Taylor Series for e^x
I found this in my math homework and had to look it up because it seemed wrong. The claim is that e^x — which is this irrational, transcendental number that goes on forever — can be approximated by a regular polynomial.
A polynomial is just: a + bx + cx² + dx³... Normal stuff. But the Maclaurin series says if you pick the coefficients just right (1, 1, 1/2, 1/6, 1/24...), your polynomial tracks e^x almost perfectly near x=0. Add more terms and the range where it works keeps growing.
The slider starts at degree 1 (just a straight line). By degree 5, you can barely see the gap near x=0. Watch the error graph at the bottom — the area shrinks as you add terms.
Polynomial Approximation
Absolute Error |e^x - P1(x)|
Why Calculators Can Actually Compute e^x
Your calculator can't compute e^x directly — transcendental functions don't have exact finite representations. So what does it actually do? It uses a polynomial approximation like this. With enough terms, the error is smaller than the display precision, so it looks exact.
The coefficients (1, 1, 1/2, 1/6, 1/24, 1/120...) are 1/n! — they come from the derivatives of e^x, which are all also e^x. That self-referential property is what makes e^x special and why this particular series converges so cleanly.
Physics simulations, signal processing, financial models — anything involving exponential growth or decay is computing an approximation like this underneath. The degree slider you're moving literally controls how accurate the computation would be.