Engineering Mathematics Multiple Choice Questions on “Taylor Mclaurin Series – 1”.
1. The Mclaurin Series expansion of sin(ex) 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=ex we have
sin(ex)=(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(ex)=(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 nth 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