Problem 11 Write down the Taylor series for... [FREE SOLUTION]

Chapter 10: Problem 11

Write down the Taylor series for \(\left(e^{x}-1\right) / x,\) based on the series for \(e^{x}\). At \(x=0\) the function is \(0 / 0\). Evaluate the series at \(x=0\). Check by l'H么pital's Rule on \(\left(e^{x}-1\right) / x\).

Short Answer

Expert verified

The series is 1 at \(x=0\), confirmed by both the series and l'H么pital's Rule.

Step by step solution

Write down the Taylor series for \(e^x\)

The Taylor series for \(e^x\) around \(x=0\) is given by:\[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!} + \cdots\]

Tailor the series to \(e^x - 1\)

Subtract 1 from the Taylor series of \(e^x\) to get:\[e^x - 1 = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!} + \cdots\]

Divide by \(x\)

Divide the whole series for \(e^x-1\) by \(x\):\[\frac{e^x - 1}{x} = 1 + \frac{x}{2!} + \frac{x^2}{3!} + \frac{x^3}{4!} + \cdots + \frac{x^{n-1}}{n!} + \cdots\]

Evaluate the series at \(x=0\)

The series \(1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots\) evaluated at \(x=0\) becomes 1, because all terms with \(x\) vanish.

Check using l'H么pital's Rule

Calculate \(\lim_{x \to 0} \frac{e^x - 1}{x}\):Apply l'H么pital's Rule, differentiating numerator and denominator:\[ \lim_{x \to 0} \frac{e^x}{1} = e^0 = 1.\]This confirms that the limit at \(x=0\) is indeed 1.

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Key Concepts

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

l'H么pital's rule

L'H么pital's Rule is a technique used in calculus to find limits of indeterminate forms like \(0/0\) or \(\infty/\infty\). In the problem, we have an indeterminate form at \(x = 0\) for \(\frac{e^x - 1}{x}\).
Here's a simpler way to understand it using l'H么pital's Rule:

If you have a limit \( \lim_{x \to c} \frac{f(x)}{g(x)}\) where both \(f(c) = 0\) and \(g(c) = 0\), you can take the derivatives: \( \lim_{x \to c} \frac{f'(x)}{g'(x)}\).
Apply this method until the limit is no longer indeterminate.

In our case, for \(\frac{e^x - 1}{x}\), differentiate the numerator \((e^x)\) and the denominator \((x)\). This gives \(\frac{e^x}{1}\), which simplifies to \(e^x\). As \(x\) approaches 0, we have \(e^0 = 1\).

series evaluation

Series evaluation is the process of working with infinite sums to derive meaningful results. When evaluating a series, especially at points like \(x = 0\), we work term by term.
Consider the series for \(\frac{e^x - 1}{x}\):

The series is \(1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots\).
Evaluate each term separately at \(x = 0\).
Only the first term, which is \(1\), remains as all other terms include \(x\) and vanish at \(x = 0\).

This approach confirms the initial calculation without using derivatives. It shows that the series evaluated at zero gives a value of 1, reinforcing the result obtained with l'H么pital's Rule.

exponential function

The exponential function \(e^x\) is a fundamental mathematical concept with a variety of applications in calculus. Understanding its Taylor series helps in manipulating expressions around it.
For \(e^x\):

The Taylor series is: \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!} + \cdots\).
When subtracting 1 (i.e., \(e^x - 1\)), you get: \(x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\).

This modified series becomes useful when divided by \(x\), forming \(\frac{e^x - 1}{x}\), which again links us back to limits and differentiability at points like \(x = 0\). Understanding this series is crucial for deeper insights into calculus problems and their solutions.

limit calculation

Limit calculation is a way to determine how a function behaves as the input value approaches a particular point. Calculating a limit involves carefully looking at the function's trend as \(x\) nears the value of interest.
In the context of \(\frac{e^x - 1}{x}\), we're interested in its behavior as \(x\) approaches zero.
Let's break it down:

Initially substitute \(x = 0\) into the series: \(1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots\).
Observe that all terms containing \(x\) vanish, leaving only 1.
The limit confirms this evaluation, as \(\lim_{x \to 0} \frac{e^x - 1}{x}\) simplifies to 1 using l'H么pital's Rule.

These steps spotlight the precision of finding limits, where direct substitutions fail. Understanding this precision is critical for applying calculus effectively to both theoretical and real-world problems.

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91影视

Write down the Taylor series for \(\left(e^{x}-1\right) / x,\) based on the series for \(e^{x}\). At \(x=0\) the function is \(0 / 0\). Evaluate the series at \(x=0\). Check by l'H么pital's Rule on \(\left(e^{x}-1\right) / x\).

Short Answer

Step by step solution

Write down the Taylor series for \(e^x\)

Tailor the series to \(e^x - 1\)

Divide by \(x\)

Evaluate the series at \(x=0\)

Check using l'H么pital's Rule

Key Concepts

l'H么pital's rule

series evaluation

exponential function

limit calculation

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