Power series recurrence relations

[IniquiTrance]

Homework Statement

In the following series':

http://image.cramster.com/answer-board/image/cramster-equation-2009410014306337491927047975008434.gif

According to my book, we only have a common range of summation here for n >= 2.

Therefore we need to treat n = 0 and n = 1 separately.

We thus write:

[2r(r-1) + 3r - 1]c0 = 0

and

[2(r+1)r + 3(r+1)-1]c1 = 0

My question is in both cases we ignored the last (4th) series to develop these two equations. How can we just ignore this power series?

I imagine it must be because its summation begins at n=2 unlike the others, but why are we allowed to ignore it up until n = 2?

Thanks!

Homework Equations

The Attempt at a Solution

 

[tiny-tim]

Hi IniquiTrance!

(try using the X² and X₂ tags just above the Reply box )

According to my book, we only have a common range of summation here for n >= 2.

Therefore we need to treat n = 0 and n = 1 separately.

We thus write:

[2r(r-1) + 3r - 1]c0 = 0

and

[2(r+1)r + 3(r+1)-1]c1 = 0

My question is in both cases we ignored the last (4th) series to develop these two equations. How can we just ignore this power series?

I imagine it must be because its summation begins at n=2 unlike the others, but why are we allowed to ignore it up until n = 2?

ah, but you're really writing:

([2r(r-1) + 3r - 1]c₀ + 0)x^r = 0

and

([2(r+1)r + 3(r+1) -1]c₁ + 0)x^r+1 = 0,

so you're not really ignoring the 4th sum at all, you're just not bothering to put in the zeros.

Alternatively, think of the 4th sum as starting at n = 0 like the others, and with c_-1 and c_-2 defined as 0

 

[IniquiTrance]

Ah, ok much clearer now. Thank you!