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Chapter 11: Problem 2

Let \(f(x)=x\) for \(0

Short Answer

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#Summary# To solve the problem, we first found the coefficients \(b_m\) using the inner product, and then we defined and calculated the mean square error R_n for several values of n. Lastly, we determined the smallest value of n for which the condition \(R_n<0.01\) is satisfied. The process involves integration by parts and working with orthogonal functions. The main goal is to find an approximation of the given function f(x) in terms of orthogonal functions with minimal mean square error.

Step by step solution

01

Find \(b_{m}\) using the Inner Product

The coefficients \(b_{m}\) can be found using the inner product of \(f(x)\) and \(\phi_{m}(x)\) with \(0<x<1\). \begin{equation} b_{m} = \langle f(x),\phi_{m}(x) \rangle = \int\limits_{0}^{1} f(x)\phi_{m}(x) dx \end{equation} Using the given functions: \begin{equation} b_{m} = \int\limits_{0}^{1} x\sqrt{2}\sin(m\pi x) dx \end{equation} To evaluate this integral, we'll use integration by parts: Let \(u = x\), \(dv = \sqrt{2}\sin(m\pi x)dx\) Then, \(du = dx\), \(v = -\frac{\sqrt{2}}{m\pi}\cos(m\pi x)\) Now, Integration by parts: \begin{equation} b_{m} = -\frac{\sqrt{2}}{m\pi}\left[x\cos(m\pi x)\right]_{0}^1 + \frac{\sqrt{2}}{m\pi}\int\limits_{0}^{1}\cos(m\pi x) dx \end{equation} Solving the integral results in: \begin{equation} b_{m} = -\frac{\sqrt{2}}{m\pi}\left[\cos(m\pi)-0\right] + \frac{2}{m^{2}\pi^{2}}\left[\sin(m\pi x)\right]_{0}^{1} \end{equation} As \(\sin(m\pi)=0\) for all integer values of \(m\): \begin{equation} b_{m} = -\frac{\sqrt{2}}{m\pi}\cos(m\pi) \end{equation}
02

Define and Calculate Mean Square Error \(R_{n}\)

The mean square error, \(R_{n}\), is given by the expression: \begin{equation} R_{n} = \int\limits_{0}^{1} \left[f(x) - \sum_{m=1}^{n} b_m \phi_{m}(x)\right]^2 dx \end{equation} To find the mean square error for different values of \(n\), calculate the approximated function \(f_n(x) = \sum_{m=1}^{n} b_m \phi_{m}(x)\). Then plug it into the expression for \(R_{n}\).
03

Determine the smallest value of \(n\) for given condition

We need to find the smallest value of \(n\) such that \(R_{n} < 0.01\). To do this, iterate through increasing values of \(n\) and keep calculating \(R_{n}\) until the condition is met.

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Key Concepts

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

Inner Product
In the context of Fourier series, the inner product plays a crucial role in determining the coefficients of the series expansion. The inner product is essentially a way of multiplying two functions to measure their "overlap". For functions defined over an interval, this is calculated as the integral of the product of these functions over that interval.

For example, if we wish to find the inner product of two functions \( f(x) \) and \( \phi(x) \), it is given by:\[\langle f, \phi \rangle = \int_{a}^{b} f(x) \phi(x) \, dx\]
  • The limits \( a \) and \( b \) denote the interval over which we integrate.
  • This concept generalizes the notion of the dot product from vectors to functions, helping us extract coefficients like \( b_m \) in a Fourier series.
By using the inner product, one can find how much of each basis function \( \phi_m(x) \) exists within \( f(x) \), which is crucial for expressing \( f(x) \) as a series sum of these basis functions. This approach significantly aids in transforming functions into easier-to-handle components.
Mean Square Error
The mean square error (MSE) is an important metric used to measure how well an approximated function represents the original function. In the context of Fourier series, we aim to approximate a function, \( f(x) \), using a finite number of terms. The MSE, denoted as \( R_n \), quantifies the error involved in using this approximation.

The formula for MSE when using \( n \) terms is given as follows:\[R_{n} = \int_{0}^{1} \left(f(x) - \sum_{m=1}^{n} b_m \phi_{m}(x)\right)^{2} \, dx\]
  • \( f(x) \) is the original function.
  • \( f_n(x) = \sum_{m=1}^{n} b_m \phi_{m}(x) \) is the approximation of \( f(x) \) using the first \( n \) terms.
  • The smaller the value of \( R_n \), the better the approximation, indicating a smaller difference between \( f(x) \) and \( f_n(x) \).
To compute the MSE for varying \( n \), we adjust the number of series terms and observe the accuracy of our representation. By decreasing \( R_n \) below a threshold like 0.01, we determine the efficiency of our approximations.
Integration by Parts
Integration by parts is a powerful technique used to solve integrals, especially when integrating the product of two functions. This method is based on the product rule for differentiation and is expressed in the self-evident formula:\[\int u \, dv = uv - \int v \, du\]
  • \( u \) and \( v \) are differentiable functions of a variable \( x \).
  • Choose \( u \) and \( dv \) such that \( du \) and \( v \) can be easily computed.

In the provided exercise, integration by parts is employed to find the expression for \( b_m \), requiring selections for \( u \) and \( dv \) that facilitate an easier evaluation. For the specific case:
  • Set \( u = x \) (differentiates simply to \( du = dx \)).
  • Choose \( dv = \sqrt{2}\sin(m\pi x)dx \), which integrates to \( v = -\frac{\sqrt{2}}{m\pi}\cos(m\pi x) \).

By applying integration by parts, complications in the integral are simplified, allowing us to obtain the necessary coefficients and further progress the Fourier series analysis.

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Most popular questions from this chapter

Consider the boundary value problem $$ r(x) u_{t}=\left[p(x) u_{x}\right]_{x}-q(x) u+F(x) $$ $$ u(0, t)=T_{1}, \quad u(1, t)=T_{2}, \quad u(x, 0)=f(x) $$ (a) Let \(v(x)\) be a solution of the problem $$ \left[p(x) v^{\prime}\right]-q(x) v=-F(x), \quad v(0)=T_{1}, \quad v(1)=T_{2} $$ If \(w(x, t)=u(x, t)-v(x),\) find the boundary value problem satisfied by \(w\), Note that this problem can be solved by the method of this section. (b) Generalize the procedure of part (a) to the case \(u\) satisfies the boundary conditions $$ u_{x}(0, t)-h_{1} u(0, t)=T_{1}, \quad u_{x}(1, t)+h_{2} u(1, t)=T_{2} $$

In the circular cylindrical coordinates \(r, \theta, z\) defined by $$ x=r \cos \theta, \quad y=r \sin \theta, \quad z=z $$ Laplace's equation is $$ u_{r r}+(1 / r) u_{r}+\left(1 / r^{2}\right) u_{\theta \theta}+u_{z z}=0 $$ (a) Show that if \(u(r, \theta, z)=R(r) \Theta(\theta) Z(z),\) then \(R, \Theta,\) and \(Z\) satisfy the ordinary differential equations $$ \begin{aligned} r^{2} R^{\prime \prime}+r R^{\prime}+\left(\lambda^{2} r^{2}-n^{2}\right) R &=0 \\ \Theta^{\prime \prime}+n^{2} \Theta &=0 \\\ Z^{\prime \prime}-\lambda^{2} Z &=0 \end{aligned} $$ (b) Show that if \(u(r, \theta, z)\) is independent of \(\theta,\) then the first equation in part (a) becomes $$ r^{2} R^{\prime \prime}+r R^{\prime}+\lambda^{2} r^{2} R=0 $$ the second is omitted altogether, and the third is unchanged.

Consider the problem $$ y^{\prime \prime}+\lambda y=0, \quad y(0)=0, \quad y^{\prime}(L)=0 $$ $$ \begin{array}{l}{\text { Show that if } \phi_{\infty} \text { and } \phi_{n} \text { are eigenfunctions, corresponding to the eigenvalues } \lambda_{m} \text { and } \lambda_{n},} \\ {\text { respectively, with } \lambda_{m} \neq \lambda_{n} \text { , then }}\end{array} $$ $$ \int_{0}^{L} \phi_{m}(x) \phi_{n}(x) d x=0 $$ $$ \text { Hint. Note that } $$ $$ \phi_{m}^{\prime \prime}+\lambda_{m} \phi_{m}=0, \quad \phi_{n}^{\prime \prime}+\lambda_{n} \phi_{n}=0 $$ $$ \begin{array}{l}{\text { Multiply the first of these equations by } \phi_{n}, \text { the second by } \phi_{m}, \text { and integrate from } 0 \text { to } L,} \\ {\text { using integration by parts. Finally, subtract one equation from the other. }}\end{array} $$

In this problem we consider a higher order eigenvalue problem. In the study of transverse vibrations of a uniform elastic bar one is led to the differential equation $$ y^{\mathrm{w}}-\lambda y=0 $$ $$ \begin{array}{l}{\text { where } y \text { is the transverse displacement and } \lambda=m \omega^{2} / E I ; m \text { is the mass per unit length of }} \\\ {\text { the rod, } E \text { is Young's modulus, } I \text { is the moment of inertia of the cross section about an }} \\ {\text { axis through the centroid perpendicular to the plane of vibration, and } \omega \text { is the frequency of }} \\ {\text { vibration. Thus for a bar whose material and geometric properties are given, the eigenvalues }} \\ {\text { determine the natural frequencies of vibration. Boundary conditions at each end are usually }} \\ {\text { one of the following types: }}\end{array} $$ $$ \begin{aligned} y=y^{\prime} &=0, \quad \text { clamped end } \\ y=y^{\prime \prime} &=0, \quad \text { simply supported or hinged end, } \\ y^{\prime \prime}=y^{\prime \prime \prime} &=0, \quad \text { free end } \end{aligned} $$ $$ \begin{array}{l}{\text { For each of the following three cases find the form of the eigenfunctions and the equation }} \\ {\text { satisfied by the eigenvalues of this fourth order boundary value problem. Determine } \lambda_{1} \text { and }} \\ {\lambda_{2}, \text { the two eigenvalues of smallest magnitude. Assume that the eigenvalues are real and }} \\ {\text { positive. }}\end{array} $$ $$ \begin{array}{ll}{\text { (a) } y(0)=y^{\prime \prime}(0)=0,} & {y(L)=y^{\prime \prime}(L)=0} \\ {\text { (b) } y(0)=y^{\prime \prime}(0)=0,} & {y(L)=y^{\prime \prime}(L)=0} \\ {\text { (c) } y(0)=y^{\prime}(0)=0,} & {y^{\prime \prime}(L)=y^{\prime \prime \prime}(L)=0 \quad \text { (cantilevered bar) }}\end{array} $$

The wave equation in polar coordinates is $$ u_{r r}+(1 / r) u_{r}+\left(1 / r^{2}\right) u_{\theta \theta}=a^{-2} u_{t t} $$ Show that if \(u(r, \theta, t)=R(r) \Theta(\theta) T(t),\) then \(R, \Theta,\) and \(T\) satisfy the ordinary differential equations $$ \begin{aligned} r^{2} R^{\prime \prime}+r R^{\prime}+\left(\lambda^{2} r^{2}-n^{2}\right) R &=0 \\ \Theta^{\prime \prime}+n^{2} \Theta &=0 \\\ T^{\prime \prime}+\lambda^{2} a^{2} T &=0 \end{aligned} $$
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