Douglas J. Bender

No, you haven't. All you've shown is that if the integral of (1/t)dt from 1 to x is interpreted as the inverse function of e^x, then x must be dimensionless (something I've not argued against). And I pointed out that one doesn't need to interpret it in that way, straight from the definition of the definite integral. Show me, from the definition of the natural logarithm (you know, "integrals" and "Riemann sums" and all that), why "x must be 'dimensionless'". If you do not, I will consider your silence to be an implicit admission of being vanquished.


In Christ,

Douglas

[ May 19, 2002: Message edited by: Douglas J. Bender ]</p>

Principia

Really. And you are disputing the MIT professor(s), the Harvard grad, and two lectures of university courses, too?

I know the answer, Douglas. Do you? You either know or you don't. If you assert you do, then there is no point for me to change your mind. You can continue being wrong, and the world could give a damn.

Remember your last failure asserting something you thought you knew the answer to. Now *that blunder* was a show of arrogance.

Scientiae

[ May 19, 2002: Message edited by: Scientiae ]</p>

MadMordigan

<img src="graemlins/banghead.gif" border="0" alt="[Bang Head]" />

katerina2

Douglas J. Bender, the only reasons that my darling Scientiae has refrained from posting in your regard are personal, one of them being that if he does he isn't going to get any from me.

Douglas J. Bender

No, as I said that I was not arguing against the reasoning that if one interprets the definite integral of (1/t)dt from 1 to x as the inverse function of e^x then x must be unitless - I thought I said exactly this in my previous post. I've said that there is no reason, strictly from the definition of that integral, that the argument ("x") needs to be unitless. You've avoided directly answering that issue for some time now.

The answer to what? To my question? If so, then I can say that I know, but that it appears that either you do not, or you do but don't want to admit the answer.

That's profound.

So, you now claim to speak for "the world"? Wow. You must really be something extra special. Whatever, however, since I speak for God (I am a Christian, and I proclaim the Gospel).

It was no "show of arrogance". Confidence misplaced, perhaps, but not "arrogance". However, as I've said now any number of times, the reason I was in error was primarily because of a lack of completeness in ardipithecus' explanation of the "isochron method".

In Christ,

Douglas

Douglas J. Bender

Get any what, katerina?


In Christ,

Douglas

katerina2

If I have to spell that out, Douglas...

Coragyps

Katerina, if you have to spell it out, you'll put me in a terrible quandry: will I send this thread to Rants&Raves or to Humo(u)r?

Coragyps

Douglas, what possible meaning could e^year or e^furlong have? You're really reaching....

Principia

Far be it for me to speak for ma cherie, but I think she would be quite happy with either, Moderator. At your discretion...

EDIT:

Well, Douglas, all definitions of ln(x) must be equivalent and consistent. If you accept the explanation I provided (along with the other sources) then the inconsistency you perceive in the integral definition is for you alone to resolve.

Dimensionless arguments are in fact consistent with all definitions of ln(x); however, as you admit, dimensional ones apparently are not. Yet everyone in the natural sciences talks about dimensional consistency. I can quite easily explain why the integral definition should also require dimensionless arguments. But it sounds like you've made up your mind again (or perhaps it is more misplaced confidence).

Scientiae

[ May 20, 2002: Message edited by: Scientiae ]</p>