Turn an approximation into a justified guarantee

Try

A polynomial estimate differs from the true value by 0.002 in one experiment. Does that observation guarantee the same error at a nearby input? A bound needs assumptions over the whole segment between center and target. Predict why knowing a derivative only at the center cannot control every intermediate curvature or higher-order change.

Predict

For a degree-n Taylor polynomial, predict whether the remainder uses the nth or the next derivative. The polynomial already matches through order n; the next order controls the first remaining contribution. Keep the distance |x−a| and factorial in your prediction rather than focusing only on the derivative magnitude.

Experiment

Live experiment: approximation with a justified remainder

Approximate eˣ using Pₙ(x)=1+x+x²/2!+…+xⁿ/n!. Predict whether the same degree gives a smaller error near zero or near an endpoint, then compare x=0.5 with x=2.

Evaluation point x Polynomial degree n 012346810Reset experiment

Polynomial: 2.5. eˣ: 2.718282. Polynomial minus eˣ: -0.218282.

Every derivative of eˣ is eˣ. Between0 and x, its magnitude is at most e^|x|. Taylor's theorem therefore gives |error| ≤ e^|x| · |x|^(n+1)/(n+1)!, bounded above by 0.453047 after upward decimal rounding.

Inspect terms and the factorial

Terms, from degree0 through degree2: 1 + 1 + 0.5. Their signs are included. The remainder uses (2+1)! = 6, not n!.

Displayed values are rounded to six decimal places, so a displayed zero error need not mean exact equality. The derivative bound guarantees the mathematical truncation error for the exact polynomial. Browser calculations also have floating-point roundoff, which this bound does not include; at very small errors roundoff can dominate. A close-looking graph alone supplies neither guarantee. This experiment concerns eˣ, whose Taylor series converges everywhere; other functions can have finite convergence intervals. Settings are temporary.

Use the e^x investigation to compare actual error with its Lagrange bound. Record a target and degree, increase the degree once, and explain why a bound can be much larger than the observed error while remaining valid. Display rounding is not the mathematical source of the guarantee.

Separate worked case	Bound inputs	Result
e^x, degree 2, target 1/2	M₃=3 on [0,1/2]	3(1/2)³/6=1/16
sin x, degree 5, targets within distance 1	M₆=1	at most 1/720

The first uses a convenient upper bound 3, not the exact maximum derivative. A larger valid derivative bound produces a looser but still correct remainder bound.

Understand

Under sufficient regularity, such as n+1 continuous derivatives on the segment between a and x, Taylor’s theorem gives a remainder f^(n+1)(ξ)(x−a)^(n+1)/(n+1)! for some intermediate ξ. Therefore a bound |f^(n+1)|≤M throughout that segment yields |f(x)−Pₙ(x)|≤M|x−a|^(n+1)/(n+1)!. The unknown ξ is why a pointwise derivative value at either endpoint is not enough.

The theorem can be derived using repeated mean-value arguments or an integral remainder. In the latter approach, repeated integration of the derivative relation gives a weighted integral of the next derivative. Bounding its magnitude by M produces the same factorial expression. The assumptions justify those derivative and integration steps; a graph that merely looks smooth is not a substitute.

For sin x near zero, all derivatives have magnitude at most 1. The degree-5 Lagrange bound at |x|≤1 is 1/720. Its alternating series also gives a smaller next-nonzero-term bound |x|⁷/7! where the magnitudes decrease. Different valid theorems can give different bounds. Neither bound is necessarily the actual error, and failure of one hypothesis calls for another argument rather than an invented guarantee.

To show e^x equals its infinite Taylor series at a fixed real x, bound every derivative on the center-to-target segment by e^|x|. The remainder is at most e^|x||x|^(n+1)/(n+1)!. The adjacent ratio of the nonconstant factor tends to zero, so the bound vanishes as n grows. This proves representation, not merely coefficient matching. The quantifier fixes x first; it does not say one finite degree achieves the same error on the entire real line.

Apply

For every practice error claim, identify the center, degree, target segment, derivative order and bound. Solve an inclusive tolerance inequality carefully. Explain when a sufficient term count is not necessarily minimal. Keep a computed actual error separate from the proof that a bound applies beyond the tested input.

Continue

Next derive new power series by differentiation and integration. These operations are justified inside the convergence radius; endpoint behavior and remainder estimates still require their own arguments.