Problem 1 Use Taylor's theorem to show tha... [FREE SOLUTION]

91影视

A Primer of Real Analysis

Dan Sloughter

$Math Studyset 91影视 Explanations$ Math

2009 Edition

Chapter 8: Problem 1

Use Taylor's theorem to show that $$ \exp (1)=e=\sum_{n=0}^{\infty} \frac{1}{n !} $$

Short Answer

Expert verified

The expansion $ e = \sum_{n=0}^{\infty} \frac{1}{n!} $ is derived from the Taylor series for $ e^x $ at $ x = 1 $.

Step by step solution

Understand Taylor's Theorem

Taylor's theorem can be used to approximate functions by polynomials. For a function $ f(x) $ at $ x = a $, Taylor's series is given by $ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2+ \frac{f'''(a)}{3!}(x-a)^3 + \cdots $.

Consider the function $ f(x) = e^x $

The function we are considering is $ f(x) = e^x $. We want to evaluate this function at $ x = 1 $, so $ a = 0 $. This will give us the series to approximate $ e = e^1 $.

Calculate Derivatives at $ x = 0 $

The derivative of $ f(x) = e^x $ is $ f'(x) = e^x $. Since $ e^x $ differentiates to $ e^x $ each time, all higher derivatives of $ f(x) $ evaluated at $ x = 0 $ are 1, i.e., $ f^{(n)}(0) = 1 $ for all $ n $.

Write down the Taylor Series at $ x = 0 $

The Taylor series for $ f(x) = e^x $ at $ x = 0 $ is $ f(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $. By substituting $ x = 1 $, we get the series $ e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots $.

Conclude the Series Sum

From the Taylor series expansion at $ x = 1 $, we have shown that $ e $ can be expressed as $ e = \sum_{n=0}^{\infty} \frac{1}{n!} $. This series is exactly the form given in the problem, confirming that it converges to $ e $.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Function

An exponential function is a mathematical function of the form $ f(x) = a^x $, where $ a $ is a positive constant. When the base, $ a $, is Euler's number $ e $, the function becomes $ f(x) = e^x $. This function is unique because it equals its own derivative, making it fundamental in the study of calculus and exponential growth.

The number $ e $ is a mathematical constant approximately equal to 2.71828. It appears in various areas like calculus, probability theory, and complex numbers.

In calculus, $ e^x $ is significant due to its simple differentiation and integration properties.
It represents the idea of growth at a constant percentage rate, often seen in real-world phenomena like population growth and radioactive decay.

Developing an understanding of $ e^x $ is essential, as it serves as a cornerstone for more advanced topics in mathematics and science.

Series Expansion

Series expansion is a method of expressing a function as an infinite sum of terms. It allows us to approximate complex functions using simpler polynomial expressions. One of the most common types is a Taylor series, where a function is represented as an infinite sum of its derivatives evaluated at a single point.

This expansion helps convert complicated functions into polynomials, which are easier to compute and understand. The general form of a Taylor series is:

$ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots $

When the point $ a $ is zero, it becomes a Maclaurin series.

Series expansions are used in several fields, including physics and engineering, for approximating functions that are otherwise hard to deal with.
They also play a crucial role in numerical analysis, enabling computers to perform precise calculations.

Maclaurin Series

A Maclaurin series is a special case of the Taylor series where the expansion is centered around $ x = 0 $. It simplifies the process of expanding functions using derivatives evaluated at zero. This series provides a straightforward way to express common functions as infinite sums.

The general form is:

$ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots $

For the exponential function $ e^x $, every derivative is $ e^x $, and when evaluated at 0, all derivatives equal 1.

Thus, its Maclaurin series is:

$ e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $

Understanding Maclaurin series is crucial for grasping how functions can be approximated easily. They also demonstrate the elegance and power of mathematical analysis in simplifying and expanding concepts.

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91影视

Use Taylor's theorem to show that $$ \exp (1)=e=\sum_{n=0}^{\infty} \frac{1}{n !} $$

Short Answer

Step by step solution

Understand Taylor's Theorem

Consider the function \( f(x) = e^x \)

Calculate Derivatives at \( x = 0 \)

Write down the Taylor Series at \( x = 0 \)

Conclude the Series Sum

Key Concepts

Exponential Function

Series Expansion

Maclaurin Series

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