08/09/2014, 07:02 PM

(08/08/2014, 11:02 PM)tommy1729 Wrote: In post 19 I wrote :

quote:

5) I would like to comment that f(x) - f(x-1) can Always be approximated by

f(x) - ( f(x) - f ' (x) + f '' (x)/2 - f "' (x)/6 + f ""(x)/24 )

(end quote)

Jay tried to approximate the Original sequence(s) by the equation

f ' (x) = f(x/2)

A better approximation is probably

f (x) - f(x-1) = f(x/2)

It seems to hard at first , however

f(x) - f(x-1) can Always be approximated by

f(x) - ( f(x) - f ' (x) + f '' (x)/2 - f "' (x)/6 + f ""(x)/24 )

Therefore we can rewrite

f ' (x) - f " (x)/2 + ... = f(x/2).

These coefficients can be solved for by using the binomium expansion of the powers in the Taylor series for f(x-1).(...)

That's actually a pretty interesting approach. I'll have to try it out. I do have one concern though.

For integer k:

A(2k) = A(2k-1) + A(k)

A(2k+1) = A(2k) + A(k)

Depending on whether x is even or odd, the functional equation would look like:

even: f(x) - f(x-1) = f(x/2)

odd: f(x+1) - f(x) = f(x/2) or f(x) - f(x-1) = f(x/2 - 1/2)

In effect, the even terms will tend to be larger than the odd terms, compared to a smooth interpolation function. The reason isn't hard to grasp. Look at the relations I posted above. A(2k) and A(2k+1) are each A(k) larger than the previous term. Assuming a smooth function with a positive curvature (which is intuitively obvious for large k). The difference A(2k+1)-A(2k) should be larger than A(2k)-A(2k-1), but it's not. The net effect is that odd terms tend to underestimate the continuous estimator, and the even terms tend to overestimate it.

But of course it gets worse. If you look at the terms by comparing the index modulo 4, you'll find that the terms with k = 3 mod 4 will tend to be smaller than the terms with k = 1 mod 4. Indeed, this pattern continues indefinitely. The terms with k = 15 mod 16 tend to be smaller than the terms with k = 7 mod 16. The "largest" odds will be the terms with k = 1 mod 2^m for arbitrary m. The smallest odds will be the terms with k = -1 mod 2^m.

The evens are more complex to analyze, but in general, they tend to be much better behaved. But they still have their "cycles" at power of 2 intervals (for arbitrary powers of 2).

So in the end, the only way to get a continuous function that will calculate every term in the sequence, is to introduce cyclic functions with periods of 2, 4, 8, 16, etc. These cyclic functions will guarantee that the function becomes unbounded with large imaginary part. Perhaps that's not harsh enough an assessment. While the sequence itself grows subexponentially on the real line, it would grow at least exponentially in the imaginary direction.

With this in mind, I've given up trying to find a continuous function that interpolates every term in the sequence. I'd rather just use a concatenation of line segments, connecting the odd terms, with the even terms at the midpoint of each line segment.

The function f(x) which I described before provides a useful first-order approximation, once you scale it by the constant 1.083063... I'm still trying to find betters ways to determine the value of this constant.

I'm also working on a second-order approximation. This would consist of two parts. The first is to fix oscillations in the error term (1/1.083...)*f(k)-A(k). These oscillations follow approximately a scaled version of f(-x). Remember the zeroes on the negative real line?

The second aspect of the second-order approximation is to come up with a function (or functions) to help approximate the even and odd terms. In essence, I want to split the (2^m)-cyclic aspects of the sequence from the smooth continuous approximation function.

Getting back to your post, I feel like your idea may provide more insights into the smooth function f(x) (and it might help me figure out the large-scale oscillations, which do not follow an exact cycle), but I don't see it helping with the (2^m)-cyclic aspects of the even/odd issue.

~ Jay Daniel Fox