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Chapter 9: Problem 1

Find the Taylor polynomials of orders \(0,1,2,\) and 3 generated by \(f\) at \(a\) $$f(x)=e^{2 x}, \quad a=0$$

Short Answer

Expert verified
\(P_0(x) = 1\); \(P_1(x) = 1 + 2x\); \(P_2(x) = 1 + 2x + 2x^2\); \(P_3(x) = 1 + 2x + 2x^2 + \frac{4}{3}x^3\).

Step by step solution

01

Understanding Taylor Polynomials

The Taylor polynomial of order n for a function \(f(x)\) centered at \(a\) is given by \(P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots + \frac{f^{(n)}(a)}{n!}(x-a)^n\). To solve this, we need to find the successive derivatives of \(f(x)=e^{2x}\) and evaluate them at \(x=0\).
02

Compute and evaluate f(a)

Calculate \(f(x)\) at \(x=0\). We know that \(f(x) = e^{2x},\) so \(f(0) = e^{0} = 1\).
03

Find f'(x) and evaluate f'(a)

Differentiate \(f(x) = e^{2x}\) with respect to \(x\). We know \(f'(x) = \frac{d}{dx}(e^{2x}) = 2e^{2x}\). Thus, \(f'(0) = 2e^0 = 2\).
04

Find f''(x) and evaluate f''(a)

Differentiate \(f'(x) = 2e^{2x}\) with respect to \(x\). We have \(f''(x) = \frac{d}{dx}(2e^{2x}) = 4e^{2x}\). Thus, \(f''(0) = 4e^0 = 4\).
05

Find f'''(x) and evaluate f'''(a)

Differentiate \(f''(x) = 4e^{2x}\) with respect to \(x\). This gives us \(f'''(x) = \frac{d}{dx}(4e^{2x}) = 8e^{2x}\). Thus, \(f'''(0) = 8e^0 = 8\).
06

Construct P_0(x)

The zeroth order Taylor polynomial is simply \(P_0(x) = f(0) = 1\).
07

Construct P_1(x)

The first-order Taylor polynomial uses the first derivative: \(P_1(x) = f(0) + f'(0)x = 1 + 2x\).
08

Construct P_2(x)

The second order Taylor polynomial uses up to the second derivative: \(P_2(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 = 1 + 2x + \frac{4}{2}x^2 = 1 + 2x + 2x^2\).
09

Construct P_3(x)

The third order Taylor polynomial is: \(P_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 = 1 + 2x + 2x^2 + \frac{8}{6}x^3 = 1 + 2x + 2x^2 + \frac{4}{3}x^3\).

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Key Concepts

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

Derivatives
Derivatives are fundamental tools in calculus, representing the rate at which a function changes. To understand Taylor polynomials, it's essential to grasp derivatives because they help describe how a function behaves at a certain point. In this exercise, we work with the function \(f(x) = e^{2x}\). To construct Taylor polynomials, we calculate derivatives up to the desired order and evaluate them at the specific point \(a = 0\). Here's a quick overview of the process:
  • First, find the function's value at \(x = 0\): \(f(0) = 1\).
  • Next, differentiate the function to find \(f'(x) = 2e^{2x}\), and evaluate \(f'(0) = 2\).
  • Then, calculate the second derivative: \(f''(x) = 4e^{2x}\), and evaluate \(f''(0) = 4\).
  • Finally, determine the third derivative: \(f'''(x) = 8e^{2x}\), with \(f'''(0) = 8\).
With these derivatives, you can build Taylor polynomials of various orders, which represent approximations of the function \(f(x)\) around \(x=0\).
Exponential Functions
Exponential functions are crucial in many areas of mathematics and science due to their unique properties. The function \(f(x) = e^{2x}\) is an example where the base \(e\) is Euler's number, approximately 2.71828. This function grows exponentially, meaning it increases at a constantly proportional rate.Some key characteristics of exponential functions like \(e^{2x}\) include:
  • The derivative of \(e^{2x}\) provides the same function up to a constant multiple. This makes exponential functions straightforward to differentiate repeatedly.
  • The exponential function is continuous and smooth, with no breaks or corners.
  • Exponential functions have a constant ratio of change rather than a constant amount of change.
Due to these properties, exponential functions like \(e^{2x}\) are easy to work with when calculating derivatives for Taylor polynomials. They inherently simplify the differentiation and evaluation process.
Maclaurin Series
The Maclaurin series is a special case of a Taylor series, where the center point \(a = 0\). It makes the expansion around \(x = 0\) particularly efficient for functions like \(e^{2x}\). The general form of a Maclaurin series for a function \(f(x)\) is:\[P_n(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots + \frac{f^{(n)}(0)}{n!}x^n\]In our exercise, the task was to find the Maclaurin series' Taylor polynomials up to the third order. Let's summarize:
  • The zeroth-order approximation or \(P_0(x)\) is simply the constant term \(f(0) = 1\).
  • The first-order polynomial or \(P_1(x)\) includes the linear term: \(1 + 2x\).
  • The second-order polynomial, \(P_2(x)\), adds a quadratic term: \(1 + 2x + 2x^2\).
  • The third-order polynomial or \(P_3(x)\) introduces the cubic term: \(1 + 2x + 2x^2 + \frac{4}{3}x^3\).
The Maclaurin series provides a powerful tool to approximate functions using polynomials, simplifying complex calculations with a few terms.

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Most popular questions from this chapter

Neither the Ratio Test nor the Root Test helps with \(p\) -series. Try them on $$ \sum_{n=1}^{\infty} \frac{1}{n^{p}} $$ and show that both tests fail to provide information about convergence.

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=3}^{\infty} \sin \left(\frac{1}{\sqrt{n}}\right)$$

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty} \frac{n 2^{n}(n+1) !}{3^{n} n !}$$

Find the series' radius of convergence. $$\sum_{n=1}^{\infty}\left(\frac{2 \cdot 4 \cdot 6 \cdots(2 n)}{2 \cdot 5 \cdot 8 \cdot \cdots(3 n-1)}\right)^{2} x^{n}$$

In Exercises 17鈥� 46, use any method to determine whether the series converges or diverges. Give reasons for your answer. \(\sum_{n=3}^{\infty} \frac{2^{n^{2}}}{n^{2^{2}}}\)
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