Maths

[AJ]

1.
The sequence (-1)^n(1+1/n), for positive integers n,
(a) has limit point 1
(b) has limit point -1
(c) has limit points 1 and -1
(d) has no limit points

[AJ]

2.

Suppose a random variable X takes values -2, 0, 1, 4 with probabilities 0.4, 0.1, 0.3, 0.2 respectively.
(a) the unique median of the distribution is 1
(b) the unique median of the distribution is 0
(c) the unique median of the distribution lies between 0 and 1
(d) the distribution has multiple medians


I got (c) .. but in key answer given is (d) ... Can we have multiple medians..??

[tim]

Hi for the 1st the answr is no limit points.

[AJ]

@tim..

ya, I too got this.. but in answer key its (c)

[tim]

May b they r takng like this...putng the the value of n..and with subsequent values of n answr cums out to b -1 and 1..
I m nt sure abt it..just givin a try..

[AJ]

Sorry, but can u explain in a bit more detail.....

[Chocolate Frog]

I'll give you a layman's definition of a limit point. For any sequence <a_n>, p is a limit point if an arbitrarily defined neighbourhood of p contains an infinite number of elements of the sequence. Or, for any ε > 0

a_n belongs to ]p-ε, p+ε[ for infinitely many values of n.

You'll see that for <(-1)^n(1+1/n)> = <-2, 3/2, -4/3, 5/4...>, both 1 and -1 satisfy the definition.

[tim]

How cn u get these values of -1 raise to power anything..?

[Chocolate Frog]

Tim, I think the way symbols and equations get rendered here creates confusion. This is what the sequence looks like:

Sequence.docx

[AJ]

@chocolate frog...

ohkayy.. so basically elements of sequence should lie in neighborhood of a point....for that point to be a limit point...

thanks a million...!!! :)

and do u have any other example of a sequence with a limit point.. I just wanna c if my concept of limit point is clear....
or may b if u have some such questions....


And, btw any thoughts on second one...??

[Chocolate Frog]

Two easiest possible examples:

<a_n> = (-1)^n has two limit points; 1 and -1.
<a_n> = (1/n) has one limit point; 0.

As for the second question, the answer should be (d). Median is the midway point in a distribution. Basically, a value x of X is the median if P(X=<x) >= 1/2 and P(X>=x) >= 1/2. This is the case for both X = 0 and X = 1.

[Badmathsboy]

Answer for Q.1 is 'No limit points'..
A seq can converge to only one point.. If there are two points, they are points of accumulation, and not point of convergence.
Answer to Q.2 is D, right explanation above. :)

[anon_econ]

badmathsboy,
limit points r not limits. they r in fact accumulation points.

[anon_econ]

ok i'm not sure if they r the same thing as accumulation points. but limit points r different from limits.

[AJ]

@ chocolate frog..

Thanks, that really helped!

@vasudha
What are accumulation points..?? :/

[anon_econ]

hehe like i said i'm not sure if they r the same thing as limit points or there is some technical difference b/w the two

[Badmathsboy]

Hey, Maybe my post was confusing. Dont confuse limits to limit of a sequence. What I meant to say was a SEQUENCE can only converge to ONE point, [LUB and HLB]. The above sequence tends to 1 and -1 and therefore does not converge. Another way is to check if this is Cauchy sequence, which it is not, not even monotonic.
Hope I am clear.
PS: Also hope i am right. :P

[Chocolate Frog]

I'm afraid you're not, Badmathsboy. Limit of a sequence, if it exists, is unique. But the above sequence tends to cluster around 1 and -1, not converge to any single number. Neither is a limit, but both are limit/cluster/accumulation points. It's a related but different concept. Please refer to the definition I've given above. It allows for multiple limit points to exist.

In contrast, the way limit of a sequence is defined precludes the existence of more than one limit:

<a_n> converges to a, or has a as its limit if for every ε > 0 there exists a number K (usually dependent on ε) such that |a_n - a| < ε for every n>=K.

[Badmathsboy]

Hey, thats exactly what i want to say, the seq clusters around -1 and 1. So these two points are points of accumulation, not limit of the seq (Limit is unique). I dont get your point. Do you argue that the answer is -1 and 1?
Edit: In my previous post, when I say [LUB AND HLB], I mean any of the two, not that BOTH will be limit of the seq.

[Chocolate Frog]

Because limit point is not the same as limit. Did you read the definitions I posted?

Limit point, cluster point and accumulation point are but different names for the same thing.