Studdy AI Logo
Math  /  Calculus

QuestionNew tal codes f 13 High Econ 10 https://www.webassign.net/web/Student/Assignment-Responses, f(x)=1(9+x)2f(x)=∑n=0∞((−1)n+1nxn−19n+1)\begin{array}{c} f(x)=\frac{1}{(9+x)^{2}} \\ f(x)=\sum_{n=0}^{\infty}\left(\frac{(-1)^{n+1 n x^{n-1}}}{9^{n+1}}\right) \end{array}
What is the radius of convergence, RR ? R=9R=9 (b) Use part (a) to find a power series for f(x)=1(9+x)3f(x)=∑n=0∞(□)\begin{array}{c} f(x)=\frac{1}{(9+x)^{3}} \\ f(x)=\sum_{n=0}^{\infty}(\square) \end{array}
What is the radius of convergence, RR ? R=R= □\square (c) Use part (b) to find a power series for f(x)=x2(9+x)3f(x)=∑n=2∞(□)\begin{array}{c} f(x)=\frac{x^{2}}{(9+x)^{3}} \\ f(x)=\sum_{n=2}^{\infty}(\square) \end{array}
What is the radius of convergence, RR ? R=R= □\square Search

Studdy Solution

STEP 1

What is this asking? Given the power series for 1(9+x)2 \frac{1}{(9+x)^2} , we need to find power series for 1(9+x)3 \frac{1}{(9+x)^3} and x2(9+x)3 \frac{x^2}{(9+x)^3} and their radii of convergence. Watch out! Don't forget the chain rule when taking derivatives, and remember the radius of convergence stays the same when differentiating.

STEP 2

1. Rewrite the initial function
2. Find the power series for 1(9+x)3 \frac{1}{(9+x)^3}
3. Find the power series for x2(9+x)3 \frac{x^2}{(9+x)^3}

STEP 3

We're given f(x)=1(9+x)2=∑n=0∞(−1)n+1nxn−19n+1 f(x) = \frac{1}{(9+x)^2} = \sum_{n=0}^{\infty} \frac{(-1)^{n+1}nx^{n-1}}{9^{n+1}} .
Let's **rewrite** the given function in a way that will help us later: 1(9+x)2=192(1+x9)2=181(1+x9)2=181(1+x9)−2 \frac{1}{(9+x)^2} = \frac{1}{9^2 (1 + \frac{x}{9})^2} = \frac{1}{81 (1 + \frac{x}{9})^2} = \frac{1}{81} (1 + \frac{x}{9})^{-2}

STEP 4

Rewriting like this helps us see the connection between the different functions we'll be working with.
It also makes it easier to apply the power rule for derivatives.

STEP 5

Notice that ddx[19+x]=−1(9+x)2 \frac{d}{dx} \left[ \frac{1}{9+x} \right] = \frac{-1}{(9+x)^2} .
Multiplying both sides by −1-1 gives us −ddx[19+x]=1(9+x)2 -\frac{d}{dx} \left[ \frac{1}{9+x} \right] = \frac{1}{(9+x)^2} .
We can also say that ddx[12(9+x)2]=−22(9+x)3=−1(9+x)3 \frac{d}{dx} \left[ \frac{1}{2(9+x)^2} \right] = \frac{-2}{2(9+x)^3} = \frac{-1}{(9+x)^3} .
Multiplying by −1-1 gives −ddx[12(9+x)2]=1(9+x)3 -\frac{d}{dx} \left[ \frac{1}{2(9+x)^2} \right] = \frac{1}{(9+x)^3} .

STEP 6

So, to find the power series for 1(9+x)3 \frac{1}{(9+x)^3} , we can **differentiate** the power series for 12(9+x)2 \frac{1}{2(9+x)^2} and multiply by −1-1. −12ddx∑n=0∞(−1)n+1nxn−19n+1=−12∑n=0∞(−1)n+1n(n−1)xn−29n+1=∑n=0∞(−1)nn(n−1)xn−22⋅9n+1 -\frac{1}{2} \frac{d}{dx} \sum_{n=0}^{\infty} \frac{(-1)^{n+1}nx^{n-1}}{9^{n+1}} = -\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n+1}n(n-1)x^{n-2}}{9^{n+1}} = \sum_{n=0}^{\infty} \frac{(-1)^{n}n(n-1)x^{n-2}}{2 \cdot 9^{n+1}} This is our **power series** for 1(9+x)3 \frac{1}{(9+x)^3} .

STEP 7

The **radius of convergence** remains the same, R=9 R = 9 .

STEP 8

Now, we just need to **multiply** the power series we found for 1(9+x)3 \frac{1}{(9+x)^3} by x2 x^2 : x2⋅∑n=0∞(−1)nn(n−1)xn−22⋅9n+1=∑n=0∞(−1)nn(n−1)xn2⋅9n+1 x^2 \cdot \sum_{n=0}^{\infty} \frac{(-1)^{n}n(n-1)x^{n-2}}{2 \cdot 9^{n+1}} = \sum_{n=0}^{\infty} \frac{(-1)^{n}n(n-1)x^n}{2 \cdot 9^{n+1}} This is our **power series** for x2(9+x)3 \frac{x^2}{(9+x)^3} .

STEP 9

The **radius of convergence** is still R=9 R = 9 .

STEP 10

The power series for 1(9+x)3 \frac{1}{(9+x)^3} is ∑n=0∞(−1)nn(n−1)xn−22⋅9n+1 \sum_{n=0}^{\infty} \frac{(-1)^{n}n(n-1)x^{n-2}}{2 \cdot 9^{n+1}} with radius of convergence R=9 R = 9 .
The power series for x2(9+x)3 \frac{x^2}{(9+x)^3} is ∑n=0∞(−1)nn(n−1)xn2⋅9n+1 \sum_{n=0}^{\infty} \frac{(-1)^{n}n(n-1)x^n}{2 \cdot 9^{n+1}} with radius of convergence R=9 R = 9 .

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord