I conjecture that there is no elementary function g (or combination of elementary functions) such that
g(x) + g(y) = xy
(i.e. the function g with argument x, added to the same function with x replaced by y, gives x times y.)
For best answer, either prove me wrong by finding such a function g or else prove me right by showing that it cannot exist.
This is nothing to do with homework or teaching, just a little question I thought of.
g(x) + g(y) = xy
(i.e. the function g with argument x, added to the same function with x replaced by y, gives x times y.)
For best answer, either prove me wrong by finding such a function g or else prove me right by showing that it cannot exist.
This is nothing to do with homework or teaching, just a little question I thought of.
-
Yes, there is no such function. Suppose, for the sake of contradiction that there is. Then,
g(x) + g(y) = xy
Since this true for all x,y, it is true for x = 0 and all y. Hence,
g(0) + g(y) = 0 . y = 0
Consequently, g(y) = -g(0).
Hence g is a constant function. We then derive using the same equation that
-2 g(0) = xy, which means that xy is a fixed value for every x and y. Contradiction. Therefore such a g cannot exist.
g(x) + g(y) = xy
Since this true for all x,y, it is true for x = 0 and all y. Hence,
g(0) + g(y) = 0 . y = 0
Consequently, g(y) = -g(0).
Hence g is a constant function. We then derive using the same equation that
-2 g(0) = xy, which means that xy is a fixed value for every x and y. Contradiction. Therefore such a g cannot exist.
-
such a relation cannot exist for independent parameters for all real values of x and y.
if y=0
g(x) + g(0) = 0
g(x) = -g(0)
likewise,
g(y) = -g(0)
This is not exactly your given parameters, but consider
ln(x) + ln(y) = ln (x*y) for all positive values of x and y.
if you have restricted domains, you may find a relation
if y=0
g(x) + g(0) = 0
g(x) = -g(0)
likewise,
g(y) = -g(0)
This is not exactly your given parameters, but consider
ln(x) + ln(y) = ln (x*y) for all positive values of x and y.
if you have restricted domains, you may find a relation
-
Put y=0 so g(x)=-g(0)=c any constant
so g is constant function equal to c.
c+c=xy
so for any xy = 2c=constant
a contradiction
hence this function does not exist.
so g is constant function equal to c.
c+c=xy
so for any xy = 2c=constant
a contradiction
hence this function does not exist.