This set of Discrete Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Discrete Probability – Power Series”.
1. The explicit formula for the geometric sequence 3, 15, 75, 375,… is _______
a) 2*6! * 3n-1
b) 3 * 5n-1
c) 3! * 8n-1
d) 7 * 4n-1
View Answer
Explanation: The initial term is 3 and each subsequent term is the product of its previous term, the common ratio is 5. Thus the formula generating this sequence is an = 3 * 5n-1.
2. The third term of a geometric progression with common ratio equal to half the initial term is 81. Determine the 12th term.
a) 312
b) 415
c) 68
d) 59
View Answer
Explanation: Let the initial term be a and the common ratio r. The 3rd term is ar2 = 27 and the initial term is a=3r so 3r3 = 81 ⇒ r=3 ⇒ a=3. The a12 = a * r11 = 3 * 311 = 312.
3. Which of the following series is called the “formal power series”?
a) b0+b1x+b2x2+…+bnxn
b) b1x+b2x2+…+bnxn
c) 1/2b0+1/3b1x+1/4b2x2+…+1/nbnxn
d) n2(b0+b1x+b2x2+…+bnxn)
View Answer
Explanation: A formal power series is also called a “formal series”, of a field F is an infinite sequence b0, b1, b2, … over F. It is a function from the set of nonnegative integers to F i.e., 0, 1, 2, 3, … → F. A formal power series can also be written as b0+b1x+b2x2+…+bnxn.
4. sec(x) has a trigonometric series that is given by _______
a) ∞∑n=0 ((-1)nE2n / (2n)!)*x2n
b) ∞∑n=0 ((-1)nE2n)
c) ((-1)nB2n / (2n)!)*x2n
d) ∞∑n=0 ((2n)!)*x2n+1
View Answer
Explanation: A trigonometric series is an example of a Maclaurin series. Here, sec(x) can be represented as ∞∑n=0 ((-1)nE2n / (2n)!)*x2n.
5. Determine the interval and radius of convergence for the power series: ∞∑n=17n/n(3x−1)n-1.
a) (2x+1)/6
b) 7|3x−1|
c) 5|x+1|
d) 3!*|4x−9|
View Answer
Explanation: Okay, let’s start off with the Ratio Test to get our hands on L = limn→ ∞∣7n+1(3x−1)n/(n+1)n7n(3x−1)n-1∣=limn→∞∣7n(3x−1)n+1∣=|3x−1|limn→∞7n/(3n-1)=7|3x−1|.
6. Determine a power series representation for the function g(x)=ln(7−x).
a) ∞∑n=0 xn+1/7n+1
b) ln(14)∞∑n=0 xn+1/7n
c) ln(7)∞∑n=0 xn+1/7n+1
d) ln∞∑n=0 x/7n+1
View Answer
Explanation: We know that ∫1/7−x dx=−ln(7−x) and there is a power series representation for 1/7−x. So, ln(7−x)=−∫1/7−xdx
=−∫ ∞∑n=0 xn/7n+1dx=C
⇒ ∞∑n=0 xn+1/7n+1
So, the answer is, ln(7−x)=ln(7)∞∑n=0 xn+1/7n+1.
7. An example of Maclaurin series is _______
a) ∞∑n=0 (xn/n!)
b) ∞∑n=0 (x/5+n!)
c) ∞∑n=0 (xn+1/(n-1)!)
d) (xn/n)
View Answer
Explanation: The exponential function ex can described as ∞∑n=0 (xn/n!) which is an example of a Maclaurin series. This series converges for all x.
8. Find the power series representation for the function f(x)=x/4−x.
a) ∞∑n=0xn+1/4n+1
b) ∞∑n=0xn+14n
c) ∞∑n=0xn4n
d) ∞∑n=0xn+1
View Answer
Explanation: So, again, we’ve got an x in the numerator. f(x)=x*1/4−x. If there is a power series representation for g(x)=1/4−x, there will be a power series representation for f(x). Suppose, g(x)=1/4*1/1−x4. To get a power series representation is to replace the x with x4. Doing this gives, g(x)=1/4 ∞∑n=0 xn/4n (xn/4 nprovided ∣x/4∣<1) ⇒ g(x) = 1/4 ∞∑n=0 xn/4n = ∞∑n=0 xn/4n+1. The interval of convergence for this series is, ∣x/4∣<1⇒1/4 |x|<1⇒|x|<4. Now, multiply g(x) by x and we have f(x)=x*1/4−x=x ⇒ ∞∑n=0 xn/4n+1 = ∞∑n=0xn+1/4n+1 and the interval of convergence will be |x|<4.
9. What is the radius of convergence and interval of convergence for the power series ∞∑n=0m!(2x-1)m?
a) 3, 12
b) 1, 0.87
c) 2, 5.4
d) 0, 1/2
View Answer
Explanation: Suppose, L=limn→∞ |(m+1)!(2x+1)m+1/m!(2x+1)m|
= limm→∞∣(m+1)m!(2x-1)/m!|
= |2x-1|limm→∞(m+1)
So, this power series will only converge if x=1/2. We know that every power series will converge for x=a and in this case a=1/2. Remember that we get a from (x−a)n. In this case, the radius of convergence is R=0 and the interval of convergence is x=1/2.
10. Determine the radius of convergence and interval of convergence for the power series: ∞∑n=0 (x−7)n+1/nn.
a) 0, −1<x<1
b) ∞, −∞<x<∞
c) 1, −2<x<2
d) 2, −1<x<1
View Answer
Explanation: So, L=limn→∞∣(x−7)n+1/nn∣
L=limn→∞∣x−7/n∣
L=|x−7|limn→∞1/n=0
So, since L=0<1 any of the value of x, this power series will converge for every x. In these cases, the radius of convergence is R=∞ and interval of convergence is −∞<x<∞.
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