Department of Math and Computer Science
Problems of the Month 1998

proposed by Dan Jurca

Find the sum of the first n terms of the following series.

1+2+4+7+11+16+22+29+37+46+56+¼

Subtracting 1 from each term we get the series

0+1+3+6+10+15+21+28+36+45+55+¼,

so that the i-th term of the given series is the (i-1)-th ``triangular number'', [((i-1)i)/2]. Hence the sum of the first n terms is

n+ n å i=1   (i-1)i 2 =  n3+5n 6 .

proposed by Professor Bill Nico

Develop a method or write a computer program to solve the following problem.

Suppose n is an integer, 3 £ n, and s₁, s₂, s₃, ¼, s_n are n positive real numbers. Determine whether there exists a circle which can be circumscribed about a polygon with sides of lengths s₁, s₂, s₃, ¼, s_n; and, if so, find (an accurate approximation of) the radius.

In particular, does there exist a circle which can be circumscribed about a polygon with sides of lengths 1, 2, 3, ¼, 10; and, if so, what is (an accurate approximation of) the radius?

proposed by Dan Jurca

Find a function f:R®R such that

0 £ n Þ  lim x®¥  f(n)(x) =¥,   and x Î R Þ  lim n®¥  f(n)(x) =0

where f⁽ⁿ⁾ is the derivative of order n of f.

Consider f:R®R by f(x)=e^x/2. Then for 0 £ n we have f⁽ⁿ⁾(x)=e^x/2/2ⁿ. Thus the function f obviously has the required properties.

Of course for each a, 0 < a < 1, the function f(x)=e^ax has the required properties.

No other solution was received.

proposed by Dan Jurca

For each continuous function f:R®R let Tf:R®R by

x Î R Þ Tf(x)= ó õ x 0  f= ó õ x 0  f(u)du,

and for n = 2, 3, ¼ let T_nf be the function obtained by applying T n times to f. For example, if g(t)=t², then

Tg(x) = ó õ x 0  g= ó õ x 0  g(u) du= ó õ x 0  u2 du=x3/3, and T2g(x) =TTg(x)= ó õ x 0  Tg= ó õ x 0  Tg(u) du= ó õ x 0  u3/3 du=x4/12.

Now let f(x)=e^x, and determine lim_n®¥T_nf.

Let C(R) be the (infinite-dimensional) real vector space of all continuous functions R®R and let T:C(R)® C(R) by

f Î C(R) Þ Tf Î C(R) by Tf(x)= ó õ x 0  f= ó õ x 0  f(u) du.

Then for n=0, 1, 2, ¼ there is a linear operator T_n:C(R)®C(R) defined as follows:

T0 =idC(R); 1 £ n Þ Tn =T°Tn-1.

If f Î C(R) by f(x)=e^x, what is lim_n®¥T_nf?

(Observe that we are asking for lim_n®¥(T_nf).)

We find at once

T0f(x) =f(x) T1f(x) = ó õ x 0  T0f= ó õ x 0  eudu=ex-1 T2f(x) = ó õ x 0  T1f= ó õ x 0  [eu-1]du=ex-x-1=ex-(1+x) T3f(x) = ó õ x 0  T2f= ó õ x 0  [eu-(1+u)]du=¼ = ex-(1+x+  x2 2 ).

Now let us write p_-1(x)=0, and, for n=0,1,2,¼,

pn(x)= n å k=0   xk k! .

Then by induction on n we find easily

0 £ n Þ Tnf(x)=ex-pn-1(x).

Now as everyone knows x Î R Þ

ex= ¥ å k=0   xk k! = lim n®¥  n å k=0   xk k! = lim n®¥  pn(x).

It follows at once that x Î R Þ

lim n®¥  Tnf(x)= lim n®¥  (ex-pn-1(x))=ex- lim n®¥  pn-1(x) = ex- lim n®¥  pn(x)=ex-ex=0.

Thus we have: x Î R Þ lim_n®¥T_nf(x)=0, and since this holds for each x, we write lim_n®¥T_nf=0.

Remark: Although the argument above holds for the particular function f(x)=e^x, in fact one can show that the same result holds for each continuous f:R®R.

Communicated by Dan Jurca

For n=0,1,2,¼ let f_n(x)=x^n-1e^1/x.

Find the n-th derivative, f_n⁽ⁿ⁾, of f_n, and prove that your answer is correct.

Let f(x)=sin(psin^-1(x)); for n=0,1,2,3,¼ evaluate the n-th derivative of f at 0; i.e., compute

f(n)(0).

Let us write y=f(x)=sin(psin^-1x); then we find

y =sin(psin-1x) sin-1y =psin-1x  y¢ Ö 1-y2 =  p Ö 1-x2 (1-x2)(y¢)2 =p2(1-y2) -2x(y¢)2+(1-x2)·2y¢y¢¢ =p2·-2yy¢ -xy¢+(1-x2)y¢¢ =-p2y,

so that y satisfies the differential equation (1-x²)y¢¢-xy¢+p²y=0. Since f is clearly an analytic function near 0, we write y=å_n=0^¥ c_nxⁿ, where c_n=[(f⁽ⁿ⁾(0))/n!]. We determine c_n using standard techniques as follows.

y = ¥ å n=0  cnxn p2y = ¥ å n=0  p2cnxn y¢ = ¥ å n=0  (n+1)cn+1xn xy¢ = ¥ å n=0  ncnxn y¢¢ = ¥ å n=0  (n+2)(n+1)cn+2xn x2y¢¢ = ¥ å n=0  n(n-1)cnxn (1-x2)y¢¢ = ¥ å n=0  [(n+2)(n+1)cn+2-n(n-1)cn]xn

Hence for n=0,1,2,¼ we have

(n+2)(n+1)cn+2-n(n-1)cn-ncn+p2cn=0.

Now one finds at once c₀=f(0)=0, c₁=f¢(0)=p, and

2 £ n Þ cn=  (n-2)2-p2 n(n-1) cn-2.

It follows that

f(n)(0)=n!·cn= ì ï ï í ï ï î 0 if n is even; p if n=1; (12-p2)(32-p2)¼[(n-2)2-p2]p if n is odd and 3 £ n.

Equivalently one may write

f(n)(0)= ì ï ï í ï ï î 0 if n is even; p ën/2û Õ i=1  [(2i-1)2-p2] if n is odd.

Communicated by Dan Jurca

Let us define a box as a subset of the cartesian plane which is the interior of a square with sides of length 1 and vertices at the integer lattice points; i.e., each corner of the square has integral coordinates. Now consider positive integers m and n, and the m×n rectangle with one corner at the origin of the cartesian plane and the diagonally opposite corner at the point (m,n). Find a formula for b(m,n), the number of boxes inside this rectangle which are intersected by the line from the point (0,0) to the point (m,n).

From the sketch below we can see that b(4,7)=10, and b(4,8)=8.

We show that b(m,n)=m+n-gcd(m,n).

First consider the case that m and n are relatively prime, so that gcd(m,n)=1. Then we show that the line segment l from (0,0) to (m,n) does not contain a point (i,j) with 0 < i < m. For otherwise the triangles with vertices (0,0),(i,0),(i,j) and (0,0),(m,0),(m,n) are similar, so that i/j=m/n. But then we have in=jm, so that m|in (since m divides jm). Since m and n are relatively prime, it follows that m|i. This is not possible if 0 < i < m. Hence l contains no other point with integer coordinates. This means that if l intersects the boundary of a box, it does so in one of precisely three ways:

1.

in the point (0,0);

2.

in the interior of an edge of the boundary;

3.

in the point (m,n).

Now consider the general case, and let g=gcd(m,n). Then there are positive integers m¢ and n¢ such that m=gm¢, n=gn¢, and gcd(m¢,n¢)=1. Again considering similar triangles the line segment l from (0,0) to (m,n) contains the point (m¢,n¢). Further, the pattern from (0,0) to (m¢,n¢) repeats g times. Hence l intersects g(m¢+n¢-1) = gm¢+gn¢-g=m+n-gcd(m,n) boxes, so b(m,n)=m+n-gcd(m,n), as asserted.

Also solved by Matthew Hubbard, Thomas Kim, and Professor Bill Nico. Thomas Kim generalized the result to higher dimensions.