Local Linearity Worksheet Linear Drafting
Lesson 1: Tangent Line Approximations & Local Linearity
Spec Sheet No.
CALC-01
Lead Engineer (Name)
Date
The Zoom-In Phenomenon
Any smooth curve, when viewed under high enough magnification at a specific point, eventually appears to be a straight line. This is the foundation of local linearity . We use the tangent line as our simplest "blueprint" for a function's behavior near a point \( c \).
1. First-Order Approximation (The Tangent Line)
The equation of the tangent line to \( f(x) \) at \( x = c \) is given by: \( L(x) = f(c) + f'(c)(x - c) \)
Task A: The Sine Wave Center
Construct the tangent line blueprint for \( f(x) = \sin(x) \) centered at \( c = 0 \).
Step 1: Find \( f(0) \)
Step 2: Find \( f'(0) \)
Final Equation \( L(x) \):
2. Error Analysis (Structural Failure)
The tangent line is only an approximation . The error \( E(x) \) is the difference between the actual function value and the blueprint value: \( E(x) = |f(x) - L(x)| \).
Task B: Testing Tolerance
Use your blueprint \( L(x) \) from Task A to approximate \( \sin(x) \) at the following values. Calculate the error (use a calculator for the true \( f(x) \)).
| Input \( x \) | Blueprint \( L(x) \) | True \( \sin(x) \) | Error \( |f-L| \) |
| --- | --- | --- | --- |
| \( 0.01 \) | | ~0.009999 | |
| \( 0.1 \) | | ~0.099833 | |
| \( 1.0 \) (approx. \( 57^\circ \)) | | ~0.841470 | |
Blueprint Debrief:
What happens to the error as you move further away from the center \( c = 0 \)? Explain why a straight line is "failing" to represent the curve at \( x = 1.0 \).
3. Future Development: Higher Order Blueprints
"The tangent line captures the slope, but ignores the bend."
To reduce error, we need our blueprint to "bend" with the curve. In the next lesson, we will match the second derivative (concavity) to create a parabolic approximation.
Linear Drafting Slides Linear Drafting
The Blueprint of Local Linearity and the Foundation of Calculus Approximations
Spec: CALC-L1
How does a calculator "know" \(\sin(0.1)\)?
Calculators don't use giant look-up tables or draw perfect circles.
They use polynomials —the simplest functions to compute—to "blueprint" complex curves.
\( y = \sin(x) \)
\( y \approx x \)
The Concept: Local Linearity
Magnify
Zoom into any point on a smooth curve.
Straighten
Under high enough zoom, the curve looks straight.
Draft
The tangent line is our 1st-order blueprint.
The Drafting Formula
\( L(x) = f(c) + f'(c)(x - c) \)
\( f(c) \)
The "Starting Point"
\( f'(c) \)
The "Slope"
\( (x - c) \)
The "Distance from Center"
Structural Failure: Error
Linear Failure
As we move away from the center \( c \), the straight line fails to account for the curve's bending .
Error = \( |f(x) - L(x)| \)
Observation Log
Close to center? Error is negligible.
Far from center? Error grows rapidly.
How do we fix the bend?
Phase 2 Preview
To capture the "bend," our blueprint needs higher dimensions .
Linear
Degree 1
Quadratic
Degree 2
Linear Drafting Answer Key Answer Key
Lesson 1: Linear Drafting (Teacher Reference)
Confidential
KEY-CALC-01
Teacher Instructions:
This document provides solutions for the Local Linearity Worksheet. Use the discussion prompts to guide students toward the realization that linear models are "local" and require curvature (the 2nd derivative) to improve accuracy over larger intervals.
1. First-Order Approximation (The Tangent Line)
Task A: The Sine Wave Center (\( f(x) = \sin(x) \), \( c = 0 \))
Step 1: Find \( f(0) \)
\( \sin(0) = 0 \)
Step 2: Find \( f'(0) \)
\( f'(x) = \cos(x) \rightarrow \cos(0) = 1 \)
Final Equation \( L(x) \):
\( L(x) = 0 + 1(x - 0) \rightarrow L(x) = x \)
2. Error Analysis (Structural Failure)
| Input \( x \) | Blueprint \( L(x) \) | True \( \sin(x) \) | Error \( |f-L| \) |
| --- | --- | --- | --- |
| \( 0.01 \) | \( 0.01 \) | ~0.009999 | \( 0.000001 \) |
| \( 0.1 \) | \( 0.1 \) | ~0.099833 | \( 0.000167 \) |
| \( 1.0 \) | \( 1.0 \) | ~0.841470 | \( 0.158530 \) |
Blueprint Debrief Solution:
As \( x \) increases, the error increases drastically. At \( x=0.01 \), the error is nearly zero (less than 0.0001%), while at \( x=1 \), the error is nearly 16%. This occurs because the sine function has "concavity" (it curves downward), while the tangent line continues straight. The linear blueprint cannot "see" the bend in the function.
Transition to Lesson 2:
Challenge students: "If a line can't match the bend, what shape could? A parabola. How do we make the parabola's 'bend' match the function's 'bend' at the center?" (This introduces matching the 2nd derivative).
Polynomial Construction Slides Polynomial Construction
Matching Derivatives to Build the Perfect Curve Approximation
Spec: CALC-L2
The "Hug" Challenge
The tangent line only matches the value and the slope .
Goal: Create a polynomial that "hugs" the function better by matching the bend.
Linear Blueprint
Matches \( f(c) \) and \( f'(c) \)
Quadratic Blueprint
Matches \( f(c) \), \( f'(c) \), AND \( f''(c) \)
Building Concavity
To match the concavity, our polynomial needs an \( x^2 \) term.
\( P(x) = a + b(x-c) + d(x-c)^2 \)
\( P(c) = a = f(c) \)
\( P'(c) = b = f'(c) \)
\( P''(c) = 2d = f''(c) \)
Structural Insight
Notice the factor of 2 !
When we take the second derivative of \( x^2 \), the power comes down. To "cancel" it out, we must divide by \( 2 \)!
\( d = \frac{f''(c)}{2} \)
The Taylor Blueprint Formula
\( P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(c)}{k!} (x-c)^k \)
\( f(c) \)
Position
\( f'(c)(x-c) \)
Slope
\( \frac{f''(c)}{2!}(x-c)^2 \)
Bend
\( \frac{f'''(c)}{3!}(x-c)^3 \)
Rate of Bend
The factorials \( k! \) account for the repeated application of the Power Rule during differentiation.
Construction Quality
Degree 1 (Linear)
Good for points very close to center.
Low Precision Range
Degree 3 (Cubic)
Captures inflection and wider range.
Medium Precision Range
Degree 7+
Virtually indistinguishable from the function near the center.
High Precision Range
Construction Site
Let's build a 3rd-degree blueprint for \( f(x) = e^x \) centered at \( c = 0 \).
Spec 1: Derivatives
Calculate \( f'(0), f''(0), f'''(0) \)
Spec 2: Assembly
Plug into the Taylor formula and simplify.
Polynomial Construction Activity Polynomial Construction
Lesson 2: Higher-Order Taylor Approximations
Site Map
MAP-CALC-02
Lead Architect
Project Date
Structural Objective
Your goal is to construct a 3rd-degree Taylor Polynomial (\( P_3(x) \)) for the function \( f(x) = \sin(x) \) centered at \( c = 0 \). This will create a cubic blueprint that "hugs" the curve better than our simple tangent line.
Phase 1: Calculating Derivatives
Fill out the derivative table for \( f(x) = \sin(x) \) at the center \( c = 0 \).
Order (\( k \)) Derivative \( f^{(k)}(x) \) Value at \( c=0 \) 0 (The Function) \( \sin(x) \) \( f(0) = \) 1 (The Slope) \( f'(0) = \) 2 (The Bend) \( f''(0) = \) 3 (The Swerve) \( f'''(0) = \)
Phase 2: Assemblage
Plug your values into the Taylor Assembly Template:
\( P_3(x) = \)
?
\( + \)
?
\( x + \)
?
\( 2! \)
\( x^2 + \)
?
\( 3! \)
\( x^3 \)
Simplified Blueprint:
\( P_3(x) = \) ________________________________________
Phase 3: Final Inspection
Comparing to Linear:
Recall the tangent line \( L(x) = x \). How does \( P_3(x) \) differ? How will this subtraction term change the shape of the line as \( x \) grows?
The Missing Link:
Notice that the \( x^2 \) term disappeared! Looking at your derivative table, explain why some terms in the Taylor series for \( \sin(x) \) become zero.
Polynomial Construction Answer Key Answer Key
Lesson 2: Polynomial Construction (Teacher Reference)
Restricted Access
KEY-CALC-02
Teacher Instructions:
This key provides the derivation for the 3rd-degree Taylor polynomial of \( \sin(x) \). Emphasize to students that the disappearing even powers (\( x^0, x^2, \dots \)) are a unique property of odd functions like sine.
Phase 1: Calculating Derivatives
Order (\( k \)) Derivative \( f^{(k)}(x) \) Value at \( c=0 \) 0 \( \sin(x) \) \( f(0) = 0 \) 1 \( \cos(x) \) \( f'(0) = 1 \) 2 \( -\sin(x) \) \( f''(0) = 0 \) 3 \( -\cos(x) \) \( f'''(0) = -1 \)
Phase 2: Assemblage
\( P_3(x) = 0 + 1x + \frac{0}{2!}x^2 + \frac{-1}{3!}x^3 \)
Simplified Blueprint:
\( P_3(x) = x - \frac{x^3}{6} \)
Phase 3: Final Inspection Solutions
Comparing to Linear:
The cubic polynomial \( x - x^3/6 \) includes a subtraction term. As \( x \) increases (positive direction), the cubic term pulls the function downward, matching the "concave down" behavior of sine in the first quadrant. This makes it more accurate than the line \( y=x \) which continues straight up.
The Missing Link:
The \( x^2 \) term disappears because the second derivative of sine at zero is \( \sin(0) = 0 \). This occurs because sine is centered at an inflection point at the origin where its concavity is momentarily zero.
Maclaurin Patterns Slides Maclaurin Cycles
Discovering the Infinite Patterns in Taylor Series Centered at Zero
Spec: CALC-L3
The Simplest Center: \( c = 0 \)
When we center a Taylor Series at the origin (\( c = 0 \)), we call it a Maclaurin Series .
\( P(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \)
No more \((x - c)\) terms to worry about—just pure powers of \( x \)!
Advantages
Calculations are significantly simpler at zero.
Patterns in cyclic derivatives emerge clearly.
The 4-Step Cycle
\( \sin(x) \)
Start
\( \cos(x) \)
Derivative 1
\( -\cos(x) \)
Derivative 3
\( -\sin(x) \)
Derivative 2
At \( x = 0 \):
\( \sin(0) = 0 \)
\( \cos(0) = 1 \)
\( -\sin(0) = 0 \)
\( -\cos(0) = -1 \)
Pattern: 0, 1, 0, -1... repeat!
The Big Three Series
\( e^x \)
\( 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \)
All powers
\( \sin(x) \)
\( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \)
Odd powers
\( \cos(x) \)
\( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \)
Even powers
Symmetry in Series
The Taylor series perfectly reflects the function's symmetry!
Odd Functions (\(\sin\)) only have odd powers in their series.Even Functions (\(\cos\)) only have even powers in their series.
EULER'S LINK?
Pattern Mission
You are about to derive the Maclaurin series for \( e^x \) and \( \cos(x) \).
Look for the Signs
Pay close attention to the signs (+/-) and which derivatives "disappear" at zero. These are the keys to the infinite patterns.
Maclaurin Discovery Worksheet Pattern Discovery
Lesson 3: Maclaurin Series Derivation
Cycle Log
LOG-CALC-03
Investigator
Cycle Date
Discovery 1: The Unchanging \( e^x \)
Derive the first 5 terms (up to \( x^4 \)) of the Maclaurin series for \( f(x) = e^x \).
Derivative Chain (\( c=0 \))
\( f(0) = \)
\( f'(0) = \)
\( f''(0) = \)
\( f'''(0) = \)
\( f^{(4)}(0) = \)
Pattern Recognition
What is unique about the derivatives of \( e^x \)? How does this make the Taylor coefficients (\( a_n \)) easy to predict?
\( e^x \approx \) __________________________________________________________________
Discovery 2: The Even Cycle of \( \cos(x) \)
Calculate the values for the Maclaurin series of \( f(x) = \cos(x) \) at \( c = 0 \).
Order \( k \) \( f^{(k)}(x) \) \( f^{(k)}(0) \) Term \( \frac{f^{(k)}(0)}{k!}x^k \) 0 \( \cos(x) \) 1 2 3 4
\( \cos(x) \approx \) __________________________________________________________________
Predicting the Infinite
Based on your discoveries, write the general term (\( \Sigma \) notation) for each series:
Exponential Series
\( \sum_{n=0}^{\infty} \frac{x^n}{?} \)
Cosine Series
\( \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{?} \)
Maclaurin Discovery Answer Key Answer Key
Lesson 3: Maclaurin Cycles (Teacher Reference)
Cycle Log Key
KEY-CALC-03
Teacher Notes:
Students often struggle with the alternating signs in trigonometric series. Remind them that \( (-1)^n \) creates the oscillation: positive when \( n \) is even, negative when \( n \) is odd.
Discovery 1: The Unchanging \( e^x \)
Derivative Chain (\( c=0 \))
\( f(0) = e^0 \) 1
\( f'(0) = e^0 \) 1
\( f''(0) = e^0 \) 1
\( f'''(0) = e^0 \) 1
\( f^{(4)}(0) = e^0 \) 1
Pattern Solution
Since the derivative of \( e^x \) is always \( e^x \), all derivatives at zero equal 1. This means the coefficient for every term \( x^n \) is simply \( 1/n! \).
\( e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \)
Discovery 2: The Even Cycle of \( \cos(x) \)
Order \( k \) \( f^{(k)}(x) \) \( f^{(k)}(0) \) Term 0 \( \cos(x) \) 1 1 1 \( -\sin(x) \) 0 0 2 \( -\cos(x) \) -1 \( -\frac{x^2}{2!} \) 3 \( \sin(x) \) 0 0 4 \( \cos(x) \) 1 \( \frac{x^4}{4!} \)
\( \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots \)
Exponential Series
\( \sum_{n=0}^{\infty} \frac{x^n}{n!} \)
Cosine Series
\( \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \)
Validity Zones Slides Validity Zones
Investigating the Radius and Interval of Convergence for Power Series
Spec: CALC-L4
When Series "Break"
Unlike polynomials, infinite series don't always represent a function everywhere.
A power series is only "valid" if the terms get small fast enough to sum to a finite value.
Case Study: \(\frac{1}{1-x}\)
\( 1 + x + x^2 + x^3 + \dots \)
Valid for \( x = 0.5 \)
Explodes for \( x = 2 \)
The Structural Integrity Test
The Ratio Test
\( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L \)
\( L < 1 \)
SAFE ZONE
Converges Absolutely
\( L > 1 \)
DANGER ZONE
Diverges
\( L = 1 \)
UNKNOWN
Test Endpoints
Mapping the Zone
Center \( c \)
Radius \( R \)
Radius of Convergence (\( R \))
The distance from the center to the edge of the safe zone.
Interval of Convergence (\( I \))
The set of all \( x \)-values where the series is valid.
Surveyor's Checklist
1
Set up the Ratio Test for the series.
2
Solve the inequality \( L < 1 \) for \( x \).
3
Identify the Radius \( R \) and the Interval \( I \).
4
Check the endpoints! (The boundary check).
Field Work
We will now find the Validity Zone for the Geometric Series:
\( \sum_{n=0}^{\infty} x^n \)
Expected result: The zone where \( |x| < 1 \).
Ratio Test Worksheet Zone Survey
Lesson 4: Radius and Interval of Convergence
Boundary Report
BND-CALC-04
Surveyor Name
Date
Safe Zone Protocol
Use the Ratio Test to find the Radius of Convergence (\( R \)) and the Interval of Convergence (\( I \)) for the given power series. Ensure you test the behavior at the boundaries.
Site 1: \( \sum_{n=1}^{\infty} \frac{x^n}{n} \)
Step 1: Ratio Test Setup
\( \lim_{n \to \infty} \left| \frac{\text{Term } n+1}{\text{Term } n} \right| = \) ____________________
Step 2: Solve for \( L < 1 \)
Step 3: Boundary Check (\( x = \pm R \))
Radius \( R \)
Interval \( I \)
Site 2: \( \sum_{n=0}^{\infty} \frac{(x-3)^n}{2^n} \)
Step 1: Ratio Test Setup
_________________________________________________
Step 2: Solve for \( L < 1 \)
Step 3: Identifying Center and Radius
Radius \( R \)
Interval \( I \)
Ratio Test Answer Key Answer Key
Lesson 4: Validity Zones (Teacher Reference)
Confidential Survey Key
KEY-CALC-04
Site 1: \( \sum_{n=1}^{\infty} \frac{x^n}{n} \)
Ratio Test Solution:
\( \lim_{n \to \infty} \left| \frac{x^{n+1}}{n+1} \cdot \frac{n}{x^n} \right| = \lim_{n \to \infty} |x| \left( \frac{n}{n+1} \right) = |x| \)
Set \( |x| < 1 \). Radius \( R = 1 \). Potential Interval: \( (-1, 1) \).
Boundary Check Solution:
At \( x = 1 \): \( \sum 1/n \) (Harmonic Series) \( \rightarrow \) Diverges .At \( x = -1 \): \( \sum (-1)^n/n \) (Alt. Harmonic Series) \( \rightarrow \) Converges .
Radius: 1
Interval: \( [-1, 1) \)
Site 2: \( \sum_{n=0}^{\infty} \frac{(x-3)^n}{2^n} \)
Ratio Test Solution:
\( \lim_{n \to \infty} \left| \frac{(x-3)^{n+1}}{2^{n+1}} \cdot \frac{2^n}{(x-3)^n} \right| = \left| \frac{x-3}{2} \right| \)
Set \( \frac{|x-3|}{2} < 1 \rightarrow |x-3| < 2 \). Center is 3. Radius is 2.
Boundary Check Solution:
At \( x = 5 \): \( \sum (2)^n/2^n = \sum 1 \) \( \rightarrow \) Diverges .At \( x = 1 \): \( \sum (-2)^n/2^n = \sum (-1)^n \) \( \rightarrow \) Diverges .
Radius: 2
Interval: \( (1, 5) \)
Precision Competition Slides Accuracy Architect
The Precision Competition: Approximating the Impossible by Hand
Spec: CALC-L5
The "Human Calculator" Challenge
How would you calculate \( e \) or \( \sqrt{2} \) to 5 decimal places if you only had a pencil and paper?
Taylor Series turn "impossible" transcendental numbers into simple arithmetic operations.
The Competition Rules
Calculate an irrational number using a Taylor Series.
Highest accuracy with the fewest terms wins.
The Architect's Trade-off
High Accuracy
Matching more derivatives (higher \( n \)) reduces the error significantly.
n=10, 20, 50...
High Cost
Each new term requires more calculation. In real-world computing, this means more time and energy.
Computation Time ++
The Power of the Center
Strategy is key! Approximating \( \sin(3) \) using a Maclaurin series (\( c=0 \)) is a nightmare.
But approximating \( \sin(3) \) using a Taylor series centered at \( c = \pi \approx 3.14 \) is incredibly fast!
Architect Tip
Choose a center \( c \) that is as close as possible to your target \( x \).
The Mission: Approximating \( \sqrt{e} \)
Target
\( e^{0.5} \)
Use the Maclaurin series for \( e^x \).
Round 1: Degree 2
Calculate \( 1 + (0.5) + \frac{(0.5)^2}{2} \)
Round 2: Degree 3
Add the cubic term: \( \frac{(0.5)^3}{6} \)
Bonus Challenge
Estimate the error without a calculator using the Remainder Theorem (preview)!
CALCULATE!
Open your Architect Logs. You have 15 minutes to reach 4 decimal places of accuracy.
Accuracy Architect Challenge Architect Log
Lesson 5: Precision Competition (Final Project)
Accuracy Rank
ACC-CALC-05
Lead Architect
Date
Mission: The Root of Precision
Irrational numbers are the ultimate test of an architect. Your task is to approximate the value of \( \sqrt{1.2} \) using a 2nd-degree Taylor Polynomial .
Target Function
\( f(x) = \sqrt{x} \)
Target Point
\( x = 1.2 \)
Best Center (\( c \))
?
Phase 1: Derivative Blueprint
Choose center \( c = 1 \) for easy square root calculation.
Order \( k \) \( f^{(k)}(x) \) Value at \( c=1 \) 0 \( \sqrt{x} = x^{1/2} \) 1 1 2
Phase 2: Assemblage and Calculation
Taylor Polynomial \( P_2(x) \) centered at \( c=1 \):
\( P_2(x) = \) ________________________________________
The Calculation: Plug in \( x = 1.2 \) and compute manually.
Show Arithmetic Work
Final Estimate
1. ________
Phase 3: Final Analysis
?
True Value (Calculator): \( \sqrt{1.2} \approx 1.095445 \dots \)
Calculate your Error:
Error = \( |1.095445 - \text{Estimate}| = \) ____________________
Were you within 0.001? If not, would adding a 3rd-degree term (the \( x^3 \) term) help or hurt your efficiency?
Accuracy Architect Answer Key Answer Key
Lesson 5: Accuracy Architect (Teacher Reference)
Confidential Key
KEY-CALC-05
Teacher Instructions:
This key provides the full manual calculation for \( \sqrt{1.2} \). The choice of center \( c=1 \) is crucial for ease of calculation. Students should notice that even with only two terms (Degree 2), the accuracy is quite high.
Phase 1: Derivative Blueprint Solutions
Order \( k \) \( f^{(k)}(x) \) Value at \( c=1 \) 0 \( x^{1/2} \) 1 1 \( \frac{1}{2}x^{-1/2} \) 1/2 (0.5) 2 \( -\frac{1}{4}x^{-3/2} \) -1/4 (-0.25)
Phase 2: Assemblage and Calculation
Taylor Polynomial \( P_2(x) \):
\( P_2(x) = 1 + \frac{1}{2}(x-1) - \frac{1}{4(2!)}(x-1)^2 = 1 + \frac{1}{2}(x-1) - \frac{1}{8}(x-1)^2 \)
Manual Calculation Solution (\( x = 1.2 \)):
\( P_2(1.2) = 1 + 0.5(0.2) - 0.125(0.2)^2 \)
\( = 1 + 0.1 - 0.125(0.04) \)
\( = 1.1 - 0.005 \)
Final Estimate: 1.095
Phase 3: Final Analysis Key
Error: \( |1.095445 - 1.095| = 0.000445 \).
The estimate is incredibly accurate for only two terms (well within the 0.001 tolerance). This demonstrates the power of Taylor series when centered near the target value. A 3rd term would improve accuracy but increase the arithmetic complexity (dividing by 3! and handling \( 0.2^3 \)).
