Chapter 10: Problem 12
Let \(f(x)=\cos x\) and let \(P_{n}(x)\) be the Taylor approximation of degree \(n\) for \(f(x)\) around \(0 .\) Explain why, for any \(x,\) we can choose an \(n\) such that $$\left|f(x)-P_{n}(x)\right|<10^{-9}$$
Short Answer
Expert verified
For any \( x \), we can choose an \( n \) to make \( \left|f(x) - P_{n}(x)\right| < 10^{-9} \) because the Taylor remainder tends to zero.
Step by step solution
01
Understand the Taylor Series
The Taylor series for a function \( f(x) \) expanded at \( x = 0 \) is given by \( T_n(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \cdots + \frac{f^{(n)}(0)x^n}{n!} \). For \( f(x) = \cos x \), the series is \( \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \). This series defines \( P_n(x) \), the Taylor polynomial of degree \( n \).
02
Consider the Convergence of the Series
The Taylor series for \( \cos x \) is an infinite series. As \( n \rightarrow \infty \), \( P_n(x) \) converges to \( \cos x \) for all \( x \). Thus, we can make \( \left|f(x) - P_{n}(x)\right| \) arbitrarily small by taking sufficiently large \( n \).
03
Use the Fact That the Remainder Goes to Zero
The error or remainder \( R_n(x) \) of the Taylor series is given by Taylor's remainder theorem as \( R_n(x) = \frac{f^{(n+1)}(c)x^{n+1}}{(n+1)!} \) for some \( c \) between \( 0 \) and \( x \). For \( \cos x \), the derivatives are bounded by \( 1 \) in magnitude. Thus, \( \left|R_n(x)\right| \leq \frac{|x|^{n+1}}{(n+1)!} \).
04
Determine \( n \) for Desired Accuracy
We want \( \left|f(x) - P_n(x)\right| < 10^{-9} \). Using the bound \( \frac{|x|^{n+1}}{(n+1)!} < 10^{-9} \), solve for \( n \). For each specific \( x \), \( n \) such that this inequality is satisfied can be determined. As \( n \) grows, \( \left|x\right|^{n+1} \) becomes small faster than \( (n+1)! \) grows.
05
Conclude the Argument
Since for a given \( x \), \( \cos x \) is bounded and the growth of \( (n+1)! \) ensures that \( \left|R_n(x)\right| = \frac{|x|^{n+1}}{(n+1)!}\) gets arbitrarily small, there exists an \( n \) such that \( \left|f(x) - P_n(x)\right| < 10^{-9} \) for any \( x \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor Polynomial
The Taylor polynomial is an essential concept in calculus, offering a powerful tool for approximation. It represents a function as a sum of its derivatives at a single point, usually near where the function behaves similarly to a polynomial. When dealing with functions like \( f(x) = \cos x \), the Taylor polynomial approximates this function around a specified point, often zero.
For instance, the Taylor polynomial of degree \( n \) for \( \cos x \) around zero is known as the Maclaurin series:\\[ T_n(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots + (-1)^n\frac{x^{2n}}{(2n)!}. \]This series is "centered" at 0, providing a polynomial that behaves similarly to \( \cos x \) when \( x \) is close to this point.
For instance, the Taylor polynomial of degree \( n \) for \( \cos x \) around zero is known as the Maclaurin series:\\[ T_n(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots + (-1)^n\frac{x^{2n}}{(2n)!}. \]This series is "centered" at 0, providing a polynomial that behaves similarly to \( \cos x \) when \( x \) is close to this point.
- Key Point: The Taylor polynomial minimizes the error between the actual function value and the polynomial approximation near the center.
Convergence
Convergence plays a critical role in understanding Taylor series. It describes how the Taylor series approaches the actual function as the polynomial’s degree increases.
Consider the Taylor series for \( \cos x \). As \( n \rightarrow \infty \), the polynomial \( P_n(x) \) increasingly resembles \( \cos x \). This is because the higher the degree, the more closely aligned are the values of the polynomial and the function across a broader range of \( x \).
Consider the Taylor series for \( \cos x \). As \( n \rightarrow \infty \), the polynomial \( P_n(x) \) increasingly resembles \( \cos x \). This is because the higher the degree, the more closely aligned are the values of the polynomial and the function across a broader range of \( x \).
- It's crucial to note that convergence means the series will approximate the function to any desired accuracy, given sufficiently high \( n \).
- This property allows us to ensure the polynomial matches the behavior of \( \cos x \) within a tolerated error margin, like \( 10^{-9} \).
Remainder Theorem
Understanding the remainder is key in determining how close a Taylor polynomial is to its original function. The Remainder Theorem offers a formula for this difference, specifically for Taylor series. It states that the error between the function \( f(x) \) and its Taylor polynomial \( P_n(x) \) can be expressed as:
\[ R_n(x) = \frac{f^{(n+1)}(c)x^{n+1}}{(n+1)!}, \] where \( c \) lies somewhere between the center and the point of evaluation, \( x \).
For \( \cos x \), all the derivatives are bounded, meaning \( \left|f^{(n+1)}(c)\right| \leq 1 \) for any \( c \). This bound helps ensure that the error \( R_n(x) \) diminishes as \( n \) increases. Thus, with a sufficiently large \( n \), the remainder can be made negligibly small, fulfilling the accuracy condition \( \left|R_n(x)\right| < 10^{-9} \) for any \( x \).
\[ R_n(x) = \frac{f^{(n+1)}(c)x^{n+1}}{(n+1)!}, \] where \( c \) lies somewhere between the center and the point of evaluation, \( x \).
For \( \cos x \), all the derivatives are bounded, meaning \( \left|f^{(n+1)}(c)\right| \leq 1 \) for any \( c \). This bound helps ensure that the error \( R_n(x) \) diminishes as \( n \) increases. Thus, with a sufficiently large \( n \), the remainder can be made negligibly small, fulfilling the accuracy condition \( \left|R_n(x)\right| < 10^{-9} \) for any \( x \).
- Key Insight: Knowing the remainder allows us to control the precision of our approximation.
Degree of Approximation
The degree of approximation is critically tied to the Taylor polynomial's degree, \( n \). This degree highlights the polynomial's capacity to closely mimic the function within an acceptable error range.
When tasked with approximating \( \cos x \) to a precision of less than \( 10^{-9} \), we adjust \( n \) until the error \( \left| f(x) - P_n(x) \right| \) is sufficiently small. This is achieved by using the error bound \( \frac{|x|^{n+1}}{(n+1)!} \) and solving for \( n \) to pinpoint what degree ensures our approximation error stays under \( 10^{-9} \).
When tasked with approximating \( \cos x \) to a precision of less than \( 10^{-9} \), we adjust \( n \) until the error \( \left| f(x) - P_n(x) \right| \) is sufficiently small. This is achieved by using the error bound \( \frac{|x|^{n+1}}{(n+1)!} \) and solving for \( n \) to pinpoint what degree ensures our approximation error stays under \( 10^{-9} \).
- The degree of the polynomial grows as \( x \) becomes larger, allowing \( \left|x\right|^{n+1} \) to reduce faster than \( (n+1)! \) increases.
- This relationship underscores the strength of Taylor polynomials in approximating functions across different values of \( x \).