Fixed Point Slides Numerical Analysis | Lesson 1
Fixed Point Logic
Discovering convergence through iterative sequences and cobweb dynamics.
ITERATION_V1.0
The Calculator Loop
Try this: Pick any number. Press cos. Then press cos again. And again.
0.739085...
The "Dottie Number"
Why does a sequence of repeated functions always lead to the same destination?
\(x = \cos(x)\)
Solving by "Doing"
What is a Fixed Point?
A value \(L\) is a fixed point of a function \(g(x)\) if:
\(g(L) = L\)
In a iterative sequence where \(x_{n+1} = g(x_n)\), if the sequence converges to \(L\), then \(L\) must be a fixed point.
Analytic View
Finding where the function intersects the line \(y = x\).
Iterative View
Feeding the output of one step as the input for the next.
Visualizing Iteration
The Cobweb Plot Algorithm
1
Start at \((x_0, 0)\) and move vertically to \((x_0, g(x_0))\).
2
Move horizontally to the line \(y = x\). You are now at \((g(x_0), g(x_0))\).
3
Repeat. This horizontal coordinate is your new \(x_1\).
The resulting path "spirals" or "staircases" toward (or away from) the fixed point.
The Convergence Test
When will \(x_{n+1} = g(x_n)\) actually find the solution?
CONVERGENCE CONDITION
\(|g'(L)| < 1\)
The "slope" at the fixed point must be shallow enough to "pull" the sequence in.
DIVERGENCE CONDITION
\(|g'(L)| > 1\)
If the function is too steep, the sequence will "fly away" or bounce infinitely.
Stable
Unstable
Fixed Point Workshop Worksheet Cobweb Workshop
Numerical Analysis | Fixed Point Iteration
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1
Stability Analysis
For each of the following iterative functions \(g(x)\), determine if the sequence \(x_{n+1} = g(x_n)\) will converge to the given fixed point \(L\). Show your work using the derivative test \(|g'(L)| < 1\).
A. \(g(x) = \frac{1}{2}x + 2\) L = 4
Converges Diverges
B. \(g(x) = x^2 - 1\) L = 1.618
Converges Diverges
2
Cobweb Mapping
Perform a cobweb iteration for the function \(g(x) = \sqrt{2x + 3}\) starting at \(x_0 = 1\). Draw the steps on the plot provided and record the first 4 terms of the sequence below.
y = x y = g(x) 0 1 2 3
Sequence Terms
x₀ = 1.0000
x₁ =
x₂ =
x₃ =
3
Algorithmic Reformulation
Suppose we want to solve the equation \(x^3 + x - 1 = 0\). We can rewrite this in multiple fixed-point forms \(x = g(x)\). Evaluate the two options below:
Option 1
\(g_1(x) = 1 - x^3\)
Starting near \(x \approx 0.68\)
Find \(g_1'(0.68)\) to check stability:
Option 2
\(g_2(x) = (1 - x)^{1/3}\)
Starting near \(x \approx 0.68\)
Find \(g_2'(0.68)\) to check stability:
Critical Reflection
Why do some rearrangements of the same equation work for iteration while others fail? What does this tell us about the "design" of numerical algorithms?
UNIT 1: ITERATIVE INSIGHTS // FIXED POINT LOGIC // WORKsheet_id: 101-FP
Newton Method Slides Lesson 2: Tangent Trail
Newton's Method
From tangent line approximations to a powerful recursive sequence for root-finding.
The Speed of Calculation
How do you solve \(x^5 + 2x - 1 = 0\)?
There is no "quintic formula." You can't isolate \(x\).
Bisection
Reliable but Slow
Newton's Method
Incredibly Fast
// NEWTON_ITERATION.LOG
n=0: x = 1.000000000
n=1: x = 0.666666667
n=2: x = 0.518518519
n=3: x = 0.487121212
n=4: x = 0.486389234
n=5: x = 0.486389035 (FIXED)
The Geometry of Roots
We want to find where \(f(x) = 0\). We start with a guess \(x_n\).
1. Find the point \((x_n, f(x_n))\).
2. Draw the tangent line at that point.
3. See where the tangent line hits the x-axis.
4. That x-intercept is our new guess, \(x_{n+1}\).
x_0 x_1
The Recursive Formula
Derived from the slope formula: \(f'(x_n) = \frac{f(x_n) - 0}{x_n - x_{n+1}}\)
\[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\]
NEWTON'S FORMULA
Fast Convergence
Usually doubles precision every step.
Initial Guess
Requires a "good" starting point \(x_0\).
Failure Modes
Fails if \(f'(x_n) = 0\) (division by zero).
Hidden Sequences
Your calculator doesn't "know" square roots. It calculates them using Newton's Method.
Example: Finding \(\sqrt{A}\)
Solve \(f(x) = x^2 - A = 0\)
\(x_{n+1} = \frac{1}{2}\left(x_n + \frac{A}{x_n}\right)\)
This is the Babylonian Method—a specific case of Newton's Method!
Embedded in every ALU (Arithmetic Logic Unit) is an iterative sequence derived from these calculus principles.
Newton Method Lab Worksheet Newton Method Lab
Numerical Analysis // Tangent Trail
LAB_ID: NM_202_TANGENT
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1. Algorithmic Derivation
Starting from the point-slope form of a line, \(y - y_0 = m(x - x_0)\), derive the Newton-Raphson iteration formula. Assume the line is tangent to \(f(x)\) at \(x_n\) and we seek the x-intercept \(x_{n+1}\).
Resulting Formula:
\(x_{n+1} = \) __________________________________
2. The "Quintic" Problem
Find the positive root of \(f(x) = x^5 + x - 1\). First, calculate the derivative: \(f'(x) = \) _____________________.
Iteration (n) Guess (\(x_n\)) \(f(x_n)\) \(f'(x_n)\) Next (\(x_{n+1}\)) 0 1.0000 1 2 3
3. Failure Modes Analysis
Newton's Method is powerful but fragile. Below is a plot of \(f(x) = \sin(x)\). If you chose a starting guess of \(x_0 = \pi/2\), what would happen in the first iteration? Explain why using the formula.
x_0 = π/2
Bonus: The Babylonian Connection
We saw that \(x_{n+1} = \frac{1}{2}(x_n + A/x_n)\) calculates \(\sqrt{A}\). Prove that this is exactly what you get when you apply Newton's Method to \(f(x) = x^2 - A\).
Convergence Speed Slides Lesson 3: Convergence Speed
Error & Efficiency
Measuring the rate at which numerical sequences approach the truth.
The Error Sequence
In numerical analysis, we don't just care that a sequence converges—we care how fast.
Absolute Error at step \(n\)
\(e_n = |x_n - L|\)
Where \(L\) is the true solution (limit).
The Goal
We want the error sequence \(\{e_n\}\) to shrink to zero as quickly as possible.
Precision = Decimal places of accuracy
Efficiency = Computation time/cost
The "Big Three" Rates
Linear (\(p=1\))
Steady
Error is reduced by a constant factor \(k\) each step.
\(e_{n+1} \approx k \cdot e_n\)
Quadratic (\(p=2\))
Explosive
Precision doubles every single step. (Newton's Method)
\(e_{n+1} \approx k \cdot e_n^2\)
Cubic (\(p=3\))
Extreme
Precision triples every step. Highly efficient but complex to compute.
\(e_{n+1} \approx k \cdot e_n^3\)
The Cost of Slowness
Algorithm Face-Off
To get 15 decimal places of \(\pi\):
Leibniz Formula
Sub-linear convergence
10,000,000,000+ Steps
Years of calculation
Newton-Raphson
Quadratic convergence
~5 Steps
Milliseconds
"Good numerical analysis doesn't just find an answer; it finds it before the heat death of the universe."
Detecting Order Graphically
How do we know the order \(p\) in the real world? We plot the error on a log-log scale.
\(\log(e_{n+1}) \approx p \log(e_n) + \log(k)\)
The slope of the log-log error plot tells you the order of convergence!
Linear (Slope 1) Quadratic (Slope 2) log(e_n) log(e_{n+1})
Efficiency Audit Worksheet Efficiency Audit
Convergence Rate Analysis
Audit Reference
REF-CONV-003
Lead Analyst
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1. Error Sequencing
A student is approximating the root \(L = 2.000000\). The table below shows the results of their iteration. Complete the "Absolute Error" column (\(e_n = |x_n - L|\)).
Step (n) Approximation (\(x_n\)) Absolute Error (\(e_n\)) 0 1.500000 0.500000 1 1.750000 2 1.875000 3 1.937500
Based on the errors above, what is the constant ratio \(k \approx e_{n+1}/e_n\)?
2. Order of Convergence Audit
Analyze the two datasets below to determine if the convergence is Linear or Quadratic. Show the ratio test that confirms your choice.
Method A Data
\(e_0 = 1.0 \times 10^{-1}\)
\(e_1 = 5.0 \times 10^{-2}\)
\(e_2 = 2.5 \times 10^{-2}\)
\(e_3 = 1.2 \times 10^{-2}\)
Perform Ratio Test:
Linear Convergence
Method B Data
\(e_0 = 1.0 \times 10^{-1}\)
\(e_1 = 1.0 \times 10^{-2}\)
\(e_2 = 1.0 \times 10^{-4}\)
\(e_3 = 1.0 \times 10^{-8}\)
Perform Ratio Test:
Quadratic Convergence
3. Executive Summary
A. The "Doubling" Phenomenon
Explain in your own words why quadratic convergence is often described as "doubling the number of correct decimal places" at each step. (Hint: Think about the exponent in \(e_{n+1} \approx e_n^2\)).
B. Algorithmic Trade-offs
If you have a quadratic method that requires twice as many operations per step as a linear method, which one is better in the long run? Justify your answer using the concept of efficiency.
OFFICIAL AUDIT DOC // CONVERGENCE_SPEED_L3
ITERATIVE INSIGHTS SERIES
Infinite Constants Slides 3.14159265... 2.71828182... 1.61803398...
Lesson 4: Infinite Constants
Infinite Constants
Calculating the building blocks of the universe through the lens of infinite sequences.
The Pi Race
In 1673, Gottfried Leibniz discovered a beautifully simple sequence for \(\pi/4\).
\(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots\)
The Leibniz Formula
The Problem:
To get just 10 digits of \(\pi\), you need over 5 billion terms.
The Modern Standard
Contrast this with Ramanujan's formula (1910), which yields 8 new digits per term.
\(\frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum_{k=0}^{\infty} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}\)
Constructing \(e\)
Approach A: Compound Interest
The Limit Sequence
\(e = \lim_{n \to \infty} (1 + \frac{1}{n})^n\)
Intuitive, but slow.
Requires \(n=1,000,000\) to get 6 decimal places.
Approach B: Power Series
The Infinite Sum
\(e = \sum_{n=0}^{\infty} \frac{1}{n!}\)
Incredibly efficient.
Requires only 10 terms to get 7 decimal places.
Sums are Sequences
Every infinite series \(\sum a_n\) is actually a sequence of partial sums \(\{S_k\}\).
S₁ = a₁
S₂ = a₁ + a₂
S₃ = a₁ + a₂ + a₃
Convergent if:
\(\lim_{k \to \infty} S_k = L\)
Calculators compute transcendental functions (sin, log, exp) by evaluating partial sums of these sequences until the desired precision is met.
"Nature does not hurry, yet everything is accomplished."
Numerical analysis is the art of balancing perfection with practicality.
Precision
Infinite processes allow us to reach any degree of truth required for engineering or science.
Limit
The sequence itself is the definition of the number. Irrationals are "becoming" through iteration.
Constant Constructor Workshop Worksheet π
Constant Constructor
Numerical Analysis // Lesson 4
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1. Evaluating the Efficiency of \(e\)
Compare the two standard definitions of Euler's constant. Calculate the partial sums up to \(n=5\) for each method and compare their accuracy to the true value \(e \approx 2.71828\).
Limit Method
\(x_n = (1 + 1/n)^n\)
\(x_1 =\) 2.00000
\(x_2 =\)
\(x_3 =\)
\(x_4 =\)
\(x_5 =\)
Series Method
\(S_n = \sum_{k=0}^{n} \frac{1}{k!}\)
\(S_1 = 1 + 1 =\) 2.00000
\(S_2 = 2 + 1/2 =\)
\(S_3 = S_2 + 1/6 =\)
\(S_4 = S_3 + 1/24 =\)
\(S_5 = S_4 + 1/120 =\)
Efficiency Conclusion:
Which sequence reaches 4 correct decimal places first? How many steps do you estimate the "Limit Method" would need to reach the accuracy that \(S_5\) achieved in just 5 steps?
2. The Slowness of Leibniz
The Leibniz series \(\pi = 4 \times (1 - 1/3 + 1/5 - 1/7 + \dots)\) is an alternating sequence. While visually elegant, it is notoriously slow to converge.
Calculate the partial sums for \(\pi\):
n=1: 4.0000...
n=2: 4(1 - 1/3) = 2.6666...
n=3:
n=4:
n=5:
Error Analysis
Sketch a plot of the partial sums. How do they "oscillate" around the value of \(\pi\)?
Convergence Challenge
Modern computers use Ramanujan-type sequences that give 14 digits per term. If you were tasked with calculating \(\pi\) to 1 million digits, why would choosing the Leibniz series be considered an "engineering failure"?
© 2026 ITERATIVE INSIGHTS // UNIT_4_CONSTANTS SEQUENCE COMPLETED
Order to Chaos Slides Lesson 5: The End of Predictability
Order to Chaos
Exploring the Logistic Map and the thin line between stability and mathematical chaos.
The Simple Equation
Originally a model for population growth, the Logistic Map is a quadratic iterative sequence:
\(x_{n+1} = r x_n (1 - x_n)\)
Where \(0 \le x_n \le 1\) is population and \(r\) is growth rate.
It looks harmless. But as we increase the parameter r, the behavior of the sequence changes fundamentally.
Parameter Sensitivity
r = 2.5: Converges to a single stable point.
r = 3.2: Bounces between 2 values (Period-2).
r = 3.5: Bounces between 4 values (Period-4).
r = 3.9: Complete Chaos. No repeating pattern.
The Tree of Life (and Chaos)
A Bifurcation Diagram plots the long-term stable values (attractors) of the sequence for every possible value of \(r\).
Period Doubling
As \(r\) increases, the fixed point splits into two, then four, then eight... this cascade leads directly into the chaotic regime.
This "self-similarity" is the birth of fractals.
r →
Fig 5.1: The Bifurcation Cascade
When "Equal" is No Longer Equal
Sensitive Dependence
In the chaotic regime, a difference of 0.000000001 in the initial value \(x_0\) leads to completely different results after just a few dozen iterations.
This is the "Butterfly Effect."
"Chaos is not randomness. It is order that hasn't been recognized yet."
Numerical Implications
In chaotic systems, long-term numerical approximation is impossible, no matter how fast your computer is. Rounding error eventually destroys all predictive power.
The Iterative Insight
Lesson 1-2
Sequences find roots and constants with infinite precision.
Lesson 3-4
Efficiency and error analysis allow us to master the infinite.
Lesson 5
Iteration also reveals the boundaries of human knowledge.
You have moved from Fixed Points to Infinite Constants to Universal Chaos.
Chaos Map Simulation Worksheet Chaos Map simulation
Numerical Analysis | The Logistic Map
Researcher:
DOC_ID: CHAOS_LOG_505
1
Stability Phase (r < 3.0)
Evaluate the sequence \(x_{n+1} = 2.8 x_n (1 - x_n)\) starting with \(x_0 = 0.5\). Calculate the first 5 terms.
x₀ 0.5000
x₁
x₂
x₃
x₄
Does the sequence seem to be converging to a single value? If so, what is it?
2
The Oscillation Phase (3.0 < r < 3.4)
Increase the growth rate to \(r = 3.2\). Using \(x_0 = 0.5\), calculate the terms until you notice a pattern.
<table class="w-full border-collapse mb-6"><tbody><tr class="bg-slate-100 font-mono text-xs"><td class="border border-slate-200 p-2 text-center w-12">n=5</td><td class="border border-slate-200 p-2 text-center w-12">n=6</td><td class="border border-slate-200 p-2 text-center w-12">n=7</td><td class="border border-slate-200 p-2 text-center w-12">n=8</td><td class="border border-slate-200 p-2 text-center w-12">n=9</td><td class="border border-slate-200 p-2 text-center w-12">n=10</td></tr><tr class="h-12 bg-white"><td class="border border-slate-200 p-2"></td><td class="border border-slate-200 p-2"></td><td class="border border-slate-200 p-2"></td><td class="border border-slate-200 p-2"></td><td class="border border-slate-200 p-2"></td><td class="border border-slate-200 p-2"></td></tr></tbody></table>
Observation:
Describe the steady-state behavior. How many distinct values does the sequence bounce between? (This is called Period-2).
3
The Chaotic Regime (r = 4.0)
When \(r=4.0\), the map becomes chaotic. Compare two sequences that start almost identically.
Sequence A (\(x_0 = 0.5\))
x₀: 0.5000
x₁: 1.0000
x₂: 0.0000
x₃: 0.0000
Sequence B (\(x_0 = 0.5001\))
x₀: 0.5001
x₁: 0.9999
x₂: 0.0004
x₃: 0.0016
Critical Analysis
Given what you see above, explain why "long-term weather forecasting" or "stock market prediction" using iterative models is inherently limited, even if our initial measurements are 99.9% accurate.
ITERATIVE INSIGHTS // THE LOGISTIC MAP // FINAL LAB END OF SEQUENCE
