# Power series representation

1. Dec 15, 2009

### rent981

1. Find a power series for F(x)= 3/4x^3-5, where c=1
2. power series = 1/a-r

3. What I did was take a derivative to get a similar function that was easier to solve. I used 1/x^3-1. Then I found a series for that function. Which I got $$\sum$$(x^3)^n. Then I added back from my orginal problem and got $$\sum$$(4/3x)^3n does this look at all right?

2. Dec 15, 2009

### Staff: Mentor

Without parentheses it's difficult to tell what your function F is. It should be interpreted as (3/4)*x^3 - 5, but I don't think that's what you had in mind.

3. Dec 15, 2009

### phsopher

Doesn't seem right to me. How does the 5 affect your result for instance?

From the geometrical series you know that

$$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$$

$$\frac{3}{4x^3 - 5}$$

is already almost in the same form, all you have to do is manipulate it a little bit.

4. Dec 15, 2009

### rent981

Ok if I rewrite I need to make the first term of the denominator 1. So I divide by -5. So then does the sum from 0 to infinity of (-3/5)(x)^3n look better?

5. Dec 15, 2009

### phsopher

Where did the 4 go?

6. Dec 15, 2009

### rent981

oops. Im meant to write [infinity]\sum[/n=0] (-3/5)(4x)^3n.

7. Dec 15, 2009

### rent981

$$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n (-3/5)(4x)^3n$$

8. Dec 15, 2009

### rent981

does that look ok?

9. Dec 15, 2009

### phsopher

When you divide by five you should do that to the whole denominator. And you have too much x's in there though I suspect that's a typo.

10. Dec 15, 2009

### rent981

Yes that was a typo. So.....$$\sum$$ (from 0-infinty) (-3/5)(4/5X)^3n would be ok? Sorry this is a new topic for me. Thanks you so much for helping!!!

11. Dec 15, 2009

### phsopher

Almost, except that 4/5 is not cubed so it's power is n:

$$\frac{3}{4x^3 - 5} = -\frac{3}{5} \frac{1}{1-\frac{4x^3}{5}} = -\frac{3}{5}\sum_{n=0}^{\infty} (\frac{4x^3}{5})^n$$