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Taylor Series Introduction

Monday, October 18, 2010 - 14:00 - 15:00

Math 2300 Section 005 – Calculus II – Fall 2010

Notes on Taylor Series – October 18, 2010

Definition of a Taylor Series

The Taylor series for f(x) centered at x=0 is

f(x)=f(0)+f^{{\prime}}(0)x+\frac{f^{{\prime\prime}}(0)}{2!}x^{2}+\frac{f^{{(3)}}(0)}{3!}x^{3}+\cdots=\sum _{{k=0}}^{\infty}\frac{f^{{(k)}}(0)}{k!}x^{k}.

Similarly, the Taylor series for f(x) centered at x=a is

f(x)=f(a)+f^{{\prime}}(a)(x-a)+\frac{f^{{\prime\prime}}(a)}{2!}(x-a)^{2}+\cdots=\sum _{{k=0}}^{\infty}\frac{f^{{(k)}}(a)}{k!}(x-a)^{k}.


Since a Taylor series is the same as a Taylor polynomial, but is taken to infinite degree, we already know many Taylor series. From the last section, we know the following.

\displaystyle\sin x \displaystyle=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\cdots=\sum _{{k=0}}^{\infty}\frac{(-1)^{k}}{(2k+1)!}x^{{2k+1}}
\displaystyle\cos x \displaystyle=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\cdots=\sum _{{k=0}}^{\infty}\frac{(-1)^{k}}{(2k)!}x^{{2k}}
\displaystyle e^{x} \displaystyle=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots=\sum _{{k=0}}^{\infty}\frac{x^{k}}{k!}
\displaystyle(1+x)^{p} \displaystyle=1+px+\frac{p(p-1)}{2!}x^{2}+\frac{p(p-1)(p-2)}{3!}x^{3}+\cdots=\sum _{{k=0}}^{\infty}\binom{p}{k}x^{k}
\displaystyle\ln(x+1) \displaystyle=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots=\sum _{{k=1}}^{\infty}\frac{(-1)^{{k+1}}}{k}x^{k}



There is a bit of a technicality here. In section 10.1, we wrote \approx between the functions and their Taylor polynomials. Now, we're writing = instead. (\approx means “approximately equal to,” while = means “equal to.”) The main point to understand here is detailed in the next portion of these notes.

Taylor Polynomials

Monday, October 11, 2010 - 14:00 - Friday, October 15, 2010 - 15:00

Math 2300 Section 005 – Calculus II – Fall 2010

Notes on Taylor Polynomials – October 11 & 15, 2010

Definition of a Taylor Polynomial

The Taylor polynomial of degree n approximating f(x) near x=0 is

P_{n}(x)=f(0)+f^{{\prime}}(0)x+\frac{f^{{\prime\prime}}(0)}{2!}x^{2}+\frac{f^{{(3)}}(0)}{3!}x^{3}+\cdots+\frac{f^{{(n)}}(0)}{n!}x^{n}=\sum _{{k=0}}^{n}\frac{f^{{(k)}}(0)}{k!}x^{k}

Similarly, the Taylor polynomial of degree n approximating f(x) near x=a is

\displaystyle P_{n}(x) \displaystyle=f(a)+f^{{\prime}}(a)(x-a)+\frac{f^{{\prime\prime}}(a)}{2!}(x-a)^{2}+\frac{f^{{(3)}}(a)}{3!}(x-a)^{3}+\cdots+\frac{f^{{(n)}}(a)}{n!}(x-a)^{n}
\displaystyle=\sum _{{k=0}}^{n}\frac{f^{{(k)}}(a)}{k!}(x-a)^{k}

This may not at first seem obvious, but if you have calculated the nth Taylor Polynomial, you can find the (n+1)th Taylor polynomial by adding one more term. That is


Book Assignment 8

Due Date: 
Monday, October 18, 2010 - 16:00

Section 10.1 Problems 28, 31 and 32

Midterm 2 Solutions

Solutions to Midterm 2 are attached.

Midterm 2 Review Solutions

Math 2300 – Calculus II – University of Colorado

Fall 2010 – Review for Midterm II

1. Rotating the ellipse \displaystyle\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 about the x-axis generates an ellipsoid. Compute its volume.

Solution: In this situation, the a and b would be numbers. But, in general, we have that

\displaystyle\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \displaystyle=1
\displaystyle\frac{y^{2}}{b^{2}} \displaystyle=1-\frac{x^{2}}{a^{2}}
\displaystyle y^{2} \displaystyle=b^{2}-\frac{b^{2}}{a^{2}}x^{2}
\displaystyle y \displaystyle=\sqrt{b^{2}-\frac{b^{2}}{a^{2}}x^{2}}

This object is symmetric with respect to the x-axis and y-axis, meaning that we only need to find one of the limits of integration. Setting y=0 in the above equation implies

\sqrt{b^{2}-\frac{b^{2}}{a^{2}}x^{2}}=0\qquad\Leftrightarrow\qquad b^{2}-\frac{b^{2}}{a^{2}}x^{2}=0\qquad\Leftrightarrow\qquad b^{2}\left(1-\frac{x^{2}}{a^{2}}\right)=0.

When we create such objects, we never have b=0 and so we must have 1=x^{2}/a^{2} and a^{2}=x^{2}. So, the x-value where the ellipsoid has a zero y-value must be a. (This makes perfect sense from the definition, if you should know it or look it up.)

So, we consider slices in the ellipsoid revolved around the x-axis, where the radius is given as


and the width of a Riemann disk is dx. So, we find that the volume is

\displaystyle\int _{{-a}}^{a}\pi r^{2}\, dx \displaystyle=\int _{{-a}}^{a}\pi\left(b^{2}-\frac{b^{2}}{a^{2}}x^{2}\right)\, dx
\displaystyle=\pi\int _{0}^{a}\left(b^{2}-\frac{b^{2}}{a^{2}}x^{2}\right)\, dx
\displaystyle=\frac{2\pi ab^{2}}{3}

Midterm 2 Review Quiz

Due Date: 
Tuesday, October 12, 2010 - 16:00


Math 2300 Section 005 – Calculus II – Fall 2010

Quiz – Due Tuesday, October 12, 2010

This quiz is intended to be a review, so it contains more problems than usual. I won't be grading each problem individually, but the quiz will be graded as a whole. Solutions will be posted on Tuesday.

  1. Find the sum

    \sum _{{n=3}}^{{12}}\frac{3^{n}+6}{4^{n}}

    Hint: Write this as two geometric series.

Radius of Convergence

Friday, October 8, 2010 - 14:00 - 15:00

Math 2300 Section 005 – Calculus II – Fall 2010

Power Series and Radius of Converges – October 8, 2010

Power Series

Definition: A power series is an infinite series of the form

f(x)=\sum _{{n=0}}^{\infty}c_{n}(x-a)^{n}=c_{0}+c_{1}(x-a)^{1}+c_{2}(x-a)^{2}+\cdots


  • c_{n} represents the coefficient of the nth term.

  • a is some (fixed) real number.

  • x varies around a, and so we sometimes say that the power series is “centered” at a.

Series Convergence Notes

Math 2300 Section 005 – Calculus II – Fall 2010

Review on Series Convergence – October 6, 2010

§ Summary of Convergence of Series

§ Some General Notes

  • Pulling a finite number of terms off from a series will not affect convergence / divergence.

  • You can pull series apart like integrals (in terms of sums/difference and multiplication by a constant).

  • Series notation is shorthand, and in most cases series can be rewritten in any number of different ways.

    \sum _{{n=1}}^{\infty}\frac{(-1)^{n}}{n^{2}}=\sum _{{n=1}}^{\infty}\frac{(-1)^{{n+2}}}{n^{2}}=\sum _{{n=5}}^{\infty}\frac{(-1)^{{n-2}}}{(n-4)^{2}}=\sum _{{n=5}}^{\infty}\frac{(-1)^{n}}{(-1)^{2}(n-4)^{2}}=\sum _{{n=5}}^{\infty}\frac{\cos(n\pi)}{n^{2}-8n+16}

Book Assignment 7

Due Date: 
Monday, October 11, 2010 - 16:00

Section 9.3 Problem 50
Section 9.4 Problems 60, 84
Section 9.5 Problems 34, 42

Limit Comparison Test

Wednesday, October 6, 2010 - 14:00

Math 2300 Section 005 – Calculus II – Fall 2010

9.4 Limit Comparison Test Examples – October 6, 2010

  1. Determine if the given series converges or diverges using the limit comparison test.

    \sum _{{n=1}}^{\infty}\frac{1}{2^{n}-1}

    Solution: The dominant terms here are 1 in the numerator and 2^{n} in the denominator. So, we compare the series terms

    a_{n}=\frac{1}{2^{n}-1}\qquad\text{and}\qquad b_{n}=\frac{1}{2^{n}}.

    The terms b_{n} form a convergent geometric series. And, we find that

    \lim _{{n\rightarrow\infty}}\frac{a_{n}}{b_{n}}=\lim _{{n\rightarrow\infty}}\frac{\displaystyle\frac{1}{2^{n}-1}}{\displaystyle\frac{1}{2^{n}}}=\lim _{{n\rightarrow\infty}}\frac{2^{n}}{2^{n}-1}=\lim _{{n\rightarrow\infty}}\frac{1}{1-1/2^{n}}=1.

    Since we obtain a finite positive number using the limit comparison test, we know that both of the series either diverge or they both converge. We already knew that the series of b_{n} terms converges, so the series in question must also converge.

© 2011 Jason B. Hill. All Rights Reserved.