37. Q. What is wrong with the following argument?
Let R e the real numbers
(1) "For all x,y in R, f(x)+f(y)=f(xy)
is equivalent to
(2) For all x,y in R, f(x) +f(y)= f((x)y)
which is equivalent to
(3) For all x,y in R, f(x) + f(y) = f((x)y)=f(x(y))= f(x)+ f(y)
From this for y=0, we make the conclusion that
(4) For all x in R, f(x)=f(x)
Since the steps are reversible, any function with property (4) has property (1). Therefore, for all x,y in R cos x+ cos y =cos(xy)
(A) (2) does not imply (1)
(B) (3) does not imply (2)
(C) (3) does not imply (4)
(D) (4) does not imply (3)
(E) (4) is not true for f = cos
I thought (2) implies (1), and (3) implies (2). What's wrong with this question
50. In a game two players take turn tossing a fair coin; the winner is the first one to toss a head. The probability that the player who makes the first toss wins the game is
The answer is 2/3 but I thought it's 3/4.
53 Let V be the vector space, under the usual operations, of real polynomials that are of degree at most 3. Let W be the subspace of all polynomials p(x) in V such that P(0)=P(1)=P(1)=0. Then dim V + dim W is
The answer is 5.
Can anyone help me? Thanks a lot
GR8767 #37, #50 #53
Re: GR8767 #37, #50 #53
37. If you reverse the steps and go from 4 to 3 you have to start with the fact that y=0 then there is no way that you can continue to work backwards as you will have f(x)+f(0)=f((x)0)=f(x(0)) which doesn't help
50. You are looking for the value of the following series
$$\sum_{k=0}^\infty (1/2)^{2k+1}$$
you can break that up into two series then use the formula for the geometric series to solve and get your answer
$$\sum_{k=1}^\infty (1/2)^k\sum_{k=1}^\infty (1/2)^{2k}$$
the value of the first series is equal to 1 and the second series becomes
$$\sum_{k=1}^\infty (1/4)^{k}$$
the value of this series is equal to 1/3 so 1(1/3)=2/3
53. Since the polynomials are of a degree at most three the dim V is equal to 4 since there are the coefficients for the x^3, x^2, x terms and coefficients with no x. It then tells us that the nullity is 3 so using rank plus nullity thorem dim W=43=1. So 4+1=5
50. You are looking for the value of the following series
$$\sum_{k=0}^\infty (1/2)^{2k+1}$$
you can break that up into two series then use the formula for the geometric series to solve and get your answer
$$\sum_{k=1}^\infty (1/2)^k\sum_{k=1}^\infty (1/2)^{2k}$$
the value of the first series is equal to 1 and the second series becomes
$$\sum_{k=1}^\infty (1/4)^{k}$$
the value of this series is equal to 1/3 so 1(1/3)=2/3
53. Since the polynomials are of a degree at most three the dim V is equal to 4 since there are the coefficients for the x^3, x^2, x terms and coefficients with no x. It then tells us that the nullity is 3 so using rank plus nullity thorem dim W=43=1. So 4+1=5

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Re: GR8767 #37, #50 #53
wow, toughluck, you got the game sewed up, as they say.

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Re: GR8767 #37, #50 #53
For 37, I think the "let y=0" was part of the argument for 3=> 4 but not part of 3 or 4.
1. f(x) + f(y) = f(xy)
2. f(x) + f(y) = f((x)y)
3. f(x) + f(y) = f(x) + f(y)
4. f(x) = f(x)
1=> 2=> 3=> 4 is clear.
Working backwards, 4=> 3 is in fact true:
f(x) + f(y) = f(x) + f(y) = f(x) + f(y)
The trouble is with 3=> 2.
This is clear is you let f = cos(x). Clearly 3 holds. But for x = 0, y = 0, we have cos(x) + cos(y) = 1 + 1 = 2, which is not cos((x)y) = 1.
1. f(x) + f(y) = f(xy)
2. f(x) + f(y) = f((x)y)
3. f(x) + f(y) = f(x) + f(y)
4. f(x) = f(x)
1=> 2=> 3=> 4 is clear.
Working backwards, 4=> 3 is in fact true:
f(x) + f(y) = f(x) + f(y) = f(x) + f(y)
The trouble is with 3=> 2.
This is clear is you let f = cos(x). Clearly 3 holds. But for x = 0, y = 0, we have cos(x) + cos(y) = 1 + 1 = 2, which is not cos((x)y) = 1.
Re: GR8767 #37, #50 #53
4 => 3 is not true. The fact that f(x) = f(x) does not imply that f(x) + f(y) = f((x)y), ie that f(x) + f(y) = f(xy). And the counter example is given in the problem, cos(x) satisfies cos(x) = cos(x), but it's not necessarily true that cos(x) + cos(y) = cos(xy), as well all know. The fact that 3 => 2 is very trivial. The fact that f(x) + f(y) = f((x)y) = f(x(y)) = f(x) + f(y), has the equality f(x) + f(y) = f((x)y) as the very first two terms.

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Re: GR8767 #37, #50 #53
You're right. I totally disregarded the middle two terms in the equality of 3.

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Re: GR8767 #37, #50 #53
For problem #53, I was able to solve it without using the RankNullity Theorem, which I find a bit confusing here.
dim(V) is clearly 4 like toughluck pointed out, but then for the subspace W it seems easier to me to note that this is just the space of all polynomials of the form:
c[x(x1)(x+1)]=0
You could multiply that all out and get c(x^3x)=0, but no matter what you only have one coefficient c. This means that dim(W)=1.
dim(V) is clearly 4 like toughluck pointed out, but then for the subspace W it seems easier to me to note that this is just the space of all polynomials of the form:
c[x(x1)(x+1)]=0
You could multiply that all out and get c(x^3x)=0, but no matter what you only have one coefficient c. This means that dim(W)=1.