Adapting 2021 BC 5

Posted on August 10, 2021 by Lin McMullin

Eight of nine. We continue our study of the 2021 free-response questions. We will look at ways to adapt, expand, and explore this question to help students better understand it and look at other questions that can be asked based on a similar stem.

2021 BC 5

This is a Differential Equation (Type 6) with a Sequence and Series (Type 10) question included. It contains topics from Units 7 and 10 of the current Course and Exam Description (CED). It is not unusual for AP Calculus exam question to include several of the types in my classification and from several of the units from the CED (here units 7 and 10). In addition, the usual solving an initial value problem and a Euler’s Method approximation are included.

The stem for 2021 BC 5 is:

Part (a): Students were asked to write the second-degree Taylor polynomial for the function centered at x = 1 and then use it to approximate f(2). Students should stop after substituting 2 into their polynomial; no arithmetic or simplification is required, and a simplifying mistake will lose a point.

Discussion and ideas for adapting this question:

Ask students to find an expression for the second derivative (implicit differentiation).
Verify that $\displaystyle {f}''\left( 1 \right)=4$
Ask students to find the third-degree polynomial and use it to approximate f(2)

Part (b): Students were required to approximate f(2) using Euler’s method with two steps of equal size.

Discussion and ideas for adapting this question:

After you solve the equation in part (c), ask students to compare the approximations from parts (a) and (b) with the exact value. Neither approximation is very close to the exact value. Discuss why this is so. Consider the slope of the graph near x = 2.
Find a more accurate approximation using 3, 4, or more smaller steps. There are graphing calculator programs that will do the arithmetic. Do not hesitate to use them. Students have already shown they know how to do a Euler’s Method approximation; the point is to understand the situation.

Part (c): Finding the solution of the differential equation by separating the variables is expected in this kind of question. The added twist is that the method of integration by parts is necessary to find one of the antiderivatives.

Discussion and ideas for adapting this question:

Be sure not to skip over removing the absolute value signs. The most efficient way is to realize that at (and near) the initial condition y > 0, so |y| = y. What do you do if y < 0?
There is not much you can do differently here. One thing is to change the initial condition. Try a negative value such as f(1) = –4.
As suggested in (b), compare, and discuss the approximations with the exact value.

Next week we will conclude this series of posts with a look at 2021 BC 6.

I would be happy to hear your ideas for other ways to use this question. Please use the reply box below to share your ideas.

Infinite Sequences and Series – Unit 10

Posted on January 19, 2021 by Lin McMullin

Unit 10 covers sequences and series. These are BC only topics (CED – 2019 p. 177 – 197). These topics account for about 17 – 18% of questions on the BC exam.

Topics 10.1 – 10.2

Topic 10.1: Defining Convergent and Divergent Series.

Topic 10. 2: Working with Geometric Series. Including the formula for the sum of a convergent geometric series.

Topics 10.3 – 10.9 Convergence Tests

The tests listed below are tested on the BC Calculus exam. Other methods are not tested. However, teachers may include additional methods.

Topic 10.3: The n^th Term Test for Divergence.

Topic 10.4: Integral Test for Convergence. See Good Question 14

Topic 10.5: Harmonic Series and p-Series. Harmonic series and alternating harmonic series, p-series.

Topic 10.6: Comparison Tests for Convergence. Comparison test and the Limit Comparison Test

Topic 10.7: Alternating Series Test for Convergence.

Topic 10.8: Ratio Test for Convergence.

Topic 10.9: Determining Absolute and Conditional Convergence. Absolute convergence implies conditional convergence.

Topics 10.10 – 10.12 Taylor Series and Error Bounds

Topic 10.10: Alternating Series Error Bound.

Topic 10.11: Finding Taylor Polynomial Approximations of a Function.

Topic 10.12: Lagrange Error Bound.

Topics 10.13 – 10.15 Power Series

Topic 10.13: Radius and Interval of Convergence of a Power Series. The Ratio Test is used almost exclusively to find the radius of convergence. Term-by-term differentiation and integration of a power series gives a series with the same center and radius of convergence. The interval may be different at the endpoints.

Topic 10.14: Finding the Taylor and Maclaurin Series of a Function. Students should memorize the Maclaurin series for $\displaystyle \frac{1}{{1-x}}$ , sin(x), cos(x), and e^x.

Topic 10.15: Representing Functions as Power Series. Finding the power series of a function by, differentiation, integration, algebraic processes, substitution, or properties of geometric series.

Timing

The suggested time for Unit 9 is about 17 – 18 BC classes of 40 – 50-minutes, this includes time for testing etc.

Previous posts on these topics:

Before sequences

Amortization Using finite series to find your mortgage payment. (Suitable for pre-calculus as well as calculus)

A Lesson on Sequences An investigation, which could be used as early as Algebra 1, showing how irrational numbers are the limit of a sequence of approximations. Also, an introduction to the Completeness Axiom.

Everyday Series

Convergence Tests

Reference Chart

Which Convergence Test Should I Use? Part 1 Pretty much anyone you want!

Which Convergence Test Should I Use? Part 2 Specific hints and a discussion of the usefulness of absolute convergence

Good Question 14 on the Integral Test

Sequences and Series

Graphing Taylor Polynomials Graphing calculator hints

Introducing Power Series 1

Introducing Power Series 2

Introducing Power Series 3

New Series from Old 1 substitution (Be sure to look at example 3)

New Series from Old 2 Differentiation

New Series from Old 3 Series for rational functions using long division and geometric series

Geometric Series – Far Out An instructive “mistake.”

A Curiosity An unusual Maclaurin Series

Synthetic Summer Fun Synthetic division and calculus including finding the (finite)Taylor series of a polynomial.

Error Bounds

Error Bounds Error bounds in general and the alternating Series error bound, and the Lagrange error bound

The Lagrange Highway The Lagrange error bound.

What’s the “Best” Error Bound?

Review Notes

Type 10: Sequences and Series Questions

Sequences and Series (Type 10)

Posted on March 27, 2020 by Lin McMullin

AP Questions Type 10: Sequences and Series (BC Only)

The last BC question on the exams usually concerns sequences and series. The question may ask students to write a Taylor or Maclaurin series and to answer questions about it and its interval of convergence, or about a related series found by differentiating or integrating. The topics may appear in other free-response questions and in multiple-choice questions. Questions about the convergence of sequences may appear as multiple-choice questions. With about 8 multiple-choice questions and a full free-response question this is one of the largest topics on the BC exams.

Convergence tests for series appear on both sections of the BC Calculus exam. In the multiple-choice section, students may be asked to say if a sequence or series converges or which of several series converge.

The Ratio test is used most often to determine the radius of convergence and the other tests to determine the exact interval of convergence by checking the convergence at the end points. Click here for a convergence test chart students should be familiar with; this list is also on the resource page.

Students should be familiar with and able to write several terms and the general term of a Taylor or Maclaurin series. They may do this by finding the derivatives and constructing the coefficients from them, or they may produce the series by manipulating a known or given series. They may do this by substituting into a series, differentiating it, or integrating it.

The general form of a Taylor series is $\displaystyle \sum\limits_{n=0}^{\infty }{\frac{{{f}^{\left( n \right)}}\left( a \right)}{n!}{{\left( x-a \right)}^{n}}}$ ; if a = 0, the series is called a Maclaurin series.

What Students Should be Able to Do

Use the various convergence tests to determine if a series converges. The test to be used is rarely given so students need to know when to use each of the common tests. For a summary of the tests click: Convergence test chart. and the posts “What Convergence Test Should I use?” Part 1 and Part 2
Understand absolute and conditional convergence. If the series of the absolute values of the terms of a series converges, then the original series is said to absolutely convergent (or converges absolutely). If a series is absolutely convergent, then it is convergent. If the series of absolute values diverges, then the original series may or may not converge; if it converges it is said to be conditionally convergent.
Write the terms of a Taylor or Maclaurin series by calculating the derivatives and constructing the coefficients of each term.
Distinguish between the Taylor series for a function and the function. DO NOT say that the Taylor polynomial is equal to the function (this will lose a point); say it is approximately equal.
Determine a specific coefficient without writing all the previous coefficients.
Write a series by substituting into a known series, by differentiating or integrating a known series, or by some other algebraic manipulation of a series.
Know (from memory) the Maclaurin series for sin(x), cos(x), e^x and $\displaystyle \tfrac{1}{1-x}$ and be able to find other series by substituting into one of these.
Find the radius and interval of convergence. This is usually done by using the Ratio test to find the radius and then checking the endpoints.
Be familiar with geometric series, its radius of convergence, and be able to find the number to which it converges, $\displaystyle {{S}_{\infty }}=\frac{{{a}_{1}}}{1-r}$ . Re-writing a rational expression as the sum of a geometric series and then writing the series has appeared on the exam.
Be familiar with the harmonic and alternating harmonic series. These are often useful series for comparison.
Use a few terms of a series to approximate the value of the function at a point in the interval of convergence.
Determine the error bound for a convergent series (Alternating Series Error Bound and Lagrange error bound). See my posts on Error Bounds and the Lagrange Highway
Use the coefficients (the derivatives) to determine information about the function (e.g. extreme values).

This list is quite long, but only a few of these items can be asked in any given year. The series question on the free-response section is usually quite straightforward. Topics and convergence test may appear on the multiple-choice section. As I have suggested before, look at and work as many past exam questions to get an idea of what is asked and the difficulty of the questions. Click on Power Series in the “Posts by Topic” list on the right side of the screen to see previous posts on Power Series or any other topic you are interested in.

Free-response questions:

2004 BC 6 (An alternate approach, not tried by anyone, is to start with $\displaystyle \sin \left( {5x+\tfrac{\pi }{4}} \right)=\sin (5x)\cos \left( {\tfrac{\pi }{4}} \right)+\cos (5x)\sin \left( {\tfrac{\pi }{4}} \right)$ )
2011 BC 6 (Lagrange error bound)
2016 BC 6
2017 BC 6
2019 BC 6

Multiple-choice questions from non-secure exams:

2008 BC 4, 12, 16, 20, 23, 79, 82, 84
2012 BC 5, 9, 13, 17, 22, 27, 79, 90,

These question come from Unit 10 of the 2019 CED.

Revised March 12, 2021

A Curiosity

Posted on February 4, 2020 by Lin McMullin

Thoughts on the power series for $f\left( x \right)=\cos \left( {\sqrt{x}} \right),x\ge 0$ , which I found curious.

Last week someone asked me a question about the Maclaurin series for $f\left( x \right)=\cos \left( {\sqrt{x}} \right),x\ge 0$ . Finding the Maclaurin series is straightforward:

$\displaystyle \cos \left( x \right)=1-\frac{{{{x}^{2}}}}{{2!}}+\frac{{{{x}^{4}}}}{{4!}}-\frac{{{{x}^{6}}}}{{6!}}+\cdots +{{\left( {-1} \right)}^{n}}\frac{{{{x}^{{2n}}}}}{{\left( {2n} \right)!}}+\cdots$

Substituting $\displaystyle \sqrt{{x}}$ for x gives

$\displaystyle \cos \left( {\sqrt{x}} \right)=R\left( x \right)=1-\frac{{{{{\left( {\sqrt{x}} \right)}}^{2}}}}{{2!}}+\frac{{{{{\left( {\sqrt{x}} \right)}}^{4}}}}{{4!}}-\frac{{{{{\left( {\sqrt{x}} \right)}}^{6}}}}{{6!}}+\cdots {{\left( {-1} \right)}^{n}}\frac{{{{{\left( {\sqrt{x}} \right)}}^{{2n}}}}}{{\left( {2n} \right)!}}+\cdots$

$\displaystyle \cos \left( {\sqrt{x}} \right)=R\left( x \right)=1-\frac{x}{{2!}}+\frac{{{{x}^{2}}}}{{4!}}-\frac{{{{x}^{3}}}}{{6!}}\cdots +{{\left( {-1} \right)}^{n}}\frac{{{{x}^{n}}}}{{\left( {2n} \right)!}}+\cdots$

We can find the radius and interval of convergence by using the Ratio test:

$\displaystyle \underset{{x\to \infty }}{\mathop{{\lim }}}\,\left| {\frac{{\frac{{{{x}^{{n+1}}}}}{{(2(n+1))!}}}}{{\frac{{{{x}^{n}}}}{{\left( {2n} \right)!}}}}} \right|=\underset{{x\to \infty }}{\mathop{{\lim }}}\,\left| {\frac{x}{{\left( {2x+2} \right)\left( {2n+1} \right)}}} \right|=0$

This indicates that Maclaurin series converges for all Real numbers. However, the original function $f\left( x \right)=\cos \left( {\sqrt{x}} \right),x\ge 0$ is not defined for negative numbers, but the series is. This can be accounted for by the fact that the series contains only even powers of x, and for all Real numbers x, $\displaystyle {{\left( {\sqrt{x}} \right)}^{2}}$ is a Real number. In addition, since the function ends at x = 0, how can the Maclaurin series be centered there? Since it is not defined to the left of zero, how can it have derivatives at zero?

This is in conformance with the graph. You can see the graph and experiment with here on Desmos. Use the slider and note the exaggerated scales. Also note that the power series extends steeply up to the left from the point (0, 1).

$\displaystyle \cos \left( {\sqrt{x}} \right)$ in red largely covered by its Maclaurin series (with n = 14) in blue.

The curiosity is that $\displaystyle \cos \left( {\sqrt{x}} \right)$ is not defined for negative numbers and is not differentiable at x = 0 (because the two-sided limit defining the derivative does not exist to the left of x = 0. But, but the Maclaurin series is continuous and differentiable for all Real numbers. The Maclaurin series is a good approximation for $f\left( x \right)=\cos \left( {\sqrt{x}} \right),x\ge 0$ but approximates a larger function to the left of x = 0.

The explanation is that there is a larger function (that is, one defined for all Real numbers with the appropriate derivatives) that includes $latex \displaystyle \cos \left( {\sqrt{x}} x > 0 as part of it. The series is

$\displaystyle R\left( x \right)=\left\{ {\begin{array}{*{20}{c}} {\cos \left( {\sqrt{x}} \right)} & {x\ge 0} \\ {\cosh \left( {\sqrt{{-x}}} \right)} & {x<0} \end{array}} \right.$

(Note: $\displaystyle \cosh \left( {\sqrt{{-x}}} \right)=1+\frac{x}{{2!}}+\frac{{{{x}^{2}}}}{{4!}}+\frac{{{{x}^{3}}}}{{6!}}+\cdots =\frac{{{{e}^{{\sqrt{{-x}}}}}+{{e}^{{-\sqrt{{-x}}}}}}}{2}$ )

I wish to thank Louis A. Talman, Ph.D,,Emeritus Professor of Mathematics Metropolitan State University of Denver for helping me understand this function better and correcting some of my early ideas. He is the one who Developed the piecewise defined series above. An explanation of the reasoning and a longer discussion of this series can be found in this note “On $f\left( x \right)=\cos \left( {\sqrt{x}} \right),x\ge 0$ .” click here. In that note he shows that the Maclaurin series R(x) approximates this piecewise defined function. The two pieces form a function that is continuous and differentiable everywhere including at x = 0. (The pieces join smoothly the point (0, 1).

A similar curious situation, where a series, but not a Taylor/Maclaurin series, approximates a function is discussed in Geometric Series – Far Out.

What’s the “Best” Error Bound?

Posted on January 28, 2020 by Lin McMullin

A know a lot of people like mathematics because there is only one answer, everything is exact. Alas, that’s not really the case. Numbers written as non-terminating decimals are not “exact;” they must be rounded or truncated somewhere. Even things like $\sqrt{7},\pi ,$ and 5/17 may look “exact,” but if you ever had to measure something to those values, you’re back to using decimal approximations.

There are many situations in mathematics where it is necessary to find and use approximations. Two if these that are usually considered in introductory calculus courses are approximating the value of a definite integral using the Trapezoidal Rule and Simpson’s Rule and approximating the value of a function using a Taylor or Maclaurin polynomial.

If you are using an approximation, you need and want to know how good it is; how much it differs from the actual (exact) value. Any good approximation technique comes with a way to do that. The Trapezoidal Rule and Simpson’s Rule both come with expressions for determining how close to the actual value they are. (Trapezoidal approximations, as opposed to the Trapezoidal Rule and Simpson’s Rule per se, are tested on the AP Calculus Exams. The error is not tested.) The error approximation using a Taylor or Maclaurin polynomial is tested on the exams.

The error is defined as the absolute value of the difference between the approximated value and the exact value. Since, if you know the exact value, there is no reason to approximate, finding the exact error is not practical. (And if you could find the exact error, you could use it to find the exact value.) What you can determine is a bound on the error; a way to say that the approximation is at most this far from the actual value. The BC Calculus exams test two ways of doing this, the Alternating Series Error Bound (ASEB) and the Lagrange Error Bound (LEB). These two techniques are discussed in my previous post, Error Bounds. The expressions used below are discussed there.

Examining Some Error Bounds

We will look at an example and the various ways of computing an error bound. The example, which seems to come up this time every year, is to use the third-degree Maclaurin polynomial for sin(x) to approximate sin(0.1).

Using technology to twelve decimal places sin(0.1) = 0.099833416647

The Maclaurin (2n – 1)^th-degree polynomial for sin(x) is

$\displaystyle x-\frac{1}{{3!}}{{x}^{3}}+\frac{1}{{5!}}{{x}^{5}}-+\cdots \frac{1}{{\left( {2n-1} \right)!}}{{x}^{{2n-1}}}$

So, using the third degree polynomial the approximation is

$\sin \left( {0.1} \right)\approx 0.1-\frac{1}{6}{{\left( {0.1} \right)}^{3}}=0.099833333333...$

The error to 12 decimal places is the difference between the approximation and the 12 place value. The error is:

$\displaystyle 0.00000008331349=8.331349\times {{10}^{{-8}}}=Error$

Using the Alternating Series Error Bound:

Since the series meets the hypotheses for the ASEB (alternating, decreasing in absolute value, and the limit of the n^th term is zero), the error is less than the first omitted term. Here that is

$\displaystyle \frac{1}{{5!}}{{\left( {0.1} \right)}^{5}}\approx 0.0000000833333\approx 8.33333\times {{10}^{-8}}={{B}_{1}}$

The actual error is less than B₁ as promised.

Using the Legrange Error Bound:

For the Lagrange Error Bound we must make a few choices. Nevertheless, each choice gives an error bound larger than the actual error, as it should.

For the third-degree Maclaurin polynomial, the LEB is given by

$\displaystyle \left| {\frac{{\max {{f}^{{(4)}}}\left( z \right)}}{{4!}}{{{(0.1)}}^{4}}} \right|$ for some number z between 0 and 0.1.

The fourth derivative of sin(x) is sin(x) and its maximum absolute value between 0 and 0.1 is |sin(0.1)|. So, the error bound is

However, since we’re approximating sin(0.1) we really shouldn’t use it. In a different example, we probably won’t know it.

What to do?

The answer is to replace it with something larger. One choice is to use 0.1 since 0.1 > sin(0.1). This gives

$\displaystyle \left| {\frac{{0.1}}{{4!}}{{{(0.1)}}^{4}}} \right|\approx 4.166666666\times {{10}^{{-7}}}={{B}_{3}}$

The usual choice for sine and cosine situations is to replace the maximum of the derivative factor with 1 which is the largest value of the sine or cosine.

$\displaystyle \left| {\frac{1}{{4!}}{{{(0.1)}}^{4}}} \right|\approx 4.166666666\times {{10}^{{-6}}}={{B}_{4}}$

Since the 4^th degree term is zero, the third-degree Maclaurin Polynomial is equal to the fourth-degree Maclaurin Polynomial. Therefore, we may use the fifth derivative in the error bound expression, $\displaystyle \left| {\frac{{\max {{f}^{{(5)}}}\left( z \right)}}{{5!}}{{{(0.1)}}^{5}}} \right|$ to calculate the error bound. The 5^th derivative of the sin(x) is cos(x) and its maximum value in the range is cos(0) =1.

$\displaystyle \left| {\frac{{\max {{f}^{{(4)}}}\left( z \right)}}{{4!}}{{{(0.1)}}^{4}}} \right|$ for some number z between 0 and 0.1.

I could go on ….

Since B₁, B₂, B₃, B₄, and B₅are all greater than the error, which should we use? Or should we use something else? Which is the “best”?

The error is what the error is. Fooling around with the error bound won’t change that. The error bound only assures you your approximation is, or is not, good enough for what you need it for. If you need more accuracy, you must use more terms, not fiddle with the error bound.

2019 CED Unit 10: Infinite Sequences and Series

Posted on January 14, 2020 by Lin McMullin

Unit 10 covers sequences and series. These are BC only topics (CED – 2019 p. 177 – 197). These topics account for about 17 – 18% of questions on the BC exam.