## 0568 #56 #57 #63

Forum for the GRE subject test in mathematics.
sjin
Posts: 10
Joined: Sat Sep 19, 2009 5:55 pm

### 0568 #56 #57 #63

56 Which of the following does not define a metric on the set of all real members?
(A) F(x,y)= 0 if x=y and 2 if x!=y

(b) F(x,y)= min{ abs(x-y),1}

(c) F(x,y)= abs(x-y)/3

(d) F(x,y)= abs(x-y)/(abs(x-y)+1)

(e) F(x,y)= (x-y)^2

I can understand (a) (c) and (d) but how to know (b) is a metric but (e) is not.

57 The set of real numbers x for which the series Sum ( n!*x^(2n)/n^(n)*(1+x^(2n))
converges is
(a) {0} (b) {x:-1<x<1} (c) {x:-1<=x<=1} (d) {x:-sqrt(e)<=x<=sqrt(e)} (e) R

I used the ratio test to get the limit but there should be a trick in doing this and I can't figure it out.

63 If f is the function defined by f(x)= x*exp(-x^2-x^(-2) if x!=0 and =0 if x=0, at how many values of x does the graph of f have a horizontal tangent line?

diogenes
Posts: 73
Joined: Sun Aug 31, 2008 9:31 pm

### Re: 0568 #56 #57 #63

sjin wrote:56 Which of the following does not define a metric on the set of all real members?

(b) F(x,y)= min{ abs(x-y),1}

I can understand (a) (c) and (d) but how to know (b) is a metric but (e) is not.

Hmm...this is way too long, but here we go. Note that:

$F(x,y)\geq 0$ when $|x-y|\geq 0.$

Now, when x=y |x-y|=0 implies that F(x,y)=0 and if F(x,y) =0 , then x=y.
So, $F\geq 0$ and F =0 when x=y.

|x-y| = |y-x| implies that F(x,y) = F(y,x).

Finally, we need to prove the triangle inequality.

Let x,y,z be in the the universe of discourse X.

Note $F(x,y)\leq 1$ for all x,y, in X.

So,if $F(x,z)=1, or F(z,y)=1$, then $F(x,y)\leq F(x,z)+F(z,y)$.

However, if F(x,z)<1, or F(z,y)<1, then F(x,z)=|x-z| and F(z,y)= |z-y|.

So, $F(x,y) = min(1,|x-y|)\leq |x-y|\leq |x-z|+|z-y| = F(x,z)+F(z,y)$.

Edit: Clearly, not everyone taking this test will have had a, e.g., functional analysis class. So, I am curious as to what "meta-level" concept would have unlocked this one.
Last edited by diogenes on Thu Oct 01, 2009 8:08 pm, edited 1 time in total.

Posts: 9
Joined: Tue Sep 29, 2009 1:07 pm

### Re: 0568 #56 #57 #63

In question 56 the function in (e) is not a metric because it doesn't satisfy the triangle inequality. For instance, if we take x=1, y=0 and z=-1, then F(x,z) = 4 while F(x,y) + F(y,z) = 2.

In question 57, your attempt to use the ratio test is correct. After you do some algebra to simplify a_{n+1}/a_n, try considering 3 cases: 1) |x|<1; (2) |x|=1 and (3) |x|>1. You'll see that in the 3 cases the desired limit is less than 1, which implies that the series converges for all real numbers.

In question 63 all you need to do is to compute f' and find where it is zero. These points are -1,0,1. However, you must be careful with the derivative at 0 because of the definition of f, so you must use the definition of f'(0) as a limit.

Swimguy112
Posts: 7
Joined: Wed Aug 12, 2009 11:42 am

### Re: 0568 #56 #57 #63

I tried the ratio test for 57 but it didn't give me anything easy to work with. I just tried the following argument instead:

We want to find when the Sum[n!*x^(2n)]/[(n^n)*(1+x^(2n))] converges. In order to eliminate some of the possible answers you can use the comparison test.

Since [n!*x^(2n)]/[(n^n)*(1+x^(2n))] < [n!*x^(2n)]/[(n^n)*(x^(2n))] for all n in the natural numbers

( I basically removed the one from the denominator)

that means that

if SUM of [n!*x^(2n)]/[(n^n)*(x^(2n))] converges, so must Sum[n!*x^(2n)]/[(n^n)*(1+x^(2n))].

When you simplify the new sum you'll see that it will converge regardless of the value of x. Therefore if the sum Sum[n!*x^(2n)]/[(n^n)*(x^(2n))] converges for all real values of x then Sum[n!*x^(2n)]/[(n^n)*(1+x^(2n))] must also converge for all real values of x.

Hope this makes sense. Good Luck!

jayre
Posts: 11
Joined: Wed Nov 04, 2009 10:30 pm

### Re: 0568 #56 #57 #63

conrado.math wrote:In question 63 all you need to do is to compute f' and find where it is zero. These points are -1,0,1.

I've got f'(0) = 0 from the definition of a derivative at x = 0. I'm having trouble finding the other two roots of f'(x). I have f'(x) = Exp(-x^2-x^-2)*(1+2x^2-2x^-2), so if f'(x) = 0, (1+2x^2-2x^-2) = 0, but I can't find the roots to this. In particular, -1 and 1 don't seem to work. I don't think I'm calculating the derivative incorrectly. Any ideas?

YKM
Posts: 45
Joined: Mon Aug 08, 2011 10:25 am

### Re: 0568 #56 #57 #63

jayre wrote:
conrado.math wrote:In question 63 all you need to do is to compute f' and find where it is zero. These points are -1,0,1.

I've got f'(0) = 0 from the definition of a derivative at x = 0. I'm having trouble finding the other two roots of f'(x). I have f'(x) = Exp(-x^2-x^-2)*(1+2x^2-2x^-2), so if f'(x) = 0, (1+2x^2-2x^-2) = 0, but I can't find the roots to this. In particular, -1 and 1 don't seem to work. I don't think I'm calculating the derivative incorrectly. Any ideas?

I have problem with this too, f'(x) = 0 iff (1 - 2x^2 + 2x^(-2) ). Can anyone show me how to do this problem?

YKM
Posts: 45
Joined: Mon Aug 08, 2011 10:25 am

### Re: 0568 #56 #57 #63

Sorry, I mean f'(x) = 0 iff (1 - 2x^2 + 2x^(-2)) = 0.

yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

### Re: 0568 #56 #57 #63

have problem with this too, f'(x) = 0 iff (1 - 2x^2 + 2x^(-2) ). Can anyone show me how to do this problem?

Hi, I was a bit puzzled by this question but I think I figured it out now.

(1 - 2x^2 + 2x^(-2)) = 0
=> $x^2-2x^4+2=0$, this is a quadratic equation in x^2, solve it and we would have two roots.

The third point where f'(x)=0 is 0, since $\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=0$ due to the $e^{(-x^2-x^{-2})}\to e^{-\infty} \to 0}$,

fulufhelo
Posts: 1
Joined: Wed Feb 20, 2013 4:30 pm

### Re: 0568 #56 #57 #63

pls cn u pls help me with the following question:does d(x,y)=(x-y)^2 define a metric on the set of real numbers

DDswife
Posts: 116
Joined: Thu Aug 14, 2014 5:29 pm

### Re: 0568 #56 #57 #63

No. As someone already posted above, the triangular property is the problem