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.

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- May 9th 2009, 10:55 AMpberardisolving 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. - May 9th 2009, 11:11 AMrunning-gag
- May 9th 2009, 11:18 AMpberardi
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. - May 9th 2009, 03:10 PMthe_doc
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! - May 9th 2009, 11:50 PMCaptainBlack