(a) Use differentiation to find a power series representation for f(x) =
1
(6 + x)2
.
f(x) =
∞
leftparen1.gif
(−1)n(n+1)xn6n+2
rightparen1.gif
sum.gif
n = 0
What is the radius of convergence, R?
R =
The Correct Answer and Explanation is :
To find a power series representation for
[
f(x) = \frac{1}{(6 + x)^2}
]
Step 1: Consider a Related Geometric Series
We start with the standard geometric series representation:
[
\frac{1}{1 – r} = \sum_{n=0}^{\infty} r^n, \quad \text{for } |r| < 1.
]
To apply this, rewrite ( f(x) ) in a suitable form. First, express the denominator as:
[
\frac{1}{6+x} = \frac{1}{6} \cdot \frac{1}{1 + \frac{x}{6}}.
]
Using the geometric series for ( \frac{1}{1 – (-x/6)} ), we obtain:
[
\frac{1}{6 + x} = \frac{1}{6} \sum_{n=0}^{\infty} \left( -\frac{x}{6} \right)^n, \quad \text{for } |x| < 6.
]
Step 2: Differentiate Both Sides
Since we need ( \frac{1}{(6+x)^2} ), differentiate both sides with respect to ( x ):
[
\frac{d}{dx} \left[ \frac{1}{6+x} \right] = \frac{d}{dx} \left[ \frac{1}{6} \sum_{n=0}^{\infty} \left( -\frac{x}{6} \right)^n \right].
]
Using the derivative of ( \frac{1}{6 + x} ), we get:
[
-\frac{1}{(6+x)^2} = \frac{1}{6} \sum_{n=1}^{\infty} n \left( -\frac{x}{6} \right)^{n-1} \cdot \left( -\frac{1}{6} \right).
]
Simplifying,
[
\frac{1}{(6+x)^2} = \sum_{n=1}^{\infty} \frac{n}{6^{n+1}} (-x)^{n-1}.
]
Rewriting the index by setting ( n \to n+1 ), we obtain:
[
f(x) = \sum_{n=0}^{\infty} (n+1) (-1)^n \frac{x^n}{6^{n+2}}.
]
Step 3: Radius of Convergence
The radius of convergence ( R ) is determined by the original geometric series, which converges for ( |x/6| < 1 ), implying:
[
|x| < 6.
]
Thus, the radius of convergence is:
[
R = 6.
]
Now, let’s generate a visual representation of the power series.
Here is a visual representation of the function ( f(x) = \frac{1}{(6+x)^2} ) alongside its power series approximation using 10 terms. The dashed red line represents the series approximation, while the solid blue line is the exact function. The dotted vertical lines at ( x = \pm 6 ) indicate the radius of convergence ( R = 6 ).