Taylor's Series MCQ Quiz - Objective Question with Answer for Taylor's Series - Download Free PDF

Concept:

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

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

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

Calculation:

For option (1)

f(0) =e0 ⋅ cos θ = 1

⇒ f’(x) = ex (- sin x) + cos x ⋅ ex

⇒ f’(x) = f(x) - ex ⋅ sin x

⇒ f’’(x) = f’(x) - ex ⋅ x - ex sin x

⇒ f''(x) = f'(x) - f(x) - ex sin x

⇒ f'''(x) = f''(x) - f'(x) - ex cos x - ex sin x

⇒ f'''(x) = f''(x) - f'(x) - f(x) - ex sin x

Now,

f’(0) = 1 - 0 = 1

f’’(0) = f’(0) - e0 (1) - 0 = 1 - 1 = 0

f’’’(0) = f’’(0) f’(0) - 1 - 0

= 1 - 1 - 1 = -2

Taylor series expansion at x = 0 is

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

Concept:

The taylor’s series expansion of (1 + x)^-1 is given by

(1 + x)^-1 = 1 – x + x² – x³ + …

Calculation:

Given complex function is (z – 1)/(z + 1);

To expand about the point z = 1, let us assume t = z – 1;

Now the function will be

\(f\left( z \right) = \frac{{z - 1}}{{z + 1}} = \frac{t}{{t + 2}} = 1 - \frac{1}{{\frac{t}{2} + 1}} = 1 - {\left( {1 + \frac{t}{2}} \right)^{ - 1}}\)

Using standard Taylor's series expansion,

\(f\left( z \right) = 1 - \left[ {1 - \frac{t}{2} + \frac{{{t^2}}}{{{2^2}}} - \frac{{{t^3}}}{{{2^3}}} + \ldots } \right]\)

\( \Rightarrow f\left( z \right) = \frac{t}{2} - \frac{{{t^2}}}{{{2^2}}} + \frac{{{t^3}}}{{{2^3}}} - \ldots \)

The third term in the expansion is \(\frac{{{t^3}}}{8} = \frac{{{{\left( {z - 1} \right)}^3}}}{8}\)

Explanation:

Maclaurian Series is  f(0)+xf'(0)+(x²/2!)f''(0)+ _______

The basic idea is to be able to express a function as a polynomial, because polynomials are easy to deal with.

Some times a Maclaurin series can make it easier to do complicated operations like integration or differentiation.

Some times Maclaurin series can be used for things like proving that e^(ix)=cos(x)+isin(x) by relating sines, cosines, and exponentials in the same “form” to establish equality.

Another application is numerical computation. 

Explanation:

A Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. These series are advantageous because they can provide approximations for more complex functions using polynomials, which are easier to calculate.

The second-order Taylor's polynomial about the point 'a' is an approximation of an original function f(x) near this point 'a'. This polynomial only involves the function's value and first two derivatives at the point 'a'.

It's given by the formula:

f(x) ≈ f(a) + f'(a)(x - a) + (1/2)f''(a)(x - a)²

Option 1:

f(a) + f'(a) (x - a)³ This is not correct. The term (x - a) should be squared, not cubed. Also, it's missing the second derivative term.

Option 2:

f(a) + f'(a) (x - a) + (1/2)f''(a) (x - a)² This is the correct form of the second-order Taylor's polynomial. It represents the value of the function at 'a', plus the first derivative at 'a' times (x - a), plus one-half times the second derivative at 'a' times (x - a) squared. This gives an approximation for f(x), specifically near 'a'.

Option 3:

f(a) + f'(a) (x - a)² This is not correct. The first derivative term should be multiplied by (x-a), not (x-a)², and it's missing the second derivative term.

Option 4:

f(a) + f'(a) (x - a) + (1/2)f''(a) (x - a)³ This is not correct. The second derivative term should be multiplied by (x - a)², not (x - a)³.

Hence, option 2 is the correct form for a second order Taylor's polynomial.

Explanation:

The Maclaurin series expansion for Sin (x) is given by the summation from n=0 to infinity of (-1)ⁿ ×  x⁽²ⁿ⁺¹⁾/(2n+1)!

The Expansion of Sinx = x -x³/3! + x⁵/5! -............

Hence, The Expansion of Sinx contains all the odd powers of x.

Hence, The Correct option is 4.

Concept:

Taylor series:

\(f(z)=f(a)+\frac{f'(a)}{1!}(z-a)+\frac{f''(a)}{2!}(z-a)^2+...\)

Sum of GP: \(f\left( z \right) = \frac{a}{{1 - r}} \)

Calculation:

f(z) = 1 + (1 – z) + (1 – z)² + ...∞ 

The above series is in the form of G.P.

a = 1, r = (1 – z)

\(f\left( z \right) = \frac{a}{{1 - r}} = \frac{1}{{1 - \left( {1 - z} \right)}} = \frac{1}{z}\)

Explanation:

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-

\(f(z)=f(a)\;+\;(z-a)f'(a)\;+\;\frac{(z-a)^2}{2!}f''(a)\;+\;........\frac{(z-a)^n}{n!}f^n(a)\)

\(\therefore f(z)=\sum_{n=0}^{\infty}\frac{(z-a)^n}{n!}f^n(a)\)

\(\therefore \sum_{n=0}^{\infty}a_n(z-a)^n\) where \(a_n=\frac{f^n(a)}{n!}\)

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).

\(f(z)= a_0\;+\;a_1(z-a)\;+\;a_2(z-a)^2\;+\;.....\;+\;a_n(z-a)^n\;+\;a_{-1}(z-a)^{-1}\;+\;a_{-2}(z-a)^{-2}\;+\;....\;+\;a_{-n}(z-a)^{-n}\)

\(\therefore \sum_{n=0}^{\infty}a_n(z-a)^n\;+\;\sum_{n=1}^{\infty}a_{-n}(z-a)^{-n}\)

\(\therefore \sum_{-\infty}^{\infty}a_n(z-a)^n\;where\;a_n=\frac{1}{2\pi i}\oint \frac{f(z)}{(z-a)^{n+1}}dz\)

The negative part of Laurent's series i.e \(\sum_{n=1}^{\infty}a_{-n}(z-a)^{-n}\) is called the singular part, and if that vanishes the terms that remain will be \( \sum_{n=0}^{\infty}a_n(z-a)^n\) , which is nothing but Taylor series.

Important Points

Maclaurin's Series:

When a = 0 in Taylor's series i.e. about origin or powers of z, the series formed is Maclaurin's series.

\(f(z)=f(0)\;+\;(z)f'(0)\;+\;\frac{(z)^2}{2!}f''(0)\;+\;........\frac{(z)^n}{n!}f^n(0)\)

Cauchy's Theorem:

If f(z) is single-valued and an analytic function of z and f'(z) is continuous at each point within and on the closed curve c, then according to the theorem, \(\mathop \oint \limits_C f\left( z \right)dz = 0\).

Taylor series expansion about x = a is given by:

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

about x = 0 i.e a = 0

\(f\left( x \right) = f\left( 0 \right) + xf'\left( 0 \right) + \frac{{{x^2}}}{{2!}}f''\left( 0 \right) + \ldots \)

Comparing the coefficients, we get:

\({a_2} = \frac{{f''\left( 0 \right)}}{{2!}}\)

With \(f\left( x \right) = \mathop \smallint \limits_0^x {e^{ - {t^2}/2}}dt\),

\(f'\left( x \right) = \frac{d}{{dx}}\left[ {\mathop \smallint \limits_0^x {e^{ - {t^2}/2}}dt} \right]\)

Using the Leibnitz rule, we can write:

\(f'\left( x \right) = {e^{ - {x^2}/2}}\)

Now, \(f''\left( x \right) = {e^{ - \frac{{{x^2}}}{2}}}\left( {\frac{{ - 2x}}{2}} \right)\)

\(f''\left( x \right) = {e^{ - \frac{{{x^2}}}{2}}}\left( { - x} \right)\)

at x = 0

f’’(0) = 0

Concept:

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

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

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

Calculation:

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

\(3\sin x = 3x - \frac{{{x^3}}}{{2!}} + \frac{{{x^5}}}{{40}} - \ldots \)          ---(1)

Similarly,

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

 \(2\cos x = 2 - {x^2} + \frac{{{x^4}}}{{12}} - \ldots \)        ---(2)

Adding equation (1) and equation (2), we get:

\(3\sin x + \cos x = 2 + 3x - {x^2} - \frac{{{x^3}}}{2} + \ldots \)

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:

\(f\left( x \right) = f\left( a \right) + \frac{{f'\left( a \right)}}{{1!}}\left( {x - a} \right) + \frac{{f''\left( a \right)}}{{2!}}{\left( {x - a} \right)^2} + \frac{{f'''\left( a \right)}}{{3!}}{\left( {x - a} \right)^3} + \; \ldots \ldots \)

Calculation:

Given:

We have to find the series expansion of \(\frac{{\sin x}}{x}\) 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:

\(sinx=x - \frac{{{x^3}}}{{3!}} + \ldots \)

Therefore the series expansion of \(\frac{{\sin x}}{x}\) near origin will be:

\(\frac{sinx}{x}=1 - \frac{{{x^2}}}{{6}} + \ldots \)

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

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

\(f(x)=f(a)\;+\;(x-a)f'(x)\;+\;\frac{(x-a)^2}{2!}f''(x)\;+\;........\infty\)

Calculation:

Given:

\(f\left( x \right) = \frac{{\sin \left( x \right)}}{{x - 54}}\)

f(100)(54) = ?

Using Taylor series expansion for Sin x at a = 54

\(f(x)=f(a)\;+\;(x-a)f'(x)\;+\;\frac{(x-a)^2}{2!}f''(x)\;+\;........\infty\)

\(\sin x~=~\sin (54)~+~\frac{(x~-~54)~\times~\cos (54)}{1!}~-~\frac{(x~-~54)^2~\times~\cos (54)}{2!}~-~.............\infty\)

Now the function transforms into:

f(x) = \(\frac{\sin (54)}{x~-~54}~+~\frac{\cos (54)}{1!}~-~\frac{(x~-~54)~\times~\cos (54)}{2!}~-~.............\infty\)

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

\(\frac{dy}{dx}=x^2y-1\), y(0) = 1, y(0, 1) = ?

Here, x₀ = 0, y₀ = 1, h = 0.1 so y(0, 1) = y₁ = ?

Here, y^I(x) = x²y - 1

⇒ y^I(0) = -1

y^II(x) = 2xy + x²y^I

⇒ y^II(0) = 0

y^III(x) = 2y + 2xy^I + 2xy^I + x²y^II

⇒ y^III(0) = 2

y^IV(x) = 2y^I + 4(xy^II + y^I) + 2xy^II + x2y^III

⇒ y^IV(0) = -6 and so on......

Hence by Taylor series:

\(y_1=y_0+hy^I(0)+\frac{h^2}{2!}y^{II}(0)+\frac{h^3}{3!}y^{III}(0)+\frac{h^4}{4!}y^{IV}(0)+....\)

\(1+0.1(-1)+\frac{(0.1)^2}{2!}(0)+\frac{(0.1)^3}{3!}(2)+\frac{(0.1)^4}{4!}(-6)+.....\)

1 - 0.1 + 0 + 0.00033 + ......

= 0.90031 ≈ 0.90033

f(x) = e^{sin x} ⇒ f(0) = e⁰ = 1

f'(x) = esin x (cos x) ⇒ f'(0) = 1

f''(x) = esin x cos²x + esin x (-sin x)

f''(0) = 1

= esin x [cos²x - sin x]

f'''(x) = esin x [-sin 2x - cos x] + esin x  [cos³x - sin x cos x]

f'''(x) = -1 + 1 = 0

f^iv(x) = esin x  [-2 cos 2x + sin x] + esin x  cos x [-sin 2x - cos x]

\(+ e^{\sin x } \left[ 3 \cos^2 x (-\sin x) - \frac{\cos 2x}{2} \times 2 \right]\)

+ esin x [cos⁴ x - sin x cos² x]

f^iv(0) = -2 -1 -1 + 1 = -3

Substitue in Maclaurin Series

\(e^{\sin x} = 1 + x + \frac{x^2}{2 } + \frac{x^3}{3!} (0) + \frac{x^4}{4!} \times (-3) + ....\)

\(= 1+ x + \frac{x^2}{2} - \frac{x^4}{8} + ....\)

Concept:

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

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

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

Calculation:

For option (1)

f(0) =e0 ⋅ cos θ = 1

⇒ f’(x) = ex (- sin x) + cos x ⋅ ex

⇒ f’(x) = f(x) - ex ⋅ sin x

⇒ f’’(x) = f’(x) - ex ⋅ x - ex sin x

⇒ f''(x) = f'(x) - f(x) - ex sin x

⇒ f'''(x) = f''(x) - f'(x) - ex cos x - ex sin x

⇒ f'''(x) = f''(x) - f'(x) - f(x) - ex sin x

Now,

f’(0) = 1 - 0 = 1

f’’(0) = f’(0) - e0 (1) - 0 = 1 - 1 = 0

f’’’(0) = f’’(0) f’(0) - 1 - 0

= 1 - 1 - 1 = -2

Taylor series expansion at x = 0 is

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