Back to Finding and using Taylor series

Exercise 8

Avg difficulty 2.5

The fact that we can differentiate or integrate a Taylor series reveals other important ways we can think about functions such as exe^x.

  1. Perhaps the most important property of the function h(x)=exh(x) = e^x is that h′(x)=exh'(x) = e^x; that is, the function exe^x is its own derivative. Suppose that we didn't yet know the coefficients of the Taylor series expansion for exe^x, so we just said ex=a0+a1x+a2x2+a3x3+⋯+anxn+⋯\begin{gathered} e^x = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots + a_n x^n + \cdots \end{gathered}. Let x=0x = 0 in Equation; what does this tell us about the value of a0a_0?

  2. Take the derivative of both sides of Equation and call your resulting equation for exe^x “Equation 2”. Why do Equation and Equation 2 together tell us that a1=a0a_1 = a_0? Combine this observation with your conclusion in (b) and note that you now know the numerical value of both a0a_0 and a1a_1.

  3. Why do Equation and Equation 2 together tell us that a2=12a1a_2 = \frac{1}{2}a_1?

  4. Continue reasoning similarly to find the value of a3a_3, a4a_4, and a5a_5. What do you observe?

Boelkins, Active Calculus, §8.5 Finding and using Taylor series, Exercise 8

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Adapted from Active Calculus © 2012–2025 Matthew Boelkins. Licensed under CC BY-SA 4.0. Source: Active Calculus. Imported at revision dc86ba3.