Re: Solving for X ?? Stumpes

joe <joeycook@xxxxxxxx> wrote:
On Jul 31, 12:51 pm, "G. A. Edgar" <ed...@xxxxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
In article <1185899859.570131.76...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, joe

<joeyc...@xxxxxxxx> wrote:
13 = 2^T + 3^T,

[...]

Is there another numerical method you can suggest ?

[...]>

Thanks for that answer. Still dissapointed there isn't some calculus
method to obtain an exact answer, but this will do.

Here's something else which is not really what you want; it shows how
an equation like that can be solved using infinite series.

First, note that the equation you suggested, 2^T + 3^T = 13, can be easily
solved for its only real root by inspection, giving T = 2.
(There are, however, infinitely many complex roots. One of them
is 2.57466... + i 5.99551...)
So let's try to solve a closely equation which _can't_ be done
by inspection, say,

2^T + 3^T = 13.1

We can guess (correctly) that its real root is a little bit larger than 2.
So let's expand 2^T + 3^T in a series about T = 2, invert that series (by a
process normally called series reversion), and then just plug 13.1 into
that inverse series to get the solution! Easier said than done, of course,
but using a computer algebra system makes it fairly easy...

The first three terms of the inverse series are
[Please view in a fixed-width font.]

| 1 4 log(2)^2 + 9 log(3)^2
| 2 + ------------------- (x - 13) - ------------------------ (x -13)^2
| 4 log(2) + 9 log(3) 2(4 log(2) + 9 log(3))^3

Terms quickly get messy after that, but let's see what we can do using just
those three terms to approximate. Setting x = 13.1, the numerical value of
the three terms is 2.00786733...; compare that with the equation's exact
solution, 2.00786749... The approximation was not bad considering
that just three terms were used.

I suspect that the infinite inverse series converges at x = 13.1; if that
suspicion is correct, then one may think of that series evaluated at 13.1
as being the precise solution. (But I don't know the coefficient of the
nth term of the inverse series...)

BTW, in case you're interested in the nonreal complex solutions to your
original equation, 2^T + 3^T = 13, it appears that their real parts are
always a bit larger than 2. As to their imaginary parts, I'll just note
that some of them are close to integer multiples of pi (such as 9pi, 11pi,
18pi, 20pi), which is perhaps not accidental.

David
.

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