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Infinitely Continued III (Posted on 2010-02-04) Difficulty: 3 of 5
Evaluate:

√(6+2√(7+3√(8+4√(9+5√(10+ ....)))))

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 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 youíre right Steve, f(x) = x, and Iíve 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 itís possible to show that all derivatives higher than the first are zero at integer values of x, so Taylorís expansion would show f(x) = x for all x >= 0.

Perhaps computerers can explain this..
Iíve used the following Maple code to work out Jerís 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)+ ..Ö))))

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

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