find max,min,sup,inf..

[evinda]

Hey!!!
I am given the following exercise:
Find max,min,sup,inf of the set:
$$B=\{\frac{(-1)^{n}m}{n+m},n,m=1,2,....\}$$
I thought the following:
$$supB=\frac{1}{3},maxB=\frac{1}{3},infB=0, \nexists minB$$

Are the above right?? If yes,how can I prove that it is actually like that??

 

[Klaas van Aarsen]

Hey!!!
I am given the following exercise:
Find max,min,sup,inf of the set:
$$B=\{\frac{(-1)^{n}m}{n+m},n,m=1,2,....\}$$
I thought the following:
$$supB=\frac{1}{3},maxB=\frac{1}{3},infB=0, \nexists minB$$

Are the above right?? If yes,how can I prove that it is actually like that??

Hola!!

Suppose we pick $m=n=2$, then I get \(\displaystyle B=\frac 2 4\).
Isn't that greater than your $\sup B$?

And suppose we pick $m=n=1$, isn't $B$ smaller than your $\inf B$ then?

 

[evinda]

Hola!!

Suppose we pick $m=n=2$, then I get \(\displaystyle B=\frac 2 4\).
Isn't that greater than your $\sup B$?

And suppose we pick $m=n=1$, isn't $B$ smaller than your $\inf B$ then?

Oh,yes!!!Right!!! So:
$$ supB=\frac{1}{2},maxB=\frac{1}{2},infB=\frac{-1}{2},minB=\frac{-1}{2}$$

Is there a way to prove that it is like that?

 

[Klaas van Aarsen]

Oh,yes!!!Right!!! So:
$$ supB=\frac{1}{2},maxB=\frac{1}{2},infB=\frac{-1}{2},minB=\frac{-1}{2}$$

Is there a way to prove that it is like that?

Good!

... but suppose we pick $m=3, n=2$, then I get \(\displaystyle B=\frac 3 5\).
Isn't that greater than your new $\sup B$?

Let's take a look at B.
$$B=\frac{(-1)^n m}{n+m}$$

To find the largest value, we need a numerator that is a big as possible, and a denominator that is as small as possible.
Oh, and we have the additional condition that the fraction should be positive.

Suppose we pick a "large" value for the number in the numerator. Say $m=1000$.
What should you pick for $n$ to make $B$ as big as possible?

 

[evinda]

Good!

... but suppose we pick $m=3, n=2$, then I get \(\displaystyle B=\frac 3 5\).
Isn't that greater than your new $\sup B$?

Let's take a look at B.
$$B=\frac{(-1)^n m}{n+m}$$

To find the largest value, we need a numerator that is a big as possible, and a denominator that is as small as possible.
Oh, and we have the additional condition that the fraction should be positive.

Suppose we pick a "large" value for the number in the numerator. Say $m=1000$.
What should you pick for $n$ to make $B$ as big as possible?

I understand..I think that we should pick then $n=2$..Right?

 

[Klaas van Aarsen]

I understand..I think that we should pick then $n=2$..Right?

Yep.
Care to make a new guess about the largest value?
And what about the smallest value?

 

[evinda]

Yep.
Care to make a new guess about the largest value?
And what about the smallest value?

Hmm..I don't really know.. If we want the numerator to be as big as possible,we have to take $ m\to \infty$ and $n=2$..But then we will have $\frac{\infty}{\infty}$ .. Could you give me a hint how I can find it?

 

[Klaas van Aarsen]

Hmm..I don't really know.. If we want the numerator to be as big as possible,we have to take $ m\to \infty$ and $n=2$..But then we will have $\frac{\infty}{\infty}$ .. Could you give me a hint how I can find it?

What is the following limit?
$$\lim_{m \to \infty} \frac m {2+m} =\lim_{m \to \infty} \frac {1} {\frac 2 m + 1}$$

 

[evinda]

What is the following limit?
$$\lim_{m \to \infty} \frac m {2+m} =\lim_{m \to \infty} \frac {1} {\frac 2 m + 1}$$

It is equal to $1$!!! So,this is the supremum,right??
And,to find the infimum,we pick $n=1$ and calculate the limit $\lim_{m \to \infty}\frac{-m}{1+m}=-1$,so the infimum is $-1$,or am I wrong?
So,$supB=1$ and $infA=-1$ ?

 

[Klaas van Aarsen]

It is equal to $1$!!! So,this is the supremum,right??
And,to find the infimum,we pick $n=1$ and calculate the limit $\lim_{m \to \infty}\frac{-m}{1+m}=-1$,so the infimum is $-1$,or am I wrong?
So,$supB=1$ and $infA=-1$ ?

Yes!

 

[evinda]

Yes!

And... there is no min and max of the set,right?

 

[Klaas van Aarsen]

And... there is no min and max of the set,right?

Right!

 

[evinda]

Right!

And how can I prove it?

 

[Klaas van Aarsen]

And how can I prove it?

Note that since $m,n \ge 1$, we have that:
$$-1 < \frac{(-1)^n m}{n+m} < 1$$
This is true because the absolute value of the numerator is always smaller than the denominator.

Furthermore, we have already found sequences approaching both $-1$ and $1$.

Therefore $\sup B = 1, \inf B = -1$, and $\min B, \max B$ do not exist. $\qquad \blacksquare$

 

[evinda]

Note that since $m,n \ge 1$, we have that:
$$-1 < \frac{(-1)^n m}{n+m} < 1$$
This is true because the absolute value of the numerator is always smaller than the denominator.

Furthermore, we have already found sequences approaching both $-1$ and $1$.

Therefore $\sup B = 1, \inf B = -1$, and $\min B, \max B$ do not exist. $\qquad \blacksquare$

Great!!!Thank you very much!!!