Category: Engineering Mathematics Objective Questions

Fourier Analysis and Partial Differential Equations Multiple Choice Questions on “Derivation and Solution of Two-dimensional Wave Equation”.

1. Who discovered the one-dimensional wave equation?
a) Jean d’Alembert
b) Joseph Fourier
c) Robert Boyle
d) Isaac Newton
View Answer

Answer: a
Explanation: Jean-Baptiste le Rond d’Alembert (16 November 1717 – 29 October 1783) was a French mathematician, mechanician, physicist, philosopher, and music theorist. Until 1759 he was co-editor with Denis Diderot of the Encyclopédie. He was the person responsible for the discovery of wave equation.

2. Wave equation is a third-order linear partial differential equation.
a) True
b) False
View Answer

Answer: b
Explanation: The wave equation is a second-order linear partial differential equation which is developed for the description of waves (water waves, sound waves, seismic waves, light waves), acoustics, electromagnetics, and fluid dynamics.

3. In which of the following fields, does the wave equation not appear?
a) Acoustics
b) Electromagnetics
c) Pedology
d) Fluid Dynamics
View Answer

Answer: c
Explanation: The wave equation arises in fields like acoustics, electromagnetics, and fluid dynamics. Historically, the problem of a vibrating string such as that of a musical instrument was studied by Jean le Rond d’Alembert, Leonhard Euler, Daniel Bernoulli, and Joseph-Louis Lagrange. In 1746, d’Alembert discovered the one-dimensional wave equation, and within ten years Euler discovered the three-dimensional wave equation.

4. The wave equation is known as d’Alembert’s equation.
a) True
b) False
View Answer

Answer: a
Explanation: D’Alembert’s formula for obtaining solutions to the wave equation is named after him. The wave equation is sometimes referred to as d’Alembert’s equation.

5. Which of the following statements is false?
a) Equations that describe waves as they occur in nature are called wave equations
b) The problem of having to describe waves arises in fields like acoustics, electromagnetics, and fluid dynamics
c) Jean d’Alembert discovered the three-dimensional wave equation
d) Jean d’Alembert discovered the one-dimensional wave equation
View Answer

Answer: c
Explanation: The one-dimensional wave equation was discovered by d’Alembert in 1746 and by 1756, the three-dimensional wave equation was discovered by Euler.

6. What is the order of the partial differential equation, (frac{∂z}{∂x}-(frac{∂z}{∂y})^3=0)?
a) Order-5
b) Order-1
c) Order-4
d) Order-2
View Answer

Answer: b
Explanation: The order of an equation is defined as the highest derivative present in the equation. Hence, in the given equation, (frac{∂z}{∂x}-(frac{∂z}{∂y})^5=0,) the order is 1.

7. The half-interval method in numerical analysis is also known as __________
a) Newton-Raphson method
b) Regula Falsi method
c) Taylor’s method
d) Bisection method
View Answer

Answer: d
Explanation: The Bisection method, also known as binary chopping or half-interval method, is a starting method which is used, where applicable, for few iterations, to obtain a good initial value.

8. Wave equation is a linear elliptical partial differential equation.
a) False
b) True
View Answer

Answer: a
Explanation: Wave equation is a linear second order hyperbolic partial differential equation whereas a Laplace equation is a linear elliptical partial differential equation.

9. Which of the following is the condition for a second order partial differential equation to be hyperbolic?
a) b²-ac < 0
b) b²-ac=0
c) b²-ac>0
d) b²-ac= < 0
View Answer

Answer: c
Explanation: For a second order partial differential equation to be hyperbolic, the equation should satisfy the condition, b²-ac>0.

10. Which of the following statements is true?
a) Hyperbolic equations have three families of characteristic curves
b) Hyperbolic equations have one family of characteristic curves
c) Hyperbolic equations have no families of characteristic curves
d) Hyperbolic equations have two families of characteristic curves
View Answer

Answer: d
Explanation: The canonical variables ξ and η for a hyperbolic pde satisfy the equations,
(aζ_x+(b+sqrt{b^2-ac}) ζ_y=0 , and , aη_x+(b+sqrt{b^2-ac}) η_y=0 )
The families of curves ξ = constant and η = constant are the characteristic curves. Hence, hyperbolic equations have two families of characteristic curves.

Global Education & Learning Series – Fourier Analysis and Partial Differential Equations.
here is complete set of 1000+ Multiple Choice Questions and Answers.

 

Engineering Mathematics Multiple Choice Questions on “Taylor Mclaurin Series – 1”.

1. The Mclaurin Series expansion of sin(e^x) is?
a) sin(1)+(frac{x.cos(1)}{1!}+sum_{n=2}^{infty}sum_{a=0}^{infty}frac{x^n.(-1)^a}{n!}timesfrac{(2a+1)^n}{(2a+1)!})
b) (frac{e^x}{1!}+frac{e^{3x}}{3!}+frac{e^{5x}}{5!}…infty)
c) (-frac{e^x}{1!}+frac{e^{3x}}{3!}-frac{e^{5x}}{5!}…infty)
d) (sum_{n=2}^{infty}sum_{a=0}^{infty}frac{x^n.(-1)^a}{n!}times frac{(2a+1)^n}{(2a+1)!})
Answer: a
Explanation: We know the series expansion for sin(x) is
sin(t)=(frac{t}{1!}-frac{t^3}{3!}+frac{t^5}{5!}…infty)
Substituting t=e^x we have
sin(e^x)=(frac{e^x}{1!}-frac{e^3x}{3!}+frac{e^5x}{5!}…infty)
Now using
(e^x=1+frac{x}{1!}+frac{x^2}{2!}+frac{x^3}{3!}+…infty)
We have
sin(e^x)=(frac{1+frac{x}{1!}+frac{x^2}{2!}+frac{x^3}{3!}+…infty}{1!}-frac{1+frac{3x}{1!}+frac{(3x)^2}{2!}+frac{(3x)^3}{3!}+…infty}{3!}+frac{1+frac{5x}{1!}+frac{(5x)^2}{2!}+frac{(5x)^3}{3!}+…infty}{5!})
Grouping terms with same power we have
(=(frac{1}{1!}-frac{1}{3!}+frac{1}{5!}…infty)+x(frac{1}{1!}-frac{3}{3!}+frac{5}{5!}…infty))
+(frac{x^2}{2!}(frac{1}{1!}-frac{3^2}{3!}+frac{5^2}{5!}…infty)+…infty)
We can rewrite the last terms of the series as a double series we have
sin(1)+(frac{x.cos(1)}{1!}+sum_{n=2}^{infty}sum_{a=0}^{infty}frac{x^n.(-1)^a}{n!}timesfrac{(2a+1)^n}{(2a+1)!})

2. What is the coefficient of x¹⁰¹⁷²⁹ in the series expansion of cos(sin(x))?
a) 0
b) ¹⁄_101729!
c) ^-1⁄_101729!
d) 1
Answer: a
Explanation: We know that the series expansion of cos(x) is
cos(t)=1-(frac{t^2}{2!}+frac{t^4}{4!}….infty)
Now substituting t=sin(x) we have
cos(sin(x))=(1-frac{1}{2!}times (frac{x}{1!}-frac{x^3}{3!}+frac{x^5}{5!}+…infty)^2+frac{1}{4!}times (frac{x}{1!}-frac{x^3}{3!}+frac{x^5}{5!}+…infty)^4+..infty)
Observe that every term has odd powered series raised to an even term.
Thus, we must have only even powered terms in the above series expansion. The coefficient of any odd powered term is zero.

3. Let τ(X) be the Taylor Series expansion of f(x) = x³ + x² + 1019 centered at a = 1019, then what is the value of the expression 2(τ(1729))² + τ(1729) * f(1729) – 3(f(1729))² + 1770?
a) 1770
b) 1729
c) 0
d) 1
Answer: a
Explanation: Observe first off that the given function is a polynomial and so any other representation (Taylor Series here) which is continuous and differentiable has to be the
same polynomial. This gives us
τ(x) = f(x)
We now evaluate the expression as follows
= 2(f(1729))² + (f(1729))² – 3(f(1729))² + 1770
= 3(f(1729))² – 3(f(1729))² +1770
= 1770

4. Find the Taylor series expansion of the function cosh(x) centered at x = 0.
a) (1-frac{x^2}{2!}+frac{x^4}{4!}+….infty)
b) (frac{x}{1!}+frac{x^3}{3!}+frac{x^5}{5!}….infty)
c) (1+frac{x^2}{2!}+frac{x^4}{4!}+….infty)
d) (1+frac{x}{1!}+frac{x^2}{2!}+….infty)
Answer: c
Explanation: We know the general expression for the expansion of the Taylor series
(tau[f(x)]=f(a)+frac{x.f^{(1)}(a)}{1!}+frac{x^2.f^{(2)}(a)}{2!}+…..infty)
Given a=0 we substitute in the equation to get
(tau[f(x)]=f(0)+f^{(1)}(0)times frac{x}{1!}+f^{(2)}(0)times frac{x^2}{2!}+….infty)
Now the n^th derivatives can be calculated as
(f^{(n)}(x)=(frac{e^x+e^{-x}}{2})^{(n)})
(=frac{e^x+(-1)^ne^{-x}}{2})
Substituting x=0 yields the final expansion
(f^{(n)}(x)=frac{1+(-1)^n}{2})
We get
(tau[f(x)]=1+(0)times frac{x}{1!}+(1)times frac{x^2}{2!}+(0)times frac{x^3}{3!}+…..infty)
(tau[f(x)]=1+frac{x^2}{2!}+frac{x^4}{4!}+…infty)

5. To find the value of sin(9) the Taylor Series expansion should be expanded with center as ___________
a) 9
b) 8
c) 7
d) Some delta (small) interval around 9
Answer: d
Explanation: The Taylor series gives accurate results around some point taken as center. As we need the value of 9 the center nearer to the point should be taken.

6. f⁽¹⁾ (n) = g⁽ⁿ⁾ (0) holds good for some functions f(x) and g(x). Now let the coordinate axes containing graph g(x) be rotated by 30 degrees clockwise, then the corresponding Taylor series for the transformed g(x) is?
a) g(0)+(frac{e^x – 1}{sqrt{3}}+frac{sum_{n=1}^infty f^{(1)}(n)x^n}{n!})
b) (g(0) + frac{g^{(1)}.x}{1!} + frac{g^{(2)}(1).x^2}{2!}+…infty)
c) No unique answer exist
d) Such function is not continuous
Answer: a
Explanation: We have f⁽¹⁾ (n) = g⁽ⁿ⁾ (0)
As the coordinate axes containing f(x) is rotated the tan(30) term gets added to the derivative of f⁽¹⁾_new(n)=g⁽ⁿ⁾(0)-tan(30)
We have
g⁽ⁿ⁾(0)=f⁽¹⁾(n)+tan(30)
The Taylor expansion centered at 0 for g(x) is given by
g(x)=(g(0)+frac{g^{(1)}(0).x}{1!}+frac{g^{(2)}(0).x^2}{2!}+…..infty )
Now substituting g⁽ⁿ⁾(0)=f⁽¹⁾(n)+tan(30) we have
g(x)=(g(0)+frac{(f^{(1)}(1)+tan(30)).x}{1!}+frac{(f^{(1)}(2)+tan(30)).x^2}{2!}+….infty)
g(x)=(g(0)+(frac{f^{(1)}(1).x}{1!}+frac{f^{(1)}(2).x}{2!}+…infty)+tan(30)times(frac{x}{1!}+frac{x^2}{2!}+….infty))
=(g(0)+frac{sum_{n=1}^{infty}f^{(1)}(n).x^n}{n!}+frac{1}{sqrt{3}}times(e^x-1))

7. Let τ(f(x)) denote the Taylor series for some function f(x). Then the value of τ(τ(τ(f(1729)))) – 2τ(τ(f(1729))) + τ(f(1729)) is?
a) 1729
b) -1
c) 1
d) 0
Answer: d
Explanation: We know that the Mclaurin Series for any given function always yields a polynomial (finite OR infinite).
Further the Mclaurin series of this polynomial (i.e.τ(τ(f(x)))) is also a polynomial. Due to uniqueness of this polynomial, no matter how many nested Mclaurin series we might find, they are all equal. Thus, we have
τ(τ…….(f(x))….)) = f(x)
Substituting this into our required expression we have
= f(1729) – 2f(1729) + f(1729)
= 0.

8. Let Mclaurin series of some f(x) be given recursively, where a_n denotes the coefficient of xⁿ in the expansion. Also given a_n = a_n-1 / n and a₀ = 1, which of the
following functions could be f(x)?
a) e^x
b) e^2x
c) c + e^x
d) No closed form exists
Answer: a
Explanation: Observing the recurrence relation we have
a_n=(frac{a_{n-1}}{n}=frac{a_{n-2}}{n(n-1)})
a_n=(frac{a_0}{n(n-1)(n-2)…3times 2 times 1})
Thus one could deduce that
a_n=(frac{1}{n!})
Putting this into the Mclaurin expansion we have
f(x)=a₀+a₁x+a₂x²+a₃x³….∞
f(x)=(1+frac{x}{1!}+frac{x^2}{2!}+frac{x^3}{3!}+….infty)
Which is the well known expansion of e^x.

9. A function f(x) which is continuous and differentiable over the real domain exists such that f⁽ⁿ⁾ (x) = [f^{(n + 1)} (x)]², f(0) = a and f⁽¹⁾(0) = 1.
a) True
b) False
Answer: a
Explanation: Writing out the Mclaurin series we have
f(x)=f(0)+(frac{f^{(1)}(0).x}{1!}+frac{f^{(2)}(0).x^2}{2!}+…infty)
Now sustituting f⁽ⁿ⁺¹⁾(0) = (sqrt{f^{(n)}(0)})we have
f(x) = f(0) + (frac{f^{(1)}(0).x}{1!}+frac{sqrt{f^{(1)}(0)}.x^2}{2!}+frac{sqrt{sqrt{f^{(1)}(0)}.x^3}}{3!}…infty)
Now f⁽¹⁾(0)=1 and f(0)=a
Substituting this we have
f(x)=a+(frac{x}{1!}+frac{x^2}{2!}+frac{x^3}{3!}+…infty)
f(x)=a-1+e^x
This is a well defined function.