### Author Topic: Web Bonus Problem to Week 8 (#1)  (Read 2785 times)

#### Victor Ivrii

• Elder Member
• Posts: 2607
• Karma: 0
##### Web Bonus Problem to Week 8 (#1)
« on: November 01, 2015, 05:11:25 PM »

#### Jeremy Li 2

• Full Member
• Posts: 15
• Karma: 0
##### Re: Web Bonus Problem to Week 8 (#1)
« Reply #1 on: December 01, 2015, 06:52:43 AM »
PART A
In case the boldfacing isn't clear below, $\mathbf{k}$ and $\mathbf{x}$ are vectors.

a)
Let $\kappa$ be $\sqrt{2\pi}$. Then, applying the Plancherel theorem to each dimension:

||Fu(\mathbf{x})||=||\hat{u}(\mathbf{x})||=||u(\mathbf{x})||

Then $F^*F=I$.

This is an infinite dimensional operator, but right-multiplying the previous equation by $F^{-1}$ we get $F^*=F^{-1}$. Therefore, $FF^*=FF^{-1}=I$

b)

(F^2 f)(\mathbf{x}) = F(F(f(\mathbf{x})))=F(\hat{f}(\mathbf{k}))
=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) e^{-i\mathbf{k}\cdot\mathbf{x}}d^n\mathbf{k}
=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) e^{i\mathbf{k}\cdot(-\mathbf{x})}d^n\mathbf{k}
=f(-\mathbf{x})

c)
This is obvious from b.

(F^4f)(\mathbf{x})=(F^2F^2f)(\mathbf{x})=(F^2f)(-\mathbf{x})=f(\mathbf{x})

d)
Firstly, assume $f(\mathbf{x})$ is even,

Then

\hat{f}(\mathbf{k})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) e^{-i \mathbf{k} \cdot \mathbf{x}} d^n\mathbf{x}

Expanding the complex exponential:

\hat{f}(\mathbf{k})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) \cos(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x} - \frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) i\sin(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x}

Since $f(\mathbf{x})$ is even, we can repeat the calculation of the Fourier Transform with $f(-\mathbf{x})$ instead.

\hat{f}(\mathbf{k})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{-x}) \cos(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x} - \frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{-x}) i\sin(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x}

We can do the substitution $\mathbf{-x}=\mathbf{x}$. We get a negative factor in the differential when we do this, but the limits of integration are also reversed on each coordinate, so the negative factor cancels. Therefore:

\hat{f}(\mathbf{k})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) \cos(-\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x} - \frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) i\sin(-\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x}\\
=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) \cos(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x} + \frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) i\sin(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x}

Adding (5) and (7) and dividing by 2 we get:

\hat{f}(\mathbf{k})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} f(\mathbf{x}) \cos(\mathbf{k}\cdot\mathbf{x}) d^n\mathbf{x}

which is real if $f(\mathbf{x})$ is real.

Now assume that $\hat{f}(\mathbf{k})$ is real. We want to prove that $f(\mathbf{x})$ is even. The IFT is:

f(\mathbf{x})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x}} d^n\mathbf{k}

Expanding the complex exponential

f(\mathbf{x})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) \cos(\mathbf{k} \cdot \mathbf{x}) d^n\mathbf{k} + \frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) i\sin(\mathbf{k} \cdot \mathbf{x}) d^n\mathbf{k}

If $\hat{f}(\mathbf{k})$ is real, than the integrand of the term on the right is purely imaginary, whereas the term on the left is purely real. With the given assumption that $f(\mathbf{x})$ is real, the second term must necessarily be zero. Then:

f(\mathbf{x})=\frac{1}{\sqrt{2\pi}} \iiint_{\mathbb{R}^n} \hat{f}(\mathbf{k}) \cos(\mathbf{k} \cdot \mathbf{x}) d^n\mathbf{k}

Therefore $f(\mathbf{x})$ is even.

The case of odd $f(\mathbf{x})$ can be done in the same way.
« Last Edit: December 01, 2015, 06:56:27 AM by Jeremy Li 2 »

#### Jeremy Li 2

• Full Member
• Posts: 15
• Karma: 0
##### Re: Web Bonus Problem to Week 8 (#1)
« Reply #2 on: December 01, 2015, 07:21:31 AM »
PART B
The transformation $Q$ is orthogonal. That means $Q^T=Q^{-1}$.

Then

\det(Q)\det(Q)=\det(Q)\det(Q^T)=\det(QQ^T)=\det(QQ^{-1})=\det(I)=1

This means that $|\det(Q)|=1$. By 5.2.A Eq(1):

f(\mathbf{x})=f(Q\mathbf{x}) \iff \hat{f}(\mathbf{k})=\hat{f}(Q^T\mathbf{k})

It's evident that if $f$ is rotationally invariant (left side is true), than the right side holds for any $Q^T$ orthogonal, and so $\hat{f}$ is rotationally invariant. It's evident that if $\hat{f}$ is rotationally invariant (right side is true), than the left side holds for any $Q$ orthogonal, and so $f$ is rotationally invariant.
« Last Edit: December 01, 2015, 08:58:04 AM by Jeremy Li 2 »