A Mathematical Emergency.

[Don Blazys]

To: CraigD. P.S. In LaTex, the logarithmic exponent should assume the form:


xln(a)
--------- -1
ln(T)
----------------------
ln(a)
--------- -1
ln(T)




Please use the above to correct your initial post on this thread.

Don.

[CraigD]

Originally Posted by Don Blazys

To: CraigD. P.S. In LaTex, the logarithmic exponent should assume the form…

Mea maxima culpa! My apologies for taking so long to get it right. All those parentheses got me mixed up – embarrassing for a guy who writes parsers and other parentheses-y things for a living. Hopefully, we’ve now gotten past the electronic barrier of writing the equation (which would have taken all of 30 seconds using an old-fashioned chalkboard). Though, to spread the culpa around a bit, Don could have saved me from my folly by learning a bit of LaTeX. It’s not difficult, and offers the pleasure at seeing ugly keystrokes rendered as pretty pictures.

Here’s the equation correctly rendered:


A bit more compactly:


Dividing both sides by , this can be written

I’m not sure if this is what the Don meant in the original post by “cross out the canceled Ts”. I don’t see anything algebraically abnormal about the equation

It is a remarkable identity, and looks like fun to prove. It’s pretty easy to demonstrate, running a little MUMPS program (with a standard, 18-digit precision calculator) on my cheap 4-year-old laptop (no supercomputer required ):

f  r "a=",a," T=",T," x=",x," " s E=x*$zln(a)/$zln(T)-1/($zln(a)/$zln(T)-1) w a,"^",x,"=",a**x," ",T,"(",a,"/",T,")^(",E,")=",T*(a/T**E),!
a=1 T=2 x=3 1^3=1 2(1/2)^(1)=1
a=2 T=3 x=4 2^4=16 3(2/3)^(-4.128533874054364329)=15.99999999999999945
a=2 T=4 x=3 2^3=8 4(2/4)^(-1)=8
a=4 T=2 x=3 4^3=64 2(4/2)^(5.000000000000000001)=63.9999999999938609
a=9 T=3 x=2 9^2=81 3(9/3)^(2.999999999999999998)=80.99999999999948382
a=16 T=4 x=2 16^2=256 4(16/4)^(3)=256
a=25 T=5 x=2 25^2=625 5(25/5)^(3)=625
a=8 T=2 x=2 8^2=64 2(8/2)^(2.5)=63.99999999999999652
a=.25 T=.5 x=2 .25^2=.0625 .5(.25/.5)^(3.000000000000000001)=.0624999999999969838
a=.5 T=.6 x=.7 .5^.7=.6155722066724512082 .6(.5/.6)^(-.1405352050771790835)=.6155722066724661252
a=.5 T=3 x=2 .5^2=.25 3(.5/3)^(1.386852807234541587)=.2499999999999999888
a=2.718281828459045236 T=7.389056098930650228 x=2 2.718281828459045236^2=7.389056098930650231 7.389056098930650228(2.718281828459045236/7.389056098930650228)^(-.000000000000000002)=7.389056098930649799
a=7.389056098930650228 T=2.718281828459045236 x=3 7.389056098930650228^3=403.4287934927351227 2.718281828459045236(7.389056098930650228/2.718281828459045236)^(5.000000000000000002)=403.4287934927350995

The difference between the two sides of the equation at more than 18 significant digits are typical of rounding error when approximating transcendental functions such as logarithms, which are used not only in the natural log focution (, $zln), but also for non-integer exponentiation.

[CraigD]

Here’s a proof of


, so, for ease of writing, rewrite as


Taking the logarithm base a of both sides gives:


, , and give


And


Multiply both sides by giving


Expand, giving


, so


And simplify, giving

[Don Blazys]

To: CraigD. Thats good. Just exellent! You have more than redeemed yourself! I too apologize and take full responsibility for not being able to write it in LaTex. I work at a high school, and have now resolved to get one of the computer teachers to show me how to do it the first chance I get. As you could probably surmise, I'm quite inexperienced at working with computers, so I greatly appreciate your hard work and patience, and thank you for seeing it through rather than giving up like so many others have done.

Since we have now established that the first equation in post #12 is indeed an identity, I can promise that from here on out, that identity will make for a very interesting discussion, mainly because it is relatively unknown and not yet in any book, journal, magazine or encyclopedia. You see, I discovered it only about a decade ago, and it's consequences and ramifications have yet to be fully addressed by the math community. In fact, it's been only in the past few months that the math departments at two major universities and the editors of a major math journal have begun a serious study of this rather remarkable identity, so it is clear that we are all entering into "uncharted mathematical territory", where there is no "general consensus" on some very basic issues.

Personally, I find this all very exiting because this stuff is both "elementary", and "cutting edge", and just as the early experiments involving quantum mechanics required "intelligent interpretation", so does this relatively new identity!

For instance, since the first equation in post #12 is a proven identity, and since that identity is true for all values of the variables T={2,3,4...}, a={1,2,3...} and x={0,1,2...}, is the term on the right more suitable for representing "common factors" than the term on the left? I say yes, because "common factors" absolutely must be defined as being greater than unity in order for that concept to make any sense whatsoever, and the term on the right clearly precludes (prevents) T=1 when T is to be construed as a "common factor". What do you say?

Also, is it now proper to "cross out" the cancelled T's in the term on the left as you did in the second equation in post#12? I have a problem with that because once the T's are "crossed out", the term on the left appears as a^x, which can be construed as (1)a^x=(N/N)a^x where N does not equal T, and while the equation would remain true, it would no longer be an identity, and we would be either ambiguous (in the case of (1)a^x), or inconsistent (in the case of (N/N)a^x) as to the value of the cancelled factor or cancelled common factor. Again, what do you say?

There are other, even more serious issues, but let's discuss these two first. Maybe others will join us.

P.S. What kind of calculator do you have that is accurate to 18 decimal places, or is such accuracy available only in conjunction with a computer? Are they expensive?

Thanks again my mathematical friend!

Don.

[CraigD]

Easy questions first…

Originally Posted by Don Blazys

P.S. What kind of calculator do you have that is accurate to 18 decimal places…

I’m using the built in calculator of an implementation of the MUMPS programming language (specifically, Intersystems Caché) that I usually have a terminal session connected to. If I recall correctly, the language standard requires it give at least 16 digits precision, and most do better.

I’ve written my own calculators in MUMPS for when I must be able to, for example, do precise arithmetic with million digit numbers. MUMPS is traditionally a database-focused language, used a lot in medicine and banking, so I’m a bit of an oddball for using it for math and science, although it has interesting and fairly unappreciated, IMHO, potential in these disciplines.

MUMPS is a programming languages, so a bit much to install and manage just to have a decent calculator, though very handy if you do, and learn just a bit about using it. You may have notices in some of my posts that you can write pretty useful MUMPS programs with a very small number of keystrokes.

There are several applications that include high or arbitrary precision calculators, such as Mathematica and Maple. Apps like these do a lot more than just literal calculations – I find that they make me feel rather stupid, as in a sense, they “know” much more math than I do.

Many folk find the ruby language handy for use as a calculator Lots about it, including documentation, tutorials, and a free Windows implementation of it, can be found here. If you can’t or don’t want to install it on your computer, you can use it via a browser here.

Originally Posted by Don Blazys

... or is such accuracy available only in conjunction with a computer?

Short answer, yes, you need a computer capable of running your calculator app of choice. Though in principle you can run nice calculator apps on the average cellphone these days, in practice this is pretty tricky. Mostly, I use a computer as a calculator. I’m rarely without a computer of a sort, having a PalmOS handheld in my pocket most of the time, though my calculators on it aren’t as good as on a Windows or unix machine. I hope someday soon to have about the same resources on my handheld as on a normal PC - I’m not quite there yet, but may be soon, as there are a growing number of handhelds being brought to market that run general-purpose operating systems, typically linux.

On the subject, you might find the hypo thread What is your favorite calculator? interesting.

Originally Posted by Don Blazys

Are they expensive?

Assuming you have a PC, you can get free copies of Caché or several other implementation of MUMPS (there’s also an open source version). Ruby is also free, via the links above.

Mathematica and Maple are a bit pricey, though if you’re with a school, you can usually get a good discount.

Now for a harder question…

Originally Posted by Don Blazys

Also, is it now proper to "cross out" the cancelled T's in the term on the left as you did in the second equation in post#12? I have a problem with that because once the T's are "crossed out", the term on the left appears as a^x, which can be construed as (1)a^x=(N/N)a^x where N does not equal T, and while the equation would remain true, it would no longer be an identity, and we would be either ambiguous (in the case of (1)a^x), or inconsistent (in the case of (N/N)a^x) as to the value of the cancelled factor or cancelled common factor. Again, what do you say?

Algebraically, it’s always proper to replace a term of the form with 1. When doing so, it’s important to retain any domain constraints of the replaced term (eg: )

Usually, one tries to write math canonically, replacing any terms equal to 1, so


is preferable to


. Whether you prefer


is largely aesthetic. I like it. I like even more this variation:


I don’t think it can be written much more tersely than this.

All of these are identities, with the restriction , , , and .

[Don Blazys]

To:CraigD. If we begin with the term:

(T/T)a^x,

then "crossing out" the cancelled T's leaves us with:

a^x,

and if all we are left with is:

a^x,

then we can't possibly derive the more defined and powerfull term:

T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)).

That's one reason why I am against teaching students that they should automatically "cross out" cancelled common factors. I believe that students should be taught the truth, and the truth is that when confronted by a term such as:

(T/T)a^x,

they have a choice. They can either "lose" or "cross out" the cancelled T's and write:

a^x,

or, they can choose to retain the cancelled T's and write:

T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)),

where it is now algebraically impossible to "lose" or "cross out" the cancelled T's. Before I invented the above "cohesive term", we didn't have this choice!

Now, the question is, if a "common factor" is defined as a positive integer T>1, then which term should we use for problems that involve cancelled "common factors"...

a^x

or

T(a/T)^((xln(a)/(ln(T))-1)/(ln(a)/(ln(T))-1)) ?

I prefer the "cohesive term". How about you?

Don.

[Qfwfq]

Don, the little time I could afford when I read your post was wasted on making sure I had not mis-read, re-checking the parenthesis etc. Although I had not translated it wron onto paper, the effort in scarce time brought me to a hasty conclusion and only last evening (and because a sane person such as Craig confirmed your claim so I deemed the challenge worthwile) I scrounged the time at the risk of ruining my frugal supper and including the time to re-derive the logarithm rules after my initial senile muddle; at first I had written the ratio as the log of the difference...

So, yes it's an actual identity although not obvious and rather tricky to prove, but I don't see why it has profound implications about the basic methods of formal calculation.

Now I had at least granted you the benefit of doubt and, given my lack of time, invited you to support your own claim:

Originally Posted by Qfwfq

Don I see no way for the exponent to identically equal 1 independently of x and if there's any way of the logarithms in the exponent compensating the T's in the base, go ahead and show us.

but you did quite otherwise. This is not the way to increase your audience either; an effort to get along with people and to be more compliant would do better.


Egaaaaaaaaash!!!!! I saw it after!

Originally Posted by Don Blazys

That's one reason why I am against teaching students that they should automatically "cross out" cancelled common factors.

I agree with the word automatically and I now get what you're after but it's no new cool discovery! I don't however agree where you use the word wrong. I certainly wasn't taught to do these things automatically, indeed I was often taught that it's sometimes necessary to do the opposite of "crossing out". Formal computations can be like getting from A to B in the mountains, the way to go isn't always straight down the gradient...

That's the difficult thing about learning to do exercizes and not many teachers are good at helping students to have the best approach to finding the way. My highschool math teacher was lousy in this and it caused me difficulty through the rest of my studies, but she definitely showed us how tricks are often necessary (add and subtract the same term, multiply and divide by the same term etc.). You seem to be banging on a wide open door instead of explaining your intent and helping to explore the complex inside of the palace.

[Don Blazys]

To: Qfwfq.
I apologize for my continuous failure to post my equations using LaTex. My "computer skills" are rudimentary at best, so I beg you to remain both patient and curious and to bear with me as I am doing my best to improve those skills under somewhat difficult circumstances and conditions. There is nothing that I want more than to have other mathematicians understand what I am presenting, and I really, really appreciate it when good folks such as yourself and CraigD take the time to contribute to this discussion.

If you can find the time to visit my website (donblazys.com), then please do so. (You can also find it by doing a "Google search" on "cohesive terms".) It was put together by the computer teachers at the school where I work and contains a lot more information on the equation (identity) that we are discussing here.

Now, here's why this really is a "cool new discovery", why it does indeed have "profound implications" for mathematics, and why I named this thread "A Mathematical Emergency".

You see, at this very moment, students throughout the world are being taught that in order to render the terms in the equation:

Ta^x+Tb^y=Tc^z

relatively prime or "co-prime", we must first divide by T, then "cross out" the T's. However, doing so results in:

(T/T)a^x+(T/T)b^y=(T/T)c^z=a^x+b^y=c^z,

which is wrong and inconsistent with both the "Beal Conjecture", and all observed results, because it falsely implies that x, y, and z can all be greater than 2 after the largest possible common factor T has been cancelled and the terms are co-prime.
Now, the proper, correct and effective way to render the terms "co-prime" is as follows:
Instead of "crossing out" the T's, we actually use them to derive at least one "cohesive term" so that it can be firmly established that the cancelled common factor T>1. This gives us three possibilities, one of which is:

(T/T)a^x+(T/T)b^y=(T/T)c^z=

T(c/T)^((zln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)).

Factoring does not involve the cancelled variable T, and results in:

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c^(z/2))^2=

(T(c/T)^(((z/2)ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)))^2,

where it is now clear that before we do anything else, we must first let z=1 in the unfactored case, and z=2 in the factored case, so that we can "cross out" and eliminate the logarithms that are preventing us from letting T=c, which common sense tells us must be allowable. This gives us both:

(T/T)a^x+(T/T)b^y=(T/T)c=T(c/T),

and

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c)^2=(T(c/T))^2,

where T=c is now clearly allowable. Simplifying, we note that our co-prime equations:

a^x+b^y=c,

and

(a^(x/2))^2+(b^(y/2))^2=c^2

are now perfectly consistent with both the "Beal Conjecture", and all observed results, because the implication that x, y and z can all be greater than 2 after the largest possible common factor T has been cancelled no longer exists.
Thus, by using a "cohesive term" to guarantee that the common factor T>1, we prevent the T's from being "crossed out" prematurely, and automatically prove both the "Beal Conjecture", and "Fermat's Last Theorem" (which involves only the "special case" where x=y=z).
Had mankind learned how to properly represent and eliminate common factors to begin with, (using at least one "cohesive term" to ensure that any common factor is "non-trivial") problems such as the BC and FLT would never have surfaced!
That, to me, is profound.

Don.

[CraigD]

Originally Posted by Don Blazys

(T/T)a^x+(T/T)b^y=(T/T)c^z=

T(c/T)^((zln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)).

Factoring does not involve the cancelled variable T, and results in:

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c^(z/2))^2=

(T(c/T)^(((z/2)ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)))^2,

This step is algebraically incorrect. It should be

((T/T)a^(x/2))^2+((T/T)b^(y/2))^2=((T/T)c^(z/2))^2=

T^2 (c/T)^((((z/2)ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1)))2)

Rendered more readably


This is because . .

[Kharakov]

Originally Posted by Don Blazys

P.S. What kind of calculator do you have that is accurate to 18 decimal places, or is such accuracy available only in conjunction with a computer? Are they expensive?

If you want something cheap and simple to use, Window's powertoys has a calculator upgrade with 32 digit precision.

http://www.microsoft.com/windowsxp/d...powertoys.mspx

Click on the powercalc.exe link.

Or you can just search microsoft.com for power calculator (or powertoys). Works nice. Just make sure to switch view to scientific (from standard) to get the enhanced features.