Consider the real numbers for . Prove that
Using the identity in combination with we have:
Evaluate the integral
First of all we note that for
Hence by Parseval we get that
Find all values of such that
We’re invoking the same technique as in the problem here. Successively we have:
Obviously is differentiable in and its derivative is
It is . We distinguish cases:
- If then attains global maximum at equal to . Since it follows that .
- If then attains global minimum at equal to . In this case , however , the inequality cannot hold for all ; since is continuous its range is:
Hence this case is rejected.
- For the inequality obviously holds for all .
Summing up , .
Examine the convergence of a series
From Taylor’s theorem with integral remainder we have that
However it is known that . Hence,
Exponentiating we get
Thus the series converges.
Prove that there does not exist a rational function with real coefficients such that
where is a non constant polynomial.
Since polynomials are defined on we have that
Since tends to a finite value as it must be a constant polynomial. In particular, must be constant in the range of which is an infinite set, implying that must also be constant. This proves what we wanted.