Taylor Series - Free MCQ Practice Test with solutions, GATE MA Engineering

Concept:

Taylor’s series method:

The Taylor series can be used to calculate the value of an entire function at every point, if the value of the function, and all of its derivatives, are known at a single point.

Taylor's series expansion for f (x + h) is
f(x+h) = f(x) + hf′(x) + h²/2!f″(x) + h³/3! + f‴(x)+…∞
f(x) = f(a) + (x−α)f′(x) +(x) +........∞

Calculation:

Given:

f⁽¹⁰⁰⁾(54) = ?
Using Taylor series expansion for Sin x at a = 54

Now the function transforms into:

After Observing carefully the first term in the above infinite series, the (x - 54) term is always in the denominator, which will become zero when we put x = 54.

Every derivative will also have the same term till infinite.

So, every term will have zero in its denominator after putting x = 54.

⇒ f⁽¹⁰⁰⁾(54) is Undefined.

Concept:

Taylor series expansion

Option 1:
The standard expansion of log(1 + z) is given as 

Hence, Option 1 is true
Option 2:
Given complex function is 
→ Let’s Resolve f(z) into partial fractions

For expanding about z = 2, let z – 2 = t ⇒ z = 2 + t


Option 3:
Cauchy’s Integral Formula:

If f(z) is an analytic function within a closed curve and if a is any point within C, then

Residue Theorem:

If f(z) is analytic in a closed curve C except at a finite number of singular points within C, then

⁡f(z)dz = 2πi × [sum of residues at the singualr points with in C]

Formula to find residue:

1. If f(z) has a simple pole at z = a, then

2. If f(z) has a pole of order n at z = a, then

Given complex integral is 
where Cis the circle |z-1| = 2;

Now for the given complex function, the pole is -4 with order 2;

The pole - 4 lies outside the given circle C;

Therefore, no residue inside the circle, hence integration will be zero.

Option 3 is also correct

Option 4:

The given complex function is f(z) = 
In this function, the singularities are z = 0, +i, -i;

Therefore, the given function has 3 singularities...

Option 4 is incorrect

f''(0) = 1


fiv(0) = -2 -1 -1 + 1 = -3
Substitue in Maclaurin Series

Concept:

Taylor series:

The Taylor series of a real or complex-valued function f (x) that is infinitely differentiable at a real or complex number ‘a’ is the power series.

Expression of Taylor series is:

Calculation:

Given:

We have to find the series expansion of sin⁡x/x near origin, or a = 0.

Let f(x) = sin x

f(0) = sin (0) = 0,

f'(0) = cos (0) = 1,

f''(0) = -sin(0) = 0,

f'''(0) = -1 .... so on

Putting all the values in Taylor series expansion, we get:

Series expansion of sin x  will be:
sinx = x − x³/3!+…
Therefore the series expansion of sin⁡ x/x near origin will be:

Taylor Series:

If f(z) is analytic inside a circle 'C', centre at z = a, and radius 'r', then for all z inside 'C'; the Taylor series is given by-


Laurent Series:
If f(z) is analytic at every point inside and on the boundary of a ring shaped region 'R' bounded by two concentric circle C₁ and C₂ having centre at 'a' & respective radii r₁ and r₂ (r₁ > r₂).


The negative part of Laurent's series i.e is called the singular part, and if that vanishes the terms that remain will be , which is nothing but Taylor series.

Concept:

Taylor series:

Calculation:
f(z) = 1 + (1 – z) + (1 – z)²
The above series is in the form of G.P.
a = 1, r = (1 – z)

Concept:

Binomial expansion of (1 - x)-n is given by
(1 - x)^-n = 1 + nx + 
Formula used:

Calculation:
Given,
y = sin-1 x
....(1)
Using the above binomial expansion formula

Integrating both sides with respect to x, 

Hence, option c is the correct answer.

Hence by Taylor series:

1 - 0.1 + 0 + 0.00033 + ......
= 0.90031 ≈ 0.90033

Concept:
Taylor series expansion for sin x and cos x are respectively:

Calculation:

Similarly,

Adding equation (1) and equation (2), we get:
3sin⁡x + cos⁡x = 2 + 3x −