I have a differential equation and I am solving it using power series. I am now at this point where I have got:
The reason I have got it like that was because my original equation read:
and so the minus 1 represents the x on the denominator, the y is obviosuly represented by the power series on the right and I have represented e^(ax) as a power series as well.
How would I go about simplifying this? So far, I haven't said what n or lambda are so could I simply swap the n's in the exponential power series for lambda and be left with
and then try and simplify like this?
Although, whilst I'm writing this, I'm thinking if theres a at the bottom, how would I shift the index to help me find the reccurance relation (which I think is the indicial equation?).

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 17042012 15:08

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 17042012 15:25
It's not true in general that , as you seem to be implying. What's the full differential equation you're trying to solve?

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 17042012 15:36
As Nuodai remarked  you are confused about how to multiply sums. In technical terms  you are using the fallacy of (a variation on) the Freshman's Dream. Never mind power series, consider the finite sums
and
Now
which clearly isn't equal to
...Anyway once you have dealt with that; as to (at least how I interpret what you are asking in your) final question:
Firstly reindex your power series expansion for by writing
where
Then
If you prefer to express this in terms of the original power series coefficients for then write:
Last edited by Jake22; 17042012 at 15:51. 
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 18042012 13:58
(Original post by nuodai)
It's not true in general that , as you seem to be implying. What's the full differential equation you're trying to solve?
Where h, m, A < 0 and a and E are constants.
Because its a Frobenius method question I used . I use 'x' rather than 'x  x_0' as x_0 = 0 because thats the singular point (I'm just writing this out to make sure I understand the reason why I've done it).
So when I worked it out, I out in that power series and got:
Now I'm trying to get rid of the e^x and so I thought about using the maclaurin series of e^x and then simplifying. 
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 18042012 14:48
(Original post by claret_n_blue)
Now I'm trying to get rid of the e^x and so I thought about using the maclaurin series of e^x and then simplifying. 
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 18042012 15:44
(Original post by claret_n_blue)
I don't get what you've done here. How did you get that on the rhs?
Look at the example of a power series where all of the coefficients except for the first two terms are zero:
where for all .
Thus both series are just finite sums of two terms.
Lets just expand:
out term by term. We get
where the is all zero since the terms are all zero by definition.
For each, we sum over all combinations of powers of from each term that multiply to give . So for example, for , we can multiply the constant term from the first sum and the term from the second and also the term from the first and the cconstant term from the second. We do things this way to collect like powers  nothing more and nothing less.Last edited by Jake22; 18042012 at 15:53. 
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 18042012 15:52
(Original post by Jake22)
Do you now understand how to reindex and multiply power series?
so
Is that correct? 
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 18042012 16:08
Forgetting about the differential equation for a moment*  it is not true that
Try to understand the summation notation for power series. Until you take the time to understand it you are just plugging things randomly into formulae you don't understand and are making mistakes. In particular  seperate your coefficients from your powers of .
So, rewrite as and try again...
*You have to forget about the differential equation for the time being. You are making the classical undergraduate mistake of trying to solve a problem x by using y without understanding what y is or how to manipulate it. In other words get comfortable with basic manipulation of power series before you use them to solve problems. In particular, be able to add, multiply, divide and reindex them. 
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 18042012 16:48
(Original post by Jake22)
You have to forget about the differential equation for the time being. You are making the classical undergraduate mistake of trying to solve a problem x by using y without understanding what y is or how to manipulate it. In other words get comfortable with basic manipulation of power series before you use them to solve problems. In particular, be able to add, multiply, divide and reindex them.
So
Now the bit I am stuck on is the actual multiplication. I understand how every term in the first sum must be multiplied by every term in the second term but I don't get how to write that?
I'm trying to use your very first point to help me write it like this but I don't get what 'i' represents and how you got it. 
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 18042012 17:32
(Original post by claret_n_blue)
Now the bit I am stuck on is the actual multiplication. I understand how every term in the first sum must be multiplied by every term in the second term but I don't get how to write that?
I'm trying to use your very first point to help me write it like this but I don't get what 'i' represents and how you got it.
i is just an index  it doesn't matter which letter you take.
Again, let and be power series. Then if we multiply both together, we do the following:
multiply the first term of the first series by everything in the second:
then multiply the second term of the first series by everything in the second:
and so on and on. We then have to add all of these expressions together.
More generally, when we multiply the th term of the first series by everything in the second  we get
so the whole sum is therefore
in summation notation, we could write this as:
but this isn't that useful to us since we want to present the answer in the standard form of a power series. Instead of adding all those series like that  we collect like terms:
Question: Which terms in the previous sum are all just some coefficient times, say, ?
Answer: The general term is . This is 'coefficient times ' precisely when . Thus the relevant terms are . If we write down the sum of all those terms, we just have:
Now, similarly, all the terms of the form 'coefficient times ' are just the sum:
Obviously, the total product is the sum of all the terms 'coefficient times x^0', 'coefficient times x^1', 'coefficient times x^2' etc. i.e.
or in summation notation:
I cannot emphasise enough: your problem is not understanding the summation notation. You must do examples with finite sums where you write out the same thing in series notation and in conventional sum notation if you want to understand how to manipulate infinite sums. e.g.
Now multiply this by the series
presenting your answer as a sum of powers of where you collect together like terms writing it out both ways.Last edited by Jake22; 18042012 at 17:42. 
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 22042012 19:35
(Original post by ttoby)
.(Original post by Jake22)
.
From here, I said my indicial equation is (k)(k1) = 0 and I need to use k = 1 to solve the equation because the question says "By direct substitution, use the larger root of the indicial equation for ...."
So I subbed in k = 1 into my reccurance relation and ended up with
Where A, , E and a are constants.
Do you think this is correct? I'm not sure how to go from here and find my solution. I have a solution that I need to work to so I can show that if it helps.
EDIT: I just retried it and it's clearly wrong because I've rearranged it to get my reccurance relation wrong. What do I do from:
I know you do:
but I don't get how to reaarange from here.Last edited by claret_n_blue; 22042012 at 19:53.
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Updated: April 22, 2012
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