Taylor Polynomial Of Product Fg: Proof And Explanation

Nov 1, 2025 by SLV Team 55 views

Proving the Taylor Polynomial of a Product: $fg$

Hey guys! Let's dive into a fascinating topic in real analysis: Taylor polynomials. Specifically, we're going to tackle the proof that the nth Taylor polynomial of a product of two functions, let's say fg, is actually the product of their individual Taylor polynomials, but only up to the nth power of (x - x₀). This might sound a bit complex at first, but trust me, we'll break it down step by step so it's super clear. This problem comes from Trench's "Introduction to Real Analysis" (exercise 2.5.10a), and it's a classic example that really solidifies your understanding of Taylor expansions.

Understanding Taylor Polynomials

Before we jump into the proof, let's make sure we're all on the same page about what a Taylor polynomial actually is. Imagine you have a function, let's call it f(x), and you want to approximate its value near a specific point, x₀. A Taylor polynomial does exactly that! It gives you a polynomial approximation of your function around that point. The more terms you include in the polynomial, the better the approximation becomes – at least, close to x₀. The nth Taylor polynomial, denoted as Fₙ(x), is given by:

Fₙ(x) = f(x₀) + f'(x₀)(x - x₀) + (f''(x₀)/2!)(x - x₀)² + ... + (fⁿ(x₀)/n!)(x - x₀)ⁿ

Where:

f(x₀) is the value of the function at x₀.
f'(x₀), f''(x₀), ..., fⁿ(x₀) are the first, second, and nth derivatives of the function evaluated at x₀.
n! is the factorial of n (i.e., n! = n × (n-1) × (n-2) × ... × 2 × 1).

So, you can see that the Taylor polynomial is built using the function's derivatives at a single point. This is what allows it to approximate the function's behavior locally. Now, with that in mind, let's think about what happens when we have two functions, f(x) and g(x), and we multiply them together. What about the Taylor polynomial of their product? That's the puzzle we're going to solve today.

Setting up the Problem: Taylor Polynomials of f, g, and fg

Okay, let's get formal for a moment and set up the problem properly. We're given two functions, f(x) and g(x), and we're interested in their Taylor polynomials around a point x₀. As the problem suggests, let's denote the nth Taylor polynomials of f(x) and g(x) as Fₙ(x) and Gₙ(x), respectively. Similarly, let Hₙ(x) represent the nth Taylor polynomial of the product function h(x) = f(x)g(x). Remember, our goal is to prove that Hₙ(x) is equal to the product of Fₙ(x) and Gₙ(x), but only when we truncate the product to include terms up to the nth power of (x - x₀). In other words, we're not keeping any terms with powers higher than n.

To make things even clearer, let's write out the general forms of these Taylor polynomials:

Fₙ(x) = ∑[k=0 to n] (f⁽ᵏ⁾(x₀) / k!) * (x - x₀)ᵏ
Gₙ(x) = ∑[k=0 to n] (g⁽ᵏ⁾(x₀) / k!) * (x - x₀)ᵏ
Hₙ(x) = ∑[k=0 to n] (h⁽ᵏ⁾(x₀) / k!) * (x - x₀)ᵏ

Where f⁽ᵏ⁾(x₀), g⁽ᵏ⁾(x₀), and h⁽ᵏ⁾(x₀) represent the kth derivatives of f, g, and h, respectively, evaluated at x₀. The summation notation simply means we're adding up terms for k = 0, 1, 2, ..., n. Now, the core of the problem lies in understanding how the derivatives of the product h(x) = f(x)g(x) relate to the derivatives of f(x) and g(x) individually. This is where the Leibniz rule comes into play. This rule is crucial for expanding the derivatives of a product, and it's the key to unlocking this proof. So, let's take a look at the Leibniz rule and see how it helps us.

The Leibniz Rule: Derivatives of a Product

The Leibniz rule (sometimes called the general product rule) is a powerful tool for finding the nth derivative of a product of two functions. It's like the product rule you learned in basic calculus, but extended to higher-order derivatives. The rule states that if we have two functions, u(x) and v(x), then the nth derivative of their product is given by:

(uv)⁽ⁿ⁾ = ∑[k=0 to n] (n choose k) * u⁽ᵏ⁾ * v⁽ⁿ⁻ᵏ⁾

Where:

(uv)⁽ⁿ⁾ denotes the nth derivative of the product uv.
u⁽ᵏ⁾ and v⁽ⁿ⁻ᵏ⁾ are the kth and (n-k)th derivatives of u and v, respectively.
(n choose k) is the binomial coefficient, which is calculated as n! / (k! * (n-k)!).

This formula might look a bit intimidating, but let's break it down. Essentially, it says that the nth derivative of the product is a sum of terms. Each term involves a product of derivatives of u and v, and a binomial coefficient. The binomial coefficient tells you how many ways you can choose k derivatives from u and (n-k) derivatives from v. The key is that the sum of the orders of the derivatives in each term always adds up to n. For example, let's consider the first few terms:

When k = 0: (n choose 0) * u⁽⁰⁾ * v⁽ⁿ⁾ = 1 * u * v⁽ⁿ⁾
When k = 1: (n choose 1) * u⁽¹⁾ * v⁽ⁿ⁻¹⁾ = n * u' * v⁽ⁿ⁻¹⁾
When k = 2: (n choose 2) * u⁽²⁾ * v⁽ⁿ⁻²⁾ = (n(n-1)/2) * u'' * v⁽ⁿ⁻²⁾

And so on. You can see how the derivatives shift between u and v, and how the binomial coefficients come into play. Now, how does this help us with our Taylor polynomial problem? Well, we can use the Leibniz rule to find the derivatives of h(x) = f(x)g(x), which are needed to construct the Taylor polynomial Hₙ(x). This is the crucial link that connects the Taylor polynomials of f, g, and their product. Let's apply the Leibniz rule to our problem and see what happens!

Applying the Leibniz Rule to h(x) = f(x)g(x)

Alright, let's put the Leibniz rule into action! We want to find the nth derivative of h(x) = f(x)g(x), so we can plug it into the formula for the Taylor polynomial Hₙ(x). Using the Leibniz rule, we have:

h⁽ⁿ⁾(x) = ∑[k=0 to n] (n choose k) * f⁽ᵏ⁾(x) * g⁽ⁿ⁻ᵏ⁾(x)

This is the nth derivative of h(x) in terms of the derivatives of f(x) and g(x). Now, remember that we need to evaluate this derivative at the point x₀ to construct the Taylor polynomial. So, let's plug in x₀:

h⁽ⁿ⁾(x₀) = ∑[k=0 to n] (n choose k) * f⁽ᵏ⁾(x₀) * g⁽ⁿ⁻ᵏ⁾(x₀)

This gives us the nth derivative of h(x) evaluated at x₀. We can now substitute this into the formula for the Taylor polynomial Hₙ(x):

Hₙ(x) = ∑[n=0 to n] (h⁽ⁿ⁾(x₀) / n!) * (x - x₀)ⁿ
Hₙ(x) = ∑[n=0 to n] (1/ n!) * [∑[k=0 to n] (n choose k) * f⁽ᵏ⁾(x₀) * g⁽ⁿ⁻ᵏ⁾(x₀)] * (x - x₀)ⁿ

Whoa, that looks like a mouthful! But don't worry, we're getting closer. This equation expresses Hₙ(x) in terms of the derivatives of f and g at x₀. Now, let's take a step back and think about what we're trying to prove. We want to show that Hₙ(x) is the product of Fₙ(x) and Gₙ(x), truncated to the nth power of (x - x₀). This means we need to compare this expression for Hₙ(x) with the product Fₙ(x)Gₙ(x) and see if they match up to the terms with (x - x₀)ⁿ. So, let's calculate the product Fₙ(x)Gₙ(x) and then carefully compare the coefficients.

Calculating Fₙ(x)Gₙ(x) and Comparing Coefficients

Okay, let's roll up our sleeves and calculate the product of the Taylor polynomials Fₙ(x) and Gₙ(x). We have:

Fₙ(x) = ∑[i=0 to n] (f⁽ⁱ⁾(x₀) / i!) * (x - x₀)ⁱ
Gₙ(x) = ∑[j=0 to n] (g⁽ʲ⁾(x₀) / j!) * (x - x₀)ʲ

So, their product is:

Fₙ(x)Gₙ(x) = [∑[i=0 to n] (f⁽ⁱ⁾(x₀) / i!) * (x - x₀)ⁱ] * [∑[j=0 to n] (g⁽ʲ⁾(x₀) / j!) * (x - x₀)ʲ]

When we multiply these two sums, we're going to get a lot of terms. We need to be organized to keep track of them. Think about it like multiplying two polynomials – you multiply each term in the first polynomial by each term in the second polynomial. The key thing to remember is that we're only interested in terms up to the nth power of (x - x₀). Any terms with higher powers will be truncated (i.e., we'll ignore them). To find the coefficient of (x - x₀)ⁿ in the product, we need to consider all pairs of terms where the powers of (x - x₀) add up to n. That is, we need to consider all pairs (i, j) such that i + j = n. So, let's write out the coefficient of (x - x₀)ⁿ in the product Fₙ(x)Gₙ(x). It will be a sum of terms of the form:

(f⁽ⁱ⁾(x₀) / i!) * (g⁽ʲ⁾(x₀) / j!)

Where i + j = n. We can rewrite j as n - i, so the coefficient of (x - x₀)ⁿ becomes:

∑[i=0 to n] (f⁽ⁱ⁾(x₀) / i!) * (g⁽ⁿ⁻ⁱ⁾(x₀) / (n - i)!)

Now, let's multiply this by n!:

n! * ∑[i=0 to n] (f⁽ⁱ⁾(x₀) / i!) * (g⁽ⁿ⁻ⁱ⁾(x₀) / (n - i)!) = ∑[i=0 to n] [n! / (i! * (n - i)!)] * f⁽ⁱ⁾(x₀) * g⁽ⁿ⁻ⁱ⁾(x₀)

Notice anything familiar? The term n! / (i! * (n - i)!) is just the binomial coefficient (n choose i)! So, we can rewrite the coefficient as:

∑[i=0 to n] (n choose i) * f⁽ⁱ⁾(x₀) * g⁽ⁿ⁻ⁱ⁾(x₀)

Guess what? This is exactly the same expression we got for h⁽ⁿ⁾(x₀) when we applied the Leibniz rule! This is the crucial step in the proof. We've shown that the coefficient of (x - x₀)ⁿ in the Taylor polynomial Hₙ(x) is the same as the coefficient of (x - x₀)ⁿ in the product Fₙ(x)Gₙ(x). This holds true for all powers of (x - x₀) up to n. Therefore, we've proven that the nth Taylor polynomial of the product fg is indeed the product of the Taylor polynomials of f and g, truncated to the nth power of (x - x₀).

Conclusion: We Did It!

Woohoo! We made it! We've successfully proven that the nth Taylor polynomial of the product of two functions is the product of their individual Taylor polynomials, truncated to the nth power. This is a pretty neat result, and it highlights the power of Taylor polynomials in approximating functions and their products. The key takeaways from this proof are:

Understanding the definition of a Taylor polynomial.
Knowing how to apply the Leibniz rule for finding derivatives of products.
Carefully comparing coefficients to show the equality.

This problem from Trench's "Introduction to Real Analysis" is a great exercise for solidifying your understanding of these concepts. So, next time you're working with Taylor polynomials, remember this result – it can be a real time-saver! Keep exploring the fascinating world of real analysis, guys, and you'll uncover even more amazing mathematical insights. This stuff might seem abstract now, but it forms the foundation for many applications in physics, engineering, and computer science. So, keep up the great work!