#1 | Benson #2 | #3 | #4 | #5 | #6 | #7 |
#8 | #9 | #10 | #11 | Oleynick #12 | #13 | #14 |
2.Let f:[0,3]->R be a continuous function. Suppose
f(0)-f(1)=f(2)-f(3).
Prove that there exists x in the interval [0,1] such that
f(x)+f(x+2)=2f(x+1).
Let g(x)=f(x)+f(x+2)-2f(x+1). We will try to show that g has a root
in the interval [0,1], which will show that for some x in [0,1]
0=f(x)+f(x+2)-2f(x+1) which implies f(x)+f(x+2)=2f(x+1). Note that
g(0)=f(0)+f(2)-2f(1) and g(1)=f(1)+f(3)-2f(2). But from our given
information,
f(3)=f(2)+f(1)-f(0). Thus g(1)=f(1)+[f(2)+f(1)-f(0)]-2f(2) so that
g(1)=-f(0)-f(2)+2f(1). We see that g(0)=-g(1). By trichotomy either
g(0) is positive, negative, or zero. If g(0) is 0, then 0 is a root
of g and we are done. If g(0) is postive, we have that g(0)>0>g(1),
and if g(0) is negative we have that g(0)<0 |
12.Let c be a cluster point of set A. Let L be a real number. Let f: A
-> R be a function such that lim x->c f(x) = L and f(x) not = L for each
x in A. Prove that L is a cluster point of { f(x) : x in A }.
To show that L is a cluster point of { f(x) : x in A }, we must show that for all delta_{B}>0, there exists a y in { f(x) : x in A }, y != L, such that |y - L| < delta-sub-B. We will demonstrate how to find a value y, given any value of delta_{B}>0. Because c is a cluster point of A and lim x->c f(x) = L, by the definition of the limit of a function, for all epsilon>0, there exists a delta_{F(epsilon)}>0 such that |f(x) - L|<epsilon when x in A and 0<|x - c|<delta_{F(epsilon)}. Because c is a cluster point of A, by the definition of cluster point, for all delta_{A}>0, there exists an x in A, x not= c, such that |x - c|<delta_{A}. For any value delta_{B}, if we take delta_{F(deltaB)} as the value of delta_{A}, then we know that there exists a z in A such that z not= c and |z - c|<delta_{F(deltaB)}. But, because lim x->c f(x) = L, and z in A, and 0<|z - c|<delta_{F(deltaB)}, we also know that |f(z) - L|<delta_{B}. Because f(x) not= L for all x in A, we also know that f(z) not= L. So, given any delta_{B}, take the value f(z) as y. y = f(z) so y is in { f(x) : x in A }, y not= L, and |y - L|<delta_{B}. Therefore, L is a cluster point of { f(x) : x in A }. Sent by John Oleynick on Sat, 12 Apr 2003 18:52:11 |