Question 1 |
The Fourier series to represent x-x^2 for - \pi \leq x\leq \pi is given by
x-x^2=\frac{a_0}{2}+\sum_{n=1}^{\infty }a_n \cos nx +\sum_{n=1}^{\infty }b_n \sin nx
The value of a_0 (round off to two decimal places), is _______.
x-x^2=\frac{a_0}{2}+\sum_{n=1}^{\infty }a_n \cos nx +\sum_{n=1}^{\infty }b_n \sin nx
The value of a_0 (round off to two decimal places), is _______.
-1.52 | |
2.12 | |
-6.58 | |
5.23 |
Question 1 Explanation:
\begin{aligned} a_0&=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx\\ &=\frac{1}{\pi}\int_{-\pi}^{\pi} (x-x^2)dx\\ &=\frac{-1}{\pi}\int_{0}^{\pi}2x^2 dx\\ &=\frac{-1}{\pi}\left ( \frac{2x^3}{3} \right )_0^{\pi}\\ &=-\frac{2}{3 \pi}[\pi^3]=\frac{-2 \pi^2}{3}\\ &=-6.58 \end{aligned}
Question 2 |
The Laplace transform of \sinh (at) is
\frac{a}{s^2-a^2} | |
\frac{a}{s^2+a^2} | |
\frac{s}{s^2-a^2} | |
\frac{s}{s^2+a^2} |
Question 2 Explanation:
L(\sinh (at))=\frac{a}{s^2-a^2}
Question 3 |
Consider a two-dimensional flow through isotropic soil along x direction and z direction. If h is the hydraulic head, the Laplace's equation of continuity is expressed as
\frac{\partial h}{\partial x}+\frac{\partial h}{\partial z}=0 | |
\frac{\partial h}{\partial x}+\frac{\partial h}{\partial x}\frac{\partial h}{\partial z}+\frac{\partial h}{\partial z}=0 | |
\frac{\partial^2 h}{\partial x^2}+\frac{\partial^2 h}{\partial z^2}=0 | |
\frac{\partial^2 h}{\partial x^2}+\frac{\partial^2 h}{\partial x \partial z}+\frac{\partial^2 h}{\partial z^2}=0 |
Question 3 Explanation:
The Laplace's equation of continuity for two dimensional flow in a soil is expressed as:
k_x\frac{\partial^2 h}{\partial x^2}+k_z\frac{\partial^2 h}{\partial z^2}=0... for anisotropic soil [k_x\neq k_z]
:
\frac{\partial^2 h}{\partial x^2}+\frac{\partial^2 h}{\partial z^2}=0... for isotropic soil [k_x = k_z]
k_x\frac{\partial^2 h}{\partial x^2}+k_z\frac{\partial^2 h}{\partial z^2}=0... for anisotropic soil [k_x\neq k_z]
:
\frac{\partial^2 h}{\partial x^2}+\frac{\partial^2 h}{\partial z^2}=0... for isotropic soil [k_x = k_z]
Question 4 |
The laplace transform F(s) of the exponential function ,f(t)=e^{at}\; when \; t\geq 0, where a is a constant and (s-a)\gt 0, is
\frac{1}{s+a} | |
\frac{1}{s-a} | |
\frac{1}{a-s} | |
\infty |
Question 4 Explanation:
\begin{aligned} L(e^{at}) &=\frac{1}{s-a} \\ L(e^{at}) &=\int_{0}^{\infty }e^{-st}e^{at}dt \\ &= \int_{0}^{\infty }e^{-(s-a)t}dt\\ &= \left.\begin{matrix} \frac{e^{-(s-a)t}}{-(s-a)} \end{matrix}\right|_{0}^{\infty }\\ &=-\frac{1}{s-a}(0-1)\\ &=\frac{1}{s-a} \end{aligned}
Question 5 |
The Fourier series of the function,
\begin{matrix} f(x) & =0 & -\pi \lt x \leq 0 \\ f(x) &=\pi-x & 0 \lt x \lt \pi \end{matrix}
in the interval [-\pi ,\pi ] is
f(x)=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{\cos x}{1^{2}}+\frac{\cos 3x}{3^{2}}+\cdots\: \cdots \ \cdots \right ] + \left [ \frac{\sin x}{1}+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+\cdots \: \cdots\: \cdot \right ]
The convergence of the above Fourier series at x = 0 gives
\begin{matrix} f(x) & =0 & -\pi \lt x \leq 0 \\ f(x) &=\pi-x & 0 \lt x \lt \pi \end{matrix}
in the interval [-\pi ,\pi ] is
f(x)=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{\cos x}{1^{2}}+\frac{\cos 3x}{3^{2}}+\cdots\: \cdots \ \cdots \right ] + \left [ \frac{\sin x}{1}+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+\cdots \: \cdots\: \cdot \right ]
The convergence of the above Fourier series at x = 0 gives
\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6} | |
\sum_{n=1}^{\infty }\frac{(-1)^{n+1}}{n^{2}}=\frac{\pi ^{2}}{12} | |
\sum_{n=1}^{\infty }\frac{1}{(2n-1)^{2}}=\frac{\pi ^{2}}{8} | |
\sum_{n=1}^{\infty }\frac{(-1)^{n+1}}{(2n-1)}=\frac{\pi}{4} |
Question 5 Explanation:
The function is f(x)=0
-p\lt x\leq 0
=p-x,\, 0 \lt x \lt \pi
And Fourier series is,
f\left ( x \right )=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{\cos x}{1^{2}}+\frac{\cos 3x}{3^{2}}+\frac{\cos 5x}{5^{2}}+... \right ]+\left [ \frac{\sin x}{1}+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+... \right ] ...\left ( i \right )
At x=0, (a point of discontinuity), the fourier series converges to \frac{1}{2}\left [ f\left ( 0^{-1} \right )+f\left ( 0^{+} \right ) \right ]
where f\left ( 0^{-} \right )=\lim_{x\rightarrow 0}\left ( \pi -x \right )=\pi
Hence, eq. (i), we get,
\frac{\pi }{2}=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{1}{1^{2}}+\frac{1}{3^{2}}+... \right ]
\Rightarrow \;\; \frac{1}{1}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+...\frac{\pi ^{2}}{8}
-p\lt x\leq 0
=p-x,\, 0 \lt x \lt \pi
And Fourier series is,
f\left ( x \right )=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{\cos x}{1^{2}}+\frac{\cos 3x}{3^{2}}+\frac{\cos 5x}{5^{2}}+... \right ]+\left [ \frac{\sin x}{1}+\frac{\sin 2x}{2}+\frac{\sin 3x}{3}+... \right ] ...\left ( i \right )
At x=0, (a point of discontinuity), the fourier series converges to \frac{1}{2}\left [ f\left ( 0^{-1} \right )+f\left ( 0^{+} \right ) \right ]
where f\left ( 0^{-} \right )=\lim_{x\rightarrow 0}\left ( \pi -x \right )=\pi
Hence, eq. (i), we get,
\frac{\pi }{2}=\frac{\pi }{4}+\frac{2}{\pi }\left [ \frac{1}{1^{2}}+\frac{1}{3^{2}}+... \right ]
\Rightarrow \;\; \frac{1}{1}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+...\frac{\pi ^{2}}{8}
Question 6 |
The infinite series 1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+... corresponds to
sec x | |
e^{x} | |
cos x | |
1+sin^{2}x |
Question 6 Explanation:
e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}...
(By McLaurin's series expansion)
(By McLaurin's series expansion)
Question 7 |
Laplace transform for the function f (x) = cosh(ax) is
\frac{a}{s^{2}-a^{2}} | |
\frac{s}{s^{2}-a^{2}} | |
\frac{a}{s^{2}+a^{2}} | |
\frac{s}{s^{2}+a^{2}} |
Question 7 Explanation:
It is a standard result that
L\left ( \cosh at \right )=\frac{s}{s^{2}-a^{2}}
L\left ( \cosh at \right )=\frac{s}{s^{2}-a^{2}}
Question 8 |
The summation of series S=2+\frac{5}{2}+\frac{8}{2^{2}}+\frac{11}{2^{3}}+....\infty is
4.5 | |
6 | |
6.75 | |
10 |
Question 8 Explanation:
S is an arithmetico-geometric series and can be summed up as follows,
\begin{aligned} S&=2+\left ( \frac{5}{2}+\frac{8}{2^{2})}+\frac{11}{2^{3}}+...\infty \right ) \\ &=2+x\;\; (say)\\ \text{where } x&=\frac{5}{2}+\frac{8}{2^{2}}+\frac{11}{2^{3}}+...\infty \;\; ...(i)\\ \text{Multiplying by }&\text{1/2, we get, } \\ \frac{1}{2}x&=\frac{5}{2^{2}}+\frac{8}{2^{3}}+\frac{11}{2^{4}}+...\infty \;\;...(ii) \\ \text{Substracting (ii) } & \text{from (i), we get, }\\ x-\frac{1}{2}x&=\frac{5}{2}+\frac{\left ( 8-5 \right )}{2^{2}} \\ &+\frac{\left ( 11-8 \right )}{2^{3}} +\frac{\left ( 13-11 \right )}{2^{4}}+...\infty \\ \frac{x}{2}&=\frac{5}{2}+\left ( \frac{3}{2^{2}}+\frac{3}{2^{3}}+\frac{3}{2^{4}}+...\infty \right ) \\ \frac{x}{2}&=\frac{5}{2}+\frac{\frac{3}{2^{2}}}{\left [ 1-\frac{1}{2} \right ]} \\ \frac{x}{2}&=\frac{5}{2}+\frac{3}{2}=4 \\ \therefore \;\; x&=8\\ \text{and } S&=2+x=10\end{aligned}
\begin{aligned} S&=2+\left ( \frac{5}{2}+\frac{8}{2^{2})}+\frac{11}{2^{3}}+...\infty \right ) \\ &=2+x\;\; (say)\\ \text{where } x&=\frac{5}{2}+\frac{8}{2^{2}}+\frac{11}{2^{3}}+...\infty \;\; ...(i)\\ \text{Multiplying by }&\text{1/2, we get, } \\ \frac{1}{2}x&=\frac{5}{2^{2}}+\frac{8}{2^{3}}+\frac{11}{2^{4}}+...\infty \;\;...(ii) \\ \text{Substracting (ii) } & \text{from (i), we get, }\\ x-\frac{1}{2}x&=\frac{5}{2}+\frac{\left ( 8-5 \right )}{2^{2}} \\ &+\frac{\left ( 11-8 \right )}{2^{3}} +\frac{\left ( 13-11 \right )}{2^{4}}+...\infty \\ \frac{x}{2}&=\frac{5}{2}+\left ( \frac{3}{2^{2}}+\frac{3}{2^{3}}+\frac{3}{2^{4}}+...\infty \right ) \\ \frac{x}{2}&=\frac{5}{2}+\frac{\frac{3}{2^{2}}}{\left [ 1-\frac{1}{2} \right ]} \\ \frac{x}{2}&=\frac{5}{2}+\frac{3}{2}=4 \\ \therefore \;\; x&=8\\ \text{and } S&=2+x=10\end{aligned}
Question 9 |
The Fourier series expansion of a symmetric and even function, f(x) where \begin{matrix} f(x) &=1+(2x/\pi), & \; -\pi \lt x \lt 0 \\ &=1-(2x/\pi), & \; 0 \lt x \lt \pi \end{matrix}
will be
will be
\sum_{n=1}^{\infty } \frac{4}{\pi ^{2}n^{2}}(1+cos n\pi ) | |
\sum_{n=1}^{\infty } \frac{4}{\pi ^{2}n^{2}}(1-cos n\pi ) | |
\sum_{n=1}^{\infty } \frac{4}{\pi ^{2}n^{2}}(1-sin n\pi ) | |
\sum_{n=1}^{\infty } \frac{4}{\pi ^{2}n^{2}}(1+sin n\pi ) |
Question 9 Explanation:
Since given function is symmetric and even, its Fourier series contain only cosine terms.
\begin{aligned} a_{0}&=\frac{2}{c}\int_{0}^{c}f\left ( x \right )dx\\ \text{and } a_{n}&=\frac{2}{c}\int_{0}^{c}f\left ( x \right )\cos \frac{n\pi x}{c}dx\\ &\text{Using above result and using,}\\ f\left ( x \right )&=1-\frac{2x}{\pi } \text{ we get}\\ a_{0}&=\frac{2}{\pi }\int_{0}^{\pi }\left ( 1-\frac{2x}{\pi } \right )dx \\ &=\frac{2}{\pi }\left ( x-\frac{x^{2}}{\pi } \right )_{0}^{\pi }=0 \\ \text{and } a_{n}&=\frac{2}{\pi }\int_{0}^{\pi }\left ( 1-\frac{2x}{\pi } \right )\cos nx\: dx \\ &=\frac{2}{\pi }\left [ \left ( 1-\frac{2x}{\pi } \right )\frac{\sin nx}{n}-\left ( -\frac{2}{\pi } \right )\left ( -\frac{\cos nx}{n^{2}} \right ) \right ]_{0}^{\pi } \\ &=\frac{2}{\pi }\left [ -\frac{2\cos n\pi }{\pi n^{2}}+\frac{2}{\pi n^{2}} \right ] \\ &=\frac{4}{n^{2}\pi ^{2}}\left [ 1-\cos n\pi \right ]\end{aligned}
\begin{aligned} a_{0}&=\frac{2}{c}\int_{0}^{c}f\left ( x \right )dx\\ \text{and } a_{n}&=\frac{2}{c}\int_{0}^{c}f\left ( x \right )\cos \frac{n\pi x}{c}dx\\ &\text{Using above result and using,}\\ f\left ( x \right )&=1-\frac{2x}{\pi } \text{ we get}\\ a_{0}&=\frac{2}{\pi }\int_{0}^{\pi }\left ( 1-\frac{2x}{\pi } \right )dx \\ &=\frac{2}{\pi }\left ( x-\frac{x^{2}}{\pi } \right )_{0}^{\pi }=0 \\ \text{and } a_{n}&=\frac{2}{\pi }\int_{0}^{\pi }\left ( 1-\frac{2x}{\pi } \right )\cos nx\: dx \\ &=\frac{2}{\pi }\left [ \left ( 1-\frac{2x}{\pi } \right )\frac{\sin nx}{n}-\left ( -\frac{2}{\pi } \right )\left ( -\frac{\cos nx}{n^{2}} \right ) \right ]_{0}^{\pi } \\ &=\frac{2}{\pi }\left [ -\frac{2\cos n\pi }{\pi n^{2}}+\frac{2}{\pi n^{2}} \right ] \\ &=\frac{4}{n^{2}\pi ^{2}}\left [ 1-\cos n\pi \right ]\end{aligned}
Question 10 |
If L defines the Laplace Transform of a function, L\left [ \sin \left ( at \right ) \right ] will equal to
\frac{a}{s^{2}-a^{2}} | |
\frac{a}{s^{2}+a^{2}} | |
\frac{s}{s^{2}+a^{2}} | |
\frac{s}{s^{2}-a^{2}} |
Question 10 Explanation:
\begin{aligned} L\left [ f\left ( t \right ) \right ]&=\int_{0}^{\infty }e^{-st}f\left ( t \right )dt \\ \therefore \;\; L\left [ \sin \left ( at \right ) \right ]&=\int_{0}^{\infty }e^{-st}\sin \left ( at \right )dt \\ &= \frac{a}{s^{2}+a^{2}} \end{aligned}
There are 10 questions to complete.
