Discrete Mathematics Questions and Answers – Discrete Probability – Power Series

«
»

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

Answer: b
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.
advertisement

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

Answer: a
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

Answer: a
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

Answer: a
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

Answer: b
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|.
advertisement

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

Answer: c
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

Answer: a
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

Answer: a
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

Answer: d
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.
advertisement

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

Answer: b
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<∞.

Sanfoundry Global Education & Learning Series – Discrete Mathematics.

To practice all areas of Discrete Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
advertisement
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn