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Old 02-11-2009   #135 (permalink)
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Don Blazys
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Re: A Mathematical Emergency.

To: Modest,

Quoting Modest:
Quote:
As I said before,
it’s just as possible as canceling the T’s in this term:
\frac{e^{\ln(T)}}{T}\ * a

The above term, as is, allows us to divide each and every T by T,
which results in:

\frac{e^{\ln(T/T)}}{T/T}\ * a=a

However, the "Blazys term" as is does not allow us to divide each and every
T by T. That's a big difference!

Quoting Modest:
Quote:
\sin^2x+\cos^2x=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
.

\sin^2x=1-\cos^2x

is how I prefer to view it.

Quoting Modest:
Quote:
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.
If:

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

then letting x=1 results in:

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

where the entire exponent on the left "cancels out" using the "white out" method!
Thus, at x=1, all we are left with is:

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

Now, and only now can we let T=a.

You see, if T\neq a were indeed true for all values of x,
then we would have the greatest mystery ever on our hands because
we would then be required to explain exactly why the properties of logarithms
should insanely preclude T=a when both the term on the right
and common sense tell us that T=a must be possible
under certain conditions such as x=1

For me, it's a pretty safe bet that the properties of logarithms are not insane!

In short, if there is absolutely no condition under which my "Blazys terms" will allow T=a,
then I have discovered a huge flaw in the properties of logarithms because clearly,
T=a must be allowable!
After all Mr. Napier was a great mathematician. He was not a "nut" , and his invention,
when used properly, could not possibly result in "nutty" preclusions!

So you see, it absolutely must be the case that
T=a is indeed possible if and only if x=1.

Don.

Last edited by Don Blazys; 02-11-2009 at 08:39 PM..
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