Math Challenge - May 2019

[Periwinkle]

Your on the right track and it has to do with $x$. But the proof is technically correct. It is the same limit, since $\xi$ is chosen accordingly. It simply uses an assumption which is not explicitly mentioned, which one?

I've been thinking so much about this question. If we want to be so precise, then the conditions of the theorem could be completed by that $a \lt b$.

 

[fresh_42]

I've been thinking so much about this question. If we want to be so precise, then the conditions of the theorem could be completed by that $a \lt b$.

No, it's more subtle than this and uses a 'tool' you wouldn't have expected in calculus.

 

[Periwinkle]

No, it's more subtle than this and uses a 'tool' you wouldn't have expected in calculus.

Then this is just the Axiom of Choice. Then I'll explain it.

 

[fresh_42]

Then this is just the Axiom of Choice. Then I'll explain it.

Yes, $\xi$ is selected from an interval which depends on $x$, and since this is done for all $x$ we have silently used a selection function $x \longmapsto \xi(x)$.

Do you also have an idea how it can be avoided?

 

[Periwinkle]

Yes, $\xi$ is selected from an interval which depends on $x$, and since this is done for all $x$ we have silently used a selection function $x \longmapsto \xi(x)$.

Do you also have an idea how it can be avoided?

The question described above is written on pages 74-77 of this book. The reverse is questionable.

 

[fresh_42]

The question described above is written on pages 74-77 of this book. The reverse is questionable.

O.k. but we don't have to fall back on rationals here. Epsilontic and continuity will do.

 

[Periwinkle]

O.k. but we don't have to fall back on rationals here. Epsilontic and continuity will do.

$\xi(x) = (\min\{x,x_0\}+\max\{x,x_0\})/2$

That's fair. There is no need for exact equality.

 

[fresh_42]

$\xi(x) = (\min\{x,x_0\}+\max\{x,x_0\})/2$

I can't see why this avoids AC. Formally we did this:
$$
\Lambda(x):=\left\{ \xi \in (\min\{x,x_0\},\max\{x,x_0\}) \, : \, \dfrac{f(x)-f(x_0)}{x-x_0}=f\,'(\xi)\right\}
$$
The mean value theorem guarantees us that all $\Lambda(x)\neq \emptyset$, but we need more: namely a function $$\xi \, : \, [a,b]-\{x_0\}\longrightarrow \bigcup_{x\in [a,b]-\{x_0\}} \Lambda(x)$$
Narrowing the interval doesn't change the argument.

 

[Periwinkle]

I can't see why this avoids AC. Formally we did this:
$$
\Lambda(x):=\left\{ \xi \in (\min\{x,x_0\},\max\{x,x_0\}) \, : \, \dfrac{f(x)-f(x_0)}{x-x_0}=f\,'(\xi)\right\}
$$
The mean value theorem guarantees us that all $\Lambda(x)\neq \emptyset$, but we need more: namely a function $$\xi \, : \, [a,b]-\{x_0\}\longrightarrow \bigcup_{x\in [a,b]-\{x_0\}} \Lambda(x)$$
Narrowing the interval doesn't change the argument.

Exist the limit $c:=\lim_{x \to x_0}f\,'(x)\,.$

Therefore, for all $\epsilon$, there is a $\delta$ that if $|x-x_0| \lt \delta$, then $|f\,'(x)-c| \lt \epsilon$.

Based on mean value theorem $\dfrac{f(x)-f(x_0)}{x-x_0}\,$ equal to one of $f\,'(\xi)$, where $\left| \xi-x_0 \right| \lt \delta$.

Therefore without choosing $\xi$ we know $\left| \dfrac{f(x)-f(x_0)}{x-x_0}\ -c \right| \lt \epsilon$ if $|x-x_0| \lt \delta$.

 

[fresh_42]

We have chosen $\xi$, but only one element from one non-empty set $\Lambda(x)$ and use $|\xi -x_0|<|x-x_0|< \delta\,.$

 

[Couchyam]

Spoiler: (Problem 9)

Let $n:=i_\ell i_{\ell-1}\dots i_0$ be the decimal representation of $n\in\mathbb N$ (that is, $n=\sum_{j=0}^\ell i_j 10^j$.) Since $n\geq 10^\ell$, $\frac{1}{n}\leq 10^{-\ell}$. Now, for a fixed $\ell$, there are no more than $9^{\ell+1}$ numbers between $10^\ell$ and $10^{\ell+1}$ whose decimal expansions do not contain the digit $9$ (this is because each number in this range has a unique decimal expansion with at most $\ell+1$ digits, and the set of length $\ell+1$ strings composed of the digits $\{0,...,8\}$ has size $9^{\ell+1}$). Hence,
\begin{align*}
\sum_{n=1}^\infty \frac{\epsilon_n}{n}=\sum_{\ell=0}^\infty \bigg(\sum_{10^\ell\leq n<10^{\ell+1}}\frac{\epsilon_n}{n}\bigg)<\sum_{\ell=0}^\infty \frac{9^{\ell+1}}{10^\ell}=90<\infty
\end{align*}
(Heuristically, the convergence of the series is related to the Cantor-set-like support of the coefficients $\frac{\epsilon_n}{n}$, in the sense that as $n$ grows, the gaps between regions where $\epsilon_n$ is nonzero expand proportionally.)

 

[SpinFlop]

Attempt at Problem 4

Part a) follows directly from the fundamental property of self adjoint operators.
$$ (\hat T{\mathbf a})\cdot{\mathbf b} = {\mathbf a}\cdot(\hat T{\mathbf b})$$
where $\hat T$ is a self-adjoint linear operator and ${\mathbf a},{\mathbf b}$ are two vectors in a given vector space.
In the following proof ${\mathbf a},{\mathbf b}$ will be taken as eigenvectors of $\hat T$ with eigenvalues:
$$\hat T {\mathbf a} = \alpha{\mathbf a}\\
\hat T {\mathbf b} = \beta{\mathbf b}$$
First show that eigenvalues of a self-adjoint operator are real:
$$
(\hat T{\mathbf a})\cdot{\mathbf a} = {\mathbf a}\cdot(\hat T{\mathbf a})\\
(\alpha{\mathbf a})\cdot{\mathbf a} = {\mathbf a}\cdot(\alpha{\mathbf a})\\
\alpha^{*}({\mathbf a}\cdot{\mathbf a}) = \alpha({\mathbf a}\cdot{\mathbf a})\\
\alpha^{*} = \alpha
$$
Thus $\alpha$ is real.
Now show that eigenvectors of $\hat T$ with distinct eigenvalues are orthogonal
$$
(\hat T{\mathbf a})\cdot{\mathbf b} = {\mathbf a}\cdot(\hat T{\mathbf b})\\
(\alpha{\mathbf a})\cdot{\mathbf b} = {\mathbf a}\cdot(\beta{\mathbf b})\\
\alpha^{*}({\mathbf a}\cdot{\mathbf b}) = \beta({\mathbf a}\cdot{\mathbf b})\\
\alpha({\mathbf a}\cdot{\mathbf b}) = \beta({\mathbf a}\cdot{\mathbf b})
$$
where the last line follows from the fact that the eigenvalues must be real.
Since the eigenvectors ${\mathbf a},{\mathbf b}$ have distinct eigenvalues this means $\alpha\neq\beta$ and so necessarily ${\mathbf a}\cdot{\mathbf b} = 0$.
Thus, by definition ${\mathbf a},{\mathbf b}$ are orthogonal.

Part b)
If I understand correctly, for a given function $f(t)$ in the Hilbert space $\mathcal {H} =L_{2}([0,1])$ the linear operator $T_g$ simply multiplies $f(t)$ by the function $g(t)$:
$$T_{g}(f)(t):=g(t)f(t)$$
If so, then the eigenvalue problem requires
$$g(t)f(t) = \lambda_{g}f(t)$$
The trivial possibility is that the function $g(t)$ is a constant and so we get $\lambda_{g} = m = M$
If $g(t)$ is not a constant, then I am not so sure I know how to proceed, the only path forward that I see is to define our eigenfunctions as delta functions, in which case our eigenvalue spectrum is continuous over the interval $[M,m]$.

 

[fresh_42]

Attempt at Problem 4

Part a) follows directly from the fundamental property of self adjoint operators.
$$ (\hat T{\mathbf a})\cdot{\mathbf b} = {\mathbf a}\cdot(\hat T{\mathbf b})$$
where $\hat T$ is a self-adjoint linear operator and ${\mathbf a},{\mathbf b}$ are two vectors in a given vector space.
In the following proof ${\mathbf a},{\mathbf b}$ will be taken as eigenvectors of $\hat T$ with eigenvalues:
$$\hat T {\mathbf a} = \alpha{\mathbf a}\\
\hat T {\mathbf b} = \beta{\mathbf b}$$
First show that eigenvalues of a self-adjoint operator are real:
$$
(\hat T{\mathbf a})\cdot{\mathbf a} = {\mathbf a}\cdot(\hat T{\mathbf a})\\
(\alpha{\mathbf a})\cdot{\mathbf a} = {\mathbf a}\cdot(\alpha{\mathbf a})\\
\alpha^{*}({\mathbf a}\cdot{\mathbf a}) = \alpha({\mathbf a}\cdot{\mathbf a})\\
\alpha^{*} = \alpha
$$
Thus $\alpha$ is real.
Now show that eigenvectors of $\hat T$ with distinct eigenvalues are orthogonal
$$
(\hat T{\mathbf a})\cdot{\mathbf b} = {\mathbf a}\cdot(\hat T{\mathbf b})\\
(\alpha{\mathbf a})\cdot{\mathbf b} = {\mathbf a}\cdot(\beta{\mathbf b})\\
\alpha^{*}({\mathbf a}\cdot{\mathbf b}) = \beta({\mathbf a}\cdot{\mathbf b})\\
\alpha({\mathbf a}\cdot{\mathbf b}) = \beta({\mathbf a}\cdot{\mathbf b})
$$
where the last line follows from the fact that the eigenvalues must be real.
Since the eigenvectors ${\mathbf a},{\mathbf b}$ have distinct eigenvalues this means $\alpha\neq\beta$ and so necessarily ${\mathbf a}\cdot{\mathbf b} = 0$.
Thus, by definition ${\mathbf a},{\mathbf b}$ are orthogonal.

Part b)
If I understand correctly, for a given function $f(t)$ in the Hilbert space $\mathcal {H} =L_{2}([0,1])$ the linear operator $T_g$ simply multiplies $f(t)$ by the function $g(t)$:
$$T_{g}(f)(t):=g(t)f(t)$$
If so, then the eigenvalue problem requires
$$g(t)f(t) = \lambda_{g}f(t)$$
The trivial possibility is that the function $g(t)$ is a constant and so we get $\lambda_{g} = m = M$
If $g(t)$ is not a constant, then I am not so sure I know how to proceed, the only path forward that I see is to define our eigenfunctions as delta functions, in which case our eigenvalue spectrum is continuous over the interval $[M,m]$.

What you wrote is correct so far. But the question was to determine the spectrum, i.e. the complement of the resolvent set, not the point spectrum of eigenvalues.

 

[Periwinkle]

Spoiler: Question 8

$$ (a(x),(b(x)) =\int_{-\pi /2}^{\pi/2} (11\sin(x) + 8\cos(x))\cdot (4\sin(x) + 13\cos(x)) \, dx =

\\ \int_{-\pi /2}^{\pi/2} (44\sin^2(x) + 175 \cos(x) \sin(x) + 104 \cos^2(x)) \, dx =
44 \pi /2 +104 \pi/2 = 148 \pi /2. $$

$$ \left|a \right| ^2 = (a(x),(a(x)) =
\\ \int_{-\pi /2}^{\pi/2} (121\sin^2(x) + 176 \cos(x) \sin(x) + 64 \cos^2(x)) \, dx = 121 \pi /2 +64 \pi/2 = 185 \pi /2.$$

$$ \left|b \right| ^2 = (b(x),(b(x)) =
\\ \int_{-\pi /2}^{\pi/2} (16\sin^2(x) + 104 \cos(x) \sin(x) + 169 \cos^2(x)) \, dx = 16 \pi /2 + 169 \pi/2 = 185 \pi /2.$$

$$ \cos(\phi) = \frac {(a(x),(b(x))} {\left|a \right| \left|b \right| } = \frac {148 \pi /2} { 185 \pi /2 }= 0.8. $$

$$ \phi = 0.6435. $$

 

[fresh_42]

Spoiler: Question 8

$$ (a(x),(b(x)) =\int_{-\pi /2}^{\pi/2} (11\sin(x) + 8\cos(x))\cdot (4\sin(x) + 13\cos(x)) \, dx =

\\ \int_{-\pi /2}^{\pi/2} (44\sin^2(x) + 175 \cos(x) \sin(x) + 104 \cos^2(x)) \, dx =
44 \pi /2 +104 \pi/2 = 148 \pi /2. $$

$$ \left|a \right| ^2 = (a(x),(a(x)) =
\\ \int_{-\pi /2}^{\pi/2} (121\sin^2(x) + 176 \cos(x) \sin(x) + 64 \cos^2(x)) \, dx = 121 \pi /2 +64 \pi/2 = 185 \pi /2.$$

$$ \left|b \right| ^2 = (b(x),(b(x)) =
\\ \int_{-\pi /2}^{\pi/2} (16\sin^2(x) + 104 \cos(x) \sin(x) + 169 \cos^2(x)) \, dx = 16 \pi /2 + 169 \pi/2 = 185 \pi /2.$$

$$ \cos(\phi) = \frac {(a(x),(b(x))} {\left|a \right| \left|b \right| } = \frac {148 \pi /2} { 185 \pi /2 }= 0.8. $$

$$ \phi = 0.6435. $$

That will take me awhile, since one of us has made a mistake and I have to figure out who and where.

Correction: We were both right. I forgot a square root in the denominator.

How do you find my solution:

We define $f(x)=\sin(x)-6\cos(x)\; , \;g(x)=6\sin(x)+\cos(x)$ and observe, that $\{\,f,g\,\}$ is a orthogonal basis for a two dimensional subspace of $L^2\left( \left[ -\frac{\pi}{2},+\frac{\pi}{2} \right] \right)$ with $\gamma :=|f|=|g|=\sqrt{\dfrac{37 \pi}{2}}$. As we are interested in an angle, we won't have to bother the length of our coordinate vectors, i.e. we do not need to normalize them. Now we have $a=-f+2g\, , \,b=-2f+g$ and
\begin{align*}
\cos \varphi &= \cos (\sphericalangle (a,b))\\
&= \cos(\sphericalangle (-f+2g,-2f+g))\\
&= \dfrac{\langle -f+2g,-2f+g \rangle}{|-f+2g|\cdot |-2f+g|}\\
&= 2 \; \dfrac{\langle f,f\rangle + \langle g,g \rangle}{\sqrt{\left( |f|^2+4|g|^2\right)} \cdot \sqrt{\left( 4|f|^2+|g|^2 \right)}}\\
&= 2\; \dfrac{\gamma^2+\gamma^2}{\sqrt{5\gamma^2 \cdot 5\gamma^2}}\\
&= \dfrac{4}{5}
\end{align*}
and $\varphi \approx 36.87° \approx 0.2 \pi$

 

[Periwinkle]

That will take me awhile, since one of us has made a mistake and I have to figure out who and where.

Correction: We were both right. I forgot a square root in the denominator.

How do you find my solution:

We define $f(x)=\sin(x)-6\cos(x)\; , \;g(x)=6\sin(x)+\cos(x)$ and observe, that $\{\,f,g\,\}$ is a orthogonal basis for a two dimensional subspace of $L^2\left( \left[ -\frac{\pi}{2},+\frac{\pi}{2} \right] \right)$ with $\gamma :=|f|=|g|=\sqrt{\dfrac{37 \pi}{2}}$. As we are interested in an angle, we won't have to bother the length of our coordinate vectors, i.e. we do not need to normalize them. Now we have $a=-f+2g\, , \,b=-2f+g$ and
\begin{align*}
\cos \varphi &= \cos (\sphericalangle (a,b))\\
&= \cos(\sphericalangle (-f+2g,-2f+g))\\
&= \dfrac{\langle -f+2g,-2f+g \rangle}{|-f+2g|\cdot |-2f+g|}\\
&= 2 \; \dfrac{\langle f,f\rangle + \langle g,g \rangle}{\sqrt{\left( |f|^2+4|g|^2\right)} \cdot \sqrt{\left( 4|f|^2+|g|^2 \right)}}\\
&= 2\; \dfrac{\gamma^2+\gamma^2}{\sqrt{5\gamma^2 \cdot 5\gamma^2}}\\
&= \dfrac{4}{5}
\end{align*}
and $\varphi \approx 36.87° \approx 0.2 \pi$

I was only the inner product of the three-dimensional Euclidean space before my eyes, which has the same meaning in Hilbert space.

 

[Periwinkle]

Question 4

​

Self-adjoint linear operator's eigenvalues are real ​

$$ \lambda (x,x) = (\lambda x, x) = (Ax,x) = (x, Ax) = (x,\lambda x,) = \bar \lambda (x,x) $$ However, in the Euclidean space $(x,x) =0$ follows $x = 0$, so $\lambda = \bar \lambda$.

The eigenvectors belonging to different eigenvalues are orthogonal ​

$$ (Ax, Ay) = (AAx, y) = (\lambda _1 Ax, y) = \lambda_1^2(x,y) $$ $$ (Ax, Ay) = (x, AAy) = (x, \lambda _2 Ay) = \lambda_2^2(x,y) $$ Because $\lambda_1$ and $\lambda_2$ is different, so $(x,y)$ is equal to $0$.

The spectrum of the $T_g$ operator is the interval $[m,M]$​

The regular values of the operator $T_g$ are the $\lambda$ numbers for which the $(T_g-\lambda I)^{-1}$ operator has a value in the whole space. The other values of $\lambda$ are the spectrum of the operator. The inverse operator is defined by the following formula: $$ (T_g-\lambda I)^{-1} f(t) = \frac 1 {g(t) -\lambda} f(t). $$ It must be proved that for each number $\lambda$ in interval $[m,M]$, there is an $s(t)$element of space $L^2([0,1])$, that is not mapped to an element of space $L^2([0,1])\,$ by the above inverse operator. It follows from the continuity of the $g(t)$ function that if $\lambda$ is a point in the $[m, M]$ interval, then $g (t) = \lambda$ on some $t_{\lambda}$. Also, provided that $\lambda$ is different from $m$, there is in the $[0,1]$ interval a $t_1, t_2, \dots, t_i, \dots$ monotone growing sequence that if $t_i \leq t \lt t_{\lambda}$, then $$\frac 1 {|g(t) - \lambda|} \gt i.$$ If $\lambda = m$, there is a similar monotone descending sequence. The appropriate $s(t)\,$element of $L^2([0,1])\,$ is constructed as follows.

The value of the function $s(t)$ on the $[t_i, t_{i+1})$ interval is $$ \frac 1 {i\sqrt {t_{i+1} - t_i}}. $$ At points outside all $(t_i, t_{i+1})$ intervals $s(t)$ is $0$ everywhere. The integral of the $s^2 (t)$ function is equal $$ \sum_{i=1}^\infty {(t_{i+1} - t_i)} {\left( \frac 1 {i \sqrt {t_{i+1} - t_i}} \right)^2} = \sum_{n=1}^\infty \frac 1 {i^2}.$$ However, for each $(t_i, t_{i+1})$ interval, the $$S(t) = \frac 1 {g(t) -\lambda} s(t) $$ function is greater than $i s(t)$, therefore, therefore $S^2(t)$ integral is divergent, thus the transformed function cannot belong to space $L^2([0,1])\,$.

At points outside $[m,M]$ interval $\lambda$ values are regular. The values of $$\left|\frac 1 {g(t) -\lambda}\right|$$ in this case have an $K\,$upper bound. However, if there is an integral of $s^2(t)$, then there is also an integral of $K^2s^2(t)$, so the smaller $$\frac 1 {(g(t) -\lambda)^2} s^2(t)$$ also has an integral. So in this case, the $(T_g-\lambda I)^{-1}$ operator has a value in the whole space.

 

[fresh_42]

Question 4

​

Self-adjoint linear operator's eigenvalues are real ​

$$ \lambda (x,x) = (\lambda x, x) = (Ax,x) = (x, Ax) = (x,\lambda x,) = \bar \lambda (x,x) $$ However, in the Euclidean space $(x,x) =0$ follows $x = 0$, so $\lambda = \bar \lambda$.

The eigenvectors belonging to different eigenvalues are orthogonal ​

$$ (Ax, Ay) = (AAx, y) = (\lambda _1 Ax, y) = \lambda_1^2(x,y) $$ $$ (Ax, Ay) = (x, AAy) = (x, \lambda _2 Ay) = \lambda_2^2(x,y) $$ Because $\lambda_1$ and $\lambda_2$ is different, so $(x,y)$ is equal to $0$.

The spectrum of the $T_g$ operator is the interval $[m,M]$​

...

Correct. It could be said a bit shorter if we don't specify a potential inverse:

From the boundaries of $g$ we get that $m,M$ are a lower, resp. upper bound of $T_g\,.$ Hence $\sigma(T_g) \subseteq [m,M]$. According to the mean value theorem for continuous functions we know, that $g$ takes every value in $[m,M]$ at least once, i.e for every $\mu \in [m,M]$ there is a real number $t_\mu \in [0,1]$ such that $g(t_\mu)=\mu\,.$ Thus $$T_g(f)(t_\mu)=g(t_\mu)f(t_\mu)=\mu\cdot f(t_\mu)$$
and $T-\mu$ isn't bounded invertible, hence $\mu \in \sigma(T_g)$ and $\sigma(T_g)=[m,M]\,.$

 

[Periwinkle]

I noticed my own mistake. Correctly: $$ |S(t)| = \left| \frac 1 {g(t) -\lambda} s(t) \right|$$ function is greater than $i |s(t)|.$

Tomorrow I will consider the above solution.