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modest · 02-10-2009

Quote:

Originally Posted by Don Blazys

Thanks for the constructive criticism.
I sincerely apologize for any of my posts that come off as "arrogant" or "demeaning".

Thank you, no problem, and I did indeed mean it as constructive criticism and it’s wonderful you took it as such. A respected member of our forums here sums up nicely where I was coming from in this post. It’s excellent advice. In fact, I'll quote it:

Quote:

Originally Posted by Buffy

When introducing revolutionary breakthroughs, probably the worst approach is to say "everything you know is wrong, and I'm about to prove it to you." It may sound more sneaky, but you'll find that you have a much easier time if you introduce what you're talking about as a "slight improvement" on existing theory--something that you almost do up until you hit the excerpted quote!

Let someone *else* characterize your astounding breakthrough as such: when you say it yourself, it pretty much guarantees that it will be dismissed out of hand, even by those who should know better.

Successful Physics requires good Marketing,
Buffy

Quote:

Originally Posted by Don Blazys

Quote:

Originally Posted by modest

Yes, you are allowed to cancel the T's on the right. It is basic arithmetic. At most you are offering a reason to choose not to cancel the T's which is a valid choice. But, your statements about the "algebraic impossibility" of canceling the T's in your term or canceling (T/T) in the lhs of your identity are wrong.

"Crossing out" the $T$ 's on both sides of the equation:

$T\left(\frac{a}{T}\right)^{\left(\frac{\frac{x\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1}\right)}=\left(\frac{T}{T}\right)a^x$

would result in:

$\left(a\right)^{\left(\frac{\frac{x\ln(a)}{\ln()}-1}{\frac{\ln(a)}{\ln()}-1}\right)}=a^x$

Nice

We can call this the whiteout method of canceling common factors... you just erase any mention of ‘em

Ok... seriously though... I think you’re probably making a play on the words “crossing out”, but you’re the only one using that phrase so there’s really no point in exaggerating its meaning. I’m saying “cancel” which is the appropriate mathematical term for eliminating a product that equals one. It is not algebraically impossible to cancel (T/T) in the RHS of this identity:

$T\left(\frac{a}{T}\right)^{\left(\frac{\frac{x\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1}\right)} = \left(\frac{T}{T}\right)a^x$

$\forall \ a,x,T \in \Bbb{R} \ \colon T>1, a>0, T \neq a$

Every reason you give for leaving (T/T) is a reason to choose to leave it and not a reason to prohibit the operation. By the way, I believe the line following the identity properly sets your domain which is necessary to include for reasons I gave in my last post.

As far as the other thing I said “canceling the T's in your term”. I’m referring to this term:

$T\left(\frac{a}{T}\right)^{\left(\frac{\frac{x\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1}\right)}$

In particular—the story you told about some student winning a bet because T could not be eliminated from this term using algebra. I contend (and I think I’ve shown) that T can be eliminated from the term so long at T isn’t zero. I realize that doing so goes against what you’re using this term for, but that does not make it algebraically impossible. It is possible. As I said before, it’s just as possible as canceling the T’s in this term:

$\frac{e^{\ln(T)}}{T} \ a$

I’m not saying you would do this to your identity. That would defeat any purpose of having the identity. I’m saying by normal rules of algebra, it can be done.

Quote:

Originally Posted by Don Blazys

Now, as for the term on the right, if we cross out those $T$ 's only,
then we will be left with:

$T\left(\frac{a}{T}\right)^{\left(\frac{\frac{x\ln(a)}{\ln(T)}-1}{\frac{\ln(a)}{\ln(T)}-1}\right)}=a^x$

which is "problematic" in several respects. For one thing, the term without the $T$ 's
shows no evidence that the greatest possible factor $T$ was extracted, then cancelled,
which means that it does not necessarily represent a prime as does the term on the left.

Possibly a fine reason to *choose* to leave (T/T).

Quote:

Originally Posted by Don Blazys

Also, there is now no evidence that the term on the left was derived from the term on the right,
which means that the equation can no longer be viewed as an "identity".

No. Notice the Pythagorean trigonometric identity,

$\sin^2 x + \cos^2 x = 1$

The right side of this identity shows no evidence that it was derived from the left yet it keeps its title “identity”. Examples abound.

Quote:

Originally Posted by Don Blazys

Most importantly however, the $T$ 's on the right tell us that it must be possible
to let $T=a$ , which litterally forces us to let $x=1$
which in turn allows us to "cross out" the expressions involving logarithms,
so that all we are left with is:

$T\left(\frac{a}{T}\right)=\left(\frac{T}{T}\right)a$

where we may now let $T=a$ , or simply "cross out" the remaining $T$ 's so as to have:

$a=a$

But T cannot equal a in your identity—it results in division by zero (if x=1 or not). So, I’m not sure where you’re coming from with that. Similarly the rest of your post goes more into that topic which I’ll have to think about to a greater extent.

~modest

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