Mixing
Recurrence Relations Defined In Terms of Each Other to Closed Form
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I am doing some research into the movement of robots executing a given algorithm, and have come up with the following recursive equations that describe the movement of the robots at each iteration:
beginequation
theta_i = frac34theta_i-1
endequation
beginequation
L_i=frac12sqrtL_i-1^2+A_i-1^2
endequation
beginequation
A_i=L_itantheta_i
endequation
Where the initial conditions are $theta_1=45^circ$, $;L_1=frac12d$, and $frac12d$.
It is clear that the first recurrence relation can simply be replaced with the following closed-form solution:
beginequation
theta_i = big(frac34big)^i-1;theta_1
endequation
But my problem is trying to obtain closed-form solutions to the other two equations that are defined in terms of each other. Is there a way to do this?
sequences-and-series algorithms recurrence-relations closed-form recursion
asked Jul 15 at 2:57
RoryHector
9012
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up vote
0
down vote
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I am doing some research into the movement of robots executing a given algorithm, and have come up with the following recursive equations that describe the movement of the robots at each iteration:
beginequation
theta_i = frac34theta_i-1
endequation
beginequation
L_i=frac12sqrtL_i-1^2+A_i-1^2
endequation
beginequation
A_i=L_itantheta_i
endequation
Where the initial conditions are $theta_1=45^circ$, $;L_1=frac12d$, and $frac12d$.
It is clear that the first recurrence relation can simply be replaced with the following closed-form solution:
beginequation
theta_i = big(frac34big)^i-1;theta_1
endequation
But my problem is trying to obtain closed-form solutions to the other two equations that are defined in terms of each other. Is there a way to do this?
sequences-and-series algorithms recurrence-relations closed-form recursion
asked Jul 15 at 2:57
RoryHector
9012
$L_i=frac12sqrtL_i-1^2+A_i-1^2=frac12 L_i-1sqrt1+tantheta_i-1$, ...
– Jens Schwaiger
Jul 15 at 3:09
You can eliminate $,A_i,$ and get $displaystyle L_i^2 = fracL_i-1^24 cos^2 theta_i-1$ which telescopes, but you still need to evaluate the trigonometric product that ensues in the denominator to get a truly closed form.
– dxiv
Jul 15 at 3:13
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am doing some research into the movement of robots executing a given algorithm, and have come up with the following recursive equations that describe the movement of the robots at each iteration:
beginequation
theta_i = frac34theta_i-1
endequation
beginequation
L_i=frac12sqrtL_i-1^2+A_i-1^2
endequation
beginequation
A_i=L_itantheta_i
endequation
Where the initial conditions are $theta_1=45^circ$, $;L_1=frac12d$, and $frac12d$.
It is clear that the first recurrence relation can simply be replaced with the following closed-form solution:
beginequation
theta_i = big(frac34big)^i-1;theta_1
endequation
But my problem is trying to obtain closed-form solutions to the other two equations that are defined in terms of each other. Is there a way to do this?
sequences-and-series algorithms recurrence-relations closed-form recursion
asked Jul 15 at 2:57
RoryHector
9012
I am doing some research into the movement of robots executing a given algorithm, and have come up with the following recursive equations that describe the movement of the robots at each iteration:
beginequation
theta_i = frac34theta_i-1
endequation
beginequation
L_i=frac12sqrtL_i-1^2+A_i-1^2
endequation
beginequation
A_i=L_itantheta_i
endequation
Where the initial conditions are $theta_1=45^circ$, $;L_1=frac12d$, and $frac12d$.
It is clear that the first recurrence relation can simply be replaced with the following closed-form solution:
beginequation
theta_i = big(frac34big)^i-1;theta_1
endequation
But my problem is trying to obtain closed-form solutions to the other two equations that are defined in terms of each other. Is there a way to do this?
sequences-and-series algorithms recurrence-relations closed-form recursion
asked Jul 15 at 2:57
RoryHector
9012
asked Jul 15 at 2:57
RoryHector
9012
asked Jul 15 at 2:57
RoryHector
9012
asked Jul 15 at 2:57
RoryHector
9012
9012
$L_i=frac12sqrtL_i-1^2+A_i-1^2=frac12 L_i-1sqrt1+tantheta_i-1$, ...
– Jens Schwaiger
Jul 15 at 3:09
You can eliminate $,A_i,$ and get $displaystyle L_i^2 = fracL_i-1^24 cos^2 theta_i-1$ which telescopes, but you still need to evaluate the trigonometric product that ensues in the denominator to get a truly closed form.
– dxiv
Jul 15 at 3:13
add a comment |Â
$L_i=frac12sqrtL_i-1^2+A_i-1^2=frac12 L_i-1sqrt1+tantheta_i-1$, ...
– Jens Schwaiger
Jul 15 at 3:09
You can eliminate $,A_i,$ and get $displaystyle L_i^2 = fracL_i-1^24 cos^2 theta_i-1$ which telescopes, but you still need to evaluate the trigonometric product that ensues in the denominator to get a truly closed form.
– dxiv
Jul 15 at 3:13
$L_i=frac12sqrtL_i-1^2+A_i-1^2=frac12 L_i-1sqrt1+tantheta_i-1$, ...
– Jens Schwaiger
Jul 15 at 3:09
$L_i=frac12sqrtL_i-1^2+A_i-1^2=frac12 L_i-1sqrt1+tantheta_i-1$, ...
– Jens Schwaiger
Jul 15 at 3:09
You can eliminate $,A_i,$ and get $displaystyle L_i^2 = fracL_i-1^24 cos^2 theta_i-1$ which telescopes, but you still need to evaluate the trigonometric product that ensues in the denominator to get a truly closed form.
– dxiv
Jul 15 at 3:13
You can eliminate $,A_i,$ and get $displaystyle L_i^2 = fracL_i-1^24 cos^2 theta_i-1$ which telescopes, but you still need to evaluate the trigonometric product that ensues in the denominator to get a truly closed form.
– dxiv
Jul 15 at 3:13
add a comment |Â
1 Answer
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I believe if you replace the $A_i-1$ with your formula for $A_i$ you should come to the conclusion that $$L_i = (frac12)^i-1prod_k=1^i-1sec(θ_k)L_1$$ and that consequently,
$$A_i = (frac12)^i-1prod_k=1^i-1sec(θ_k)L_1tan(θ_i)$$
where you have already given a closed form formula for $θ_i$.
answered Jul 15 at 3:14
BelowAverageIntelligence
339212
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I believe if you replace the $A_i-1$ with your formula for $A_i$ you should come to the conclusion that $$L_i = (frac12)^i-1prod_k=1^i-1sec(θ_k)L_1$$ and that consequently,
$$A_i = (frac12)^i-1prod_k=1^i-1sec(θ_k)L_1tan(θ_i)$$
where you have already given a closed form formula for $θ_i$.
answered Jul 15 at 3:14
BelowAverageIntelligence
339212
add a comment |Â
up vote
1
down vote
accepted
I believe if you replace the $A_i-1$ with your formula for $A_i$ you should come to the conclusion that $$L_i = (frac12)^i-1prod_k=1^i-1sec(θ_k)L_1$$ and that consequently,
$$A_i = (frac12)^i-1prod_k=1^i-1sec(θ_k)L_1tan(θ_i)$$
where you have already given a closed form formula for $θ_i$.
answered Jul 15 at 3:14
BelowAverageIntelligence
339212
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I believe if you replace the $A_i-1$ with your formula for $A_i$ you should come to the conclusion that $$L_i = (frac12)^i-1prod_k=1^i-1sec(θ_k)L_1$$ and that consequently,
$$A_i = (frac12)^i-1prod_k=1^i-1sec(θ_k)L_1tan(θ_i)$$
where you have already given a closed form formula for $θ_i$.
answered Jul 15 at 3:14
BelowAverageIntelligence
339212
I believe if you replace the $A_i-1$ with your formula for $A_i$ you should come to the conclusion that $$L_i = (frac12)^i-1prod_k=1^i-1sec(θ_k)L_1$$ and that consequently,
$$A_i = (frac12)^i-1prod_k=1^i-1sec(θ_k)L_1tan(θ_i)$$
where you have already given a closed form formula for $θ_i$.
answered Jul 15 at 3:14
BelowAverageIntelligence
339212
answered Jul 15 at 3:14
BelowAverageIntelligence
339212
answered Jul 15 at 3:14
BelowAverageIntelligence
339212
answered Jul 15 at 3:14
BelowAverageIntelligence
339212
339212
add a comment |Â
add a comment |Â
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$L_i=frac12sqrtL_i-1^2+A_i-1^2=frac12 L_i-1sqrt1+tantheta_i-1$, ...
– Jens Schwaiger
Jul 15 at 3:09
You can eliminate $,A_i,$ and get $displaystyle L_i^2 = fracL_i-1^24 cos^2 theta_i-1$ which telescopes, but you still need to evaluate the trigonometric product that ensues in the denominator to get a truly closed form.
– dxiv
Jul 15 at 3:13