Re: Circle Laying On Curves

"Ioannis" <morpheus@xxxxxxxxxxxx> wrote in message
news:1172687240.48149@xxxxxxxxxxx

"Narek Saribekyan" <narek.saribekyan@xxxxxxxxx> wrote in message
news:1172681916.585957.261320@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

Consider curves e^(-x) and ln(x) and a tangent unit circle. How can I
fins the center coordinates?

The set of points at distance 1 above the curve exp(-x), is exp(-x) + 1.
Similarly, the set of points at distance 1 below the curve ln(x), is ln(x) -
1.

Therefore the center of the circle (x,y) will lie at the intersection of the
curves:
exp(-x) + 1 and ln(x) - 1.

x then must satisfy the equation:

exp(-x) + 1 = ln(x) - 1 (1)

(1) cannot be solved exactly, so you have to use numerical methods. Here I
use
a Hyper-Lambert HW function to solve this equation numerically as:

ln(x) - exp(-x) = 2 =>
x*(ln(x)/x - exp(-x)/x) = 2 =>
x*exp(ln(ln(x)/x - exp(-x)/x)) = 2 =>
x = HW(ln(ln(x)/x - exp(-x)/x);2) =>
x =~ 7.393603058

Therefore the circle's origin will be at: (x, ln(x) - 1)

Hmmmm. Upon closer inspection, the above is *almost* right, but not quite
right.

We need actually the intersection of the loci which are formed by all the
points equidistant from both curves, exp(-x) and ln(x), but measuring distance
in the direction of the normal to the tangent at every point of both curves,
both above and below.

These loci are not the functions ln(x) - 1 and exp(-x) + 1, because these are
at distance 1 only vertically from the original functions.

If we call these loci f(x) and g(x), then the solution must indeed satisfy the
equation f(x) = g(x), and there should be two solutions, one to the left and
one to the right of the intersection of the original curves. One of the
solutions will likely be very close to x ~= 7.393603058.

I have no idea what these loci are, or whether they will even be functions.

Apologies for the confusion...
--
I.N. Galidakis
http://ioannis.virtualcomposer2000.com/

.

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