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Re: [Phys-L] solve equation w/o calculus



On 12/15/2016 07:02 PM, Carl Mungan wrote:

y(x) = sqrt(4+4x) - sqrt(x).

Using calculus, one finds x=1/3 and thus y=sqrt(3).

The question is: Is there a neat way to find this solution without using
calculus? I don't want an approximation, which you can do graphically or
numerically. Nor do I think a Taylor expansion around x=1/3 is really fair.

I'm seeking a geometric solution. Rearrange the equation into a parabola or
some such

Well, I don't know how to do it in strictly geometric terms,
but I can arrange it to look like a parabola ... locally.
It is only /locally/ a parabola.

Use graphical methods or whatever to generate the hypothesis that
Q = (x, y^2) = (1/3, 3) is a point worthy of further investigation.
Plot y^2 as a function of 1/x to make this even more plausible.
Nothing has been proved at this point; it's just a hypothesis.

Substitute y^2 --> z. Minimizing z is the same as minimizing
y, at least in the neighborhood of the special point Q. There
will be spurious roots, just not near Q.

Substitute x -> 1/3 + dx and z --> 3 + dz, to focus attention
where we want it.

More-or-less straightforward algebra [see below] gets rid of the
square roots and leaves us with a relatively simple polynomial:

3 dz^2 + ((- 16) - 30 dx) dz + 27 dx^2 = 0 [1]

Confirm by direct evaluation that (dx,dz) = (0,0) is a solution.
So this is a point on the curve. It remains to be shown that it
is a minimum.

Now if (0,0) is a point on the curve, there must be other points
where dx and dz are reeeally small, but not quite zero. There
must be some points where dz is less than, say, 1e-50000.

In equation [1], throw away the dz^2 term as being "small squared"
compared to the dz term that we plan to keep. Similarly, throw
away 30 dx since it is negligible compared to 16. What remains
is a parabola: dz is proportional to dx squared, in the vicinity
of the special point Q.

This uses a notion of expansion to lowest order /without/ using
calculus directly or indirectly. That includes not using Taylor
series directly or indirectly.

Actually this is a slightly tricky version of expansion to lowest
order, because it's first order in dz and (equivalently) second
order in dx. In terms of dx, we kept the second order terms dx^2
and dz, while throwing away the third-order term dx dz and the
fourth-order term dz^2.

You can then estimate the size of the terms that were dropped,
so that this becomes a /controlled/ approximation. That is to
say, you have a handle on how accurate it is. The is more
professional than the usual Taylor series, where it can be
difficult to ascertain the radius of convergence, let alone
the rate of convergence.

You can even pick off the curvature of the parabola, by inspection.
Add the parabola to the aforementioned graph for comparison.

Here's the algebra, with help from the maxima program:

(%i2) step1 : y = sqrt(4*x+4)-sqrt(x)

(%o2) y = sqrt(4 x + 4) - sqrt(x)

(%i3) step2 : ratsimp(step1*step1)

2
(%o3) y = (- 2 sqrt(x) sqrt(4 x + 4)) + 5 x + 4

(%i4) step3 : (-4)-5*x+step2

2
(%o4) y - 5 x - 4 = - 2 sqrt(x) sqrt(4 x + 4)

(%i5) step4 : step3*step3

2 2
(%o5) ((- 4) - 5 x + y ) = 4 x (4 x + 4)

(%i6) step5 : ev(step4, y = sqrt(z))

2
(%o6) ((- 4) - 5 x + z) = 4 x (4 x + 4)

(%i7) step6 : ratsimp(step5-rhs(step5))

2 2
(%o7) z + ((- 8) - 10 x) z + 9 x + 24 x + 16 = 0


(%i8) ratsimp(ev(step6*3, x = dx+1/3, z = dz+3))

2 2
(%o8) 3 dz + ((- 16) - 30 dx) dz + 27 dx = 0


And in case you were wondering, yes, for me it is easier to use
maxima than to try to do 7 lines of algebra without making a mistake.
This does not mean maxima is easy to use; it just means I am lousy
at algebra.