Elise T
2/26/2016 08:44:16 am
Remember that, in terms of converging, a power series will do 1 of 3 things:
Nathan Rao
2/26/2016 08:50:17 am
Section 9.1 Exercise #7
Abigail Chase
2/26/2016 09:10:13 am
If a power series is differentiated term by term the new series will converge on the same integral of the derivative of the function represented by the original series.
Michelle H
3/8/2016 09:25:07 pm
interval* not integral 2/26/2016 02:20:36 pm
If the common ratio of a geometric series is 1, then its sum will oscillate between its first term and 0, and its infinite sum will not exist.
Rachel Willy
2/26/2016 03:50:29 pm
Don't forget!
Michelle H
2/26/2016 09:51:30 pm
A harmonic series is:
Ananth Putcha
2/26/2016 10:08:27 pm
In order to take the sum of a geometric series from n=1 to infinity, you can take the integral of the function of the series from 1 to infinity. If it starts at n=0 then you take the integral from 0 and so on.
Shawn Park
2/26/2016 10:54:04 pm
Section 9.1 #9
Juan Guevara
2/26/2016 11:00:40 pm
______________________________________________________
Taylor Garcia
2/26/2016 11:01:50 pm
Section 9.1 #15
Elise T
2/29/2016 08:47:32 am
A harmonic series is a Pseries where p=1;
Nathan Rao
2/29/2016 05:11:13 pm
Section 9.1 Exercise #17
Michelle H
2/29/2016 05:21:51 pm
A power series centered at x=0 will be:
Rachel Willy
2/29/2016 05:33:04 pm
Just a refresher of geometric series because all of the series in chapter 9 are geometric!
Allison Y
2/29/2016 06:08:49 pm
9.1 #25 2/29/2016 06:40:56 pm
P.466 9.1 #19
Abigail Chase
2/29/2016 08:10:47 pm
9.1 #42. for 1/ (1+3x) find the integral of convergence:
Rumi Venkatesh
2/29/2016 08:26:32 pm
9.1 #21
Kavya Anjur
2/29/2016 10:09:27 pm
The two important theorems of 9.1 are Theorem 1, which states that if a function (the summation from n =0 to infinity of c sub n * (xa)^n) converges for (the absolute value of xa) is less than R, then the series which is the derivative of the original function (obtained by differentiating f(x) term by term) also converges for all (absolute value of xa) is less than R. Theorem 2 states the same thing but for the integral of f(x). Thus, the integral of each term of a series constitutes the integral from a to x of f(t)dt, which is equal to the summation from n=0 to infinity of c sub n * [(xa)^(n1)] / (n+1). They both converge to the same interval.
Christopher Glenn
2/29/2016 10:41:34 pm
9.1 #3
Herven
2/29/2016 10:54:54 pm
sum(ar^(n1))
Herven
2/29/2016 10:55:20 pm
9.1 #27
Tanmayi K
2/29/2016 10:57:10 pm
9.1 #27 part a
Shawn Park
2/29/2016 11:03:55 pm
HW 9.1 #11
Tanmayi K
2/29/2016 11:14:36 pm
9.1 #28
Kevin Knox
2/29/2016 11:16:10 pm
9.1 #15
Kevin Knox
2/29/2016 11:17:35 pm
This is actually #17, which appears to already have been posted.
Kevin Knox
2/29/2016 11:26:17 pm
9.1 #10
Juan Guevara
2/29/2016 11:57:37 pm
______________________________________________________
Nathan Rao
3/1/2016 08:33:54 am
Telescoping Series:
Elise T
3/1/2016 01:04:38 pm
While the example you gave would converge, not all telescoping series are the same. When there is a series of equal but opposite integers (3+33+33) would diverge because there is no result of the summation, there is also ∞Σ n [n(n+1)] which would diverge because the sum would ultimately be 1, and any infinite sum with a constant term diverges.
Michelle H
3/1/2016 05:34:16 pm
A harmonic series is a type of p series that has an exponent of 1. Since it has an exponent of 1, it will always diverge. While you may be able to say that all harmonic series are p series, you may not say that all p series are harmonic.
Allison Y
3/1/2016 06:41:04 pm
9.1 #12
Kavya Anjur
3/1/2016 06:47:19 pm
For 9.1, problem #23, one can see that the partial sums tend to go towards infinity for part A as none of them approach zero and they all have positive values. For part b, one can observe that the partial sums alternate between being the value of 1 and being the value of 0 from the graph. Furthermore, it helps to use the integral test on functions to identify what value the series converges to. For part c, one can see that the partial sums alternate between positive and negative numbers while their overall value, as could be represented by their magnitude, actually increases towards infinity.
Abigail Chase
3/1/2016 08:07:17 pm
9.1 #48
Christopher Glenn
3/1/2016 08:16:49 pm
Just so no one (including me) forgets my flashdrivewinning analogy:
Christopher Glenn
3/2/2016 10:30:45 pm
The analogy also applies to Taylor Series and Maclaurin Series. 3/1/2016 08:53:01 pm
p.467 9.1 #26
Kevin Knox
3/1/2016 09:29:12 pm
9.1 #8
Juan Guevara
3/1/2016 11:53:17 pm
______________________________________________________
Ananth
3/1/2016 11:54:51 pm
______________________________________________________
Abigail Chase
3/2/2016 01:50:13 pm
9.2 #30
Elise T
3/2/2016 02:34:23 pm
There are 4 main Maclaurin Series that we need to make sure to know:
Kavya Anjur
3/2/2016 07:00:01 pm
Three additional Maclaurin Series include
Allison Y
3/2/2016 07:20:50 pm
9.2 #1
Brandon C
3/2/2016 07:42:18 pm
9.2 #3
Rumi Venkatesh
3/2/2016 08:03:31 pm
9.2 #8
Nathan Rao
3/2/2016 08:21:54 pm
9.2 Exercise #9
Ananth Putcha
3/2/2016 10:59:06 pm
In order to find the general equation for the nth derivative of a function, first take the first 3 or 4 derivatives and try to find a pattern.
Juan Guevara
3/2/2016 11:44:11 pm
______________________________________________________
Kevin Knox
3/2/2016 11:54:25 pm
9.1 #4
Nathan Rao
3/3/2016 08:49:18 am
Section 9.3 # 5
Abigail Chase
3/3/2016 01:18:17 pm
9.3
Juan Guevara
3/3/2016 01:25:29 pm
______________________________________________________
Elise
3/3/2016 05:18:09 pm
The higher the degree of a Taylor Polynomial Approximation, the more accurate the approximation will be. (more derivatives=more accurate) 3/3/2016 05:42:39 pm
Theorem 9: The Ratio Test
Allison Y
3/3/2016 09:04:24 pm
9.2 #26
Kavya Anjur
3/3/2016 09:09:13 pm
For 9.2 #4, you start with the function 7xe^x. You then break it apart into 7x(1 + x + (x^2)/(2!) + ... (x^n)/(n!). After using the Taylor series equation, you see how the approximation equals 7x + 7x^2 + 7x^3/ (2!) +... (7x^[n+1])/(n!). Thus, it converges for all real x.
Kavya Anjur
3/3/2016 09:14:54 pm
9.2 #6
Kavya Anjur
3/3/2016 09:19:40 pm
9.2 #7
Kavya Anjur
3/3/2016 09:27:03 pm
9.2 #18
Tanmayi K
3/3/2016 09:32:40 pm
Remember the analogy of a square and a rectangle and apply it to Taylor Series and MacLaurin series. While all squares are rectangles, not all rectangles are squares. Just like that, while all MacLaurin Series are Taylor Series, not all Taylor Series are MacLaurin Series. MacLaurin series are Taylor Series centered at x=0.
Brandon C
3/3/2016 10:34:44 pm
We use Taylor Series and MacLaurin Series to approximate a graph using polynomials. Think of it like you are using Linearization at a certain point x=a. But in this case a is the center. Also, maybe just a coincidence, the first two terms of the Polynomial Series Formula is the formula for linearization. f(a)+f'(a)(xa).
Kevin Knox
3/3/2016 11:51:04 pm
9.1 #43 3/3/2016 11:56:40 pm
Absolute Convergence:
Kevin Knox
3/3/2016 11:57:37 pm
9.1 #44
Juan Guevara
3/3/2016 11:57:48 pm
When we use part of a Taylor series to estimate the value of a function, the end of the series that we do not use is called the remainder, R_n(x). If we know the size of the remainder, M (max size), then we know how close our estimate is. 3/3/2016 11:58:57 pm
Theorem 7 The Direct Comparison Test
Kevin Knox
3/4/2016 07:15:46 pm
9/1 #36
Allison Y
3/4/2016 08:03:09 pm
9.2 #27
Nathan Rao
3/4/2016 08:17:10 pm
9.3 #19
Kavya Anjur
3/4/2016 09:25:23 pm
9.3 #9
Juan Guevara
3/4/2016 11:37:14 pm
______________________________________________________
Juan Guevara
3/4/2016 11:57:37 pm
______________________________________________________
Abigail Chase
3/7/2016 02:05:42 pm
The summation of a sub n as n goes to infinity diverges if the lim as n goes to infinity of a sub n fails to exist or is different from zero. (The nth term test for divergence)
Michelle H
3/7/2016 05:38:00 pm
However, just because the limit as n goes to infinity equals 0, it does not imply convergence
Elise T
3/7/2016 02:17:02 pm
REMEMBER
Shreyas Mohan
3/7/2016 06:25:33 pm
Alright guys this is really interesting so read this:
Rachel Willy
3/7/2016 06:28:49 pm
A sequence converges if its limit at infinity exists. Otherwise, it diverges.
Shreyas Mohan
3/8/2016 07:09:19 pm
Watch out I think you know the right thing but typed it out wrong. If the limit to infinity is not equal to 0, it diverges. Otherwise, more tests are needed to determine if it converges. 3/7/2016 07:00:26 pm
The remainder R(b) at point b on a Taylor approximation can be estimated with the inequality:
Allison Y
3/7/2016 08:13:57 pm
9.1 #32
Kevin Knox
3/7/2016 08:35:30 pm
9.2 #16 Part A
Kevin Knox
3/7/2016 09:45:11 pm
I forgot to include the factorials in the Taylor series. It should be
Nathan Rao
3/7/2016 10:02:40 pm
9.2 #11
Kavya Anjur
3/7/2016 11:15:56 pm
9.2 #5
Kavya Anjur
3/7/2016 11:32:26 pm
Theorem 3
Juan Guevara
3/7/2016 11:37:49 pm
The direct comparison test is a tool we can use to determine convergence for complicated, positive series. Often a given series closely resembles a pseries, but doesn't exactly match up termbyterm to apply the direct comparison test. If this is the case, there is a second comparison test called the Limit Comparison Test, which works well when comparing algebraic series with a pseries. To choose the appropriate pseries, match it up with the highest power of the complicated series.
Kevin Knox
3/8/2016 12:09:29 am
9.2 # 18 Part A
Nathan Rao
3/8/2016 08:29:27 am
The nthTerm Test for Divergence
Elise T
3/8/2016 10:33:22 am
For example: you can use the nth term divergence test for ∞ Σ n=1 of (1+3n^2+n^3)/(4n^35n+2) by taking (n^3)/(4n^3) because those are the terms to the highest degree, and then determining that their leading coefficients equals 1/4, meaning that the function is divergent. On the other side, you cannot use the nth sum divergence test for ∞ Σ n=1 of 1/(x^2), because this function is a Pseries, in which case you need to use the Pseries test, and the function would converge because the pvalue is greater than 1.
Abigail Chase
3/8/2016 10:29:36 am
In order to find a power series representation you must rewrite the function in the form 1/(1___)
Allison Y
3/8/2016 06:34:35 pm
9.1 #16 3/8/2016 06:57:23 pm
Calculator Shortcuts:
Shreyas Mohan
3/8/2016 06:59:00 pm
When faced with finding the interval of convergence set the absolute value of r less than 1. This is really written like 1<r<1.
Michelle H
3/8/2016 07:29:15 pm
Procedure for determining convergence (p. 505)
Juan Guevara
3/8/2016 10:53:02 pm
*** There are three possibilities for power series convergence. ***
Elise T
3/9/2016 10:59:00 am
9.2 #15
Abigail Chase
3/9/2016 01:32:01 pm
Does the function lnx/ (x^2) converge or diverge?
Tanmayi K
3/9/2016 02:28:19 pm
Here's the extra credit from the quiz:
Nathan Rao
3/9/2016 08:23:33 pm
9.3 Exercise #3
Michelle H
3/9/2016 08:24:18 pm
One way to tell if a telescoping series is converging or diverging is by analyzing each partial sum.
Ananth Putcha
3/9/2016 10:58:34 pm
To find the power series representation, you have to rewrite it in the form: 1/(1x)
Kevin Knox
3/9/2016 11:22:00 pm
9.2 #13
Juan Guevara
3/9/2016 11:48:07 pm
The radius of convergence is always half of the interval of convergence. Here is an example.
Nathan Rao
3/10/2016 09:06:10 am
Section 9.4 Exercise #1
Elise T
3/10/2016 11:55:05 am
The Ratio Test:
Abigail Chase
3/10/2016 02:07:27 pm
Remember that this test cannot be used it the series is a Pseries.
Michelle H
3/10/2016 05:40:58 pm
Example:
Shreyas Mohan
3/10/2016 03:28:35 pm
I want to take some time to talk about Seki Kowa (16421708). He was a child prodigy, a brilliant mathematician, and a great teacher. He was born into a samurai warrior family in Fujioka, Kozuke, Japan. A few of his contributions were a method of solving higherdegree equations, using determinants to solve simultaneous equations, and a form of calculus known in Japan as yenri. It's difficult to know the extent of his contributions because the Samurai code demanded great modesty. Pretty cool guy. For more information you can go to p. 487 in your textbook. 3/10/2016 04:40:56 pm
I want to take some time to talk about Srinivasa Ramanujan (18871920). He was a largely self taught mathematician from Southern India who wrote on a wide range of topics such as infinite series, prime and composite numbers, function theory, and combinatorics. His theorems have influeced medical research and statistical mechanics, and one of his identities has been used by programmers to calculate pi to millions of decimal digits. He was so involved with mathematics in fact, that he ended up neglecting his other schoolwork and was unable to graduate from college. Pretty cool guy. For more information, you can go to p.487 in your textbook.
Rachel Willy
3/10/2016 06:23:25 pm
Just to clear this up!
Christopher Glenn
3/10/2016 06:46:35 pm
9.4 # 12
Kavya Anjur
3/10/2016 06:51:43 pm
9.4 #7
Rumi Venkatesh
3/10/2016 08:27:04 pm
9.4 #15
Brandon C
3/10/2016 10:41:20 pm
9.4 #2
Kevin Knox
3/11/2016 12:00:19 am
9.2 #19 part A
Nathan Rao
3/11/2016 08:33:33 am
9.4 Exercise #11
Abbey Chase
3/11/2016 02:01:00 pm
Integral test: the function must be positive, decreasing, and continuous. when comparing a summation of a function to the integral of that function.if the integral converges then so does the summation. If the integral diverges then so does the summation.
Michelle H
3/11/2016 02:11:55 pm
In order to use the integral test, the function must be positive, decreasing, and continuous.
Christopher Glenn
3/11/2016 08:17:00 pm
To use the comparison test, it is best to use a series that has a known limit. For example to find the convergence of n/(n^2) +3 you could compare it to the limit of 1/n.
Kavya Anjur
3/11/2016 09:35:56 pm
9.4 #13
Elise T
3/11/2016 10:15:00 pm
Remember PARTINGC  it'll make our lives a lot easier when remembering all of our options for testing.
Juan Guevara
3/11/2016 10:20:17 pm
______________________________________________________
Juan Guevara
3/11/2016 10:46:53 pm
______________________________________________________
Christopher Glenn
3/14/2016 02:43:40 am
9.5 #16
Christopher Glenn
3/14/2016 02:46:33 am
EDIT:This is #6 not 16 whoops
Rachel Willy
3/14/2016 04:19:21 pm
Find f(1/4) of f(x) = 5  15x + 45x^2  135x^3 ...
Allison Y
3/14/2016 05:16:13 pm
9.4 #4
Elise T
3/14/2016 05:21:34 pm
Alternating Series Test
Abigail Chase
3/16/2016 01:40:44 pm
Alternating series estimation theorem: error is no more than the magnitude of the next term
Kevin Knox
3/14/2016 08:01:26 pm
Chapter 9 review: #23
Kavya Anjur
3/14/2016 09:32:00 pm
9.4 #14
Kavya Anjur
3/14/2016 09:38:15 pm
9.4 #16
Kavya Anjur
3/14/2016 09:44:21 pm
9.4 #21
Ananth Putcha
3/14/2016 09:47:35 pm
In a geometric series, when r is negative, you know the the series is oscillating.
Nathan Rao
3/14/2016 11:25:07 pm
Chapter 9 Review # 37
Juan Guevara
3/14/2016 11:33:34 pm
What will be the sum of infinite geometric series 2/3 + 1/3 + 1/6..... up to 8 term? And what does that mean up to 8 term?
Abigail Chase
3/14/2016 11:39:41 pm
9.5
Tanmayi K
3/14/2016 11:46:02 pm
In 9.2, we used Taylor Series to solve problems in the homework... but what exactly were we doing?
Kevin Knox
3/15/2016 12:03:02 am
Chapter 9 review: # 19
Allison Y
3/15/2016 10:17:30 am
9.4 #3
Abigail Chase
3/15/2016 03:38:34 pm
The summation as n goes to zero to infinity of (lnx)^n converges when the absolute value of lnx <1 which is the same as when 1/e<x < e 3/15/2016 08:28:50 pm
Slightly late for pi day, but here are some ways pi can be expressed as a sum of infinite series:
Christopher Glenn
3/15/2016 09:31:11 pm
9.5 #12
Ananth Putcha
3/15/2016 10:57:03 pm
The Taylor series can be thought of as the summation of multiple tangent lines of a function.
Nathan Rao
3/15/2016 11:09:25 pm
Chapter 9 Review # 39
Brandon C
3/15/2016 11:12:44 pm
Practice Problem from Khan Academy taken from this test:
Juan Guevara
3/15/2016 11:38:48 pm
Unlike stronger convergence tests, the nthterm test for divergence cannot prove by itself that a series converges. The assumption that the series converges means that it passes Cauchy's convergence test.
Kavya Anjur
3/16/2016 07:32:22 am
Review #20
Elise T
3/16/2016 07:53:28 am
Ch. 9 Review #3
Nathan Rao
3/16/2016 08:41:13 am
Chapter 9 Review #21
Allison Y
3/16/2016 08:55:46 am
9.4 #10
Juan Guevara
3/16/2016 12:58:34 pm
______________________________________________________
Michelle H
3/16/2016 04:36:08 pm
Here is a quick summary of the direct comparison test: 3/16/2016 05:01:44 pm
The Integral Test:
Rachel Willy
3/16/2016 08:27:48 pm
P. 477 Quick Review #8
Shreyas Mohan
3/16/2016 09:06:39 pm
When you deal with alternating series, they can either diverge, converge conditionally, or converge absolutely. If the series is decreasing with time a(n+1)>a(n) then the function will converge. To determine if it's conditionally or absolutely, the the absolute value of the series. If it still converges it converges absolutely, if it diverges, then it converges conditionally.
Christopher Glenn
3/16/2016 09:07:37 pm
CH 9 Review
Brandon Cruz
3/16/2016 09:23:24 pm
If the radius of convergence is x<4. Then the interval of convergence is 4<x<4. Basically without the absolute value.
Rumi Venkatesh
3/16/2016 09:53:19 pm
Alternating Series Test Example
Rumi Venkatesh
3/16/2016 09:55:28 pm
Alternating Series Test Example (Formatting fixed)
Ananth Putcha
3/16/2016 10:27:49 pm
9.2 #22
Tanmayi Kwhen
3/16/2016 10:41:50 pm
Out of the many tests we know how to use to determine convergence and divergence, there are many instances where the tests can be inconclusive
Shawn Park
3/16/2016 11:38:48 pm
9.19.3 Quiz #2
Kevin Knox
3/16/2016 11:41:19 pm
9.5 #3
Kavya Anjur
3/17/2016 08:46:40 pm
Review #9
Allison Y
3/17/2016 09:12:10 pm
9.4 #6
Allison Y
3/17/2016 09:13:35 pm
9.4 #8
Christopher Glenn
3/17/2016 10:27:40 pm
9.19.3 Quiz
Brandon C
3/17/2016 11:15:23 pm
Write the Taylor Polynomial with these specific coefficients.
Juan Guevara
3/17/2016 11:32:01 pm
It is important to note that no matter what else is happening in the power series we are guaranteed to get convergence for x = a. The series may not converge for any other value of x, but it will always converge for x = a.
Juan Guevara
3/18/2016 11:39:18 pm
Taylor series are important because they allow us to compute functions that cannot be computed directly. While the Taylor polynomial for the sine function looks complicated and is annoying to evaluate by hand, it is just the sum of terms consisting of exponents and factorials, so the Taylor polynomial can be reduced to the basic operations of addition, subtraction, multiplication, and division. We can obtain an approximation by truncating the infinite Taylor series into a finitedegree Taylor polynomial, which we can evaluate.
Michelle H
3/21/2016 08:27:03 pm
Let f be the function given by f(x) = sin(5x + pi/4) and let P(x) be the 3rd degree Taylor Polynomial for f about x = 0
Brandon C
3/21/2016 10:29:14 pm
When you do the Limit Comparison Test or the Direct Comparison Test, think of it as using the sandwich theorem. Knowing how to use the sandwich theorem will help you know which functions to use when doing the comparison tests. For the two tests, you need to find functions that are greater or less than the function you are finding the sum of. So, this is where the Sandwich Theorem comes in handy and is good for a tool box that will help you succeed in chapter 9!
Brandon C
3/24/2016 07:59:51 am
3. Chapter 9 Quiz Comments are closed.

AuthorMrs. Johnson's 20152016 BC Calculus Center for Review. By participating in this blog, you are indicating that the work that you submit is your own. If found to be otherwise true, you will not receive credit. Happy blogging!
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