above by 1, in particular*y*_{n}converges to some positive number ≤1. It follows that*y*_{n}converges to 1.

- Let
*y*∈[0, 1] be such that (*g*(*y*) +*uf*(*y*)) =*u*. Let us suppose we do have constants*A*and*B*such that*F*(*x*) =*Ag*(*x*)/(*f*(*x*) +*B*) is a continuous function on [0.1] with max_{0≤x≤1}*F*(*x*) =*u*. We will guess that the maximum occurs when*x*=*y*, so*u*=*Ag*(*y*)/(*f*(*y*) +*B*). Then*A*=*B*= -1 satisfies these equations, so*F*(*x*) =*g*(*x*)/(1 -*f*(*x*)).So let us prove that

*F*(*x*) =*g*(*x*)/(1 -*f*(*x*)) has the required properties. Certainly*F*(*x*) is continuous because*f*(*x*) < 1 for all*x*∈[0, 1], and*F*(*y*) =*u*from above. Finally max_{0≤x≤1}(*g*(*x*) +*uf*(*x*)) =*u*, so*g*(*x*)≤*u*(1 -*f*(*x*)) for all*x*and we conclude that*F*(*x*)≤*u*. The result is proven. - Suppose we can disconnect
*F*by removing only 8 points. Then the resulting framework will consist of two nonempty frameworks*A*,*B*such that there is no segment joining a point of*A*to a point of*B*. Let*a*be the number of points in*A*. Then there are 9 -*a*points in*B*, at most*a*(*a*- 1)/2 line segments joining the points of*A*, and at most (10 -*a*)(10 -*a*- 1)/2 line segments joining the points of*B*. It follows that the resulting framework has at most 45 - 10*a*+*a*^{2}. Since 10*a*-*a*^{2}> 8 for 1≤*a*≤9, the result follows.

Peter Linnell 2010-06-02