Formation of the General Equation for a Power Series

[WhiteTrout]

So I've gotten into the Method of Frobenius and all; Solved a few questions, however the most inconvenient part would be the formulation of the general equations for the final answer.

Granted, the lecturer told us to not spend so much time on that segment due to its minimal weightage, but I prefer to know.

So, here's one which I am on and about right now;

C_(k+1) = -2Ck/(k+1)(2k-1)

k=0; C1= -2Co/(1)(-1)
k=1; C2= -2C1/(2)(1)
k=2; C3= -2C2/(3)(3)
k=3; C4= -2C3/(4)(5)

Their equivalent in terms of Co being
-2Co/(1)(-1)
(-2)^2 Co/(1)(-1)(2)(1)
(-2)^3 Co/(1)(-1)(2)(1)(3)(3)
(-2)^3 Co/(1)(-1)(2)(1)(3)(3)(4)(5)
respectively.

What I have tried doing was
y= Co [Summation of](-2)^(n+1)/(n+1)!(<missing link>)

I cannot complete it because I don't know any function that allows me to pile up previous values, so that the current will be multiplied by the previous.
However, I am pretty sure that this isn't the correct method, so any help given will be very appriciated.

Also, I would love to hear advice on what to look for when creating the general equation.
Thank you.

 

[LCKurtz]

So I've gotten into the Method of Frobenius and all; Solved a few questions, however the most inconvenient part would be the formulation of the general equations for the final answer.

Granted, the lecturer told us to not spend so much time on that segment due to its minimal weightage, but I prefer to know.

So, here's one which I am on and about right now;

C_(k+1) = -2Ck/(k+1)(2k-1)

k=0; C1= -2Co/(1)(-1)
k=1; C2= -2C1/(2)(1)
k=2; C3= -2C2/(3)(3)
k=3; C4= -2C3/(4)(5)

Their equivalent in terms of Co being
-2Co/(1)(-1)
(-2)^2 Co/(1)(-1)(2)(1)
(-2)^3 Co/(1)(-1)(2)(1)(3)(3)
(-2)^3 Co/(1)(-1)(2)(1)(3)(3)(4)(5)
respectively.

What I have tried doing was
y= Co [Summation of](-2)^(n+1)/(n+1)!(<missing link>)

I cannot complete it because I don't know any function that allows me to pile up previous values, so that the current will be multiplied by the previous.
However, I am pretty sure that this isn't the correct method, so any help given will be very appriciated.

Also, I would love to hear advice on what to look for when creating the general equation.
Thank you.

You can't always write them in closed forms with factorials. Sometimes you need to use the ellipsis (...) to indicate "and so forth". When looking for a pattern is is a good idea not to multiply them out (which you apparently already know) and keep things arising from the same factor together. You have an apparent (1)(2)(3)... sequence and a (1)(3)(5)... sequence. So write it like this (Use the X₂ icon for subscripts) for n = 3:

(-2)³c₀/(-1){(1)(2)(3)(4)}{(1)(3)(5)}

You have almost got it, grouping the -1 with the powers of -2 and the factorial. So the general term for c_n would be written

c_n=(-2)ⁿ⁺¹c₀/((n+1)!(1)(3)(5)...(2n-1))

Sometimes there is no avoiding the ellipsis (the three dots).

 

[vela]

There's also double factorial notation where $n! = n(n-2)(n-4)\cdots 1$ when n is odd and $n! = n(n-2)(n-4)\cdots 2$ when n is even. 0! and (-1)! are both defined to be 1 to make writing general formulas easier.

You can also write, for example,
$$1\cdot 3\cdot 5 \cdots (2k+1) = \frac{1\cdot 2 \cdot 3 \cdots (2k+1)}{2\cdot 4\cdots 2k} = \frac{(2k+1)!}{2^k k!}$$

 

[WhiteTrout]

Hello,

LCKurtz: Alright I'll keep that in mind. I had not much prior encounters with ellipsis because I try not to use them; but this and the other advices would be very very helpful to remember in a pinch, thank you very much.

Vela: So that was what I was missing. I have thought about it before, but I didn't know how to express it correctly. Thanks for the insight, I know what to expect now; And for the heads up about double factorials too. The extra knowledge should be useful.


Thank you both for your time.
It has really helped me a great deal.

<edit> I just did some reading into double factorials and I must say that the identities would be useful;
In such case, the above can also be written as (2n+1)! right?
And its even numbered sibling would be (2n)!

I would really like to know more, but I am not around that area yet so I'll just keep it at this point.

Thank you.