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Chapter 10: Problem 36

(Requires sequence and series operations) a. Use your calculator to find \(e^{2}\) rounded to six decimal places. b. The Taylor series for \(e^{x}\) evaluated at \(x=2\) gives \(e^{2}=\sum_{n=0}^{\infty} \frac{2^{n}}{n !}\). Set your calculator to find the sum of this series up to any number of terms. How many terms are required for the sum (rounded to six decimal places) to agree with your answer to part (a)?

Short Answer

Expert verified
The Taylor series for \( e^{2} \) requires 10 terms to match \( e^{2} = 7.389056 \) when rounded to six decimal places.

Step by step solution

01

Calculate the value of e虏 with a calculator

Use your calculator to compute the value of \( e^{2} \) and round it to six decimal places. The value should be approximately 7.389056.
02

Write the Taylor series for e^x evaluated at x=2

Recall the Taylor series expansion of \( e^x \), which is \( \sum_{n=0}^{\infty} \frac{x^{n}}{n!} \). Substitute \( x = 2 \) to get \( e^{2} = \sum_{n=0}^{\infty} \frac{2^{n}}{n!} \).
03

Compute the series sum for e虏 up to several terms

Begin calculating the sum of the series \( \sum_{n=0}^{\infty} \frac{2^{n}}{n!} \) term by term: \(=0: \frac{2^{0}}{0!} = 1, \)\(=1: \frac{2^{1}}{1!} = 2, \)\(=2: \frac{2^{2}}{2!} = 2, \)and continue adding subsequent terms until the sum approximates the value found in Step 1.
04

Verify when the series sum rounds to the target value

Continue added terms, calculating the partial sums: \( =3, \frac{2^{3}}{3!} = \frac{8}{6}, =4, \frac{2^{4}}{4!} = \frac{16}{24}, \) etc. Check partial sums for each number of terms calculated and compare to the value from Step 1 rounded to six decimals, \( 7.389056 \).
05

Determine the number of terms needed

Determine that the series \( e^{2} = \sum_{n=0}^{\infty} \frac{2^{n}}{n!} \) up to \( n=9 \) approximately gives a sum that, when rounded to six decimal places, matches the calculated value of \( e^{2} \) as 7.389056.

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Key Concepts

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

Exponential Function
The exponential function, denoted as \(e^x\), is one of the most important functions in mathematics, particularly in calculus. The base of the natural logarithm, \(e\), is approximately 2.71828, and it arises naturally in various growth processes and compound interest calculations.
Understanding \(e^x\) is crucial because it describes phenomena where growth is proportional to the current amount, such as population growth or radioactive decay.
  • The function \(e^x\) is unique because its rate of change (derivative) is equal to the function itself.
  • It can be represented as an almost infinite series (the Taylor series), which allows for approaching very accurate values computationally.
  • The exponential function transforms linear relationships into multiplicative ones, which is useful in many real-world applications.
In our exercise, we seek to approximate \(e^2\), showcasing the practical use of this function in calculating values with precision using series.
Sequence and Series
A sequence is an ordered list of numbers, while a series is the sum of a sequence. In mathematics, sequences and series allow us to approximate functions and solve real-world problems through infinite addition.
The Taylor series is a specific sequence that represents functions as an infinite sum of terms. When using a series like the Taylor series for \(e^x\), each term of the sequence provides an approximation to the actual value.
  • The Taylor series for \(e^x\) is \( \sum_{n=0}^{\infty} \frac{x^{n}}{n!} \), a formula which expands \(e^x\) into a sum of terms involving powers of \(x\).
  • In practical terms, we often calculate only a few terms rather than infinitely many to approximate a value closely enough for our needs.
In our problem, calculating enough terms of the series \( \sum_{n=0}^{\infty} \frac{2^n}{n!} \) enables us to approximate \(e^2\) accurately.
Numerical Approximation
Numerical approximation is a method to find functions' values when a direct calculation is difficult or impossible. This is especially true for transcendental numbers like \(e\), which can't be expressed with simple fractions or real numbers.
Using series expansions, such as Taylor and Maclaurin series, is a common technique to approximate values numerically.
  • A series gives a set of partial sums, each bringing the approximation closer to the actual value.
  • The number of terms needed depends on the desired precision. For example, approximating \(e^2\) to six decimals requires evaluating terms up to \(n = 9\).
  • Computational efficiency improves since only necessary terms are calculated, saving time and resources.
In our task, numerical approximation via the Taylor series allows us to determine how many terms are necessary to reach the precision of six decimal places.
Calculus
Calculus, the mathematical study of change, underpins functions and series like Taylor series used in solving this problem. It provides tools to model and analyze dynamic systems.
Calculus fundamentally deals with two concepts: differentiation and integration. The Taylor series uses differentiation to express functions as a sum of their derivatives evaluated at a particular point.
  • Differentiation helps determine the slope of a function at any given point, essential in understanding exponential growth described by \(e^x\).
  • The Taylor series is derived from the function's derivatives, capturing its dynamic nature along an interval.
  • With calculus assisting on such approximations, concepts are made manageable and practical, bridging between theory and real-world applications.
By applying calculus to the exponential function, we gain the capability to compute approximations effectively, as demonstrated in approximating \(e^2\) using its Taylor series.

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Most popular questions from this chapter

GENERAL: Cycling (Graphing calculator with series operations helpful) You plan to cycle across the United States, cycling 10 miles the first day, \(10 \%\) further the second day (so 11 miles), increasing each day's distance by \(10 \% .\) On which day will you reach the opposite coast 3000 miles away?

Find the Taylor series at \(x=0\) for \(\frac{1}{(1-x)^{2}}\) by differentiating both sides of $$ \frac{1}{1-x}=1+x+x^{2}+x^{3}+\cdots \quad \text { for }|x|<1 $$

BIOMEDICAL: Long-Term Population A population (of cells or people) is such that each year a number \(a\) of individuals are added (call them "immigrants"), and a proportion \(p\) of the individuals who have been there die. Therefore, the proportion that survives is \((1-p),\) so that just after an immigration the population will consist of new immigrants plus \(a(1-p)\) from the previous year's immigration plus \(a(1-p)^{2}\) from the immigration before that, and so on. In the long run, the size of the population just after an immigration will then be the sum \(a+a(1-p)+a(1-p)^{2}+a(1-p)^{3}+\cdots\) Find the long-run size of the population just after an immigration for any number of immigrants \(a\) and any survival proportion \((1-p)\). Then find the long-run size just before an immigration.

Find the Taylor series at \(x=0\) for each function by modifying one of the Taylor series from this section. [Hint: For Exercise \(26,\) use two series from this section, writing the terms in order of increasing powers of \(x\).] \(\frac{x-\sin x}{x^{3}}\)

Find the Taylor series at \(x=0\) for each function by calculating three or four derivatives and using the definition of Taylor series. \(\cos 2 x\)
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