# Introducing Power Series

The posts for the next several weeks will be on topics tested only on the BC Calculus exams. Continuing with some posts on introducing power series (the Taylor and Maclaurin series)

Introducing Power Series 1 Two examples to lead off with.

Introducing Power Series 2 Looking at the graph of a power series foreshadows the idea of the interval of convergence.

Introducing Power Series 3 The Taylor Approximating Polynomial with examples of using a series to approximate.

Graphing Taylor Polynomials Graphing calculator hints.

# Synthetic Summer Fun

Today, for some summer fun, let’s look at synthetic division a/k/a synthetic substitution. I’ll assume you all know how to do that since it is a pretty common pre-calculus topic and even comes up again in calculus.

Why Does Synthetic Division Work?

An example: consider the polynomial

$P(x)=2{{x}^{4}}-3{{x}^{3}}-11{{x}^{2}}+14x-1$.

This can be written in nested form like this

$P(x)=((((2x-3)x-11)x+14)x-1)$

To evaluate this last expression at, say x = 2, we do the arithmetic as follows:

1.   2 x 2 – 3 = 1
2.   2 x 1 – 11 = –9
3.   2 x (–9) + 14 = –4
4.   2 x (–4) – 1 = – 9 = f(2)

Notice that this requires only multiplication and addition or subtraction, no raising to powers. More to the point, this is the same arithmetic, in the same order when you do the evaluation by synthetic division, and the work is a little easier to keep track of.

$\begin{matrix} {} & 2 & -3 & -11 & 14 & -1 \\ 2) & 2 & 1 & -9 & -4 & -9 \\ \end{matrix}\begin{matrix} {} \\ \,=P(2) \\ \end{matrix}$

Synthetic division has another advantage: the other numbers in the second row are the coefficients of a quotient polynomial, a polynomial of one less degree that the original. So,

$\displaystyle \frac{P(x)}{x-2}=2{{x}^{3}}+{{x}^{2}}-9x-4+\frac{-9}{x-2}$

The Remainder Theorem and the Factor Theorem

In general, a polynomial of degree n, divided by a linear factor (x – a) gives a polynomial Q(x) of degree n – 1 and a remainder R

$\displaystyle \frac{P(x)}{x-a}=Q(x)+\frac{R}{x-a}$

Or

$P(x)=Q(x)(x-a)+R$

From here it is easy to see that $P(a)=R$. This is called the remainder theorem. It has a corollary called the factor theorem: If R = 0, then (x – a) is a factor of P(x).

Calculus

But wait there is more: differentiating the equation above using the product rule gives

${P}'(x)=Q(x)(1)+Q(x)(x-a)+0$ and substituting x = a  gives

${P}'(a)=Q(a)$. The value of the quotient polynomial at a is the derivative of the original polynomial at a.

Of course, we could also rewrite the same equation as $\displaystyle \frac{P(x)-P(a)}{x-a}=Q(x)$ . Then

$\displaystyle {P}'(a)=\underset{x\to a}{\mathop{\lim }}\,\frac{P(x)-P(a)}{x-a}=\underset{x\to a}{\mathop{\lim }}\,Q(x)=Q(a)$

Taylor Series

But wait, there’s even more.

A polynomial is a Maclaurin series in which all the terms after the nth term are zero. When you students are first learning how to write a Taylor series, by finding all the derivatives and substituting in the general term, a good exercise is to have them write the Taylor series for a polynomial centered away from the origin. For the example above:

$P(x)=-9-2\left( x-2 \right)+19{{\left( x-2 \right)}^{2}}+13{{\left( x-2 \right)}^{3}}+2{{\left( x-2 \right)}^{4}}$

Then ask them to expand the expression above and collect term etc. They should get the original polynomial again (and have some great practice expand powers of a binomial).

Can synthetic division help us? Yes, of course.  Here, is the original computation again:

$\begin{matrix} {} & 2 & -3 & -11 & 14 & -1 \\ 2) & 2 & 1 & -9 & -4 & -9 \\ \end{matrix}\begin{matrix} {} \\ \,=P(2) \\ \end{matrix}$

If we ignore the –9 and divide the quotient numbers by 2 we get

$\begin{matrix} {} & 2 & 1 & -9 & -4 \\ 2) & 2 & 5 & 1 & -2 \\ \end{matrix}\begin{matrix} {} \\ \ ={P}'(2) \\ \end{matrix}$

$\begin{matrix} {} & 2 & 5 & 1 \\ 2) & 2 & 9 & 19 \\ \end{matrix}\begin{matrix} {} \\ \ =\frac{{{P}'}'(2)}{2} \\ \end{matrix}$

And again

$\begin{matrix} {} & 2 & 9 \\ 2) & 2 & 13 \\ \end{matrix}\begin{matrix} {} \\ \ =\frac{{{{P}'}'}'\left( 2 \right)}{3!} \\ \end{matrix}$

One more time

$\begin{matrix} {} & 2 \\ 2) & 2 \\ \end{matrix}\begin{matrix} {} \\ \ =\frac{{{P}^{(4)}}\left( 2 \right)}{4!} \\ \end{matrix}$

What do you see? Right, the last numbers in each computation, –9, –2, 19, 13, and 2, are the coefficients of the Taylor polynomial!

If you really want to dive this home and have some more summer fun here’s the start of a proof (at least for n = 4). Let

$P(x)={{c}_{4}}{{x}^{4}}+{{c}_{3}}{{x}^{3}}+{{c}_{2}}{{x}^{3}}+{{c}_{1}}x+{{c}_{0}}$ and divide this by a:

$\begin{matrix} {} & {{c}_{4}} & {{c}_{3}} & {{c}_{2}} & {{c}_{1}} & {{c}_{0}} \\ a) & {{c}_{4}} & {{c}_{4}}a+{{c}_{3}} & {{c}_{4}}{{a}^{2}}{{+}_{{}}}{{c}_{3}}a+{{c}_{2}} & {{c}_{4}}{{a}^{3}}+{{c}_{3}}{{a}^{2}}+{{c}_{a}}a+{{c}_{1}} & {{c}_{4}}{{a}^{4}}+{{c}_{3}}{{a}^{3}}+{{c}_{a}}{{a}^{2}}+{{c}_{1}}a+{{c}_{0}}=P(a) \\ \end{matrix}$

Again

$\begin{matrix} {} & {{c}_{4}} & {{c}_{4}}a+{{c}_{3}} & {{c}_{4}}{{a}^{2}}{{+}_{{}}}{{c}_{3}}a+{{c}_{2}} & {{c}_{4}}{{a}^{3}}+{{c}_{3}}{{a}^{2}}+{{c}_{2}}a+{{c}_{1}} \\ a) & {{c}_{4}} & 2{{c}_{4}}a+{{c}_{3}} & 3{{c}_{4}}{{a}^{2}}+2{{c}_{3}}a+{{c}_{2}} & 4{{c}_{4}}{{a}^{3}}+3{{c}_{3}}{{a}^{2}}+2{{c}_{2}}a+{{c}_{1}} \\ \end{matrix}\begin{matrix} {} \\ \ ={P}'(a) \\ \end{matrix}$

And I’ll leave the rest to you.  Really, why should I have all the fun?

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# Sequences and Series (Type 10 for BC only)

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.

On the free-response section there is usually one full question devoted to sequences and series. This question usually involves writing a Taylor or Maclaurin polynomial for a series.

Students should be familiar with and able to write several terms and the general term of a 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.
• 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), ex and $\displaystyle \tfrac{1}{1-x}$ and be able to find other series by substituting into them.
• Find the radius and interval of convergence. This is usually done by using the Ratio test and 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 post of  February 22, 2013.
• Use the coefficients (the derivatives) to determine information about the function (e.g. extreme values).
• 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 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.

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.

The concludes the series of posts on the type questions in review for the AP Calculus exams.

Next Post

Friday April 7, 2017 The Domain of the solution of a differential equation.

# Sequences and Series

AP Type Question 10

Sequences and Series – for BC only

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.

On the free-response section there is usually one full question devoted to sequences and series. This question usually involves writing a Taylor or Maclaurin polynomial for a series.

Students should be familiar with and able to write several terms and the general term of a 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.

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.
• 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; 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), ex and $\displaystyle \tfrac{1}{1-x}$ and be able to find other series by substituting into them.
• Find the radius and interval of convergence. This is usually done by using the Ratio test and 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.
• 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 post of  February 22, 2013.
• 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 exam is usually quite straightforward. 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.

# Inrtoducing Power Series 2

In our last post we found that we could produce better and better polynomial approximations to a function. That is, we produced a set of polynomials of increasing degree that had the same value for the functions and its derivatives at a given point. To see what is going on I suggest we graph these approximating polynomials along with the given function.

We found that the polynomials $\left( x-1 \right)$, $\left( x-1 \right)+\left( -\tfrac{1}{2} \right){{\left( x-1 \right)}^{2}}$, $\left( x-1 \right)+\left( -\tfrac{1}{2} \right){{\left( x-1 \right)}^{2}}+\left( \tfrac{1}{6} \right){{\left( x-1 \right)}^{3}}$, and $\left( x-1 \right)+\left( -\tfrac{1}{2} \right){{\left( x-1 \right)}^{2}}+\left( \tfrac{1}{6} \right){{\left( x-1 \right)}^{3}}+\left( -\tfrac{1}{4!} \right){{\left( x-1 \right)}^{4}}$ produced approximations to the natural logarithm function at the point (1, 0). To see how this works, graph each of these polynomials, one after the other. See the figure below.

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Notice that each polynomial comes closer to the graph of the graph of y = ln(x), the black graph, in the figures.

You students can do this on their graphing calculators or with a graphing program. More on how to do  this below.

Now do the same thing with the polynomials found for the sine function.

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However, there is a difference. The sine polynomials seem to hug the sine graph over increasingly wider intervals while the logarithm polynomials do not. This may not be a surprise since the logarithm function has no values for $x\le 0$ while the polynomials do. The polynomials cannot come close to the graph if there is no graph.

Students should notice these things:

• Successively higher degree polynomials seem to come closer to the graph of the function than the previous one.
• The polynomials may exist outside the domain of the function (outside of $x>0$ for ln(x) for example).
• The interval where the graphs are near the function is limited.

Taken together these two examples suggest several questions (which you can perhaps draw out of your class):

1. If there were an infinite number of terms would the Polynomial be the same as the function?
2. How do you add an infinite number of terms?
3. Over what interval is the approximation “good”? Is the interval the same for all functions? How do you find the interval?
4. How good is the approximation?
5. Is there an easier way to build the polynomial? Do you have to figure out and evaluate all of the derivatives?

These questions will be the topic of my next post.

________________________________________

How to Graph these Polynomials using Winplot

You can enter each polynomial separately of course, but here is an easier way.

1. After setting your viewing window (CTRL+V), push [F1] to get the explicit equation entry window and enter sin(x) (or the function you are interested in) and click [OK] to graph y = sin(x).
2. Then push [F1] again and enter Sum( (-1)^(n+1)x^(2n-1)/(2n-1)! ,n,1,A) and click [OK]. The underlined part may be changed to the general term of any series.  The n identifies the variable, the 1 is the starting value of n and the A will be the final value which we will change.
3. Next click on [ANIM] > [Individual] > [A]. This will bring up a slider. Enter 100 in the box and click [Set R] and then enter 0 and click [Set L]. This will make the A values change by exactly 1 allowing you to look at A = 1, 2, 3, 4, … in order.
4. Click the tab on the “A” slider window box and see the various approximating polynomials “hug” the graph

# Introducing Power Series 1

The next few posts will discuss a way to introduce Taylor and Maclaurin series to students. We will kind of sneak up on the idea without mentioning where we are going or using any special terms. In this post we will find a way of approximating a function with a polynomial of any degree we choose. In the next post we will look at the graph of these polynomials and finally suggest some questions for further thought.

Making Better Approximations

Students already know and have been working with the tangent line approximation of a function at a point (a, f(a)):

$f(x)\approx f\left( a \right)+{f}'\left( a \right)\left( x-a \right)$

ln(x):

For the function $f\left( x \right)=\ln \left( x \right)$ at the point (1, 0) ask your students to write the tangent line approximation: $y=0+(1)(x-1)$ .Point out that this line has the same value as  ln(xand its derivative as at (1, 0).

Then suggest that maybe having a polynomial that has the same value, first derivative and second derivative might be a better approximation. Suggest they start with $y=a+b\left( x-1 \right)+c{{\left( x-1 \right)}^{2}}$ and see if they can find values of a, b and c that will make this happen.

Since $f\left( 1 \right)=0,{f}'\left( 1 \right)=1\text{ and }{{f}'}'\left( x \right)=-1$ we can write

$y=a+b\left( x-1 \right)+c{{\left( x-1 \right)}^{2}};\quad y\left( 1 \right)=a+0+0=0;\quad a=0$

${y}'=b+2c\left( x-1 \right);\quad {y}'\left( 1 \right)=b+0=\tfrac{1}{1};\quad b=1$

${{y}'}'=2c;\quad {{y}'}'\left( 1 \right)=c=-\tfrac{1}{{{1}^{2}}}=-1;\quad c=-\tfrac{1}{2}$

$y=0+\left( x-1 \right)-\tfrac{1}{2}{{\left( x-1 \right)}^{2}}$

Then suggest they try a third degree polynomial starting with $y=a+b\left( x-1 \right)+c{{\left( x-1 \right)}^{2}}+d{{\left( x-1 \right)}^{3}}$. Proceeding as above, all the numbers come out the same and we find that

$\ln \left( x \right)\approx 0+\left( x-1 \right)+\left( -\tfrac{1}{2} \right){{\left( x-1 \right)}^{2}}+\left( \tfrac{1}{3} \right){{\left( x-1 \right)}^{3}}$

Then go for a fourth- and fifth-degree polynomial until they discover the patterns. (The signs alternate, and the denominators are the factorial of the exponent.)

See if the class can write a general polynomial of degree N :

$\displaystyle \ln \left( x \right)\approx \sum\limits_{k=1}^{N}{\frac{{{\left( -1 \right)}^{k+1}}}{k}{{\left( x-1 \right)}^{k}}}$

sin(x):

Then have the class repeat all this for a new function such as $f\left( x \right)=\sin \left( x \right)$ at the point (0, 0). This could be assigned as homework or group work. Ask them to do enough terms until they see the pattern. There will be patterns similar to ln(x ) and every other term (the even powers) will have a coefficient of zero.

$\sin \left( x \right)\approx x-\tfrac{1}{3!}{{x}^{3}}+\tfrac{1}{5!}{{x}^{5}}-\tfrac{1}{7!}{{x}^{7}}+\tfrac{1}{9!}{{x}^{9}}$

or in general the polynomial of degree N is

$\displaystyle \sin \left( x \right)\approx \sum\limits_{k=1}^{N}{\frac{{{\left( -1 \right)}^{k+1}}}{\left( 2k-1 \right)!}{{x}^{2k-1}}}$

How good is this approximation? Using only the first three terms of the polynomial above you will tell you that. Pretty close: correct to 5 decimal places.  Using four terms gives correct to 7 decimal places when rounded.

Finally, see if they can generalize this idea to any function f at any point on the function $\left( {{x}_{0}},f\left( {{x}_{0}} \right) \right)$. This time you will not have the various derivatives as numbers, rather they will be expressions like . Work through the powers one at a time to go from $y=a+b\left( x-{{x}_{0}} \right)+c{{\left( x-{{x}_{0}} \right)}^{2}}+d{{\left( x-{{x}_{0}} \right)}^{3}}+e{{\left( x-{{x}_{0}} \right)}^{4}}$

and so on, until you get to

$f\left( x \right)\approx f\left( {{x}_{0}} \right)+\frac{{f}'\left( {{x}_{0}} \right)}{1!}\left( x-{{x}_{0}} \right)+\frac{{{f}'}'\left( {{x}_{0}} \right)}{2!}{{\left( x-{{x}_{0}} \right)}^{2}}$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad +\cdots +\frac{{{f}^{\left( N \right)}}\left( {{x}_{0}} \right)}{N!}{{\left( x-{{x}_{0}} \right)}^{N}}$

For example the third derivative computation would look like this:

${{{y}'}'}'=3\cdot 2\cdot 1d+4\cdot 3\cdot 2e\left( x-{{x}_{0}} \right)$

${{{y}'}'}'\left( {{x}_{0}} \right)=3\cdot 2\cdot 1d+4\cdot 3\cdot 2e(0)={{{f}'}'}'\left( {{x}_{0}} \right)$

$d=\frac{{{{f}'}'}'\left( {{x}_{0}} \right)}{3!}$

The computations here are perhaps a little different than what students have seen, so take your time doing this. Two or even three class days may be necessary.

Notice these things:

• The first two terms are the tangent line approximation.
• The various derivatives are numbers that must be calculated.
• All the terms of any degree are the same as the terms of the previous degree with one additional term.

Next post in this series: Looking at all this graphically.

(Typos in an earlier version of this post have been corrected – LMc)