# Thread: Pulling out n=0 and n=1 terms...

1. ## Pulling out n=0 and n=1 terms...

I am following the example from this website to learn: An additional example of a series solution

How does he pull out the n=0 and n=1 terms to get: -2a2-6a3x+a1x-a0-a1x?

It's driving me insane, please help. If anyone could listen step for step, how he got that number from the summations I would greatly appreciate it.

2. ## Re: Pulling out n=0 and n=1 terms...

He wants to match the index of summations, to make them all start from n = 2.The the first one starts from n =2, so leave that; the second one and the fourth one start from n = 0, so write down the first two terms of each; and, finally, the third one starts from n =1, so write down its first term. That's basically where your expression came from (check it). It seems to me, though, in the next two latex lines, he had forgotten that he had matched the index of the summations. In any case, websites aren't that good a source of learning -- it's far better to learn from a book, in my opinion.

3. ## Re: Pulling out n=0 and n=1 terms...

Originally Posted by Planko
I am following the example from this website to learn: An additional example of a series solution

How does he pull out the n=0 and n=1 terms to get: -2a2-6a3x+a1x-a0-a1x?
The equation to that point is
$\displaystyle \sum_{n=2}^\infty n(n-1)a_nx^n- \sum_{n= 0}^\infty (n+1)(n+2)a_{n+2}x^n+ \sum_{n=1}^\infty na_nx^n- \sum_{n=0}^\infty a_nx^n= 0$

The sums have been adjusted so they each have "$\displaystyle x^n$" but the first sum does not start until n= 2 and the third sum does not start until n= 1.
We can write the second sum as
$\displaystyle \sum_{n= 0}^\infty (n+1)(n+2)a_{n+2}x^n$$\displaystyle = (0+ 1)(0+ 2)a_{0+2}x^0+ (1+ 1)(1+ 2)a_{1+ 2}x+ \sum_{n= 2}^\infty (n+1)(n+2)a_{n+2}x^n$$\displaystyle = 2a_2+ 6a_3x+\sum_{n=2}^\infty(n+1)(n+2)a_{n+2}x^n$

Similarly, we can write third sum as
$\displaystyle \sum_{n=1}^\infty na_nx^n= (1)a_1x+ \sum_{n= 2}^\infty na_nx^n$

and the fourth sum as
$\displaystyle \sum_{n= 0}^\infty a_nx^n= a_0+ a_1x+ \sum_{n=2}^\infty a_nx^n$

Now, all of those sums start at n= 2 so we can combine them into a single sum. The parts are not in those sums are
$\displaystyle -(2a_2+ 6a_3x)+ a_1x- (a_0+ a_1x)$

4. ## Re: Pulling out n=0 and n=1 terms...

Your name made me laugh. I had been staring at my books and websites for while... too long. So my brain is essentially mush... Anyways, I had a cup of coffee and took a break.

After coming back, I saw how he obtained the numbers finally. Thanks for the response.

5. ## Re: Pulling out n=0 and n=1 terms...

Thanks HallsofIvy, you gave a better explanation than any of my books or websites I've visited all day.

6. ## Re: Pulling out n=0 and n=1 terms...

Originally Posted by Planko
I am following the example from this website to learn: An additional example of a series solution

How does he pull out the n=0 and n=1 terms to get: -2a2-6a3x+a1x-a0-a1x?

It's driving me insane, please help. If anyone could listen step for step, how he got that number from the summations I would greatly appreciate it.

$\displaystyle \sum_{n=2}^{\infty}n(n-1)a_nx^n-\sum_{n=0}^{\infty}(n+1)(n+2)a_{n+2}x^n+\sum_{n=1} ^{\infty}na_nx^n- \sum_{n=0}^{\infty}a_nx^n=0$

$\displaystyle \sum_{n=2}^{\infty}n(n-1)a_nx^n-\{2a_2+6a_3x+\sum_{n=2}^{\infty}(n+1)(n+2)a_{n+2}x ^n\}+\{a_1x+\sum_{n=2}^{\infty}na_nx^n\}-\{a_0+a_1x+ \sum_{n=2}^{\infty}a_nx^n\}=0$

$\displaystyle -2a_2-6a_3x+a_1x-a_0-a_1x+\{\sum_{n=2}^{\infty}n(n-1)a_nx^n-\sum_{n=2}^{\infty}(n+1)(n+2)a_{n+2}x^n+\sum_{n=2} ^{\infty}na_nx^n-\sum_{n=2}^{\infty}a_nx^n\}=0$

$\displaystyle -2a_2-6a_3x-a_0+\{\sum_{n=2}^{\infty}[n(n-1)a_nx^n-(n+1)(n+2)a_{n+2}x^n+na_nx^n-a_nx^n] \}=0$

$\displaystyle -2a_2-6a_3x-a_0+\{\sum_{n=2}^{\infty}[n(n-1)a_n-(n+1)(n+2)a_{n+2}+na_n-a_n]x^n \}=0$