Finding Taylor Series
Use substitution (as in Formula (7)) to find the Taylor series at x = 0 of the functions in Exercises 1–12.
e⁻ˣ/²

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Recall the Taylor series expansion of the exponential function \(e^x\) centered at \(x=0\), which is given by: \[e^x = \sum_{n=0}^\infty \frac{x^n}{n!}\]

In the given function, we have \(e^{-x^2}\). To use substitution, let’s replace \(x\) in the exponential series with \(-x^2\). This gives: \[e^{-x^2} = \sum_{n=0}^\infty \frac{(-x^2)^n}{n!}\]

Simplify the powers inside the summation: \[e^{-x^2} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{n!}\]

This series is already centered at \(x=0\), so this is the Taylor series expansion of \(e^{-x^2}\) at \(x=0\).

To write out the first few terms explicitly, substitute \(n=0,1,2,3,\ldots\) into the summation to see the pattern of coefficients and powers of \(x\).

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Taylor Series Expansion

A Taylor series represents a function as an infinite sum of terms calculated from the derivatives at a single point, usually around x = 0 (Maclaurin series). It approximates functions using polynomials, making complex functions easier to analyze and compute.

Substitution Method in Series

Substitution involves replacing the variable in a known Taylor series with another expression to find the series of a related function. For example, substituting x² into the series for e^x helps find the series for e^(x²). This method simplifies finding series for composite functions.

Exponential Function Series

The exponential function e^x has a well-known Taylor series: sum of x^n/n! for n from 0 to infinity. Understanding this base series is essential because many functions, like e^(-x²/2), can be expressed by modifying this series through substitution and algebraic manipulation.

Finding Taylor SeriesUse substitution (as in Formula (7)) to find the Taylor series at x = 0 of the functions in Exercises 1–12.e⁻ˣ/²