solving trig equation for homework(?) Re: (no subject)

On 6/15/2013 7:37 AM, Rupert Swarbrick wrote:
> ?????? ????? <ya.natasha42 at yandex.ru> writes:
>> Hello!
>>
>> Please help with the equation: cos 3x-sin x=sqrt(3)(cos x-sin 3x)
>>
>> I can not bring it to the form in order to find the real roots in the
>> interval [0,2*%pi].

Rubert's method is maybe too much unnecessary work.
  I suspect the original homework assignment
was to use a plotting program.  Anyway...

eq:   (%o39) cos(3*x)-sin(x)-  (sqrt(3)*(cos(x)-sin(3*x)));

%,exponentialize;
ratsubst(y,exp(%i*x),%);

/*notice that when y=0 there may be a problem from denominator. Also 
that y=infinity
might be a solution, though closer examination via limit shows it is not.*/
ratnumer(%);

/* at this point solve() can be called to find exact values for y, but 
the answer is
complicated.  Trying numerics ... */

%,numer;    /*  allroots doesn't like exact rationals. */
allroots(%,y);

/* cleaning up presentation etc */
subst(y=exp(%i*x), %);
map (lambda([r],rectform(-%i*log(r))),%o63);

returns
[x=0.19541387185512*%i+0.5235987755983,
x=2.094395102393195,
x=0.19541387185512*%i-2.617993877991494,
x=-1.047197551196598,
x=0.5235987755983-0.19541387185512*%i,
x=-0.19541387185512*%i-2.617993877991489]

Two real roots.




> One way to solve this sort of question is to substitute log(y) for %i*x
> into its exponential form. For example:
>
>    eq: cos(3*x) - sin(x) = sqrt(3)*(cos(x)-sin(3*x));
>    eq1: exponentialize(eq);
>    eq2: subst([%i*x = log(y), 3*%i*x = 3*log(y)], eq1);
>
> (the strange form for subst is because subst works by matching
> syntactically and %i*x doesn't appear as a subexpression of
> 3*%i*x. ratsubst deals with this correctly, but jumps straight to eq3,
> which makes it less obvious what's going on)
>
> eq2 is then:
>
>         3   1                               3   1
>        y  + --           1             %i (y  - --)       1
>              3   %i (y - -)                      3    y + -
>             y            y                      y         y
>        ------- + ---------- = sqrt(3) (------------ + -----)
>           2          2                      2           2
>
> This is a sixth order polynomial equation. One way to write it as a
> polynomial is to apply rat and then multiply through by the denominator:
>
>    eq3: rat(eq2);
>    eq4: multthru(2*y^3, eq3);
>
>            6       4       2                   6            4
>    (%o26) y  + %i y  - %i y  + 1 = sqrt(3) %i y  + sqrt(3) y
>                                                  2
>                                       + sqrt(3) y   - sqrt(3) %i
>
>
> This is promising, since it's a cubic in y^2. So let's tidy up further:
>
>    (%i31) eq5: ratsubst(z, y^2, eq4);
>              3        2                         3                  2
>    (%o31)   z  + %i (z  - z) + 1 = sqrt(3) %i (z  - 1) + sqrt(3) (z  + z)
>
>
> Now, if u is a root of eq5, then exp(log(sqrt(u))/%i) will be a root of
> your original equation. Very promising... But the result of
>
>    solve(eq5, z);
>
> is pretty horrible.
>
> Notice that %solve, from the to_poly_solve package, seems to get the
> same answer, starting with the original equation. It also takes quite a
> long time to run. I'm writing up the stuff above, since it explains
> what's going on (and I never believe a long calculation that gives a
> horrible answer...)
>
> Incidentally, plotting eq with
>
>    plot2d(lhs(eq)-rhs(eq), [x,0,2*%pi]);
>
> convinced me that there are six roots. A relief, since each of the three
> roots of eq5 is the square of two roots of eq4. Numerically, you can do:
>
>    (%i45) zroots: allroots(eq5), numer;
>    (%o45) [z = .7071067811865129 %i + .7071067811865259,
>    z = - .7071067811865503 %i - .7071067811865429,
>    z = .5000000000000374 %i + .8660254037844556]
>
> (the numer flag is to put in a numerical value for the sqrt(3)).
>
> AHAH! If you recognise the decimal expansion of 1/sqrt(2), this looks
> very familiar! Now, there are two things we can do. Firstly, we can
> continue with the numerical calculations. I confess that I didn't spot
> the magic numbers the first time around, so that's what I did
> first. Secondly, you can factor the cubic by eye.
>
> The numerical approach:
>
> First, we want the square roots:
>
>    (%i48) yroots: flatten(map(lambda([x], [sqrt(rhs(x)), -sqrt(rhs(x))]), zroots));
>    (%o48) [.3826834323650779 %i + .9238795325112702,
>    - .3826834323650779 %i - .9238795325112702,
>    .3826834323650919 - .9238795325112852 %i,
>    .9238795325112852 %i - .3826834323650919,
>    .2588190451025366 %i + .9659258262890813,
>    - .2588190451025366 %i - .9659258262890813]
>
> And to convert from y to x:
>
>    (%i51) almost_x_roots: map(lambda([y], rectform(log(y)/%i)), yroots);
>    (%o51) [1.983135877736725e-14 %i + .3926990816987195,
>    1.983135877736725e-14 %i - 2.748893571891074,
>    6.24500451351651e-16 %i - 1.17809724509617,
>    6.24500451351651e-16 %i + 1.963495408493623,
>    .2617993877991614 - 1.66811009449927e-14 %i,
>    - 1.66811009449927e-14 %i - 2.879793265790632]
>
> Woo! These all have tiny imaginary parts! I haven't got it horribly
> wrong! Let's throw away the imaginary parts, which we think are just
>  From numerical error, to get
>
>    (%i53) x_roots: subst(0, %i, almost_x_roots);
>    (%o53) [.3926990816987195, - 2.748893571891074, - 1.17809724509617,
>                         1.963495408493623, .2617993877991614, - 2.879793265790632]
>
> Oh, and these are sometimes in the wrong 2*%pi-long interval, so let's
> fix that:
>
>    (%i57) x_roots_in_interval: map(lambda([x], mod(x, float(2*%pi))), x_roots);
>    (%o57) [.3926990816987195, 3.534291735288512, 5.105088062083416,
>                           1.963495408493623, .2617993877991614, 3.403392041388954]
>
> Incidentally, I made a slight mistake when working that out the first
> time, and wrote:
>
>    (%i58) map(lambda([x], mod(x, float(%pi))), x_roots);
>    (%o58) [.3926990816987195, .3926990816987193, 1.963495408493623,
>                           1.963495408493623, .2617993877991614, .2617993877991611]
>
> Given that the pairs of square roots were next to each other in our
> list, this shows that the roots are actually %pi-periodic and the
> positive and negative square roots pick out the solutions on the odd and
> even intervals. Rather neat, considering that the function itself
> doesn't have any obvious symmetries (looking at a graph). Anyone know a
> nice mathematical explanation for this?
>
>
> Now let's go back to the bit where we spotted the magic numbers. Notice
> that
>
>    (%i75) map(lambda([z0], expand(subst(z0, z, lhs(eq5)-rhs(eq5)))),
>               [(1 + %i)/sqrt(2), -(1 + %i)/sqrt(2), sqrt(3)/2+%i/2]);
>    (%o75)                             [0, 0, 0]
>
> So we actually have the solutions. In fact, they have very nice
> numerical answers. Taking logarithms, we get the following six answers:
>
>    (%i105) y_roots: flatten(map(lambda([x], [sqrt(x), -sqrt(x)]),
>                                [(1+%i)/sqrt(2), -(1+%i)/sqrt(2), sqrt(3)/2+%i/2]))$
>
>    (%i106) x_roots: map(lambda([y], rectform(log(y)/%i)), y_roots);
>                      %pi    7 %pi    3 %pi  5 %pi  %pi    11 %pi
>    (%o106)          [---, - -----, - -----, -----, ---, - ------]
>                       8       8        8      8    12       12
>
> and we can force them into the expected intervals with
>
>    (%i108) x_roots_in_interval: map(lambda([x], mod(x, 2*%pi)), x_roots);
>                        %pi  9 %pi  13 %pi  5 %pi  %pi  13 %pi
>    (%o108)            [---, -----, ------, -----, ---, ------]
>                         8     8      8       8    12     12
>
> Incidentally, taking the original numerical answers as num_xrii, you can
> do the following:
>
>    (%i125) num_xrii / float(%pi);
>    (%o125) [.1249999999999985, 1.124999999999998, 1.625000000000001,
>                     .6250000000000009, .08333333333333715, 1.083333333333337]
>
> and at least the first four are obviously recognizable. Phew!
>
> Questions:
> ==========
>
>   (1) Natasha, was your original question a homework question? If you
>       waded through this whole email, I don't feel too guilty for having
>       answered it. But this probably wasn't the approach the homework
>       setter expected you to follow. (Spotting that 0.707... is a
>       numerical solution kind of requires a computer...) If so, was there
>       some sort of hint?
>
>   (2) I'm still mystified by the fact that the two roots in each square
>       root give the roots over the two length-%pi intervals. Anyone got
>       any idea why?
>
>   (3) Why did algsys give such a horrible answer on the cubic, I wonder?
>
> Anyway, Natasha, I hope this answered your original question. I may have
> written slightly more than you were after, but I was enjoying myself :-)
>
> Rupert
>
>
> PS: Natasha, your email was held in a moderation queue because you're
>      not subscribed to the mailing list. If you want your emails to
>      appear more quickly, I recommend subscribing.
>
>
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