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Topic: new tutor here
Replies: 151   Last Post: Jul 17, 2013 4:09 AM

 Messages: [ Previous | Next ]
 Joe Niederberger Posts: 4,657 Registered: 10/12/08
Re: new tutor here
Posted: Jun 24, 2013 10:57 AM

R. Hansen says:
>Ok, so I think I see what you are complaining about. My statement that the function was ever increasing and would only intersect at one point.

No, (though that's a tantalizing statement and figures in other ways of seeing matters.)

Let's go back to your attached PDF proof. Your reasoning gets tangled and confused between function and their derivatives. (And you said you weren't using calculus!!!)

You have a blue rectangle (BR), as a function of x
[1] BR(x) = x.
You have a red rectangle also a function of x.
[2] RR(x) = e log(x)

You say (see your own quote below) that BR increases at a constant rate. True, in fact BR' = 1. You say RR increases at an ever decreasing rate. This is also true if it is translated as follows: RR is ever increasing while RR' is ever decreasing.

The problem is the conclusion you draw, which doesn't follow from those observed facts alone. BR(e) = RR(e)
is true, but you haven't *said anything* thing about
the value of RR' at x=e! It could be the case that RR'(e) >> BR'(e) (which is still and always 1). It could even be the case that RR'(x) is always strictly greater than 1, even though it is *ever decreasing*. If those things were true (and you haven't shown or even mentioned that they are not!) then RR would always be larger than BR, for all x > e!

([There is a bit I snipped out of the quote below (because transcribing was a pain. It had to do with inverse functions, call them BR2 and RR2].
I also note that you make a claim that BR, *as a function of y*, call it BR2(y) is increasing at an ever increasing rate (vis-a-vis y), and that a similar inverse of RR, call it RR2, has constant derivative RR2'(y) = 1.
These inverse relationship don't get you any pertinent facts that I can see, other than to bloat and confuse the argument. The value of BR2' at y=1 is similarly unknown.)

So your conclusion about the areas in completely unfounded. The red rectangle could be larger than the blue at x=pi, and you haven't brought forth any reasons why it cannot be. The facts you do bring up simply don't warrant the desired conclusion.

I think its always a bad idea to say "we are proving such and such" *without calculus*, and then launch into rates of change arguments. Calculus was invented to make clear thinking about those situations possible; abandon it and confusions like the one I explain above are easy to fall into.

The key fact that I used from calculus was that the derivative of log is 1/e at x=e. In terms of your argument (mine was given earlier) this means RR'(e) = 1.

[Interpretation: at x=e, RR' = BR', (after which, RR' is strictly decreasing while BR' = 1.) This is *key* to making your argument work, but is *missing* from your argument. I happen to think its pretty clear you confused BR(e)=RR(e) as carrying the day, where you *really* needed BR'(e)=RR'(e). ]

So, RR'(e) = BR'(e), plus RR' is ever decreasing, means *intuitively*, that RR(pi) has to be < BR(pi) = pi. To really write a "proof", though, one would need to invoke something like the mean value theorem (as I mentioned in my solution).

[Note on "non-calculus" proofs: I would gladly accept arguments based on pre-calculus "principles" such as Cavalieri's, if they are clearly stated and the reasoning sticks to just that principle. Something like that *may* be possible here (I'm doubtful), and it will necessarily involve at least one special fact about e, that would normally be considered as "calculus" derived fact.]

In closing my post here, I'd like to reiterate that I've always maintained a distinction in this thread between mathematical visualizations, and proof that may be *based* on those. One can get the picture correct but the proof wrong. They are not the same, but there is intimate connection.

Cheers,
Joe N

R. Hansen says:
- --------------------------------
>We can see that as x increases, the blue rectangle?s width (and its area) increases at a constant rate while the red rectangle's height (and its area) increases at
an ever decreasing rate (because of the log(x) relationship).
<snip>
Our proof is done. When one function increases at a
constant rate and a second function increases at an
ever increasing rate then if these two functions are
ever equal they will only be equal once. The areas of
the red and blue rectangles are equal at point P, thus
they will never be equal again and since the blue rectangle?s area is increasing at an ever increasing
rate, it will always be greater than the area of the red
rectangle, at all points after P and at point Q, thus
e^pi > pi^e.
- ------------------

------- End of Forwarded Message

Date Subject Author
6/12/13 Michael Mossey
6/12/13 Wayne Bishop
6/12/13 Michael Mossey
6/13/13 Wayne Bishop
6/13/13 Dave L. Renfro
6/13/13 Joe Niederberger
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