lectures

warning: Creating default object from empty value in /home/sidehop/math.jasonbhill.com/modules/taxonomy/taxonomy.pages.inc on line 34.

Separation of Variables

Section: 
11.4
Date: 
Wednesday, October 27, 2010 - 14:00 - 15:00

Math 2300 Section 005 – Calculus II – Fall 2010

Notes on Separation of Variables – October 27, 2010

Definition: We call a differential equation separable if it can be written in the form

\frac{dy}{dx}=g(x)f(y)

for f a function of y and g a function of x.

The Formal Approach to Solving Separable Differntial Equations

If f(y)\neq 0 then we can write

\frac{1}{f(y)}\frac{dy}{dx}=g(x).

Integrating both sides with respect to the variable x gives

\displaystyle\int\frac{1}{f(y)}\frac{dy}{dx}\, dx \displaystyle=\int g(x)\, dx
\displaystyle\int\frac{1}{f(y)}\, dy \displaystyle=\int g(x)\, dx

What will usually happen is that this introduces a natural log in terms of y. Exponentiating and solving for y then provides a solution in terms of x.

Slope Fields

Section: 
11.2
Date: 
Monday, October 25, 2010 - 14:00 - Tuesday, October 26, 2010 - 15:00

Math 2300 Section 005 – Calculus II – Fall 2010

Notes on Slope Fields – October 25–26, 2010

Slope Field Examples

  1. Find an infinite collection of solutions to the differential equation

    y^{{\prime}}=-\frac{x}{y}.

    Solution: It's nearly impossible to ask for even a single solution to this differential equation without knowing more about it. You can graph out the slope field yourself (perhaps not quite as neatly as a computer or calculator, but you can do it). You get a slope field like this.

    Now it appears a bit more obvious that we're dealing with circles centered at the origin. Let's see if that is actually correct by taking a generic circle, centered at the origin, and determining if it satisfies the differential equation. In what follows, we use implicit differentiation to view y as a function of x.

    \displaystyle x^{2}+y^{2} \displaystyle=r^{2}
    \displaystyle\frac{d}{dx}\left[x^{2}+y^{2}\right] \displaystyle=\frac{d}{dx}r^{2}
    \displaystyle 2x+2y\frac{dy}{dx} \displaystyle=0
    \displaystyle 2y\frac{dy}{dx} \displaystyle=-2x
    \displaystyle\frac{dy}{dx} \displaystyle=-\frac{2x}{2y}
    \displaystyle\frac{dy}{dx} \displaystyle=-\frac{x}{y}
    \displaystyle y^{{\prime}} \displaystyle=-\frac{x}{y}.

    So, each circle centered at the origin satisfies the given differential equation.

Taylor Polynomial Error Bounds

Section: 
10.4
Date: 
Wednesday, October 20, 2010 - 14:00 - 15:00

Math 2300 Section 005 – Calculus II – Fall 2010

Notes on Taylor Polynomial Error Bounds – October 20, 2010

We'll start by discussing the formal error bound for Taylor polynomials. (I.e., how badly does a Taylor polynomial approximate a function?) Then, we'll see some standard examples. Finally, we'll see a powerful application of the error bound formula.

Lagrange Error Bound for P_{n}(x)

We know that the nth Taylor polynomial is P_{n}(x), and we have spent a lot of time in this chapter calculating Taylor polynomials and Taylor Series. The question is, for a specific value of x, how badly does a Taylor polynomial represent its function? We define the error of the nth Taylor polynomial to be

E_{n}(x)=f(x)-P_{n}(x).

That is, error is the actual value minus the Taylor polynomial's value. Of course, this could be positive or negative. So, we force it to be positive by taking an absolute value.

|E_{n}(x)|=|f(x)-P_{n}(x)|.

The following theorem tells us how to bound this error. That is, it tells us how closely the Taylor polynomial approximates the function. Essentially, the difference between the Taylor polynomial and the original function is at most |E_{n}(x)|. At first, this formula may seem confusing. I'll give the formula, then explain it formally, then do some examples. You may want to simply skip to the examples.

Theorem 10.1 Lagrange Error Bound  Let f be a function such that it and all of its derivatives are continuous. If P_{n}(x) is the nth Taylor polynomial for f(x) centered at x=a, then the error is bounded by

|E_{n}(x)|\le\frac{M}{(n+1)!}|x-a|^{{n+1}}

where M is some value satisfying |f^{{(n+1)}}(x)|\le M on the interval between a and x.

New Taylor Series From Old

Section: 
10.3
Date: 
Monday, October 25, 2010 - 14:00 - 15:00

Math 2300 Section 005 – Calculus II – Fall 2010

Notes on Obtaining New Taylor Series from Old – October 19, 2010

Taylor Series We Currently Know

\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
\displaystyle\frac{1}{1+x} \displaystyle=(1+x)^{{-1}}=1-x+x^{2}-x^{3}+x^{4}-\cdots=\sum _{{k=0}}^{\infty}(-1)^{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 are several ways we could go about obtaining new series from these ones. First, we can substitute inside an existing series. Then, we could differentiate or integrate existing series. We'll consider these with some examples.

Taylor Series Introduction

Section: 
10.2
Date: 
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}.

Examples

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}

where

\binom{p}{k}=\frac{p!}{k!(p-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

Section: 
10.1
Date: 
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

P_{{n+1}}(x)=P_{n}(x)+\frac{f^{{(n+1)}}(x)}{(n+1)!}(x-a)^{{n+1}}.

Radius of Convergence

Section: 
9.5
Date: 
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

where

  • 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.

Limit Comparison Test

Section: 
9.4
Date: 
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.

Ratio Test

Section: 
9.4
Date: 
Wednesday, October 6, 2010 - 14:00 - 15:00

Math 2300 Section 005 – Calculus II – Fall 2010

9.4 Ratio Test Examples – October 6, 2010

  1. What does the ratio test say about the series?

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

    Solution:

    \lim _{{n\rightarrow\infty}}\frac{\displaystyle\frac{1}{n+1}}{\displaystyle\frac{1}{n}}=\lim _{{n\rightarrow\infty}}\frac{n}{n+1}=1

    and so the ratio test says nothing about the series. Of course, we already know that the harmonic series is divergent.

Comparison Test

Section: 
9.4
Date: 
Tuesday, October 5, 2010 - 14:00 - Wednesday, October 6, 2010 - 15:00

Math 2300 Section 005 – Calculus II – Fall 2010

9.4 Comparison Test Examples – October 5, 2010

  1. Use the comparison test to determine if the given series converges or diverges.

    \sum _{{n=1}}^{\infty}\frac{1}{n^{4}+1}

    Solution: Since

    0\le\frac{1}{n^{4}+1}\le\frac{1}{n^{4}}

    and we know the p-series

    \sum _{{n=1}}^{\infty}\frac{1}{n^{4}}

    converges, the given series also converges.

© 2011 Jason B. Hill. All Rights Reserved.