Understanding and Calculating e^x Using Taylor Series: An In-Depth Guide

To solve or calculate the value of e^x using a partial sum of its Taylor series expansion, you need to understand the underlying series and how to manipulate it. In this article, we will explore how the Taylor series expansion is used to approximate the value of e^x, and how to determine the precision of the approximation.

Introduction to the Taylor Series Expansion for e^x

The Taylor series expansion for the exponential function, e^x, is given by the infinite series:

t

tt(e^x sum_{k0}^{infty} frac{x^k}{k!} 1 frac{x}{1!} frac{x^2}{2!} ldots) t

The right side of your equation represents the partial sum of this series up to the nth term:

t

tt1 frac{x}{1!} frac{x^2}{2!} ldots frac{x^n}{n!} t

By using this partial sum, you can approximate the value of e^x with a specified degree of accuracy. This is particularly useful when you need a quick and reasonably accurate estimate of the exponential function.

Step-by-Step Guide to Calculating e^x with the Taylor Series

Step 1: Understand the Series

The full Taylor series expansion offers an exact representation of e^x as an infinite sum. However, in practical applications, it's often more feasible to work with a finite number of terms. The series can be truncated to yield a more manageable polynomial expression:

t

tt(e^x approx 1 frac{x}{1!} frac{x^2}{2!} ldots frac{x^n}{n!}) t

Step 2: Rearrange the Equation

Let's rearrange the equation to isolate terms on one side:

t

tt(e^x - left(1 frac{x}{1!} frac{x^2}{2!} ldots frac{x^n}{n!}right) 0) t

Step 3: Define a Function

Define a function (f(x)) as follows:

t

tt(f(x) e^x - left(1 frac{x}{1!} frac{x^2}{2!} ldots frac{x^n}{n!}right)) t

Step 4: Analyze the Function

Analyze the behavior of (f(x)) for different values of x:

tAt x 0: t

tt

ttt(f(0) e^0 - 1 0) tt

t

t

So, x 0 is a solution.

tBehavior as x increases: t

As x increases, e^x grows much faster than the polynomial expansion on the right side. Therefore, for large values of x, (f(x) approx 0).

tBehavior as x decreases: t

As x decreases significantly, for example, when x to -infty, e^x to 0, while the polynomial still yields finite values. Hence, (f(x) approx 0).

Step 5: Finding Other Solutions

To determine if there are any other solutions, analyze the derivative (f'(x)):

t

tt(f'(x) e^x - left(frac{1}{1!}frac{2x}{2!} ldots frac{nx^{n-1}}{n!}right)) t

Since (e^x) grows exponentially while the polynomial derivative grows polynomially, (f'(x)) will be positive for sufficiently large x. This indicates that the function is strictly increasing after a certain point.

t

ttTherefore, the only solution in the vicinity of x 0 is x 0. t

There are no other solutions for x 0 or x 0 due to the growth rates of the exponential function and the polynomial.

Conclusion:

tThe only solution in the vicinity of x 0 is x 0. tThere are no other solutions for x 0 or x 0. tThe equation is satisfied when x 0.

Practical Example: Calculating e^x Using the Taylor Series

This isn’t a thing you have to solve. Instead, it is an expression you use to calculate something. For example, if you want to know what (sqrt{e}) is, you can put x 0.5 into the expansion you provided and get:

t

tt(e^{0.5} 1 frac{0.5}{1!} frac{0.5^2}{2!} frac{0.5^3}{3!} ldots) t

Simply add as many terms as you need and crunch the numbers to approximate the value of (e^{0.5}) as accurately as required.