- The answer is no. One example is
*A*=( 0 1 0 0 ) *B*=( 0 0 1 0 ) *C*=( 1 0 0 0 ). Then det

*M*=/= 0 (expand by the fourth row), whereas det*N*= 0 (fourth row consists entirely of 0's). Therefore*M*is invertible and*N*is not invertible, as required. - For
*n*a non-negative integer, as*n*increases from*n*to*n*+ 6, we add 3 twice, 1 twice and 2 twice to*f*(*n*); in other words*f*(*n*+ 6) =*f*(*n*) + 12. We deduce that*f*(*n*) = 2*n*when*n*= 0 (mod 6). - Let
*s*_{n}denote the number of strings of length*n*with no three consecutive*A's*. Thus*s*_{1}= 3,*s*_{2}= 9 and*s*_{3}= 26. We claim that we have the following recurrence relation:*s*_{n}= 2*s*_{n-1}+2*s*_{n-2}+2*s*_{n-3}(*n*__>__3).*B*or*C*; the second term indicates the number of such strings which begin with*AB*of*AC*, and the third term indicates the number of such strings which begin with*AAB*or*AAC*. Using this recurrence relation, we find that*s*_{4}= 76,*s*_{5}= 222 and*s*_{6}= 648. Since the total number of strings is 3^{6}, we conclude that the probability of a string on 6 symbols not containing 3 successive*A*'s is 648/3^{6}= 8/9. - The answer is no. Let us color the chess board in the usual way with
alternating black and white squares, say the corners are colored with
black squares. Then by determining the number of black squares in
each row, working from top to bottom, we see that the number of black
squares is
4 + 4 + 5 + 4 + 4 + 4 + 5 + 4 + 4 = 38.Since there are 78 squares in all, we see that the number of white squares is 40. Now each domino cover 1 white and 1 black square, so if the board could be covered by dominoes, then there would be an equal number of black and white squares, which is not the case.
- Expanding the sine, we have
*f*(*x*) = cos*x*sin(*t*^{2}-*t*) d*t*+ sin*x*cos(*t*^{2}-*t*) d*t*.*f'*(*x*)= cos *x*sin(*x*^{2}-*x*) - sin*x*sin(*t*^{2}-*t*) d*t*+sin *x*cos(*x*^{2}-*x*) + cos*x*cos(*t*^{2}-*t*) d*t*,*f''*(*x*)= - sin *x*sin(*x*^{2}-*x*) + (2*x*- 1)cos*x*cos(*x*^{2}-*x*)-sin *x*sin(*x*^{2}-*x*) - cos*x*sin(*t*^{2}-*t*) d*t*+cos *x*cos(*x*^{2}-*x*) + (1 - 2*x*)sin*x*sin(*x*^{2}-*x*)-sin *x*sin(*t*^{2}-*t*) d*t*+ cos*x*sin(*x*^{2}-*x*).

We deduce that*f*(*x*) +*f''*(*x*)= (2 *x*+ 1)(cos*x*cos(*x*^{2}-*x*) - sin*x*sin(*x*^{2}-*x*))= (2 *x*+ 1)cos*x*^{2}.

Set

*g*(*x*) =*f''*(*x*) +*f*(*x*). To find*f*^{(12)}(0) +*f*^{(10)}(0), we need to compute*g*^{(10)}(0), which we can find by considering the coefficient of*x*^{10}in the Maclaurin series expansion for (2*x*+ 1)cos*x*^{2}. Since cos*x*^{2}= 1 -*x*^{4}/2! +*x*^{8}/4! -*x*^{12}/6! + ..., we see that this coefficient is 0. Therefore*g*^{(10)}(0) = 0 and the result follows. - Suppose first that there is an infinite subset
*S*such that each person only knows a finite number of people in*S*. Then pick a person*A*_{1}in*S*. Then there is an infinite subset*S*_{1}of*S*containing*A*_{1}such that*A*_{1}knows nobody in*S*_{1}. Now choose a person*A*_{2}other than*A*_{1}in*S*_{1}. Then there is an infinite subset*S*_{2}of*S*_{1}containing {*A*_{1},*A*_{2}} such that nobody in*S*_{2}knows*A*_{2}. Of course nobody in*S*_{2}will know*A*_{1}either. Now choose a person*A*_{3}in*S*_{2}other than*A*_{1}and*A*_{2}. Then nobody from {*A*_{1},*A*_{2},*A*_{3}} knows each other. Clearly we can continue this process indefinitely to obtain an arbitrarily large number of people who don't know each other.Therefore we may assume in any infinite subset of people there is a person who knows an infinite number of people. So we can pick a person

*B*_{1}who knows infinitely many people*T*_{1}. Then we can pick a person*B*_{2}in*T*_{1}who knows an infinite number of people*T*_{2}of*T*_{1}, because we are assuming in any infinite subset of people, there is somebody who knows infinitely many of them. Of course,*B*_{1}and*B*_{2}know all the people in*T*_{2}. Now choose a person*B*_{3}in*T*_{2}who knows an infinite number of people in*T*_{2}. Then {*B*_{1},*B*_{2},*B*_{3}} know each other. Clearly we can continue this process indefinitely to obtain an arbitrarily large number of people who know each other.**Remark**A simple application of the axiom of choice shows that we can find an infinite number of people in the party such that either they all know each other or they all don't know each other. - Set
*b*_{n}= 1 -*a*_{n+1}/*a*_{n}. Let us suppose to the contrary that S|*b*_{n}| is convergent. Then lim_{n--> }*b*_{n}= 0, so may assume that |*b*_{n}| < 1/2 for all*n*. Now*a*_{n+1}= *a*_{1}(1 -*b*_{1})(1 -*b*_{2})...(1 -*b*_{n}),

henceln( *a*_{n+1})= ln *a*_{1}+ ln(1 -*b*_{1}) + ln(1 -*b*_{2}) + ... + ln(1 -*b*_{n}).

Since lim_{n--> }*a*_{n}= 0, we see that lim_{n--> }ln(*a*_{n}) = - . Now ln(1 -*b*)__>__- 2|*b*| for |*b*| < 1/2; one way to see this is to observe that 1/(1 -*x*)__<__2 for 0__<__*x*__<__1/2 and then to integrate between 0 and |*b*|. Therefore lim_{n--> }(- 2|*b*_{1}| - ... -2|*b*_{n}|) = - . This proves that S|*b*_{n}| is divergent and the result follows.

Peter Linnell 2004-10-29