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Chapter 11: Problem 29

Use the Taylor series for \(e^{x}\) to show that \(\frac{d}{d x} e^{x}=e^{x}\).

Short Answer

Expert verified
The Taylor series of \(e^{x}\), when differentiated term-by-term, results in the same series: \(e^{x}\).

Step by step solution

01

- Write the Taylor Series for \(e^{x}\)

The Taylor series for \(e^{x}\) around \(x = 0\) is given by: \[ e^{x} = \sum_{n=0}^\infty \frac{x^n}{n!} \]
02

- Write the General Term

The general term in the Taylor series for \(e^{x}\) can be written as: \(\frac{x^n}{n!}\)
03

- Derive Each Term

Differentiate the general term \( \frac{x^n}{n!} \) with respect to \(x\). \[ \frac{d}{dx} \left( \frac{x^n}{n!} \right) = \frac{n x^{n-1}}{n!} = \frac{x^{n-1}}{(n-1)!} \]
04

- Re-index the Series

Rewrite the derived terms by re-indexing the series: \[ \sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!} \] Replace \(n-1\) with \(m\), so \(m = n - 1\): \[ \sum_{m=0}^\infty \frac{x^m}{m!} \]
05

- Recognize the Result

Notice that the re-indexed series is the same as the original Taylor series: \[ e^{x} = \sum_{m=0}^\infty \frac{x^m}{m!} \]
06

- Conclude the Derivative

Therefore, differentiating \(e^{x}\) using its Taylor series yields: \[ \frac{d}{dx} e^{x} = e^{x} \]

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Key Concepts

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

Taylor Series
Taylor series are an important tool in calculus. They allow us to express functions as infinite sums of their derivatives at a single point. This representation can often simplify complex functions and make them easier to understand. The general form of a Taylor series for a function \(f(x)\) centered around \(x=a\) is given by:

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Most popular questions from this chapter

Use the integral test to determine whether the infinite series is convergent or divergent. (You may assume that the hypotheses of the integral test are satisfied.) $$\sum_{k=1}^{\infty} \frac{1}{(3 k)^{2}}$$

It can be shown that \(\lim b e^{-b}=0 .\) Use this fact and the integral test to show that \(\sum_{k=1}^{\infty} \frac{k}{e^{k}}\) is convergent.

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