What!?! Spring Fair is this weekend. As such, I'll consider reserving a room and leaving a pile of tests on a table for those who want to come in and sit to take the test. The goal is to let you guys come in freely on Friday 3-7 PM. (4/20/2012). Though I recommend when you come, you commit to staying for an hour block. Perhaps I will come in and out to check. But I hope everybody enjoys Spring Fair, minus the one hour they come.
Then Friday 3-7 PM, we will do it again (4/27/2012). I will make the exam available online for those who can't make it.
In addition, I will make a test that you can work on before each Friday. That's 4 practice tests!
Hint 1: It will be based on your homework. Why? 1) That's most likely how Dr. Brown will write his test. 2) If you get a problem wrong, or don't know how to do a problem, then you'll know which problems you'll need to do more of. 3) In order to write more, I have to somehow make it easy to write.
Hint 2 (for first two practice tests): There will be 8 problems. One of the problems will try to ask something from the first half of the course (in minor preparation for the final). One of the problems will involve polar coordinates (Section 10.4). One of the problems will be an improper integral. The remaining five problems will come from Section 11.1 to 11.7.
Advice: Do all the problems from Section 11.7.
Thank you so much Tim for setting up four practice exams! I really appreciate it!
ReplyDeleteI really appreciate all of this! Thank you!
ReplyDeleteI understand the idea of an interval of convergence and testing the endpoints, but I am confused on how a series can be convergent only at a single point. Doesn't convergence require a pattern over more than one x value?
ReplyDeleteA power series is a function of $x$. Every value of $x$ gives us a series. We ask if this series converges or diverges. Thus, it is possible for a power series to be convergent at only one point.
DeleteExample:
$$f(x)=\sum_{n=1}^\infty n!x^n$$
For $a > 0$, $a>\frac{1}{N}$ for some large positive integer $N$.
For $n>N$, $n! \left(\frac{1}{N}\right)^n= \frac{n}{N}\frac{n-1}{N}\cdots \frac{N+1}{N}\frac{N}{N}\frac{N-1}{N}\cdots\frac{1}{N}$.
Thus $n! \left(\frac{1}{N}\right)^n > \frac{N}{N}\frac{N-1}{N}\cdots\frac{1}{N}$.
We conclude that $\lim _{n\to \infty} n! a^n \neq 0$ for $a>0$.
Convince yourself that $\lim _{n\to \infty} n! a^n \neq 0$ for $a<0$.
Thus, $f(a)$ is divergent for $a\neq 0$. Of course $f(0)=\sum 0=0$. Thus $f(x)$ is convergent only at a single point.
Thanks Tim! That definitely clarifies it for me!
DeleteHi
ReplyDeleteIs today possible to talk about the ratio and root test. I'm having troubles with some problems from the hw due this week
nahyr
Could we go over some common forms of problems that require (or are most likely to be solved) a specific test. For example what test is usually the most effective when we see a factorial in the problem.
ReplyDeletethanks
When there's a factorial in the series I think it's helpful to use the ratio test because the factorial always cancels!
DeleteCan a series converge on two separate intervals (for example: (0,1) and (2,4) or does it always converge on a single interval/point?
ReplyDeleteAnswer:
DeleteSuppose you had a power series centered at $a$ with radius of convergence $R$: $$\sum_{n=1}^\infty c_n (x-a)^n$$. Then the series will converge absolutely on $(a-R,a+R)$, and diverge on $(-\infty,a-R)\cup (a+R,\infty)$. The convergence at the endpoints must be determined on a case-to-case basis.
Reasoning:
The above conclusion comes from either the ratio or the root test.
Root Test:
Here I write $\limsup$ which is a special kind of limit called the limit superior. In the problems you encounter, taking the limit will give the same result as the taking the $\limsup$.
$\limsup _{n\to \infty} \sqrt[n]{\left| c_n (x-a)^n \right|}=\left|(x-a) \right| \limsup _{n\to \infty} \sqrt[n]{\left| c_n \right|}$
The computation of $\limsup _{n\to \infty} \sqrt[n]{\left| c_n \right|}$ always exists and is some non-negative real number or $\infty$, call it $\frac{1}{R}$.
Let fix an $x\in \mathbb{R}$. The root test says $\left| x-a \right| \frac{1}{R} <1 $ means $\sum_{n=1}^\infty c_n (x-a)^n$ is absolutely convergent. $\left| x-a \right| \frac{1}{R} >1 $ means is $\sum_{n=1}^\infty c_n (x-a)^n$ is divergent. The test is inconclusive when $\left| x-a \right| \frac{1}{R} =1 $.
Thus we summarize and conclude (multiplying both sides by $R$), that the power series converges absolutely for all $x$, such that $\left| x-a \right|R$.
A similar argument happens for the ratio test.
I believe most of this in some for is in your book.
Oh, the direct answer to your question is, it will only converge on one interval, and that interval is necessarily symmetric about $a$, the center of the power series. Clearly the power series centered at $a$ converges at least at $x=a$.
DeleteThank you! This was so helpful!
DeleteHi,
ReplyDeleteI'm having trouble with problem 25. i'm using the Ratio test but it is not working because the answer is that it converges absolutely, but I'm not getting that answer.
Nahyr
This may be a bit late, but: the Ratio test should give Lim n-> infinity for |n+1^100(100^n+1)/n+1!|/|n^100(100^n)/n!|, which after cancelling out gives Lim n-> infinity n+1^100(100)/n+1 or Lim n-> infinity 100(n+1)^99. Which would appear to diverge. Are you sure that this is supposed to converge?
DeleteWhile it might appear to diverge, after cancelling we get:
Delete$$\lim_{n\to \infty} \frac{(n+1)^{100} 100}{n^{100} n}$$
This goes to zero.
Reason:
Both $$\lim _{n\to\infty} \frac {(n+1)^{100}}{n^{100}}$$ and $$\lim _{n\to\infty} \frac {100}{n}$$ exist. The first equals 1 and the second equals 0. As a property of limits, the limit of their product exists and equals 0.
Note that one way to show the first one exists is by considering the continuous function $f(x)=x^{100}$. This allows us to bring the limit inside. We get $$\left(\lim_{n\to\infty} \frac{n+1}{n}\right)^{100}$$.
As a remark, the quotient of polynomials portion of the series under the ratio and/or root test always goes to 1. This is mentioned in your book.
DeleteExplicitly, if your series is of the form $\sum p(n)b(n)$ where $p(x)$ is a quotient of polynomials, then $\lim_{n\to \infty} \left|\frac{p(n+1)b(n+1)}{p(n)b(n)}\right| = \lim _{n\to\infty}\left|\frac{b(n+1)}{b(n)}\right|$.
HI Tim,
ReplyDeleteI remember you showing us during section something like this,but I was wondering if you could post the proof on how:
lim as n--->infinity of (1+x/n)^n =e^(x)
Thank you!
This comment has been removed by the author.
DeleteThank you this is really helpful!
DeleteOriginally posted April 18, 2012 8:34 PM with an error. Correct answer below:
DeleteWe find that the base goes to 1 and the exponent goes to $\infty$.
Rewrite the expression as $$e^{n\cdot \ln \left(1+\frac{x}{n}\right)}$$.
Because $e^y$ is a continuous function on $\mathbb{R}$, $$\lim_{n\to \infty} e^{a_n}=e^{\lim_{n\to\infty} a_n}$$
The limit of the exponent is of the type $\infty \cdot 0$.
Rewrite the expression as $$\frac{\ln\left(1+\frac{x}{n}\right)}{\frac{1}{n}}$$.
To use L'Hospital's rule, we consider the associated function: $$\frac{\ln\left(1+\frac{x}{y}\right)}{\frac{1}{y}}$$.
After using the rule we get $$\lim _{y\to\infty} \frac{x}{1+\frac{x}{y}}=x$$.
We conclude the limit is $e^{x}$.
Check out related material here: http://en.wikipedia.org/wiki/E_(mathematical_constant)
DeleteHey Tim, thanks for the practice exams. I just want to say that I'm unlikely to be taking it friday, the 20th, on account of other work due in the near future. I will however be there on the 27th.
ReplyDeleteThank you for the practice exam! Will you be posting answers online?
ReplyDeleteAlso, I'm having problems finding the constant of integration for series. I'm given a function and told to represent it as a power series and find the radius of convergence. The original function can't be put into power series form right away, but its derivative can. So then I integrate the derivative. I don't understand how to figure out what C is in this case.
For reference, problem 11.9.14 a is a good example of this: f(x) = ln(1-x).
Thanks
I will get around to posting answers online. Hopefully you've tacked all the problems in one form or another.
DeleteAs for the constant of integration, the easiest scenario is if you're given $f(0)$ and the power series is centered around $0$. In that case, $C=f(0)$.
Every time you take the derivative of a function to make the function a power series, you have to then integrate the function of the power series? If so why?
ReplyDeleteBecause integration and differentiation are inverse functions.
DeleteWe see the use of inverse functions quite a bit in mathematics. It allows us to change the look of an expression and take advantage of different parts of the expression.
Example: Addition and subtraction
$x^2+2x-4$
We can add 1 as long as we subtract 1.
$=x^2+2x+1-1-4$
Then we get
$=(x+1)^2-5$
This new expression is the same as the first, but we can read off various properties, like it has a minimum at $(-1,-5)$.
Example: Multiplication and division.
$\lim _{n\to \infty} \frac{1}{\sqrt{n}-\sqrt{2n}}$
We can multiply by $\sqrt{n}+\sqrt{2n}$ as long as we divide by $\sqrt{n}+\sqrt{2n}$
Then we get
$=\lim _{n\to \infty} \frac{\sqrt{n}+\sqrt{2n}}{n-2n}$
$=- \lim _{n\to \infty} \frac{\sqrt{n}+\sqrt{2n}}{n}$
This new expression is the same as the first, but we see it goes to $0$.
Example: Exponentiation and logarithm
There is an example above. CTRL+F "Rewrite the"
Example: Differentiation and integration
$\frac{1}{(1-x)^2}$
Integrate and differentiate
$=\frac{d}{dx}\left[ \int \left( \frac{1}{(1-x)^2} \right) dx\right]$
$=\frac{d}{dx}\left[ \frac{1}{1-x}\right]$
$=\frac{d}{dx}\left[ \sum_{n=0}^\infty x^n \right]$
How does this apply to real life? Abstractly, instead of forcing the object to change, try changing how you view it. It's a glass half-full/half-empty type of mindset. Perhaps we can't change the volume of the water we're given, but we can change it's shape by pouring it into different containers. Do something, and yet do nothing.
hi! Can tomorrow during section, can we do more power series problems? sometimes I get confused on what to do or how to even start.
ReplyDeleteNahyr S. Lugo-Fagundo
For homework problems concerning power series do we need to prove convergency of the endpoints?
ReplyDeleteIf the question asks for the "interval of convergence" then you have to determine convergence/divergence of the endpoints.
DeleteTanks for the practice exams!
ReplyDeleteI'm confused about the homework problem 11.8.38. These kind of problems in general (the problems towards the end of each problem set) are really annoying because IDK how the book expects us to be able to figure these kind of problems out. Maybe you can guide me on how I can approach this problem?
ReplyDeleteI like to think of the tougher problems as being designed to test our true understanding of the material. Sometimes we learn by following examples, but we have the ability to learn by using what we know and applying it to new situations.
DeleteOn that note, I suggest you approach the problem by simplifying or changing it. Can you do a simpler version of the problem? Can you do it if you actually choose some numbers for $c_n$?
Fortunately, the book actually asks you a simpler version of the problem. See #37.
All of the example problems in section 11.9 (Representing Power Functions as Power Series), they don't test the endpoints for convergence/divergence - they always just assume the interval of convergence is (-R, R). Do we need to test the endpoints for convergence for the homework problems for this section? Is there a reason that the book doesn't test the endpoints?
ReplyDeleteThe book actually gives a reason in each example why the interval happens to be an open interval.
DeleteMany examples in that section converge on the open interval because $$\frac{1}{1-x}=\sum _{n=0}^\infty x^n$$ only converges on $\left| x \right| < 1$.
In general, if you are asked for the interval of convergence, then you need to know the behavior at the endpoints.
I was puzzled by the same facts that Jackson just mentioned.
ReplyDeleteI was wondering if we could discuss Taylor Series for Tuesday's section if that was possible?
ReplyDeletethanks
Hello,
ReplyDeleteI was a little confused about the Taylor and Maclaurin Series and what kind of questions would appear on the test. Could we maybe go over this in section on Tuesday?
Thank you!!!
Michelle Bohrson
In general, homework questions or questions like homework questions are great candidates for test questions.
DeleteI thought of some questions for Practice Exam 4
Tim, where can we find practice test for the final?
ReplyDeleteThanks a lot!
Practice Test = Midterm 1 + Midterm 2 + my four practice tests + homework questions + make your own!
DeleteMy four practice tests and Dr. Brown's midterms are often simple modifications of your homework exercises. You guys can alter the numbers and test yourselves or test each other!
I was recently reminded of the saying "Give a man a fish and you feed him for a day. Teach a man to fish and you feed him for a lifetime."
So take a look at all the exams you've been given and ask yourself, what can I change about the problem and how that would change the problem? If you have any difficulties coming up with problems, then you can ask me.
As I've mentioned from time to time, math has the effect of teaching you how to think. You might not use the actual mathematics, but it's the thought process.
Understanding concepts related to infinity is not a simple feat. How can adding infinitely many things be finite? How can adding infinitely many things, even though the added increments are going to zero, be infinite? Being able to perceive these various concepts expands the way you think. Gets you to stretching your mind's capacity.
If I'm up for it, then I'll give you another fish. For now, try catching your own.