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'Spiky Periodic Things' - Do these objects have a name, and is there a method for finding the boundary...
$begingroup$
This question was originally about evaluating the sum $sum_{n=0}^infty e^{nix}$, but I figured out the answer about half way through writing it. So instead, I decided to ask a slightly different question.
Now, obviously, the sum $sum_{n=0}^infty e^{nix}$ does not converge - however, the closure of the set of points $(x,z)$ given by $z=sum_{n=0}^omega e^{nix}$ is bounded for arbitrarily large $omega$; the object formed by these points is well defined as $omegatoinfty$.
This becomes clear when you break the sum into a real and imaginary part:
$$sum_{n=0}^infty e^{nix}=sum_{n=0}^infty cos{nx}+isum_{n=0}^infty sin{nx}$$
Both sums form bounded sets, and the boundary is clearly another periodic function with vertical asymptotes at integer multiples of $2pi$.
The resulting object can be expressed (admittedly awkwardly) as a piece-wise set-valued function of a single real variable $x$:
$$r_1(x)=-frac{1}{2}csc{frac{x}{2}}+frac{1}{2}qquad r_2(x)=frac{1}{2}csc{frac{x}{2}}+frac{1}{2}$$
$$m_1(x)=-frac{1}{2}tan{frac{x}{4}}qquad m_2(x)=frac{1}{2}cot{frac{x}{4}}$$
$$f(x)=begin{cases}[r_1(x),r_2(x)]+i[m_1(x),m_2(x)]&cot{frac{x}{4}}>0\ mathbb{C}&cot{frac{x}{4}}=0^{pm1}\ [r_2(x),r_1(x)]+i[m_2(x),m_1(x)]&cot{frac{x}{4}}<0end{cases}$$
Using $0^{-1}=pminfty$, $quad Y_1+Y_2=left{y_1+y_2mid y_1in Y_1land y_2in Y_2right}$, and $iY=left{iymid yin Yright}$. (If there is a more elegant way to write this, please tell me).
Naturally my next question was whether or not other 'spiky summations' exist, so I played around with different periodic functions, trying to get the summation to 'converge' to a bounded shape. After experimenting for a while, it seems that there is an entire class of these objects - which have interesting geometric and statistical properties.
Do these objects have a name? And is there a general method for finding the bounding curves given the summation used to generate them?
sequences-and-series curves periodic-functions
asked Jan 24 at 20:16
R. BurtonR. Burton
638110
$endgroup$
add a comment |
$begingroup$
This question was originally about evaluating the sum $sum_{n=0}^infty e^{nix}$, but I figured out the answer about half way through writing it. So instead, I decided to ask a slightly different question.
Now, obviously, the sum $sum_{n=0}^infty e^{nix}$ does not converge - however, the closure of the set of points $(x,z)$ given by $z=sum_{n=0}^omega e^{nix}$ is bounded for arbitrarily large $omega$; the object formed by these points is well defined as $omegatoinfty$.
This becomes clear when you break the sum into a real and imaginary part:
$$sum_{n=0}^infty e^{nix}=sum_{n=0}^infty cos{nx}+isum_{n=0}^infty sin{nx}$$
Both sums form bounded sets, and the boundary is clearly another periodic function with vertical asymptotes at integer multiples of $2pi$.
The resulting object can be expressed (admittedly awkwardly) as a piece-wise set-valued function of a single real variable $x$:
$$r_1(x)=-frac{1}{2}csc{frac{x}{2}}+frac{1}{2}qquad r_2(x)=frac{1}{2}csc{frac{x}{2}}+frac{1}{2}$$
$$m_1(x)=-frac{1}{2}tan{frac{x}{4}}qquad m_2(x)=frac{1}{2}cot{frac{x}{4}}$$
$$f(x)=begin{cases}[r_1(x),r_2(x)]+i[m_1(x),m_2(x)]&cot{frac{x}{4}}>0\ mathbb{C}&cot{frac{x}{4}}=0^{pm1}\ [r_2(x),r_1(x)]+i[m_2(x),m_1(x)]&cot{frac{x}{4}}<0end{cases}$$
Using $0^{-1}=pminfty$, $quad Y_1+Y_2=left{y_1+y_2mid y_1in Y_1land y_2in Y_2right}$, and $iY=left{iymid yin Yright}$. (If there is a more elegant way to write this, please tell me).
Naturally my next question was whether or not other 'spiky summations' exist, so I played around with different periodic functions, trying to get the summation to 'converge' to a bounded shape. After experimenting for a while, it seems that there is an entire class of these objects - which have interesting geometric and statistical properties.
Do these objects have a name? And is there a general method for finding the bounding curves given the summation used to generate them?
sequences-and-series curves periodic-functions
asked Jan 24 at 20:16
R. BurtonR. Burton
638110
$endgroup$
2
$begingroup$
The more elegant way to write $left{y_1inmathbb{R}mid r_1(x)leq y_1leq r_2(x)right}$ is $[r_1(x),r_2(x)]$.
$endgroup$
– Chris Culter
Jan 24 at 20:33
1
$begingroup$
Also, instead of bounding the limit set of the series by a rectangle in the complex plane, it would be even more elegant to say that it's a subset of a certain circle. You can find the center and radius of this circle using the geometric series formula.
$endgroup$
– Chris Culter
Jan 24 at 20:37
$begingroup$
@ChrisCulter How do I use the geometric series formula?
$endgroup$
– R. Burton
Jan 24 at 23:21
add a comment |
$begingroup$
This question was originally about evaluating the sum $sum_{n=0}^infty e^{nix}$, but I figured out the answer about half way through writing it. So instead, I decided to ask a slightly different question.
Now, obviously, the sum $sum_{n=0}^infty e^{nix}$ does not converge - however, the closure of the set of points $(x,z)$ given by $z=sum_{n=0}^omega e^{nix}$ is bounded for arbitrarily large $omega$; the object formed by these points is well defined as $omegatoinfty$.
This becomes clear when you break the sum into a real and imaginary part:
$$sum_{n=0}^infty e^{nix}=sum_{n=0}^infty cos{nx}+isum_{n=0}^infty sin{nx}$$
Both sums form bounded sets, and the boundary is clearly another periodic function with vertical asymptotes at integer multiples of $2pi$.
The resulting object can be expressed (admittedly awkwardly) as a piece-wise set-valued function of a single real variable $x$:
$$r_1(x)=-frac{1}{2}csc{frac{x}{2}}+frac{1}{2}qquad r_2(x)=frac{1}{2}csc{frac{x}{2}}+frac{1}{2}$$
$$m_1(x)=-frac{1}{2}tan{frac{x}{4}}qquad m_2(x)=frac{1}{2}cot{frac{x}{4}}$$
$$f(x)=begin{cases}[r_1(x),r_2(x)]+i[m_1(x),m_2(x)]&cot{frac{x}{4}}>0\ mathbb{C}&cot{frac{x}{4}}=0^{pm1}\ [r_2(x),r_1(x)]+i[m_2(x),m_1(x)]&cot{frac{x}{4}}<0end{cases}$$
Using $0^{-1}=pminfty$, $quad Y_1+Y_2=left{y_1+y_2mid y_1in Y_1land y_2in Y_2right}$, and $iY=left{iymid yin Yright}$. (If there is a more elegant way to write this, please tell me).
Naturally my next question was whether or not other 'spiky summations' exist, so I played around with different periodic functions, trying to get the summation to 'converge' to a bounded shape. After experimenting for a while, it seems that there is an entire class of these objects - which have interesting geometric and statistical properties.
Do these objects have a name? And is there a general method for finding the bounding curves given the summation used to generate them?
sequences-and-series curves periodic-functions
asked Jan 24 at 20:16
R. BurtonR. Burton
638110
$endgroup$
This question was originally about evaluating the sum $sum_{n=0}^infty e^{nix}$, but I figured out the answer about half way through writing it. So instead, I decided to ask a slightly different question.
Now, obviously, the sum $sum_{n=0}^infty e^{nix}$ does not converge - however, the closure of the set of points $(x,z)$ given by $z=sum_{n=0}^omega e^{nix}$ is bounded for arbitrarily large $omega$; the object formed by these points is well defined as $omegatoinfty$.
This becomes clear when you break the sum into a real and imaginary part:
$$sum_{n=0}^infty e^{nix}=sum_{n=0}^infty cos{nx}+isum_{n=0}^infty sin{nx}$$
Both sums form bounded sets, and the boundary is clearly another periodic function with vertical asymptotes at integer multiples of $2pi$.
The resulting object can be expressed (admittedly awkwardly) as a piece-wise set-valued function of a single real variable $x$:
$$r_1(x)=-frac{1}{2}csc{frac{x}{2}}+frac{1}{2}qquad r_2(x)=frac{1}{2}csc{frac{x}{2}}+frac{1}{2}$$
$$m_1(x)=-frac{1}{2}tan{frac{x}{4}}qquad m_2(x)=frac{1}{2}cot{frac{x}{4}}$$
$$f(x)=begin{cases}[r_1(x),r_2(x)]+i[m_1(x),m_2(x)]&cot{frac{x}{4}}>0\ mathbb{C}&cot{frac{x}{4}}=0^{pm1}\ [r_2(x),r_1(x)]+i[m_2(x),m_1(x)]&cot{frac{x}{4}}<0end{cases}$$
Using $0^{-1}=pminfty$, $quad Y_1+Y_2=left{y_1+y_2mid y_1in Y_1land y_2in Y_2right}$, and $iY=left{iymid yin Yright}$. (If there is a more elegant way to write this, please tell me).
Naturally my next question was whether or not other 'spiky summations' exist, so I played around with different periodic functions, trying to get the summation to 'converge' to a bounded shape. After experimenting for a while, it seems that there is an entire class of these objects - which have interesting geometric and statistical properties.
Do these objects have a name? And is there a general method for finding the bounding curves given the summation used to generate them?
sequences-and-series curves periodic-functions
sequences-and-series curves periodic-functions
asked Jan 24 at 20:16
R. BurtonR. Burton
638110
asked Jan 24 at 20:16
R. BurtonR. Burton
638110
edited Jan 24 at 22:21
R. Burton
asked Jan 24 at 20:16
R. BurtonR. Burton
638110
asked Jan 24 at 20:16
R. BurtonR. Burton
638110
asked Jan 24 at 20:16
R. BurtonR. Burton
638110
638110
2
$begingroup$
The more elegant way to write $left{y_1inmathbb{R}mid r_1(x)leq y_1leq r_2(x)right}$ is $[r_1(x),r_2(x)]$.
$endgroup$
– Chris Culter
Jan 24 at 20:33
1
$begingroup$
Also, instead of bounding the limit set of the series by a rectangle in the complex plane, it would be even more elegant to say that it's a subset of a certain circle. You can find the center and radius of this circle using the geometric series formula.
$endgroup$
– Chris Culter
Jan 24 at 20:37
$begingroup$
@ChrisCulter How do I use the geometric series formula?
$endgroup$
– R. Burton
Jan 24 at 23:21
add a comment |
2
$begingroup$
The more elegant way to write $left{y_1inmathbb{R}mid r_1(x)leq y_1leq r_2(x)right}$ is $[r_1(x),r_2(x)]$.
$endgroup$
– Chris Culter
Jan 24 at 20:33
1
$begingroup$
Also, instead of bounding the limit set of the series by a rectangle in the complex plane, it would be even more elegant to say that it's a subset of a certain circle. You can find the center and radius of this circle using the geometric series formula.
$endgroup$
– Chris Culter
Jan 24 at 20:37
$begingroup$
@ChrisCulter How do I use the geometric series formula?
$endgroup$
– R. Burton
Jan 24 at 23:21
2
2
$begingroup$
The more elegant way to write $left{y_1inmathbb{R}mid r_1(x)leq y_1leq r_2(x)right}$ is $[r_1(x),r_2(x)]$.
$endgroup$
– Chris Culter
Jan 24 at 20:33
$begingroup$
The more elegant way to write $left{y_1inmathbb{R}mid r_1(x)leq y_1leq r_2(x)right}$ is $[r_1(x),r_2(x)]$.
$endgroup$
– Chris Culter
Jan 24 at 20:33
1
1
$begingroup$
Also, instead of bounding the limit set of the series by a rectangle in the complex plane, it would be even more elegant to say that it's a subset of a certain circle. You can find the center and radius of this circle using the geometric series formula.
$endgroup$
– Chris Culter
Jan 24 at 20:37
$begingroup$
Also, instead of bounding the limit set of the series by a rectangle in the complex plane, it would be even more elegant to say that it's a subset of a certain circle. You can find the center and radius of this circle using the geometric series formula.
$endgroup$
– Chris Culter
Jan 24 at 20:37
$begingroup$
@ChrisCulter How do I use the geometric series formula?
$endgroup$
– R. Burton
Jan 24 at 23:21
$begingroup$
@ChrisCulter How do I use the geometric series formula?
$endgroup$
– R. Burton
Jan 24 at 23:21
add a comment |
1 Answer
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$begingroup$
This behavior is very easy to explain when you view the sum as a geometric series. We have a closed formula $$sum_{n=0}^omega e^{nix}=frac{e^{(omega+1)ix}-1}{e^{ix}-1}$$ and the bounding curves you are finding are just what you get by maximizing and minimizing the real and imaginary parts of $$frac{w-1}{e^{ix}-1}$$ where $w$ is allowed to range over the unit circle. When $omega$ is large, $e^{(omega+1)ix}$ is cycling through the entire unit circle very quickly as $x$ changes (while the denominator $e^{ix}-1$ is changing slowly), and so the real and imaginary parts of the sum quickly oscillate between the maximum and minimum values where $w$ ranges over the unit circle.
Explicitly, to find the real and imaginary parts we write $w=a+bi$ and find $$frac{w-1}{e^{ix}-1}=frac{a-1+bi}{cos x-1+isin x}=frac{a(cos x-1)+bsin x+1-cos x}{(cos x-1)^2+sin^2 x}+ifrac{-asin x+b(cos x-1)+sin x}{(cos x-1)^2+sin^2 x}.$$ By Cauchy-Schwarz, given the constraint $a^2+b^2=1$, we maximize or minimize $a(cos x-1)+bsin x$ by making $(a,b)$ a scalar multiple of $(cos x-1,sin x)$, so that $a=pmfrac{cos x-1}{sqrt{(cos x-1)^2+sin^2 x}}$ and $b=pmfrac{sin x}{sqrt{(cos x-1)^2+sin^2 x}}$. This tells us that when $2-2cos x$ is positive, the real part is bounded by $$frac{1}{2}pmfrac{(cos x-1)^2+sin^2 x}{((cos x-1)^2+sin^2 x)^{3/2}}=frac{1}{2}pmfrac{1}{2}csc(x/2).$$
Similarly, the imaginary part is bounded by $$frac{1}{2}frac{sin x}{1-cos x}pmfrac{1}{2}csc(x/2)$$ which is $frac{1}{2}cot(x/4)$ or $-frac{1}{2}tan(x/4)$ depending on the sign chosen.
The tangent-like appearance of the imaginary part graph (with half the period of the bounding curves) is explained by the fact that the average value between the oscillations is $frac{1}{2}frac{sin x}{1-cos x}=frac{1}{2}cot(x/2)$. In other words, the imaginary part is the graph of $frac{1}{2}cot(x/2)$ superimposed with oscillations whose peaks behave like $frac{1}{2}csc(x/2)$.
The bounding curves for other similar summations can be found in the same way, by writing everything in terms of complex exponentials and then summing as a geometric series. In particular, the same method should work for three of the four examples you presented at the end (the exception being $sum (-1)^n cos^2(n/(x-n))$, which is really not the same sort of sum at all and has very different behavior).
answered Jan 24 at 23:55
Eric WofseyEric Wofsey
189k14216347
$endgroup$
$begingroup$
Ah, now I see what ChrisCulter meant about the geometric series. You might want to double-check your answer, by the way.
$endgroup$
– R. Burton
Jan 25 at 3:48
$begingroup$
Oops, I missed some square roots when optimizing. Your bounding curves were correct. :)
$endgroup$
– Eric Wofsey
Jan 25 at 5:03
$begingroup$
Also, in this case it works out to be the same, but shouldn't the geometric series give $frac{1-e^{(omega+1)ix}}{1-e^{ix}}$ (with everything after that having the same sign change)?
$endgroup$
– R. Burton
Jan 25 at 15:59
1
$begingroup$
That's the same: you just multiply the numerator and denominator by $-1$.
$endgroup$
– Eric Wofsey
Jan 25 at 16:05
add a comment |
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$begingroup$
This behavior is very easy to explain when you view the sum as a geometric series. We have a closed formula $$sum_{n=0}^omega e^{nix}=frac{e^{(omega+1)ix}-1}{e^{ix}-1}$$ and the bounding curves you are finding are just what you get by maximizing and minimizing the real and imaginary parts of $$frac{w-1}{e^{ix}-1}$$ where $w$ is allowed to range over the unit circle. When $omega$ is large, $e^{(omega+1)ix}$ is cycling through the entire unit circle very quickly as $x$ changes (while the denominator $e^{ix}-1$ is changing slowly), and so the real and imaginary parts of the sum quickly oscillate between the maximum and minimum values where $w$ ranges over the unit circle.
Explicitly, to find the real and imaginary parts we write $w=a+bi$ and find $$frac{w-1}{e^{ix}-1}=frac{a-1+bi}{cos x-1+isin x}=frac{a(cos x-1)+bsin x+1-cos x}{(cos x-1)^2+sin^2 x}+ifrac{-asin x+b(cos x-1)+sin x}{(cos x-1)^2+sin^2 x}.$$ By Cauchy-Schwarz, given the constraint $a^2+b^2=1$, we maximize or minimize $a(cos x-1)+bsin x$ by making $(a,b)$ a scalar multiple of $(cos x-1,sin x)$, so that $a=pmfrac{cos x-1}{sqrt{(cos x-1)^2+sin^2 x}}$ and $b=pmfrac{sin x}{sqrt{(cos x-1)^2+sin^2 x}}$. This tells us that when $2-2cos x$ is positive, the real part is bounded by $$frac{1}{2}pmfrac{(cos x-1)^2+sin^2 x}{((cos x-1)^2+sin^2 x)^{3/2}}=frac{1}{2}pmfrac{1}{2}csc(x/2).$$
Similarly, the imaginary part is bounded by $$frac{1}{2}frac{sin x}{1-cos x}pmfrac{1}{2}csc(x/2)$$ which is $frac{1}{2}cot(x/4)$ or $-frac{1}{2}tan(x/4)$ depending on the sign chosen.
The tangent-like appearance of the imaginary part graph (with half the period of the bounding curves) is explained by the fact that the average value between the oscillations is $frac{1}{2}frac{sin x}{1-cos x}=frac{1}{2}cot(x/2)$. In other words, the imaginary part is the graph of $frac{1}{2}cot(x/2)$ superimposed with oscillations whose peaks behave like $frac{1}{2}csc(x/2)$.
The bounding curves for other similar summations can be found in the same way, by writing everything in terms of complex exponentials and then summing as a geometric series. In particular, the same method should work for three of the four examples you presented at the end (the exception being $sum (-1)^n cos^2(n/(x-n))$, which is really not the same sort of sum at all and has very different behavior).
answered Jan 24 at 23:55
Eric WofseyEric Wofsey
189k14216347
$endgroup$
$begingroup$
Ah, now I see what ChrisCulter meant about the geometric series. You might want to double-check your answer, by the way.
$endgroup$
– R. Burton
Jan 25 at 3:48
$begingroup$
Oops, I missed some square roots when optimizing. Your bounding curves were correct. :)
$endgroup$
– Eric Wofsey
Jan 25 at 5:03
$begingroup$
Also, in this case it works out to be the same, but shouldn't the geometric series give $frac{1-e^{(omega+1)ix}}{1-e^{ix}}$ (with everything after that having the same sign change)?
$endgroup$
– R. Burton
Jan 25 at 15:59
1
$begingroup$
That's the same: you just multiply the numerator and denominator by $-1$.
$endgroup$
– Eric Wofsey
Jan 25 at 16:05
add a comment |
$begingroup$
This behavior is very easy to explain when you view the sum as a geometric series. We have a closed formula $$sum_{n=0}^omega e^{nix}=frac{e^{(omega+1)ix}-1}{e^{ix}-1}$$ and the bounding curves you are finding are just what you get by maximizing and minimizing the real and imaginary parts of $$frac{w-1}{e^{ix}-1}$$ where $w$ is allowed to range over the unit circle. When $omega$ is large, $e^{(omega+1)ix}$ is cycling through the entire unit circle very quickly as $x$ changes (while the denominator $e^{ix}-1$ is changing slowly), and so the real and imaginary parts of the sum quickly oscillate between the maximum and minimum values where $w$ ranges over the unit circle.
Explicitly, to find the real and imaginary parts we write $w=a+bi$ and find $$frac{w-1}{e^{ix}-1}=frac{a-1+bi}{cos x-1+isin x}=frac{a(cos x-1)+bsin x+1-cos x}{(cos x-1)^2+sin^2 x}+ifrac{-asin x+b(cos x-1)+sin x}{(cos x-1)^2+sin^2 x}.$$ By Cauchy-Schwarz, given the constraint $a^2+b^2=1$, we maximize or minimize $a(cos x-1)+bsin x$ by making $(a,b)$ a scalar multiple of $(cos x-1,sin x)$, so that $a=pmfrac{cos x-1}{sqrt{(cos x-1)^2+sin^2 x}}$ and $b=pmfrac{sin x}{sqrt{(cos x-1)^2+sin^2 x}}$. This tells us that when $2-2cos x$ is positive, the real part is bounded by $$frac{1}{2}pmfrac{(cos x-1)^2+sin^2 x}{((cos x-1)^2+sin^2 x)^{3/2}}=frac{1}{2}pmfrac{1}{2}csc(x/2).$$
Similarly, the imaginary part is bounded by $$frac{1}{2}frac{sin x}{1-cos x}pmfrac{1}{2}csc(x/2)$$ which is $frac{1}{2}cot(x/4)$ or $-frac{1}{2}tan(x/4)$ depending on the sign chosen.
The tangent-like appearance of the imaginary part graph (with half the period of the bounding curves) is explained by the fact that the average value between the oscillations is $frac{1}{2}frac{sin x}{1-cos x}=frac{1}{2}cot(x/2)$. In other words, the imaginary part is the graph of $frac{1}{2}cot(x/2)$ superimposed with oscillations whose peaks behave like $frac{1}{2}csc(x/2)$.
The bounding curves for other similar summations can be found in the same way, by writing everything in terms of complex exponentials and then summing as a geometric series. In particular, the same method should work for three of the four examples you presented at the end (the exception being $sum (-1)^n cos^2(n/(x-n))$, which is really not the same sort of sum at all and has very different behavior).
answered Jan 24 at 23:55
Eric WofseyEric Wofsey
189k14216347
$endgroup$
$begingroup$
Ah, now I see what ChrisCulter meant about the geometric series. You might want to double-check your answer, by the way.
$endgroup$
– R. Burton
Jan 25 at 3:48
$begingroup$
Oops, I missed some square roots when optimizing. Your bounding curves were correct. :)
$endgroup$
– Eric Wofsey
Jan 25 at 5:03
$begingroup$
Also, in this case it works out to be the same, but shouldn't the geometric series give $frac{1-e^{(omega+1)ix}}{1-e^{ix}}$ (with everything after that having the same sign change)?
$endgroup$
– R. Burton
Jan 25 at 15:59
1
$begingroup$
That's the same: you just multiply the numerator and denominator by $-1$.
$endgroup$
– Eric Wofsey
Jan 25 at 16:05
add a comment |
$begingroup$
This behavior is very easy to explain when you view the sum as a geometric series. We have a closed formula $$sum_{n=0}^omega e^{nix}=frac{e^{(omega+1)ix}-1}{e^{ix}-1}$$ and the bounding curves you are finding are just what you get by maximizing and minimizing the real and imaginary parts of $$frac{w-1}{e^{ix}-1}$$ where $w$ is allowed to range over the unit circle. When $omega$ is large, $e^{(omega+1)ix}$ is cycling through the entire unit circle very quickly as $x$ changes (while the denominator $e^{ix}-1$ is changing slowly), and so the real and imaginary parts of the sum quickly oscillate between the maximum and minimum values where $w$ ranges over the unit circle.
Explicitly, to find the real and imaginary parts we write $w=a+bi$ and find $$frac{w-1}{e^{ix}-1}=frac{a-1+bi}{cos x-1+isin x}=frac{a(cos x-1)+bsin x+1-cos x}{(cos x-1)^2+sin^2 x}+ifrac{-asin x+b(cos x-1)+sin x}{(cos x-1)^2+sin^2 x}.$$ By Cauchy-Schwarz, given the constraint $a^2+b^2=1$, we maximize or minimize $a(cos x-1)+bsin x$ by making $(a,b)$ a scalar multiple of $(cos x-1,sin x)$, so that $a=pmfrac{cos x-1}{sqrt{(cos x-1)^2+sin^2 x}}$ and $b=pmfrac{sin x}{sqrt{(cos x-1)^2+sin^2 x}}$. This tells us that when $2-2cos x$ is positive, the real part is bounded by $$frac{1}{2}pmfrac{(cos x-1)^2+sin^2 x}{((cos x-1)^2+sin^2 x)^{3/2}}=frac{1}{2}pmfrac{1}{2}csc(x/2).$$
Similarly, the imaginary part is bounded by $$frac{1}{2}frac{sin x}{1-cos x}pmfrac{1}{2}csc(x/2)$$ which is $frac{1}{2}cot(x/4)$ or $-frac{1}{2}tan(x/4)$ depending on the sign chosen.
The tangent-like appearance of the imaginary part graph (with half the period of the bounding curves) is explained by the fact that the average value between the oscillations is $frac{1}{2}frac{sin x}{1-cos x}=frac{1}{2}cot(x/2)$. In other words, the imaginary part is the graph of $frac{1}{2}cot(x/2)$ superimposed with oscillations whose peaks behave like $frac{1}{2}csc(x/2)$.
The bounding curves for other similar summations can be found in the same way, by writing everything in terms of complex exponentials and then summing as a geometric series. In particular, the same method should work for three of the four examples you presented at the end (the exception being $sum (-1)^n cos^2(n/(x-n))$, which is really not the same sort of sum at all and has very different behavior).
answered Jan 24 at 23:55
Eric WofseyEric Wofsey
189k14216347
$endgroup$
This behavior is very easy to explain when you view the sum as a geometric series. We have a closed formula $$sum_{n=0}^omega e^{nix}=frac{e^{(omega+1)ix}-1}{e^{ix}-1}$$ and the bounding curves you are finding are just what you get by maximizing and minimizing the real and imaginary parts of $$frac{w-1}{e^{ix}-1}$$ where $w$ is allowed to range over the unit circle. When $omega$ is large, $e^{(omega+1)ix}$ is cycling through the entire unit circle very quickly as $x$ changes (while the denominator $e^{ix}-1$ is changing slowly), and so the real and imaginary parts of the sum quickly oscillate between the maximum and minimum values where $w$ ranges over the unit circle.
Explicitly, to find the real and imaginary parts we write $w=a+bi$ and find $$frac{w-1}{e^{ix}-1}=frac{a-1+bi}{cos x-1+isin x}=frac{a(cos x-1)+bsin x+1-cos x}{(cos x-1)^2+sin^2 x}+ifrac{-asin x+b(cos x-1)+sin x}{(cos x-1)^2+sin^2 x}.$$ By Cauchy-Schwarz, given the constraint $a^2+b^2=1$, we maximize or minimize $a(cos x-1)+bsin x$ by making $(a,b)$ a scalar multiple of $(cos x-1,sin x)$, so that $a=pmfrac{cos x-1}{sqrt{(cos x-1)^2+sin^2 x}}$ and $b=pmfrac{sin x}{sqrt{(cos x-1)^2+sin^2 x}}$. This tells us that when $2-2cos x$ is positive, the real part is bounded by $$frac{1}{2}pmfrac{(cos x-1)^2+sin^2 x}{((cos x-1)^2+sin^2 x)^{3/2}}=frac{1}{2}pmfrac{1}{2}csc(x/2).$$
Similarly, the imaginary part is bounded by $$frac{1}{2}frac{sin x}{1-cos x}pmfrac{1}{2}csc(x/2)$$ which is $frac{1}{2}cot(x/4)$ or $-frac{1}{2}tan(x/4)$ depending on the sign chosen.
The tangent-like appearance of the imaginary part graph (with half the period of the bounding curves) is explained by the fact that the average value between the oscillations is $frac{1}{2}frac{sin x}{1-cos x}=frac{1}{2}cot(x/2)$. In other words, the imaginary part is the graph of $frac{1}{2}cot(x/2)$ superimposed with oscillations whose peaks behave like $frac{1}{2}csc(x/2)$.
The bounding curves for other similar summations can be found in the same way, by writing everything in terms of complex exponentials and then summing as a geometric series. In particular, the same method should work for three of the four examples you presented at the end (the exception being $sum (-1)^n cos^2(n/(x-n))$, which is really not the same sort of sum at all and has very different behavior).
answered Jan 24 at 23:55
Eric WofseyEric Wofsey
189k14216347
edited Jan 25 at 5:03
answered Jan 24 at 23:55
Eric WofseyEric Wofsey
189k14216347
answered Jan 24 at 23:55
Eric WofseyEric Wofsey
189k14216347
answered Jan 24 at 23:55
Eric WofseyEric Wofsey
189k14216347
189k14216347
$begingroup$
Ah, now I see what ChrisCulter meant about the geometric series. You might want to double-check your answer, by the way.
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– R. Burton
Jan 25 at 3:48
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Oops, I missed some square roots when optimizing. Your bounding curves were correct. :)
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– Eric Wofsey
Jan 25 at 5:03
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Also, in this case it works out to be the same, but shouldn't the geometric series give $frac{1-e^{(omega+1)ix}}{1-e^{ix}}$ (with everything after that having the same sign change)?
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– R. Burton
Jan 25 at 15:59
1
$begingroup$
That's the same: you just multiply the numerator and denominator by $-1$.
$endgroup$
– Eric Wofsey
Jan 25 at 16:05
add a comment |
$begingroup$
Ah, now I see what ChrisCulter meant about the geometric series. You might want to double-check your answer, by the way.
$endgroup$
– R. Burton
Jan 25 at 3:48
$begingroup$
Oops, I missed some square roots when optimizing. Your bounding curves were correct. :)
$endgroup$
– Eric Wofsey
Jan 25 at 5:03
$begingroup$
Also, in this case it works out to be the same, but shouldn't the geometric series give $frac{1-e^{(omega+1)ix}}{1-e^{ix}}$ (with everything after that having the same sign change)?
$endgroup$
– R. Burton
Jan 25 at 15:59
1
$begingroup$
That's the same: you just multiply the numerator and denominator by $-1$.
$endgroup$
– Eric Wofsey
Jan 25 at 16:05
$begingroup$
Ah, now I see what ChrisCulter meant about the geometric series. You might want to double-check your answer, by the way.
$endgroup$
– R. Burton
Jan 25 at 3:48
$begingroup$
Ah, now I see what ChrisCulter meant about the geometric series. You might want to double-check your answer, by the way.
$endgroup$
– R. Burton
Jan 25 at 3:48
$begingroup$
Oops, I missed some square roots when optimizing. Your bounding curves were correct. :)
$endgroup$
– Eric Wofsey
Jan 25 at 5:03
$begingroup$
Oops, I missed some square roots when optimizing. Your bounding curves were correct. :)
$endgroup$
– Eric Wofsey
Jan 25 at 5:03
$begingroup$
Also, in this case it works out to be the same, but shouldn't the geometric series give $frac{1-e^{(omega+1)ix}}{1-e^{ix}}$ (with everything after that having the same sign change)?
$endgroup$
– R. Burton
Jan 25 at 15:59
$begingroup$
Also, in this case it works out to be the same, but shouldn't the geometric series give $frac{1-e^{(omega+1)ix}}{1-e^{ix}}$ (with everything after that having the same sign change)?
$endgroup$
– R. Burton
Jan 25 at 15:59
1
1
$begingroup$
That's the same: you just multiply the numerator and denominator by $-1$.
$endgroup$
– Eric Wofsey
Jan 25 at 16:05
$begingroup$
That's the same: you just multiply the numerator and denominator by $-1$.
$endgroup$
– Eric Wofsey
Jan 25 at 16:05
add a comment |
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$begingroup$
The more elegant way to write $left{y_1inmathbb{R}mid r_1(x)leq y_1leq r_2(x)right}$ is $[r_1(x),r_2(x)]$.
$endgroup$
– Chris Culter
Jan 24 at 20:33
1
$begingroup$
Also, instead of bounding the limit set of the series by a rectangle in the complex plane, it would be even more elegant to say that it's a subset of a certain circle. You can find the center and radius of this circle using the geometric series formula.
$endgroup$
– Chris Culter
Jan 24 at 20:37
$begingroup$
@ChrisCulter How do I use the geometric series formula?
$endgroup$
– R. Burton
Jan 24 at 23:21