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find max,min,sup,inf..

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
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

MHB Seeker
Staff member
Mar 5, 2012
8,774
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!! :D

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

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
Hola!! :D

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!!! :eek: 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? (Thinking)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Oh,yes!!!Right!!! :eek: 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? (Thinking)
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

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
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

MHB Seeker
Staff member
Mar 5, 2012
8,774
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

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
Yep. :)
Care to make a new guess about the largest value?
And what about the smallest value?
Hmm..I don't really know.. (Thinking) 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 $\frac{\infty}{\infty}$ (Worried).. Could you give me a hint how I can find it? :eek:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Hmm..I don't really know.. (Thinking) 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 $\frac{\infty}{\infty}$ (Worried).. Could you give me a hint how I can find it? :eek:
What is the following limit?
$$\lim_{m \to \infty} \frac m {2+m} =\lim_{m \to \infty} \frac {1} {\frac 2 m + 1}$$
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
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?? :rolleyes:
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

MHB Seeker
Staff member
Mar 5, 2012
8,774
It is equal to $1$!!! So,this is the supremum,right?? :rolleyes:
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! (Mmm)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,718

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
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

Well-known member
MHB Site Helper
Apr 13, 2013
3,718
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!!! :)