Does #int fg = int f int g# ?
3 Answers
Proof by contradiction
Explanation:
This is an interesting question, but can be simply prooved via contradiction:
Let
This is the statement we are trying to disprove:
Let
and
Hence we see that
=>
Hence we were able to select
See explanation...
Explanation:
There are some functions for which:
#int fg = int f int g#
The easiest examples to find are functions which are mostly
For example:
#f(x) = g(x) = { (1 " if " x " is rational"), (0 " if " x " is irrational") :}#
Then:
#int fg = 0+C = int f int g#
If however
#int fg != int f int g#
For example suppose
Then the lowest degree non-zero term of
#(a_m b_n)/((m+1)(n+1)) x^(m+n+2)#
while that of
#(a_m b_n)/(m+n+1) x^(m+n+1)#
So:
#int fg != int f int g#
If at least one of
Explanation:
To show that the equality does not hold universally, it suffices to provide a counterexample.
That is: to counter the claim that the integrals are equal, we must show that for some
(We do not need to show that, for all
Rhys has provided a counterexample using one of my favorite integrals --
My favorite counterexample to this "multiplication rule" has always been