# Thread: solving by power series...just seeing the pattern

1. ## solving by power series...just seeing the pattern

If I have a denominator that goes 2 4 6 8...2k ( all multiplied) what is that pattern and how do I find it?

Similarly if I have a denominator for the odds 3 5 7...2k+1 ( all multiplied ) what is the pattern for this and how do I get it?

Thanks.

2. Originally Posted by pberardi
If I have a denominator that goes 2 4 6 8...2k ( all multiplied) what is that pattern and how do I find it?

Similarly if I have a denominator for the odds 3 5 7...2k+1 ( all multiplied ) what is the pattern for this and how do I get it?

Thanks.
Hi

Could you be more precise ?

3. This was the diffeq. y'' +xy' + y = 0. I have to find a power series solution. After getting a general solution, differentiating, plugging in, I get a formula. Then I have to start plugging in different values of n to see a pattern so that I can get a power series. When I start plugging in n values for the formula my denominator looks like this for the even values of a: 2*4*6*8*10

Similarly for the odd values they are 3*5*7*9.

The numerator is simple so I need not mention it.

I need a formula for these two denominators. After some research I know that for 2k the formula is 2^k(k!) and for 2k+1 it is (2k+1)! however I do no know how yet to reach that conclusion except for sheer memorization which doesn't help much.

Thanks and I hope this is clear. If not please continue to ask if you don't mind. Thank you.

4. I did the problem and think I see what you mean! You can write the coefficients using a number of different notational methods:

$\displaystyle a_{2n+1} = \frac{(-1)^n}{\prod_{k=1}^{k=n}(2k +1) } \, a_1 = \frac{(-1)^n \, n! \, 2^n}{(2n +1)! } \, a_1 = \frac{(-1)^n }{(2n +1)!! } \, a_1$

and similarly

$\displaystyle a_{2n} = \frac{(-1)^n}{\prod_{k=1}^{k=n}(2k) } \, a_0 = \frac{(-1)^n }{n! \, 2^n } \, a_0 = \frac{(-1)^n }{(2n)!! } \, a_0$ .

So the three notations used were the product notation (capital Pi), the factorial and the double factorial respectively. Hope this is of some help to you!

5. Originally Posted by pberardi
If I have a denominator that goes 2 4 6 8...2k ( all multiplied) what is that pattern and how do I find it?

Similarly if I have a denominator for the odds 3 5 7...2k+1 ( all multiplied ) what is the pattern for this and how do I get it?

Thanks.
2 4 6 8...2k = 2^k k!

3 5 7...(2k+1) = (2k+1)!/(2^k k!)

CB