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Magazine Chapter 7: Problem 5
Using the first 6 terms of the series for \(e^{x}\), determine approximate values of \(e^{2}\) and \(\sqrt{e}\) to 4 sig fig.
Short Answer
Expert verified
\( e^{2} \approx 7.267 \), \( \sqrt{e} \approx 1.649 \).
Step by step solution
01
Know the Series for \( e^{x} \)
The Taylor series for \( e^{x} \) around \( x = 0 \) is given by \( e^{x} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \ldots \). We will use the first 6 terms of this series.
02
Apply the Series to \( e^{2} \)
Substitute \( x = 2 \) into the series: \( e^{2} \approx 1 + \frac{2}{1!} + \frac{2^2}{2!} + \frac{2^3}{3!} + \frac{2^4}{4!} + \frac{2^5}{5!} \). Calculate each term and sum them to get an approximate value for \( e^{2} \).
03
Calculate Each Term for \( e^{2} \)
Calculate: 1. \( 1 = 1 \)2. \( \frac{2}{1!} = 2 \)3. \( \frac{2^2}{2!} = 2 \)4. \( \frac{2^3}{3!} = \frac{8}{6} \approx 1.3333 \)5. \( \frac{2^4}{4!} = \frac{16}{24} \approx 0.6667 \)6. \( \frac{2^5}{5!} = \frac{32}{120} \approx 0.2667 \)Now sum these: \( 1 + 2 + 2 + 1.3333 + 0.6667 + 0.2667 = 7.2667 \).
04
Approximate \( e^{2} \)
The approximate value of \( e^{2} \) is 7.2667. Rounding to 4 significant figures gives us 7.267.
05
Determine Series for \( \sqrt{e} \)
Knowing \( sqrt{e} = e^{0.5} \), we use the series: \( e^{0.5} \approx 1 + \frac{0.5}{1!} + \frac{0.5^2}{2!} + \frac{0.5^3}{3!} + \frac{0.5^4}{4!} + \frac{0.5^5}{5!} \).
06
Calculate Each Term for \( \sqrt{e} \)
Calculate:1. \( 1 = 1 \)2. \( \frac{0.5}{1!} = 0.5 \)3. \( \frac{0.5^2}{2!} = 0.125 \)4. \( \frac{0.5^3}{3!} = \frac{0.125}{6} \approx 0.020833 \)5. \( \frac{0.5^4}{4!} = \frac{0.0625}{24} \approx 0.002604 \)6. \( \frac{0.5^5}{5!} = \frac{0.03125}{120} \approx 0.000260 \)Sum these: \( 1 + 0.5 + 0.125 + 0.020833 + 0.002604 + 0.000260 \approx 1.648697 \).
07
Approximate \( \sqrt{e} \)
The approximate value of \( \sqrt{e} \) is 1.648697. Rounding to 4 significant figures gives us 1.649.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Function in Taylor Series
The exponential function, often symbolized as \( e^{x} \), plays a crucial role in many areas of mathematics and science. It's known for its unique properties, such as having its own derivative and integral. When we expand it using a Taylor series, we make it easier to approximate the value of \( e^{x} \) without the need for a calculator that carries the constant "e" to infinite precision. This series is expressed as:
- \( e^{x} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \ldots \)
Approximation Methods Using Taylor Series
Approximation is a key strategy in both mathematics and science when exact solutions are either impossible or impractical to obtain. The Taylor series provides a powerful method for approximating functions like \( e^{x} \) by summing its series expansion terms.
- By truncating the series after a few terms (like the first six terms mentioned), we can get an approximate value of \( e^{x} \).
- This approximation holds especially true when \( x \) is small, as higher powers of \( x \) contribute less to the sum.
- The accuracy of the approximation depends on how many terms we include: the more terms, the more accurate the approximation.
Understanding Significant Figures in Approximations
Significant figures are a method of expressing the precision of a numerical result. When we use approximation methods like Taylor series to find values such as \( e^{2} \) or \( \sqrt{e} \), it's crucial to represent our results with an appropriate amount of significant figures to convey a proper level of accuracy.
- Significant figures reflect the certainty of a measurement or calculation. In our calculation, we approximate values to 4 significant figures.
- Rounding helps us determine which digits are meaningful in the context of the measurement. For instance, approximating \( e^{2} = 7.2667 \) gives us 7.267 when limited to 4 significant figures.
- This precision ensures that small errors in our approximation don't mislead us into thinking we have more accurate results than we actually do.
