If f(m, n) = (1-x^m)(1-x^{m-1})...(1-x^{m-n+1}) / (1-x)(1-x²)...(1-xⁿ) where m and n are positive in

Q: Why can't we just expand the products and cancel terms? It seems like a lot of work to factor things out.

That is a classic trap! While you *could* expand everything, you would be drowning in a sea of terms. The beauty of this problem lies in recognizing the recursive structure. By factoring, you are essentially 'peeling the onion' layer by layer, which is much more elegant and less prone to calculation errors.

Q: What is a $q$-analog, and why does it matter?

A $q$-analog is a mathematical generalization that recovers the original object when a parameter (in this case, $x$) approaches 1. As $x \to 1$, $f(m, n)$ becomes the standard binomial coefficient $\binom{m}{n}$. Understanding this helps you predict the answer before you even start the algebra!

Q: Wait, why do we reject the negative root here? Or is there no root involved?

You are absolutely right to be cautious! In this specific problem, we are dealing with a functional identity, not a quadratic equation, so there are no roots to reject. We are simply proving an equality between two expressions. Always check if you are solving for a variable or proving an identity—it changes your entire strategy!

Question

If $f(m, n) = \frac{(1-x^m)(1-x^{m-1})\dots(1-x^{m-n+1})}{(1-x)(1-x^2)\dots(1-x^n)}$ where $m$ and $n$ are positive integers ($n \le m$), show that $f(m, n+1) = f(m-1, n+1) + x^{m-n-1} f(m-1, n)$.

Sigma Percentile · Accepted Answer

### The Elegance of the Gaussian Binomial Coefficient

Welcome, future engineers. Today, we are not just solving an algebraic identity; we are stepping into the world of combinatorics and $q$-calculus.

The function $f(m, n)$ you see before you is known as the Gaussian Binomial Coefficient. It is a profound generalization of the standard binomial coefficient $\binom{m}{n}$.

When you see expressions like

$$f(m, n) = \frac{(1-x^m)(1-x^{m-1})\dots(1-x^{m-n+1})}{(1-x)(1-x^2)\dots(1-x^n)}$$

do not let the product notation intimidate you. It is merely a structured way of counting, and today, we will prove its recursive nature.

### Phase 1: Defining the Target

Our objective is to prove the identity $f(m, n+1) = f(m-1, n+1) + x^{m-n-1} f(m-1, n)$. To begin, we must clearly define our target, $f(m, n+1)$.

By substituting $n 	o n+1$ into our definition, we get:

$$f(m, n+1) = \frac{(1-x^m)(1-x^{m-1})\dots(1-x^{m-n})}{(1-x)(1-x^2)\dots(1-x^{n+1})}$$

Notice the denominator now extends to $(1-x^{n+1})$. This is our destination. We need to manipulate the Right Hand Side (RHS) until it matches this form perfectly.

### Phase 2: The Decomposition of the RHS

Now, let us look at the RHS: $f(m-1, n+1) + x^{m-n-1} f(m-1, n)$. This looks like a daunting sum of two complex fractions.

However, the secret to JEE Advanced problems is often finding the 'hidden' commonality. Let us expand these terms individually.

For $f(m-1, n+1)$, we substitute $m 	o m-1$ and $n 	o n+1$:

$$f(m-1, n+1) = \frac{(1-x^{m-1})(1-x^{m-2})\dots(1-x^{m-n-1})}{(1-x)(1-x^2)\dots(1-x^{n+1})}$$

For $f(m-1, n)$, we simply substitute $m 	o m-1$:

$$f(m-1, n) = \frac{(1-x^{m-1})(1-x^{m-2})\dots(1-x^{m-n})}{(1-x)(1-x^2)\dots(1-x^n)}$$

### Phase 3: The Algebraic Dance

Here is where the magic happens. If you compare $f(m-1, n+1)$ and $f(m-1, n)$, you will notice they are almost identical.

Specifically, $f(m-1, n+1)$ contains an extra factor in the numerator $(1-x^{m-n-1})$ and an extra factor in the denominator $(1-x^{n+1})$. This gives us a beautiful relationship:

$$f(m-1, n+1) = f(m-1, n) \cdot \frac{1-x^{m-n-1}}{1-x^{n+1}}$$

Now, substitute this into our RHS expression. We can factor out the common term $f(m-1, n)$:

$$	ext{RHS} = f(m-1, n) \left[ \frac{1-x^{m-n-1}}{1-x^{n+1}} + x^{m-n-1} ight]$$

### Phase 4: The Grand Unification

We are in the home stretch. Let us simplify the expression inside the brackets by taking the Least Common Multiple (LCM), which is $(1-x^{n+1})$:

$$	ext{Bracket} = \frac{1-x^{m-n-1} + x^{m-n-1}(1-x^{n+1})}{1-x^{n+1}}$$

Expanding the numerator, we get $1 - x^{m-n-1} + x^{m-n-1} - x^{(m-n-1) + (n+1)}$. Notice how the exponents add up: $(m-n-1) + (n+1) = m$.

The terms $-x^{m-n-1}$ and $+x^{m-n-1}$ cancel out perfectly, leaving us with:

$$	ext{Bracket} = \frac{1-x^m}{1-x^{n+1}}$$

Finally, we multiply this back with our common factor $f(m-1, n)$:

$$	ext{RHS} = f(m-1, n) \cdot \frac{1-x^m}{1-x^{n+1}}$$

When you write out the full product for $f(m-1, n)$ and multiply by this fraction, the terms align perfectly to reconstruct $f(m, n+1)$. We have arrived at our destination.

The identity is **proven**. Remember, in mathematics, as in life, the most complex problems often yield to the simplest, most elegant truths if you are patient enough to find the pattern.

Visualized Solution

Understanding the Function $f (m, n)$

Defining the Target Term $f (m, n + 1)$

Expanding the First RHS Term

Expanding the Second RHS Term

Relating the RHS Terms

Setting Up the RHS Expression

Factoring Out the Common Term

Taking the LCM Inside the Bracket

Expanding the Numerator

Simplifying the Bracket

Recombining with the Common Factor

Final Proof and Conclusion

Key Takeaways and Pascal's Identity

The Sigma Insight: Properties of Binomial Coefficients

The Elegance of the Gaussian Binomial Coefficient

Phase 1

Phase 2

Phase 3

Phase 4

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