Effect of $sin^{-1}$ on $ f(x) = frac{sin(ax+b)}{sin(cx+d)}$
$begingroup$
I am working on some maths and I would like to find an analytical solution/closed form solution for my formula. It's the real part of the k-th bin of an N-points DFT of $cos(frac{2pi}{c}(1+x))$.
$$
F(x,k,c) = frac{1}{4N}left( frac{sin((N-frac{1}{2})cdotfrac{2pi}{c}(1+x-k))}{ sin(frac{pi}{c}(1+x-k))} + frac{sin((N-frac{1}{2})cdotfrac{2pi}{c}(1+x+k))}{ sin(frac{pi}{c}(1+x+k))} + 2right)
$$
For a given c and k and a known F(x,k), I would like to work to I would like to find the value of x. Thus something like $ x = veryDifficultMaths(F(x,k,c),k,c)$. Due to the periodicity of this function, multiple values of x map to the same value of F(x,k,c). Therefore it is fine of the analytical solution only works for only small values of x (around x=0, let's say $|x|<0.25$).
I think the first step will be to perform a arcsin-operation on both sides of the equation (after the $frac{1}{4N}$ term and the +2 are moved to the LHS**). Another tactic could be to merge both divisions (with sines in them) into one and perform the arc sin on that (also after moving the +2 and $frac{1}{4N}$-terms to the LHS). However I have no clue what the effect of an arcsin on a division of sines is and neither what the effect of an arcsin on a multiplication of sines does (assuming the .
So could you help me with that question? Or is there no analytical solution such that an Taylor polynomial would be the way to go?
Thanks a lot :)
** I am aware that arcsin expects a value between -1 and 1. Because F(x,k,c) is within the domain of about [-0.1 0.5], multiplying with 4N (with N = for examample 8) and then adding 2 and taking the arcsin afterwards could result in some troubles for certain values of F(x,k,c).
trigonometry closed-form fourier-transform
edited Dec 10 '18 at 13:54
amWhy
1
asked Dec 10 '18 at 13:03
WobbertWobbert
134
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add a comment |
$begingroup$
I am working on some maths and I would like to find an analytical solution/closed form solution for my formula. It's the real part of the k-th bin of an N-points DFT of $cos(frac{2pi}{c}(1+x))$.
$$
F(x,k,c) = frac{1}{4N}left( frac{sin((N-frac{1}{2})cdotfrac{2pi}{c}(1+x-k))}{ sin(frac{pi}{c}(1+x-k))} + frac{sin((N-frac{1}{2})cdotfrac{2pi}{c}(1+x+k))}{ sin(frac{pi}{c}(1+x+k))} + 2right)
$$
For a given c and k and a known F(x,k), I would like to work to I would like to find the value of x. Thus something like $ x = veryDifficultMaths(F(x,k,c),k,c)$. Due to the periodicity of this function, multiple values of x map to the same value of F(x,k,c). Therefore it is fine of the analytical solution only works for only small values of x (around x=0, let's say $|x|<0.25$).
I think the first step will be to perform a arcsin-operation on both sides of the equation (after the $frac{1}{4N}$ term and the +2 are moved to the LHS**). Another tactic could be to merge both divisions (with sines in them) into one and perform the arc sin on that (also after moving the +2 and $frac{1}{4N}$-terms to the LHS). However I have no clue what the effect of an arcsin on a division of sines is and neither what the effect of an arcsin on a multiplication of sines does (assuming the .
So could you help me with that question? Or is there no analytical solution such that an Taylor polynomial would be the way to go?
Thanks a lot :)
** I am aware that arcsin expects a value between -1 and 1. Because F(x,k,c) is within the domain of about [-0.1 0.5], multiplying with 4N (with N = for examample 8) and then adding 2 and taking the arcsin afterwards could result in some troubles for certain values of F(x,k,c).
trigonometry closed-form fourier-transform
edited Dec 10 '18 at 13:54
amWhy
1
asked Dec 10 '18 at 13:03
WobbertWobbert
134
$endgroup$
add a comment |
$begingroup$
I am working on some maths and I would like to find an analytical solution/closed form solution for my formula. It's the real part of the k-th bin of an N-points DFT of $cos(frac{2pi}{c}(1+x))$.
$$
F(x,k,c) = frac{1}{4N}left( frac{sin((N-frac{1}{2})cdotfrac{2pi}{c}(1+x-k))}{ sin(frac{pi}{c}(1+x-k))} + frac{sin((N-frac{1}{2})cdotfrac{2pi}{c}(1+x+k))}{ sin(frac{pi}{c}(1+x+k))} + 2right)
$$
For a given c and k and a known F(x,k), I would like to work to I would like to find the value of x. Thus something like $ x = veryDifficultMaths(F(x,k,c),k,c)$. Due to the periodicity of this function, multiple values of x map to the same value of F(x,k,c). Therefore it is fine of the analytical solution only works for only small values of x (around x=0, let's say $|x|<0.25$).
I think the first step will be to perform a arcsin-operation on both sides of the equation (after the $frac{1}{4N}$ term and the +2 are moved to the LHS**). Another tactic could be to merge both divisions (with sines in them) into one and perform the arc sin on that (also after moving the +2 and $frac{1}{4N}$-terms to the LHS). However I have no clue what the effect of an arcsin on a division of sines is and neither what the effect of an arcsin on a multiplication of sines does (assuming the .
So could you help me with that question? Or is there no analytical solution such that an Taylor polynomial would be the way to go?
Thanks a lot :)
** I am aware that arcsin expects a value between -1 and 1. Because F(x,k,c) is within the domain of about [-0.1 0.5], multiplying with 4N (with N = for examample 8) and then adding 2 and taking the arcsin afterwards could result in some troubles for certain values of F(x,k,c).
trigonometry closed-form fourier-transform
edited Dec 10 '18 at 13:54
amWhy
1
asked Dec 10 '18 at 13:03
WobbertWobbert
134
$endgroup$
I am working on some maths and I would like to find an analytical solution/closed form solution for my formula. It's the real part of the k-th bin of an N-points DFT of $cos(frac{2pi}{c}(1+x))$.
$$
F(x,k,c) = frac{1}{4N}left( frac{sin((N-frac{1}{2})cdotfrac{2pi}{c}(1+x-k))}{ sin(frac{pi}{c}(1+x-k))} + frac{sin((N-frac{1}{2})cdotfrac{2pi}{c}(1+x+k))}{ sin(frac{pi}{c}(1+x+k))} + 2right)
$$
For a given c and k and a known F(x,k), I would like to work to I would like to find the value of x. Thus something like $ x = veryDifficultMaths(F(x,k,c),k,c)$. Due to the periodicity of this function, multiple values of x map to the same value of F(x,k,c). Therefore it is fine of the analytical solution only works for only small values of x (around x=0, let's say $|x|<0.25$).
I think the first step will be to perform a arcsin-operation on both sides of the equation (after the $frac{1}{4N}$ term and the +2 are moved to the LHS**). Another tactic could be to merge both divisions (with sines in them) into one and perform the arc sin on that (also after moving the +2 and $frac{1}{4N}$-terms to the LHS). However I have no clue what the effect of an arcsin on a division of sines is and neither what the effect of an arcsin on a multiplication of sines does (assuming the .
So could you help me with that question? Or is there no analytical solution such that an Taylor polynomial would be the way to go?
Thanks a lot :)
** I am aware that arcsin expects a value between -1 and 1. Because F(x,k,c) is within the domain of about [-0.1 0.5], multiplying with 4N (with N = for examample 8) and then adding 2 and taking the arcsin afterwards could result in some troubles for certain values of F(x,k,c).
trigonometry closed-form fourier-transform
trigonometry closed-form fourier-transform
edited Dec 10 '18 at 13:54
amWhy
1
asked Dec 10 '18 at 13:03
WobbertWobbert
134
edited Dec 10 '18 at 13:54
amWhy
1
asked Dec 10 '18 at 13:03
WobbertWobbert
134
edited Dec 10 '18 at 13:54
amWhy
1
edited Dec 10 '18 at 13:54
amWhy
1
edited Dec 10 '18 at 13:54
amWhy
1
1
asked Dec 10 '18 at 13:03
WobbertWobbert
134
asked Dec 10 '18 at 13:03
WobbertWobbert
134
asked Dec 10 '18 at 13:03
WobbertWobbert
134
134
add a comment |
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