[QUOTE=davar55;375056]Don't know about the insult, but my post was just a comment.
How can a seventh degree polynomial be flat over an interval?
So perhaps a larger view would show up the actual non-flat curvature?[/QUOTE]

It's not that it is "flat" (i.e. has zero first derivative) over that *entire* interval, but that it is *nearly* "flat" (i.e. has a very small magnitude derivative) over that interval. You would readily see this behavior more clearly on a zoomed-in view (try graphing the function on a calculator and playing with the window size).

All odd polynomial functions exhibit this behavior, in fact. Take [TEX]f(x)=x^3[/TEX]. Near [TEX]x = 0[/TEX], the graph of this function appears "flat". Why? Well, [TEX]f'(x)=3x^2[/TEX]. So, you see that at [TEX]x=0[/TEX], indeed, the function has zero derivative, and hence is "flat". But even as far out as, say, [TEX]x=0.25[/TEX], the derivative is still near enough to 0 for the function to *appear* flat in a larger-range window.

No insult intended, by the way. I suppose I was just taken somewhat aback at your elementary question, given that (IIRC) you have some degree of mathematical credentials. Call it a "Bob moment"...

Also, what I thought might be inaccurate was not the flatness
but the program that generated the graph. But I see now it
was just a matter of the size of the graph outputted.
In this situation, size matters.

[QUOTE=davar55;375033]In that graph, for x in (-.25,+.25) the y-value looks flat at -1.0.

Could this be wrong?[/QUOTE]

True, I spoke too hastily - clearly f`(x) = 1701*x^6 + 96*x^5 + 5*x^4 = 0 at x = 0. This leads to our next pair of fun homework problems:
[i]
True or False: Under the assumption of infinite numerical precision, Newton-Raphson on f(x) always converges to the root as long as the initial iterate x0 != 0. [Hint: consider whether an iterative-update sequence can start at x != 0 and land at x = 0 at some later iterate.]

True or False: Under the assumption of infinite numerical precision, Halley's method applied to f(x) always converges to the root irrespective of the choice of the initial iterate.[/i]

C'mon, you guys - at least the 2nd part of the above problem is trivially easy....

[spoiler]just examine the denominator which appears in Halley's iterative-update formula and see if it can = 0 or not for the function in question.[/spoiler]

On that graph, 4 squares is 1, so a square is 0.25, and 0.25^7 is 0.000061, or 1/4096 of a square. So, it is "totally flat" :razz: for plus and minus a square around x=0.

Well, if you have a monitor with 4096x8192 resolution, or 6 hi-res monitors, or about 24 "normal monitors" connected in such a way that the image "spreads" on all of them (each shows a different part of the image, like on stadiums), and enlarge (magnify) that graphic SO MUCH until ONE SINGLE square of it is shown on your (assembly of) monitor(s), and look to the interval from (-0.25, 0.25) on horizontal and (-1.5, -0.5) on vertical, then you may be able to see the graph of the line raising ONE SINGLE PIXEL, in the rightmost part of your (assembly of) screen (s). :rofl:. And of course, the line will drop one pixel, on the leftmost part of the screen.

First derivative near x=0:

I am not arguing with you, my post was for davar&co. About your derivative graph, it still has exactly the same "aspect", no matter how much you zoom in. If you change the Ox labels from +/-0.25 to +/-0.000001, and modify accordingly on Oy, it will still look exactly the same.

[QUOTE=ewmayer;375082][I]
True or False: Under the assumption of infinite numerical precision, Newton-Raphson on f(x) always converges to the root as long as the initial iterate x0 != 0. [Hint: consider whether an iterative-update sequence can start at x != 0 and land at x = 0 at some later iterate.]
[/I][/QUOTE]
For anyone interested in the theory behind this, the key is the Banach fixed-point theorem
[URL]http://en.wikipedia.org/wiki/Contraction_mapping_theorem[/URL]
In practice, this can take a while to converge, and Newton-Raphson is an optimization.

Applied to spaces of continuous functions instead, the Banach fixed-point theorem underlies many proofs of the existence (and uniqueness) of solutions to differential equations.

Thanks for the link, but note that the kinds of situations which are problematic for Newton-style iterative schemes are just those which break the assumption of a contraction mapping which underpins the BFPT. Depending on the details of the scheme, function to which applied and initial iterate one can have iterates which shoot off arbitrarily far toward +-oo, or which bounce around either deterministically (limit cycle) or chaotically.

The BFPT is - roughly speaking - essentially a proof of convergence once one is within the "contiguous basin of attraction" of a fixed point.

Re the OP:

[QUOTE]
...
I have an approximation for the real root, but I want an exact formula for it, even if it includes an infinite series:
...
[/QUOTE]Since the particular seventh degree polynomial equation is irreducible,
by Galois it has no closed form (by radicals) solution. I wonder what
is meant by an exact formula solution that includes an infinite series
whose convergent is either expressible in radicals (reducing to the
reducible case) or not (in which case not exact.) Or is it me who's
a bit confused?

By exact, I mean something like this:

[TEX]\sum\limits_{i=a}^\infty f(i)[/TEX]

where f(i) is a function of i such that the above sum converges to the root. If you need to, use nested sums.