Tuesday, March 27, 2012

Class Discussion due 20120401

Again never a due date. Keep asking questions.

Things are getting busy for me, but that doesn't mean I won't be there to help! With that being said, consider the many other useful options. Graduate students in the math help room should usually be able to help with conceptual questions as well as homework questions. Dr. Brown also has office hours. I rarely see anybody during mine, so do drop by!

But I like these tips a high school teacher once told the class:
"College is NOT a big high school."
: He literally emphasized the word 'not'.
"Work smart, not hard."
: Don't take this literally. You should work hard, but you should work hard on the right things. In other words, you don't go into the forest and cut down a bunch of trees if you want to build a brick house. For example, pay attention to concepts which Dr. Brown emphasizes.

53 comments:

  1. I was completing the homework for 7.8 and came across an interesting question. Problem 32 asks to find if the integral of 1/(1-x^2)^(1/2) is convergent or divergent from 0 to 1. I know that for 1 the integral is 1/0... but when you integrate you get sin^-1(x) from 1 to 0 so the problem works out well. I guess my question here is that do we determine whether or not the problem is improper by the original integral or its integration value.

    in this case
    1/(1-x^2)^(1/2)
    or
    sin^-1(x)

    Thanks :)

    ReplyDelete
    Replies
    1. I answered this for you in class, but as a reference, then the integral is improper, because of the original function. Whether or not it converges depends on after we take the integral.

      We must keep in mind that before coming to improper integrals, the construction of the integral was only valid for bounded functions on a finite interval. Section 7.8 allows us to give meaning to other cases we might come across.

      Delete
    2. Looking at the book, I have a slight correction to my statement. The definition of the definite interval (in the book you have) does not require the function to be bounded. Even so, section 7.8 helps develop what it means to integrate over infinite intervals and to integrate a function defined only on a half-open or open interval.

      Delete
  2. Tim,
    When trying to solve 7.8.38, I am having trouble to integrate the function, therefore, I can't find the limit value. Could you go over it in section?

    Thanks,

    Marcelo

    ReplyDelete
    Replies
    1. I went over this with Marcelo after the end of class, if anybody else needs help with it let me know. But also state the problem, because I left my book at the office.

      Delete
  3. Tim,
    I was trying to solve 7.8.56, and I'm having trouble stating if its divergent or convergent.

    thank you,
    Nahyr

    ReplyDelete
    Replies
    1. Oh no, I left my book at the office, what is the statement of the problem?

      Delete
    2. You'll first want to compute the indefinite integral. Whether or not it is divergent or convergent depends on whether the limits exist.

      How should we evaluate the integral?

      The form of the integrand suggests we make a trigonometric substitution. Do you agree?

      Don't forget that after choosing $x=2\sec(\theta)$, we should also have $dx = 2 \sec(\theta) \tan(\theta) d\theta$.

      Delete
  4. Are there any polar curves with periods other than $\pi$ and $2\pi$?

    ReplyDelete
    Replies
    1. Prove that a polar curve of the form $r(\theta)$, excluding $r=0$, which is periodic (repeats itself) will have a period which is a multiple of $\pi$.

      If you believe that result, then convince yourself that the smallest multiple of $\pi$ for which a polar curve can repeat itself is $\pi$. Argue that such a curve satisfies the property $r(\theta+\pi)=-r(\theta)$.

      Delete
  5. Hi Tim,

    I thought it was interesting today how Professor Brown showed us how to represent a repeating decimal in terms of a geometric series and I was wondering if this method can be used to approximate decimals, like e?

    -Erica

    ReplyDelete
    Replies
    1. The decimal representation of a rational number either stops or eventually repeats itself. A decimal number that either stops or eventually repeats itself is a rational number. $e$, however, is an irrational number. Thus $e$ cannot be represented as a geometric series.

      Delete
    2. However, when we get to section 11.8, we'll have a series for the function $e^x$ and plugging in $x=1$, we can use the terms in that series to approximate $e$.

      Delete
  6. I believe there was a question similar to this posted earlier, but it hadn't been answered. When you're determining if an integral is improper or not, do you take the integral first and then see if you can evaluate the resulting function at those values, or do the limits of integration need to work in the original integral before you actually integrate that function. I'm at problem 32 which has this issue.

    ReplyDelete
    Replies
    1. I got around to answering this question above. CTRL+F "Sophia".

      Delete
  7. So just to make sure, is it ok to solve the indefinite integral first, and then apply the limit to solve the improper integral?
    Tanks.

    ReplyDelete
    Replies
    1. Yes. But as we saw in class, it might actually be useful to change the limiting variable instead of going back to the original variable.

      Yu're Wlcome. Aytime.

      Delete
  8. I have another question. Basically, if you're given an improper integral and told to evaluate it using the Comparison Theorem (like 7.8 problem 52), what do you compare it to? One of the examples in the book (and I think Prof. Brown might have done in class) uses 1/x to define the integral of (1+e^-x)/x from 1 to infinity as divergent. But that choice 1/x seemed kind of arbitrary? Is that the comparison we should usually go to? How do we know what to pick to compare our function to?

    ReplyDelete
    Replies
    1. I suppose it depends on your point of view. To me it's the obvious choice.

      In discussion section, we had the following:
      $$\int_a^\infty \frac{1}{x^p}dx$$ converges for $p>1$ and diverges for $p\leq 1$, where $a>0$.

      $$\int_a^\infty e^{-x}dx$$ converges, where $a\in \mathbb{R}$.

      If we want the integral of a function from $a$ to $\infty$ to converge, we should try to show it is less than $e^{-x}$ or $\frac{1}{x^p}$ for some $p>1$.

      If we want the integral of a function from $a$ to $\infty$ to diverge, we should try to show it is greater than $\frac{1}{x^p}$ for some $p\leq 1$.

      These don't exhaust the choices of functions to compare to, but they are fundamental when it comes to using the comparison theorem.

      Delete
    2. We also have
      $$\int_0^a \frac{1}{x^p}dx$$ converges for $p<1$ and diverges for $p\geq 1$, where $a>0$.

      Delete
  9. Just clarifying, just because the improper integrand converges does not mean that the improper integral as a whole converges on the designated interval, right?

    ReplyDelete
    Replies
    1. What do you mean by the integrand converging?

      Delete
    2. Dr. Brown responded to this in class today: It is a comment asking whether or not the function F(x) inside the integral of F(x)has any effect on the convergence of the integral, just because F(x) converges as it approaches infinity. The answer was that it does not have any effect, because the integral is an entirely different function from the original F(x).

      Delete
    3. Thank you Joshua! That's exactly what I was trying to clarify!

      Delete
  10. Today Professor Brown mentioned that you can add series, subtract series, etc. because they behave like limits in some sense. I was wondering, can we take derivatives and antiderivatives of series?

    Thanks!

    -Erica

    ReplyDelete
    Replies
    1. You must be talking about power series. Your answer is on page 748.

      Delete
  11. I was wondering if there is an easy way to spot a pattern for geometric series, in the case where a and r are not already in the correct form. Could we do some recognition in section for figuring out if a series is geometric?

    ReplyDelete
    Replies
    1. Part of this reply is a subset of CTRL+F "Mario".

      When you see a series, you should be able to identify if it is a geometric series. Read over example 2 on page 705. Then see if you can spot all the geometric series on page 711, numbers 5 through 42. Post your answer here and I'll confirm. (Anybody can do this exercise).

      Numbers 57 through 63 are all geometric series.

      The remainder of this reply is a rewording of CTRL+F "Summarize the strategy".

      I will do recognition with y'all in class (evens on page 740). I recommend you do the odds on your own. At first read page 739. Try to memorize and/or learn as much of the strategy outlined as you can. Get help on the ones that give you difficulty.

      Delete
  12. Professor Brown has mentioned in class before about using L'Hospital's rule in sequences and series by associating them with a function? Is this right? and how do you do this?

    Michelle Bohrson

    ReplyDelete
    Replies
    1. We can associate a sequence with a function and use L'Hospital's rule as follows:

      Suppose we have a sequence ${a_1, a_2 ,\dots,a_n,\dots}$ which is given by some formula $f(n)$, for $n=1,2,\dots$.

      Our goal is to compute the limit of $f(n)$, where $n$ is a positive integer, as $n$ goes to $\infty$.

      To use L'Hospital's rule, instead of $f(n)$, we extend the function $f(n)$ to the real numbers. So now we have $f(x)$, where $x$ is a real number.

      We needed the function on the real line in order to take derivatives.

      Example:
      Compute $$\lim _{n\to \infty} \frac{\log(n)}{n}$$.

      Then we consider the function $\log(x)/x$, which is only defined for $x>0$, but we are taking the limit, so we're okay.

      Using L'Hospital's rule, we get
      $$\lim _{x\to \infty} \frac{\log(x)}{x}=\lim_{x\to\infty} \frac{1/x}{1}=0$$.

      Delete
    2. The analogue to a series is an integral. There are theorems in your book that relate the two. We can use integrals to determine whether a series converges or diverges, and when it converges, estimate the value of the series.

      Example:
      Determine whether $$\sum _{n=1}^\infty \frac{1}{n^2}$$ converges or diverges. If it converges, estimate it's value.

      So we consider the function $\frac{1}{x^2}$. This is a decreasing function, and looking at the graph, we'd see that $\int _1 ^\infty \frac{1}{x^2}dx<\sum _{n=1}^\infty \frac{1}{n^2}<1+\int _1 ^\infty \frac{1}{x^2}dx$.

      Now $\int_1 ^\infty \frac{1}{x^2}dx=-\lim_{a\to \infty}\left[ \frac{1}{3x^3}\right] _1^a=\frac{1}{3}$

      We conclude $\sum _{n=1}^\infty \frac{1}{n^2}$ converges and is a number between $\frac{1}{3}$ and $\frac{4}{3}$

      April 2, 2012 11:15 AM

      Delete
  13. I have learned the difference between series and sequences. I also learned there are many techniques to finding the sum of a series and determining whether it converges or diverges. As for figuring out which method to use it is sort of like determining the different methods of integrating, you just have to identify key features in the problem to know which method to use, and practice to figure it out?

    ReplyDelete
    Replies
    1. Exactly. Convergence for sequences is more or less a review of limits of functions as $x$ goes to $\infty$. You'll want to look at section 11.7 on page 739 for practicing and identifying convergence and divergence of series. Also, CTRL+F "Summarize".

      Delete
  14. I think Professor Brown's approach to teaching us different methods of determining convergence is thorough and helpful. I learned a lot of the convergence tests in high school, but never understood where they came from, so I really appreciate the derivation of each test we've learned so far! I think the Integral Test is among the most helpful on exams (if the series has an easy antiderivative).

    -Erica

    ReplyDelete
  15. Hey, this doesn't have a lot to do with what we've been covering. I was just wondering where we are on the practice test idea? we still have 3 weeks till the next test but have any decisions been made?

    ReplyDelete
  16. Does Theorem 6 (the one containing absolute values) still apply even if there are multiple values in the numerator?

    ReplyDelete
    Replies
    1. Short answer, yes.

      Long answer, why wouldn't it?

      The statement in words: If the absolute value of a sequence goes to zero, then the sequence itself goes to zero.

      Delete
  17. Also how do we begin to think about a problem such as # 44 in Section 11.1? Do we just do this theoretically because solving for the limit seems almost impossible?

    ReplyDelete
    Replies
    1. First, it's typically useful to convert the $n$th root symbol to a rational exponent. Here we get $$\left(2^{1+3n}\right)^{1/n}$$. Second, we make use of theorem 7 on page 695.

      Theorem
      If $\lim _{n\to \infty} a_n =L$ and the function $f$ is continuous at $L$, then $$\lim_{n\to \infty} f(a_n)=f(L)$$

      Delete
  18. Regarding question 11.1.82, intuition tells me that the sequence goes to 0 as n goes to infinity, but I can't find a way to mathematically express this in a satisfying way. Is there a good way to find the limits of recursive sequences?

    ReplyDelete
    Replies
    1. Try to use the method of induction to show that $0 < a_n \leq 2$. The same suggestion (use induction) goes to show that the sequence is decreasing.

      The monotonic sequence theorem on page 698 implies the sequence converges.

      As we did in class, and you can follow in example 14, take the limit of both sides of the recursive definition.

      In combination with theorem 7 on page 695, we obtain $$L = \frac{1}{3-L}$$. Solve for $L$ and figure out which solution is the correct one.

      Delete
  19. I had a question about theorem 6 in chapter 11.1.
    It states If lim as n goes to infinity of the absolute value of |an|=0 then the lim of an=0.
    My question here is that does this work for other numbers. say lim|an|=1 does the lim of an=1?

    Problem 34 has a problem that converges to 1 if the absolute value is taken, but if this is not true then then since the values alternate form -1 to 1 this is divergent?

    ReplyDelete
    Replies
    1. Answering your first question:
      In general, no.

      We have the three possible cases if $\lim _{n\to \infty} \left| a_n \right|$ exists (finite) and not equal to $0$.

      Case 1:
      If the sequence has infinitely many positive and negative terms, then $\lim _{n\to \infty} \left| a_n \right| = c>0$ implies $\lim_{n\to \infty} a_n$ does not exist.

      Case 2:
      If the sequence has finitely many negative terms, then $\lim _{n\to\infty} a_n = \lim _{n\to \infty} \left| a_n \right| =c$.

      Case 3:
      If the sequence has finitely many positive terms, then $\lim _{n\to\infty} a_n = -\lim _{n\to \infty} \left| a_n \right| = -c$.

      Remark:
      There are finitely many terms equal to $0$.

      Answering your second question:
      Problem 34 falls under the first case.

      Delete
  20. Is there a hint when we see an equation to know if it will be geometric or for example as Professor Brown said in class a telescoping series?

    ReplyDelete
    Replies
    1. When you see a series, you should be able to identify if it is a geometric series. Read over example 2 on page 705. Then see if you can spot all the geometric series on page 711, numbers 5 through 42. Post your answer here and I'll confirm. (Anybody can do this exercise).

      Note that numbers 43 through 48 are all telescoping sums. I can't say there's a clear method to recognizing when a series is a telescoping sum. However, if you're asked to find the sum and it's not a geometric series, then it's probably a telescoping sum.

      Numbers 57 through 63 are all geometric series.

      Delete
  21. Can we go over finding the limit when trigonometric functions are involved? I don't think it's addressed in any of the examples in the book, and I just want to clarify and make sure I'm not doing something wrong.

    ReplyDelete
    Replies
    1. It's easy to miss how to deal with the trigonometric functions. Ultimately, this is captured in the first two or so lines on page 694 and/or the theorem on page 693.

      Theorem
      If $\lim _{x\to \infty} f(x)=L$ and $f(n)=a_n$, when $n$ is an integer, then $\lim_{n\to \infty} a_n =L$.

      The Squeeze Theorem
      The Squeeze Theorem can also be adapted for sequences as follows.

      If $a_n \leq b_n \leq c_n $ for $n\geq n_0$ and $\lim _{n\to \infty} a_n = \lim _{n\to \infty} c_n =L$, then $\lim _{n\to \infty} b_n =L$.

      Thus, read example 11 on page 105.

      Remark:
      In general, the consequence of the above theorem is that you can apply your knowledge of limits of functions to limits of sequences.

      Delete
  22. I'm having trouble with some homework problem with series. I was wondering if it was possible if you could explain how to determine if it converges or diverges using series.

    thank you
    nahyr

    ReplyDelete
    Replies
    1. Please read section 11.7 thoroughly (page 739). Then try to test yourself with problems there. In between every problem, memorize or learn the strategy suggested by the book. Put the strategy into your own words. Write a short-list of the strategy provided. Summarize the strategy.

      I will go through evens in class. I suggest you go through odds and get help with the ones that give you trouble.

      Delete
  23. Could we go over an induction proof question in section today like 11.1.82 where you have to find the limit of the sequence?

    ReplyDelete
  24. I thought the derivation of the Limit Comparison Test today in class was so interesting. It seems a lot of these tests involve comparing the given series to an arbitrarily chosen series. I was wondering if we could go through several examples in section and focus on what test to use and which series would be optimal for comparison.

    Thanks!
    -Erica

    ReplyDelete