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- Thread starter 57Strat
- Start date
Lee
Well-Known Member
- Joined
- Feb 11, 2002
- Messages
- 6,076
Devin,
Got your package Saturday! Man, what a treat! TheLars is.........well, different and the Cohiba is a work of art!
I will definately smoke the Lars soon and post my review! It will be humorous, I'm sure! BTW, it's not in the humi, smells too wierd!
As for the Cohiba, I'll save that for a special day!
Thanks Bro!
Lee
Got your package Saturday! Man, what a treat! TheLars is.........well, different and the Cohiba is a work of art!
I will definately smoke the Lars soon and post my review! It will be humorous, I'm sure! BTW, it's not in the humi, smells too wierd!
As for the Cohiba, I'll save that for a special day!
Thanks Bro!
Lee
57Strat
...................
- Joined
- Jan 14, 2001
- Messages
- 2,815
A short weekend......sigh
Glad to hear everyone is up for another contest!!!
Kenny, I play a little.
Pete, having the Robert Johnson does count! And yes, I have a copy...
Lee, I know what you mean as I had to keep them in the garage. ??? Enjoy those cigars!
Glad to hear everyone is up for another contest!!!
Kenny, I play a little.
Pete, having the Robert Johnson does count! And yes, I have a copy...
Lee, I know what you mean as I had to keep them in the garage. ??? Enjoy those cigars!
57Strat
...................
- Joined
- Jan 14, 2001
- Messages
- 2,815
A sample of the new contest.........
Please determine whether these series converge or diverge, and all logarithms are natural logarithms to the base e=2.71828...
It is well-known that the series
1/1p + 1/2p + 1/3p + ...
1/3(ln 3)p + 1/4(ln 4)p + 1/5(ln 5)p + ...
1/16(ln 16)((ln(ln 16))p + 1/17(ln 17)((ln(ln 17))p + 1/18(ln 18)((ln(ln 18))p + ...
etc. all converge if p>1 and diverge if p<1. The sums are taken from the appropriate starting points to infinity.
In this problem we consider series that lie on the edge between the case where p=1 and these series diverge, and p>1 where these series converge. Specifically, we wish to consider the case where the product of the logarithms becomes arbitrarily deep, (ln x) (ln ln x) (ln ln ln x) ... However, there is a difficulty as the nesting of the logarithms gets deep. Suppose we start with a fairly large number like B=10000000001000000000. Then ln B = 20723265837, ln ln B =23.755, ln ln ln B = 3.1678, ln ln ln ln B = 1.1530, ln ln ln ln ln B = .14239, ln ln ln ln ln ln B = .14239, ln ln ln ln ln ln ln B = -1.9492, and ln ln ln ln ln ln ln ln B is undefined, since negative numbers do not have logarithms.
This means that to get beyond 6 levels of logarithms we would have to start at a number much larger than B. Two different approaches will be used to avoid this problem with the starting point. The first approach is to step up the depth of the logarithms very slowly, so slowly that all of the terms in each product are greater than 1. Specifically, we consider the series
1/f(1) + 1/f(2) + 1/f(3) + 1/f(4) + ...
where f(n) is defined by
prod = n
term = n
while term > e do
term = ln(term)
prod = prod * term
end
f(n) = prod
Does this first, converge or diverge?
The second approach to the starting point difficulty is to substitute the function (1+ln n) for the function (ln n). The function (1+ln n) can be nested arbitrarily deeply when n>1. As before, the series
1/1p + 1/2p + 1/3p + ...
1/1(1+ln 1)p + 1/2(1+ln 2)p + 1/3(1+ln 3)p + ...
1/1(1+ln 1)((1+ln(1+ln 1))p + 1/2(1+ln 2)((1+ln(1+ln 2))p + 1/3(1+ln 3)((1+ln(1+ln 3))p + ...
etc, all converge when p>1 and diverge when p<1. The sums can now be taken from n=1 to infinity.
Now consider the following series which sits on the border between p=1 and p>1.
1/1 + 1/2(1+ln 2) + 1/3(1+ln 3)((1+ln(1+ln 3)) + 1/4(1+ln 4)((1+ln(1+ln 4))(1+ln(1+ln(1+ln 4))) + ...
Does this second, converge or diverge?
Please determine whether these series converge or diverge, and all logarithms are natural logarithms to the base e=2.71828...
It is well-known that the series
1/1p + 1/2p + 1/3p + ...
1/3(ln 3)p + 1/4(ln 4)p + 1/5(ln 5)p + ...
1/16(ln 16)((ln(ln 16))p + 1/17(ln 17)((ln(ln 17))p + 1/18(ln 18)((ln(ln 18))p + ...
etc. all converge if p>1 and diverge if p<1. The sums are taken from the appropriate starting points to infinity.
In this problem we consider series that lie on the edge between the case where p=1 and these series diverge, and p>1 where these series converge. Specifically, we wish to consider the case where the product of the logarithms becomes arbitrarily deep, (ln x) (ln ln x) (ln ln ln x) ... However, there is a difficulty as the nesting of the logarithms gets deep. Suppose we start with a fairly large number like B=10000000001000000000. Then ln B = 20723265837, ln ln B =23.755, ln ln ln B = 3.1678, ln ln ln ln B = 1.1530, ln ln ln ln ln B = .14239, ln ln ln ln ln ln B = .14239, ln ln ln ln ln ln ln B = -1.9492, and ln ln ln ln ln ln ln ln B is undefined, since negative numbers do not have logarithms.
This means that to get beyond 6 levels of logarithms we would have to start at a number much larger than B. Two different approaches will be used to avoid this problem with the starting point. The first approach is to step up the depth of the logarithms very slowly, so slowly that all of the terms in each product are greater than 1. Specifically, we consider the series
1/f(1) + 1/f(2) + 1/f(3) + 1/f(4) + ...
where f(n) is defined by
prod = n
term = n
while term > e do
term = ln(term)
prod = prod * term
end
f(n) = prod
Does this first, converge or diverge?
The second approach to the starting point difficulty is to substitute the function (1+ln n) for the function (ln n). The function (1+ln n) can be nested arbitrarily deeply when n>1. As before, the series
1/1p + 1/2p + 1/3p + ...
1/1(1+ln 1)p + 1/2(1+ln 2)p + 1/3(1+ln 3)p + ...
1/1(1+ln 1)((1+ln(1+ln 1))p + 1/2(1+ln 2)((1+ln(1+ln 2))p + 1/3(1+ln 3)((1+ln(1+ln 3))p + ...
etc, all converge when p>1 and diverge when p<1. The sums can now be taken from n=1 to infinity.
Now consider the following series which sits on the border between p=1 and p>1.
1/1 + 1/2(1+ln 2) + 1/3(1+ln 3)((1+ln(1+ln 3)) + 1/4(1+ln 4)((1+ln(1+ln 4))(1+ln(1+ln(1+ln 4))) + ...
Does this second, converge or diverge?
The answer is (D) all of the above and yes, it's my final answer! LOL!
Do I qualify for the consolation prize? ROTF!
Brother Devin... sorry your weekend was cut short but I hope you and Wendy had a great time!
Aloha Bro...
Wade
LOL!Do I qualify for the consolation prize? ROTF!
Brother Devin... sorry your weekend was cut short but I hope you and Wendy had a great time!
Aloha Bro...
Wade
Horse
Member 200
- Joined
- Aug 1, 2001
- Messages
- 2,981
It's as easy as elephants and bugs:
In the Infinite Seriesian Jungle, there are only two kinds of animals, elephants (divergent series) and bugs (convergent series). There are two basic ways to recognize elephants, i.e., to show that series diverge. Other tests for divergence are really special cases of the comparison test.
If the parts of the beast aren't small, it's an elephant. (If an does not tend to 0, the series does not converge.)
(comparison test) If a beast in the jungle is bigger than an elephant, it's another elephant!
Just as there are two kinds of elephants, there are two ways a series can fail to converge. The partial sums may tend to infinity, but another kind of divergence ("by oscillation") happens if they simply wander around finitely without ever committing themselves
Nice one Devin...pulling it from Someone else's contest
Mathematical Series
The Contest Center
59 DeGarmo Hills Road
Wappingers Falls, NY 12590
Even thought it's been (gulp) 20 years, I can still remember my t-shirt that said 'Math 318's a Bitch'
At A&M, that would be Differential Equations....and yes, it's a Bitch.
In the Infinite Seriesian Jungle, there are only two kinds of animals, elephants (divergent series) and bugs (convergent series). There are two basic ways to recognize elephants, i.e., to show that series diverge. Other tests for divergence are really special cases of the comparison test.
If the parts of the beast aren't small, it's an elephant. (If an does not tend to 0, the series does not converge.)
(comparison test) If a beast in the jungle is bigger than an elephant, it's another elephant!
Just as there are two kinds of elephants, there are two ways a series can fail to converge. The partial sums may tend to infinity, but another kind of divergence ("by oscillation") happens if they simply wander around finitely without ever committing themselves
Nice one Devin...pulling it from Someone else's contest
Mathematical Series
The Contest Center
59 DeGarmo Hills Road
Wappingers Falls, NY 12590
Even thought it's been (gulp) 20 years, I can still remember my t-shirt that said 'Math 318's a Bitch'
At A&M, that would be Differential Equations....and yes, it's a Bitch.
Allofus123
Here ducky, ducky, ducky!
- Joined
- Aug 25, 2001
- Messages
- 3,869
From pure reasoning with beyond a basic knowledge of Relativity, and theory of. Time, is in essence A measurement used by humanity to establish order, not only in daily life where reality exists, but to trace historically the past, which is not real, or impossible to recall from memory without method of evidence (verifiable or not) to impart facts of what occurred then, which is unknown to us in our personal life' experience. Those considerations, are paramount to examine how it is possible for Eternity and Eternal to exist in our perception of inductive reasoning, when we have no verifiable or provable evidence to support Eternal or Eternity exists. Eternal or Eternity is metaphysical only, and to cross the boundary of time without physical experience by observation, is demanding another IT exists, or the premise is not possible to reason if no knowledge of it, is present. Moreover, with no history for us to study or learn from, we have no past within which to dwell; therefore, Eternal and Eternity are INNATE, or they would not exist in with comprehension of A Potential, without time present.
So the answer is...... Who cares! 
ROFLMAO!
So the answer is...... Who cares!
ROFLMAO!
D
DMC
Guest
Gentlemen, I have changed my screen name... No more, Devin Cole.
Thanks Rod!
BTW, thanks to a great CP member, I have your addy now...
No more, Devin Cole.Thanks Rod!
BTW, thanks to a great CP member, I have your addy now...
Allofus123
Here ducky, ducky, ducky!
- Joined
- Aug 25, 2001
- Messages
- 3,869
I think you opened up a can of worms here......
A concert starts in seventeen minutes and Devin along with 3 other chupas must all cross a bridge to get there. All four men begin on the same side of the bridge. You must help them across to the other side. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses (either one or two people) must have the flashlight with them. The flashlight must be walked back and forth. It cannot be thrown etc. Each person walks at a different speed. A pair must walk together at the rate of the slower man's pace.
Time to cross the bridge
Devin 1 minute
Matt 2 minutes
Mark 5 minutes
Ryan 10 minutes
For example, if Ryan and Devin walk across first, ten minutes have elapsed when they get to the other side of the bridge. If Ryan then returns with the flashlight, a total of twenty minutes have passed and you have failed the mission. Note: there are two known answers to this problem.
I'll be back with an answer or two after I finish BBQing.....
A concert starts in seventeen minutes and Devin along with 3 other chupas must all cross a bridge to get there. All four men begin on the same side of the bridge. You must help them across to the other side. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses (either one or two people) must have the flashlight with them. The flashlight must be walked back and forth. It cannot be thrown etc. Each person walks at a different speed. A pair must walk together at the rate of the slower man's pace.
Time to cross the bridge
Devin 1 minute
Matt 2 minutes
Mark 5 minutes
Ryan 10 minutes
For example, if Ryan and Devin walk across first, ten minutes have elapsed when they get to the other side of the bridge. If Ryan then returns with the flashlight, a total of twenty minutes have passed and you have failed the mission. Note: there are two known answers to this problem.
I'll be back with an answer or two after I finish BBQing.....
D
DMC
Guest
No, not running.....LMAO Just got tired of seeing my name, so I am using my initials.
Kenny,
1. Devin crosses with Matt (2 min)
2. Devin returns (3 min)
3. Mark crosses with Ryan (13 min)
4. Matt returns (15 min)
5. Devin crosses with Matt (17 min)
Total time for all to cross, 17 minutes.
ROFLMAO!!!
Kenny,
1. Devin crosses with Matt (2 min)
2. Devin returns (3 min)
3. Mark crosses with Ryan (13 min)
4. Matt returns (15 min)
5. Devin crosses with Matt (17 min)
Total time for all to cross, 17 minutes.
ROFLMAO!!!
7
74-1021679779
Guest
DMC I'm not in this pass,but how did you change your name?
I did not mean to put my whole name for post and I'd like to have a cool name on it.
I did not mean to put my whole name for post and I'd like to have a cool name on it.
D
DMC
Guest
Jason, just send Rod a message. I am sure he would be more than happy to help.