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Expansion of $x$ in powers of $u$
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Given:
$$sin(x) = u sin(x+a),qquad u<1$$
How do I expand $x$ in powers of $u$?
I tried using Taylor series but it failed to proceed.
trigonometry complex-numbers power-series trigonometric-series
edited Jul 18 at 14:45
user7530
33.4k558109
asked Jul 18 at 14:23
David
267
add a comment |Â
up vote
2
down vote
favorite
Given:
$$sin(x) = u sin(x+a),qquad u<1$$
How do I expand $x$ in powers of $u$?
I tried using Taylor series but it failed to proceed.
trigonometry complex-numbers power-series trigonometric-series
edited Jul 18 at 14:45
user7530
33.4k558109
asked Jul 18 at 14:23
David
267
And you want to expand about $u=0$? I.e. $x(u) = sum alpha_i u^i$?
– user7530
Jul 18 at 15:09
@user7530 yes I need to expand in ascending powers of u. No condition for u=0 is given in statement. It is just asking me to develop an infinite series.
– David
Jul 18 at 15:25
@lab bhattacharjee That doesn't help. The answer is: usin(a) + (1/2) u^2 sin(2a) and so on......
– David
Jul 18 at 15:53
@labbhattacharjee your form doesn't gets simplified
– David
Jul 18 at 15:54
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given:
$$sin(x) = u sin(x+a),qquad u<1$$
How do I expand $x$ in powers of $u$?
I tried using Taylor series but it failed to proceed.
trigonometry complex-numbers power-series trigonometric-series
edited Jul 18 at 14:45
user7530
33.4k558109
asked Jul 18 at 14:23
David
267
Given:
$$sin(x) = u sin(x+a),qquad u<1$$
How do I expand $x$ in powers of $u$?
I tried using Taylor series but it failed to proceed.
trigonometry complex-numbers power-series trigonometric-series
edited Jul 18 at 14:45
user7530
33.4k558109
asked Jul 18 at 14:23
David
267
edited Jul 18 at 14:45
user7530
33.4k558109
edited Jul 18 at 14:45
user7530
33.4k558109
edited Jul 18 at 14:45
user7530
33.4k558109
33.4k558109
asked Jul 18 at 14:23
David
267
asked Jul 18 at 14:23
David
267
asked Jul 18 at 14:23
David
267
267
And you want to expand about $u=0$? I.e. $x(u) = sum alpha_i u^i$?
– user7530
Jul 18 at 15:09
@user7530 yes I need to expand in ascending powers of u. No condition for u=0 is given in statement. It is just asking me to develop an infinite series.
– David
Jul 18 at 15:25
@lab bhattacharjee That doesn't help. The answer is: usin(a) + (1/2) u^2 sin(2a) and so on......
– David
Jul 18 at 15:53
@labbhattacharjee your form doesn't gets simplified
– David
Jul 18 at 15:54
add a comment |Â
And you want to expand about $u=0$? I.e. $x(u) = sum alpha_i u^i$?
– user7530
Jul 18 at 15:09
@user7530 yes I need to expand in ascending powers of u. No condition for u=0 is given in statement. It is just asking me to develop an infinite series.
– David
Jul 18 at 15:25
@lab bhattacharjee That doesn't help. The answer is: usin(a) + (1/2) u^2 sin(2a) and so on......
– David
Jul 18 at 15:53
@labbhattacharjee your form doesn't gets simplified
– David
Jul 18 at 15:54
And you want to expand about $u=0$? I.e. $x(u) = sum alpha_i u^i$?
– user7530
Jul 18 at 15:09
And you want to expand about $u=0$? I.e. $x(u) = sum alpha_i u^i$?
– user7530
Jul 18 at 15:09
@user7530 yes I need to expand in ascending powers of u. No condition for u=0 is given in statement. It is just asking me to develop an infinite series.
– David
Jul 18 at 15:25
@user7530 yes I need to expand in ascending powers of u. No condition for u=0 is given in statement. It is just asking me to develop an infinite series.
– David
Jul 18 at 15:25
@lab bhattacharjee That doesn't help. The answer is: usin(a) + (1/2) u^2 sin(2a) and so on......
– David
Jul 18 at 15:53
@lab bhattacharjee That doesn't help. The answer is: usin(a) + (1/2) u^2 sin(2a) and so on......
– David
Jul 18 at 15:53
@labbhattacharjee your form doesn't gets simplified
– David
Jul 18 at 15:54
@labbhattacharjee your form doesn't gets simplified
– David
Jul 18 at 15:54
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Here is what I did. I started with $, x_0 := O(u). ,$ Define by recursion
$, x_n+1 := sin^-1(u sin(x_n + a)). ,$ The first few values are
$, x_1 = sin(a), u + O(u^2), , x_2 = sin(a), u + sin(2a),u^2/2 + O(u^3). ,$
Using the pattern and taking it to the limit, I found that
$, x = log( (1 - u exp(-ia)) / (1 - uexp(ia)) )/(2i). ,$ In the limit
$, x = sum_n=1^infty sin(na),u^n/n . ,$
Another method uses exponentials. Let
$, X := exp(i x), , A := exp(i a) ,$
and substitute them in equation
$, sin(x) = u sin(x+a) ,$ to get
$, (X - 1/X) = u (X A - 1/(X A)) ,$
and solving for $, X ,$ gives
$ X^2 = (1 - u/A) / (1 - A u). ,$
Using $, log(1-x) = - sum_n>0 x^n/n ,$
now gives the power series in $, u.$
answered Jul 18 at 17:30
Somos
11.6k1933
Don't know if this is solution is feasible, but it is a brute force way to find the answer
– David
Jul 18 at 22:46
Ok thank you for helping
– David
Jul 19 at 2:55
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
Here is what I did. I started with $, x_0 := O(u). ,$ Define by recursion
$, x_n+1 := sin^-1(u sin(x_n + a)). ,$ The first few values are
$, x_1 = sin(a), u + O(u^2), , x_2 = sin(a), u + sin(2a),u^2/2 + O(u^3). ,$
Using the pattern and taking it to the limit, I found that
$, x = log( (1 - u exp(-ia)) / (1 - uexp(ia)) )/(2i). ,$ In the limit
$, x = sum_n=1^infty sin(na),u^n/n . ,$
Another method uses exponentials. Let
$, X := exp(i x), , A := exp(i a) ,$
and substitute them in equation
$, sin(x) = u sin(x+a) ,$ to get
$, (X - 1/X) = u (X A - 1/(X A)) ,$
and solving for $, X ,$ gives
$ X^2 = (1 - u/A) / (1 - A u). ,$
Using $, log(1-x) = - sum_n>0 x^n/n ,$
now gives the power series in $, u.$
answered Jul 18 at 17:30
Somos
11.6k1933
Don't know if this is solution is feasible, but it is a brute force way to find the answer
– David
Jul 18 at 22:46
Ok thank you for helping
– David
Jul 19 at 2:55
add a comment |Â
up vote
1
down vote
accepted
Here is what I did. I started with $, x_0 := O(u). ,$ Define by recursion
$, x_n+1 := sin^-1(u sin(x_n + a)). ,$ The first few values are
$, x_1 = sin(a), u + O(u^2), , x_2 = sin(a), u + sin(2a),u^2/2 + O(u^3). ,$
Using the pattern and taking it to the limit, I found that
$, x = log( (1 - u exp(-ia)) / (1 - uexp(ia)) )/(2i). ,$ In the limit
$, x = sum_n=1^infty sin(na),u^n/n . ,$
Another method uses exponentials. Let
$, X := exp(i x), , A := exp(i a) ,$
and substitute them in equation
$, sin(x) = u sin(x+a) ,$ to get
$, (X - 1/X) = u (X A - 1/(X A)) ,$
and solving for $, X ,$ gives
$ X^2 = (1 - u/A) / (1 - A u). ,$
Using $, log(1-x) = - sum_n>0 x^n/n ,$
now gives the power series in $, u.$
answered Jul 18 at 17:30
Somos
11.6k1933
Don't know if this is solution is feasible, but it is a brute force way to find the answer
– David
Jul 18 at 22:46
Ok thank you for helping
– David
Jul 19 at 2:55
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Here is what I did. I started with $, x_0 := O(u). ,$ Define by recursion
$, x_n+1 := sin^-1(u sin(x_n + a)). ,$ The first few values are
$, x_1 = sin(a), u + O(u^2), , x_2 = sin(a), u + sin(2a),u^2/2 + O(u^3). ,$
Using the pattern and taking it to the limit, I found that
$, x = log( (1 - u exp(-ia)) / (1 - uexp(ia)) )/(2i). ,$ In the limit
$, x = sum_n=1^infty sin(na),u^n/n . ,$
Another method uses exponentials. Let
$, X := exp(i x), , A := exp(i a) ,$
and substitute them in equation
$, sin(x) = u sin(x+a) ,$ to get
$, (X - 1/X) = u (X A - 1/(X A)) ,$
and solving for $, X ,$ gives
$ X^2 = (1 - u/A) / (1 - A u). ,$
Using $, log(1-x) = - sum_n>0 x^n/n ,$
now gives the power series in $, u.$
answered Jul 18 at 17:30
Somos
11.6k1933
Here is what I did. I started with $, x_0 := O(u). ,$ Define by recursion
$, x_n+1 := sin^-1(u sin(x_n + a)). ,$ The first few values are
$, x_1 = sin(a), u + O(u^2), , x_2 = sin(a), u + sin(2a),u^2/2 + O(u^3). ,$
Using the pattern and taking it to the limit, I found that
$, x = log( (1 - u exp(-ia)) / (1 - uexp(ia)) )/(2i). ,$ In the limit
$, x = sum_n=1^infty sin(na),u^n/n . ,$
Another method uses exponentials. Let
$, X := exp(i x), , A := exp(i a) ,$
and substitute them in equation
$, sin(x) = u sin(x+a) ,$ to get
$, (X - 1/X) = u (X A - 1/(X A)) ,$
and solving for $, X ,$ gives
$ X^2 = (1 - u/A) / (1 - A u). ,$
Using $, log(1-x) = - sum_n>0 x^n/n ,$
now gives the power series in $, u.$
answered Jul 18 at 17:30
Somos
11.6k1933
edited Jul 18 at 23:31
answered Jul 18 at 17:30
Somos
11.6k1933
answered Jul 18 at 17:30
Somos
11.6k1933
answered Jul 18 at 17:30
Somos
11.6k1933
11.6k1933
Don't know if this is solution is feasible, but it is a brute force way to find the answer
– David
Jul 18 at 22:46
Ok thank you for helping
– David
Jul 19 at 2:55
add a comment |Â
Don't know if this is solution is feasible, but it is a brute force way to find the answer
– David
Jul 18 at 22:46
Ok thank you for helping
– David
Jul 19 at 2:55
Don't know if this is solution is feasible, but it is a brute force way to find the answer
– David
Jul 18 at 22:46
Don't know if this is solution is feasible, but it is a brute force way to find the answer
– David
Jul 18 at 22:46
Ok thank you for helping
– David
Jul 19 at 2:55
Ok thank you for helping
– David
Jul 19 at 2:55
add a comment |Â
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And you want to expand about $u=0$? I.e. $x(u) = sum alpha_i u^i$?
– user7530
Jul 18 at 15:09
@user7530 yes I need to expand in ascending powers of u. No condition for u=0 is given in statement. It is just asking me to develop an infinite series.
– David
Jul 18 at 15:25
@lab bhattacharjee That doesn't help. The answer is: usin(a) + (1/2) u^2 sin(2a) and so on......
– David
Jul 18 at 15:53
@labbhattacharjee your form doesn't gets simplified
– David
Jul 18 at 15:54