The equation can be written in more convenient form as...
(1)
... i.e in the form , and it can be solved iteratively using the Newton-Raphson method...
(2)
... starting from an arbitrary ...
Kind regards
can i solve it without using any numerical method? also, when you solve using this method, -sqrt(2) (which is a solution) will be eliminated.The equation can be written in more convenient form as...
that's true.I am positive that there are only two real solutions:
x_1=sqrt(2)
x_2=-sqrt(2)
The Newton's sequence for solving the equation...
(1)
... is expressed as...
(2)
The function is represented here...
If x has to be consider positive real, there is only one attractive fixed point at and for any the sequence converges to the solution. For x negative or in general complex the task is much more complex and the 'solution' found by Zarathustra, though perfectly correct, is symply among the complex solutions the only that has the imaginary part equal to 0...
Kind regards
and we can continue with as many
times as we want for to appear in those powers (an even number of times, though).
Taking limit we have . Therefore so
Generalize: for a constant number of times x appears in the power.
In case you want it more rigorously:
Let be defined by , where appears times in those powers
(where in the given equation we have appear times in the powers).
It's easy to show that for all .
is a sub-sequence of which is defined just like but for .
so it is also the limit of its sub-sequence .
Therefore is the solution to , which would be and possibly the negative of it, if it can be substituted in the original equation.
Note an interesting point here. From what we proved it follows that if we're talking in the domain then
converges only if is a solution to for some .
But using the usual invetigation methods of functions it follows that the maximal value of in
is when . That is, the maximal value is , so converges for only if
We see that is about as we would expect, as a result of the original problem.