Construct Taylor models from derivative evidence

Try

A mathematical model supplies local derivative data but no formula for its function. At the center a=2, its exact values are listed below. Inputs and outputs are dimensionless. Construct a cubic polynomial that agrees with all four entries. Do not assume these numbers are polynomial coefficients: differentiating a cubic multiplies its coefficients by factorials.

Derivative order	Supplied value at 2	Required coefficient of (x−2) to that order
0	f(2)=3	3
1	f′(2)=−2	−2
2	f″(2)=4	4/2!=2
3	f‴(2)=6	6/3!=1

Write h=x−2 so that h=0 means the center. Predict whether evaluating the local polynomial at x=5/2 should use h=5/2 or h=1/2. The distinction is essential when the center is not zero.

Predict

Suppose a second, independent statement says f has a continuous fourth derivative on [2,5/2] and its absolute value never exceeds 12 there. Predict which evidence determines the polynomial and which evidence controls its approximation error. Could a different smooth function share every table entry while violating that bound?

Also predict how the guaranteed error changes if the distance from the center halves while the same derivative bound remains valid. For a cubic approximation the first uncontrolled derivative is the fourth, so the distance enters to the fourth power.

Experiment

The polynomial is P₃(x)=3−2h+2h²+h³, with h=x−2. Verify it by differentiating three times and substituting h=0. Then compute each contribution at h=1/2.

Contribution	Value at x=5/2
Constant	3
Linear	−1
Quadratic	1/2
Cubic	1/8
Total	21/8

Now use the independent bound. Taylor’s theorem gives an absolute remainder no larger than 12(1/2)⁴/4!=1/32. Therefore the actual value lies in [83/32,85/32], centered at 21/8. Draw this interval on a number line with its midpoint labeled; distinguish the single polynomial estimate from the guaranteed enclosure.

Repeat the bound at x=9/4. The radius becomes 12(1/4)⁴/24=1/512, one sixteenth of the previous radius. This comparison assumes the same bound on the shorter interval, which follows here because it is a subset of the supplied interval.

Understand

For a function with sufficient derivatives near a, the degree-n Taylor polynomial is the sum of f⁽ᵏ⁾(a)(x−a)ᵏ/k! for integer k from zero to n. The notation f⁽ᵏ⁾ means derivative order k, with order zero meaning the original function. The factorial divides out the repeated factors introduced by differentiating a power. This formula explains the table’s coefficients rather than merely naming a construction.

Under the stated continuity of the next derivative, a bound M for its absolute value throughout the interval between a and x gives remainder at most M times the distance to a raised to n+1, divided by (n+1)!. A derivative value at a single point is not such an interval bound. Here M=12 was separately supplied; it was not estimated by looking at the table.

The family P₃(x)+C(x−2)⁴ demonstrates the limitation. Every real C gives the same first four entries at 2. At 5/2, however, the added value is C/16, which is unrestricted without additional information. The fourth derivative is 24C, so the bound of 12 rules out large choices. Smoothness and matching a finite derivative table alone do not guarantee a small error.

Known infinite series support another kind of transfer. Starting from the exponential series, substitute u=−3x to obtain e^(−3x)=1−3x+(9/2)x²−(9/2)x³+⋯. The original exponential series converges for every real u, so this substitution is valid for every real x. At x=1/6, the argument is −1/2. The cubic approximation is 1−1/2+1/8−1/48=29/48, and the next alternating term has magnitude 1/384. Terms decrease in magnitude at this input, so that next term bounds the error. The alternating-series justification is separate from the derivative-table construction.

A substitution can change powers and hence the meaning of degree. In e^(−2x²), four nonzero terms extend through x⁶, not x³. Integration also changes coefficients and convergence questions. A geometric-series identity first holds on an open interval; if an integrated series converges at an endpoint, its equality with the original integral there needs a justified boundary argument, not a silent extension of an interior theorem.

Apply

Build the fresh practice polynomial from its own center and derivative table. Show a derivative-match check and label the separate remainder assumption. Report an interval when a guaranteed enclosure is requested, rather than presenting the approximation as exact.

For the transformed-series task, retain the source series, the substitution, the first coefficients and the validity interval. For an endpoint, test convergence separately. The final mixed motion task integrates a series to approximate one coordinate; that coordinate is not the particle’s traveled distance.

Change the center and audit the guarantee

At center a=1, suppose f(1)=2, f′(1)=3 and f″(1)=−4. The quadratic Taylor polynomial is P₂(x)=2+3(x−1)−2(x−1)². At x=1.2 it gives 2.52. Its quadratic coefficient is −4/2!, not −4, and the displacement is 0.2, not 1.2. Differentiate to check the three local entries before estimating any error.

Now separately assume f has a continuous third derivative with absolute value at most six on [1,1.2]. Taylor’s theorem gives radius 6(0.2)³/3!=0.008, hence the enclosure [2.512,2.528]. This is a guaranteed interval under the stated hypothesis, not an exact evaluation of f. The control changes the displacement only within the interval where the bound is supplied. Halving the displacement divides this cubic error bound by eight.

Remove the interval bound while retaining the derivative table. Every function P₂(x)+C(x−1)³ shares those same three entries, but at 1.2 it differs by 0.008C. Since C is arbitrary, there is no finite uniform guarantee from the table alone. Each such function is smooth, so saying “the function is smooth” does not supply the missing quantitative control.

For the changed retry, annotate which information constructs the polynomial and which information certifies its error. Retain the interval and derivative order in the written explanation. Neither a successful numerical example nor a public practice answer validates an unseen remainder claim outside those hypotheses.

A coefficient is not an error certificate

Review criterion: Keep local derivative data separate from an interval remainder bound and justify any transferred bound.

Predict, then change one input

Displacement h from center 1:

0.2

Polynomial P₂(1+h)

2.52

Guaranteed radius, assuming |f‴|≤6

0.008

Lower enclosure endpoint

2.512

Upper enclosure endpoint

2.528

This bound uses the supplied third-derivative bound on [1,1.2]. Moving beyond that interval is deliberately disabled; the local derivative table alone gives no uniform guarantee.

Explain the changed case: A changed center

At a=1, f(a)=2, f′(a)=3 and f″(a)=−4. A continuous third derivative satisfies |f‴|≤6 on [1,1.2]. Which statement is justified at x=1.2?

• P₂(x)=2+3(x−1)−2(x−1)²
• Taylor radius: 6·(0.2)³/3!

Write a reason before selecting a claim. Feedback is public practice, not a proof grader.

f(1.2)=2.52 exactlyThe error is at most 0.2 because the displacement is 0.2.f(1.2) lies in [2.512,2.528].

Switch to changed retryReset this practice

Inputs and choices are temporary. Nothing is saved, no lesson completion or score is changed, and repeating a public case is not unseen assessment.

Continue

Return to Taylor error bounds for the theorem behind the remainder estimate, or series operations for justified transformations. The synthesis assessment asks you to combine these actions on fresh cases. Completing its public prompts records practice, not an independently verified examination result.