1. Cauchy’s Remainder for Maclaurin’s Theorem is given by \(\frac{x^n(1-θ)^{n-1}}{(n-1)!}f^{(n)}(θx)\).
a) True
b) False
Explanation: Schlomilch’s Remainder for Maclaurin’s Theorem is given by, \(\frac{x^n(1-θ)^{n-p}}{(n-1)!p} f^{(n)}(θx).\) To obtain Cauchy’s Remainder for Maclaurin’s Theorem, we put p=1, which gives us,
\(\frac{x^n(1-θ)^{n-1}}{(n-1)!}f^{(n)}(θx).\)
2. What is the Taylor series expansion of f(x)= x2-x+1 about the point x=-1?
a) f(x) = -3-3(x+1)+(x+1)2
b) f(x) = -3-(x+1)+(x+1)2
c) f(x) = -3-3(x+1)+2(x+1)2
d) f(x) = -1-3(x+1)+2(x+1)2
Explanation: Given, f(x)= x2-x+1 , f(-1)=1+1+1=3
f'(x) = 2x-1,f'(-1)=2(-1)-1= -3
f”(x) = 2, f”(-1)=2
fn (x)=0, for n > 2
The Taylor series expansion of f(x) about x=a is,
f(x) = f(a)+(x-a) f'(a)+(x-a)2 \(\frac{f”}{2!}\) (a)+⋯
Here a=-1,
f(x) = f(-1)+(x+1) f'(-1)+(x+1)2 \(\frac{f”}{2!}\) (-1) since,for n > 2, fn (x) = 0.
f(x) = -3-3(x+1)+(x+1)2
3. What is the first term in the Taylor series expansion of f(x) = 8x5-3x2-5x about x=2?
a) 232
b) 244
c) 234
d) 222
Explanation: Given, f(x) = 8x5-3x2-5x
First term in the Taylor series for f(x) is f(a). Here, a=2. Therefore,
f(2) = 8(2)5-3(2)2-5(2)=8(32)-3(4)-10=234.
4. In recurrence relation, each further term of a sequence or array is defined as a function of its succeeding terms.
a) True
b) False
Explanation: A recurrence relation is an equation that recursively defines a sequence or multidimensional array of values, once one or more initial terms are given; each further term of the sequence or array is defined as a function of its preceding terms.
5. The initial condition for the recurrence relation of a factorial is ___________
a) 0!=0
b) 0!=1
c) 1!=1
d) 1!=0
Explanation: Factorial of a number is defined by the recurrence relation as, n!=n(n-1)! for n>0 and the initial condition is 0!=1.
6. For the power series of the form, \(∑_{i=0}^∞ a_i z^i,\) which one of the following may not be true?
a) The series converges only for z=0
b) The series converges absolutely for all z
c) The series converges absolutely for all z in some finite open interval (-R, R) and diverges if z<-R or z>R
d) At the points z=R and z=-R, the series will diverge
Explanation: A power series in a variable z is in the form of an infinite sum as given, i.e., \(∑_{i=0}^∞ a_i z^i,\), where ai are integers, real numbers, complex numbers, or any other quantities of a given type.
For any power series, one of the following is true:
i. The series converges only for z=0.
ii. The series converges absolutely for all z.
iii. The series converges absolutely for all z in some finite open interval (-R, R) and diverges if z<-R or z>R. At the points z=R and z=-R, the series may converge absolutely, converge conditionally, or diverge.
7. Maclaurin Series is named after ______________
a) Colin Maclaurin
b) Normand Maclaurin
c) Ian Maclaurin
d) Richard Cockburn Maclaurin
Explanation:
Colin Maclaurin (February 1698 – 14 June 1746), was a Scottish Mathematician. The Maclaurin Series was named after him.
Normand Maclaurin (10 December 1835 – 24 August 1914) was a Scottish-born physician, company director, Australian politician and university administrator.
Ian Maclaurin, born 30 March 1937, is a British businessman, who has been chairman of Vodafone and chairman and chief executive of Tesco.
Richard Cockburn Maclaurin (June 5, 1870 – January 15, 1920) was a Scottish-born U.S. educator and mathematical physicist.
8. Maclaurin’s Theorem is a special type of Taylor’s Theorem.
a) False
b) True
Explanation: The Taylor series expansion of f(x) about x=a is,
f(x)= f(a)+(x-a) f'(a)+(x-a)2\(\frac{f”}{2!}\) (a)+⋯
If we put x=0, we get the Maclaurin series which is given by, f(x)= f(0)+xf'(0)+x2 \(\frac{f”}{2!}\) (0)+⋯
9. The Mclaurin Series expansion of sin(ex) is?
a) sin(1)+\(\frac{x.cos(1)}{1!}+\sum_{n=2}^{\infty}\sum_{a=0}^{\infty}\frac{x^n.(-1)^a}{n!}\times\frac{(2a+1)^n}{(2a+1)!}\)
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) \(\sum_{n=2}^{\infty}\sum_{a=0}^{\infty}\frac{x^n.(-1)^a}{n!}\times \frac{(2a+1)^n}{(2a+1)!}\)
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)+\(\frac{x.cos(1)}{1!}+\sum_{n=2}^{\infty}\sum_{a=0}^{\infty}\frac{x^n.(-1)^a}{n!}\times\frac{(2a+1)^n}{(2a+1)!}\)
10. What is the coefficient of x101729 in the series expansion of cos(sin(x))?
a) 0
b) 1⁄101729!
c) -1⁄101729!
d) 1
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.