Mixing
Laplace transform of square root of a trigonometric function
$begingroup$
Need help with this question from my university paper.
My question : Find Laplace Transform of the following: $sqrt{1 + sin(4t)}$
I do know how to solve $sqrt{1 + sin(t)}$
By taking $1 = sin^2 left(frac{t}{2}right) + cos^2 left(frac{t}{2}right)$
Further taking $sin(t) = 2sin left(frac{t}{2}right) cos left(frac{t}{2}right)$
But I'm a bit confused about the above question.
Thanks in advance.
trigonometry laplace-transform
edited Jan 1 at 14:31
Moo
5,53131020
asked Jan 1 at 11:14
Tapan LathiyaTapan Lathiya
84
$endgroup$
add a comment |
$begingroup$
Need help with this question from my university paper.
My question : Find Laplace Transform of the following: $sqrt{1 + sin(4t)}$
I do know how to solve $sqrt{1 + sin(t)}$
By taking $1 = sin^2 left(frac{t}{2}right) + cos^2 left(frac{t}{2}right)$
Further taking $sin(t) = 2sin left(frac{t}{2}right) cos left(frac{t}{2}right)$
But I'm a bit confused about the above question.
Thanks in advance.
trigonometry laplace-transform
edited Jan 1 at 14:31
Moo
5,53131020
asked Jan 1 at 11:14
Tapan LathiyaTapan Lathiya
84
$endgroup$
$begingroup$
@MariuszIwaniuk You're right, thanks!
$endgroup$
– caverac
Jan 1 at 12:39
$begingroup$
If you know that the transform of $sqrt{1 + sin t}$ is $F(s)$, then $sqrt{1 + sin 4 t}$ gives $F(s/4)/4$.
$endgroup$
– Maxim
Jan 2 at 3:08
add a comment |
$begingroup$
Need help with this question from my university paper.
My question : Find Laplace Transform of the following: $sqrt{1 + sin(4t)}$
I do know how to solve $sqrt{1 + sin(t)}$
By taking $1 = sin^2 left(frac{t}{2}right) + cos^2 left(frac{t}{2}right)$
Further taking $sin(t) = 2sin left(frac{t}{2}right) cos left(frac{t}{2}right)$
But I'm a bit confused about the above question.
Thanks in advance.
trigonometry laplace-transform
edited Jan 1 at 14:31
Moo
5,53131020
asked Jan 1 at 11:14
Tapan LathiyaTapan Lathiya
84
$endgroup$
Need help with this question from my university paper.
My question : Find Laplace Transform of the following: $sqrt{1 + sin(4t)}$
I do know how to solve $sqrt{1 + sin(t)}$
By taking $1 = sin^2 left(frac{t}{2}right) + cos^2 left(frac{t}{2}right)$
Further taking $sin(t) = 2sin left(frac{t}{2}right) cos left(frac{t}{2}right)$
But I'm a bit confused about the above question.
Thanks in advance.
trigonometry laplace-transform
trigonometry laplace-transform
edited Jan 1 at 14:31
Moo
5,53131020
asked Jan 1 at 11:14
Tapan LathiyaTapan Lathiya
84
edited Jan 1 at 14:31
Moo
5,53131020
asked Jan 1 at 11:14
Tapan LathiyaTapan Lathiya
84
edited Jan 1 at 14:31
Moo
5,53131020
edited Jan 1 at 14:31
Moo
5,53131020
edited Jan 1 at 14:31
Moo
5,53131020
5,53131020
asked Jan 1 at 11:14
Tapan LathiyaTapan Lathiya
84
asked Jan 1 at 11:14
Tapan LathiyaTapan Lathiya
84
asked Jan 1 at 11:14
Tapan LathiyaTapan Lathiya
84
84
$begingroup$
@MariuszIwaniuk You're right, thanks!
$endgroup$
– caverac
Jan 1 at 12:39
$begingroup$
If you know that the transform of $sqrt{1 + sin t}$ is $F(s)$, then $sqrt{1 + sin 4 t}$ gives $F(s/4)/4$.
$endgroup$
– Maxim
Jan 2 at 3:08
add a comment |
$begingroup$
@MariuszIwaniuk You're right, thanks!
$endgroup$
– caverac
Jan 1 at 12:39
$begingroup$
If you know that the transform of $sqrt{1 + sin t}$ is $F(s)$, then $sqrt{1 + sin 4 t}$ gives $F(s/4)/4$.
$endgroup$
– Maxim
Jan 2 at 3:08
$begingroup$
@MariuszIwaniuk You're right, thanks!
$endgroup$
– caverac
Jan 1 at 12:39
$begingroup$
@MariuszIwaniuk You're right, thanks!
$endgroup$
– caverac
Jan 1 at 12:39
$begingroup$
If you know that the transform of $sqrt{1 + sin t}$ is $F(s)$, then $sqrt{1 + sin 4 t}$ gives $F(s/4)/4$.
$endgroup$
– Maxim
Jan 2 at 3:08
$begingroup$
If you know that the transform of $sqrt{1 + sin t}$ is $F(s)$, then $sqrt{1 + sin 4 t}$ gives $F(s/4)/4$.
$endgroup$
– Maxim
Jan 2 at 3:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The function $sqrt{1+sin at}$ is periodic with period $T = 2pi/a$ and the first zero at $3pi/2a$. Following your steps, we can write this as
$$ sqrt{1+sin at} = sqrt{cos^{2}frac{at}{2} + sin^{2}frac{at}{2} + 2sinfrac{at}{2}cosfrac{at}{2}} = left|cosfrac{at}{2}+sinfrac{at}{2}right|. $$
Using the Laplace transform of a periodic function, we split the integral into two pieces. This is necessary because of the absolute value sign. To get rid of it in the region $[3pi/2a, 2pi/a]$, we have to take the negative of the function.
$$begin{aligned} I &= int_{0}^{infty}sqrt{1+sin at},e^{-st},mathrm{d}t = frac{1}{1-e^{-2pi s/a}}int_{0}^{2pi/a}left|cosfrac{at}{2} + sinfrac{at}{2}right|e^{-st},mathrm{d}t \
&= frac{1}{1-e^{-2pi s/a}}left[int_{0}^{3pi/2a}left(cosfrac{at}{2} + sinfrac{at}{2}right)e^{-st},mathrm{d}t - int_{3pi/2a}^{2pi/a}left(cosfrac{at}{2} + sinfrac{at}{2}right)e^{-st},mathrm{d}t right]end{aligned}$$
These integrals can be done by considering the integral
$$ int_{0}^{3pi/2a}e^{iat/2}e^{-st},mathrm{d}t $$
and summing the real and imaginary contributions. One should then obtain
$$ mathcal{L}[sqrt{1+sin at}] = frac{1}{s^{2}+a^{2}/4}left(s + frac{a}{2} + frac{sqrt{2},ae^{-3pi s/2a}}{1-e^{-2pi s/a}}right). $$
For $a = 4$, we have
$$ boxed{mathcal{L}[sqrt{1+sin 4t}] = frac{1}{s^{2}+4}left(s + 2 + frac{4sqrt{2},e^{-3pi s/8}}{1-e^{-pi s/2}}right)}.$$
answered Jan 1 at 18:36
IninterrompueIninterrompue
63519
$endgroup$
$begingroup$
Beautiful solution, Thanks :)
$endgroup$
– Tapan Lathiya
Jan 1 at 20:58
$begingroup$
I was thinking along the lines of convolution, but wow, that solution is quite neat and compact. Great job!
$endgroup$
– bjcolby15
Jan 1 at 21:32
add a comment |
$begingroup$
With CAS help and for general $a$ and $aneq 0$ we have:
$mathcal{L}_tleft[sqrt{1+sin (a t)}right](s)=frac{e^{-frac{pi s}{2 a}} left(-a left(sqrt{2}-2 cosh left(frac{pi s}{2
a}right)right)+frac{left(a^2+4 s^2right) cosh left(frac{pi s}{2 a}right) , _3F_2left(-frac{1}{4},frac{1}{4},1;1-frac{i s}{2
a},1+frac{i s}{2 a};1right)}{s}right) left(1+tanh left(frac{pi s}{2 a}right)right)}{a^2+4 s^2}$
Mathematica code:
HoldForm[LaplaceTransform[Sqrt[1 + Sin[a t]], t, s] == (E^(-(([Pi] s)/(
2 a))) (-a (Sqrt[2] - 2 Cosh[([Pi] s)/(2 a)]) + ((a^2 +
4 s^2) Cosh[([Pi] s)/(
2 a)] HypergeometricPFQ[{-(1/4), 1/4, 1}, {1 - (I s)/(2 a),
1 + (I s)/(2 a)}, 1])/s) (1 + Tanh[([Pi] s)/(2 a)]))/(
a^2 + 4 s^2)] // TeXForm
For $a=4$:
(E^(-(([Pi] s)/8)) (-s (Sqrt[2] - 2 Cosh[([Pi] s)/8]) + (4 + s^2) Cosh[([Pi] s)/
8] HypergeometricPFQ[{-(1/4), 1/4, 1}, {1 - (I s)/8,
1 + (I s)/8}, 1]) (1 + Tanh[([Pi] s)/8]))/(s (4 + s^2))
$mathcal{L}_tleft[sqrt{1+sin (4 t)}right](s)=frac{e^{-frac{1}{8} (pi s)} left(-s left(sqrt{2}-2 cosh left(frac{pi
s}{8}right)right)+left(4+s^2right) cosh left(frac{pi s}{8}right) , _3F_2left(-frac{1}{4},frac{1}{4},1;1-frac{i s}{8},1+frac{i
s}{8};1right)right) left(1+tanh left(frac{pi s}{8}right)right)}{s left(4+s^2right)}$
answered Jan 1 at 12:48
Mariusz IwaniukMariusz Iwaniuk
1,9721718
$endgroup$
$begingroup$
Thanks for the answer. Is there any way to get rid of the square root? If so then it will be much easier to solve it further.
$endgroup$
– Tapan Lathiya
Jan 1 at 12:57
$begingroup$
@TapanLathiya Possible,yes,maybe so: $sqrt{1+sin (a t)}=sum _{j=0}^{infty } frac{left((-1)^{1+j} 4^{-j} (2 j)!right) sin ^j(a t)}{(-1+2 j) (j!)^2}$ or another way.
$endgroup$
– Mariusz Iwaniuk
Jan 1 at 13:02
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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$begingroup$
The function $sqrt{1+sin at}$ is periodic with period $T = 2pi/a$ and the first zero at $3pi/2a$. Following your steps, we can write this as
$$ sqrt{1+sin at} = sqrt{cos^{2}frac{at}{2} + sin^{2}frac{at}{2} + 2sinfrac{at}{2}cosfrac{at}{2}} = left|cosfrac{at}{2}+sinfrac{at}{2}right|. $$
Using the Laplace transform of a periodic function, we split the integral into two pieces. This is necessary because of the absolute value sign. To get rid of it in the region $[3pi/2a, 2pi/a]$, we have to take the negative of the function.
$$begin{aligned} I &= int_{0}^{infty}sqrt{1+sin at},e^{-st},mathrm{d}t = frac{1}{1-e^{-2pi s/a}}int_{0}^{2pi/a}left|cosfrac{at}{2} + sinfrac{at}{2}right|e^{-st},mathrm{d}t \
&= frac{1}{1-e^{-2pi s/a}}left[int_{0}^{3pi/2a}left(cosfrac{at}{2} + sinfrac{at}{2}right)e^{-st},mathrm{d}t - int_{3pi/2a}^{2pi/a}left(cosfrac{at}{2} + sinfrac{at}{2}right)e^{-st},mathrm{d}t right]end{aligned}$$
These integrals can be done by considering the integral
$$ int_{0}^{3pi/2a}e^{iat/2}e^{-st},mathrm{d}t $$
and summing the real and imaginary contributions. One should then obtain
$$ mathcal{L}[sqrt{1+sin at}] = frac{1}{s^{2}+a^{2}/4}left(s + frac{a}{2} + frac{sqrt{2},ae^{-3pi s/2a}}{1-e^{-2pi s/a}}right). $$
For $a = 4$, we have
$$ boxed{mathcal{L}[sqrt{1+sin 4t}] = frac{1}{s^{2}+4}left(s + 2 + frac{4sqrt{2},e^{-3pi s/8}}{1-e^{-pi s/2}}right)}.$$
answered Jan 1 at 18:36
IninterrompueIninterrompue
63519
$endgroup$
$begingroup$
Beautiful solution, Thanks :)
$endgroup$
– Tapan Lathiya
Jan 1 at 20:58
$begingroup$
I was thinking along the lines of convolution, but wow, that solution is quite neat and compact. Great job!
$endgroup$
– bjcolby15
Jan 1 at 21:32
add a comment |
$begingroup$
The function $sqrt{1+sin at}$ is periodic with period $T = 2pi/a$ and the first zero at $3pi/2a$. Following your steps, we can write this as
$$ sqrt{1+sin at} = sqrt{cos^{2}frac{at}{2} + sin^{2}frac{at}{2} + 2sinfrac{at}{2}cosfrac{at}{2}} = left|cosfrac{at}{2}+sinfrac{at}{2}right|. $$
Using the Laplace transform of a periodic function, we split the integral into two pieces. This is necessary because of the absolute value sign. To get rid of it in the region $[3pi/2a, 2pi/a]$, we have to take the negative of the function.
$$begin{aligned} I &= int_{0}^{infty}sqrt{1+sin at},e^{-st},mathrm{d}t = frac{1}{1-e^{-2pi s/a}}int_{0}^{2pi/a}left|cosfrac{at}{2} + sinfrac{at}{2}right|e^{-st},mathrm{d}t \
&= frac{1}{1-e^{-2pi s/a}}left[int_{0}^{3pi/2a}left(cosfrac{at}{2} + sinfrac{at}{2}right)e^{-st},mathrm{d}t - int_{3pi/2a}^{2pi/a}left(cosfrac{at}{2} + sinfrac{at}{2}right)e^{-st},mathrm{d}t right]end{aligned}$$
These integrals can be done by considering the integral
$$ int_{0}^{3pi/2a}e^{iat/2}e^{-st},mathrm{d}t $$
and summing the real and imaginary contributions. One should then obtain
$$ mathcal{L}[sqrt{1+sin at}] = frac{1}{s^{2}+a^{2}/4}left(s + frac{a}{2} + frac{sqrt{2},ae^{-3pi s/2a}}{1-e^{-2pi s/a}}right). $$
For $a = 4$, we have
$$ boxed{mathcal{L}[sqrt{1+sin 4t}] = frac{1}{s^{2}+4}left(s + 2 + frac{4sqrt{2},e^{-3pi s/8}}{1-e^{-pi s/2}}right)}.$$
answered Jan 1 at 18:36
IninterrompueIninterrompue
63519
$endgroup$
$begingroup$
Beautiful solution, Thanks :)
$endgroup$
– Tapan Lathiya
Jan 1 at 20:58
$begingroup$
I was thinking along the lines of convolution, but wow, that solution is quite neat and compact. Great job!
$endgroup$
– bjcolby15
Jan 1 at 21:32
add a comment |
$begingroup$
The function $sqrt{1+sin at}$ is periodic with period $T = 2pi/a$ and the first zero at $3pi/2a$. Following your steps, we can write this as
$$ sqrt{1+sin at} = sqrt{cos^{2}frac{at}{2} + sin^{2}frac{at}{2} + 2sinfrac{at}{2}cosfrac{at}{2}} = left|cosfrac{at}{2}+sinfrac{at}{2}right|. $$
Using the Laplace transform of a periodic function, we split the integral into two pieces. This is necessary because of the absolute value sign. To get rid of it in the region $[3pi/2a, 2pi/a]$, we have to take the negative of the function.
$$begin{aligned} I &= int_{0}^{infty}sqrt{1+sin at},e^{-st},mathrm{d}t = frac{1}{1-e^{-2pi s/a}}int_{0}^{2pi/a}left|cosfrac{at}{2} + sinfrac{at}{2}right|e^{-st},mathrm{d}t \
&= frac{1}{1-e^{-2pi s/a}}left[int_{0}^{3pi/2a}left(cosfrac{at}{2} + sinfrac{at}{2}right)e^{-st},mathrm{d}t - int_{3pi/2a}^{2pi/a}left(cosfrac{at}{2} + sinfrac{at}{2}right)e^{-st},mathrm{d}t right]end{aligned}$$
These integrals can be done by considering the integral
$$ int_{0}^{3pi/2a}e^{iat/2}e^{-st},mathrm{d}t $$
and summing the real and imaginary contributions. One should then obtain
$$ mathcal{L}[sqrt{1+sin at}] = frac{1}{s^{2}+a^{2}/4}left(s + frac{a}{2} + frac{sqrt{2},ae^{-3pi s/2a}}{1-e^{-2pi s/a}}right). $$
For $a = 4$, we have
$$ boxed{mathcal{L}[sqrt{1+sin 4t}] = frac{1}{s^{2}+4}left(s + 2 + frac{4sqrt{2},e^{-3pi s/8}}{1-e^{-pi s/2}}right)}.$$
answered Jan 1 at 18:36
IninterrompueIninterrompue
63519
$endgroup$
The function $sqrt{1+sin at}$ is periodic with period $T = 2pi/a$ and the first zero at $3pi/2a$. Following your steps, we can write this as
$$ sqrt{1+sin at} = sqrt{cos^{2}frac{at}{2} + sin^{2}frac{at}{2} + 2sinfrac{at}{2}cosfrac{at}{2}} = left|cosfrac{at}{2}+sinfrac{at}{2}right|. $$
Using the Laplace transform of a periodic function, we split the integral into two pieces. This is necessary because of the absolute value sign. To get rid of it in the region $[3pi/2a, 2pi/a]$, we have to take the negative of the function.
$$begin{aligned} I &= int_{0}^{infty}sqrt{1+sin at},e^{-st},mathrm{d}t = frac{1}{1-e^{-2pi s/a}}int_{0}^{2pi/a}left|cosfrac{at}{2} + sinfrac{at}{2}right|e^{-st},mathrm{d}t \
&= frac{1}{1-e^{-2pi s/a}}left[int_{0}^{3pi/2a}left(cosfrac{at}{2} + sinfrac{at}{2}right)e^{-st},mathrm{d}t - int_{3pi/2a}^{2pi/a}left(cosfrac{at}{2} + sinfrac{at}{2}right)e^{-st},mathrm{d}t right]end{aligned}$$
These integrals can be done by considering the integral
$$ int_{0}^{3pi/2a}e^{iat/2}e^{-st},mathrm{d}t $$
and summing the real and imaginary contributions. One should then obtain
$$ mathcal{L}[sqrt{1+sin at}] = frac{1}{s^{2}+a^{2}/4}left(s + frac{a}{2} + frac{sqrt{2},ae^{-3pi s/2a}}{1-e^{-2pi s/a}}right). $$
For $a = 4$, we have
$$ boxed{mathcal{L}[sqrt{1+sin 4t}] = frac{1}{s^{2}+4}left(s + 2 + frac{4sqrt{2},e^{-3pi s/8}}{1-e^{-pi s/2}}right)}.$$
answered Jan 1 at 18:36
IninterrompueIninterrompue
63519
answered Jan 1 at 18:36
IninterrompueIninterrompue
63519
answered Jan 1 at 18:36
IninterrompueIninterrompue
63519
answered Jan 1 at 18:36
IninterrompueIninterrompue
63519
63519
$begingroup$
Beautiful solution, Thanks :)
$endgroup$
– Tapan Lathiya
Jan 1 at 20:58
$begingroup$
I was thinking along the lines of convolution, but wow, that solution is quite neat and compact. Great job!
$endgroup$
– bjcolby15
Jan 1 at 21:32
add a comment |
$begingroup$
Beautiful solution, Thanks :)
$endgroup$
– Tapan Lathiya
Jan 1 at 20:58
$begingroup$
I was thinking along the lines of convolution, but wow, that solution is quite neat and compact. Great job!
$endgroup$
– bjcolby15
Jan 1 at 21:32
$begingroup$
Beautiful solution, Thanks :)
$endgroup$
– Tapan Lathiya
Jan 1 at 20:58
$begingroup$
Beautiful solution, Thanks :)
$endgroup$
– Tapan Lathiya
Jan 1 at 20:58
$begingroup$
I was thinking along the lines of convolution, but wow, that solution is quite neat and compact. Great job!
$endgroup$
– bjcolby15
Jan 1 at 21:32
$begingroup$
I was thinking along the lines of convolution, but wow, that solution is quite neat and compact. Great job!
$endgroup$
– bjcolby15
Jan 1 at 21:32
add a comment |
$begingroup$
With CAS help and for general $a$ and $aneq 0$ we have:
$mathcal{L}_tleft[sqrt{1+sin (a t)}right](s)=frac{e^{-frac{pi s}{2 a}} left(-a left(sqrt{2}-2 cosh left(frac{pi s}{2
a}right)right)+frac{left(a^2+4 s^2right) cosh left(frac{pi s}{2 a}right) , _3F_2left(-frac{1}{4},frac{1}{4},1;1-frac{i s}{2
a},1+frac{i s}{2 a};1right)}{s}right) left(1+tanh left(frac{pi s}{2 a}right)right)}{a^2+4 s^2}$
Mathematica code:
HoldForm[LaplaceTransform[Sqrt[1 + Sin[a t]], t, s] == (E^(-(([Pi] s)/(
2 a))) (-a (Sqrt[2] - 2 Cosh[([Pi] s)/(2 a)]) + ((a^2 +
4 s^2) Cosh[([Pi] s)/(
2 a)] HypergeometricPFQ[{-(1/4), 1/4, 1}, {1 - (I s)/(2 a),
1 + (I s)/(2 a)}, 1])/s) (1 + Tanh[([Pi] s)/(2 a)]))/(
a^2 + 4 s^2)] // TeXForm
For $a=4$:
(E^(-(([Pi] s)/8)) (-s (Sqrt[2] - 2 Cosh[([Pi] s)/8]) + (4 + s^2) Cosh[([Pi] s)/
8] HypergeometricPFQ[{-(1/4), 1/4, 1}, {1 - (I s)/8,
1 + (I s)/8}, 1]) (1 + Tanh[([Pi] s)/8]))/(s (4 + s^2))
$mathcal{L}_tleft[sqrt{1+sin (4 t)}right](s)=frac{e^{-frac{1}{8} (pi s)} left(-s left(sqrt{2}-2 cosh left(frac{pi
s}{8}right)right)+left(4+s^2right) cosh left(frac{pi s}{8}right) , _3F_2left(-frac{1}{4},frac{1}{4},1;1-frac{i s}{8},1+frac{i
s}{8};1right)right) left(1+tanh left(frac{pi s}{8}right)right)}{s left(4+s^2right)}$
answered Jan 1 at 12:48
Mariusz IwaniukMariusz Iwaniuk
1,9721718
$endgroup$
$begingroup$
Thanks for the answer. Is there any way to get rid of the square root? If so then it will be much easier to solve it further.
$endgroup$
– Tapan Lathiya
Jan 1 at 12:57
$begingroup$
@TapanLathiya Possible,yes,maybe so: $sqrt{1+sin (a t)}=sum _{j=0}^{infty } frac{left((-1)^{1+j} 4^{-j} (2 j)!right) sin ^j(a t)}{(-1+2 j) (j!)^2}$ or another way.
$endgroup$
– Mariusz Iwaniuk
Jan 1 at 13:02
add a comment |
$begingroup$
With CAS help and for general $a$ and $aneq 0$ we have:
$mathcal{L}_tleft[sqrt{1+sin (a t)}right](s)=frac{e^{-frac{pi s}{2 a}} left(-a left(sqrt{2}-2 cosh left(frac{pi s}{2
a}right)right)+frac{left(a^2+4 s^2right) cosh left(frac{pi s}{2 a}right) , _3F_2left(-frac{1}{4},frac{1}{4},1;1-frac{i s}{2
a},1+frac{i s}{2 a};1right)}{s}right) left(1+tanh left(frac{pi s}{2 a}right)right)}{a^2+4 s^2}$
Mathematica code:
HoldForm[LaplaceTransform[Sqrt[1 + Sin[a t]], t, s] == (E^(-(([Pi] s)/(
2 a))) (-a (Sqrt[2] - 2 Cosh[([Pi] s)/(2 a)]) + ((a^2 +
4 s^2) Cosh[([Pi] s)/(
2 a)] HypergeometricPFQ[{-(1/4), 1/4, 1}, {1 - (I s)/(2 a),
1 + (I s)/(2 a)}, 1])/s) (1 + Tanh[([Pi] s)/(2 a)]))/(
a^2 + 4 s^2)] // TeXForm
For $a=4$:
(E^(-(([Pi] s)/8)) (-s (Sqrt[2] - 2 Cosh[([Pi] s)/8]) + (4 + s^2) Cosh[([Pi] s)/
8] HypergeometricPFQ[{-(1/4), 1/4, 1}, {1 - (I s)/8,
1 + (I s)/8}, 1]) (1 + Tanh[([Pi] s)/8]))/(s (4 + s^2))
$mathcal{L}_tleft[sqrt{1+sin (4 t)}right](s)=frac{e^{-frac{1}{8} (pi s)} left(-s left(sqrt{2}-2 cosh left(frac{pi
s}{8}right)right)+left(4+s^2right) cosh left(frac{pi s}{8}right) , _3F_2left(-frac{1}{4},frac{1}{4},1;1-frac{i s}{8},1+frac{i
s}{8};1right)right) left(1+tanh left(frac{pi s}{8}right)right)}{s left(4+s^2right)}$
answered Jan 1 at 12:48
Mariusz IwaniukMariusz Iwaniuk
1,9721718
$endgroup$
$begingroup$
Thanks for the answer. Is there any way to get rid of the square root? If so then it will be much easier to solve it further.
$endgroup$
– Tapan Lathiya
Jan 1 at 12:57
$begingroup$
@TapanLathiya Possible,yes,maybe so: $sqrt{1+sin (a t)}=sum _{j=0}^{infty } frac{left((-1)^{1+j} 4^{-j} (2 j)!right) sin ^j(a t)}{(-1+2 j) (j!)^2}$ or another way.
$endgroup$
– Mariusz Iwaniuk
Jan 1 at 13:02
add a comment |
$begingroup$
With CAS help and for general $a$ and $aneq 0$ we have:
$mathcal{L}_tleft[sqrt{1+sin (a t)}right](s)=frac{e^{-frac{pi s}{2 a}} left(-a left(sqrt{2}-2 cosh left(frac{pi s}{2
a}right)right)+frac{left(a^2+4 s^2right) cosh left(frac{pi s}{2 a}right) , _3F_2left(-frac{1}{4},frac{1}{4},1;1-frac{i s}{2
a},1+frac{i s}{2 a};1right)}{s}right) left(1+tanh left(frac{pi s}{2 a}right)right)}{a^2+4 s^2}$
Mathematica code:
HoldForm[LaplaceTransform[Sqrt[1 + Sin[a t]], t, s] == (E^(-(([Pi] s)/(
2 a))) (-a (Sqrt[2] - 2 Cosh[([Pi] s)/(2 a)]) + ((a^2 +
4 s^2) Cosh[([Pi] s)/(
2 a)] HypergeometricPFQ[{-(1/4), 1/4, 1}, {1 - (I s)/(2 a),
1 + (I s)/(2 a)}, 1])/s) (1 + Tanh[([Pi] s)/(2 a)]))/(
a^2 + 4 s^2)] // TeXForm
For $a=4$:
(E^(-(([Pi] s)/8)) (-s (Sqrt[2] - 2 Cosh[([Pi] s)/8]) + (4 + s^2) Cosh[([Pi] s)/
8] HypergeometricPFQ[{-(1/4), 1/4, 1}, {1 - (I s)/8,
1 + (I s)/8}, 1]) (1 + Tanh[([Pi] s)/8]))/(s (4 + s^2))
$mathcal{L}_tleft[sqrt{1+sin (4 t)}right](s)=frac{e^{-frac{1}{8} (pi s)} left(-s left(sqrt{2}-2 cosh left(frac{pi
s}{8}right)right)+left(4+s^2right) cosh left(frac{pi s}{8}right) , _3F_2left(-frac{1}{4},frac{1}{4},1;1-frac{i s}{8},1+frac{i
s}{8};1right)right) left(1+tanh left(frac{pi s}{8}right)right)}{s left(4+s^2right)}$
answered Jan 1 at 12:48
Mariusz IwaniukMariusz Iwaniuk
1,9721718
$endgroup$
With CAS help and for general $a$ and $aneq 0$ we have:
$mathcal{L}_tleft[sqrt{1+sin (a t)}right](s)=frac{e^{-frac{pi s}{2 a}} left(-a left(sqrt{2}-2 cosh left(frac{pi s}{2
a}right)right)+frac{left(a^2+4 s^2right) cosh left(frac{pi s}{2 a}right) , _3F_2left(-frac{1}{4},frac{1}{4},1;1-frac{i s}{2
a},1+frac{i s}{2 a};1right)}{s}right) left(1+tanh left(frac{pi s}{2 a}right)right)}{a^2+4 s^2}$
Mathematica code:
HoldForm[LaplaceTransform[Sqrt[1 + Sin[a t]], t, s] == (E^(-(([Pi] s)/(
2 a))) (-a (Sqrt[2] - 2 Cosh[([Pi] s)/(2 a)]) + ((a^2 +
4 s^2) Cosh[([Pi] s)/(
2 a)] HypergeometricPFQ[{-(1/4), 1/4, 1}, {1 - (I s)/(2 a),
1 + (I s)/(2 a)}, 1])/s) (1 + Tanh[([Pi] s)/(2 a)]))/(
a^2 + 4 s^2)] // TeXForm
For $a=4$:
(E^(-(([Pi] s)/8)) (-s (Sqrt[2] - 2 Cosh[([Pi] s)/8]) + (4 + s^2) Cosh[([Pi] s)/
8] HypergeometricPFQ[{-(1/4), 1/4, 1}, {1 - (I s)/8,
1 + (I s)/8}, 1]) (1 + Tanh[([Pi] s)/8]))/(s (4 + s^2))
$mathcal{L}_tleft[sqrt{1+sin (4 t)}right](s)=frac{e^{-frac{1}{8} (pi s)} left(-s left(sqrt{2}-2 cosh left(frac{pi
s}{8}right)right)+left(4+s^2right) cosh left(frac{pi s}{8}right) , _3F_2left(-frac{1}{4},frac{1}{4},1;1-frac{i s}{8},1+frac{i
s}{8};1right)right) left(1+tanh left(frac{pi s}{8}right)right)}{s left(4+s^2right)}$
answered Jan 1 at 12:48
Mariusz IwaniukMariusz Iwaniuk
1,9721718
edited Jan 1 at 13:46
answered Jan 1 at 12:48
Mariusz IwaniukMariusz Iwaniuk
1,9721718
answered Jan 1 at 12:48
Mariusz IwaniukMariusz Iwaniuk
1,9721718
answered Jan 1 at 12:48
Mariusz IwaniukMariusz Iwaniuk
1,9721718
1,9721718
$begingroup$
Thanks for the answer. Is there any way to get rid of the square root? If so then it will be much easier to solve it further.
$endgroup$
– Tapan Lathiya
Jan 1 at 12:57
$begingroup$
@TapanLathiya Possible,yes,maybe so: $sqrt{1+sin (a t)}=sum _{j=0}^{infty } frac{left((-1)^{1+j} 4^{-j} (2 j)!right) sin ^j(a t)}{(-1+2 j) (j!)^2}$ or another way.
$endgroup$
– Mariusz Iwaniuk
Jan 1 at 13:02
add a comment |
$begingroup$
Thanks for the answer. Is there any way to get rid of the square root? If so then it will be much easier to solve it further.
$endgroup$
– Tapan Lathiya
Jan 1 at 12:57
$begingroup$
@TapanLathiya Possible,yes,maybe so: $sqrt{1+sin (a t)}=sum _{j=0}^{infty } frac{left((-1)^{1+j} 4^{-j} (2 j)!right) sin ^j(a t)}{(-1+2 j) (j!)^2}$ or another way.
$endgroup$
– Mariusz Iwaniuk
Jan 1 at 13:02
$begingroup$
Thanks for the answer. Is there any way to get rid of the square root? If so then it will be much easier to solve it further.
$endgroup$
– Tapan Lathiya
Jan 1 at 12:57
$begingroup$
Thanks for the answer. Is there any way to get rid of the square root? If so then it will be much easier to solve it further.
$endgroup$
– Tapan Lathiya
Jan 1 at 12:57
$begingroup$
@TapanLathiya Possible,yes,maybe so: $sqrt{1+sin (a t)}=sum _{j=0}^{infty } frac{left((-1)^{1+j} 4^{-j} (2 j)!right) sin ^j(a t)}{(-1+2 j) (j!)^2}$ or another way.
$endgroup$
– Mariusz Iwaniuk
Jan 1 at 13:02
$begingroup$
@TapanLathiya Possible,yes,maybe so: $sqrt{1+sin (a t)}=sum _{j=0}^{infty } frac{left((-1)^{1+j} 4^{-j} (2 j)!right) sin ^j(a t)}{(-1+2 j) (j!)^2}$ or another way.
$endgroup$
– Mariusz Iwaniuk
Jan 1 at 13:02
add a comment |
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@MariuszIwaniuk You're right, thanks!
$endgroup$
– caverac
Jan 1 at 12:39
$begingroup$
If you know that the transform of $sqrt{1 + sin t}$ is $F(s)$, then $sqrt{1 + sin 4 t}$ gives $F(s/4)/4$.
$endgroup$
– Maxim
Jan 2 at 3:08