How do you solve #7^(x - 2) = 5x#?
The difference between the RHS and LHS changes sign in (0. .5).
0 appears to be closer to the zero of the equation.
Rearrange to the form
befits application of an iterative method, for successive
approximations, with a starter guess-value
Now, use the discrete analogue
This is of the form
The sequence of approximations is
0.00408... 0.00414... 0.004114 0.0041144..., for n = 1, 2, 3, 4, ., with
For x = 0.004144,
Despite that we get a good approximation to the solution of the given equation, the sequence tends to the solution of the discrete analogue and not the solution of the given equation, in mathematical exactitude. Yet, we get closer, when we advance, and the difference is bounded, with a limit that is numerically
This analysis is exclusive, for every such discrete analogue, in respect of every such equation..