 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Infinitely Continued III (Posted on 2010-02-04) Evaluate:

√(6+2√(7+3√(8+4√(9+5√(10+ ....)))))

 See The Solution Submitted by K Sengupta Rating: 4.5000 (2 votes) Comments: ( Back to comment list | You must be logged in to post comments.) re(3): Is my reasoning valid? | Comment 6 of 7 | (In reply to re(2): Is my reasoning valid? Spoiler by Steve Herman)

Yes, I think you�re right Steve, f(x) = x, and I�ve probably been expecting too much of my Maple plot function.

In fact, the Jer function is clearly continuous and differentiable (x>0), and I believe it�s possible to show that all derivatives higher than the first are zero at integer values of x, so Taylor�s expansion would show f(x) = x for all x >= 0.

Perhaps computerers can explain this..
I�ve used the following Maple code to work out Jer�s function as far as the 20th radical as an approximation to f(x).

f(x) = sqrt((x+2)+(x-2)sqrt((x+3)+(x-1)sqrt((x+4)+xsqrt((x+5)+ ..�))))

Code
f := proc (x) local n, p, s, a;
n := 20; s := 0;
for a from x+n by -1 to x+2 do
p := sqrt(a+s);
s := (a-5)*p;
end do;
p
end proc;

For every value of x (>0) that I try, this gives a value very close to x (for example, f(3.8)=3.799998630), supporting the theory that f(x) = x.
So why does plot(f,0..) show a graph that is jumping between the lines y = x and y = x + 1? The jumps are at irregular intervals and the plot remains on each line for a while before jumping back again.

 Posted by Harry on 2010-02-06 13:00:27 Please log in:

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