The figure below depicts a function f(x) and the Taylor polynomials P₂₀(x) and P₂₁(x) which approximate it.

• The Taylor polynomials are expanded about x = a.
• If the radius of convergence of the Taylor series is R, then higher and higher-order Taylor polynomials will more and more closely match f(x) for | x - a | < R.
• Outside this region the Taylor polynomials will tend to diverge, often in the manner indicated.

We cannot integrate the function e^-.04t2. We therefore cannot find an exact solution to the differential equation shown at the top of the figure below.

• We can, however, integrate any given Taylor series, which is essentially just an infinite polynomial.
• In the third and forth lines we show the Taylor series of our function. The nth term is (-.04 t²)ⁿ / n! = (-.04ⁿ)/n! * t⁽²ⁿ⁾.
• We then proceed to integrate the series term by term.
• The integral of the nth term is (-.04ⁿ)/n! * t⁽²ⁿ⁺¹⁾ / ((2n+1)n!).
• The power series for the integral is thus the one shown in the last line.

The power series representing the integral in the figure below is the one obtained above. We test the convergence of this power series.

• The reciprocal ratio of coefficients shown in the third line simplifies as indicated in that line.
• Since the limit of .04ⁿ / .04^(n+1) = 1/.04, the limit of (2n+3)/(2n+1) is 1 and the limit of (n+1)! / n! is n+1, we see in the fourth line that the limit of the reciprocal ratios is infinite.
• We conclude that radius of convergence is infinite, and that the series therefore converges for all x.

We find limits on the error of the first-degree Taylor polynomial of e^(-.04x2) on the interval from -1 to 1.

• The error term for P₁ involves the maximum value of f''(x) on the interval in question.
• We must therefore find the limits over this interval for the function f''(x) indicated in the last line below.

At x = 0, .0032 x² - .08 takes its minimum value and e^(-.04 x²) takes its maximum value. Since the minimum value of the polynomial is negative, the entire function must therefore have a minimum at x = 0. The minimum is therefore L = -.08.

• Furthermore the polynomial turns out to be negative for all x between -5 and 5, so it is certainly negative on the interval -1 < x < 1.
• On this interval both factors are getting closer and closer to zero so that f''will reach its maximum value as x -> 1.

More specifically we see that the exponential function decreases toward a limit of 0 as we move away from x = 0.

• The polynomial function has a parabolic graph symmetric about the y axis.
• The quadratic polynomial is negative for x = 0, positive for large x and zero when x = 5 (the latter is easily seen by solving the indicated equation).
• It follows that the maximum, which occurs at either 1 or -1, is approximately -.073 and is less than -.07.

It follows that, since f''(x) lies between L and M for -1 < x < 1, f(x) - P₁(x) lies between 1/2 L x² and 1/2 M x², as indicated in the last line for L = -.08 and M = -.07.