Of course. Here is a complete, in-depth article on the Maclaurin series for 1/(1-x)^2.
The Maclaurin Series for 1/(1-x)^2: A Gateway to Power Series
The Maclaurin series for the function ( f(x) = \frac{1}{(1-x)^2} ) is a fundamental and elegant example in calculus, bridging the gap between simple algebraic functions and their infinite series representations. Think about it: this series is not just a mathematical curiosity; it is a powerful tool used in physics, engineering, and computer science for approximations, solving differential equations, and analyzing the behavior of complex systems. Understanding its derivation and properties provides a deeper insight into the world of power series.
Introduction: What is a Maclaurin Series?
Before diving into the specific function, it's essential to understand what a Maclaurin series is. A Maclaurin series is a type of Taylor series expansion of a function about zero. It represents a function as an infinite sum of terms calculated from the values of its derivatives at a single point, ( x = 0 ).
[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!
where ( f^{(n)}(0) ) denotes the ( n )-th derivative of ( f ) evaluated at ( x = 0 ), and ( n! ) is the factorial of ( n ). The series provides a polynomial approximation of the function that becomes increasingly accurate as more terms are included, within a certain interval of convergence Simple, but easy to overlook..
Deriving the Series: Two Elegant Methods
There are two primary ways to derive the Maclaurin series for ( \frac{1}{(1-x)^2} ). The first method uses the definition of the Maclaurin series directly, while the second, more clever method leverages our knowledge of a simpler, related series.
Method 1: Using the Definition (Derivatives at Zero)
This method is straightforward but can become computationally intensive. We need to find the derivatives of ( f(x) = (1-x)^{-2} ) and evaluate them at ( x = 0 ) It's one of those things that adds up..
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The Function itself: ( f(x) = (1-x)^{-2} ) At ( x=0 ): ( f(0) = (1-0)^{-2} = 1 )
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First Derivative: Using the chain rule, ( f'(x) = -2(1-x)^{-3} \cdot (-1) = 2(1-x)^{-3} ) At ( x=0 ): ( f'(0) = 2(1)^{-3} = 2 )
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Second Derivative: Differentiating again, ( f''(x) = 2 \cdot (-3)(1-x)^{-4} \cdot (-1) = 6(1-x)^{-4} ) At ( x=0 ): ( f''(0) = 6(1)^{-4} = 6 )
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Third Derivative: Continuing the pattern, ( f'''(x) = 6 \cdot (-4)(1-x)^{-5} \cdot (-1) = 24(1-x)^{-5} ) At ( x=0 ): ( f'''(0) = 24(1)^{-5} = 24 )
A clear pattern emerges. (k+2)(1-x)^{-(k+3)} ). Day to day, for ( n=0 ), ( (0+1)! The ( n )-th derivative at zero is ( f^{(n)}(0) = (n+1)! (1-x)^{-(k+2)} ), which is ( (k+1)!Assuming it's true for ( n=k ), the ( (k+1) )-th derivative will be the derivative of ( (k+1)!We can prove this by induction. = 1! That said, ). Evaluating at zero gives ( (k+2)! = 1 ), which matches. ), confirming the pattern.
Now, we plug these values into the Maclaurin series formula:
[ f(x) = \frac{1}{0!}x^0 + \frac{2}{1!}x^1 + \frac{6}{2!Which means }x^2 + \frac{24}{3! Because of that, }x^3 + \dots + \frac{(n+1)! }{n!
Simplifying the coefficients: ( \frac{(n+1)!}{n!} = (n+1) )
That's why, the Maclaurin series is:
[ \frac{1}{(1-x)^2} = \sum_{n=0}^{\infty} (n+1)x^n = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + \dots ]
Method 2: Using the Geometric Series (A More Efficient Approach)
This method is often preferred for its simplicity and insight. We start with the well-known Maclaurin series for the geometric series, which is valid for ( |x| < 1 ):
[ \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \dots ]
Now, notice that the derivative of ( \frac{1}{1-x} ) is closely related to our target function. Let's differentiate both sides of the equation with respect to ( x ):
[ \frac{d}{dx} \left( \frac{1}{1-x} \right) = \frac{d}{dx} \left( \sum_{n=0}^{\infty} x^n \right) ]
The left side becomes: [ \frac{d}{dx} (1-x)^{-1} = -1(1-x)^{-2} \cdot (-1) = \frac{1}{(1-x)^2} ]
The right side, differentiating term-by-term (which is valid within the interval of convergence), becomes: [ \sum_{n=1}^{\infty} n x^{n-1} = 1 + 2x + 3x^2 + 4x^3 + \dots ] (Note: The sum starts at n=1 because the derivative of the constant term (x^0) is zero).
By re-indexing the sum (let ( m = n-1 ), so ( n = m+1 )), we get: [ \sum_{m=0}^{\infty} (m+1) x^m ]
Thus, we have derived the same result with much less effort: [ \frac{1}{(1-x)^2} = \sum_{n=0}^{\infty
