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Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\ln \left(1+x e^{x}\right)$$

Short Answer

Expert verified
The first few terms are f(x) = ... . They match the computed series.

Step by step solution

01

Understand the Maclaurin Series

The Maclaurin series is a special case of the Taylor series centered at zero. It expands a function into a sum of terms that are derivatives of that function at zero: ... ].
02

Find the first derivative

Compute the first derivative of the function. Let \(f(x) = \ln(1 + x e^x)\). To find \(f'(x)\), use the chain rule: [ ... ]
03

Calculate higher-order derivatives

Find the second and third derivatives: f''(x) = \displaystyle \frac{d}{dx} f'(x) = ( ... f'''(x) = \displaystyle ( ... Terms evaluation at \(x=0\) for the first few fits
04

Evaluate the derivatives at x = 0

Compute the derivatives at \(x = 0\): . Therefore,
05

Form the Maclaurin series

Using the evaluated derivatives, collect the first few terms of the series: f(x) = ... First few terms of the Maclaurin series:
06

Verify using a computer

Use computational tools like Python/Mathematica/WolframAlpha to find the series. ... The computed series matches the analytically determined one.

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Key Concepts

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

Maclaurin Series
The Maclaurin series is a special kind of Taylor series centered at zero. For any function, the Maclaurin series expresses it as an infinite sum of its derivatives evaluated at zero.
The general form of a Maclaurin series for a function \(f(x)\) is:
\[ f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \text{...} \]
Each term includes a derivative of the function at zero and a factor of \(x\) raised to the power of the term index. You obtain these derivatives step by step and plug them into this formula.
For instance, in this exercise, if we calculate the Maclaurin series for \( \text{ln}(1 + x e^x)\), you would:
1. Define \(f(x)\) as \(\text{ln}(1 + x e^x)\)
2. Compute the derivatives \(f'(x)\), \(f''(x)\), \(f'''(x)\) at \(x = 0\)
3. Substitute these derivatives into the Maclaurin series formula.
Taylor Series
The Taylor series generalizes the Maclaurin series to any point, not just zero.
The Taylor series for a function \(f(x)\) around a point \(a\) is:
\[ f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \text{...} \]
Like the Maclaurin series, the Taylor series converts a function into an infinite sum of terms, each involving derivatives. However, the center point \(a\) provides flexibility in approximating the function around different points. In practice, you often truncate these infinite series after a few terms for simplicity, leading to polynomial approximations of the function. Imagine expanding \(\text{ln}(1 + x e^x)\) at different points to see how the series form changes.
Derivatives
Derivatives are essential for forming both Taylor and Maclaurin series. A derivative of a function gives us the slope or rate of change at a point.
For a function \(f(x)\), the first derivative \(f'(x)\) describes the slope, the second derivative \(f''(x)\) describes the concavity, and so on. In our example with \(\text{ln}(1 + x e^x)\), derivatives are crucial:
- First derivative \(f'(x)\): Using the chain rule, you find the rate of change at any \(x\)
- Higher-order derivatives \(f''(x)\), \(f'''(x)\): These describe deeper levels of changes and are obtained by differentiating repeatedly.
Evaluating all derivatives at \(x=0\) lets us write out the Maclaurin series.
Computational Tools
In modern mathematics, computational tools play a huge role in simplifying and verifying derivatives and series.
Tools like Python, Mathematica, or WolframAlpha can automatically compute derivatives, series expansions, and more. Python offers libraries like SymPy to symbolically compute the Maclaurin series with ease.
Here's an example in Python:
```python
import sympy as sp
x = sp.symbols('x')
f = sp.ln(1 + x * sp.exp(x))
series = sp.series(f, x, 0, 4)
print(series)
```
This level of automation ensures accuracy and saves time, making it easier for students to learn and cross-check their manual solutions.

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Most popular questions from this chapter

A tall tower of circular cross section is reinforced by horizontal circular disks (like large coins), one meter apart and of negligible thickness. The radius of the disk at height \(n\) is \(1 /(n \ln n)(n \geq 2)\) Assuming that the tower is of infinite height: (a) Will the total area of the disks be finite or not? Hint: Can you compare the series with a simpler one? (b) If the disks are strengthened by wires going around their circumferences like tires, will the total length of wire required be finite or not? (c) Explain why there is not a contradiction between your answers in (a) and (b). That is, how is it possible to start with a set of disks of finite area, remove a little strip around the circumference of each, and get an infinite total length of these strips? Hint: Think about units-you can't compare area and length. Consider two cases: (1) Make the width of each strip equal to one percent of the radius of the disk from which you cut it. Now the total length is infinite but what about the total area? (2) Try to make the strips all the same width; what happens? Also see Chapter 5, Problem 3.31(b).

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

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: $$\sum_{n=1}^{\infty} \frac{(x+2)^{n}}{(-3)^{n} \sqrt{n}}$$

Assume that the Maclaurin series converges to the function. Show that \(1-\cos x=x^{2} / 2\) with an error less than 0.003 for \(|x|<\frac{1}{2}\).

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