Problem 7 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Assume that the sixth degree polynomials, \(S(x), C(x),\) and \(E(x)\) that match, respectively, \(\sin x, \cos x,\) and \(e^{x},\) are $$ \begin{array}{l} S(x)=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !} \\ C(x)=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\frac{x^{6}}{6 !} \\ E(x)=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\frac{x^{5}}{5 !}+\frac{x^{6}}{6 !} \end{array} $$ a. Compute \(S^{\prime}(x)\) and compare it with \(C(x)\). b. Compute \(C^{\prime}(x)\) and compare it with \(S(x)\). c. Compute \(E^{\prime}(x)\) and compare it with \(E(x)\). d. (Only for the adventurous.) Let \(i=\sqrt{-1}\). Note that \(i^{2}=-1, i^{3}=i^{2} \cdot i=-i,\) and \(i^{4}=i^{2} \cdot i^{2}=1,\) and continue this sequence. Compute \(E(i \cdot x)\) and write it in terms of \(S(x)\) and \(C(x)\).

Step-by-Step Solution

Answer

a. \(S'(x) = C(x)\); b. \(C'(x) = -S(x)\); c. \(E'(x) = E(x)\); d. \(E(ix) = C(x) + iS(x)\).

Step 1: Differentiate S(x)

Find the derivative of the polynomial function \(S(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}\). Differentiate each term individually. The derivative of \(x\) is \(1\).The derivative of \(-\frac{x^3}{3!}\) is \(-3 \frac{x^2}{3!} = -\frac{x^2}{2!}\).The derivative of \(\frac{x^5}{5!}\) is \(5 \frac{x^4}{5!} = \frac{x^4}{4!}\).Combining, \(S'(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!}\), which is equal to \(C(x)\).

Step 2: Differentiate C(x)

Find the derivative of the polynomial function \(C(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!}\). Differentiate each term individually. The derivative of \(1\) is \(0\).The derivative of \(-\frac{x^2}{2!}\) is \(-2 \frac{x}{2!} = -x\).The derivative of \(\frac{x^4}{4!}\) is \(4 \frac{x^3}{4!} = \frac{x^3}{3!}\).The derivative of \(-\frac{x^6}{6!}\) is \(-6 \frac{x^5}{6!} = -\frac{x^5}{5!}\).Combining, \(C'(x) = -x + \frac{x^3}{3!} - \frac{x^5}{5!}\), which is equal to \(-S(x)\).

Step 3: Differentiate E(x)

Find the derivative of \(E(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!}\). Differentiate each term:The derivative of \(1\) is \(0\).The derivative of \(x\) is \(1\).The derivative of \(\frac{x^2}{2!}\) is \(2 \frac{x}{2!} = x\).The derivative of \(\frac{x^3}{3!}\) is \(3 \frac{x^2}{3!} = \frac{x^2}{2!}\).The derivative of \(\frac{x^4}{4!}\) is \(4 \frac{x^3}{4!} = \frac{x^3}{3!}\).The derivative of \(\frac{x^5}{5!}\) is \(5 \frac{x^4}{5!} = \frac{x^4}{4!}\).The derivative of \(\frac{x^6}{6!}\) is \(6 \frac{x^5}{6!} = \frac{x^5}{5!}\).Combining, \(E'(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!}\), which is equal to \(E(x)\).

Step 4: Evaluate E(ix)

Substitute \(ix\) into \(E(x)\):\[ E(ix) = 1 + (ix) + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} \]Simplify using powers of \(i\): \(i^2 = -1\) gives \(-\frac{x^2}{2!}\).\(i^3 = -i\) gives \(-\frac{ix^3}{3!}\).\(i^4 = 1\) gives \(\frac{x^4}{4!}\).\(i^5 = i\) gives \(\frac{ix^5}{5!}\).\(i^6 = -1\) gives \(-\frac{x^6}{6!}\).Combining terms:\[ E(ix) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + i \left ( x - \frac{x^3}{3!} + \frac{x^5}{5!} \right) \]Notice that the real part is \(C(x)\) and the imaginary part is \(iS(x)\).

Key Concepts

Differentiation

Differentiation is a key concept in calculus used to determine the rate at which a function is changing at any point. This is done by finding the derivative, which is the function that gives the slope of the tangent line to the curve of the original function at any point. In the exercise, the aim is to differentiate sixth degree polynomials like \( S(x) \), \( C(x) \), and \( E(x) \). The differentiation process involves applying simple rules to each term of the polynomial separately. For instance:

• The derivative of a constant is zero.
• The derivative of \(x^n\) is \(nx^{n-1}\).

Using these rules, we can differentiate any polynomial by applying them to each term and adding up the results. For example, differentiating \( S(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} \) involves handling each term: - \( x \) becomes 1,- \( -\frac{x^3}{3!} \) becomes \(-\frac{x^2}{2!}\),- and \( \frac{x^5}{5!} \) becomes \( \frac{x^4}{4!}\).Differentiation not only tells us how functions change, but also helps in comparing functions like \( S'(x) \) and \( C(x) \) as shown in this exercise.

Complex Numbers

Complex numbers extend real numbers by including an imaginary unit \(i\), where \(i\) satisfies \(i^2 = -1\). They are crucial for solving polynomial equations that have no real solutions. A complex number has the form \(a + bi\), where \(a\) and \(b\) are real numbers.In the given problem, we explore complex numbers by substituting \(ix\) into \(E(x)\), the polynomial matching \(e^x\), transforming it to \(E(i \cdot x)\). When simplifying, we use the powers of \(i\):

• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\)

This substitution shows why complex numbers are useful. Substituting \(ix\) produces imaginary terms alongside real ones. These imaginary terms give further insights, like showing the relation between complex exponentials and trigonometric functions through Euler's formula: \(e^{ix} = \cos(x) + i\sin(x)\). Thus, the real part corresponds to \(C(x)\) and the imaginary part to \(iS(x)\), linking exponential functions to trigonometric functions.

Polynomial Approximation

Polynomial approximation is a powerful mathematical tool used to approximate more complicated functions with polynomials. The Taylor Polynomial is a useful method for approximating functions within a certain interval.For example, the function \(\sin x\) can be approximated by its Taylor polynomial of degree six, \(S(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}\). This approximates \(\sin x\) near zero with a high degree of accuracy for small values of \(x\). Similarly, \(\cos x\) and \(e^x\) are approximated by their respective polynomials \(C(x)\) and \(E(x)\).The accuracy of polynomial approximation generally increases with the degree of the polynomial. The more terms included in the polynomial, the closer it will match the original function over a range of values. In practical applications, this allows for both easier computation and a deeper understanding of the behavior near significant points.

Trigonometric Functions

Trigonometric functions such as \(\sin x\) and \(\cos x\) are foundational in mathematics, particularly in geometry and analyzing periodic phenomena. They describe relationships in right triangles and circles.In the exercise, \(\sin x\) and \(\cos x\) are approximated by sixth degree polynomials \(S(x)\) and \(C(x)\). These polynomials are derived using the Taylor series expansion, capturing the oscillating nature of trigonometric functions effectively over specific intervals.An intriguing feature of trigonometric functions is their relationship through differentiation. The derivative of \(\sin x\) is \(\cos x\), and for \(\cos x\), it's \(-\sin x\). Hence, in our exercise, differentiating \(S(x)\) yields \(C(x)\) and differentiating \(C(x)\) results in \(-S(x)\). This interchange through differentiation illustrates the derivative patterns of trigonometric functions, providing insight into how these functions evolve using calculus.