Predict Digits of Irrational Numbers with Modular Arithmetic Summation?

[Chrono G. Xay]

Would it be possible to write an equation utilizing a summation of a modular function of a Cartesian function, whose degree is dependent upon the index of the root, in that it predicts the digits less than 1 of the root, that when summed equals the computed value sqrt( n )?

I already have what feels like a basis upon which to start,

[Summation; x=1, (inf)] \frac{(integer(f(x))mod10)}{10^x}

in that it would return such values as...

0.1 + 0.02 + 0.003 + 0.0004

but am having trouble coming up with an algorithm which would strip away, for example, the 1.0 from 1.732... in sqrt( 3 ) .

It seems as though even if such a thing were doable, it would have been nice to be able to say that, with the appropriate formulae for f(x), for one, it could compute the digits of any irrational number, except that for non-root irrational numbers, such as e, θ, φ, etc. the index is arguably one (given π is obtained from a quotient, and e is obtained from the limit of an exponential...)[/sup]

 

[Mark44]

but am having trouble coming up with an algorithm which would strip away, for example, the 1.0 from 1.732... in sqrt( 3 ) .

Use the floor function, also called the greatest integer function, and written as $\lfloor x \rfloor$. The greatest integer in x is the largest integer that is less than or equal to x. For example, $\lfloor 1.732 \rfloor = 1$.

To strip off the integer part, use $x -\lfloor x \rfloor$. Some programming languages have floor() and ceil() as part of their standard libraries.

 

[Chrono G. Xay]

Ok. I thought that might have been it, but at the same time it felt like it might have been incorrect because I would be using a term of sqrt( x ), similar to

y = 2x - | y | ...

Ok, so now we have

n^(1/m) = floor(n^(1/m)) + (Summation; x=1, (inf)] \frac{(integer(f(x))mod10)}{10^x} ...

Then again, shouldn't the greatest integer also be predictable by such an equation?