Power Series Convergence Slides Power Series
The Calculus of Convergence
Lesson 01
How does a calculator work?
When you press sin(2), your calculator isn't drawing a triangle and measuring the opposite side.
Instead, it uses a **Power Series**—an infinite polynomial that mimics the function's behavior.
\(\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots\)
Transcendental functions are transformed into basic arithmetic: Addition and Multiplication.
What is a Power Series?
A power series centered at \(c\) is an infinite series of the form:
\[ \sum_{n=0}^{\infty} a_n (x - c)^n = a_0 + a_1(x-c) + a_2(x-c)^2 + \dots \]
Variable
x
Center
c
Coefficients
a_n
Think of it as a polynomial with "infinite degree."
The Radius of Convergence
For any power series \(\sum a_n(x-c)^n\), there are exactly three possibilities:
1
Single Point
Converges only at \(x = c\).
R = 0
2
Interval
Converges if \(|x-c| < R\) and diverges if \(|x-c| > R\).
R > 0
3
Always
Converges for all real values of \(x\).
R = \infty
The Ratio Test: Our Primary Tool
To find the radius \(R\), analyze the limit of the absolute ratio of consecutive terms:
\[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}(x-c)^{n+1}}{a_n(x-c)^n} \right| \]
Set \(L < 1\) to force convergence.
The Ratio Test is **inconclusive** when \(L = 1\).
Warning: Check Your Endpoints
The Ratio Test tells us what happens inside the interval. It says nothing about the boundaries.
Protocol:
Solve \(L < 1\) to get the open interval \((c-R, c+R)\).
Plug \(x = c-R\) into the original series. Test for convergence.
Plug \(x = c+R\) into the original series. Test for convergence.
Possible Interval Shapes
( c-R , c+R ) Open
[ c-R , c+R ] Closed
[ c-R , c+R ) Half-Open
( c-R , c+R ] Half-Open
Let's Try One
Find the radius and interval of convergence for:
\[ \sum_{n=1}^{\infty} \frac{(-1)^n (x-3)^n}{n \cdot 2^n} \]
Step 1
Set up Ratio Test limit.
Step 2
Solve \(L < 1\) for \(|x-3|\).
Step 3
Identify Radius \(R\).
Step 4
Test \(x=1\) and \(x=5\).
Convergence Workshop Worksheet Convergence Workshop
Subject: Calculus II / Power Series Analysis
Name:
Date:
Instructions
For each of the following power series, determine the **Radius of Convergence (R)** and the **Interval of Convergence (I)**. You must show the full limit process using the Ratio Test and provide a rigorous analysis of the endpoints.
1
\(\sum_{n=1}^{\infty} \frac{(x-2)^n}{n \cdot 3^n}\)
Show Ratio Test Limit & Find R
Test Endpoint 1
Test Endpoint 2
Radius \(R\):
Interval \(I\):
2
\(\sum_{n=0}^{\infty} \frac{(-1)^n n! x^n}{10^n}\)
Analysis Area
Radius \(R\):
Interval \(I\):
3
\(\sum_{n=1}^{\infty} \frac{x^n}{n^2}\)
Analysis Area (Include p-series or comparison test for endpoints)
Radius \(R\):
Interval \(I\):
Conceptual Challenge
Suppose that the power series \(\sum c_n x^n\) converges when \(x=4\) and diverges when \(x=-6\). What can be said about the convergence or divergence of the following series? Explain your reasoning.
a) \(\sum c_n 2^n\)
b) \(\sum c_n (-4)^n\)
Convergence Teacher Guide Teacher Guide: Convergence
Lesson 01 | Answer Key & Facilitation Notes
Workshop Problem Solutions
Problem 1: \(\sum \frac{(x-2)^n}{n \cdot 3^n}\)
Ratio Test: \(\lim_{n \to \infty} |\frac{(x-2)^{n+1}}{(n+1)3^{n+1}} \cdot \frac{n 3^n}{(x-2)^n}| = \frac{|x-2|}{3} \lim \frac{n}{n+1} = \frac{|x-2|}{3}\).Radius: Set \(\frac{|x-2|}{3} < 1 \implies |x-2| < 3 \implies R = 3\).Open Interval: \(-3 < x-2 < 3 \implies -1 < x < 5\).Endpoint \(x = -1\): \(\sum \frac{(-3)^n}{n 3^n} = \sum \frac{(-1)^n}{n}\) (Alternating Harmonic Series) → **Converges**.Endpoint \(x = 5\): \(\sum \frac{3^n}{n 3^n} = \sum \frac{1}{n}\) (Harmonic Series) → **Diverges**.Final Interval: \(I = [-1, 5)\).
Problem 2: \(\sum \frac{(-1)^n n! x^n}{10^n}\)
Ratio Test: \(\lim_{n \to \infty} |\frac{(n+1)! x^{n+1}}{10^{n+1}} \cdot \frac{10^n}{n! x^n}| = \lim_{n \to \infty} \frac{(n+1)|x|}{10} = \infty\) (for all \(x \neq 0\)).Radius: \(R = 0\).Interval: \(I = \{0\}\). Only converges at the center.
Problem 3: \(\sum \frac{x^n}{n^2}\)
Ratio Test: \(\lim |x| \frac{n^2}{(n+1)^2} = |x|\). Convergence for \(|x| < 1\).Radius: \(R = 1\).Endpoint \(x = -1\): \(\sum \frac{(-1)^n}{n^2}\) → Converges (AST).Endpoint \(x = 1\): \(\sum \frac{1}{n^2}\) → Converges (p-series, \(p=2\)).Final Interval: \(I = [-1, 1]\).
Facilitation & Misconceptions
Key Questions
"Why does the factorial in Problem 2 make the radius zero?" (Fast growth of numerator)
"What happens to the radius if we square the \(n\) in the denominator?" (R stays the same, endpoints might change).
Common Errors
Forgetting to include the absolute value in the Ratio Test.
Mistaking the center for the radius.
Neglecting to plug endpoints back into the **original** series.
Representing Functions Slides Series Alchemy
Transforming Functions into Power Series
Lesson 02
The Unsolvable Problem
Try to evaluate this integral using standard techniques (Substitution, Parts, Trig Sub):
\[ \int e^{-x^2} dx \]
It is impossible in terms of elementary functions.
"Wait... if we can turn \(e^{-x^2}\) into a polynomial, we can integrate it term-by-term! Simple Power Rule!"
This is the power of Series Manipulation.
The "Grandfather" Series
Everything starts with the Geometric Series representation:
Master Formula
\[ \frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n \]
Valid only for \( |x| < 1 \)
Through substitution, differentiation, and integration, we can turn this single formula into thousands of others.
Rule 1: Substitution
"Replace the x with a block of code"
Find a series for \(\frac{1}{1 + x^2}\):
Treat as \(\frac{1}{1 - (-x^2)}\)
\[ \sum (-x^2)^n = \sum (-1)^n x^{2n} \]
Important Note on Convergence:
The new interval of convergence is found by substituting the inner block into the original condition:
\( |-x^2| < 1 \implies |x| < 1 \)
Rule 2: Term-by-Term Calculus
Differentiation
We can take the derivative of a power series just like a polynomial:
\[ \frac{d}{dx} \sum a_n x^n = \sum n a_n x^{n-1} \]
Radius of convergence stays the **same**. Endpoints may change.
Integration
Similarly, we can integrate term-by-term:
\[ \int \sum a_n x^n dx = C + \sum \frac{a_n x^{n+1}}{n+1} \]
Essential for functions like \(\arctan(x) = \int \frac{1}{1+x^2} dx\).
Practice: Building \(\ln(1+x)\)
1
Start with \(\frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n\)
2
Recognize that \(\int \frac{1}{1+x} dx = \ln(1+x)\)
3
Integrate the series term-by-term:
\[ \ln(1+x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1} \]
Series Alchemy Activity Series Alchemy
Inquiry: Manipulating Geometric Series
Name:
01
The Reciprocal Function
Find a power series representation for \(f(x) = \frac{1}{x}\) centered at \(c=1\).
Hint: Rewrite the function as \(\frac{1}{1 - (1-x)}\) or \(\frac{1}{1 + (x-1)}\).
Work Area
Final Series:
02
The Arctangent Bridge
Use the series for \(\frac{1}{1+x^2}\) to derive the series for \(f(x) = \arctan(x)\). Evaluate the constant of integration \(C\) by using \(f(0) = \arctan(0) = 0\).
Derivation Area
\(\arctan(x) =\)
03
The Gaussian Approximation
This integral is vital in statistics and physics but impossible with elementary algebra. Use series to solve it.
Part A: Find the series for \(e^{-t^2}\) using substitution into the series for \(e^x = \sum \frac{x^n}{n!}\).
Part B: Integrate your result from Part A to find the general antiderivative.
Part C: Evaluate the definite integral \(\int_{0}^{0.5} e^{-t^2} dt\) using the first three non-zero terms of your series.
Approximate Value:
Taylor Construction Slides The Taylor Formula
Custom Polynomial Construction
The Matching Game
How do we build a polynomial \(P(x)\) that looks exactly like a function \(f(x)\) at a point \(x=c\)?
We match the behavior:
P(c) = f(c) [Value]
P'(c) = f'(c) [Slope]
P''(c) = f''(c) [Concavity]
...and so on for every derivative.
As we add more derivatives...
The "shadow" of the polynomial covers more of the function's curve.
The Taylor Series Definition
If \(f\) has derivatives of all orders at \(c\), then:
\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n \]
Taylor Series
Centered at any point \(c\).
Maclaurin Series
A Taylor Series centered at \(c=0\).
Case Study: \(\sin(x)\) at \(c=0\)
n \(f^{(n)}(x)\) \(f^{(n)}(0)\) 0 \(\sin(x)\) 0 1 \(\cos(x)\) 1 2 \(-\sin(x)\) 0 3 \(-\cos(x)\) -1
The Resulting Series:
\[ \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \]
"Only odd powers remain because the sine function is odd, and even-order derivatives are zero at the origin."
The Essential Library
Exponential
\[ e^x = \sum \frac{x^n}{n!} \]
R = \infty
Sine
\[ \sin x = \sum \frac{(-1)^n x^{2n+1}}{(2n+1)!} \]
R = \infty
Cosine
\[ \cos x = \sum \frac{(-1)^n x^{2n}}{(2n)!} \]
R = \infty
Geometric
\[ \frac{1}{1-x} = \sum x^n \]
R = 1
Expansion Workshop Worksheet Expansion Workshop
Deriving Taylor and Maclaurin Series
Name:
Taylor's General Form:
\( f(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \dots + \frac{f^{(n)}(c)}{n!}(x-c)^n + \dots \)
1
The Natural Logarithm
Construct the Taylor series for \(f(x) = \ln(x)\) centered at \(c=1\).
n \(f^{(n)}(x)\) \(f^{(n)}(1)\) Term Coefficient: \(\frac{f^{(n)}(1)}{n!}\) 0 1 2 3
Resulting Series:
2
Exponential Base-Change
Construct the Maclaurin series (\(c=0\)) for \(f(x) = 3^x\). Use the relation \(3^x = e^{x \ln 3}\).
Derivation Area
Maclaurin Series:
3
Manipulation Relay
Use existing series for \(\cos(x)\) to find the first four non-zero terms of \(f(x) = x \cos(x^2)\).
Manipulation Steps
\(x \cos(x^2) \approx\)
Workshop #3: Power Series Construction
\(n! = n(n-1)\dots(1)\)
\(0! = 1\)
Error Bound Slides Quantifying Error
The Lagrange Error Bound
Lesson 04
How many terms are enough?
In 1962, a missing hyphen in code caused the Mariner 1 probe to self-destruct.
Approximation error in engineering isn't just a math problem—it's a safety problem.
"If we need \(\pi\) to 15 decimal places for a Mars landing, how many terms of the series must we calculate?"
We need a way to find the **maximum possible error** without knowing the actual value of the function.
The Remainder Term
Every function can be split into a Polynomial and a Remainder:
\[ f(x) = P_n(x) + R_n(x) \]
Approximation
P_n(x)
Error (Remainder)
R_n(x)
The goal is to show that \(|R_n(x)|\) is very small.
Taylor's Inequality (Lagrange Bound)
If \(|f^{(n+1)}(t)| \leq M\) for all \(t\) between \(x\) and \(c\), then:
\[ |R_n(x)| \leq \frac{M}{(n+1)!} |x - c|^{n+1} \]
M: The "Max Value" of the \((n+1)\)-th derivative on the interval.
n: The degree of the polynomial being used.
How to find "M"
Sine & Cosine
Since \(|\sin(t)| \leq 1\) and \(|\cos(t)| \leq 1\), you can almost always set **\(M = 1\)**.
Exponential (\(e^x\))
For \(e^x\) on \([0, x]\), the derivative is always \(e^t\). Since \(e^t\) is increasing, the max value is **\(e^x\)**.
The Strategy
Identify the interval between the center \(c\) and evaluation point \(x\).
Find the \((n+1)\)-th derivative of the function.
Determine its maximum absolute value on that interval (this is \(M\)).
Plug into the bound formula.
Example: Estimating \(\sin(0.2)\)
Approximate \(\sin(0.2)\) using a 3rd degree Maclaurin polynomial. Find the error bound.
1. Parameters
\(n=3, c=0, x=0.2\)
\(f^{(4)}(t) = \sin(t)\)
2. Finding M
Max value of \(|\sin(t)|\) on \([0, 0.2]\) is at \(0.2\), but we use the global max **\(M=1\)** for simplicity.
\[ |R_3(0.2)| \leq \frac{1}{4!} |0.2|^4 = \frac{0.0016}{24} \approx 0.000067 \]
Uncertainty Report Worksheet Uncertainty Report
Case Study: The Lagrange Error Bound
Project Code: NASA-SERIES-X
Analyst:
Lagrange Error Bound:
\( |R_n(x)| \leq \frac{M}{(n+1)!} |x - c|^{n+1} \)
Case A
The Square Root Approximation
We want to approximate \(\sqrt{4.2}\) using a 2nd degree Taylor polynomial for \(f(x) = \sqrt{x}\) centered at \(c=4\).
Step 1: Find derivatives up to \(n+1\)
f'(x) =
f''(x) =
f'''(x) =
Step 2: Find M (Max of \(|f'''(t)|\) on \([4, 4.2]\))
Step 3: Calculate the bound
Calculation Area
Error \(\leq\)
Case B
High-Precision Requirements
Find the degree \(n\) needed to approximate \(e^{0.5}\) using a Maclaurin polynomial such that the error is less than \(0.0001\). (Use \(e^1 < 3\) to find your \(M\)).
Iterative Solving Space (Test values of n)
Parameters
\(x=0.5, c=0, M=3\)
Final Recommendation
Required Degree \(n\):
...to ensure accuracy to 4 decimal places.
"Everything is a polynomial if you have enough terms and a little bit of tolerance."
Physics Series Slides Series in Motion
Physics & Engineering Applications
Pendulums
Dynamics
Computation
The Engineer's Secret
In physics textbooks, you often see this substitution for small angles \(\theta\):
\(\sin(\theta) \approx \theta\)
Where does this come from? Is it "cheating"?
It's a Taylor Polynomial!
The Maclaurin series for \(\sin(\theta)\) is:
\[ \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \dots \]
If \(\theta\) is small (e.g., \(0.1\) radians), then \(\theta^3\) is **\(0.001\)**. The higher terms become negligible instantly.
Series as Solutions
When a differential equation is too hard for integration factors, we "guess" that the solution is a power series:
\[ y = \sum_{n=0}^{\infty} a_n x^n \]
The Procedure:
Differentiate the series term-by-term for \(y'\) and \(y''\).
Plug them into the Diff Eq.
Shift indices to combine the sums.
Solve for the coefficients \(a_n\).
Case Example: Airy's Equation
\[ y'' - xy = 0 \]
This equation models light diffraction. It has no solution in terms of sines, cosines, or logs.
"The solution is purely defined by its series coefficients. These are known as the **Airy Functions**."
Physics Spotlight: The Pendulum
The exact differential equation for a pendulum is:
\[ \frac{d^2\theta}{dt^2} + \frac{g}{L} \sin(\theta) = 0 \]
This is "Non-linear" and very hard to solve.
Apply the 1st-degree Taylor polynomial (\(\sin\theta \approx \theta\)):
\[ \frac{d^2\theta}{dt^2} + \frac{g}{L} \theta = 0 \]
This is a Simple Harmonic Oscillator! We can solve this in seconds:
\(\theta(t) = \theta_0 \cos(\sqrt{g/L}t)\)
Final Challenge
How would an engineer approximate the energy lost in a system where resistance is modeled by \(e^{-x^2}\)?
"They don't look for a closed form. They integrate the series until the Lagrange bound is smaller than their measurement precision."
Series are not just tools; they are the solution.
Physics Pendulum Lab Activity Field Lab: Pendulum
Subject: Series Approximations in Mechanics
Engineer:
The Small Angle Validity
Engineers use the approximation \(\sin(\theta) \approx \theta\) to linearize the pendulum equation. In this lab, you will use Taylor series to determine exactly how "small" an angle must be for this approximation to be valid within a given error tolerance.
1
Taylor Polynomial Comparison
Write the 3rd-degree Maclaurin polynomial for \(f(\theta) = \sin(\theta)\) and calculate the percent error when using \(P_1(\theta) = \theta\) at \(\theta = 15^\circ\) (Convert to radians first!).
A) \(\theta\) in Radians (\(15^\circ\)):
B) Value of \(P_1(\theta) = \theta\):
C) Value of \(P_3(\theta) = \theta - \frac{\theta^3}{6}\):
D) Difference (Error):
2
Lagrange Verification
Use the Lagrange Error Bound to prove that the error in using \(\sin(\theta) \approx \theta\) for any angle \(|\theta| < 0.2\) radians is less than \(0.002\).
Proof Area
3
Solving the "Unsolvable"
Find the first four non-zero coefficients of the power series solution for the differential equation \(y' = y + x\) with initial condition \(y(0) = 1\).
Step A: Write the series for \(y\) and \(y'\).
Step B: Substitute into the equation and solve for coefficients \(a_0, a_1, a_2, a_3\).
\(a_0 =\)
\(a_1 =\)
\(a_2 =\)
\(a_3 =\)
"Theoretical Physics is just Calculus with a bigger budget."
Small Angle Exit Ticket Final Check
Exit Ticket: The Small Angle Approximation
Question: In engineering, we often replace \(\cos(\theta)\) with the quadratic approximation \(1 - \frac{\theta^2}{2}\) for small oscillations.
1. Use the Maclaurin series for \(\cos(\theta)\) to explain why this is a valid 2nd-degree approximation.
2. If \(\theta = 0.5\) radians, find the maximum error of this approximation using the Lagrange Error Bound (hint: look at the next term in the series).
Error Bound \(\leq\)
Conceptual Reflection
How does the concept of Power Series change your view of complex functions like \(\sin(x)\) or \(e^x\)?
Name:
Lesson 05 | Applied Series
