Approximate functions with Taylor/Maclaurin polynomials. Add terms incrementally and see convergence radius visually.
The plotted curve is your function; the overlaid curve is a Taylor polynomial that gains one term at a time. Pick a function such as e^x or sin(x) and set a center; every added term is the next derivative order of the expansion, so the polynomial hugs the function near the center and falls away as you move off it. The tool shades the interval of convergence, making the radius visible at a glance. The usual surprise: the interval does not depend on how many terms you add - outside it, extra terms make the error worse.
A concrete input and expected output from the current implementation.
Input
f(x) = e^x, center a = 0, degree 4; check the value at x = 0.5
Expected output
P4(x) = 1 + x + x^2/2 + x^3/6 + x^4/24. At x = 0.5 the polynomial gives 1.6484375 against e^0.5 = 1.6487213, an error of about 2.8e-4. The plotted curves are visually overlapping between x = -1 and x = 1 and separate visibly past x = 2; the convergence interval covers the whole plot, which is right for e^x.
The Maclaurin coefficients are 1/n! because every derivative of e^x equals 1 at the center 0. Four terms already match e^0.5 to three decimals, and the remaining gap is the tail of the series, dominated by the first omitted term 0.5^5/120 = 0.000260, which is the size of the error the plot shows.
Each added term is the next order of the expansion: order n contributes f^(n)(a)(x-a)^n/n!, where a is the center, and the degree-n polynomial is the running sum through that order. Coefficients come from derivatives of f at the center only. The patterns are simple for common functions: 1/n! for e^x, alternating odd powers divided by factorials for sin(x), all ones for 1/(1-x).
The series converges on an interval about the center; the radius is the distance from the center to the nearest point where f misbehaves. That point can be off the real axis, which is why 1/(1+x^2) looks smooth everywhere real yet has radius 1: its complex singularities sit at +/- i, one unit from 0. Inside the interval the partial sums approach f; outside, they grow without bound.
Because the Maclaurin series of 1/(1+x^2) has radius of convergence 1. The radius is the distance from the center to the nearest singularity of the complex function, and 1/(1+x^2) has singularities at +/- i, one unit from 0. The real graph never reveals them, so the divergence at x = 2 is invisible in the function but visible in the series.
It depends on the function and on x. For alternating series like sin(x), the error of the degree-n polynomial is no larger than the first omitted term (Leibniz's bound), so the needed degree follows directly. Near the center a few terms suffice; near the edge of the radius the terms shrink slowly and you may need dozens. Trust the omitted-term bound, not the plotted overlap, when the value matters.
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