Hi experts!
I want to resolve the equation: $A.sin(x)=B.(C-x)$
where A, C and B are constants, $0\leq x\leq \pi $, $A$ and $B$ are fixed, and $0\leq C\leq \pi $.
I want to obtain the analytic solution, $x=f(C)$, (if exist) for different values of $C$ and if only implicit solution exist, obtain the curve and and table with the values of:
$ x$ vs. $C$
How can I do that?
Waiting for your answer.
Thanks a lot!
One option, and this might be suboptimal, is to replace sin(x) with its taylor series.
Because
sin(x) = x - (1/6)x^3 + (1/120)x^5 - (1/5040)x^7 + (1/362880)x^9 - O(x^11)
we would change
Asin(x) = BC - B*x
into instead
Ax - (A/6)x^3 + (A/120)x^5 - (A/5040)x^7 + (A/362880)x^9 - O(A x^11) = BC - Bx
giving you then
-BC + (A+B)x - (A/6)x^3 + (A/120)x^5 - (A/5040)x^7 + (A/362880)x^9 - O(A x^11) = 0
or if you want a few more terms
-BC + (A+B)x - (A/3!)x^3 + (A/5!)x^5 - (A/7!)x^7 + (A/9!)x^9 - (A/11!)x^11 + (A/13!)x^13 + O(A x^15) = 0
This does reveal that varying C will only affect the answer by a constant.
If A and B were fixed constants, known in advance, then you could use "for" and "find_root" to find a value of x for a long sequence of C values, forming a table of values.
For example:
Thanks. It can't be done wothout using Taylor expansion?
I'm guessing that you need either the Taylor expansion or some competitor of it, like the Chebychev polynomials. The main issue is that you have both sin x and a linear polynomial in x. Therefore, I really can't see how you would "solve for x." I'm about 98% sure but I don't want to say 100%, because mathematics is always full of surprises.
Asked: 2014-07-19 15:53:05 +0200
Seen: 647 times
Last updated: Jul 20 '14
Copyright Sage, 2010. Some rights reserved under creative commons license. Content on this site is licensed under a Creative Commons Attribution Share Alike 3.0 license.
