In_Training

 Seems our pilots are training for the incoming F-35As. Below:

https://www.msn.com/en-ca/money/topstories/canada-s-military-is-preparing-for-f-35s-despite-government-review/ss-AA1MJUQt?ocid=winp2fptaskbarhover&cvid=68ef91ab1bd1498293955f7beffe3445&ei=23#image=1

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Decided to get my own visuals for my cubic spline problem. With help on the equations from

Gemini, and help with the Desmos interface from Copilot:

                                                                                 

In reverse contrast:

                                                                                       

Analysis of Your Equations and Target Conditions

1. The Given Points

The points your spline should pass through are:

P₁ = (1, 2)

P₂ = (2, 3)

P₃ = (3, 5)

2. Your Defined Segments

You have two segments, f₁(x) for and f₂(x) for .

Segment 1 (f₁):

f(x) = 2 + 0.75(x − 1) + 0.25(x − 1)³, for 1 ≤ x ≤ 2

Segment 2 (f₂):

f(x) = 3 + 1.5(x − 2) + 0.75(x − 2)² − 0.25(x − 2)³, for 2 ≤ x ≤ 3

Checking the Spline Conditions

For a smooth cubic spline, these conditions must hold at the interior knot x = 2:

Continuity (C₀): f₁(2) = f₂(2)

First derivative (C₁): f₁′(2) = f₂′(2)

Second derivative (C₂): f₁″(2) = f₂″(2)

A. Point Interpolation Check

At x = 1:

f₁(1) = 2 + 0.75(0) + 0.25(0)³ = 2 ✓ (passes through (1, 2))

At x = 2 (using f₁):

f₁(2) = 2 + 0.75(1) + 0.25(1)³ = 3 ✓

At x = 2 (using f₂):

f₂(2) = 3 + 1.5(0) + 0.75(0)² − 0.25(0)³ = 3 ✓

At x = 3:

f₂(3) = 3 + 1.5(1) + 0.75(1)² − 0.25(1)³ = 5 ✓ (passes through (3, 5))

Result: The function is continuous and interpolates all three points.

B. First Derivative Check

Derivative of f₁(x):

f₁′(x) = 0.75 + 0.75(x − 1)²

Derivative of f₂(x):

f₂′(x) = 1.5 + 1.5(x − 2) − 0.75(x − 2)²

At x = 2:

f₁′(2) = 1.5

f₂′(2) = 1.5

Result: f₁′(2) = f₂′(2) = 1.5 ✓ (C₁ continuity satisfied)

C. Second Derivative Check

Second derivative of f₁(x):

f₁″(x) = 1.5(x − 1)

Second derivative of f₂(x):

f₂″(x) = 1.5 − 1.5(x − 2)

At x = 2:

f₁″(2) = 1.5

f₂″(2) = 1.5

Result: f₁″(2) = f₂″(2) = 1.5 ✓ (C₂ continuity satisfied)

Conclusion on the Overlap

Both C₁ and C₂ continuity conditions are met at x = 2.

The first and second derivatives match at 1.5.

Any visual discrepancy in your plot likely stems from the displayed range of the derivative axes rather than the spline’s definition.

Note on the Equations

For a general cubic spline:

f(x) = aᵢ + bᵢ(x − xᵢ) + cᵢ(x − xᵢ)² + dᵢ(x − xᵢ)³

Then:

f(xᵢ) = aᵢ

f′(xᵢ) = bᵢ

f″(xᵢ) = 2cᵢ

f‴(xᵢ) = 6dᵢ

Segment coefficients:

f₁(x): a₁ = 2, b₁ = 0.75, c₁ = 0, d₁ = 0.25 → f₁′(1) = 0.75

f₂(x): a₂ = 3, b₂ = 1.5, c₂ = 0.75, d₂ = −0.25 → f₂′(2) = 1.5

Thus, f₁′(2) = f₂′(2) = 1.5, verifying full smoothness across the spline.

 Copilot:

🧠 What Do Derivatives Tell Us?

🔹 First Derivative: Rate of Change

Meaning: Measures how fast a function is changing.

Graphically: It gives the slope of the tangent line to the curve.

Physically: If the function represents position, the first derivative is velocity.

🔹 Second Derivative: Change of the Rate of Change

Meaning: Measures how the rate of change itself is changing.

Graphically: It tells us about the concavity of the graph (curving up or down).

Physically: If the function is position, the second derivative is acceleration.

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Decided to get clear on the relationship between the DeepSeek answer to the

cubic spline problem and the ChatGpt one.

These are the proposed DeepSeek equations:

S₁(x) = (1/2)(x - 1)² + (1/2)(x - 1) + 2

S₂(x) = (1/2)(x - 2)² + (3/2)(x - 2) + 3

Turns out that the two equations DeepSeek produced for a cubic approximation simplify

to the initial quadratic DSeek wanted to give me all along. 

                                                                                  

On questioning Gemini on the matter

of possible identity between the two approaches, Gemini admitted they were identical at the 

data points given; BUT they are not identical overall. 

What is the relationship between the function (1/2)x^2 - (1/2)x +2 between 1 and 3, and the adjacent functions in my image?

                                                                            

                                                                                    

Could you generate data points other than those given that show the inequality?

Yes, I can generate data points to demonstrate that the quadratic function and the cubic spline approximation are not identical.

The two functions were specifically designed to be equal only at the "node" points x=1, x=2, and x=3. At any other point between these nodes, they will have different values, which proves the inequality.

Here are the values at two intermediate points, x=1.5 and x=2.5: