Mathematical Proof of the Stable swap Equation

We start with two equations. Equation 1: $\quad \chi D^{n-1} \sum x_i + \prod x_i = \chi D^n + \left(\frac{D}{n}\right)^n$ Equation 2: $\quad \chi = \frac{A \prod x_i}{(D/n)^n}$

Step 1: Simplify Equation 2

Before plugging the messy $\chi$ fraction into Equation 1, let’s simplify it first. The denominator is $(D/n)^n$. Distributing the exponent gives $\frac{D^n}{n^n}$. Since this quantity is in the denominator of a larger fraction, we can invert it and move $n^n$ to the numerator:

$\chi = \frac{A \cdot (\prod x_i) \cdot n^n}{D^n}$

Now we have a cleaner expression for $\chi$ to use going forward.

Step 2: Plug $\chi$ into the left side of Equation 1

Let’s start with the left side of Eq. 1: $\chi \dot D^{n-1} \sum x_i + \prod x_i$. Substitute our simplified $\chi$:

$\left[ \frac{A \cdot (\prod x_i) \cdot n^n}{D^n} \right] \cdot D^{n-1} \sum x_i + \prod x_i$

Now group the two $D$ terms so we can simplify them:

$\frac{D^{n-1}}{D^n} \cdot \left( A \cdot (\prod x_i) \cdot n^n \sum x_i \right) + \prod x_i$

To simplify$\frac{D^{n-1}}{D^n}$, subtract exponents: $(n-1) - n = -1$. That gives $D^{-1} = \frac{1}{D}$. So the $D$ terms collapse into a single $D$ in the denominator.

New left side: $\frac{A \cdot (\prod x_i) \cdot n^n}{D} \sum x_i + \prod x_i$

Step 3: Plug $\chi$ into the right side of Equation 1

Start with the right side of Eq 1: $\chi D^n + \left(\frac{D}{n}\right)^n$. (To save a step, rewrite $(D/n)^n$ as $\frac{D^n}{n^n}$.

Right Side: $\chi D^n + \frac{D^n}{n^n}$

Substitute our clean $\chi$:

$\left[ \frac{A \cdot (\prod x_i) \cdot n^n}{D^n} \right] \cdot D^n + \frac{D^n}{n^n}$

This is straightforward: the $D^n$ in the denominator cancels with the $D^n$ we are multiplying by.

New Right Side: $A \cdot (\prod x_i) \cdot n^n + \frac{D^n}{n^n}$

Step 4: Add the Equals Sign back

Let's glue our New Left Side to our New Right Side.

$\frac{A \cdot (\prod x_i) \cdot n^n}{D} \sum x_i + \prod x_i \quad = \quad A \cdot (\prod x_i) \cdot n^n + \frac{D^n}{n^n}$

This equation is perfectly accurate! To get the final whitepaper formula, the author just cleans it up in the next two steps so there isn't a giant messy fraction on the left side.

Step 5: Clean-up (Multiply the whole equation by D)

To get rid of the $D$ on the bottom of the Left Side, let's multiply the entire equation by $D$.

Multiply the Left Side by D:

The $D$ on the bottom of the first part cancels out. The second part becomes $D \prod x_i$. Result: $A\cdot(\prod x_i) \cdot n^n \sum x_i + D \prod x_i$

Multiply the Right Side by $D$:

The first part just gets a $D$ added to the front. In the second part, $D^n \times D$= $D^{n+1}$. Result: $A \cdot D \cdot (\prod x_i) \cdot n^n + \frac{D^{n+1}}{n^n}$

Our current equation looks like this:

$A \cdot (\prod x_i) \cdot n^n \sum x_i + D \prod x_i \quad = \quad A \cdot D \cdot (\prod x_i) \cdot n^n + \frac{D^{n+1}}{n^n}$

Step 6: Final Clean-up (Divide the whole equation by $\prod x_i$)

Notice how there is a messy $(\prod x_i)$ floating around in almost every single term? Let's get rid of it by dividing the entire equation by $\prod x_i$.

Divide the Left Side by $\prod x_i$:

It perfectly cancels out in the first term, leaving $A n^n \sum x_i$. It perfectly cancels out in the second term, leaving just $D$. Final Left Side: $A n^n \sum x_i + D$

Divide the Right Side by $\prod x_i$:

It perfectly cancels out in the first term, leaving $ADn^n$. Because the second term doesn't have a $\prod x_i$ to cancel, it just drops to the bottom of the fraction, leaving $\frac{D^{n+1}}{n^n \prod x_i}$. Final Right Side: $ADn^n + \frac{D^{n+1}}{n^n \prod x_i}$

Glue those two final results together, and you have exactly derived the holy grail Stableswap equation:

$A n^n \sum x_i + D = ADn^n + \frac{D^{n+1}}{n^n \prod x_i}$