# null sequences

Watch
Announcements
#1
Hi

I've been looking at the definition of a null sequence and don't quite understand it.

My understanding of it leads to the notion that every sequence is a null sequence which is obviously false.

I don't quite understand the mathematical statement with the three quantifiers is the issue I think
0
6 years ago
#2
(Original post by Jam')
...
A null sequence is one that converges to zero.

In terms of the definition. Choose any distance from zero (that's our epsilon), then there is a point in the sequence, N, beyond which (n>N) all values, f(n), in the sequence lie within epsilon of zero. |f(n)| < epsilon.
1
6 years ago
#3
(Original post by Jam')
I don't quite understand the mathematical statement with the three quantifiers is the issue I think
Yes, this is the problem.

The definition says:

.

Rewriting this to be a bit wordier, we get: "for all natural numbers n, the statement "" implies |f(n)| < epsilon.

Your interpretation is more like: "the statement "for all natural numbers n, " implies |f(n)| < epsilon". (If you wrote this with quantifies it wouldn't look like the definition, it would be more like: ).

As far as understanding the definitions goes, it's probably best not to think too much in terms of the quantifier symbols - they save writing, but they don't help clarity.

f(n) is null if given we can find N such that whenever n is an integer > N, we have .
2
#4
(Original post by DFranklin)
Yes, this is the problem.

The definition says:

.

Rewriting this to be a bit wordier, we get: "for all natural numbers n, the statement "" implies |f(n)| < epsilon.

Your interpretation is more like: "the statement "for all natural numbers n, " implies |f(n)| < epsilon". (If you wrote this with quantifies it wouldn't look like the definition, it would be more like: ).

As far as understanding the definitions goes, it's probably best not to think too much in terms of the quantifier symbols - they save writing, but they don't help clarity.

f(n) is null if given we can find N such that whenever n is an integer > N, we have .
Ohhhh thanks I think I sort of get this. Is it saying therefore that as epsilon tends to 0, f(n) n-> infinity?

And because |f(n)| < Epsilon => lim n-> infinity f(n) = 0? Hence why the sequence converges?
0
#5
(Original post by ghostwalker)
A null sequence is one that converges to zero.

In terms of the definition. Choose any distance from zero (that's our epsilon), then there is a point in the sequence, N, beyond which (n>N) all values, f(n), in the sequence lie within epsilon of zero. |f(n)| < epsilon.
Thanks the picture is much clearer thanks to you and Dfranklin
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (68)
14.81%
I'm not sure (16)
3.49%
No, I'm going to stick it out for now (154)
33.55%
I have already dropped out (8)
1.74%
I'm not a current university student (213)
46.41%