Need Help

[nasi112]

Question 53 in the video. blackpenredpen says if we take the limit it doesn't work. If I use distribution it works. Why he didn't use the distribution?

 

[Dr.Peterson]

Question 53 in the video. blackpenredpen says if we take the limit it doesn't work. If I use distribution it works. Why he didn't use the distribution?

Please be kind enough to state the problem (or at least give a link to the appropriate place in the video). You should know better by now.

This time I took the time to locate it (3:13:31)and copy the problem, which appears to be to determine convergence of this:

I haven't watched what he does, so I can try to solve it on my own.

Now show what you did; what distribution?? And what limit did he take?

Presumably, you are making a mistake somewhere (unless he turns out to be wrong). But we can't know until we see what you did.

 

[nasi112]

Thanks and sorry. I promise next time I'll be more helpful.

[math]\lim_{n\to \infty} 1 - \cos \frac{1}{n} = \lim_{n\to \infty} 1 - \lim_{n\to \infty} \cos \frac{1}{n} = [/math] div - div

The limit is not zero if I use distribution.

 

[blamocur]

Thanks and sorry. I promise next time I'll be more helpful.

[math]\lim_{n\to \infty} 1 - \cos \frac{1}{n} = \lim_{n\to \infty} 1 - \lim_{n\to \infty} \cos \frac{1}{n} = [/math] div - div

The limit is not zero if I use distribution.

Which will, potentially, make us more helpful too

 

[nasi112]

Thanks. I'm not careless. I just thought the video is there and it doesn't take seconds to go to question 53.

 

[Dr.Peterson]

Thanks and sorry. I promise next time I'll be more helpful.

[math]\lim_{n\to \infty} 1 - \cos \frac{1}{n} = \lim_{n\to \infty} 1 - \lim_{n\to \infty} \cos \frac{1}{n} = [/math] div - div

The limit is not zero if I use distribution.

What does "div - div" mean?

Are you claiming that the difference of two divergent series is divergent? Show me a theorem that says that!

What you've really shown is that the limit is 0; and that makes it possible for the series to converge (but doesn't prove anything). It's if the limit of the terms was not zero that you would know it is divergent.

And what you show is not distribution!

Also, this is not the limit he takes (at least not at the start, which is as far as I have looked so far).

You really must use more words to explain your thinking. Bad thinking is not revealed by minimal communication (and is often associated with it).

Thanks. I'm not careless. I just thought the video is there and it doesn't take seconds to go to question 53.

I didn't say you were careless, but that you were not kind. Wouldn't it be easy enough for you to do that little bit of work (which also included finding your previous question, and so on)? When you ask people for help, it's appropriate make it easier for them.

 

[nasi112]

I forgot to state my last post was a reply to blamocur. Anyway it doesn't matter what I did is called distribution or something else. My point is when blackpenredpen wrote question 53 on the board, he compared it with question 52. He said unfortunately if we have cosine, [imath]\cos \frac{1}{\infty} = 1[/imath] and \(\displaystyle 1 - 1 = 0\). It's like he took the limit without writing it on the board. If he used distribution, I'm sorry I don't know what else to call it, he would end up with each limit not being zero. Then the sum would be div - div. I thought div - div means the series is divergent. div means diverges or divergent.

 

[Dr.Peterson]

My point is when blackpenredpen wrote question 53 on the board, he compared it with question 52. He said unfortunately if we have cosine, [imath]\cos \frac{1}{\infty} = 1[/imath] and \(\displaystyle 1 - 1 = 0\). It's like he took the limit without writing it on the board. If he used distribution, I'm sorry I don't know what else to call it, he would end up with each limit not being zero. Then the sum would be div - div. I thought div - div means the series is divergent. div means diverges or divergent.

Yes, he took the limit of terms in his head, and got 0. By the "divergence test", if the limit of the terms is non-zero, as I mentioned, then that series diverges.

It's true that each of the two series into which you split series 53 diverges. But the theorem about adding two series applies only when they both converge. It does not say that if the two series that are combined diverge, then their sum diverges.

It's essential that you learn theorems to know what you can properly do. Both the divergence test and the summation of series are covered here:

Calculus II - Convergence/Divergence of Series

In this section we will discuss in greater detail the convergence and divergence of infinite series. We will illustrate how partial sums are used to determine if an infinite series converges or diverges. We will also give the Divergence Test for series in this section.

tutorial.math.lamar.edu

In particular, it says this, which refers back to the previous section, here:

​

An example of the latter fact is if [imath]a_n=1[/imath] and [imath]b_n=-1[/imath].

So you are thinking wrongly when you say,

I thought div - div means the series is divergent. div means diverges or divergent.

This is a very important thing to learn!

Are you learning this subject from a textbook?

 

[nasi112]

Thanks Doctor for clearing up my thought. I'm studying infinite series along with other subjects from youtube. Do you think these 100 series are enough to understand the subject or I have to open my calculus's book again?

 

[Dr.Peterson]

Thanks Doctor for clearing up my thought. I'm studying infinite series along with other subjects from youtube. Do you think these 100 series are enough to understand the subject or I have to open my calculus's book again?

I doubt that he explains the theorems thoroughly enough that you could learn well from this alone. Some sort of textbook or equivalent is needed, at least as a supplement.

 

[blamocur]

For this particular problem I would try limit comparison test.

Also, I want to use the occasion to second @Dr.Peterson's advice to use a textbook. I'd also suggest spending some time on simpler problems than the ones in this YouTube video. Once you feel confident with exercises in some basic calculus book you could revisit the video if you are still interested in the topic.

Good luck and have fun!

 

[Dr.Peterson]

Looking at the work in the video, he first tries the limit comparison test with 1/x, and finds that it doesn't work, because that limit is 0, which doesn't help when comparing to a divergent series.

Now, I first considered what I should compare to by using the Taylor approximation to cos(x), and seeing that the term behaves like 1/x^2; that suggests both that it is likely to converge, and that comparison to 1/x^2 makes sense.

And in fact, that's what he does next, without as good a reason for that choice.

I do notice that he speeds through certain parts, assuming that you know all the theorems, and does not show the work as carefully as a textbook would.

 

[nasi112]

I understood how he applied the limit comparison test. It makes sense that he first used [imath]\frac{1}{n}[/imath] since it's included in the expression [imath]\cos \frac{1}{n}[/imath]. Using [imath]\frac{1}{n^2}[/imath] as the second comparison also makes sense, as it's related to the derivative of the first term. You Doctor have a good vision to introduce the Taylor series of the cosine, I don't. Even if the methods used in this video don't make sense and don't fit, I will try them. You told me in the previous thread of mine, "Usually. But sometimes, when you don't see that any of the familiar methods exactly fits, you can try using methods in slightly different ways. And they may work." It's true that [imath]\frac{1}{n}[/imath] and [imath]\frac{1}{n^2}[/imath] were used within the same method, but the expressions themselves are slightly different. I studied calculus in Hebrew, and it's challenging to follow how the theorems are explained. I think I need a calculus textbook in English.

 

[Dr.Peterson]

Deciding what to compare to can be tricky; you do just have to look for ideas, sometimes based on your first impression of whether it might converge or not, and sometimes on the form of the term. And, yes, if the first thing you try doesn't work, that can give you ideas for what to try next, as he did. That's one benefit of a video like that, where he is just showing you his own thinking (sometimes even slowly enough that you can follow it!).

But there's a reason I gave you links to a decent online textbook (in English)! It might be worth looking at, to see if his style fits you. I've recommended it for a long time.

 

[blamocur]

I studied calculus in Hebrew, and it's challenging to follow how the theorems are explained. I think I need a calculus textbook in English.

IMHO, the quality of the textbook is more important than the language it is written in. Translating online is real easy these days, so I'd look for a book with a bunch of exercises and answers. But it is true that a textbook in English would make it easier to ask questions on an English-only forum.