Hint for Spivak's Calculus Problem 2-4-a

Question

(note: this is very similar to a related question but as I'm trying to solve it without looking at the answer yet, I hope the gods may humor me anyways)

I'm self-learning math, and an answer to an /r/math post about self-learning was finally enough to motivate me to try getting feedback on this site :) . After reading throughout the internet, I've decided to start with Spivak's Calculus. I'm loving the book thankfully, but I'm stuck on this problem and I don't want to look at the answer quite yet.

4 . (a) Prove that $$\sum_{k=0}^l \binom{n}{k} \binom{m}{l-k} = \binom{n+m}{l}$$. Hint: Apply the binomal theorem to $(1+x)^n(1+x)^m$ .

I've done all the prior problems, including Problem 3 (proving the Binomial Theorem) which is obviously closely tied to this, but even with the hint it feels like there's too much I don't know. I applied the hint and found:

$$\left(\sum_{i=0}^n \binom{n}{i} x^i\right)\left(\sum_{j=0}^m \binom{m}{j} x^j\right) = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k$$

And, setting $x=1$ got something even more interesting:

$$\left(\sum_{i=0}^n \binom{n}{i}\right)\left(\sum_{j=0}^m \binom{m}{j}\right) = \sum_{k=0}^{n+m} \binom{n+m}{k}$$

However, I don't know where to go after that...is there some property of multiplying sums that I need to prove first, to relate the multiplication of two sums of binomials with the sum of the multiplication of two binomials?

Thank you!

Answer 1

Hint:

Think about the sum as the coefficient of $x^l$ in the expansion ${(1+x)}^n{(x+1)}^m$

Answer 2

The key to using the below identity is to look at the coefficient of $x^l$ on both sides. Since the polynomials in $x$ are equal, their coefficients must be as well. $$\left(\sum_{i=0}^n \binom{n}{i} x^i\right)\left(\sum_{j=0}^m \binom{m}{j} x^j\right) = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k$$ The coefficient of $x^l$ of the summation on the right hand side is obviously $\binom{n+m}{l}$. The question is, what is the coefficient of the $x^l$ in the product of summations on the left?

It might help to look at a smaller example: say $n=2,m=3,l=3$. The left hand side of the above equation becomes $$ \left(\binom{2}0x^0+\binom21x^1+\binom22x^2\right) \left(\binom{3}0x^0+\binom31x^1+\binom32x^2+\binom33x^3\right) $$ Now, expand out this product, and collect terms with equal powers of $x$. The the $x^3$ term of the result is $$ \left(\binom22\binom31+\binom21\binom32+\binom20\binom33\right)x^3 $$ Therefore, this summation must be equal to the coefficient of $x^3$ in $\sum_{k=0}^{2+3}\binom{2+3}{k}x^k$, which is $\binom{2+3}3$.

Answer 3

Instead of setting $x=1$ in $$ \left(\sum_{i=0}^n \binom{n}{i} x^i\right)\left(\sum_{j=0}^m \binom{m}{j} x^j\right) = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k, $$ expand the LHS and collect terms of degree $k$ together to obtain $$ \sum_{k=0}^{n+m} \Bigg( \sum_{\substack{0\le i\le n\\0\le j\le m\\i+j=k}} \binom{n}{i} \binom{m}{j}\Bigg) x^k = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k. $$ Since $i+j=k$ determines $j=k-i$ this yields $$ \sum_{k=0}^{n+m} \Bigg( \sum_{i=0}^k \binom{n}{i} \binom{m}{k-i}\Bigg) x^k = \sum_{k=0}^{n+m} \binom{n+m}{k} x^k. $$ Comparing coefficients yields the desired identity.

Answer 4

The solution uses some facts about polynomials that, unfortunately, Spivak hasn't yet proven.

He hasn't formally introduced polynomials yet. This happens in Chapter 3 (3rd Ed.). You may wish to return to this problem after reading chapter 3.

A polynomial function $f$ (functions are also introduced in chapter 3) has the form: $$f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$$ where the $a_i$ coefficients are constants.

This polynomial is said to have degree $n$, where $n$ is the largest power of $x$ with a nonzero coefficient.

A number $x_i$ is said to be a root of the polynomial if $f(x_i) = 0$.

An $n$-th degree polynomial can have at most $n$ roots. That is, there can be no more than $n$ different numbers $x_1, x_2, \dots, x_n$ such that $f(x_i) = 0$

This fact comes from Chapter 3 exercise 7 (3rd Ed.)

Because an $n$ degree polynomial has at most $n$ distinct roots, if a polynomial is zero for all $x$ then all of its coefficients (the $a_i$'s) must be zero. (If any $a_i \neq 0$, this forces $f$ to have a finite number of roots, which contradicts $f$ being zero for all values of $x$.)

Restating, $$0 = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0 \text{ for all }x \text{ if and only if } a_n = 0, a_{n-1} = 0, \dots a_0 = 0$$

Closely related to this fact is the following: If two polynomials are equal for all values of $x$, they must have the same coefficients.

To write it out explicitly, suppose

$$f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$$ $$g(x) = b_nx^n + b_{n-1}x^{n-1} + \dots + b_1x + b_0$$ and $$f(x) = g(x) \text{ for all }x$$

Then $a_i = b_i$ for all $i = n, n-1, \dots, 1, 0.$

(You can show this by looking at $f-g$ and using what we previously learned about polynomials that are $0$ for all $x$.)

Now let's return to the exercise.

We have $$(1+x)^n(1+x)^m = (1+x)^{n+m}$$

We can expand each side using the binomial coefficients developed in preceding problems.

When we do so, we will have polynomials on the left hand side and right hand side of the equation. These polynomials will have different looking coefficients for each power of $x$.

However, using what we know about polynomials, we know that these different looking coefficients must actually be equal. The coefficient for $x^l$ on the LHS must equal the coefficient for $x^l$ on the RHS. You can use this fact to prove the desired result.

Edit: When $a < b$, $\binom{a}{b}$ is defined to be $0$. IIRC this is also used in the problem (despite not being defined in the text).