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Use Taylor's Formula to determine the number of terms of the Maclaurin series for \(e^{x}\) that should be used to estimate \(e^{0.1}\) to within \(0.00001 .\)

Short Answer

Expert verified
Use up to the \(x^4\) term in the Maclaurin series to estimate \(e^{0.1}\) within \(0.00001\).

Step by step solution

01

Understand Maclaurin Series

The Maclaurin series for a function \( f(x) \) is given by \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \). For \(e^x\), this simplifies to \( 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \), since \(f^{(n)}(0) = 1\) for all \(n\).
02

Determine Remainder Term

To ensure the error is within the desired range, we use the remainder term \( R_n = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} \). For \(e^x\), this becomes \( R_n = \frac{x^{n+1}}{(n+1)!} \) since the derivatives of \(e^x\) are all \(e^x \) evaluated at some \( c \), which is less than or equal to the maximum value, which is \(e^{0.1} \approx 1.1052\). We need \( R_n < 0.00001 \).
03

Approximate Error Bound and Solve

The approximation error \( R_n = \frac{(0.1)^{n+1}}{(n+1)!} < 0.00001 \). We need to find the smallest \( n \) such that this is true. We test successive \( n \) values:- \( n=4 \): \( R_4 = \frac{(0.1)^5}{5!} = \frac{0.00001}{120} = 0.00000008333 \), which is less than 0.00001.Therefore, we select \( n=4 \) because for \( n=3 \), the remainder exceeds \( 0.00001 \).
04

Conclusion

To ensure the desired accuracy for estimating \(e^{0.1}\) using the Maclaurin series, we should use up to the \(x^4\) term, which means using the polynomial \(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}\). This ensures the error is less than \(0.00001\).

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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 case of the Taylor series where the expansion is done around zero. This makes it a powerful tool for approximating functions like exponential, trigonometric, or logarithmic functions near the point zero. Let's break this down a bit:
  • A Maclaurin series for a function \( f(x) \) is a sum of derivatives at zero, scaled by powers of \( x \) divided by factorials. It's represented as: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \]
  • For the exponential function \( e^x \), this simplifies to:\[ 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]
Since all derivatives of \( e^x \) are \( e^x \) itself, and when evaluated at zero, they equal 1, this simplifies the Maclaurin series for \( e^x \) wonderfully.

This simplification allows for a straightforward way to approximate values of the exponential function around zero by cutting off the series at a certain number of terms to minimize computational effort while maintaining accuracy.
Error Bound
When using series like the Maclaurin series, understanding the **Error Bound** is crucial. This helps you know how accurate your approximation is. In our case, it was necessary to ensure that when estimating \( e^{0.1} \), the error remained below a very tight limit of 0.00001.How does the error bound work for Maclaurin series? It's all about analyzing the remainder term:
  • The remainder \( R_n \) of a series can be represented as:\[ R_n = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} \]where \( c \) is some value between 0 and \( x \).
  • For the exponential function \( e^x \), this simplifies to:\[ R_n = \frac{x^{n+1}}{(n+1)!} \]
  • This represents the "tail" of the series, or the summation of terms beyond the \( n^{th} \) term.
To achieve the error margin, you analyze these remainders by systematically increasing \( n \) until: \[ \frac{(0.1)^{n+1}}{(n+1)!} < 0.00001 \]For our problem, we found \( n = 4 \) to be the smallest number of terms to meet the desired accuracy. This guarantees that all subsequent terms have negligible impact on the approximation.
Exponential Function
The **Exponential Function**, particularly the base \( e \), denoted as \( e^x \), is a fundamental mathematical function with key characteristics. It's known for its constant rate of growth, a property that is crucial in many fields such as calculus, complex numbers, and financial calculations.Why is \( e^x \) special?
  • Derivative Property: The derivative of \( e^x \) is itself \( e^x \), which means that its rate of change is equal to its current value. This is unique among other functions.
  • Applications: It appears naturally in scenarios of continuous growth or decay (think populations, or radioactive decay).
  • Maclaurin Series: As we explored above, the series expansion of \( e^x \) simplifies remarkably because each derivative evaluated at zero is one, making the series easy to compute: \[ 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]
This smooth and straightforward series expansion provides a robust method to estimate values of \( e^x \) when direct calculation might be cumbersome. Whether for simple approximations or deeper mathematical models, the function's properties make it immensely useful.

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