Problem 13 - Calculus for Biology and Medicine

Question

Determine the values of the sequence \(\left\\{a_{n}\right\\}\) for \(n=0,1,2, \ldots, 5\) $$ f(n)=e^{\sqrt{n}} $$

Step-by-Step Solution

Answer

The sequence values are approximately: \(a_0 = 1\), \(a_1 = e\), \(a_2 \approx 4.113\), \(a_3 \approx 5.642\), \(a_4 \approx 7.389\), \(a_5 \approx 9.487\).

Step 1: Understanding the Sequence Formula

The sequence is defined by the function \(f(n) = e^{\sqrt{n}}\). The task requires us to calculate the values of the sequence for \(n = 0, 1, 2, 3, 4, 5\).

Step 2: Calculate for n=0

Substitute \(n = 0\) into the formula: \(f(0) = e^{\sqrt{0}} = e^{0} = 1\). So, \(a_0 = 1\).

Step 3: Calculate for n=1

Substitute \(n = 1\) into the formula: \(f(1) = e^{\sqrt{1}} = e^{1} = e\). So, \(a_1 = e\).

Step 4: Calculate for n=2

Substitute \(n = 2\) into the formula: \(f(2) = e^{\sqrt{2}}\). Calculate the numerical value: \(\sqrt{2} \approx 1.414\), so \(e^{1.414} \approx 4.113\). So, \(a_2 \approx 4.113\).

Step 5: Calculate for n=3

Substitute \(n = 3\) into the formula: \(f(3) = e^{\sqrt{3}}\). Calculate the numerical value: \(\sqrt{3} \approx 1.732\), so \(e^{1.732} \approx 5.642\). So, \(a_3 \approx 5.642\).

Step 6: Calculate for n=4

Substitute \(n = 4\) into the formula: \(f(4) = e^{\sqrt{4}} = e^{2}\). Calculate the numerical value: \(e^{2} \approx 7.389\). So, \(a_4 \approx 7.389\).

Step 7: Calculate for n=5

Substitute \(n = 5\) into the formula: \(f(5) = e^{\sqrt{5}}\). Calculate the numerical value: \(\sqrt{5} \approx 2.236\), so \(e^{2.236} \approx 9.487\). So, \(a_5 \approx 9.487\).

Key Concepts

Understanding Sequences

Sequences are a set of numbers arranged in a specific order according to a given rule. In mathematics, each element in a sequence is called a term, usually denoted by variables like \(a_n\). In our case, the sequence is defined by the function \(f(n) = e^{\sqrt{n}}\). This means each term is determined by this formula where \(n\) represents the position in the sequence.

When working with sequences, it’s important to note:

• The position \(n\) is usually a non-negative integer starting from 0, which is why we start calculating from \(n = 0\).
• The rule or function, like our given exponential function, dictates the form of the sequence. This dictates how each term is derived from its preceding or succeeding terms.
• Calculating sequences involves substituting integers into the formula one at a time, allowing you to generate specific numbers in the sequence.

Exploring Natural Logarithms

The natural logarithm, represented as \(\ln\), is the inverse of the exponential function where the base is the number \(e\). The constant \(e\) is approximately equal to 2.718. In our sequence, \(e\) is raised to the power of the square root of \(n\), creating a specific exponential form.

Understanding natural logarithms involves some key ideas:

• It helps to solve equations involving exponential calculations since it undoes the exponential effect. For example, if \(e^x = y\), then \(x = \ln(y)\).
• Using natural logs simplifies complex calculations by transforming multiplicative relationships into additive ones.
• Natural logarithms are closely related to exponential growth. They describe rates at which processes like growth or decay occur, useful in many real-world applications from finance to science.

Demystifying Square Roots

A square root of a number \(x\) is a number \(y\) such that \(y^2 = x\). In our sequence, \(\sqrt{n}\) affects how the exponential function grows as sections of \(n\) increase.

Key points to understand square roots include:

• Square roots reduce the power of numbers. For instance, \(\sqrt{9} = 3\) because \(3^2 = 9\).
• They are particularly useful in simplifying quadratic equations or understanding geometric dimensions.
• Square roots are often approximated in practical calculations, especially for non-perfect squares like \(\sqrt{2} \approx 1.414\) and \(\sqrt{3} \approx 1.732\), which are essential for evaluating our sequence.
• Square roots form the basis of the radical expressions and the hypotenuse formula in trigonometry, showcasing their broad usefulness beyond simple arithmetic.