Solution of equation in $k$, $sin k$ and $cos k$Solution set of $cos(cos(cos(cos(x)))) =...
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Solution of equation in $k$, $sin k$ and $cos k$
Solution set of $cos(cos(cos(cos(x)))) = sin(sin(sin(sin(x))))$Solving $ sin x + sqrt 3 cos x = 1 $ - is my solution correct?Solving $sqrt{3cos^2 x - sin 2x} = - sin x$Trigonometry equation solutionSum of $sin$ and $cos$solution of the equation $2sintheta/2=cos(theta(n-1/2))?$Is the solution of $cos(sin(x))-sin(cos(x))=x$ rational, algebraic irrational or transcendental?Number of solutions of $3sin^2 x+cos^2 x+sqrt{3} sin x+cos x+1=sqrt{3}sin xcos x$Determining $A+B$, given $sin A + sin B = sqrt{frac{3}{2}}$ and $cos A - cos B = sqrt{frac12}$. Different approaches give different answers.General solution to $sin alpha + sin beta$ and $cos alpha + cos beta$?
$begingroup$
Some physical problem reduces to some equation that looks pretty transcendental:
$$
k sinmathopen{}left(sqrt{k^2+2delta},pi/2right)mathclose{}cos(kpi/2) + sqrt{k^2+2delta}cosmathopen{}left(sqrt{k^2+2delta},pi/2right)mathclose{}sin(kpi/2) =0.
$$
with $delta < k in Bbb R^+$.
For $delta=0$ the solutions are $kinBbb N$. What could one do for $deltane 0$?
real-analysis trigonometry transcendental-numbers
asked yesterday
Rudi_BirnbaumRudi_Birnbaum
625819
$endgroup$
add a comment |
$begingroup$
Some physical problem reduces to some equation that looks pretty transcendental:
$$
k sinmathopen{}left(sqrt{k^2+2delta},pi/2right)mathclose{}cos(kpi/2) + sqrt{k^2+2delta}cosmathopen{}left(sqrt{k^2+2delta},pi/2right)mathclose{}sin(kpi/2) =0.
$$
with $delta < k in Bbb R^+$.
For $delta=0$ the solutions are $kinBbb N$. What could one do for $deltane 0$?
real-analysis trigonometry transcendental-numbers
asked yesterday
Rudi_BirnbaumRudi_Birnbaum
625819
$endgroup$
2
$begingroup$
If $delta << k$, you can try Taylor-expanding the sines and cosines (as well as the square roots). Grouping the terms order-by-order of $delta$ will give your a series expansion for the solution $k$.
$endgroup$
– D.B.
yesterday
add a comment |
$begingroup$
Some physical problem reduces to some equation that looks pretty transcendental:
$$
k sinmathopen{}left(sqrt{k^2+2delta},pi/2right)mathclose{}cos(kpi/2) + sqrt{k^2+2delta}cosmathopen{}left(sqrt{k^2+2delta},pi/2right)mathclose{}sin(kpi/2) =0.
$$
with $delta < k in Bbb R^+$.
For $delta=0$ the solutions are $kinBbb N$. What could one do for $deltane 0$?
real-analysis trigonometry transcendental-numbers
asked yesterday
Rudi_BirnbaumRudi_Birnbaum
625819
$endgroup$
Some physical problem reduces to some equation that looks pretty transcendental:
$$
k sinmathopen{}left(sqrt{k^2+2delta},pi/2right)mathclose{}cos(kpi/2) + sqrt{k^2+2delta}cosmathopen{}left(sqrt{k^2+2delta},pi/2right)mathclose{}sin(kpi/2) =0.
$$
with $delta < k in Bbb R^+$.
For $delta=0$ the solutions are $kinBbb N$. What could one do for $deltane 0$?
real-analysis trigonometry transcendental-numbers
real-analysis trigonometry transcendental-numbers
asked yesterday
Rudi_BirnbaumRudi_Birnbaum
625819
asked yesterday
Rudi_BirnbaumRudi_Birnbaum
625819
edited 22 hours ago
Rudi_Birnbaum
asked yesterday
Rudi_BirnbaumRudi_Birnbaum
625819
asked yesterday
Rudi_BirnbaumRudi_Birnbaum
625819
asked yesterday
Rudi_BirnbaumRudi_Birnbaum
625819
625819
2
$begingroup$
If $delta << k$, you can try Taylor-expanding the sines and cosines (as well as the square roots). Grouping the terms order-by-order of $delta$ will give your a series expansion for the solution $k$.
$endgroup$
– D.B.
yesterday
add a comment |
2
$begingroup$
If $delta << k$, you can try Taylor-expanding the sines and cosines (as well as the square roots). Grouping the terms order-by-order of $delta$ will give your a series expansion for the solution $k$.
$endgroup$
– D.B.
yesterday
2
2
$begingroup$
If $delta << k$, you can try Taylor-expanding the sines and cosines (as well as the square roots). Grouping the terms order-by-order of $delta$ will give your a series expansion for the solution $k$.
$endgroup$
– D.B.
yesterday
$begingroup$
If $delta << k$, you can try Taylor-expanding the sines and cosines (as well as the square roots). Grouping the terms order-by-order of $delta$ will give your a series expansion for the solution $k$.
$endgroup$
– D.B.
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $delta leq 0$, the equation LHS gives complex values when $|k| < sqrt{-2delta}$ and the only solution is $|k| = sqrt{-2delta}$.
The matter is much more interesting for $delta > 0$. There is a set of discrete values of $delta$ for which the equation can be solved exactly for $k$. Namely, if $delta = 2n+2$ with $n in Bbb N$ then $k=n$ solves the equation, because then $sqrt{k^2+2delta} = k+2$ so the LHS becomes
$$
k sin left(k frac{pi}2 + pi right) cos left(k frac{pi}2 right)
+ (k+2) cos left(k frac{pi}2 + pi right) sin left(k frac{pi}2 right) \=
(-k-(k+2)) sin left(k frac{pi}2 right) cos left(k frac{pi}2 right) =
-(2k+2) sin kpi = 0
$$
For $delta = 2n+2+epsilon$, with $epsilon$ small,we can do perturbation expansion, and the results are intriguing. When $n$ is even, it seems the function is zero at
$$k = frac{n-3}{2n+2} epsilon + O(epsilon^2)$$
But when $n$ is odd, it seems the solution value of $k$ moves away from its integer value much more rapidly as $delta$ moves from its even integer value. The behavior is
$$
k = frac{n+2}{2} sqrt{epsilon} + frac{n}{4n+4} epsilon + O(epsilon^{3/2})$$
and this covers the solutions on both sides of the except that there are no solutions for $0 < delta < 4$.
This lack of solutions must have some physical meaning; I suspect that for these values of $delta$ the energy is too small and the solution to the left of the boundary is a decaying exponential rather than a sine wave.
answered yesterday
Mark FischlerMark Fischler
33.2k12452
$endgroup$
1
$begingroup$
"and this covers the solutions on both sides of the except that there are no solutions for 0<𝛿<4. " Numerically this is not confirmed. $delta=0.1$ gives about $0.947$ for $k$, which makes sense, as well.
$endgroup$
– Rudi_Birnbaum
22 hours ago
1
$begingroup$
Physically 𝛿<0 is completely equivalent to 𝛿>0, though the problem was formulated bearing in mind that both 𝛿 and 𝑘 are in ℝ+, that was what was intended in my clumsy formulation 𝛿<𝑘∈ℝ+. Integer k are the solutions for 𝛿=0, since 𝛿 should be smaller 𝑘 the most interesting values are 𝛿<<1. The discrete spectrum otherwise just reflects the quantization. To put in in another way, its exactly the $varepsilon$ dependence which is of interest.
$endgroup$
– Rudi_Birnbaum
22 hours ago
1
$begingroup$
Due to the symmetry of the problem (symmetric with respect to reflection at $x=pi/4$ in the arguments of the trigonometric functions) only one in $n$ even or odd should produce valid solutions. The other ones should be like "artefacts" from the calculation.
$endgroup$
– Rudi_Birnbaum
22 hours ago
add a comment |
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$begingroup$
For $delta leq 0$, the equation LHS gives complex values when $|k| < sqrt{-2delta}$ and the only solution is $|k| = sqrt{-2delta}$.
The matter is much more interesting for $delta > 0$. There is a set of discrete values of $delta$ for which the equation can be solved exactly for $k$. Namely, if $delta = 2n+2$ with $n in Bbb N$ then $k=n$ solves the equation, because then $sqrt{k^2+2delta} = k+2$ so the LHS becomes
$$
k sin left(k frac{pi}2 + pi right) cos left(k frac{pi}2 right)
+ (k+2) cos left(k frac{pi}2 + pi right) sin left(k frac{pi}2 right) \=
(-k-(k+2)) sin left(k frac{pi}2 right) cos left(k frac{pi}2 right) =
-(2k+2) sin kpi = 0
$$
For $delta = 2n+2+epsilon$, with $epsilon$ small,we can do perturbation expansion, and the results are intriguing. When $n$ is even, it seems the function is zero at
$$k = frac{n-3}{2n+2} epsilon + O(epsilon^2)$$
But when $n$ is odd, it seems the solution value of $k$ moves away from its integer value much more rapidly as $delta$ moves from its even integer value. The behavior is
$$
k = frac{n+2}{2} sqrt{epsilon} + frac{n}{4n+4} epsilon + O(epsilon^{3/2})$$
and this covers the solutions on both sides of the except that there are no solutions for $0 < delta < 4$.
This lack of solutions must have some physical meaning; I suspect that for these values of $delta$ the energy is too small and the solution to the left of the boundary is a decaying exponential rather than a sine wave.
answered yesterday
Mark FischlerMark Fischler
33.2k12452
$endgroup$
1
$begingroup$
"and this covers the solutions on both sides of the except that there are no solutions for 0<𝛿<4. " Numerically this is not confirmed. $delta=0.1$ gives about $0.947$ for $k$, which makes sense, as well.
$endgroup$
– Rudi_Birnbaum
22 hours ago
1
$begingroup$
Physically 𝛿<0 is completely equivalent to 𝛿>0, though the problem was formulated bearing in mind that both 𝛿 and 𝑘 are in ℝ+, that was what was intended in my clumsy formulation 𝛿<𝑘∈ℝ+. Integer k are the solutions for 𝛿=0, since 𝛿 should be smaller 𝑘 the most interesting values are 𝛿<<1. The discrete spectrum otherwise just reflects the quantization. To put in in another way, its exactly the $varepsilon$ dependence which is of interest.
$endgroup$
– Rudi_Birnbaum
22 hours ago
1
$begingroup$
Due to the symmetry of the problem (symmetric with respect to reflection at $x=pi/4$ in the arguments of the trigonometric functions) only one in $n$ even or odd should produce valid solutions. The other ones should be like "artefacts" from the calculation.
$endgroup$
– Rudi_Birnbaum
22 hours ago
add a comment |
$begingroup$
For $delta leq 0$, the equation LHS gives complex values when $|k| < sqrt{-2delta}$ and the only solution is $|k| = sqrt{-2delta}$.
The matter is much more interesting for $delta > 0$. There is a set of discrete values of $delta$ for which the equation can be solved exactly for $k$. Namely, if $delta = 2n+2$ with $n in Bbb N$ then $k=n$ solves the equation, because then $sqrt{k^2+2delta} = k+2$ so the LHS becomes
$$
k sin left(k frac{pi}2 + pi right) cos left(k frac{pi}2 right)
+ (k+2) cos left(k frac{pi}2 + pi right) sin left(k frac{pi}2 right) \=
(-k-(k+2)) sin left(k frac{pi}2 right) cos left(k frac{pi}2 right) =
-(2k+2) sin kpi = 0
$$
For $delta = 2n+2+epsilon$, with $epsilon$ small,we can do perturbation expansion, and the results are intriguing. When $n$ is even, it seems the function is zero at
$$k = frac{n-3}{2n+2} epsilon + O(epsilon^2)$$
But when $n$ is odd, it seems the solution value of $k$ moves away from its integer value much more rapidly as $delta$ moves from its even integer value. The behavior is
$$
k = frac{n+2}{2} sqrt{epsilon} + frac{n}{4n+4} epsilon + O(epsilon^{3/2})$$
and this covers the solutions on both sides of the except that there are no solutions for $0 < delta < 4$.
This lack of solutions must have some physical meaning; I suspect that for these values of $delta$ the energy is too small and the solution to the left of the boundary is a decaying exponential rather than a sine wave.
answered yesterday
Mark FischlerMark Fischler
33.2k12452
$endgroup$
1
$begingroup$
"and this covers the solutions on both sides of the except that there are no solutions for 0<𝛿<4. " Numerically this is not confirmed. $delta=0.1$ gives about $0.947$ for $k$, which makes sense, as well.
$endgroup$
– Rudi_Birnbaum
22 hours ago
1
$begingroup$
Physically 𝛿<0 is completely equivalent to 𝛿>0, though the problem was formulated bearing in mind that both 𝛿 and 𝑘 are in ℝ+, that was what was intended in my clumsy formulation 𝛿<𝑘∈ℝ+. Integer k are the solutions for 𝛿=0, since 𝛿 should be smaller 𝑘 the most interesting values are 𝛿<<1. The discrete spectrum otherwise just reflects the quantization. To put in in another way, its exactly the $varepsilon$ dependence which is of interest.
$endgroup$
– Rudi_Birnbaum
22 hours ago
1
$begingroup$
Due to the symmetry of the problem (symmetric with respect to reflection at $x=pi/4$ in the arguments of the trigonometric functions) only one in $n$ even or odd should produce valid solutions. The other ones should be like "artefacts" from the calculation.
$endgroup$
– Rudi_Birnbaum
22 hours ago
add a comment |
$begingroup$
For $delta leq 0$, the equation LHS gives complex values when $|k| < sqrt{-2delta}$ and the only solution is $|k| = sqrt{-2delta}$.
The matter is much more interesting for $delta > 0$. There is a set of discrete values of $delta$ for which the equation can be solved exactly for $k$. Namely, if $delta = 2n+2$ with $n in Bbb N$ then $k=n$ solves the equation, because then $sqrt{k^2+2delta} = k+2$ so the LHS becomes
$$
k sin left(k frac{pi}2 + pi right) cos left(k frac{pi}2 right)
+ (k+2) cos left(k frac{pi}2 + pi right) sin left(k frac{pi}2 right) \=
(-k-(k+2)) sin left(k frac{pi}2 right) cos left(k frac{pi}2 right) =
-(2k+2) sin kpi = 0
$$
For $delta = 2n+2+epsilon$, with $epsilon$ small,we can do perturbation expansion, and the results are intriguing. When $n$ is even, it seems the function is zero at
$$k = frac{n-3}{2n+2} epsilon + O(epsilon^2)$$
But when $n$ is odd, it seems the solution value of $k$ moves away from its integer value much more rapidly as $delta$ moves from its even integer value. The behavior is
$$
k = frac{n+2}{2} sqrt{epsilon} + frac{n}{4n+4} epsilon + O(epsilon^{3/2})$$
and this covers the solutions on both sides of the except that there are no solutions for $0 < delta < 4$.
This lack of solutions must have some physical meaning; I suspect that for these values of $delta$ the energy is too small and the solution to the left of the boundary is a decaying exponential rather than a sine wave.
answered yesterday
Mark FischlerMark Fischler
33.2k12452
$endgroup$
For $delta leq 0$, the equation LHS gives complex values when $|k| < sqrt{-2delta}$ and the only solution is $|k| = sqrt{-2delta}$.
The matter is much more interesting for $delta > 0$. There is a set of discrete values of $delta$ for which the equation can be solved exactly for $k$. Namely, if $delta = 2n+2$ with $n in Bbb N$ then $k=n$ solves the equation, because then $sqrt{k^2+2delta} = k+2$ so the LHS becomes
$$
k sin left(k frac{pi}2 + pi right) cos left(k frac{pi}2 right)
+ (k+2) cos left(k frac{pi}2 + pi right) sin left(k frac{pi}2 right) \=
(-k-(k+2)) sin left(k frac{pi}2 right) cos left(k frac{pi}2 right) =
-(2k+2) sin kpi = 0
$$
For $delta = 2n+2+epsilon$, with $epsilon$ small,we can do perturbation expansion, and the results are intriguing. When $n$ is even, it seems the function is zero at
$$k = frac{n-3}{2n+2} epsilon + O(epsilon^2)$$
But when $n$ is odd, it seems the solution value of $k$ moves away from its integer value much more rapidly as $delta$ moves from its even integer value. The behavior is
$$
k = frac{n+2}{2} sqrt{epsilon} + frac{n}{4n+4} epsilon + O(epsilon^{3/2})$$
and this covers the solutions on both sides of the except that there are no solutions for $0 < delta < 4$.
This lack of solutions must have some physical meaning; I suspect that for these values of $delta$ the energy is too small and the solution to the left of the boundary is a decaying exponential rather than a sine wave.
answered yesterday
Mark FischlerMark Fischler
33.2k12452
answered yesterday
Mark FischlerMark Fischler
33.2k12452
answered yesterday
Mark FischlerMark Fischler
33.2k12452
answered yesterday
Mark FischlerMark Fischler
33.2k12452
33.2k12452
1
$begingroup$
"and this covers the solutions on both sides of the except that there are no solutions for 0<𝛿<4. " Numerically this is not confirmed. $delta=0.1$ gives about $0.947$ for $k$, which makes sense, as well.
$endgroup$
– Rudi_Birnbaum
22 hours ago
1
$begingroup$
Physically 𝛿<0 is completely equivalent to 𝛿>0, though the problem was formulated bearing in mind that both 𝛿 and 𝑘 are in ℝ+, that was what was intended in my clumsy formulation 𝛿<𝑘∈ℝ+. Integer k are the solutions for 𝛿=0, since 𝛿 should be smaller 𝑘 the most interesting values are 𝛿<<1. The discrete spectrum otherwise just reflects the quantization. To put in in another way, its exactly the $varepsilon$ dependence which is of interest.
$endgroup$
– Rudi_Birnbaum
22 hours ago
1
$begingroup$
Due to the symmetry of the problem (symmetric with respect to reflection at $x=pi/4$ in the arguments of the trigonometric functions) only one in $n$ even or odd should produce valid solutions. The other ones should be like "artefacts" from the calculation.
$endgroup$
– Rudi_Birnbaum
22 hours ago
add a comment |
1
$begingroup$
"and this covers the solutions on both sides of the except that there are no solutions for 0<𝛿<4. " Numerically this is not confirmed. $delta=0.1$ gives about $0.947$ for $k$, which makes sense, as well.
$endgroup$
– Rudi_Birnbaum
22 hours ago
1
$begingroup$
Physically 𝛿<0 is completely equivalent to 𝛿>0, though the problem was formulated bearing in mind that both 𝛿 and 𝑘 are in ℝ+, that was what was intended in my clumsy formulation 𝛿<𝑘∈ℝ+. Integer k are the solutions for 𝛿=0, since 𝛿 should be smaller 𝑘 the most interesting values are 𝛿<<1. The discrete spectrum otherwise just reflects the quantization. To put in in another way, its exactly the $varepsilon$ dependence which is of interest.
$endgroup$
– Rudi_Birnbaum
22 hours ago
1
$begingroup$
Due to the symmetry of the problem (symmetric with respect to reflection at $x=pi/4$ in the arguments of the trigonometric functions) only one in $n$ even or odd should produce valid solutions. The other ones should be like "artefacts" from the calculation.
$endgroup$
– Rudi_Birnbaum
22 hours ago
1
1
$begingroup$
"and this covers the solutions on both sides of the except that there are no solutions for 0<𝛿<4. " Numerically this is not confirmed. $delta=0.1$ gives about $0.947$ for $k$, which makes sense, as well.
$endgroup$
– Rudi_Birnbaum
22 hours ago
$begingroup$
"and this covers the solutions on both sides of the except that there are no solutions for 0<𝛿<4. " Numerically this is not confirmed. $delta=0.1$ gives about $0.947$ for $k$, which makes sense, as well.
$endgroup$
– Rudi_Birnbaum
22 hours ago
1
1
$begingroup$
Physically 𝛿<0 is completely equivalent to 𝛿>0, though the problem was formulated bearing in mind that both 𝛿 and 𝑘 are in ℝ+, that was what was intended in my clumsy formulation 𝛿<𝑘∈ℝ+. Integer k are the solutions for 𝛿=0, since 𝛿 should be smaller 𝑘 the most interesting values are 𝛿<<1. The discrete spectrum otherwise just reflects the quantization. To put in in another way, its exactly the $varepsilon$ dependence which is of interest.
$endgroup$
– Rudi_Birnbaum
22 hours ago
$begingroup$
Physically 𝛿<0 is completely equivalent to 𝛿>0, though the problem was formulated bearing in mind that both 𝛿 and 𝑘 are in ℝ+, that was what was intended in my clumsy formulation 𝛿<𝑘∈ℝ+. Integer k are the solutions for 𝛿=0, since 𝛿 should be smaller 𝑘 the most interesting values are 𝛿<<1. The discrete spectrum otherwise just reflects the quantization. To put in in another way, its exactly the $varepsilon$ dependence which is of interest.
$endgroup$
– Rudi_Birnbaum
22 hours ago
1
1
$begingroup$
Due to the symmetry of the problem (symmetric with respect to reflection at $x=pi/4$ in the arguments of the trigonometric functions) only one in $n$ even or odd should produce valid solutions. The other ones should be like "artefacts" from the calculation.
$endgroup$
– Rudi_Birnbaum
22 hours ago
$begingroup$
Due to the symmetry of the problem (symmetric with respect to reflection at $x=pi/4$ in the arguments of the trigonometric functions) only one in $n$ even or odd should produce valid solutions. The other ones should be like "artefacts" from the calculation.
$endgroup$
– Rudi_Birnbaum
22 hours ago
add a comment |
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2
$begingroup$
If $delta << k$, you can try Taylor-expanding the sines and cosines (as well as the square roots). Grouping the terms order-by-order of $delta$ will give your a series expansion for the solution $k$.
$endgroup$
– D.B.
yesterday