# Calculus of Finite Differences Part-1

*Achyut Bharadwaj, Saee Patil, Valentio Iverson, Danny Li*

##### August 2021

# Acknowledgements #

This work was carried out as part of a research project at PROMYS-2021. I am a grateful recipient of the Mehta Fellowship to the PROMYS programs in 2021 and 2022.

# Preliminary Definitions #

## Finding Patterns #

Here’s a harder one:

## Notations #

What is $\Delta\left(f\right)$ when $f$ is $1,4,7,10,\cdots$?

What is $\Delta\left(f\right)$ when $f$ is $3,4,6,9,13\cdots$?

What is $\Delta^2(f)$ when $f$ is $3,4,6,9,13\cdots$?

# Some Results #

$\Delta\left(f\right)$ is a polynomial iff $f$ is a polynomial.

Let

## $\Delta^{-1}$ #

We define $\Delta^{-1}(f(x))$ as a function $g(x)$ such that $\Delta\left(g(x)\right) = f(x)$.

## Patterns again! #

## Generalizing #

Observe that

The above becomes:

So, we have

Hence,

# Calculus vs Finite Differences #

## Addition and multiplication under a finite difference #

## Multiplication of finite difference #

We now try to find a generic formula for the product of two functions under a finite difference.

## Division under a finite difference #

Before trying to find a formula for division, we try to find one for the multiplicative inverse, i.e. finding $\Delta\left(\frac{1}{f(x)}\right)$ in terms of $f(x)$

## Chain Rule #

Evaluate $\Delta\left(f(x)\right)$ when $f(x)=(2x)^2$

$(2x)^2 = 4x^2$. Now, $\Delta\left(4x^2\right) = 4\Delta\left(x^2\right) = 4(2x+1) = 8x+4$.

We compare this with the calculus chain rule. Is it possible to let $u=2x$, find $\Delta\left(f(u)\right)$, and multiply it by $\Delta u$ to get the result as we do in calculus?

We think of the function $\Delta$ as finding the slope of a curve between two integral points.

So, we can conclude that

We need

We know that

In calculus however, the difference is not finite. Hence, the above holds.

## Some fun with $\Delta^{-1}$ #

Find $\Delta^{-1}(x + 1)$. Can you find $\Delta^{-1}(f(x)+g(x))$ in terms of $f(x),g(x)$?

We know

## Addition under $\Delta^{-1}$ #

We know that

Also,

Equating the LHS of the above two equations, we get:

# Generalizing #

## A motivating question #

We are interested in the following question.

*Can we find a polynomial that fits a certain given data?*

Let us consider the following example. Can we find a quadratic polynomial $f$ such that

A motivating question indeed, and the answer is **no**. Suppose there exists
such quadratic polynomial $f$, then

Consider a variant of the question before: Can we find a cubic polynomial $f$ such that

Since

Can we generalize this? Therefore, we began to ask the following interesting question:

Given the values of $f(0), f(1), \dots, f(n)$, can we find a polynomial $f$ of degree not greater than $n$ that fits this data? Is this polynomial unique?

And the answer is yes to both questions.

## Finding a polynomial that passes through $n$ points #

**Claim.** For any
$n,k, i \le k$, we have

We will prove this by induction on $i$. Indeed, this is true for $i = 1$, since

Indeed,

Every polynomial $P(x)$ can be uniquely represented as

Using this fact, note that if $P(x) \not= 0$, then

The previous argument proves the existence of such a polynomial passing through $f(0), f(1), f(2), \dots, f(n)$, ensuring the degree is not greater than $n$. In fact, we claim an even more generalized result.

Let $x_1, x_2, \dots, x_n$ be $n$ distinct real values and
$y_1, y_2, \dots, y_n$ be $n$ real numbers. Then there exists a *unique*
polynomial $f \in \mathbb{R}[x]$ of degree less than $n$ such that
$f(x_i) = y_i$ for $1 \le i \le n$.

## Proof #

*Proof.* We’ll first prove that there exists such polynomial.
Consider the polynomial