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Chapter 1: Problem 24

Use series you know to show that: $$\ln 3+\frac{(\ln 3)^{2}}{2 !}+\frac{(\ln 3)^{3}}{3 !}+\cdots=2$$

Short Answer

Expert verified
The given series sums up to 2.

Step by step solution

01

Identify the Series

Recall the Taylor series expansion for \( e^x \) around \( x = 0 \): \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]
02

Rewrite the Series in Terms of the Exponential Function

Express the given series as \( e^{\ln 3} - 1 \). The reason for subtracting 1 is because the given series starts with \( x = \ln 3 \) instead of 1.So the given series \( \ln 3 + \frac{(\ln 3)^2}{2!} + \cdots \) becomes \( e^{\ln 3} - 1 \)
03

Simplify the Expression

Since \( e^{\ln a} = a \) for any \( a > 0 \), simplify \( e^{\ln 3} = 3 \).Then, subtract 1 to match the given series: \[ e^{\ln 3} - 1 = 3 - 1 = 2 \]

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Key Concepts

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

exponential function
The exponential function, usually denoted as \( e^x \), is one of the most important functions in mathematics. It is defined as a function where the independent variable is in the exponent. This function grows very rapidly and is widely used in various fields like mathematics, physics, and engineering. A key property of the exponential function is that its rate of change (the derivative) is proportional to the value of the function itself:
differentiation:
d\( e^x \)/dx = \( e^x \).
natural logarithm
The natural logarithm (ln) is the inverse of the exponential function. Its base is Euler's number, \( e \). So, if \( e^y = x \), then \( \text{ln}(x) = y \). Natural logarithms are useful for simplifying expressions involving exponentiation. They convert products into sums, powers into products, and so on. This makes them invaluable in calculus and algebra.
The most notable property of ln is that \( \text{ln}(e) = 1 \) and \( \text{ln}(1) = 0 \).
series simplification
Series simplification involves using algebraic techniques to make a series easier to understand or evaluate. The Taylor series expansion is one common tool for doing this. For the exponential function \( e^x \), the Taylor series around \( x = 0 \) is:
e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots.
In simplification, converting a complex series into a known form makes it easier to handle. For example, the given series \( \text{ln 3} + \frac{(\text{ln 3})^2}{2!} + \frac{(\text{ln 3})^3}{3!} + \cdots \) in the exercise can be simplified to \( e^{\text{ln 3}} - 1 \).
euler's number
Euler's number, denoted as \( e \), is approximately equal to 2.71828. It is the base of the natural logarithm and a fundamental constant in mathematics. Euler's number has many important properties, including:
  • \( e^{\text{ln} \, a} = a \) for any positive number \( a \)
  • The derivative of \( e^x \) is \( e^x \)
  • The limit definition: e= \lim_{{n \to \infty }} \left ( 1 + \frac{1}{n} \right )^n
. Euler's number appears in many areas of mathematics, including calculus, complex analysis, and number theory.

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Most popular questions from this chapter

The series \(\sum_{n=1}^{\infty} 1 / n^{s}, s>1,\) is called the Riemann Zeta function, \(\zeta(s) .\) \((\text { a })\) you found \(\zeta(2)=\pi^{2} / 6 .\) When \(n\) is an even integer, these series can be summed exactly in terms of \(\pi .\) ) By computer or tables, find $$\text { (a) } \quad \zeta(4)=\sum_{n=1}^{\infty} \frac{1}{n^{4}}$$ $$\text { (b) } \quad \zeta(3)=\sum_{n=1}^{\infty} \frac{1}{n^{3}}$$ $$\text { (c) } \quad \zeta\left(\frac{3}{2}\right)=\sum_{n=1}^{\infty} \frac{1}{n^{3 / 2}}$$

The velocity \(v\) of electrons from a high energy accelerator is very near the velocity \(c\) of light. Given the voltage \(V\) of the accelerator, we often want to calculate the ratio \(v / c .\) The relativistic formula for this calculation is (approximately, for \(V \gg 1\) ) $$\frac{v}{c}=\sqrt{1-\left(\frac{0.511}{V}\right)^{2}}, \quad V=\text { number of million volts. }$$ Use two terms of the binomial series (13.5) to find 1 \(-v / c\) in terms of \(V\). Use your result to find \(1-v / c\) for the following values of \(V\). Caution: \(V=\) the number of million volts. (a) \(\quad V=100\) million volts (b) \(\quad V=500\) million volts (c) \(\quad V=25,000\) million volts (d) \(\quad V=100\) gigavolts \(\left(100 \times 10^{9} \text { volts }=10^{5}\) million volts) \right.

Find the interval of convergence, including end-point tests: $$\sum_{n=1}^{\infty} \frac{(x+2)^{n}}{(-3)^{n} \sqrt{n}}$$

Test for convergence: $$\sum_{n=2}^{\infty} \frac{1}{n \ln \left(n^{3}\right)}$$

Find the Maclaurin series for the following functions. $$e^{1-\sqrt{1-x^{2}}}$$
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