Fake math stuff

Question

Fake math stuff

kainui · Answer

Basically the point is since $\{x\} < 1$ this means larger and larger terms are less important as k increases and thought maybe this would be an interesting way to approximate functions.

ganeshie8 · Answer

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kainui · Answer

I was looking at some power series and was playing around with expanding it in terms of the integer and fractional part of a number, $x = \lfloor x \rfloor + \{x\}$.

So first I start here with the power series:

$$f(x) = \sum_{n=0}^\infty a_n x^n$$

Then I wonder, hey what will this look like? So I expand the binomial terms inside:

$$f(x) = \sum_{n=0}^\infty a_n (\lfloor x \rfloor + \{x\})^n$$

Looking at the terms individually  (less scary)

$$(\lfloor x \rfloor + \{x\})^n = \sum_{k=0}^n \binom{n}{k} \lfloor x \rfloor^k \{x\}^{n-k}$$

A little playing around and rearranging this sum in the other sum gets us:

$$f(x) = \sum_{k=0}^\infty \{x\}^k \sum_{n=0}^\infty a_{n+k} \binom{n+k}{k} \lfloor x \rfloor^n$$

kainui · Answer

This is where it gets interesting cause if you pull out the k=0 term we end up with this:

$$f(x) = f(\lfloor x \rfloor) +\sum_{k=1}^\infty \{x\}^k \sum_{n=0}^\infty a_{n+k} \binom{n+k}{k} \lfloor x \rfloor^n$$

So I got to thinking, if $g(x) = f(\lfloor x\rfloor )$ was really a number theory function then this would what it looks like when it's interpolated between points. 

We can tell that when $\{x\} = 0$ that the entire right part is 0 so that we have $f(x) = f(\lfloor x \rfloor)$ when x is an integer.

So I thought that's interesting, if I define this polynomial here:
$$p(x) = f(x) - f(\lfloor x \rfloor) =\sum_{k=1}^\infty \{x\}^k \sum_{n=0}^\infty a_{n+k} \binom{n+k}{k} \lfloor x \rfloor^n$$

Then we automatically know that p(x) has this form:

$$p(x) = q(x)*\prod_{r=1}^\infty (x-r)$$

since we require for all positive numbers r that they're roots $p(r)=0$ since that's where the interpolating function and the number theory function are equal.

kainui · Answer

>>> into the trash

astrophysics · Answer

Looks neat, but not sure what I'm looking at yet haha, I do want to understand it though, by the way what does {x} mean, I see the other one is a floor function I believe but the squiggly brackets imply what?

kainui · Answer

Haha yeah well it's fake math cause it's definitely divergent if I write that product of roots, so something's fishy.

You can take that to be the definition of {x} actually,

$$x-\lfloor x \rfloor = \{x \}$$