Math Challenge - May 2019

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  • #37
Periwinkle
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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.

 
  • #38
fresh_42
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##\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.
 
  • #39
Periwinkle
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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##.
 
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  • #40
fresh_42
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We have chosen ##\xi##, but only one element from one non-empty set ##\Lambda(x)## and use ##|\xi -x_0|<|x-x_0|< \delta\,.##
 
  • #41
Couchyam
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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.)
 
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  • #42
SpinFlop
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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]##.
 
  • #43
fresh_42
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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.
 
  • #44
Periwinkle
63
13
$$ (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. $$

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

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

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

$$ \phi = 0.6435. $$
 
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  • #45
fresh_42
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$$ (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##
 
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  • #46
Periwinkle
63
13
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.
 
  • #47
Periwinkle
63
13
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.
 
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  • #48
fresh_42
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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]\,.##
 
  • #49
Periwinkle
63
13
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.
 

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