Power Series and Taylor Series MCQ Quiz - Objective Question with Answer for Power Series and Taylor Series - Download Free PDF
Last updated on May 19, 2025
Latest Power Series and Taylor Series MCQ Objective Questions
Power Series and Taylor Series Question 1:
\(\rm \frac{3}{1!}+\frac{5}{3!}+\frac{7}{5!}+\frac{9}{7!}+...\) is equal to
Answer (Detailed Solution Below)
Power Series and Taylor Series Question 1 Detailed Solution
Explanation:
ex = \(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...\)
Putting x = 1 and x = -1
e = \(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...\)
and
e-1 = \(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...\)
So,
3e - e-1 = (\(3+\frac{3}{1!}+\frac{3}{2!}+\frac{3}{3!}+...\)) - (\(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...\))
=
Power Series and Taylor Series Question 2:
Let f(x) = √x + αx, x > 0 and
g(x) = a0 + a1(x − 1) + a2(x − 1)2
be the sum of the first three terms of the Taylor series of f(x) around x = 1. If g(3) = 3, then α is .
Answer (Detailed Solution Below) 0.5
Power Series and Taylor Series Question 2 Detailed Solution
Explanation:
The general formula for the Taylor series of a function f(x) around x=a is:
f(x) = f(a) + f'(a)(x-a) + (f''(a)/2!)(x-a)² + ⋯
⇒ f(x) = √x + αx
⇒ \(f'(x) = (1/2)x^{-1/2} + α \)
⇒ \(f''(x) = -(1/4)x^{-3/2} \)
Now, evaluate these derivatives at x=1:
f(1) = 1 + α
⇒ f'(1) = 1/2 + α
⇒ f''(1) = -1/4
Substitute these values into the Taylor series formula:
g(x) = (1 + α) + (1/2 + α)(x-1) - (1/8)(x-1)²
Substitute x=3 into the expression for g(x):
g(3) = (1 + α) + (1/2 + α)(2) - (1/8)(2)²
⇒ g(3) = 1 + α + 1 + 2α - 1/2
⇒ g(3) = 3/2 + 3α
Set g(3) = 3 and solve for α:
3/2 + 3α = 3
⇒ 3α = 3 - 3/2
⇒ 3α = 3/2
⇒ α = 1/2
Therefore, the value of α is 1/2
Hence 0.5 is the Correct Answer.
Top Power Series and Taylor Series MCQ Objective Questions
Power Series and Taylor Series Question 3:
\(\rm \frac{3}{1!}+\frac{5}{3!}+\frac{7}{5!}+\frac{9}{7!}+...\) is equal to
Answer (Detailed Solution Below)
Power Series and Taylor Series Question 3 Detailed Solution
Explanation:
ex = \(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...\)
Putting x = 1 and x = -1
e = \(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...\)
and
e-1 = \(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...\)
So,
3e - e-1 = (\(3+\frac{3}{1!}+\frac{3}{2!}+\frac{3}{3!}+...\)) - (\(1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...\))
=
Power Series and Taylor Series Question 4:
Let f(x) = √x + αx, x > 0 and
g(x) = a0 + a1(x − 1) + a2(x − 1)2
be the sum of the first three terms of the Taylor series of f(x) around x = 1. If g(3) = 3, then α is .
Answer (Detailed Solution Below) 0.5
Power Series and Taylor Series Question 4 Detailed Solution
Explanation:
The general formula for the Taylor series of a function f(x) around x=a is:
f(x) = f(a) + f'(a)(x-a) + (f''(a)/2!)(x-a)² + ⋯
⇒ f(x) = √x + αx
⇒ \(f'(x) = (1/2)x^{-1/2} + α \)
⇒ \(f''(x) = -(1/4)x^{-3/2} \)
Now, evaluate these derivatives at x=1:
f(1) = 1 + α
⇒ f'(1) = 1/2 + α
⇒ f''(1) = -1/4
Substitute these values into the Taylor series formula:
g(x) = (1 + α) + (1/2 + α)(x-1) - (1/8)(x-1)²
Substitute x=3 into the expression for g(x):
g(3) = (1 + α) + (1/2 + α)(2) - (1/8)(2)²
⇒ g(3) = 1 + α + 1 + 2α - 1/2
⇒ g(3) = 3/2 + 3α
Set g(3) = 3 and solve for α:
3/2 + 3α = 3
⇒ 3α = 3 - 3/2
⇒ 3α = 3/2
⇒ α = 1/2
Therefore, the value of α is 1/2
Hence 0.5 is the Correct Answer.