Manipulating the Geometric Series

Calculus Level 3

Determine the power series representation of the function f ( x ) = 1 2 x + 3 f(x) = \frac{1}{2x + 3}

∑ n = 0 ∞ ( − 1 ) n 2 n x n 3 n + 1 \sum_{n = 0}^{\infty}(-1)^n \frac{2^n x^n}{3^{n+1}} ∑ n = 0 ∞ ( − 1 ) n 2 n x n 3 n \sum_{n = 0}^{\infty}(-1)^n \frac{2^n x^n}{3^{n}} ∑ n = 0 ∞ ( − 1 ) n x n 3 n + 1 \sum_{n = 0}^{\infty}(-1)^n \frac{x^n}{3^{n+1}} ∑ n = 0 ∞ ( − 1 ) n x n 3 n \sum_{n = 0}^{\infty}(-1)^n \frac{x^n}{3^{n}}

View wiki
Andrew Ellinor by Andrew Ellinor , 33, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Andrew Ellinor
Aug 31, 2015

For a function like f ( x ) = 1 2 x + 3 f(x) = \frac{1}{2x + 3} , there's one small hurdle to clear before applying the infinite geometric sum formula. In the function's current form, the denominator does not contain the 1 that is prescribed in the formula a 1 − r \frac{a}{1 - r} . We will start by dividing both numerator and denominator by 3 to resolve this issue. 1 2 x + 3 = 1 3 2 x 3 + 1 = 1 3 1 + 2 x 3 \frac{1}{2x + 3} = \frac{\frac{1}{3}}{\frac{2x}{3} + 1} = \frac{\frac{1}{3}}{1 + \frac{2x}{3}}

Now it is much easier to see that a = 1 3 a = \frac{1}{3} and r = − 2 x 3 r = -\frac{2x}{3} . Again using the formula for the infinite geometric summation, 1 2 x + 3 = ∑ n = 0 ∞ ( 1 3 ) ( − 2 x 3 ) n = ∑ n = 0 ∞ ( − 1 ) n 2 n x n 3 n + 1 \frac{1}{2x + 3} = \sum_{n = 0}^{\infty}(\frac{1}{3})(-\frac{2x}{3})^n = \sum_{n = 0}^{\infty}(-1)^n \frac{2^n x^n}{3^{n+1}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Very lucid explanation! :)

Pranshu Gaba - 5 years, 9 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...