Browsed by
Tag: calculus

Epsilon-N Proof of a Limit of a Sequence: Lim[2n/(3n+2)] = 2/3 – Article by G. Stolyarov II

Epsilon-N Proof of a Limit of a Sequence: Lim[2n/(3n+2)] = 2/3 – Article by G. Stolyarov II

Comments 1 comment
The New Renaissance Hat
G. Stolyarov II
July 12, 2014
******************************
Note from the Author: This proof was originally published on Associated Content (subsequently, Yahoo! Voices) in 2007.  The article earned over 17,250 page views on Associated Content/Yahoo! Voices, and I seek to preserve it as a valuable resource for readers, subsequent to the imminent closure of Yahoo! Voices. Therefore, this essay is being published directly on The Rational Argumentator for the first time.  ***
***
~ G. Stolyarov II, July 12, 2014
***

This is the first in a series of formal mathematical proofs I intend to present in order to assist anybody who has ventured into the challenging but fascinating world of advanced calculus and real analysis. I start with a fairly basic proof: the limit of the nth term of a sequence as n becomes increasingly large. This is an epsilon-N proof, which uses the following definition: lim(n approaches ∞)xn= L iff for each real number ε>0, there exists a positive integer N(ε) such that if n ≥ N(ε), then │xn-L│< ε, i.e., L- ε < xn< L+ ε.

The epsilon-N proof has two component steps; first, we assume that our ε>0 is given and work backward to transform the inequality │xn-L│< ε in order to find an appropriate value of N(ε) that corresponds to a given value of ε. Then, using the value of N(ε) we found, we work forward to show that if n ≥ N(ε), then │xn-L│< ε.

The proof also uses the Archimedean Property, which states that the set of positive integers is not bounded above. There are actually four equivalent conditions that are known as the Archimedean Property:

1. If a,b are in R, a>0, b>0, then there is a positive integer n such that na>b

2. The set of positive integers is not bounded above.

3. For each real number x, there exists an integer n such that n ≤x < n+1

4. For each positive real number x, there exists a positive integer n such that 1/n ≤ x

Prove: lim(n approaches ∞)[2n/(3n+2)] = 2/3

Proof: Let ε>0 be given. Find N(ε)є Z+ such that if N(ε) < n, then │2n/(3n+2)-2/3│< ε.

Working backward to transform the inequality: │2n/(3n+2)-2/3│< ε

│6n/[3(3n+2)]-2(3n+2)/[3(3n+2)]│< ε

│[6n-2(3n+2)]/[3(3n+2)]│< ε

│[6n-6n-4]/[3(3n+2)]│< ε

│-4/[3(3n+2)]│< ε

Since ε>0, (3n+2)>0, the above inequality is the same as

4/[3(3n+2)] < ε

4/(3ε) < (3n+2)

4/(3ε)- 2 < 3n

4/(9ε)- 2/3 < n

Now I work forward to prove the original result:

Let N(ε)є Z+ э 4/(9ε)- 2/3 < N(ε).

Since 4/(9ε)- 2/3 is a real number, by the Archimedean Property it must be the case that some integer exists which is greater than that real number-since the set of positive integers is not bounded above. Here, we call that integer N(ε).

For all n>N, if 4/(9ε)- 2/3 < N < n, then:

4/(3ε)- 2 < 3n

4/(3ε) < (3n+2)

4/[3(3n+2)] < ε

│-4/[3(3n+2)]│< ε

│6n/[3(3n+2)]-2(3n+2)/[3(3n+2)]│< ε

│2n/(3n+2)-2/3│< ε

I have demonstrated the above inequality, which is sufficient to demonstrate that lim(n approaches ∞)[2n/(3n+2)] = 2/3.

I have hence proved what was desired. Another way to express that the desired objective has been obtained (which I shall use in future proofs of this sort) is the abbreviation

“Q. E. D.” of the Latin “Quod Erat Demonstraterum,” which means “that which was to be demonstrated.”

How to Solve Problems of Integration by Parts Using the Tabular Method (2007) – Article by G. Stolyarov II

How to Solve Problems of Integration by Parts Using the Tabular Method (2007) – Article by G. Stolyarov II

Comments 2 comments
The New Renaissance Hat
G. Stolyarov II
July 6, 2014
******************************
Note from the Author: This essay was originally written and published on Associated Content (subsequently, Yahoo! Voices) in 2007. The tabular method is one of my favorite mathematical techniques, capable of saving tremendous amounts of time in completing certain integration problems. While the method is not commonly emphasized in conventional calculus curricula, the enthusiastic reception of this article showed that there is significant interest among students in this concise, elegant approach. This article earned over 30,000 page views on Associated Content/Yahoo! Voices, and I seek to preserve it as a valuable resource for readers, subsequent to the imminent closure of Yahoo! Voices. Therefore, this essay is being published directly on The Rational Argumentator for the first time. 
***
~ G. Stolyarov II, July 6, 2014
***
Doing calculus problems which require integration by parts can be a lengthy and tedious process, even for someone with experience in finding the integrals of functions. Fortunately, for many of the most common types of integration by parts, there is fast, simple shortcut available. I have used the tabular method to great advantage on exams and math contests; this technique was one of the tools that enabled me to earn a perfect score of 40 on the Continental Mathematics League’s nationwide calculus competition in 2005.
***

The tabular method can be applied to any function which is the product of two expressions, where one of the expressions has some nth derivative equal to zero. For instance, the tabular method can be used to find the indefinite integral of x4e3x, but not of sin(x)e3x. This is because the 5th derivative of x4 is equal to zero, whereas sin(x) does not have any nth derivative which always exhibits zero values.

The tabular method uses a convenient table with three columns. We can call the first column “Signs,” the second column “u” and the third column “dv.” Under the column called “Signs,” we list positive and negative signs in alternating order for as many times as the problem requires. The first entry in the column labeled “u” will be the part of the function we want to integrate which can be reduced to zero through successive differentiation. In integrating our sample function, x4e3x, we will put x4 in the column labeled “u.” The subsequent entries in the “u” column will be the successive derivatives of the first entry — all the way to zero.

The first entry in the column labeled “dv” includes the other part of the function we want to integrate. The subsequent entries in this column will be the successive integrals of the first entry. For our sample problem, the first entry in the “dv” column will be e3x.

This is how the table for finding the integral of x4e3x will look:

Signs u dv
+ x4 e3x
– 4x3 (1/3)e3x
+ 12x2 (1/9)e3x
– 24x (1/27)e3x
+ 24 (1/81)e3x
– 0 (1/243)e3x

Using this table to find the indefinite integral of the function requires taking the sign from each row in the column except the last, applying it to the entry for “u” in the same row, and multiplying the result by the entry for “dv” in the next row. Doing this problem on paper, you would simply draw arrows between the following expressions:

+ x4 and (1/3)e3x
– 4x3 and (1/9)e3x
+ 12x2 and (1/27)e3x
– 24x and (1/81)e3x
+ 24 and (1/243)e3x

Now multiply each of the expressions linked by “and” (or the arrows on paper) and add them together to get the final indefinite integral:

(1/3)x4e3x – (4/9)x3e3x + (4/9)x2e3x – (8/27)xe3x + (8/81)e3x + C

Remember to include the constant C if you wish to leave the integral in indefinite form without evaluating it.

The great advantage of using the tabular method is the ability to do integration by parts mechanically, without needing to exert a large amount of thinking about the special circumstances of the problem. The procedure is fairly quick to memorize and easy to retain. After you learn it once, it will always be at your disposal as a tool for quickly and easily determining indefinite integrals that would otherwise take an immense amount of time to find.