Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 2 / 10G / i

Table of contents
- Solution i

Solution

The integral is linear in its first argument and conjugate-linear in its second, and complex conjugation of the integrand shows

⟨ g, f ⟩ = \overline{⟨ f, g ⟩}

. Thus it is a Hermitian form.

f (z) = \sum_{j = 0}^{N} a_{j} z^{j}

, Fourier orthogonality gives

⟨ f, f ⟩ = \sum_{j, k} a_{j} \overline{a_{k}} \frac{1}{2 π} \int_{0}^{2 π} e^{i (j - k) θ} d θ = \sum_{j = 0}^{N} ∣ a_{j} ∣^{2} .

(1)

This is nonnegative and vanishes only when every coefficient vanishes, that is, only for the zero polynomial. Hence the Hermitian form is positive definite.

 Ancestors (10)

10G
Paper 2
ib
2007
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
List of universities
 Home

 View article source

 Discussion (0) New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version

Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 2 / 10G / i

Solution

 Ancestors (10)

 Discussion (0)  Subscribe (1) New discussion

 Articles by others on the same topic (0)

 Discussion (0) New discussion