God Himself cannot solve this problem?

Question

Let the function f(x,n,j) equal 1 when for a sequence of integers x of length n there is also a sequence of the same integer of length n+1 contained in some real number j and zero when this is not true. (For example, f(9,2,pi)=1 because there is a sequence of 2 9s at about the hundreth decimal place.)

Let the function f(x,n,j) be given in the form of a triple infinte series:
__________ n n n
f(x,n,j) = lim ∑ ∑ ∑a(x,q) + b(n,w) + c(j,e),
______ n →∞ q=1 w=1 e=1

where the varibles q, w and e are indicies of summation. Let the functions a, b and c be continuous everywhere.

(a) Is it true that

lim ∂f/∂j = k, k < ∞ ?
j→∞

(b) Is it true that for intervals (0, j) , (j+1, j+2), etc, that the following inequalities are always true?
j
∫f(x,j,n) dj <
0

2j
∫f(x,j,n) dj < ... n→∞
j

i had to use "______" over and under f(x,n,j) to keep the indicies in place when i submitted the problem.

Answer 1

You need to 'tighten up' your 'definition' of the function
f(x,n,j) since you don't show how its definition causes it to have any of the characteristics you've given it.

And, Hey! Maybe it *is* Gödel Undecideable. In which case no, not even God Himself could solve it ☺


Doug