Maclaurin Series Of E^-x Sinx A Step-by-Step Guide
Hey guys! Today, we're diving deep into the fascinating world of Maclaurin series, and we're going to tackle a particularly interesting function: f(x) = e^{-x}sin(x). If you're scratching your head wondering how to represent this function as an infinite sum of terms, you've come to the right place. We'll break down the process step-by-step, exploring different approaches and highlighting key concepts along the way. Whether you're a calculus whiz or just starting your journey, this guide will equip you with the knowledge and confidence to conquer Maclaurin series like a pro.
Understanding Maclaurin Series: The Foundation
Before we jump into the specifics of f(x) = e^{-x}sin(x), let's lay a solid foundation by understanding what Maclaurin series actually are. In essence, a Maclaurin series is a special type of Taylor series, which provides a way to represent a function as an infinite sum of terms involving its derivatives evaluated at a single point. Specifically, the Maclaurin series is centered at x = 0. Think of it as expressing a function as a polynomial with an infinite number of terms. This representation can be incredibly powerful, allowing us to approximate function values, solve differential equations, and gain deeper insights into the function's behavior. The general formula for a Maclaurin series is given by:
f(x) = f(0) + f'(0)x + (f''(0)x^2)/2! + (f'''(0)x^3)/3! + ... = Σ[n=0 to ∞] (f^(n)(0)x^n)/n!
Where f^(n)(0) represents the n-th derivative of f(x) evaluated at x = 0, and n! denotes the factorial of n. Now, why is this so useful, you might ask? Well, many functions that are difficult to work with directly can be easily approximated by their Maclaurin series, especially for values of x close to 0. This is because the higher-order terms in the series become increasingly smaller as n increases, allowing us to truncate the series after a certain number of terms and still obtain a good approximation. But before blindly applying the formula, let's consider our function f(x) = e^{-x}sin(x). Directly calculating higher-order derivatives can become quite cumbersome. So, we'll explore some clever techniques to make our lives easier.
Methods for Finding the Maclaurin Series of e^{-x}sin(x)
Alright, let's get our hands dirty with finding the Maclaurin series for f(x) = e^{-x}sin(x). There are a couple of main approaches we can take, each with its own strengths and weaknesses. We'll start by discussing the method of directly calculating derivatives, and then we'll explore the elegant approach using Cauchy products. The method of direct differentiation involves finding the first few derivatives of f(x), evaluating them at x = 0, and then plugging these values into the Maclaurin series formula. This is a straightforward approach, but it can become quite tedious as the order of the derivatives increases. However, it's a good way to understand the underlying principles and identify patterns. On the other hand, the Cauchy product method leverages the known Maclaurin series of simpler functions, like e^{-x} and sin(x), and combines them to obtain the series for their product. This method often requires less computation, but it demands a solid understanding of series manipulation and convergence properties. The user mentioned attempting the Cauchy product, which is a great approach! However, let's first make sure we have a firm grasp of both methods before proceeding. Remember, the key to mastering Maclaurin series is understanding the core concepts and choosing the most efficient method for each specific problem. Let's delve deeper into each of these methods in the following sections.
Method 1: Direct Differentiation
Let's start with the direct differentiation method. This involves finding the derivatives of f(x) = e^{-x}sin(x) and evaluating them at x = 0. Buckle up, because this can get a little messy! First, we find the first derivative, f'(x), using the product rule:
f'(x) = (e^{-x})'(sin(x)) + (e^{-x})(sin(x))' = -e^{-x}sin(x) + e^{-x}cos(x)
Evaluating at x = 0, we get f'(0) = -e^(0)sin(0) + e^(0)cos(0) = 0 + 1 = 1. Next, we find the second derivative, f''(x), by differentiating f'(x):
f''(x) = (-e^{-x}sin(x) + e^{-x}cos(x))' = e^{-x}sin(x) - e^{-x}cos(x) - e^{-x}cos(x) - e^{-x}sin(x) = -2e^{-x}cos(x)
Evaluating at x = 0, we get f''(0) = -2e^(0)cos(0) = -2. Now, let's find the third derivative, f'''(x):
f'''(x) = (-2e^{-x}cos(x))' = 2e^{-x}cos(x) + 2e^{-x}sin(x)
Evaluating at x = 0, we get f'''(0) = 2e^(0)cos(0) + 2e^(0)sin(0) = 2. We can continue this process to find higher-order derivatives, but you might already be noticing a pattern emerging. The derivatives involve combinations of e^{-x}sin(x) and e^{-x}cos(x), and their values at x = 0 seem to alternate in sign and magnitude. Now, let's plug these values into the Maclaurin series formula:
f(x) = f(0) + f'(0)x + (f''(0)x^2)/2! + (f'''(0)x^3)/3! + ... = 0 + 1x + (-2x^2)/2! + (2x^3)/3! + ... = x - x^2 + (x^3)/3 - ...
This gives us the first few terms of the Maclaurin series. However, finding a general formula for the n-th derivative and the n-th term of the series can be challenging with this method. This is where the Cauchy product method comes in handy, which we'll explore next.
Method 2: Cauchy Product
The Cauchy product method offers a more elegant way to find the Maclaurin series of f(x) = e^{-x}sin(x). This method leverages the known Maclaurin series of e^{-x} and sin(x) and combines them to obtain the series for their product. This approach is often less computationally intensive than direct differentiation, especially for higher-order terms. The Maclaurin series for e^{-x} is given by:
e^{-x} = 1 - x + (x^2)/2! - (x^3)/3! + (x^4)/4! - ... = Σ[n=0 to ∞] (-1)^n (x^n)/n!
And the Maclaurin series for sin(x) is given by:
sin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ... = Σ[n=0 to ∞] (-1)^n (x^(2n+1))/(2n+1)!
The Cauchy product of two series, Σ[n=0 to ∞] a_n and Σ[n=0 to ∞] b_n, is a new series Σ[n=0 to ∞] c_n, where the coefficients c_n are given by:
c_n = Σ[k=0 to n] a_k * b_(n-k)
In our case, a_n = (-1)^n / n! (coefficients of e^{-x}) and b_n are the coefficients of sin(x) (which are 0 for even powers of x and (-1)^n / (2n+1)! for odd powers). Applying the Cauchy product formula, we can find the coefficients of the Maclaurin series for e^{-x}sin(x). Let's calculate the first few coefficients:
- c_0 = a_0b_0 = (1)(0) = 0
- c_1 = a_0b_1 + a_1b_0 = (1)(1) + (-1)(0) = 1
- c_2 = a_0b_2 + a_1b_1 + a_2b_0 = (1)(0) + (-1)(1) + (1/2)(0) = -1
- c_3 = a_0b_3 + a_1b_2 + a_2b_1 + a_3b_0 = (1)(-1/6) + (-1)(0) + (1/2)(1) + (-1/6)(0) = 1/3
Therefore, the Maclaurin series for e^{-x}sin(x) is:
f(x) = 0 + 1x - 1x^2 + (1/3)x^3 + ... = x - x^2 + (x^3)/3 - ...
This matches the first few terms we obtained using direct differentiation. The Cauchy product method provides a more systematic way to find the coefficients, but it requires careful attention to detail and a good understanding of series manipulation. The user correctly identified the Cauchy product as a potential method. The crucial point they mentioned is the absolute convergence requirement. The Cauchy product theorem states that if Σ a_n and Σ b_n are absolutely convergent series, then their Cauchy product Σ c_n also converges to the product of the sums of the original series. Both the Maclaurin series for e^{-x} and sin(x) are absolutely convergent for all real numbers x, so we can confidently apply the Cauchy product method in this case. Let's recap and further refine our approach in the next section.
Refining the Cauchy Product Approach and Finding the General Term
Now that we've explored both direct differentiation and the Cauchy product method, let's focus on refining the Cauchy product approach and see if we can find a general term for the Maclaurin series of f(x) = e^-x}sin(x)*. We've already established that the Maclaurin series for e^{-x} and sin(x) are absolutely convergent, making the Cauchy product a valid and efficient method. We also calculated the first few terms of the resulting series, and b_(n-k) are the coefficients of sin(x). The tricky part is that the coefficients b_(n-k) are zero for even values of n-k. This means that the summation will only involve terms where n-k is odd. Let's analyze this more closely. Suppose n is even. Then n-k is odd only when k is odd. Similarly, if n is odd, then n-k is odd only when k is even. This alternating pattern adds complexity to finding a closed-form expression for c_n. One approach to tackle this is to write out the summation explicitly for different values of n and try to identify a pattern. For example:
- For n = 4, c_4 = Σ[k=0 to 4] a_k * b_(4-k) = a_1b_3 + a_3b_1 = (-1/1!)(-1/3!) + (-1/3!)(1/1!) = 1/6 - 1/6 = 0
- For n = 5, c_5 = Σ[k=0 to 5] a_k * b_(5-k) = a_0b_5 + a_2b_3 + a_4b_1 = (1)(1/5!) + (1/2!)(-1/3!) + (1/4!)(1/1!) = 1/120 - 1/12 + 1/24 = -1/30
As you can see, the calculations become quite involved. However, by carefully tracking the terms and simplifying, we can start to discern a pattern. An alternative approach is to use complex exponentials. Recall Euler's formula: e^(ix) = cos(x) + isin(x). We can rewrite sin(x) as the imaginary part of e^(ix): sin(x) = Im(e^(ix)). Therefore, e^{-x}sin(x) = Im(e{-x}e(ix)) = Im(e^(x(i-1))). Now, we can find the Maclaurin series for e^(x(i-1)) and then take the imaginary part. This approach can sometimes simplify the calculations. Let's explore this further.
Utilizing Complex Exponentials: A Clever Trick
As we discussed earlier, using complex exponentials can be a clever trick to find the Maclaurin series for f(x) = e^{-x}sin(x). This method leverages Euler's formula, e^(ix) = cos(x) + isin(x), to express sin(x) as the imaginary part of a complex exponential. This allows us to combine the exponential terms and work with a single exponential function, which is often easier to handle. We've already established that e^{-x}sin(x) = Im(e{-x}e(ix)) = Im(e^(x(i-1))). Now, let's find the Maclaurin series for e^(x(i-1)). We know that the Maclaurin series for e^u is Σ[n=0 to ∞] (u^n)/n!. Substituting u = x(i-1), we get:
e^(x(i-1)) = Σ[n=0 to ∞] ((x(i-1))^n)/n! = Σ[n=0 to ∞] ((i-1)^n (x^n))/n!
Now, we need to find the imaginary part of this series. Let's write out the first few terms:
e^(x(i-1)) = 1 + x(i-1) + (x^2(i-1)^2)/2! + (x^3(i-1)^3)/3! + (x^4(i-1)^4)/4! + ...
We need to simplify the powers of (i-1). Let's calculate the first few:
- (i-1)^0 = 1
- (i-1)^1 = i-1
- (i-1)^2 = i^2 - 2i + 1 = -1 - 2i + 1 = -2i
- (i-1)^3 = (i-1)(-2i) = -2i^2 + 2i = 2 + 2i
- (i-1)^4 = ((i-1)2)2 = (-2i)^2 = -4
- (i-1)^5 = (i-1)(-4) = -4i + 4
- (i-1)^6 = (i-1)(-4i+4) = -4i^2 +4i + 4i -4 = 4 + 8i -4 = 8i
Substituting these values back into the series, we get:
e^(x(i-1)) = 1 + x(i-1) + (x^2(-2i))/2! + (x^3(2+2i))/3! + (x^4(-4))/4! + (x^5(4-4i))/5! + (x^6(8i))/6! + ...
Now, we take the imaginary part of each term:
Im(e^(x(i-1))) = Im(1) + Im(x(i-1)) + Im((x^2(-2i))/2!) + Im((x^3(2+2i))/3!) + Im((x^4(-4))/4!) + Im((x^5(4-4i))/5!) + Im((x^6(8i))/6!) + ...
= 0 + x - x^2 + (x^3)/3 - (x^5)/30 - (x^6)/90 + ...
This gives us the Maclaurin series for e^{-x}sin(x). Notice that this matches the first few terms we obtained using the Cauchy product method and direct differentiation. This method, while involving complex numbers, provides a systematic way to find the coefficients and can be more manageable than direct differentiation for higher-order terms. It showcases the power of complex analysis in solving real-valued problems. Let's summarize our findings and discuss the convergence of the series in the final section.
Conclusion: Maclaurin Series and Convergence
Alright, guys, we've journeyed through the fascinating process of finding the Maclaurin series for f(x) = e^{-x}sin(x)! We explored multiple methods, including direct differentiation, the Cauchy product, and a clever technique using complex exponentials. Each method provided valuable insights and highlighted different aspects of series manipulation and calculus. We found that the Maclaurin series for e^{-x}sin(x) is given by:
f(x) = x - x^2 + (x^3)/3 - (x^5)/30 - (x^6)/90 + ...
While we didn't find a simple closed-form expression for the general term c_n, we successfully calculated the first few terms using various techniques. Now, a crucial question remains: For what values of x does this Maclaurin series converge? To determine the convergence, we can use the ratio test. The ratio test states that a series Σ a_n converges if the limit lim[n→∞] |a_(n+1) / a_n| < 1. Applying the ratio test to the Maclaurin series is a bit tricky without a general term, but we can analyze the convergence of the individual Maclaurin series for e^{-x} and sin(x). The Maclaurin series for e^{-x} converges for all real numbers x, and the Maclaurin series for sin(x) also converges for all real numbers x. Since the Cauchy product of two absolutely convergent series is also absolutely convergent, we can conclude that the Maclaurin series for e^{-x}sin(x) converges for all real numbers x. This means that we can use this series to approximate the value of e^{-x}sin(x) for any value of x, with the accuracy of the approximation increasing as we include more terms in the series. In conclusion, finding the Maclaurin series of f(x) = e^{-x}sin(x) is a rewarding exercise that demonstrates the power of calculus and series manipulation. We've successfully navigated the challenges, explored different approaches, and gained a deeper understanding of Maclaurin series and their applications. Keep practicing, and you'll become a Maclaurin series master in no time!
