Hello folks,
I'm supposed to solve the equation
k * (cos(x)^2 - sin(x)^2) = sin(x)^(2-e)
for x. k and e are symbolic constants.
Can you tell me if this is possible?
Thanks a lot!
Is this equation solvable?
Quote:Original post by Desperado
Hello folks,
I'm supposed to solve the equation
k * (cos(x)^2 - sin(x)^2) = sin(x)^(2-e)
for x. k and e are symbolic constants.
What is the variable you are solving for?
Unfortunately for a case like this (mainly because of the constant in the exponent of the sine function), you will need to use an approximating method like Newton's.
This method zeroes in on an approximate answer to f(x) = 0. So in this case in order to find an x that satisfies a(x) = b(x), we find an x that satisfies f(x) = 0 = a(x) - b(x)
f(x) = sin(x)^(2-e) - k * (cos(x)^2 - sin(x)^2)
f'(x) = (2-e)cos(x)sin(x)^(1-e) + 4ksin(x)cos(x))
Make an initial guess. Best to use a graphing calculator or software of some kind to get an idea of where f(x) = 0, make your initial guess x_0, then continue improving your guess using this formula:
x_(n+1) = x_n - f(x_n) / f'(x_n)
This method zeroes in on an approximate answer to f(x) = 0. So in this case in order to find an x that satisfies a(x) = b(x), we find an x that satisfies f(x) = 0 = a(x) - b(x)
f(x) = sin(x)^(2-e) - k * (cos(x)^2 - sin(x)^2)
f'(x) = (2-e)cos(x)sin(x)^(1-e) + 4ksin(x)cos(x))
Make an initial guess. Best to use a graphing calculator or software of some kind to get an idea of where f(x) = 0, make your initial guess x_0, then continue improving your guess using this formula:
x_(n+1) = x_n - f(x_n) / f'(x_n)
@alvaro: He's solving for x.
@Desperado:
You can transform the problem into two easier-looking problems. First you'd solve
s^(2-e) + 2 k s^2 - k = 0
for s, and then
sin(x) = s
for x.
The second of these is easy, so we just need to worry about the first.
How easy or hard it is to solve depends entirely on what 'e' is. If 2-e is a positive integer less than five, closed-form solutions are possible (0 <= e <= 2 are the easiest cases of course; then it's quadratic). If e is noninteger (or worse, irrational), I doubt there's a closed-form solution.
That said, closed-form solutions are overrated IMO. One nice thing about the polynomial you've got there is that there are nice root-finders that are guaranteed to return all the roots of polynomials.
Newton's method is also a (fast, easy) option, though you don't generally have global convergence guarantees.
Personally, if using Newton's method, I'd actually solve the related system of equations,
k * (c^2 - s^2) - s^(2-e) = 0
s^2 + c^2 - 1 = 0
for s and c. Then you can recover 'x' with a two-argument arctangent. The Jacobian of this system, by the way, is,
(which you'll need for Newton's method).
@Desperado:
You can transform the problem into two easier-looking problems. First you'd solve
s^(2-e) + 2 k s^2 - k = 0
for s, and then
sin(x) = s
for x.
The second of these is easy, so we just need to worry about the first.
How easy or hard it is to solve depends entirely on what 'e' is. If 2-e is a positive integer less than five, closed-form solutions are possible (0 <= e <= 2 are the easiest cases of course; then it's quadratic). If e is noninteger (or worse, irrational), I doubt there's a closed-form solution.
That said, closed-form solutions are overrated IMO. One nice thing about the polynomial you've got there is that there are nice root-finders that are guaranteed to return all the roots of polynomials.
Newton's method is also a (fast, easy) option, though you don't generally have global convergence guarantees.
Personally, if using Newton's method, I'd actually solve the related system of equations,
k * (c^2 - s^2) - s^(2-e) = 0
s^2 + c^2 - 1 = 0
for s and c. Then you can recover 'x' with a two-argument arctangent. The Jacobian of this system, by the way, is,
[ 2 k c (-2 k - 2 + e) s ][ 2 c 2 s ](which you'll need for Newton's method).
Folks,
Please review the Forum FAQ discussion of homework policy for this forum. I know that the way people learn has evolved thanks to the Internet, and lately I have been more tolerant of people asking this sort of question as a result. That said, please do be careful in your replies. I really don't want to see detailed answers that someone could copy/plagiarize and turn in as their own work. Offer advice and hints/suggestions, that's fine.
Thank you.
Please review the Forum FAQ discussion of homework policy for this forum. I know that the way people learn has evolved thanks to the Internet, and lately I have been more tolerant of people asking this sort of question as a result. That said, please do be careful in your replies. I really don't want to see detailed answers that someone could copy/plagiarize and turn in as their own work. Offer advice and hints/suggestions, that's fine.
Thank you.
Is e small? If so, you might want to solve it for e=0, then use that as your initial guess for the iterative method you use (like Newtons method mentioned above).
Wolfram Alpha provided closed form formulas for the solutions when e=0, but I can't seem to get the links to work right. You can just go there:
http://www.wolframalpha.com/
and type in:
k * (cos(x)^2 - sin(x)^2) = sin(x)^2
and it will give you the solutions.
Wolfram Alpha provided closed form formulas for the solutions when e=0, but I can't seem to get the links to work right. You can just go there:
http://www.wolframalpha.com/
and type in:
k * (cos(x)^2 - sin(x)^2) = sin(x)^2
and it will give you the solutions.
Quote:Original post by Maze Master
[...]
Wolfram Alpha provided closed form formulas for the solutions when e=0, but I can't seem to get the links to work right. You can just go there:
http://www.wolframalpha.com/
and type in:
k * (cos(x)^2 - sin(x)^2) = sin(x)^2
and it will give you the solutions.
That one you can solve by hand: Substitute cos(x)^2 = 1 - sin(x)^2, solve the equation treating sin(x) as the variable, and then take arcsin at the end.
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