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Gambler’s ruin is Darwin’s ruin

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The same day I first watched “Expelled” in theaters, I also watched the movie “21”. The movie “21” is based on the true story of MIT students who made a fortune in Las Vegas casinos through the use of mathematics.

The real story behind the movie began with an associate of Claude Shannon by the name of Dr. Edward O. Thorp of MIT. In the Early 60’s, Thorp published a landmark mathematical treatise on how to beat casinos. His research was so successful that Las Vegas casinos shut down many of their card tables for an entire year until they could devise counter measures to impede Thorp’s mathematics.

Thorp is arguably the greatest gambler of all time. He extended his gambling science to the stock market and made a fortune. His net worth is in the fractional to low billions. He is credited with some independent discoveries which were the foundation to the Black-Scholes-Merton equation relating heat transfer thermodynamics to stock option pricing. The equation won the Nobel prize and was the subject of the documentary: The Trillion Dollar Bet.

Thorp would probably be even richer today if Rudy Gulliani had not falsely implicated him in the racketeering scandal involving Michael Milken. Thorp, by the way, keeps a dartboard with Gulliani’s picture on it… 🙂

The relevance of Thorp’s math to Darwinism is that Thorp was a pioneer of risk management (which he used to create the world’s first hedge fund). In managing a hedge fund or managing the wagers in casinos, one is confronted with the mathematically defined problem of Gambler’s Ruin. The science of risk management allows a risk manager or a skilled gambler to defend against the perils gamblers ruin. Unfortunately for Darwinism, natural selection has little defense against the perils of gambler’s ruin.

Even if an individual has a statistical advantage over a casino game, it is possible the individual can lose. Let’s say a skilled player has a 1% advantage on average over the casino. He wanders into the casino, looks for a favorable opportunity and wagers $500,000.00.

If he has a 1% statistical advantage, that means he has a 50.5% chance of winning and a 49.5% chance of losing. Even though he has a slight edge, he still has a very substantial chance of losing. It would be unwise to bet $500,000.00 if that is his life savings!

The movie “21” romanticized the advantage skilled players have. The movie “21” portrayed the MIT students as people who could sit at card tables and bilk casinos like ATM machines. That’s not how it works as testified by one of the more noteworthy members of the real MIT team by the name of Andy Bloch. Bloch reported that during his tenure as manager of the MIT team, the team was once in the red for 9 months before recovering. Skilled players lose big bets not quite 50% of the time. It is not unusual, on average, to have a losing streak of 8 hands in a row every 256 rounds. Ben Mezrich reported in his book, Bringing Down the House, an incident where the Big Player of the MIT team lost 3 hands in a row in 45 seconds of play for a sum total of $120,000.00! It happens…

A skilled player with a 1% advantage might expect to play 50,000 hands before his expected value exceeds the effect of one standard deviation of bad luck. That means he might have to play a looooong time before he realizes a profit….

What does this have to do with Darwinism? Darwin argued that

Natural selection acts only by taking advantage of slight successive variations; she can never take a great and sudden leap, but must advance by short and sure, though slow steps.”

But that is complete nonsense mathematically speaking because of the problem of gambler’s ruin. It is not surprising that Darwin could not see the flaw in his argument because he could not even do high school algebra even after substantial effort. The lack of basic math and logic pervades his flawed theory.

The problem is that a selectively-advantaged traits are still subject to random events. The most basic random event is with whether a parent will even pass down a gene to a child in the first place! Added to that problem is the nature of random events in general. A genetically advantaged individual may die by accident, get consumed by a predator, etc.

And the problem gets worse. Even if selectively advantage traits get spread to a small percentage of the population, it still has a strong chance of being wiped out by the sum total of random events. The mathematics of gambler’s ruin helped clarify the effect of random “selection” on natural selection.

Without going into details, I’ll quote the experts who investigated the issues. Consider the probability a selectively advantaged trait will survive in a population a mere 7 generations after it emerges:

if a mutant gene is selectively neutral the probability is 0.79 that it will be lost from the population
….
if the mutant gene has a selective advantage of 1%, the probability of loss during the fist seven generations is 0.78. As compared with the neutral mutant, this probability of extinction [with natural selection] is less by only .01 [compared to extinction by purely random events].
….

Theoretical Aspects of Population Genetics
Motoo Kimura and Tomoko Ohta

This means is that natural selection is only slightly better than random chance. Darwin was absolutely wrong to suggest that the emergence of a novel trait will be preserved in most cases. It will not! Except for extreme selection pressures (like antibiotic resistance, pesticide resistance, anti-malaria drug resistance), selection fails to make much of an impact.

The contrast between a skilled gambler and natural selection is that a skilled player can wager small fractions of the money he sets aside for his trade. If a skilled gambler has $50,000, he might wager $100 at a time until the law of large numbers causes his statistical advantage to be asserted. He can attempt many many trials until his advantage eventually prevails. In this manner a skilled gambler can protect himself against the mathematics of gamblers ruin.

But natural selection is a blind watchmaker. It does not know how to perform risk management like a skilled player or the great math wizard, Edward Thorp. For natural selection to succeed in the way Thorp succeeded in the great casinos of Nevada and Wall Street, it has to hope the same mutant appears spontaneously many many times in many individuals. But for complex genes, this doesn’t happen. Truly novel and beneficial mutations are rare. They don’t repeat themselves very often, and when they arise, they will likely be wiped out unless there is fairly intense selection pressure (like we see in pesticide resistance or anti-biotic resistance or anti-malaria drug resistance, or malaria resistance associated with sickle cell anemia).

A further constraint on selective advantage of a given trait is the problem of selection interference and dilution of selective advantage if numerous traits are involved. If one has a population of 1000 individuals and each has a unique, novel, selectively-advantaged trait that emerged via mutation, one can see this leads to an impasse –selection can’t possibly work in such a situation since all the individuals effectively cancel out each other’s selective advantage.

This illustrates that there has to be a limit to the number of innovations appearing in a population simultaneously for selection to work. The emergence of advantageous mutations in a population has the net effect of diluting the selective advantage of all the traits.

If trait A has a large selective advantage in relation to trait B, trait A dilutes the selective advantage of trait B. Thus trait B is exposed more and more to gambler’s ruin because of the existence of trait A. For example an individual with better eyesight (trait A) might prevail over an individual with higher intelligence (trait B). An otherwise good trait (intelligence) is lost because another trait (good eyesight) interferes with the ability of that trait (intelligence) to be maintained…

Thus one can see the problem of many “slight advantageous traits” being necessarily “slight” because of the problem of interference. But “slight” implies they are subject to gambler’s ruin, and thus unlikely to be preserved as Darwin asserted. Thus Darwin was dead wrong….

John Sanford gives a more rigorous treatment in his book Genetic Entropy where he gives more exact numbers on the limits of selective advantage based on problems such as interference. Sanford shows that a 1% selective advantage is fairly generous, and is usually less than 1%. [I emphasize the word “usually”].

Most ironic is that Fisher’s analysis of the effect of gambler’s ruin essentially trashes his own theorem, Fisher’s Fundamental Theorem of Natural Selection. Fisher’s Malthusian notions of “fitness” in his fundamental theorem do not account for the effect of random events taking out selectively advantaged traits. The fundamental theorem assumes evolution is noise free with respect to fitness, that advantageous traits always result in more offspring. We know empirically and theoretically this cannot possibly be true even on the approximate model of Mendelian inheritance.

For reasons such as those I laid out, many believe molecular evolution had to be mostly invisible to selection. Attributing even 5% of molecular evolution to Darwinism would be extremely generous. See: Kimura’s Neutral Theory.

Kimura gave an obligatory salute to Darwin by claiming adaptational features (like morphology) are exempt from his math. I’ve seen nothing supporting Kimura’s obligatory salute to Darwin. It seems his neutralist ideas apply quite well to realms beyond the molecular. NAS member Masotoshi Nei has finally been bold enough to assert most everything else about evolution, not just molecular evolution, is under much less selection pressure than previously assumed. I think Nei is right.

Yesterday afternoon I showed Kimura’s books to an ID-friendly senior in biology. His jaw dropped. He had studied molecular genetics, but our conversation yesterday helped him make the connections he had not made before. The math clearly indicates Darwin couldn’t possibly be right, and by way of extension, neither can Richard Dawkins.

These fairly obvious considerations were not lost upon Michael Lynch:

the uncritical acceptance of natural selection as an explanatory force for all aspects of biodiversity (without any direct evidence) is not much different than invoking an intelligent designer

Michael Lynch
The Origins of Genome Architecture, p 368

Notes:

1. I created a Microsoft Excel Spreadsheet is provided for illustration of these concepts. I used a random number generator to simulate the progress of 10 equally skilled gamblers in a casino. Press the “F9” to redraw the graph. One can see that even “selectively” advantaged individuals can lose. The important thing to grasp is that “slight selective” advantages do not look very different from random walks except in the long run. The problem for natural selection in the wild is that there usually is no “long run” for a newly emerged trait if it suffers from gamblers ruin. The “long run” exists for skilled and intelligent risk managers like Edward Thorp, it does not exist, statistically speaking, for most selectively advantageous traits.

A copy of my spreadsheet can be accessed here.

Sometimes pressing “F9” will cause most of the gamblers to win, and other time it will cause most of them to lose. This underscores the strong effect of random events even when one possess an inherent statistical advantage such as a gambling skill or a selectively advantaged trait.

2. Here is a nice pic of Bill with a standard casino die.

In the 1970’s, casinos had to redesign their craps tables in order to foil skilled dice throwers who exploited slightly non-random behaviors of dice. Las Vegas laws were passed that prevented skilled players from using there specially designed tosses which would exhibit non-random, statistically advantageous behavior.

Some people still claim to be able to influence dice so as to create non-random outcomes in a legal way. However, even skilled crap shooters need principles of risk management and precautions against gambler’s ruin to succeed.

[UPDATE:

1. 5/5/08 World Renowned Geneticist Joe Felsenstein responds to my essay here: Gambler’s Ruin is Darwin’s Gain.

2. 5/5/08 HT: ICON-RIDS:

Natural Selection is daily and hourly scrutinising, throughout the world, the slightest variations; rejecting those that are bad, preserving and adding up all that are good.

C.DARWIN sixth edition Origin of Species — Ch#4 Natural Selection

This is an even better quote showing how wrong Darwin was in light of these discussions.

See: this comment

3. Thanks to pantrog of PT for his editorial correction about sickle cell anemia. That was my editorial mistake not seeing it in the first place. My error was pointed out here here.

4. 5/8/08 One could easily modify the spreadsheet to stop progress when zero is hit, except if I did this, one would not easily see all the lines since most of them abort early thus giving a misleading impression of large scale progress. See this comment:
Comment about Spreadsheet

5. I wrote: If he has a 1% statistical advantage, that means he has a 50.5% chance of winning and a 49.5% chance of losing. To clarify, the outcomes are complicated by double-downs, splits, and blackjacks, etc. so the notion of “win” in this thread is effective average win over time per round….I didn’t want to get into these deep specifics earlier as it was peripheral to the thread…

6. 5/31/08 In response to various comments by those at UD and PandasThumb, I created another spreadsheet with some improvements. See the improvements at: ruin_olegt_mod1.xls. The princple changes were in response to suggestions by a very fine physicist by the name of Olegt who sometimes posts at TelicThoughts and PT. The new simulation has more rounds and actually prevents a player from playing once he is ruined.

]

Comments
There are two very serious errors in this post. 1. Equating "natural variation" to mutation. 2. Ignoring time. No matter how low the probability of an outcome, if there are enough trials, the outcome is sure to occur. Related to the willingness to make such obvious errors, the core problem with the ID paradigm is that it has no explanatory power. It is fundamentally an argument from ignorance. "I thought really hard about this and I claim natural laws cannot account for it." The history of science is one long laugh at such claims. Note that this has nothing to do with theism. One could claim for example that the very existence of natural laws, or that the very persistence of the universe, reflects the continuous involvement of God in the universe.Alan
May 26, 2012
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scordova wrote: "That is a nit-pick that is also an erroneous nit-pick. See the discussion of Black-Scholes from the perspective of Brownian motion and Statistical Mechanics Here. That discussion includes thermodynamics and heat flow. Equation 10 is the famous Black-Scholes equation. Its solutions are widely adopted for financial analysis by traders, fund managers, economists, and so on. The Equation can be further transformed into standard form of the heat diffusion equation from Physics (Thermodynamics) " For the record, placing information in these comments which helps the scientifically untrained folks with scientific distinctions such as I wrote, is not nitpicking; in my view it ups the profile of these discussions by demonstrating that there are professional folks in the sciences posting here. Scordova does not need to take this as a put down. After going to the site of the link he provides I see that an economist equates heat flow with thermodynamics. Not helpful. My Holman thermo text from 1970 makes no mention of Fourier's law. The decent wikipedia article on thermodynamics makes no mention of heat flow: http://en.wikipedia.org/wiki/Thermodynamics#Related_branchesgroovamos
October 20, 2008
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Legit or crank? As so often, the case is not so clear cut. Earlier, he obviously has done some legit work, while his International Committee for Scientific Ethics and Accountability shouts "CRANK". He isn't referred to in the books with which I was introduced to Lie-theory - and that seems to be part of his problem :-)DiEb
June 25, 2008
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Yes, that is the one. What do you think? Legit or crank? Have you heard of him in the course of your work? See: Hadronic Mathematics I Hadronic Mathematics II Many thanks. I figured whatever the outcome of your evaluation you might find his ideas amusing. I'm not familiar with Lie-admissible Algebras at all... I hear the term Lie-algebra all the time in connection with advanced physics, but I've yet to run into it in formal study... Thank you again, and thanks for responding to my questions here at UD.scordova
June 24, 2008
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Ruggero Santilli?DiEb
June 24, 2008
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DiEB, Can you tell if R. Santilli is a real mathematician? He claims to have taught at some good schools. Thanks. Salvadorscordova
June 24, 2008
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Sal, so, we're talking about the Dirichlet conditions as stated in post #160.
However, it appears that condition 3 (absolute integrability of P(x)) will ensure that a normalization integral exists. So that seemed reasonable to me….is this correct?
Even for a finite period, condition 3 ("f(x) must be absolutely integrable over a period") isn't sufficient for the existence of a normalization integral, you'd need condition 4 ("f(x) must be bounded"), too. But, as I stated before, you wouldn't invoke the Dirichlet conditions in the context of L², as there is a beautiful Fourier transform on this space. I never used Apostol's "Modern Analysis" before. If someone is interested in mathematics, I'd usually hint him at books for which more current editions exist.DiEb
June 24, 2008
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Condition 3 in which enumeration? In post #159, you give condition 3 as 3. ?(x) and d?(x)/dx must be finite I don’t see there anything about absolute integrability…
I thought you meant condition 3 from the list of 4 Dirichlet conditions. I mis-interpreted your remarks.
as stated so nicely in the text of Apostol.
Do you have Apostol's book? Do you like it?
“physically realizable boundary conditions”? That I don’t know neither - I’m a mathematician, not a physicist.
I'll point you to a simple example in the Britney Spears Guide to Finite-Barrier Quantum Wells. See equation (3)... One can see the solutions to the schrodinger equation (when they are defined from -infinity to +infinity). In this case the solutions are going to be functions damped by a decaying exponential. In this case, we have two well-defined boundary conditions imposed by 2 physical boundaries. We could in priniciple create n-boundaries, for n-boundary conditions..... It's seems it would be impossible (in a universe with finite resources) to construct an infinite number physical boundaries spaced apart in such a way that we have a function Psi(x) that is not absolutely integrable. I believe if we have a finite set of boundaries, with a finite distance from each other, we'll eventually end up with functions damped with decaying exponentials on the ends that go to +/- infinity. Does that seem correct?scordova
June 23, 2008
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What I presume you meant was the series converges on points except where the the function lies outside the null-set.
What I wanted to say: If a function satisfies the Dirichlet conditions - or Jordan's test - then it has at most a countably number of discontinuities, i.e., the points of discontinuity form a null-set (regarding λ.) In this points, the series will converge to the mean of the left-hand-side and right-hand side value of the function, as stated so nicely in the text of Apostol.
It is possible in principle that there exists a Psi(x) that is not absolutely integrable over a period of infinity but Psi*(x)Psi(x) is integrable.
yes.
But I don’t know that there exists a set of physically realizable boundary conditions which would imply such a Psi(x) solution to Schrodinger’s equation!
"physically realizable boundary conditions"? That I don't know neither - I'm a mathematician, not a physicist.
However, it appears that condition 3 (absolute integrability of P(x)) will ensure that a normalization integral exists. So that seemed reasonable to me….is this correct?
Condition 3 in which enumeration? In post #159, you give condition 3 as
3. Ψ(x) and dΨ(x)/dx must be finite
I don't see there anything about absolute integrability...DiEb
June 21, 2008
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Correct me if I’m wrong (heck, you’ll correct even when I’m right…), but isn’t it possible that ? violates condition no. 3, as as ? ? L² per normalization integral?
It is possible in principle that there exists a Psi(x) that is not absolutely integrable over a period of infinity but Psi*(x)Psi(x) is integrable. But I don't know that there exists a set of physically realizable boundary conditions which would imply such a Psi(x) solution to Schrodinger's equation! However, it appears that condition 3 (absolute integrability of P(x)) will ensure that a normalization integral exists. So that seemed reasonable to me....is this correct?scordova
June 20, 2008
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Using Dirichlet conditions in the context of general Fourier transforms isn’t appropriate. Using them for Fourier series is all fine and dandy: they’ll provide you with the existence of on inverse transform, at least outside of a null-set.
I don't believe that it is customary to speak of an inverse transform when one is speaking of a fourier SERIES! :-) What I presume you meant was the series converges on points except where the the function lies outside the null-set. BUT, does there exist a function x(t) satisfying Dirichlet conditions with transform X(f), where the inverse transform of X(f) yields a function that is not in the same class as x(t) in L1? If not one might complain Dirichlet is too austere for Fourier Transforms. But then I'll counter by saying, will Jordan's Test or Dini's test for Fourier series lead to convergence (yes according to Apostol), so in that sense Dirichlet is too austere for Fourier Seiries.scordova
June 20, 2008
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upps - bounded differentiability trumps hölder-continuity...DiEb
June 20, 2008
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Sal- thanks for your surprisingly modest post. I'm afraid I have to clarify a few points... 1. DiEb:
Correct me if I’m wrong (heck, you’ll correct even when I’m right…), but isn’t it possible that ? violates condition no. 3, as Ψ ∈ L² per normalization integral? Therefore, the Dirichlet conditions wouldn’t be appropriate “in the context of the discussion…”
Sal:
That’s why I put it up to get your editorial feed back. :-) I don’t know. But recall Psi(x) must have NO discontinuities, and dPsi(x)/dx must be continuous. We can’t have any of the more extreme behaviors which motivated the generalization of Riemann integrals to Lesbegue integrals. So I don’t know. Thank you for your editorial comments.
My question was rhetorical... 2.
Dirichlet is not the approprate set of conditions for inverse Fourier transforms and series as I had supposed in December 2007 when I put my post up at YoungCosmos.
To repeat it: Using Dirichlet conditions in the context of general Fourier transforms isn't appropriate. Using them for Fourier series is all fine and dandy: they'll provide you with the existence of on inverse transform, at least outside of a null-set. But you know that I tend to neglect null-sets, as does your professor at the GMU, I presume (post #159). 3.
An appropriate conditions for inverse transforms are that the functions be Holder Continuous. I do not know if Holder Continuous functions are both necessary and sufficient. They are sufficient, I do not know if they are necessary.
Sal, in post #142, you quoted Apostol:
These integrals, which are in many ways analogous to Fourier series, are known as Fourier integrals, and the theorem which gives sufficient conditions for representing a function by such an integral is known as the Fourier integral theorem.
So, if you read on in Apostol's text, you'll find some sufficient conditions for the existence of an inverse transform (outside the null-sets of the points of not-continuity :-) )... I presented those in post #144, and in post #147, I gave an example of a function that doesn't satisfy the Dirichlet conditions and isn't hölder-continuous, but nevertheless has an inverse transform (outside the null-sets of the points of not-continuity :-) ). So, what's necessary and sufficient? On L², the Fourier transformation is an isometric isomorphism... 4.
I’m presuming Psi(x) is Holder Continuous, but I actually don’t know.
Differentiability trumps hölder-continuity. But the real question is: What's about the second (weak) derivation (is it in L²?), as the appropriate spaces for the Schrödinger equations are Sobolev :-) n.b.: on this board, I prefer L¹ and L¹ to other spaces, as I only can invoke the superscripts ¹ and ² and have not found a way to write a script-S... :-)DiEb
June 19, 2008
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Correct me if I’m wrong (heck, you’ll correct even when I’m right…), but isn’t it possible that ? violates condition no. 3, as ? ? L² per normalization integral? Therefore, the Dirichlet conditions wouldn’t be appropriate “in the context of the discussion…”
That's why I put it up to get your editorial feed back. :-) I don't know. But recall Psi(x) must have NO discontinuities, and dPsi(x)/dx must be continuous. We can't have any of the more extreme behaviors which motivated the generalization of Riemann integrals to Lesbegue integrals. So I don't know. Thank you for your editorial comments. Dirichlet is not the approprate set of conditions for inverse Fourier transforms and series as I had supposed in December 2007 when I put my post up at YoungCosmos. My physics professor at JohnsHopkins made passing reference to WavePacket formation and their likeness to FourierTransforms. Some of the math looked strikingly similar (including the use of the modulation theorem).... An appropriate conditions for inverse transforms are that the functions be Holder Continuous. I do not know if Holder Continuous functions are both necessary and sufficient. They are sufficient, I do not know if they are necessary. I'm presuming Psi(x) is Holder Continuous, but I actually don't know. :-) That's why you'll tell me, won't you, because you're eager to point out my mistakes....scordova
June 18, 2008
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-Sal are you still monitoring this thread? :-)DiEb
June 16, 2008
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Consider the context of the discussion, and the class of functions I was considering — solutions ot the Schrodinger Equations in certain contexts. From my physics textbook by Tipler and Llewellyn: For future reference, we may summarize the conditions that the wave function Psi(x) must meet in order to be acceptable: 1. Psi(x) must exist and satisfy the Schrodinger equation 2. Psi(x) and dPsi(x)/dx must be continuous 3. Psi(x) and dPsi(x)/dx must be finite 4. Psi(x) and dPsi(x)/dx must be single-valued 5.Psi(x) approaches 0 fast enough as x apporaches +or- infinity so that the normalization integral remains bounded Modern Physics by Tipler and Llewellyn page 249 It appears Psi(x)obey Dirichlet conditions with respect to x. That was the context of the discussion.
Ummm, no? In the wikipedia's article you quoted, the Dirichlet conditions are stated as: * f(x) must have a finite number of extrema in any given interval * f(x) must have a finite number of discontinuities in any given interval * f(x) must be absolutely integrable over a period. * f(x) must be bounded Correct me if I'm wrong (heck, you'll correct even when I'm right...), but isn't it possible that Ψ violates condition no. 3, as Ψ ∈ L² per normalization integral? Therefore, the Dirichlet conditions wouldn't be appropriate "in the context of the discussion..."DiEb
June 13, 2008
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Dieb wrote: Yes, you can use it sometimes, but it’s totally inappropriate in most cases.
In most cases? By the same reasoning, would you then say it is inappropriate to apply Dirichlet conditions to Fourier Series? Also, are the Dirichlet conditions more restrictive than the Jordan test or Dini's test with respect to Fourier Series? I recall you said:
the Fourier integral theorem as stated by Apostol does not necessarily satisfy the Dirichlet conditions, as the Jordan’s test is a generalization of these conditions (every function satisfying the Dirichlet conditions is - locally - of finite variation, the converse isn’t true).
[note: do you see the corner you are boxing yourself into? Which ever way you answer, you know it will lead to some embarassment to your claims about me not bein able to distinguish between a Fourier Transform and a Fourier Seires.] Where did I insist that Dirichlet conditions are necessary conditions? The link I provided to the wiki article states them as sufficient conditions, not necessary conditions. I said:
If we have an arbitrary function x(t) which obey the Dirichelet Conditions, the Fourier transform of that function is defined as ...
Does that mean that every function needs to obey Dirichlet conditions to have a Fourier transform? NO. NO. NO. You misread what I wrote, and used your misunderstanding as an argument to suggest that I didn't know the difference between and Fourier Transform and a Fourier Series. Consider the context of the discussion, and the class of functions I was considering -- solutions ot the Schrodinger Equations in certain contexts. From my physics textbook by Tipler and Llewellyn:
For future reference, we may summarize the conditions that the wave function Psi(x) must meet in order to be acceptable: 1. Psi(x) must exist and satisfy the Schrodinger equation 2. Psi(x) and dPsi(x)/dx must be continuous 3. Psi(x) and dPsi(x)/dx must be finite 4. Psi(x) and dPsi(x)/dx must be single-valued 5.Psi(x) approaches 0 fast enough as x apporaches +or- infinity so that the normalization integral remains bounded Modern Physics by Tipler and Llewellyn page 249
It appears Psi(x)obey Dirichlet conditions with respect to x. That was the context of the discussion.
Dieb wrote: And so, x1 and x2 are the same function - when we are talking about integration
They are not the same function, they merely result in the same integral and the same Fourier Transform. There is no need to put them in the same class of L1 if they were the same function. It's painful to watch a fine mind like yours try to save face when a simple admission of a mistake would suffice. But notice I stated the inverse transform as well as the transform at YoungCosmos. For this to be true, one actually needs something stronger than Dirichlet conditions, like Holder continuity [I realize now my professor at GMU made a passing remark which was in error which made me presume Dirichlet conditions are sufficient signal reconstruction via inverse transform. The topic he discussed was the phenomenon observed by Josiah Gibbs and Albert Michelson here] All this to say, your arguments that x1 and x2 are the same function fail in light of the fact I described the inverse transform as well, not to mention I was exploring the restrictions on Psi(x). There appear to be functions which would be in the same class in L1, but not result in correct reconstruction under an inverse transform. So you are clearly in error since the context of my discussions included inverse transforms. :-) PS lighten up Dieb, you Darwinists are so humorlessscordova
June 12, 2008
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Sal- 1. It's DiEb, not Dwieb, as it is Dirichlet, not Dirichelet - or was this an attempt of humour? As a moderator of this board, shouldn't be you above this kind of name-calling? 2.
By the way, Dwieb, for the reader’s benefit, tell them if functions satisfying Dirichlet conditions have Fourier Transforms.
Your usage of Dirichlet conditions in the context of general Fourier transforms raised my interest, as it was like spotting a sledgehammer in a watchmaker's workshop. Yes, you can use it sometimes, but it's totally inappropriate in most cases. 3.
me: That is a very sloppy example of equivocation!
Sal: In such case I’ll try to make my equivocations a little tighter next time. I’m surprised you’d objected, I figured such forms of argumentation are acceptable to you since I suspect you believe in Darwinian evolution (which is based on equivocation and double speak). Am I wrong to suppose you believe in Darwinian evolution?
As a general rule (in fact, as a categorical imperative), you should apply your own standard to your actions, not what you perceive/dream/imagine to be the standard of others. If your standard allows for sloppy or tight equivocations, then I've to ask: Are you a YEC? (Just kidding, I know you are...) 4. What do you call an element of L¹? I call it a function. And so, x1 and x2 are the same function - when we are talking about integration (as you phrased it somewhere else: the S-sign, not the Σ) For an engineer, that's often a surprise - for a mathematician, not so much...DiEb
June 10, 2008
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Dwieb wrote: But, you should have seen that the [omega] in my expressions is the variable of the transformed function, i.e., your renamed “f”.
Of course I saw it, that's exactly why I said above:
omega = 2 pi f,
Are you not reading what I wrote, and then accusing ME of not a reading when you are clearly the one who didn't see. By the way, Dwieb, for the reader's benefit, tell them if functions satisfying Dirichlet conditions have Fourier Transforms. PS
That is a very sloppy example of equivocation!
In such case I'll try to make my equivocations a little tighter next time. I'm surprised you'd objected, I figured such forms of argumentation are acceptable to you since I suspect you believe in Darwinian evolution (which is based on equivocation and double speak). Am I wrong to suppose you believe in Darwinian evolution?scordova
June 10, 2008
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LOL - this has to be one of the worst “gotcha” moments in history… how does consider two periodic signals x(t) and y(t) which are identical except at a single point….both x(t) and y(t) have identical Fourier series contradict anything I said before?
Hiya Dwieb, In the example discussed, are x1(t) and x2(t) different functions or not? You can't argue that merely because you can create a class of functions where both x1(t) and x2(t) are members that the two functions are identical. What you said before was:
your question doesn’t make much sense.
and my question was:
Can there be two functions say f(x) and g(x), and there exists some x where f(x) not equal to g(x), but their Fourier Transforms are equivalent?
The question made sense, and the answer was "yes". Instead you began to obfuscate and talk about L1 and say the question didn't make sense. You were wrong and you pile more obfuscations to cover your error. Did you think I wouldn't catch your error? :-)scordova
June 10, 2008
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-Sal
As I promised, I showed DiEB bungled.
LOL - this has to be one of the worst "gotcha" moments in history... how does consider two periodic signals x(t) and y(t) which are identical except at a single point….both x(t) and y(t) have identical Fourier series contradict anything I said before? From post #136 onwards, I tried to inform you about the idea of L¹ functions (and functions in similar classes)! Really, your view of mathematics is sooo 19th century :-)DiEb
June 7, 2008
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-Sal A single point is a null-set.DiEb
June 7, 2008
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My question was: Can there be two functions say f(x) and g(x), and there exists some x where f(x) not equal to g(x), but their Fourier Transforms are equivalent?
to which
Dieb responded: So, in this concept, your question doesn’t make much sense.
Now consider, Ziemer and Tranter:
consider two periodic signals x(t) and y(t) which are identical except at a single point....both x(t) and y(t) have identical Fourier series
And that is true of their Fourier transforms as well... As I promised, I showed DiEB bungled.scordova
June 7, 2008
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-Sal I bag to differ: the x1 and x2 of your example were representatives of the same function in L¹loc - and so, your question is trivial - and therefore, didn't make much sense... BTW: Are you saying that the f(x) in the Fourier Integral Theorem does satisfy the following properties: on every finite interval, f(x) is bounded and has at most a finite number of local maxima and minima and a finite number of discontinuities (i.e., perhaps you could react on my posts #143, #144, and #147) PS: Generally, there are three definitions of the Fourier transform - they differ in where to place this annoying factor of 2π (that's 2 pi). Therefore, answers to questions like yours may vary by a multiplicative constant... But, you should have seen that the ω in my expressions is the variable of the transformed function, i.e., your renamed "f".DiEb
June 7, 2008
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While f(x) is of finite variation - and it is bounded, it has an infinite number of discontinuities and local maxima. Nevertheless, the sum of the Fourier series converges to the function - at least at the points where the function is continuous.
Well then, there are functions then which do not satisfy Dirichlet conditions, but which have a Fourier Series representation. I never said that there weren't, I even cited Fejer 1904 theorems which show Dirichlet conditions are not necessary, but are sufficient conditions. However, if, even knowing Dirichlet conditions are more restrictive than Fejer's criteria, we still use them in connection with Fourier Series, why are you raising such a stink when applying them to Fourier Transforms? I never argued they were necessary conditions, and the links I provided never argued they were necessary conditions, only sufficient ones.scordova
June 7, 2008
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-Sal For both cases: F(omega) = i ? (?(? + 2 ? f) + ?(? - 2 ? f))
Why thank you DiEB. My calculations resulted in a different form, but given that omega = 2 pi f, I suppose it's the same result x1(t) = sin (2 pi f t) should be x1(f) = sin (2 pi f0 t) X1(f) =(1/2i) dirac_delta (f-f0) - (1/2i) dirac_delta(f+f0) and becasue x2 is essentially x1 except for single point of finite- valued discontinuity X1(f) = X2(f). So at least we are in agreement there. But this of course serves as a counter example to your claim:
So, in this concept, your question doesn’t make much sense.
My question was:
Can there be two functions say f(x) and g(x), and there exists some x where f(x) not equal to g(x), but their Fourier Transforms are equivalent?
Your showing that X1(f) = X2(f) is an example that my question made perfect sense. One only need to make the independent variable "t" instead of "x", and use x1 and x2 instead of f and g. note: From Signals Continuous and Discrete 2nd ed, by Ziemer, Tranter, and Fanninscordova
June 7, 2008
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arrgh, errata: f(x):= x-2^(-n) for x∈[2^(-n),2^(-n+1)[.... infinite number of ... local minima... And here's a useful hint which points to the world behind the Dirichlet conditions: http://eom.springer.de/f/f041090.htmDiEb
June 7, 2008
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-Sal For both cases: F(ω) = i π (δ(ω + 2 π f) + δ(ω - 2 π f)) Now, back to you: as what kind of function x1 (resp. x2) was treated here. Hint: x1∉L¹DiEb
June 6, 2008
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Are you saying that the f(x) in the Fourier Integral Theorem does not satisfy the following properties. on every finite interval, f(x) is bounded and has at most a finite number of local maxima and minima and a finite number of discontinuities
Yes, a function satisfying the conditionds for the Fourier integral theorem as stated by Apostol does not necessarily satisfy the Dirichlet conditions, as the Jordan's test is a generalization of these conditions (every function satisfying the Dirichlet conditions is - locally - of finite variation, the converse isn't true). Similarly for the Fourier series: Define for example, a periodic function on L²[0,1] by setting: f(x) := x-2^n for x∈[2^(-n),2^(-n+1)[ n∈N While f(x) is of finite variation - and it is bounded, it has an infinite number of discontinuities and local maxima. Nevertheless, the sum of the Fourier series converges to the function - at least at the points where the function is continuous.DiEb
June 6, 2008
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Sal- I don’t know what you want to achieve by posing your little questions (and yes, every Riemann-integrable function is Lebesgue-integrable)…
The reason I asked is because I'm math challenged, and I wanted to learn from a great master like you the answer to the following question which relates our discussion: Can you tell me what the Fourier Transform is of x1 and x2
x1(t) = sin (2 pi f t) and the Fourier Transform of x2(t) = 5 for t = 0 and sin (2 pi f t) for t everywhere else
Since it seems you perceive me as non-comprehending with respect to math, perhaps you can assist me in my understanding. Please provide the Fourier Transforms for x1 and x2.
Just out of curiosity: can you trade your three undergrad degrees for one graduate degree?
No, that's why I'm going to grad school.scordova
June 6, 2008
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