This was posted on com.sys.hp48 today. It is an excel file with new calulators for 2003 for several producers including HP.
http://www.dstewart.com/newdsc/c2es_03_orderbook/excel/Calculators_2003.XLS
Any comments?
New HP Calculators


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07092003, 07:42 PM
This was posted on com.sys.hp48 today. It is an excel file with new calulators for 2003 for several producers including HP. http://www.dstewart.com/newdsc/c2es_03_orderbook/excel/Calculators_2003.XLS Any comments? ▼
07092003, 11:41 PM
Hi, Richard; thank you for sharing the information. I saw these two references that called my attetnion: the HP33S, as a new release, and the three vage references: HP39G+, HP49G+ (more memory?) and what actualy called my attention: the HP48GII. What would these be? Weird... Luiz (Brazil)
07102003, 12:10 AM
You try to stay firm, you try not to hope too much, and then you see this! How are we good folk supposed to maintain our skepticism when such a credible looking document is presented?? The supposed MSRP of the 33S suggests that this is no Kinpo special. See? I'm doing it again... you can see real hope in that last sentence. Darn, darn, darn. ▼
07102003, 01:19 AM
Hi, Patrick; actualy, I do not know. Brand reference? Cheers. Luiz (Brazil)
07102003, 08:46 AM
Well, the 17BII + and 19BII + caught my eye. I wonder what features they will posess? I find it hard to believe that they could improve on either of those calcs, save adding trig to the 17... Maybe they added some newer bond calculations or something...more memory (not that they needed it...most 19BII users never even use the memory outside of the registers!) B. ▼
07102003, 08:54 AM
Well, key in your contact list and your RAM will not support it... A 128K RAM would be more convenient UMHO ▼
07102003, 12:04 PM
True, although with Palms and Blackberries that sync up to your Outlook/Lotus Notes calendar, very few people use it for appointments anymore. The fact that this calculator was a PDA yeaars before Palm shows just how visionary HP was!!! B.
07112003, 06:25 AM
I wonder what features they will posess? I find it hard to believe that they could improve on either of those calcs, save adding trig to the 17... Not that I think they'll do it, but my dream would be PC connectivity so Solver equations could be typed in/proofed on a computer and then d/l'd to the calc.
07102003, 04:23 PM
No, I don't expect you to identify yourselves (if you're even out there) nor break your confidentiality agreement. I just would like you to check something for me... What is the calculator's answer to 0^0? It would be great if the answer was 1 and not "Error", like it is on the 11C. I mean, 0! is calculated correctly as 1, why not 0^0? ▼
07102003, 07:25 PM
If I remember right, 0! is not calculated as 1, but is agreed to be 1. Didn't our mathematical forefathers decide it should evaluate to 1 just to keep the curve smooth? However, 1^0=1.00, 1^0=1.00, but yes, also on my 34c, 0^0 = "Error 0". And, 0^1=0.00?!?!? In a politician, such inconsistency would provoke outrage! ▼
07102003, 07:58 PM
In my professional opinion, I would have to challenge you on your distinction of calculated vs agreed. The definition of any function at any point is something that is agreed upon by the scientific community. Whether or not there is an arithmetic or other calculation involved is just a particular kind of evaluation algorithm.
So, one definition of the factorial function consistent with the scientific literature is: 0! = 1Likewise, the definition of y^x includes certain special cases: 0^x = 0 for any x != 0 There are good reasons for defining 0^0 to be 1 which I believe have to do with the continuity of the exponental function on the complex plane. I'm surprised Mr. Kahan, with his careful attention to detail in the complexities of the SOLVE and INTEGRATE functions of the 34C, and in all of its algorithms in general, let this one sneak by.
07112003, 04:05 PM
A definition of the factorial function n! = product(k) for k =1 to k greater than or equal to n (My symbols did not survive the posting process.) Note that if n = 3, we start the product with k = 1 and go up to k =3. n! = 6 If n = 0, we start the product with k = 1. At this point k is greater than n so we stop with the product equal to one. Thus 0! = 1 without treating 0 as a special case. It is a natural definition of the function as it is usually used to determine the number of possible ways to order a set of n objects. If n = 0 there is one way to order the set. This assumes that whether 0 is considered to be a number or not, it is an object in that it represents the empty set or at least a place holder.
07102003, 09:36 PM
Patrick: 0^0 is undefined by definition  like 0/0 and 1 raised to infinity. In general, x^0 is 1 because 0 = n  n, so x^(n  n) = x^n/x^n = 1. But if x is 0, the denominator would be 0^n, which is 0, and you cannot divide by zero. I am not certain if this is the correct explanation for the undefinability of 0^0, but it's a good one anyway. If your calculator reports "Error" when you attempt 0 [ENTER] 0 [y^x], the calculator is only giving you the mathematically correct answer. Other undefined expressions are ln(x) for x <= 0, tan(90), etc. Ernie ▼
07102003, 11:26 PM
I have to disagree with you, Ernie. The mathematical definition of y^x for y>0 is exp(x ln(y)). The ln(x) function is defined as the integral from 1 to x of 1/t dt and the exp(x) function is defined as the inverse of ln(x). Since the definition does not provide a value when y is zero, we are free to define it as we like. The natural definition of 0^x for x nonzero is zero. This leaves the hole at x=0, y=0. To fill that hole, one might try to take the limit as either x or y approaches zero with the other one equal to zero. The result is that there is no unique limit, meaning there is no definition of y^x which makes the function continuous. However, there are good reasons to define 0^0 as equal to 1, as it makes a number of other theorems easy to state without having to have an exception at that point. This is what actually has been done. In higher mathematics, we define the value of 0^0 to be equal to 1. The function does not end up to be continuous, of course, it cannot be, but it is the best value we can choose for a host of other reasons. As a reference, I direct you to this article written by the mathematician William H. Kahan, the numerical analyst who consulted with HP on many of their algorithms, including SOLVE and INTEGRATE. You can see on page 10 of that document a definition for the complex power function z^w, which is defined to be equal to 1 whenever w = 0, for all complex z. ▼
07112003, 01:22 AM
Patrick: I'll defer to your obviously greater expertise on the matter. From my longago college days, however, I vividly remember that my textbooks specified 0^0 as undefined  every bit as much as 0/0. I'll try to read the article you mentioned, but I doubt that I'll understand much of it. I flunked integral calculus, barely learning to integrate e^x. 8^) Ernie
07112003, 01:36 AM
Patrick: I checked page 10 of the document. Unless I'm missing something, it seems to confirm what I said:
Quote: Observation 1: There is really no need to state as an exception the case when w=0, since w*ln(z) will be 0 whenever w=0 (no matter what value z may have), and exp(0) = 1 anyway. Observation 2: "and 0^w :=0 if Re(w)>0" Notice that w is nonzero (otherwise it couldn't have a real portion greater than 0), so the definition given for z^w does _not_ cover the case when both z and w are 0. If you still disagree with me I'll take your word for it, but that's what I'm reading here. Ernie ▼
07112003, 02:45 AM
Observation 1: You can't use the w*ln(z) argument because ln(z) is undefined for z=0. You can't argue that a multiplication involving zero has to be zero when the other part of the multiplication is undefined! Observation 2: The definition does cover the case when both are zero since it says that z^0 = 1 for all z. Surely, the number z=0 would have to be included in all, would it not?
07112003, 02:54 AM
Hi Ernie, guys; There is no reason to discuss this fact. 0^0 is not defined, period. And what Patrick wrote is easily identified as conflicting.
Quote: Unfortunately, x^0 = 1 for all x (including zero) is wrong. Zero is not included. Suppose you have y^x and x^y. If y == x, then x^y == y^x, right? Based on what you wrote: 0^0 is not defined when you analise 0^x when x = 0; and 0^0 == 1 when you analise x^0 when x = 0. So, 0^0 != 0^0. Not consistent, isn't it? And that's enough reason to make it undefined. My 2¢. Luiz (Brazil)
Edited: 11 July 2003, 3:01 a.m. ▼
07112003, 05:18 AM
Luiz wrote:"There is no reason to discuss this fact. 0^0 is not defined, period." Are you sure ? It's certainly rare to see such a nice, suave person like you uttering such a bold, intransigent statement. I usually think that absolutely everything can be discussed, and that denying that possibility is akin to fundamentalism (i.e: "There is no reason to discuss the existence (or nonexistence) of God (say). God (say) exists (or doesn't), period." Get the point ?
Also, I tend to think that mathematics aren't invented, but discovered, have an existence of their own, independent of human beings or a physical universe existing or nor, and thus tend to be rather impervious to anyone's attempt to "define" or not what they should do. For the case in discussion, x^x when x is 0, look at these results: x x^xDoes that ring a bell ? Of course, one can argue that when x is exactly zero, an indefinition occurs. That's debatable but if it were true, so what ? Any calculation will show the values merrily going to 1 as x goes to 0, and if some mathematician's committee were to define that 0^0 = Pi, it would do them no good, for it would generate contradictions at once. If you are forced to accept a definite value to avoid contradictions, you're not defining anything, you're merely trying to save your face on account of the evidence.
As a further example, consider a function defined thus: f(x) = x / xNow, you evaluate it for all finite arguments, and find that f(x) is always 1. Then you try to evaluate it for x=0 and you claim that 0/0 is undefined, and so assigning the value of "1" to f(0) is a matter of definition ... Come on !! Common sense shouldn't be incompatible with mathematics !! Best regards from V.
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07112003, 06:21 AM
Thank you , Valentin; I'd remove the first two sentences in my post and convert it to an open discussion and turn it into my view of the situation: I see as not defined based on what I wrote. Back to the arena: B^} Math approaches are (always?) a matter of discussion and definition. And I teach that... Thanks. Luiz (Brazil) ▼
07112003, 08:25 AM
I'm not sure this can be done, but I simply extracted part of what's written in the below mentiond site. Particualr attention to the first sentence I set as italic.
Quote: Extracted from: The Math Forum under 0^0 (zero to zero power)[/link]. This eaddress was suggested by Juergen(CH) Please, if this cannot be posted here, remove password is 12345. Thanks.
Luiz (Brazil) Edited: 11 July 2003, 8:27 a.m.
07112003, 07:03 AM
Valentin: Your table of X and X^X values only proves that the _limit_ (not the function itself) is 1 as x approaches zero.
Quote: And mathematics shouldn't be incompatible with common sense, either. You can't ignore singularities and discontinuities with specialcase definitions and leave it at that. The function f(x) = x/x is a straight line, yes, but it gets interrupted for x=0. "Defining" f(0) as 1 goes against common sense. Ernie ▼
07112003, 08:54 AM
Hi again, Ernesto: Obviously this discussion could go on a on, and we'll probably end it "agreeing to disagree", but there's one final remark I'd like to bring to your attention: stating that x / x => 1 for all x, just means (passing the denominator to the right side) that: x = x which I'm sure you'll agree it's valid for every conceivable x, be it whatever it may, zero included. That's what the equation and the function are telling us, and trying to see "discontinuities" where there are none, and using "definitions" where none are necessary is just plain silly, a kinf of bureaucratic formalism, that goes against common sense. Also, "defining" that some function has such and such value where no other value is possible at all without inconsistency, is absurd, because if you have no choice, if your are forced, issuing a "definition" is a meaningless act on your part. That said, don't think I'm blinded to the subleties of mathematical formalism, is just that sometimes I think some of them are pretty stupid, and 0^0 or x/x for that matter just trigger that feeling for me. On the other hand, I know all too well that you can never prove anything my common sense or intuition alone; logically consistent, formal proofs are always necessary. For instance, assume that you are calculating this infinite sum: Sum(n = 1, n > inf, INT(n * tanh(Pi))/10^n) and the answer you get coincides with 1/81 to 12 decimal places. "How nice!" you think, "I didn't expect this hyperbolic tangent thing to add up to a simple rational fraction as 1/81. Maybe it's just a coincidence ?" and so you duly proceed to compute it to greater accuracy, say 30 decimals. It still agrees with 1/81 to all 30 decimals. "It's wonderful, let's try that multiprecision package I've got for my 48/49/71 !", you say, and proceed to compute the sum to 100 decimal places. It agress with 1/81. Then to 200 decimal places. It still agrees. Obviously, common sense would dictate by now that the sum equals exactly 1/81, no doubt about it, right ? 200 decimal places in agreement is more than enough evidence to settle the matter, isn't it ? Yet common sense would be wrong this time, as the sum does NOT equal 1/81. So much for common sense in mathematics ... :) Best regards from V.
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07112003, 02:02 PM
Valentin:
Quote: Yes, the discussion could go on forever. Which is another way of saying "it's undetermined." If instead of "0^0 = x" it were "2 + 2 = 4" no one would question it and there would be no discussion at all. It seems, therefore, that we need to agree to disagree.
Quote: Except for x=0, of course, since you cannot divide by zero. Ernie
07112003, 03:39 PM
Luiz wrote:"There is no reason to discuss this fact. 0^0 is not defined, period." Then Valentin responded:"Are you sure ?" Right, here we go...
0^0 is undefined. Period. Simple as that. Now, in certain applications we would like the expression x^x to be continuous. So, we define 0^0 to be 1, which is quite reasonable since lim(x>0) x^x is 1 whether x approaches zero from above or below. Soooo...the point x=0 where f(x)=x^x is a 'removeable singularity'. Quite simple really. Can we stop the argument now? PS: 0.9999...=1
ttfn ▼
07132003, 01:02 PM
Okay, What would all of you do? "Error (0)" _or_ "1" ? It would probably confuse heck out of the average user to offer a CHOICE, though this probably could be done, say, via some flag setting. The default might be "error (0)", but you could flip a bit and make the calc default to 1. Or, maybe we should have the calculator give predefined answers, but light up a "YMMV" ("Your Mileage May Vary") indicator: "I'm saying this is 1, but it's really not". I have heard some describe mathematics as "the perfect science". I believe it probably is, but the subtleties and special cases here and there seem to indicate our mathematical grammar has had a hard time perfectly expressing it in toto. So, what's a calculator to do? ;) ▼
07132003, 01:26 PM
Hi, Glynn; good reading your posts again. The title says much of what I want to express, but I'll take the benefit of doubt: 0^0 is not defined. After reading The Math Forum comments about the limits for x^y when each one tend to zero, what shows it is discontinuous because both limits tend to different values, I embrace the cause. About programming: I'd set flag 25 in an HP41/42 and test it later, or I'd use an IFERR structure if using an HPRPL model or any errortrap structure available in other languages to detect this sort of fact, but I'd keep the original math definitions (unless others appear) if I'm developing a new language or compiler. I'd not let the user interfere at this point because it would lead to the development of "conditional programs", that would behave in different ways without specific preconditioning. I think the HP49G math approach is already outstanding when allowing the use of many flags to predefine the way the user want an answer to be treated by the O.S. and how it must be shown. The HP48G already gives much control, but after reading some (far from a complete reference) of CAS resources, I can see some interesting possibilities. Nothing I can convert into words right now, but surely new perspectives. About a year ago I posted a message where I criticized CAS without knowing what was it able to do. One of the contributors called my attention and mentioned Urroz' books. I bought four of them and read parts of them in brief. Enough to change my mind about CAS and related stuff. I still keep math principles as valid and applicable, although many math enhancements have occurred and I'm not aware af them as a whole. For as long as there are math situations with doubtful answers, I'll take them as undefined. As you say, math must be precise all the times, mostly when treating its own lacks. Luiz (Brazil)
07132003, 01:55 PM
Well, I plead guilty to starting all of this brouhaha. I never anticipated this kind of response. But here is what I think would happen... If you have a company driven by engineering principles (e.g., the old HP) they might consider defining the function to be equal to 1. They would see the mathematical arguments for this, such as the rather important statement that the binomial theorem would extend naturally to the case of n=0. If you have a company driven by marketing principles (e.g., the new HP) they would no doubt leave the function undefined at (0,0), and to hell with the binomial theorem and its ilk. A marketeer would need no further justification than this thread to convince them that is the right thing to do. You never want to confuse your customers. Not with that 50% rollback in customer support coming next month. On the other hand, a company that was driven by engineering principles (i.e., the old HP) did in fact leave 0^0 undefined in their calculator logic. Earlier, I said that I was surprised that Dr. Kahan let this one slip through. I now believe him to be much wiser than I had ever thought. I now believe this was a conscious decision on his part. He was intimately aware of the good mathematical logic behind the reason to set the value to 1  he teaches this very fact in his complex analysis course for crying out loud  so he must have had a very good reason not to do it. I think I might agree with him now. ▼
07132003, 02:42 PM
If you look at the sci.math reference: ▼
07132003, 06:29 PM
The 49G gives 1 in approximation mode and ? in exact mode. Massimo ▼
07142003, 01:58 AM
Indeed. Werner
07132003, 06:00 PM
Didn't the IEEE allow for similar situations: I think you can get NAN ("not a number") as the official result from illegal (questionable?) operations such as divide by zero or exponentiations giving results that are outside the range (e.g. 10^+/99) handled by various computers and/or compilers? Maybe 0^0 deserves something similar. ▼
07132003, 11:29 PM
IEEE 854 (IEEE Standard for RadixIndependent FloatingPoint Arithmetic) only talks about +,,/,* and remainder as required Operations. Most of the rest of it is about rounding, infinity arithmetic, normalization, NaN's, etc. The other functions such as trig, exponential, log, are left up to the implementor of a particular computer. Our topic here, 0^0 isn't mentioned at all. Just for grins, I checked what Maple and Mathematica do. Maple returns 1 without any beeps or error indications. Mathematica returns "Indeterminate".
07112003, 05:59 AM
See http://www.faqs.org/faqs/scimathfaq/specialnumbers/0to0/
07112003, 05:45 AM
Seems to be an old discussion: ▼
07112003, 08:42 AM
Some place I have seen a defininiton of the fatorial function, n!, that includes 0! = 1 as a general case rather than a special case. I'll try to dig it up. As for 0^0 I don't have any answers just now, but 0 is often viewed as not being a number at all. Of course the thing about that view that bothers me is what do we put on the number line to plug the space between the smallest positive number and the smallest negative number? On the other hand the number line graphs like a function and lots of functions have discontinuities. Now I have to go read the article on Z^W. ▼
07112003, 08:33 PM
Quote: Would this one case: on digital computers, zero represents not one number but all the numbers between the positive and negative numbers with the smallest magnitudes that can be represented?
07112003, 09:20 AM
Ask Dr. Math is a valuable service and 99.99% of the time their answers are accurate and useful. But not always ... Have a look for instance at the answer they gave to this question: quoting from it:
" ... when pi appears, it is likely
Best regards from V. ▼
07112003, 11:17 AM
Valentino, you wrote :
Quote: Well, I'm not a Maths specialist, so if you could give some examples, I'd be delighted !
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07112003, 12:18 PM
At least the PI and/or SQRT(PI) which appears on normal distribution formulas (gaussian distribution) is not obviously related to any circle, but I am prepared to find someone with great math skills who may show the relation. On the other hand, I do remember a curved surface we studied in my Math II course in Electronics Engineering, which was defined by many arcs (it was kind of part of a hemispheric surface, but with a Sshaped cut). While all its features looked unavoidably very "circular", PI was remarkably absent of it's area formula, obtained as an integration exercise. I vaguely remember a possible name for such surface, which could be translated as Viviani's Vault (Bóveda de Viviani), but I may be confusing similar exercises from some 27 years ago, and I have no time now to look for more references. Undoubtedly some of the more math knowledgeable people here can offer enlightment and/or corrections to this comment. Or, I can just wait for a couple of months, because my first daughter is about to take the same course on her first Industrial Engineering year. (She is taking many of the "same" courses than I did, but her calculator is a 32Sii, while mine was a 25 at the time)
07112003, 12:40 PM
Hi, Thibauto: Thibauto posted: "Valentino, [ ... ] Well, I'm not a Maths specialist, so if you could give some examples, I'd be delighted !"
by David Blatner. Read it, as I did, and afterwards, if you still want more examples, we can discuss it, ok ? Best regards from V. ▼
07112003, 02:44 PM
Valentin, Has anyone published "The Joy of 'e'"? tm ▼
07112003, 05:26 PM
re: Has anyone published "The Joy of 'e'"? That's not quite the name, but yes. I have it  somewhere! If I find it, I'll post a proper reference. Actually, it was by the bed: "e The Story of a Number" by Eli Maor. There is also a book about zero (which was a very strange idea all by itself to folks like the ancient Greeks  let alone the above discussion about 0^0): "ZERO The Biography of a Dangerous Idea" by Charles Seife. ▼
07112003, 07:45 PM
Thanks Dave. tm
07122003, 06:28 AM
Quote: Another one from Robert Kaplan: The nothing that is: a natural history of Zero Massimo
07112003, 03:16 PM
The probability that two random integers are relatively prime (ie contain no common factors) is 6/PI^2. Werner ▼
07112003, 05:52 PM
Hi, Werner; such a well defined value, although with as many digits as a powerful computer can compute, remembered me Introduction to Stochastic Statistics. Just a feeling of mine? Or it is obtained by common statistics analysis? Just curiosity. Luiz (Brazil) (This thread is heating up everytime I touch it to sense temperature... Wow! Brainstorm, for sure... Healthy... brainstorm)
Edited: 11 July 2003, 9:44 p.m. ▼
07142003, 08:17 AM
Hi Luiz. ▼
07142003, 12:54 PM
The term "random integer" is, of course, not well defined. There is no such thing  at least none which captures the implied uniformity of the distribution. However, I believe the statistic of which you speak is obtained as the limit as N>infinity of the probability, P_{N}, of two random integers in the interval [1,N] being relatively prime. Picking two integers "at random" from a finite interval is, of course, very well defined.
07112003, 11:19 AM
Hi Valentim, guys; this is not a math forum, of course, but computing is always a related subject. If the definition for PI goes beyond the relation between the radius (or diameter) of a circle and the lenght of a complete arc (360º, 400 grads, 2PI radians), then I see what you mean. If not, if PI is only related to this particular relation, I think that what Dr. Math states is that wherever PI is used, that radius<>lenght relation is there, too. What he is proposing, in my understanding, is that: found the relation, a circle, as base defintion, is there, too. My 2¢. Luiz (Brazil) ▼
07112003, 02:38 PM
Dr. Math is not always the best source for definitions and the like. For one thing it tends to written for those who have not been exposed to much math theory or higher math. It's a bit like high school when I was told than negative numbers do not have logs. Utter nonsense, but they did not have the time to explain it to high school level kids. My calculus text says that e^(x^2) has no integral. Also utter nonsense, but they probably didn't want to spend the time on the question. Made me mad, so I went home and solved the integral myself. Of course this is the sort of exercise all students should be doing. You learn far more this way than doing things by rote. Once past calculus, the defintions angles (excepting circular angles and ordinary solid angles), trig functions, and the like do not involve circles at all. There is probably some such defintion of pi. After all, the ratio of diameter to circumferance of a circle is only pi if you are talking about plane geometry, a concept that does not exist in the real world. (At least given the thereotical physics models currewntly in vogue.) For comittee decided defintions I have always heard a story to the effect that some state legislature created a legal defintion of pi as 3.14. I find this quite believable after lookign at some of the other lawas around.
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07112003, 05:20 PM
Quote: That's nothing. According to _Asimov on Numbers_ by Isaac Asimov, "There is always the danger that some individuals, too wedded to the literal words of the Bible, may consider 3 to be the divinely ordained value of pi in consequence. I wonder if this may not have been the motive of the simple soul in some state legislature who some years back, introduced a bill which would have made pi legally equal to 3 inside the bounds of the state. Fortunately, the bill did not pass or all the wheels in that state (which would, of course, have respected the laws of the state's august legislators) would have turned hexagonal." Ernie ▼
07122003, 01:17 AM
Quote: I've read about that story: the state was Indiana, I believe. Regarding approximations as "definitions", 22/7 wouldn't have been *too* far off the mark, but the HP35 manual gives the amazinglygood example of 355/113.
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07122003, 06:56 AM
Quote: Actually this approximation predates the HP35 manual a little... ;) Tsu Ch'ungchih (430501) proposed this value versus 22/7. This was the best approximation until XV century even if the same Tsu Ch'ungchih computed 3.1415927 as an "excess" approx. and 3.1415926 as a "defect" approx. [source: A History of Mathemathics by Carl B. Boyer] Quite impressive, isn't it? Massimo
07132003, 05:24 PM
In 1896 a bill was passed in the Indiana legislature 670 to make the value of pi = 3.2. ▼
07132003, 08:16 PM
The worst part about the Indiana legislature proposing that pi be made equal to 3.2 is that 3.1 would be a better approximation. Maybe they decided 3.2, being larger than 3.1, was "better". Perhaps it's the same logic that leads modern politicians to spend more than they take in ....  Ed
07122003, 11:54 AM
Since 1 radian = arctan(b/a), for e^i written in the form a+ib, and a complete circle is 2pi radians, then it looks like pi can be related to 'e' and 'i'. ▼
07132003, 10:24 PM
Its not about 3.1 or 3.2 The definition of pi is which flavor U like the best. I think w/o question that chocolate pudding pi with whipped cream topping is absolutely unbeatable, but it needs some chocolate sprinkles to decorate the top.
07142003, 04:12 AM
1 + e^(i * pi)= 0 In a single equation you related geometry (pi), mathemathical analysis (e), complex numbers (i), 1 and 0 (whatever the last two may mean).
07132003, 11:20 PM
Hi, guys;
I read many posts related to PI as a number and its value. Unspellable posted: Quote: I still wonder about its "origin" or "definition" because, as far as I remember, PI is considered an irrational number, together with "e" ( :( I can't remember any other irrational constant). As an irrational number cannot be expressed by a relation, PI must be expressed in another way (maybe as a movie...). I'm trying to keep only "plain" (number) set theory and definitions. I know many of you will discuss what is an irrational number, that all numbers have an infinite set of digits, and all numbers are contained in Rset (real), if Euclid is right/wrong... I think it's a valuable brainstorm, and I'll read all posts. But I still wonder if PI was defined as a relation based on circle elements (plane geometry) or in some other math relation, what leads us to the fact that PI is NOT an irrational number if it is defined as a relation. By definition, of course. And concepts are mostly related to the "conceptual" world, not the real one, as unspellable stated so well. Algebra and math, in a conceptual, pure number theory, also deal with a conceptual world as well. I know it's going a bit far from what we are used to, but this is a teasing subject and I'd love going further. Thanks. Comments? Luiz (Brazil) ▼
07142003, 08:02 AM
The square root of two is an irrational number. pi (A capital PI is the symbol for a series product.) and e are transcendental numbers. Transcendental numbers, among other things, cannot be the roots of a polynomial equation. This is connected with the fact they are represented as an infinite series when evaluated.
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07142003, 07:02 PM
Let's not forget the golden ratio which is the ratio of height to width of a golden rectangle. It is frequently referred to as phi and is derived with the following series: phi = sqrt(1+sqrt(1+sqrt(1+sqrt(1+sqrt(1+...))))) phi = 1.6180339... Mark Hardman ▼
07142003, 09:36 PM
Odd, I've always known the derivation of phi as phi = (1+sqrt(5))/2 It, and its reciprocal, are the solutions to the quadratic x^{2}=x+1. It is also the limit of f_{n+1}/f_{n}, where f_{n} is the n'th Fibonnacci number. ▼
07142003, 11:50 PM
But its so much more fun to derive phi as an infinite series.
Here is another one for phi: 1
One interesting property of phi is that any positive integer can be represented as the sum of nonconsecutive powers of phi. For example: 234 = phi^{11} + phi^{7} + phi^{3} + phi + phi^{5} + phi^{9} + phi^{12}You can visit The Phi Numbering System page to try this out on any integer you want. Mark Hardman (LED)
Edited: 14 July 2003, 11:58 p.m. ▼
07152003, 05:24 AM
Phi's always been a favourite constant of mine, though when I was young and fresh, it was called 'tau' (a greek letter), not 'phi'. This URL has a lot of interesting information on phi: It seems that the Indian mathematical Genius (with a capital 'G'), Srinivasa Ramanujan was also very fond of it, as he discovered a number of outofthisworldish identities and infinite series featuring it. No wonder worldclass mathematicians Hardy and Littlewood were so impressed with him: "In 1913, the English mathematician G. H. Hardy received a strange letter from an unknown clerk in Madras, India [...]Every prominent mathematician gets letters from cranks, and at first glance Hardy no doubt put this letter in that class. But something about the formulas made him take a second look, and show it to his collaborator J. E. Littlewood. After a few hours, they concluded that the results "must be true because, if they were not true, no one would have had the imagination to invent them". (Source: Srinivasa Ramanujan)
An interesting phi fact rarely mentioned is that the increasing powers of phi are closer and closer to being integer values, something I noticed for the first time when programming a trivial loop on my recently acquired (1976 or so) HP25, which simply filled up the stack with phi and kept on multiplying it by itself in an endless loop. I noticed at once that the results approached integer values, from above and from below alternately. Look at this 'evidence': phi = 1.6180339887498948482045...Speaking of near integers, Ramanujan knew about the following striking result, where a simple looking exponential expression gives a totally unexpected, very near integer value (less than 1E12 from being an integer), called 'Ramanujan's constant': e^(pi*sqrt(163)) = 262537412640768743.999999999999 25... Certainly the number of amazing mathematical curiosities is inexhaustible, indeed ! Best regards from V.
Edited: 15 July 2003, 5:26 a.m.
07152003, 03:53 AM
...89
07142003, 09:52 PM
There is no link between transcendental numbers and infinite series. While it is true that transcendental numbers can often be expressed as the sum of an infinite series, it is possible to get even rational numbers through nontrivial infinite series. For those who have never seen it, the proof that sqrt(2) is irrational is quite short. We simply suppose the opposite is true and reach a contradition. Here goes...
If sqrt(2) is rational, then it must be expressed as a fraction of integers. Let a/b be such a fraction reduced to lowest terms (i.e., there are no common factors between a and b): sqrt(2) = a/b
Square both sides of this equation and clear denominators to get: 2b^{2} = a^{2}
Clearly, this says that the square of a is an even number. Well, if you square any odd number you get an odd number, so a cannot be odd. Therefore a must be even! This means we can write it as equal to twice some integer, k: a = 2k
Substitute into the earlier equation: 2b^{2} = (2k)^{2} = 4k^{2}
Divide both sides by 2: b^{2} = 2k^{2} Well now, lookie here... the square of b is also even. Like before, we must reason that b is even! Hold on, now. Both a and b are even, which means they have a common factor!! But we assumed right at the start that this was not the case. This contradiction proves that our starting assumption, that the square root of 2 was rational, is in error. QED
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07152003, 08:40 AM
Transcendental numbers are not "often" expressed as an infinite series, they are always expressed as an infinite series. This a property stemming from the fact that they cannot be the root of a polynomial. They are called transcendental numbers because they are always expressed as an infinite series. ▼
07152003, 01:24 PM
Every number is the sum of an infinite series whose elments are rational numbers. If you take a number, x, between 0 and 1, say, and let its decimal representation be 0.d_{1}d_{2}d_{3}... then what you are really saying is that x can be represented as the following infinite sum of rationals: x = Sum_{(n=1,2,3,...)}(d_{n}/10^{n}) By adding an extra term to the above sum to represent the integer portion of x, you can easily see how to remove the restriction that x is between 0 and 1. Of course, the sum isn't infinite if the decimal representation of x terminates, but that is not the point. The point is that, since every number has such a representation, this property hardly distinguishes the ones which are transcendental. The thing about transcendentals is that there are usually very few other ways to represent them, such as by giving a rational polynomial for which they are its roots. The motive for the original mathematician to call that class of numbers transcendental was that such numbers transcend normal arithmetic (at least, this is how it was explained to me). You cannot get such numbers from integers through addition, subtraction, multiplication, division and the taking of roots.
Edited: 15 July 2003, 1:29 p.m. 
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