Chapter 10: Problem 1
Find the Taylor polynomials of degree \(n\) approximating the functions for \(x\) near \(0 .\) (Assume \(p\) is a constant. \()\) $$\frac{1}{1-x}, \quad n=3,5,7$$
Short Answer
Expert verified
Taylor polynomials:
- n=3: \(1 + x + x^2 + x^3\)
- n=5: \(1 + x + x^2 + x^3 + x^4 + x^5\)
- n=7: \(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7\)
Step by step solution
01
Understand the Function and Taylor Series
The given function is \( f(x) = \frac{1}{1-x} \). We need to find Taylor polynomials for this function near \(x = 0\). The general formula for a Taylor series centered at \(c\) is: \[ T_n(x) = f(c) + f'(c)(x - c) + \frac{f''(c)}{2!}(x - c)^2 + \cdots + \frac{f^{(n)}(c)}{n!}(x - c)^n \] For this problem, \(c = 0\).
02
Derive the First Few Derivatives
Calculate the derivatives of \( f(x) = \frac{1}{1-x} \). The first few derivatives at \(n=0\) are:- \( f(x) = (1-x)^{-1}, \quad f(0) = 1 \)- \( f'(x) = (1-x)^{-2}, \quad f'(0) = 1 \)- \( f''(x) = 2(1-x)^{-3}, \quad f''(0) = 2 \)- \( f'''(x) = 6(1-x)^{-4}, \quad f'''(0) = 6 \)- \( f^{(4)}(x) = 24(1-x)^{-5}, \quad f^{(4)}(0) = 24 \)
03
Find the Taylor Polynomial for n=3
Use the derivatives found earlier to form the Taylor polynomial degree 3 centered at \(0\):\[ T_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 \]Substitute the derivatives:\[ T_3(x) = 1 + x + \frac{2}{2}x^2 + \frac{6}{6}x^3 = 1 + x + x^2 + x^3 \]
04
Calculate the Taylor Polynomial for n=5
Building upon the previous result for \(n=3\), we extend to \(n=5\):\[ T_5(x) = T_3(x) + \frac{f^{(4)}(0)}{4!}x^4 + \frac{f^{(5)}(0)}{5!}x^5 \]First, find the 5th derivative:\( f^{(5)}(x) = 120(1-x)^{-6}, \quad f^{(5)}(0) = 120 \)So:\[ T_5(x) = 1 + x + x^2 + x^3 + \frac{24}{24}x^4 + \frac{120}{120}x^5 = 1 + x + x^2 + x^3 + x^4 + x^5 \]
05
Extend to the Taylor Polynomial for n=7
Finally, extend to \(n=7\):\[ T_7(x) = T_5(x) + \frac{f^{(6)}(0)}{6!}x^6 + \frac{f^{(7)}(0)}{7!}x^7 \]Calculate the 6th and 7th derivatives:- \( f^{(6)}(x) = 720(1-x)^{-7}, \quad f^{(6)}(0) = 720 \)- \( f^{(7)}(x) = 5040(1-x)^{-8}, \quad f^{(7)}(0) = 5040 \)So:\[ T_7(x) = 1 + x + x^2 + x^3 + x^4 + x^5 + \frac{720}{720}x^6 + \frac{5040}{5040}x^7 = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor Series
A Taylor series is a clever way to approximate functions that are potentially complex. It does this by expanding the function into an infinite sum of terms, calculated from the function's derivatives at a particular point. For our exercise, the function is approximated around the point where \( x = 0 \). Here's a breakdown of how it works:
- Identify the function to be expanded: In this case, it's \( f(x) = \frac{1}{1-x} \).
- Determine the center for the expansion: Here, it is around \( x = 0 \).
- Calculate the derivatives of the function at the center.
- Plug these derivatives into the Taylor series formula to obtain the approximation.
Derivatives
Derivatives are fundamental in constructing a Taylor series. They indicate the rate at which a function is changing at any point. Calculating derivatives lies at the heart of finding a Taylor polynomial.
- The first step involves calculating the derivatives of the function at the expansion point. These derivatives determine the coefficients of the polynomial.
- In this exercise, for \( f(x) = \frac{1}{1-x} \), we found that \( f(x) = (1-x)^{-1} \) makes it straightforward to compute derivatives using the power rule.
- For example: The first derivative, \( f'(x) = (1-x)^{-2} \), and continues with increasing complexity, revealing the function's changing rate more precisely at \( x = 0 \).
Polynomial Approximation
Polynomial approximation is the end product of using a Taylor series. By understanding derivatives, you can approximate a function with a polynomial. Here's why this is significant:
- Polynomials are easier and quicker to compute than many functions. They simplify the process of calculation, especially in iterations and algorithms.
- This exercise illustrates polynomial approximation by reaching Taylor polynomials of different degrees: 3, 5, and 7. Each offers varying levels of precision.
- The polynomial for degree 3 is \( T_3(x) = 1 + x + x^2 + x^3 \). For degree 5, including higher derivative terms leads to \( T_5(x) = 1 + x + x^2 + x^3 + x^4 + x^5 \).