Introduction#

Umbral calculus is a scary branch of mathematics that deals with sequences.

Prerequisites#

I’ll assume you know a little bit about what the following are:

• differentiation: $$\frac{\mathrm{d}}{\mathrm{d}x}f(x)$$
• integration: $$\int f(x)\ \mathrm{d}x$$
• how they are related: $$\frac{\mathrm{d}}{\mathrm{d}x}\int f(x)\ \mathrm{d}x = f(x)$$
• the binomial theorem: $$(a+b)^x = \sum_{n=0}^\infin \binom{x}{n} a^nb^{x-n}$$

Also, I’ll be using the following notation to accentuate the main ideas:

• differantiation: $$\frac{\mathrm{d}}{\mathrm{d}x}f(x) = \mathcal{D}f(x)$$
• integration: $$\int f(x)\ \mathrm{d}x = \mathcal{I}f(x)$$

Discrete Calculus#

In discrete calculus we only care about the discrete rate of change, in other words, what do we need to add to a member of a sequence to get the next member of the sequence.

Forward Difference#

Take for example $f(x) = x^2$. We have:

x	…	0	1	2	3	4	…
x^2	…	0	1	4	9	16	…

We can find the discrete rate of change of $f(x)$ by taking the difference between two consecutive terms of the sequence:

$$ \begin{align*} \Delta x^2 &= (x+1)^2 - x^2 \\ &= x^2 + 2x + 1 - x^2 \\ &= 2x + 1 \end{align*} $$

This is called the forward difference operator, and is defined as: $$\Delta f(x) = f(x+1) - f(x)$$

Notice that it’s similar to: $$\mathcal{D}f(x) = \lim_{h\rarr 0} \frac{f(x+h)-f(x)}{h}$$

It’s important to note that the forward difference operator is linear, meaning that: $$\Delta f(x) + \Delta g(x)=\Delta (f(x) + g(x)) $$ and $$\Delta af(x) = a\Delta f(x)$$

(try proving these yourself)

Some useful stuff: $$ \begin{align*} \Delta x^n &= \sum_{k=0}^{n-1} \binom{n}{k} x^k \\ \Delta n^x &= n^x(n-1) \\ \end{align*} $$ Notice that the $e$ of forward differences is $2$, since $\Delta 2^x = 2^x$.

Summation#

Notice that: $$ \begin{align*} x^2 &= (x-1)^2 + \Delta(x-1)^2 \\ &= (x-2)^2 +\Delta (x-2)^2 +\Delta(x-1)^2 \\ & \qquad \vdots \\ &= \sum ^{x-1}_{n=0} \Delta n^2 \end{align*} $$

More generally, we have: $$f(x) - f(a)= \sum ^{x-1}_{n=a} \Delta f(n)$$

And that looks a lot like the fundamental theorem of calculus: $$f(x) - f(a)= \int ^{x}_{a} \mathcal{D}f(n)\ \mathrm{d}n$$ SO LETS CALL THIS THE FUNDEMENTAL THEOREM OF DISCRETE CALCULUS!

Let us define the indefinite summation operator $\Sigma$ as: $$\Delta \Sigma f(x)=f(x)$$

Notice again that it’s kinda like: $$\mathcal{D}\mathcal{I}f(x)=f(x)$$

Let’s try to find: $$\sum_{n=0}^{x-1} n^2=f(x) \iff \Delta f(x)=x^2$$ Knowing

$$ \begin{align*} \Delta x = 1 \\ \Delta x^2 = 2x + 1 &\implies 2x = \Delta x^2-1 \\ \Delta x^3 = 3x^2 + 3x + 1&\implies 3x^2=\Delta x^3 - 3x^2-1 \ \end{align*} $$ From where we get $$x^2 = \Delta \left(\frac{x^3}{3} - \frac{x^2}{2} + \frac{x}{6}\right)$$

So

$$\sum_{n=0}^{x-1} n^2 = \frac{x^3}{3} - \frac{x^2}{2} + \frac{x}{6}$$

Falling Factorials/Powers#

Let us define the falling power $x_n$ with $x\in\mathbb N$ as: $$x_n = x(x-1)(x-2)\cdots(x-n+1) $$ also $$x_{-n} = \frac{1}{x(x+1)(x+2)\cdots(x+n)} $$ and $$x_0 = 1$$

It’s not hard to notice that:

$$ x_n = x_{n-1} (x-n+1) \qquad \qquad (x+1) _ n=(x+1) x _ {n-1} $$

And that:

$$\frac{x_n}{n!}= \binom{x}{n}$$

So

$$\Delta x_n = (x+1)n - x_n =nx{n-1}$$

This seem quite similar to $\mathcal{D}x^n = nx^{n-1}$.

Discrete to Continuous and back (Actual Umbral Calculus)#

This is a some good insight that leads us to define the operator $\phi$ that turns a power into a falling power: $$\phi x^n = x_n$$

Notice that: $$\phi \mathcal D x^n = \Delta \phi x^n$$

Also this is a good time to define the inverse of $\phi$ as $\phi^{-1}$. Let’s try to find some values for $\phi^{-1}$ (this is similar to how we got stuff for $x^2=\Delta \text{something}$ ). We know that:

$$ \begin{align*} \phi x = x &\implies \phi^{-1} x = x \\ \phi x^2 = x(x-1) = x^2 - \phi x &\implies \phi^{-1} x^2 = x^2 - x \\ \phi x^3 = x(x-1)(x-2) = x^3 - 3\phi x^2 + 2\phi x &\implies \phi^{-1} x^3 = x^3 - 3x^2 + 2x \\ \end{align*} $$

So let’s expand on this idea and define the operator $\phi$ such that: $$\phi \mathcal D f(x) = \Delta \phi f(x)$$

Or just: $$\phi \mathcal D = \Delta \phi$$

(keep in mind that this is not multiplication, it’s applying operators)

Similarly: $$\phi \mathcal I = \Sigma \phi$$

Notice that:

$$\Delta= \phi \mathcal D \phi^{-1} $$ and $$\Sigma = \phi \mathcal I \phi^{-1}$$

Notice that $\phi$ is linear, meaning that:

$$\phi (f(x) + g(x)) = \phi f(x) + \phi g(x)$$ and $$\phi (af(x)) = a\phi f(x)$$

Let’s try to use this to find a formula for: $$\sum_{n=0}^{x-1} n^2$$

We have:

$$ \begin{align*} \sum_{n=0}^{x-1} n^2 &= \phi \int_0^x \phi^{-1} n^2 \ \mathrm{d} n \\ &= \phi \int_0^x n^2+n \ \mathrm{d} n \\ &= \phi \left(\frac{x^3}{3} + \frac{x^2}{2}\right) \\ &= \frac{x^3}{3} - \frac{x^2}{2} + \frac{x}{6} \end{align*} $$

Let’s try to find some more stuff with $\phi$.

$$ \begin{align*} \phi e^{ax} &= \phi \sum_{n=0}^\infin \frac{a^nx^n}{n!} = \sum_{n=0}^\infin a^n\frac{x_n}{n!} = \sum_{n=0}^\infin \binom{x}{n}a^n = (1+a)^x \\ \phi \sin x &= \phi \frac{-i}{2}(e^{ix}-e^{-ix}) = \frac{-i}{2}((1+i)^x-(1-i)^x) = \sqrt{2^x}\sin(\pi x/4) \\ \phi \cos x &= \sqrt{2^x}\cos(\pi x/4) \end{align*} $$

If anyone says things like “oh u cant just put complex numbers in there and expect it to work” the appropriate response is “shut up nerd”.

NOW, something slightly harder:

We will attempt to find $\phi \ln x$.

$$\ln x = \mathcal I x^{-1} \iff \phi \ln x = \phi \mathcal I x^{-1}=\Sigma \phi x^{-1}= \Sigma x^{-1}=H_{x-1} $$

uhh ill wait for you to take all that in

so there are some more stuff but im actually too lazy ill do more stuff later

Taylor Series#

Remember the general taylor series formula: (with a=0)

$$f(x) = \sum_{n=0}^\infin \binom{x}{n}\mathcal D ^ nf(0) $$

Notice that: $$\begin{align*} \Delta = \phi \mathcal D \phi^{-1} &\implies \Delta^2=\phi\mathcal D^2\phi^{-1} \\ &\implies \Delta^n=\phi\mathcal D^n\phi^{-1} \\ &\iff \Delta^n \phi = \phi \mathcal D^n \end{align*}$$

From where we get:

$$\phi f(x) = \sum_{n=0}^\infin \binom{x}{n}\Delta ^ n \phi f(0) $$ And by substituting $\phi f(x)$ as $f(x)$ we get:

$$f(x) = \sum_{n=0}^\infin \binom{x}{n}\Delta ^ n f(0) $$

AND THATS NEWTONS FORWARD DIFFERENCE FORMULA!!!!!