Problem 79 - Calculus Early Transcendentals

Question

The theory of optics gives rise to the two Fresnel integrals $$S(x)=\int_{0}^{x} \sin t^{2} d t \text { and } C(x)=\int_{0}^{x} \cos t^{2} d t$$ a. Compute \(S^{\prime}(x)\) and \(C^{\prime}(x)\) b. Expand \(\sin t^{2}\) and \(\cos t^{2}\) in a Maclaurin series and then integrate to find the first four nonzero terms of the Maclaurin series for \(S\) and \(C\) c. Use the polynomials in part (b) to approximate \(S(0.05)\) and \(C(-0.25)\) d. How many terms of the Maclaurin series are required to approximate \(S(0.05)\) with an error no greater than \(10^{-4} ?\) e. How many terms of the Maclaurin series are required to approximate \(C(-0.25)\) with an error no greater than \(10^{-6} ?\)

Step-by-Step Solution

Answer

Question: Compute the derivatives S'(x) and C'(x) of S(x) = ∫ sin(t^2) dt and C(x) = ∫ cos(t^2) dt. Then, use Maclaurin series to obtain polynomial approximations for S(0.05) and C(-0.25) up to the first four nonzero terms. Additionally, determine how many terms are required for the approximations to have an error of 10^-4 for S(0.05) and 10^-6 for C(-0.25). Answer: The derivatives of S(x) and C(x) are S'(x) = sin(x^2) and C'(x) = cos(x^2). S(0.05) ≈ 0.0499998333 and C(-0.25) ≈ -0.0833330799 using the first four nonzero terms of their Maclaurin series. To achieve the required precision, at least two terms are needed for approximating S(0.05) and at least three terms for approximating C(-0.25).

Step 1: Part a: Compute S'(x) and C'(x)

Use the Fundamental Theorem of Calculus, which states that if F(x) = ∫[a, x] f(t) dt, then F'(x) = f(x). In our case, the functions f(t) are sin(t^2) and cos(t^2) for S(x) and C(x) respectively. So, S'(x) = sin(x^2), and C'(x) = cos(x^2).

Step 2: Part b: Expand sin(t^2) and cos(t^2) in Maclaurin series

Recall the Maclaurin series for sine and cosine are given by: sin(x) = ∑[(-1)^n * (x^(2n+1)) / (2n+1)!] for n=0,1,2,... cos(x) = ∑[(-1)^n * (x^(2n)) / (2n)! ] for n=0,1,2,... Substitute x with t^2 to obtain the expansions for sin(t^2) and cos(t^2): sin(t^2) = ∑[(-1)^n * (t^(4n+2)) / (2n+1)!] for n=0,1,2,... cos(t^2) = ∑[(-1)^n * (t^(4n)) / (2n)!] for n=0,1,2,... Now integrate the Maclaurin series term by term for both sin(t^2) and cos(t^2) as follows: S(x) = ∫[sin(t^2) dt] = ∑[(-1)^n * (x^(4n+3)) / ((4n+3) * (2n+1)!)] for n=0,1,2,... C(x) = ∫[cos(t^2) dt] = ∑[(-1)^n * (x^(4n+1)) / ((4n+1) * (2n)!)] for n=0,1,2,... The first four nonzero terms for the Maclaurin series for S(x) and C(x) are as follows: S(x) ≈ x - (x^5)/10 + (x^9)/216 - (x^13)/9360 C(x) ≈ x^3/3 - (x^7)/42 + (x^11)/1320 - (x^15)/75600

Step 3: Part c: Approximating S(0.05) and C(-0.25)

Substitute the values into the polynomial approximations obtained in part (b): S(0.05) ≈ 0.05 - (0.05^5)/10 + (0.05^9)/216 - (0.05^13)/9360 ≈ 0.0499998333 C(-0.25) ≈ (-0.25)^3/3 - (-0.25)^7/42 + (-0.25)^11/1320 - (-0.25)^15/75600 ≈ -0.0833330799 So, S(0.05) ≈ 0.0499998333 and C(-0.25) ≈ -0.0833330799.

Step 4: Part d: Terms required for S(0.05) with error <= 10^-4

Let an be the nth term in the Maclaurin series expansion for S(x). By the alternating series estimation theorem, we know that the error in the approximation is less than the absolute value of the first omitted term, i.e., |(an+1)| < 10^-4. For S(x), let N be the smallest integer such that |aN+1| < 10^-4: |(0.05)^(4(N+1)+3) / ((4(N+1)+3) * (2(N+1))!)| < 10^-4 Using trial and error, we find that N = 1 satisfies this inequality. So, we can approximate S(0.05) with an error less than 10^-4 using at least two terms of the Maclaurin series.

Step 5: Part e: Terms required for C(-0.25) with error <= 10^-6

Similarly, for C(x), let M be the smallest integer such that |aM+1| < 10^-6: |(-0.25)^(4(M+1)+1) / ((4(M+1)+1) * (2(M+1))!)| < 10^-6 Using trial and error, we find that M = 2 satisfies this inequality. So, we can approximate C(-0.25) with an error less than 10^-6 using at least three terms of the Maclaurin series.

Key Concepts

Maclaurin Series

The Maclaurin series is a special case of the Taylor series and is an infinite series that represents a function as a sum of its derivatives at a single point, usually zero. It is a powerful tool in calculus for approximating functions that are difficult to manage otherwise.
\[ f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \ldots \]

For functions like \(\sin(t^2)\) and \(\cos(t^2)\), we substitute \(x\) with \(t^2\) to find their Maclaurin expansions. By doing this, you transform complex trigonometric functions into easier polynomials which are much simpler to integrate or differentiate. The first few terms of these series provide a good approximation of the function over a certain range.

• **\(\sin(t^2)\)** expands to \(t^2 - \frac{t^6}{3!} + \cdots\)
• **\(\cos(t^2)\)** expands to \(1 - \frac{t^4}{2!} + \cdots\)

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is central to calculus. It links the concept of differentiation with integration and gives us a powerful tool to solve problems in this area.
It states that if \( F(x) = \int_{a}^{x} f(t) \, dt \), then \( F'(x) = f(x) \). This theorem allows us to find the derivative of an integral function directly.

When applied to Fresnel integrals like \(S(x) = \int_{0}^{x} \sin(t^2) \, dt\) and \(C(x) = \int_{0}^{x} \cos(t^2) \, dt\), the theorem simplifies the process:

• For \( S'(x) \), you simply take \( \sin(x^2) \).
• For \( C'(x) \), just take \( \cos(x^2) \).

This method saves time and ensures accuracy, especially useful when dealing with complex or cumbersome integrals.

Error Estimation

Error estimation is crucial when using polynomial approximations, especially in mathematical calculations that demand high precision. We often need to know how far our approximation is from the true value.
In cases like approximating Fresnel integrals, the *Alternating Series Estimation Theorem* helps. It states that the error in the approximation is less than the absolute value of the first omitted term in the series.

Here's a quick guideline:

• For \( S(0.05) \), you can check terms of the series until the omitted term is less than \(10^{-4}\).
• For \( C(-0.25) \), ensure that the series term you omit is less than \(10^{-6}\).

By checking through these terms with some trial and error, you ensure your approximation is sufficiently accurate for practical use.

Polynomial Approximation

Polynomial approximation involves expressing functions in simpler polynomial terms, making calculations more feasible. Particularly for functions like sine and cosine, which can be expressed through their Maclaurin series.
This method is often used to approximate values of functions at specific points. By truncating the series, you reduce it to a polynomial, which is much easier to evaluate.

For instance:

• Compute \(S(0.05)\) using the first few terms of the Maclaurin series for \(S(x)\).
• Use the similar approach for \(C(-0.25)\), focusing on the early terms of \(C(x)\).

These shortened polynomial forms make computations of complex integrals not only manageable but quick, offering a balance between simplicity and accuracy. This is especially useful when you're dealing with small intervals or need rapid approximations in fields like physics or engineering.