Problem 15 - Advanced Engineering Mathematics

Question

We compute $$\begin{array}{l} c_{0}=\frac{1}{2} \int_{0}^{1} x P_{0}(x) d x=\frac{1}{2} \int_{0}^{1} x d x=\frac{1}{4} \\ c_{1}=\frac{3}{2} \int_{0}^{1} x P_{1}(x) d x=\frac{3}{2} \int_{0}^{1} x^{2} d x=\frac{1}{2} \\ c_{2}=\frac{5}{2} \int_{0}^{1} x P_{2}(x) d x=\frac{5}{2} \int_{0}^{1} \frac{1}{2}\left(3 x^{3}-x\right) d x=\frac{5}{16} \\ c_{3}=\frac{7}{2} \int_{0}^{1} x P_{3}(x) d x=\frac{7}{2} \int_{0}^{1} \frac{1}{2}\left(5 x^{4}-3 x^{2}\right) d x=0 \end{array}$$ $$\begin{array}{l} c_{4}=\frac{9}{2} \int_{0}^{1} x P_{4}(x) d x=\frac{9}{2} \int_{0}^{1} \frac{1}{8}\left(35 x^{5}-30 x^{3}+3 x\right) d x=-\frac{3}{32} \\ c_{5}=\frac{11}{2} \int_{0}^{1} x P_{5}(x) d x=\frac{11}{2} \int_{0}^{1} \frac{1}{8}\left(63 x^{6}-70 x^{4}+15 x^{2}\right) d x=0 \\ c_{6}=\frac{13}{2} \int_{0}^{1} x P_{6}(x) d x=\frac{13}{2} \int_{0}^{1} \frac{1}{16}\left(231 x^{7}-315 x^{5}+105 x^{3}-5 x\right) d x=\frac{13}{256}. \end{array}$$ Thus $$f(x)=\frac{1}{4} P_{0}(x)+\frac{1}{2} P_{1}(x)+\frac{5}{16} P_{2}(x)-\frac{3}{32} P_{4}(x)+\frac{13}{256} P_{6}(x)+\cdots.$$ The figure above is the graph of \(S_{5}(x)=\frac{1}{4} P_{0}(x)+\frac{1}{2} P_{1}(x)+\frac{5}{16} P_{2}(x)-\frac{3}{32} P_{4}(x)+\frac{13}{256} P_{6}(x)\).

Step-by-Step Solution

Answer

The polynomial \(S_5(x)\) approximates \(f(x)\) using calculated coefficients.

Step 1: Understanding the Problem

The task involves computing coefficients \(c_0, c_1, ..., c_6\) of the Legendre polynomials \(P_n(x)\) series expansion of some function \(f(x)\). Each coefficient is obtained by evaluating a definite integral of the form \(\frac{2n+1}{2}\int_{0}^{1} x P_n(x) dx\).

Step 2: Compute Each Coefficient

Each coefficient is computed by integrating its respective Legendre polynomial weighted by \(x\):\[c_{0} = \frac{1}{4},\ c_{1} = \frac{1}{2},\ c_{2} = \frac{5}{16},\ c_{3} = 0,\ c_{4} = -\frac{3}{32},\ c_{5} = 0,\ c_{6} = \frac{13}{256}.\]

Step 3: Construct the Series

Using the coefficients found, the function \(f(x)\) can be expressed as a series in terms of Legendre polynomials:\[f(x) = \frac{1}{4} P_{0}(x) + \frac{1}{2} P_{1}(x) + \frac{5}{16} P_{2}(x) - \frac{3}{32} P_{4}(x) + \frac{13}{256} P_{6}(x) + \cdots.\]

Step 4: Identify the Approximating Polynomial

The polynomial \(S_5(x)\) representing a truncated series up to \(P_6(x)\) is given by the same terms without the remainder parts: \[S_5(x) = \frac{1}{4} P_{0}(x) + \frac{1}{2} P_{1}(x) + \frac{5}{16} P_{2}(x) - \frac{3}{32} P_{4}(x) + \frac{13}{256} P_{6}(x).\]

Key Concepts

Series Expansion

In mathematics, series expansion is a tool used to express a function as a sum of simpler terms, often infinite. For our purpose, we are focusing on expressing some function \( f(x) \) in terms of Legendre polynomials, which are a series of orthogonal polynomials. This means they form the building blocks for approximating complex functions.

The task involves finding coefficients \( c_0, c_1, ..., c_6 \) which align with each Legendre polynomial \( P_n(x) \). These coefficients are essential as they weight the contribution of each polynomial in representing the overall function.

In our exercise, this series expansion is carried out using the formula:

• \[ c_n = \frac{2n+1}{2} \int_{0}^{1} x P_n(x) \, dx \]

This formula considers the integral of the function \( x \) multiplied by the Legendre polynomial over the defined interval \([0, 1]\). The calculation of these coefficients allows for a piecewise approximation of \( f(x) \).

Overall, series expansion simplifies complex expressions by breaking them into manageable parts dictated by an infinite or finite series, guided by these essential coefficients.

Definite Integrals

Definite integrals are a fundamental concept in calculus allowing us to calculate the exact "accumulated" value of a function over a particular interval. In our exercise, definite integrals are utilized to compute each of the coefficients \( c_n \) for the Legendre polynomial series expansion.

To compute \( c_n \), we integrate each Legendre polynomial \( P_n(x) \) multiplied by the function \( x \) over the interval \([0, 1]\). This not only helps in obtaining the necessary coefficients but also leverages the special property of Legendre polynomials' orthogonality. The integral becomes:

• \[ c_n = \frac{2n+1}{2} \int_{0}^{1} x P_n(x) \, dx \]

Using this integration technique, each polynomial's effect is isolated and scaled properly, ensuring the function approximation aligns closely with the behavior of \( f(x) \) within the given interval.

Overall, the use of definite integrals in this context provides the means to gauge the function's weight for each polynomial component, offering a precise approximation through calculus.

Approximating Polynomials

Approximating polynomials are employed to approximate a function over a particular range, making complex or otherwise formidable functions easier to analyze. In our exercise, the function is approximated through a truncated series of Legendre polynomials, where the polynomial \( S_5(x) \) is derived by excluding higher-order terms beyond \( P_6(x) \).

The approximating polynomial in our case, \( S_5(x) \), combines several weighted Legendre polynomials of lower orders:

• \[ S_5(x) = \frac{1}{4} P_{0}(x) + \frac{1}{2} P_{1}(x) + \frac{5}{16} P_{2}(x) - \frac{3}{32} P_{4}(x) + \frac{13}{256} P_{6}(x). \]

Each term in this polynomial has a coefficient that determines its contribution to the entire function. By truncating the series – essentially ignoring higher-order polynomials – we achieve a simpler function that's still quite close to \( f(x) \).

Ultimately, approximating polynomials serve to transform complex functions into something manageable and analytically friendly, using fewer terms to still retain essential characteristics of the original function within a desired accuracy.