Mathematics Question Review (2025-12-16)
Question:
Show that the line x+y=1 is the axis of symmetry of both y=e^x and y=ln(x).
Solution:
The shortest distance between y=e^x and y=ln(x) is the distance between A(1,0) and B(0,1). Since the shortest distance must be straight line cutting y=x in equal halves, forming perpendicular bisector of line AB, the slope of the required equation must be -1.
QED.
What 7
What 7
Mutant form of the question:
Proofread version:
Another quesion:
Show that e^x and ln(x) diverge for x>1.
Solution:
Since e^x is the inverse function of ln(x),
if ln(x) diverges, e^x must diverge as well.
Now we are to show that ln(x) diverges:
For a increasing function to converge,
it must have a upper bound,
and it must be monotonic increasing.
Now we show ln(x) has no upper bound:
Assuming there is a upper bound y=l,
where l is a constant.
Since ln'(x) = 1/x >0,
and ln(x) is an analytic function for x>1,
the slope of ln(x) can never be zero,
i.e. for sufficiently large value of l,
ln(l) = l, i.e. they cut each other only once,
then given the domain x>1,
ln(x) is bijective and thus diverges.
Solution:
Since e^x is the inverse function of ln(x),
if ln(x) diverges, e^x must diverge as well.
Now we are to show that ln(x) diverges:
For a increasing function to converge,
it must have a upper bound,
and it must be monotonic increasing.
Now we show ln(x) has no upper bound:
Assuming there is a upper bound y=l,
where l is a constant.
Since ln'(x) = 1/x >0,
and ln(x) is an analytic function for x>1,
the slope of ln(x) can never be zero,
i.e. for sufficiently large value of l,
ln(l) = l, i.e. they cut each other only once,
then given the domain x>1,
ln(x) is bijective and thus diverges.
i.e. for sufficiently large value of l,
ln(x) = l,
if x=l, then they must be infninty.
infinity*
Last question:
What is the use of partial differential equations (PDE) in Feng Shui?
Solution:
Differential Equations form direction fields. In Ordinary ones, the fields are in 2D. For PDE, the fields can be exhibited in 3D or beyond.
Ref: