Given the sequence of functions where such that
(i) the serieses and converge uniformly to functions .
(ii) the functions are continuous.
(iv) it holds that
(i) This is an immediate consequence of the Weierstrass M Test . We simply note that and of course
(ii) The uniform limit of continuous functions is continuous. This is enough for us to extract that both and throughout .
(iii) For a limit of differentiable functions, a sufficient condition for claiming that the derivative of the limit is the limit of the derivatives is the uniform convergence of the sequence of derivatives. Since the series for converges uniformly, we can claim that .
(iv) For the first integral we have that
and for the second integral we have
since the function is odd.
Let be two matrices with real entries. Prove that
provided all the inverses appearing on the left hand side exist.
(Vojtech Jarnik / 2nd Category/ 2015)
be elements of an arbitrary associative algebra with unit. Then:
(Vojtech Jarnik / 2nd Category/ 2011)
We are applying the classical trick, thus:
Let be defined as:
where is a sequence such that and and be an enumeration of the rationals of the interval . Prove that is Riemann integrable and that .
Apparently are Riemann integrable and . Also uniformly because coincide except of some points (). Hence , since , we have that:
Thefore is Riemann integrable and thus:
The exercise can also be found in mathematica.gr
(i) Let be the unit disk and be its positive oriented boundary. Evaluate the line integral
(ii) Can you deduce if the function
is a conservative field using the above question?
(i) We are invoking Green’s theorem. Thus:
(ii) If was a conservative field then then the line integral over all closed curves would have to be . But in the previous question we found one closed curve whose line integral is not . Thus is not a conservative field.
The exercise can also be found in the Jom Forum here.