Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

[M] Let \({f_{\bf{4}}}\) and \({f_{\bf{5}}}\) be the fourth-order and fifth order Fourier approximations in \(C\left[ {{\bf{0}},{\bf{2}}\pi } \right]\) to the square wave function in Exercise 10. Produce separate graphs of \({f_{\bf{4}}}\) and \({f_{\bf{5}}}\) on the interval \(\left[ {{\bf{0}},{\bf{2}}\pi } \right]\), and produce graph of \({f_{\bf{5}}}\) on \(\left[ { - {\bf{2}}\pi ,{\bf{2}}\pi } \right]\).

Short Answer

Expert verified

The figure below represents the graph of the curve \({f_4}\) in the interval \(\left[ {0,2\pi } \right]\).

The figure below represents the graph of the curve \({f_5}\) in the interval \(\left[ {0,2\pi } \right]\).

The figure below represents the graph of the curve \({f_5}\) in the interval \(\left[ { - 2\pi ,2\pi } \right]\).

Step by step solution

01

Find the Fourier coefficients \({a_{\bf{0}}}\)

The coefficients \({a_0}\) can be calculated as follows:

\(\begin{array}{c}\frac{{{a_0}}}{2} = \frac{1}{{2\pi }}\int_0^{2\pi } {f\left( t \right){\rm{d}}t} \\ = \frac{1}{{2\pi }}\left[ {\pi - \left( {2\pi - \pi } \right)} \right]\\ = 0\end{array}\)

02

F Find the Fourier coefficients \({a_k}\)

The coefficients \({a_k}\) can be calculated as follows:

\(\begin{array}{c}{a_k} = \frac{1}{\pi }\int_0^{2\pi } {f\left( t \right)\cos ktdt} \\ = \frac{1}{\pi }\int_0^\pi {\cos kt{\rm{d}}t} - \frac{1}{\pi }\int_\pi ^{2\pi } {\cos kt{\rm{d}}t} \\ = \frac{1}{\pi }\left\{ {\left[ {\frac{{\sin kt}}{k}} \right]_0^\pi - \left[ {\frac{{\sin kt}}{k}} \right]_\pi ^{2\pi }} \right\}\\ = 0\end{array}\)

03

Find the Fourier coefficients \({b_k}\)

The coefficients \({b_k}\) can be calculated as follows:

\(\begin{array}{c}{a_k} = \frac{1}{\pi }\int_0^{2\pi } {f\left( t \right)\sin ktdt} \\ = \frac{1}{\pi }\int_0^\pi {\sin kt{\rm{d}}t} - \frac{1}{\pi }\int_\pi ^{2\pi } {\sin kt{\rm{d}}t} \\ = \frac{1}{\pi }\left\{ {\left[ { - \frac{{\cos kt}}{k}} \right]_0^\pi + \left[ {\frac{{\cos kt}}{k}} \right]_\pi ^{2\pi }} \right\}\\ = \frac{2}{{\pi k}} - \frac{2}{{\pi k}}\cos k\pi \\ = \left\{ {\begin{array}{*{20}{c}}{\frac{4}{{k\pi }}}&{{\rm{if}}\;{\rm{k}}\,{\rm{is odd}}}\\0&{{\rm{if}}\;{\rm{k}}\,{\rm{is even}}}\end{array}} \right.\end{array}\)

04

Write the Fourier function and plot the graph of

The Fourier function can be written as follows:

\({f_4}\left( t \right) = \frac{4}{\pi }\sin t + \frac{4}{{3\pi }}\sin 3t\)

And,

\({f_5}\left( t \right) = \frac{4}{\pi }\sin t + \frac{4}{{3\pi }}\sin 3t + \frac{4}{{5\pi }}\sin 5t\)

The figure below represents the graph of the curve \({f_4}\) in the interval \(\left[ {0,2\pi } \right]\).

The figure below represents the graph of the curve \({f_5}\) in the interval \(\left[ {0,2\pi } \right]\).

The figure below represents the graph of the curve \({f_5}\) in the interval \(\left[ { - 2\pi ,2\pi } \right]\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises 13 and 14, the columns of Q were obtained by applying the Gram-Schmidt process to the columns of A. Find an upper triangular matrix R such that \(A = QR\). Check your work.

13. \(A = \left( {\begin{aligned}{{}{}}5&9\\1&7\\{ - 3}&{ - 5}\\1&5\end{aligned}} \right),{\rm{ }}Q = \left( {\begin{aligned}{{}{}}{\frac{5}{6}}&{ - \frac{1}{6}}\\{\frac{1}{6}}&{\frac{5}{6}}\\{ - \frac{3}{6}}&{\frac{1}{6}}\\{\frac{1}{6}}&{\frac{3}{6}}\end{aligned}} \right)\)

Determine which pairs of vectors in Exercises 15-18 are orthogonal.

15. \({\mathop{\rm a}\nolimits} = \left( {\begin{aligned}{*{20}{c}}8\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm b}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 3}\end{aligned}} \right)\)

Given data for a least-squares problem, \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\), the following abbreviations are helpful:

\(\begin{aligned}{l}\sum x = \sum\nolimits_{i = 1}^n {{x_i}} ,{\rm{ }}\sum {{x^2}} = \sum\nolimits_{i = 1}^n {x_i^2} ,\\\sum y = \sum\nolimits_{i = 1}^n {{y_i}} ,{\rm{ }}\sum {xy} = \sum\nolimits_{i = 1}^n {{x_i}{y_i}} \end{aligned}\)

The normal equations for a least-squares line \(y = {\hat \beta _0} + {\hat \beta _1}x\)may be written in the form

\(\begin{aligned}{{\hat \beta }_0} + {{\hat \beta }_1}\sum x = \sum y \\{{\hat \beta }_0}\sum x + {{\hat \beta }_1}\sum {{x^2}} = \sum {xy} {\rm{ (7)}}\end{aligned}\)

16. Use a matrix inverse to solve the system of equations in (7) and thereby obtain formulas for \({\hat \beta _0}\) , and that appear in many statistics texts.

In exercises 1-6, determine which sets of vectors are orthogonal.

\(\left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\0\\3\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 4}\\1\\{ - 3}\\8\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}3\\3\\5\\{ - 1}\end{array}} \right]\)

Question: In Exercises 1 and 2, you may assume that\(\left\{ {{{\bf{u}}_{\bf{1}}},...,{{\bf{u}}_{\bf{4}}}} \right\}\)is an orthogonal basis for\({\mathbb{R}^{\bf{4}}}\).

2.\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{2}}}\\{\bf{1}}\\{ - {\bf{1}}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{3}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\\{ - {\bf{1}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{4}}} = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({\bf{x}} = \left[ {\begin{aligned}{\bf{4}}\\{\bf{5}}\\{ - {\bf{3}}}\\{\bf{3}}\end{aligned}} \right]\)

Write v as the sum of two vectors, one in\({\bf{Span}}\left\{ {{{\bf{u}}_1}} \right\}\)and the other in\({\bf{Span}}\left\{ {{{\bf{u}}_2},{{\bf{u}}_3},{{\bf{u}}_{\bf{4}}}} \right\}\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free