Engineering Mathematics Test 5

The Fourier series for the function f(x) in the interval α < x < α + 2π is given by

\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos nx + \mathop \sum \limits_{n = 1}^\infty {b_n}\sin nx\)

where

\({a_o} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)dx;\;{a_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\cos nxdx;\;{b_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\sin nxdx\)

An even function is any function f such that f(-x) = f(x)

Example: cos x, sec x, x², x⁴, x⁶ …….., x^-2, x^-4 ……..

An odd function is any function f such that f(-x) = -f(x)

Example: sin x, tan x, cosec x, cot x, n, x³ ……., x^-1, x^-3 ……..

\(\mathop \smallint \limits_{ - L}^L f\left( x \right)dx = \left\{ {\begin{array}{*{20}{c}} {2\mathop \smallint \limits_0^L f\left( x \right)dx,\;\;when\;f\left( x \right)\;is\;an\;even\;function}\\ {0,\;\;when\;f\left( x \right)\;is\;an\;odd\;function} \end{array}} \right.\)

When f is an even periodic function of period 2L, then its Fourier series contains only cosine (include possibly, the constant term) terms.

\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\frac{{\cos n\pi x}}{L}\)

\({a_o} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)dx\)

\({a_n} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)\cos \frac{{n\pi x}}{L}dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)\cos \frac{{n\pi x}}{L}dx\)

When f is an odd periodic function of period 2L, then its Fourier series contains only sine terms.

\(f\left( x \right) = \mathop \sum \limits_{n = 1}^\infty {b_n}\sin \frac{{n\pi x}}{L}\)

\({b_n} = \frac{1}{L}\mathop \smallint \limits_{ - L}^L f\left( x \right)\sin \frac{{n\pi x}}{L}dx = \frac{2}{L}\mathop \smallint \limits_0^L f\left( x \right)\sin \frac{{n\pi x}}{L}dx\)

Concept:

Taylor expansion About 0 = Maclaurin series

f(x) = f(0) + xf’(0) + x²/2! f’’(0) +  …

Calculation:

To find f’’(0) /2!

f(x) = e^2x

f’(x) = 2e^2x

f’’(x) = 2²e^2x

f’’(0) = 4

Concept:

Fourier series of f(x)

\(f\left( x \right) = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos \frac{{n\pi x}}{L} + \mathop \sum \limits_{n = 1}^\infty {b_n}\sin \frac{{n\pi x}}{L}\)

\({a_0} = \frac{1}{l}\;\mathop \smallint \limits_0^{2l} f\left( x \right)dx\)

Calculation:

\(L = 3,\;{a_0} = \frac{1}{3}\mathop \smallint \limits_0^6 f\left( x \right)dx\)

\({a_0} = \frac{1}{3}\left[ {\mathop \smallint \limits_0^2 {x^2}dx + \mathop \smallint \limits_2^6 \left( {6 - x} \right)dx} \right]\)

\({a_0} = \frac{1}{3}\left[ {\frac{8}{3} + 6\left( 4 \right) - \frac{1}{2}\left( {36 - 4} \right)} \right]\)

\({a_0} = \frac{1}{3}\left[ {\frac{{32}}{3}} \right]\)

\(\therefore {a_0} = \frac{{32}}{9}\)

Explanation:

Expansion of some important function:

\(\sin \theta = \theta - \frac{{{\theta ^3}}}{{3!}} + \frac{{{\theta ^5}}}{{5!}} - \frac{{{\theta ^7}}}{{7!}} + \ldots \)

\(\cos \theta = 1 - \frac{{{\theta ^2}}}{{2!}} + \frac{{{\theta ^4}}}{{4!}} - \frac{{{\theta ^6}}}{{6!}} + \ldots \)

\(\tan \theta = \theta + \frac{{{\theta ^3}}}{2} + \frac{{{{2\theta}^5}}}{{15}} + \ldots \)

\({e^x} = 1 + x + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^4}}}{{4!}} + \ldots \)

The given series is

\(\pi - \frac{{{\pi ^3}}}{{3!}} + \frac{{{\pi ^5}}}{{5!}} - \frac{{{\pi ^7}}}{{7!}} + \ldots = \sin \pi = 0\)

Concept:

Taylor series expansion

\(f\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right) + \frac{{f'\left( a \right)}}{2}{\left( {x - a} \right)^2} + \frac{{f'''\left( a \right)}}{{3!}}{\left( {x - a} \right)^3} + \ldots \)

Calculation:

Here a = 3

\(f\left( x \right) = f\left( 3 \right) + f'\left( 3 \right)\left( {x - 3} \right) + \frac{{f'\left( 3 \right)}}{2}{\left( {x - 3} \right)^2} + \frac{{f'''\left( 3 \right)}}{{3!}}{\left( {x - 3} \right)^3} + \ldots \)

For linear approximation, take only first two terms

f(x) = f(3) + f’(3) (x-3)

f(x) = x³ - 10x²+6, f(3) = -57

f’(x) = 3x² - 20x, f’(3) = -33

∴ f(x) = -57 - 33(x-3) = 42 - 33x = 3(14 - 11x)

Explanation:

Try to solve options.

Let us take option 3

Then we can say f(x) = e^{sin x}

F(0) = 1

f’(x) = e^{sin x} cos x

f’(0) = 1

f”(x) = e^{sin x} cos² x – e^{sin x} sin x

f”(0) = 1

Similarly f’’’ (0) = 0

f^iv (0) = -3

Now from Maclaurin’s series of expansion –

\(f\left( x \right)=f\left( 0 \right)+\frac{x}{1!}~{f}'\left( 0 \right)+\frac{{{x}^{2}}}{2!}~{f}''\left( 0 \right)+\frac{{{x}^{3}}}{3!}~{f}'''\left( 0 \right)+\frac{{{x}^{4}}}{4!}f'''\left( 0 \right)\)

\(=1+x\cdot 1+\frac{{{x}^{2}}}{2!}~\left( 1 \right)+\frac{{{x}^{3}}}{3!}~\left( 0 \right)+\frac{{{x}^{4}}}{4!}\left( -3 \right)\)

Concept:

Taylor’s series expansion is given by,

\(f\left( x \right) = f\left( {{x_0}} \right) + f'\left( {{x_0}} \right)\left( {x - {x_0}} \right) + \frac{{f''\left( {{x_0}} \right)}}{{2!}}{\left( {x - {x_0}} \right)^2}\)

Calculation:

f(x) = e^x + sin x, f(π) = e^π

f’(x) = e^x + cos x, f’(π) = e^π – 1

f’’(π) = e^x – sin x, f’’(π) = e^π

\(f\left( \pi \right) = {e^\pi } + \left( {{e^\pi } - 1} \right)\left( {x - \pi } \right) + \frac{{{e^\pi }}}{{2!}}{\left( {x - \pi } \right)^2} + \ldots \)

Concept:

In the interval (-l, l) Fourier series is defined as:-

\(f\left( x \right) = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos \frac{{n\pi x}}{l} + \mathop \sum \limits_{n = 1}^\infty {b_n}\sin \frac{{n\pi x}}{l}\)

Where, \({a_0} = \frac{1}{l}\mathop \smallint \nolimits_{ - l}^l f\left( x \right)dx\)

\({a_n} = \frac{1}{l}\mathop \smallint \nolimits_{ - l}^l f\left( x \right)\cos \frac{{n\pi x}}{l}dx\)

\({b_n} = \frac{1}{l}\mathop \smallint \nolimits_{ - l}^l f\left( x \right)\frac{{\sin n\pi x}}{l}dx\)

Euler Definition: In the interval (-π, π)

\(f\left( x \right) = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos nx + \mathop \sum \limits_{n = 1}^\infty {b_n}\sin x\)

Where,

\({a_o} = \frac{1}{\pi }\mathop \smallint \nolimits_{ - \pi }^\pi f\left( x \right)dx\)

\({a_n} = \frac{1}{\pi }\mathop \smallint \nolimits_{ - \pi }^\pi f\left( x \right)\cos nx\;dx\)

\({b_n} = \frac{1}{\pi }\mathop \smallint \nolimits_{ - \pi }^\pi f\left( x \right)\sin nx\;dx\)

Calculation:

f(x) = (x – x²)

\(\therefore {a_0} = \frac{1}{l}\mathop \smallint \nolimits_{ - l}^l f\left( x \right)dx = \frac{1}{\pi }\mathop \smallint \nolimits_{ - \pi }^\pi \left( {x - {x^2}} \right)dx = \frac{1}{\pi }\left[ {\frac{{{x^2}}}{2} - \frac{{{x^3}}}{3}} \right]_{ - \pi }^\pi \)

Concept:

Taylor expansion about a:

\(f\left( {a + x} \right) = f\left( a \right) + xf'\left( a \right) + \frac{{{x^2}}}{{2!}}f''\left( a \right) + \ldots + \frac{{{x^n}}}{{n!}}{f^n}\left( a \right)\)

Calculation:

f(x) = x² + e^2x

f(1) = 1 + e²

f’(x) = 2x + 2e^2x ⇒ f’(1) = 2 + 2e²

f’’(x) = 2 + 2² e^2x ⇒ f’’(1) = 2 + 2² e²

f’’’(x) = 2³ e^2x ⇒ f’’’(1) = 2³e²

\(\therefore f\left( x \right) = \left( {1 + {e^2}} \right) + x\left( {2 + 2{e^2}} \right) + \frac{{{x^2}}}{{2!}}\left( {2 + 4{e^2}} \right) + \frac{{{x^3}}}{{3!}}\left( {8{e^3}} \right) + \ldots \)

\(\therefore f\left( x \right) = 1 + {e^2} + 2x + 2x{e^2} + \frac{{2{x^2}}}{2} + 2{x^2}{e^2} + \frac{4}{3}{x^3}{e^2} + \ldots \)

\(\therefore f\left( x \right) = \left( {1 + {e^2}} \right)\left( {1 + x} \right) + {x^2}\left( {1 + 2{e^2}} \right) + \frac{{4{x^3}}}{3}{e^2} + \ldots \)

Explanation:

When f is an even periodic function of period 2L, then its Fourier series contains only cosine (include possibly, the constant term) terms.

\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\frac{{\cos n\pi x}}{L}\)

When f is an odd periodic function of period 2L, then its Fourier series contains only sine terms.

\(f\left( x \right) = \mathop \sum \limits_{n = 1}^\infty {b_n}\sin \frac{{n\pi x}}{L}\)

\(\begin{array}{l} {b_n} = \frac{1}{\pi }\mathop \smallint \limits_{ - \pi }^\pi f\left( x \right)\sin nx\,dx\\ = \frac{1}{\pi }\mathop \smallint \limits_{ - \pi }^\pi \sin ax\,\sin nx\,dx = \frac{2}{\pi }\mathop \smallint \limits_0^\pi \sin ax\,\sin nx\,dx = \frac{1}{\pi }\mathop \smallint \limits_0^\pi 2\sin ax\,\sin nx\,dx\\ = \frac{1}{\pi }\mathop \smallint \limits_0^\pi [\cos (n - a)x - \cos (n + a)]dx = \frac{1}{\pi }\mathop \smallint \limits_0^\pi [\cos (n - a)x - \cos (n + a)]dx\\ = \frac{1}{\pi }\left[ {\frac{{\sin (n - a)x}}{{(n - a)}} - \frac{{\sin (n + a)x}}{{(n + a)}}} \right]_0^\pi = \frac{1}{\pi }\left[ {\frac{{\sin (n - a)\pi }}{{(n - a)}} - \frac{{\sin (n + a)\pi }}{{(n + a)}}} \right]\\ \end{array}\)

\(= \frac{1}{\pi }\left[ {\frac{{\sin (n\pi - a\pi )}}{{(n - a)}} - \frac{{\sin (n\pi + a\pi )}}{{(n + a)}}} \right] = \frac{1}{\pi }\left[ {\frac{{{{( - 1)}^n}( - \sin a\pi )}}{{(n - a)}} - \frac{{{{( - 1)}^n}\sin a\pi }}{{(n + a)}}} \right]\\ = - \frac{{{{( - 1)}^n}\sin a\pi }}{\pi }\left[ {\frac{1}{{(n - a)}} + \frac{1}{{(n + a)}}} \right]\\ {b_n} = - \frac{{{{( - 1)}^n}2n.\sin a\pi }}{{\pi ({n^2} - {a^2})}} = \frac{{{{( - 1)}^{n + 1}}2n.\sin a\pi }}{{\pi ({n^2} - {a^2})}}\\ {b_2} = \frac{{{{( - 1)}^3}2(2).\sin a\pi }}{{\pi ({2^2} - {a^2})}} = - \frac{{4\sin a\pi }}{{\pi ({2^2} - {a^2})}}\\ f(x) = \sin ax = \mathop \sum \limits_{n = 1}^\infty {b_n}\sin nx\\ = \frac{{2\sin a\pi }}{\pi }\mathop \sum \limits_{n = 1}^\infty \frac{{{{( - 1)}^{n + 1}}n}}{{({n^2} - {a^2})}}\sin nx\\ f(x) = \frac{{2\sin a\pi }}{\pi }\left[ {\frac{{\sin x}}{{({1^2} - {a^2})}} - \frac{{2\sin 2x}}{{({2^2} - {a^2})}} + \frac{{3\sin 3x}}{{({3^2} - {a^2})}} - ...} \right]\\ \& \,{b_2} = - \frac{{4\sin a\pi }}{{\pi ({2^2} - {a^2})}}\)