[MD-sorular] kapalı fonksiyon

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Sn Şirin Yılmaz
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Math Forum'da İngilizce olarak var
okudum anladım, belki iyi bir çeviri yapamam
diyerek Türkçe'ye çevirmedim.
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----- Özgün İleti -----
Kimden : "m.sirin yılmaz"
Kime : md-sorular at matematikdunyasi.org
Gönderme tarihi : 26/05/2010 15:47
Konu : [MD-sorular] kapalı fonksiyon
x^y=y^x bir kapalı fonksiyon kuralımı?
Eğer öyleyse tanım kümesi ile ilgili
ne söylenebilir? Saygılarımla...
M.Şirin Yılmaz
________________________________________

solving the Equation x^y = y^x
First take the log of both sides:
  log(x^y) = log(y^x)
and simplify:
  Y*log(X) = X*log(Y)
and then divide by X*Y:
  log(X)   log(Y)
  ------ = ------.
     X        Y
Now you should consider the function
         log(x)
  f(x) = ------.
            x
and simplify:
  Y*log(X) = X*log(Y)
and then divide by X*Y:
  log(X)   log(Y)
  ------ = ------.
     X        Y
Now you should consider the function
         log(x)
  f(x) = ------.
            x
Clearly, we have a solution to the last equation
if and only if
  f(X) = f(Y).
Well, this happens when X = Y,
but does it happen elsewhere?  If we graph
  y = f(x)
we will find that f increases from y = -infinity at x = 0
to y = 1/e
at x = e (that's e = 2.71828... whether you used the
common log or the natural log or the log to any other base),
and then f decreases from
y = 1/e at x = e to y = 0 at x = infinity.
Well, if X and Y are different values and
  f(X) = f(Y),
then that means that there is a horizontal line which
passes through our function at two points (namely X and Y).
Look at the function,
and you'll find that the smaller value is somewhere
between 1 and e,
and the larger value is bigger than e.
Also, the closer the smaller
value is to e, the closer the larger value is to e.
The closer the
smaller value is to 1, the bigger the larger value is.
So what you find is that if X <= 1,
then the only solution is Y = X.
Similarly, if X = e, then the only solution is Y = X.
But if
1 < X < e,
-1 power and get
X^(-Y) = Y^(-X)
and then if X is odd and Y is even or vice-versa,
then the signs don't match,
but if X and Y are both odd, then we multiply
both sides of the equation by -1 to get
(-X)^(-Y) = (-Y)^(-X).
If both X and Y are even, then we don't need to multiply,
and we still get the same equation.
So (-X, -Y) is a solution in positive integers.