- The probability of
*p*gains is the coefficient of (1/2)^{p}(1/2)^{n - p}in (1/2 + 1/2)^{n}. Therefore, without the insider trading scenario, on average the investor will have 10000(3/5 + 9/20)^{n}dollars at the end of*n*days. With the insider trading, the first term (3/5)^{n}becomes 0. Therefore on average the investor will have10000(21/20)dollars at the end of^{n}- 10000(3/5)^{n}*n*days. - We have
-ln(1 -Therefore
*x*) =*x*+*x*^{2}/2 +*x*^{3}/3 + ... = S_{i = 1}^{}(*x*^{n})/*n*.(1 - *x*)ln(1 -*x*)= - *x*+*x*^{2}/2 +*x*^{3}/6 + ...= - *x*+ S_{i = 1}^{}*x*^{n + 1}(1/*n*- 1/(*n*+ 1)) = -*x*+ S_{i = 1}^{}*x*^{n + 1}/(*n*(*n*+ 1)).

Dividing by*x*, we deduce thatSfor_{n = 1}^{}*x*^{n}/(*n*(*n*+ 1)) = 1 + ((1 -*x*)ln(1 -*x*))/*x**x*=/= 0, and the sum is 0 for*x*= 0. - Let
*I*denote the 2 by 2 identity matrix. Since*A*=*A*^{-1}, we see that*A*^{2}=*I*and hence the eigenvalues l of*A*must satisfy l^{2}= 1, so l = ±1. First consider the case det*A*= 1. Then*A*has a repeated eigenvalue ±1, and*A*is similar to( *r**s*0 *r*) where

*r*= ±1 and*s*= 0 or 1. Since*A*^{2}=*I*, we see that*s*= 0 and we conclude that*A*= ±*I*.Now suppose det

*A*= - 1. Then the eigenvalues of*A*must be 1, - 1, so the trace of*A*must be 0, which means that*A*has the form( *a**b**b*- *a*) where

*a*,*b*are complex numbers satisfying*a*^{2}+*b*^{2}= 1. Therefore*b*= (1 -*a*^{2})^{1/2}(where the exponent 1/2 means one of the two complex numbers whose square is 1 -*a*^{2}). We conclude that the matrices satisfying*A*=*A'*=*A*^{-1}are ±*I*and( *a*(1 - *a*^{2})^{1/2}(1 - *a*^{2})^{1/2}- *a*) where

*a*is any complex number. - Set
*R*=*e*^{2pi/7}= cos 2p/7 +*i*sin 2p/7. Since*R*=/= 1 and*R*^{7}= 1, we see that 1 +*R*+ ... +*R*^{6}= 0. Now for*n*an integer,*R*^{n}= cos 2*n*p/7 +*i*sin 2*n*p/7. Thus by taking the real parts and using cos(2p -*x*) = cos*x*, cos(p -*x*) = - cos*x*, we obtain1 + 2cos(2p/7) - 2cos(p/7) - 2cos(3p/7) = 0.Since cosp/7 + cos 3p/7 = 2cos(2p/7)(cosp/7), the above becomes4cos(2p/7)cos(p/7) - 2cos(2p/7) = - 1.Finally cos(2p/7) = 2cos^{2}(p/7) - 1, hence (2cos^{2}(p/7) - 1)(4cos(p/7) - 2) = - 1 and we conclude that 8cos^{3}(p/7) - 4cos^{2}(p/7) - 4cos(p/7) = - 1. Therefore the rational number required is -1/4. - Since
__/__*ABC*+__/__*PQC*= 90 and__/__*ACB*+__/__*PRB*= 90, we see that__/__*QPR*=__/__*ABC*+__/__*ACB*. Now*X*,*Y*,*Z*being the midpoints of*BC*,*CA*,*AB*respectively tells us that*AY*is parallel to*ZX*,*AZ*is parallel to*XY*, and*BX*is parallel to*YZ*. We deduce that__/__*ZXY*=__/__*BAC*and hence__/__*QPR*+__/__*ZXY*= 180. Therefore the points*P*,*Z*,*X*,*Y*lie on a circle and we deduce that__/__*QPX*=__/__*ZYX*. Using*BZ*parallel to*XY*and*BX*parallel to*ZY*from above, we conclude that__/__*ZYX*=__/__*ABC*. Therefore__/__*QPX*+__/__*PQX*=__/__*ABC*+__/__*PQX*= 90 and the result follows. - Set
*g*=*f*^{2}. Note that*g*is continuous,*g*^{3}(*x*) =*x*for all*x*, and*f*(*x*) =*x*for all*x*if and only if*g*(*x*) =*x*for all*x*. Suppose*y*e [0, 1] and*f*(*y*) =/=*y*. Then the numbers*y*,*f*(*y*),*f*^{2}(*y*) are distinct. Replacing*y*with*f*(*y*) or*f*^{2}(*y*) and*f*with*g*if necessary, we may assume that*y*<*f*(*y*) <*f*^{2}(*y*). Choose*a*e (*f*(*y*),*f*^{2}(*y*)). Since*f*is continuous, there exists*p*e (*y*,*f*(*y*)) and*q*e (*f*(*y*),*f*^{2}(*y*)) such that*f*(*p*) =*a*=*f*(*q*). Thus*f*(*p*) =*f*(*q*), hence*f*^{3}(*p*) =*f*^{3}(*q*) and we deduce that*p*=*q*. This is a contradiction because*p*<*f*(*y*) <*q*, and the result follows. - Let the tetrahedron have vertices
*A*,*B*,*C*,*D*and let*X*denote the midpoint of*BC*. Then*AX*= = /2 and we see that*ABC*has area /4. Let*R*,*S*,*T*,*U*denote the regions vertically above and distance at most 1 from*ABC*,*BCD*,*ABD*,*ACD*respectively. Then the volumes of*R*,*S*,*T*and*U*are all /4. Since these regions are disjoint, they will contribute to the volume required.Let

*Y*denote the point on*AX*which is vertically below*D*. Then*Y*is the center of*ABC*(i.e. where the medians meet), in particular__/__*YBX*= p/6 and we see that*BY*= 1/. Therefore*DY*= = and we deduce that*ABCD*has volume(1/3)*(/4)* = /12.Next consider the region which is distance 1 from

*BC*and is between*R*and*S*. We need the angle between*R*and*S*, and for this we find the angle between*DX*and*DY*. Now*DX*=*AX*= /2 and*DY*= . Therefore*XY*= = 1/(2). If q =__/__*YDX*, then sinq =*XY*/*DX*= 1/3. We deduce that the angle between*R*and*S*is p/2 + q = p/2 + sin^{-1}1/3. Therefore the region at distance 1 from*BC*and between*R*and*S*has volume p/4 + (sin^{-1}1/3)/2. There are 6 such regions, which contribute 3p/2 + 3sin^{-1}1/3 to the volume required.For the remaining volume, we shrink the sides of the tetrahedron to zero. This keeps the remaining volume constant, but the volumes above go to zero. We are left with the volume which is distance 1 from the center of the pyramid, which is 4p/3. Since 3p/2 + 4p/3 = 17p/6, we conclude that the volume of the region consisting of points which are distance at most 1 from

*ABCD*is + /12 + 17p/6 + 3sin^{-1}(1/3) 11.77. Other expressions for this are + /12 + 13p/3 - 3cos^{-1}(1/3) and + /12 + 13p/3 - 6sin^{-1}(1/).