Limits of Sequences: 8.4 |s_n||t_n| < \frac{\epsilon}{M}

  • Thread starter Thread starter Artusartos
  • Start date Start date
  • Tags Tags
    Limits Sequences
Artusartos
Messages
236
Reaction score
0
In this link:

http://people.ischool.berkeley.edu/~johnsonb/Welcome_files/104/104hw3sum06.pdf

For number 8.4...

Why don't we just say...

|s_n||t_n| &lt; \frac{\epsilon}{M} M = \epsilon?

Thanks in advance
 
Last edited by a moderator:
Physics news on Phys.org
If you mean, choose N0 such that if n> N0 then |s_n|&lt; \frac{\epsilon}{M} rather than M+1, that would, give us |s_nt_n|= \epsilon, not "<" which is required for the definition of convergence.
 
Last edited by a moderator:
HallsofIvy said:
If you mean, choose N0 such that if n> N0 then |s_n|&lt; \frac{\epsilon}{M} rather than M+1, that would, as you say, give us |s_nt_n|= \epsilon, not &quot;&lt;&quot; which is required for the definition of convergence.
<br /> <br /> No I didn&#039;t say that...<br /> <br /> I said |s_nt_n| &amp;lt; \frac{\epsilon}{M} M = \epsilon. So the first sign is an inequality.
 
Yes, you are right about what you said and I have edited my post to remove "as you said". But you are incorrect that it would be "<". You would have, instead, "=", as I said.
 
Artusartos said:
Why don't we just say...

Probably because "we" didn't think that closely when "we" wrote that paper. If you ensure that ##|s_n|<\frac\epsilon M## what you actually get is
$$
|s_nt_n| = |s_n|\cdot|t_n| < \frac\epsilon M\cdot|t_n| \leq \frac\epsilon M\cdot M = \epsilon,
$$
which is what you have yourself.

If you look at that paper again, you'll see that the author writes
$$
\begin{eqnarray*}
|s_nt_n − 0| & = & |s_n| \cdot |t_n| \\
& < & \left|\frac\epsilon{M + 1}\right| \cdot |M| \\
& < & \epsilon.
\end{eqnarray*}
$$
Why did he suddenly need absolute values in the middle line? I think he probably didn't proofread what he'd written.
 
Michael Redei said:
Probably because "we" didn't think that closely when "we" wrote that paper. If you ensure that ##|s_n|<\frac\epsilon M## what you actually get is
$$
|s_nt_n| = |s_n|\cdot|t_n| < \frac\epsilon M\cdot|t_n| \leq \frac\epsilon M\cdot M = \epsilon,
$$
which is what you have yourself.

If you look at that paper again, you'll see that the author writes
$$
\begin{eqnarray*}
|s_nt_n − 0| & = & |s_n| \cdot |t_n| \\
& < & \left|\frac\epsilon{M + 1}\right| \cdot |M| \\
& < & \epsilon.
\end{eqnarray*}
$$
Why did he suddenly need absolute values in the middle line? I think he probably didn't proofread what he'd written.

Thanks :)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
8
Views
1K
Replies
2
Views
1K
Replies
1
Views
2K
Replies
6
Views
2K
Replies
8
Views
2K
Replies
1
Views
1K
Replies
1
Views
2K
Back
Top