Nth Derivative Of Arctan(x): Finding A Closed Form Solution

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Nth Derivative of arctan(x): Finding a Closed Form Solution

Hey guys! Today, we're diving deep into the fascinating world of calculus to tackle a problem that might seem daunting at first: finding a closed form for the nth derivative of f(x) = arctan(x). This isn't just a theoretical exercise; understanding higher-order derivatives is crucial in various fields, from physics to engineering, where we often need to model complex systems and their behavior. So, grab your thinking caps, and let's get started!

Understanding the Problem

Before we jump into the solution, let's make sure we all understand the core of the problem. We're given the function f(x) = arctan(x), which is the inverse tangent function. Our goal is to find a general formula, a closed form, for its nth derivative. This means we want an expression that tells us what f(n)(x) is for any positive integer n. Now, you might be thinking, "Why can't we just keep differentiating?" Well, you certainly could, but you'd quickly find that the expressions become increasingly complex and unwieldy. A closed form, on the other hand, provides a neat and compact representation, making it much easier to work with.

Why is this important?

You might be wondering, "Okay, but why should I care about the nth derivative of arctan(x)?" Great question! Higher-order derivatives pop up in numerous applications. For example, in physics, they can describe the rate of change of acceleration (known as jerk) or even higher-order rates. In engineering, they're used in control systems, signal processing, and more. Finding a closed form makes these calculations much more manageable. Plus, the process of finding this closed form is a fantastic exercise in mathematical problem-solving, sharpening your skills in calculus and mathematical induction.

The Proposed Form and Our Strategy

The problem often gives us a hint, suggesting that the nth derivative of arctan(x) can be written in a specific form:

f(n)(x) = Pn(x) / (1 + x2)n

where Pn(x) is a polynomial. This is a crucial piece of information! It tells us that the nth derivative can be expressed as a fraction, with a polynomial in the numerator and a power of (1 + x2) in the denominator. Our strategy will be to prove this form and then figure out how to find the polynomial Pn(x).

Our Roadmap

Here's the plan of attack:

  1. Base Case: We'll start by showing that the proposed form holds for a small value of n, typically n = 1. This gives us a starting point for our proof.
  2. Inductive Step: We'll assume that the form holds for some arbitrary positive integer k, and then we'll show that it must also hold for k + 1. This is the heart of mathematical induction.
  3. Characterizing the Polynomial: Once we've proven the form, we'll need to figure out how to actually find the polynomial Pn(x). We'll explore a recursive relationship that allows us to calculate Pn+1(x) from Pn(x).

Proof by Induction

Let's dive into the mathematical details. We'll use the principle of mathematical induction to prove that f(n)(x) = Pn(x) / (1 + x2)n for all n ≥ 1.

Base Case (n = 1)

First, we need to show that the formula holds for n = 1. The first derivative of f(x) = arctan(x) is:

f'( x ) = 1 / (1 + x2)

We can rewrite this as:

f'( x ) = 1 / (1 + x2)1

This perfectly matches our proposed form, with P1(x) = 1. So, the base case holds!

Inductive Step

Now, let's assume that the formula holds for some arbitrary positive integer k. This means we're assuming that:

f(k)(x) = Pk(x) / (1 + x2)k

for some polynomial Pk(x). Our goal is to show that this implies the formula also holds for n = k + 1. In other words, we need to show that:

f(k+1)(x) = Pk+1(x) / (1 + x2)k+1

for some polynomial Pk+1(x).

To do this, we'll differentiate both sides of our inductive hypothesis (the assumption that the formula holds for k) with respect to x. This will give us an expression for f(k+1)(x). Using the quotient rule, we get:

f(k+1)(x) = [ Pk'(x) * (1 + x2)k - Pk(x) * k * (1 + x2)k-1 * (2x) ] / (1 + x2)2k

Let's simplify this expression. We can factor out (1 + x2)k-1 from the numerator:

f(k+1)(x) = [ (1 + x2)k-1 [ Pk'(x) * (1 + x2) - 2x k Pk(x) ] ] / (1 + x2)2k

Now, we can cancel out (1 + x2)k-1 from the numerator and denominator:

f(k+1)(x) = [ Pk'(x) * (1 + x2) - 2x k Pk(x) ] / (1 + x2)k+1

Notice that the denominator now has the form (1 + x2)k+1, which is exactly what we wanted! Now, we just need to show that the numerator is a polynomial. Let's define Pk+1(x) as:

Pk+1(x) = Pk'(x) * (1 + x2) - 2x k Pk(x)

Since Pk(x) is a polynomial, its derivative Pk'(x) is also a polynomial. The sum and product of polynomials are also polynomials. Therefore, Pk+1(x) is a polynomial. This means we can write:

f(k+1)(x) = Pk+1(x) / (1 + x2)k+1

This is exactly the form we wanted to show! So, we've successfully completed the inductive step.

Conclusion of the Induction

By the principle of mathematical induction, we've proven that:

f(n)(x) = Pn(x) / (1 + x2)n

for all n ≥ 1, where Pn(x) is a polynomial.

Characterizing the Polynomial P_n(x)

Okay, we've proven the general form, which is a huge step! But we're not quite done yet. We still need to figure out how to actually find the polynomial Pn(x) for a given n. Remember how we defined Pk+1(x) in the inductive step? That gives us a recursive relationship:

Pn+1(x) = Pn'(x) * (1 + x2) - 2x n Pn(x)

This is a powerful formula! It tells us that if we know Pn(x), we can find Pn+1(x) by differentiating Pn(x), multiplying by (1 + x2), and then subtracting 2x n Pn(x).

The Base Case for the Recursion

We already found P1(x) in the base case of our induction: P1(x) = 1. This is our starting point for the recursion. Now, we can use the recursive formula to find P2(x), P3(x), and so on.

Example: Finding P_2(x) and P_3(x)

Let's see this in action. To find P2(x), we use the recursive formula with n = 1:

P2(x) = P1'(x) * (1 + x2) - 2x * 1 * P1(x)

Since P1(x) = 1, its derivative P1'(x) is 0. So,

P2(x) = 0 * (1 + x2) - 2x * 1 * 1 = -2x

Now, let's find P3(x) using the recursive formula with n = 2:

P3(x) = P2'(x) * (1 + x2) - 2x * 2 * P2(x)

Since P2(x) = -2x, its derivative P2'(x) is -2. So,

P3(x) = -2 * (1 + x2) - 4x * (-2x) = -2 - 2x2 + 8x2 = 6x2 - 2

You can continue this process to find P4(x), P5(x), and so on. You'll notice that the polynomials become increasingly complex, but the recursive formula provides a systematic way to calculate them.

Summary and Conclusion

Wow, we've covered a lot! Let's recap what we've accomplished:

  • We set out to find a closed form for the nth derivative of f(x) = arctan(x).
  • We proved, using mathematical induction, that f(n)(x) = Pn(x) / (1 + x2)n, where Pn(x) is a polynomial.
  • We derived a recursive formula for calculating the polynomials: Pn+1(x) = Pn'(x) * (1 + x2) - 2x n Pn(x).
  • We calculated P2(x) and P3(x) as examples.

This journey demonstrates the power of mathematical induction and recursive relationships. Finding a closed form for the nth derivative of arctan(x) is a challenging but rewarding problem that highlights important concepts in calculus and mathematical problem-solving. Keep practicing, and you'll be amazed at what you can achieve! Now you guys know how to find the closed form of the nth derivative of arctan(x)! Go forth and conquer those derivatives!