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Multiplying power series

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    I have a differential equation and I am solving it using power series. I am now at this point where I have got:

     \displaystyle \sum_{n = 0}^{\infty} \frac{{(-ax)}^n}{n!} \cdot \sum_{\lambda = 0}^{\infty} a_{\lambda}x^{\lambda + k - 1}

    The reason I have got it like that was because my original equation read:

     \displaystyle \frac{e^{-ax}}{x} y 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

     \displaystyle \sum_{\lambda = 0}^{\infty} \left( \frac{{(-ax)}^{\lambda}}{\lambda!} \cdot a_{\lambda}x^{\lambda + k - 1} \right) and then try and simplify like this?

    Although, whilst I'm writing this, I'm thinking if theres a  \lambda ! 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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    It's not true in general that \sum a_nx^n \sum b_nx^n = \sum a_nb_nx^n, as you seem to be implying. What's the full differential equation you're trying to solve?
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    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

    \displaystyle \sum_{n=0}^1 a_n x^n = a_0 + a_1x and \displaystyle \sum_{m=0}^1 b_m x^n = b_0 + b_1x

    Now

    \displaystyle \sum_{n=0}^1 \left[ (a_n x^n) (b_n x^n)\right] = a_0b_0 + a_1b_1x^2

    which clearly isn't equal to

    \displaystyle \left(\sum_{n=0}^1 a_n x^n\right) \cdot\left(\sum_{m=0}^1 b_m x^n\right) = (a_0 + a_1x)(b_0 + b_1x) = a_0b_0 + a_0b_1x + a_1b_0x + a_1b_1x^2

    ...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 y by writing
    \displaystyle \sum_{m=0}^\infty a_m x^{m+k-1}=\sum_{m=0}^\infty b_m x^{m} where b_m = a_{m-(k-1)}

    Then \displaystyle  \left(\sum_{n=0}^\infty c_n x^n \right)\cdot \left(\sum_{m=0}^\infty b_m x^m \right) = \sum_{n=0}^\infty \left(\sum_{i=0}^n c_i b_{n-i}\right) x^n

    If you prefer to express this in terms of the original power series coefficients for y then write:

     \displaystyle \sum_{n=0}^\infty \left(\sum_{i=0}^n c_i b_{n-i}\right) x^n = \sum_{n=0}^\infty \left(\sum_{i=0}^n c_i a_{n-i-(k-1)}\right) x^n
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    (Original post by nuodai)
    It's not true in general that \sum a_nx^n \sum b_nx^n = \sum a_nb_nx^n, as you seem to be implying. What's the full differential equation you're trying to solve?
    The equation is:

     \frac{h^2}{2m} \frac{d^2y}{dx^2} + (E - \frac{Ae^{-ax}}{x}) y = 0

    Where h, m, A < 0 and a and E are constants.

    Because its a Frobenius method question I used  y = \sum^{\nifty}_{\lambda} = a_k x^{k + \lambda} . 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:

     E \cdot y - Ae^{-ax} \sum^{\nifty}_{\lambda} a_k x^{k + \lambda - 1}

    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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    (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.
    Do you now understand how to reindex and multiply power series?
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    (Original post by claret_n_blue)
    I don't get what you've done here. How did you get that on the rhs?
    As in my first post - you are basically forgetting to cross multiply and add indices. If you multiply two sums you have to multiply each term of the first sum with each term of the second sum. In your formula - you just multiply the first term with the first term, the second with the second etc. and you also forget that a_nx^n \cdot b_nx^n = a_nb_n x^{2n} \neq a_nb_n x^{n}


    Look at the example of a power series where all of the coefficients except for the first two terms are zero:

    \displaystyle \sum_{n=0}^\infty a_nx^n, \sum_{m=0}^\infty b_mx^m

    where a_i = 0 = b_i for all i \geq 2.

    Thus both series are just finite sums of two terms.

    Lets just expand:
    \displaystyle \sum_{m=0}^\infty \sum_{n=0}^m a_n b_{m-n} x^m
    out term by term. We get

    a_0b_0 + (a_0b_1 +a_1b_0)x^1 + a_1b_1 x^2 + \cdots

    where the \cdots is all zero since the terms a_2,b_2,a_3,b_3,\ldots are all zero by definition.

    For eachx^m, we sum over all combinations of powers of x from each term that multiply to give x^m. So for example, for x^1, we can multiply the constant term from the first sum and the x term from the second and also the x 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.
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    (Original post by Jake22)
    Do you now understand how to reindex and multiply power series?
    Ok, so I understand it all now, so that leaves my differential equation to read:

    \displaystyle y = \sum_{n = 0}^{\infty} a_n x^{n + k}, C = \frac{h^2}{2m}

    so

     \displaystyle C \cdot \sum_{n}^{\infty} (n + k)(n + k - 1) x ^{n + k - 1} + E \cdot \sum_{n = 0}^{\infty} a_n x^{n + k} - \sum_{n = 0}^{\infty} \left( \sum_{i = 0}^{n} \frac{(-ax)^i}{i!} \cdot a_{n - i - (k - 1)} \right)

    Is that correct?
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    Forgetting about the differential equation for a moment* - it is not true that

    \displaystyle \left(\sum_{n=0}^\infty \frac{(-ax)^n}{n!}\right) \cdot \left(\sum_{m=0}^\infty a_m x^{m+k-1}\right) = \sum_{n=0}^\infty \left(\frac{(-ax)^i}{i!}\cdot a_{n-i-(k-1)}\right)

    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 x.

    So, rewrite \displaystyle \sum_{n=0}^\infty \frac{(-ax)^n}{n!} as \displaystyle \sum_{n=0}^\infty \frac{(-a)^n}{n!}x^n 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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    (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.
    The problem I'm having is in the letter. After what you wrote in that post, what I have done is this:

    \displaystyle y = \sum_{\m = 0}^{\infty} a_m x^{m + k - 1} \, , \, e^{-ax} = \sum_{n = 0}^{\infty} \frac{(-a)^n}{n!} \cdot x^n

    So  e^{-ax} \cdot y =

     \displaystyle \sum_{n = 0}^{\infty} \frac{(-a)^n}{n!} \cdot x^n \cdoy\sum_{m = 0}^{\infty} a_m x^{m + k - 1}

    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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    (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.
    Reread my posts #3 and #6 - you need to try and see how it works with finite sums before you see how it works in general with power series. I wrote out an example there. The point about the formula I mentioned is that we collect like terms to present the final answer in the standard form for a power series i.e. we collect together all the terms involving x, collect all the terms involving x^2,... etc.

    i is just an index - it doesn't matter which letter you take.

    Again, let \displaystyle \sum_{n=0}^\infty a_n x^n and \displaystyle \sum_{m=0}^\infty b_m x^m 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:

    \displaystyle a_0x^0 \sum_{m=0}^\infty b_m x^m = \sum_{m=0}^\infty a_0b_m x^0x^m=\sum_{m=0}^\infty a_0b_m x^m

    then multiply the second term of the first series by everything in the second:

    \displaystyle a_1x^1 \sum_{m=0}^\infty b_m x^m = \sum_{m=0}^\infty a_1b_m x^1x^m = \sum_{m=0}^\infty a_1b_m x^{m+1}

    and so on and on. We then have to add all of these expressions together.

    More generally, when we multiply the nth term of the first series by everything in the second - we get

    \displaystyle a_nx^n \sum_{m=0}^\infty b_m x^m = \sum_{m=0}^\infty a_nb_m x^nx^m = \sum_{m=0}^\infty a_nb_m x^{m+n}

    so the whole sum is therefore

    \displaystyle \sum_{m=0}^\infty a_0b_m x^{m} + \sum_{m=0}^\infty a_1b_m x^{m+1} + \cdots + \sum_{m=0}^\infty a_nb_m x^{m+n} + \cdots

    in summation notation, we could write this as:

    \displaystyle \sum_{n=0}^\infty \left( \sum_{m=0}^\infty a_nb_m x^{m+n} \right)

    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, x^3?

    Answer: The general term is a_nb_m x^{m+n}. This is 'coefficient times x^3' precisely when m+n = 3. Thus the relevant terms are a_0b_3 x^{0+3}, a_1b_2 x^{1+2}, a_2b_1 x^{2+1},a_3b_0 x^{3+0}. If we write down the sum of all those terms, we just have:

    \displaystyle \sum_{m=0}^3 a_mb_{3-m}x^{m + (3-m)} = \sum_{m=0}^3 a_mb_{3-m}x^{3}

    Now, similarly, all the terms of the form 'coefficient times x^n' are just the sum:

    \displaystyle \sum_{m=0}^n a_mb_{n-m}x^{n}

    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.

    \displaystyle \sum_{m=0}^0 a_mb_{0-m}x^{0} + \sum_{m=0}^1 a_mb_{1-m}x^{1} + \cdots + \sum_{m=0}^n a_mb_{n-m}x^{n} +  \cdots

    or in summation notation:

    \displaystyle \sum_{n=0}^\infty \left(\sum_{m=0}^n a_mb_{n-m}x^{n}\right)

    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.

    \displaystyle \sum_{n=0}^3 a_n x^n = a_0 + a_1x + a_2 x^2 + a_3 x^3

    Now multiply this by the series \displaystyle \sum_{m=0}^3 b_m x^m = b_0 + b_1x + b_2 x^2 + b_3 x^3

    presenting your answer as a sum of powers of x where you collect together like terms writing it out both ways.
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    (Original post by ttoby)
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    (Original post by Jake22)
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    Hello. So I did "everything" and after I multiplied these power series, I showed it to my tutor and he told me a simpler way would be to expand the e^x power series and multiplt term by term. So I did all that and after shifting the index I ended up with:

    \left( \frac{h^2}{2m}(k^2 - k)b_0\right)x^{k - 2} +\left(\frac{h^2}{2m}(k^2 +k) b_0 - Ab_1 \right)x^{k -1} + \sum_{\lambda = 2}^{\infty} \left[ \frac{h^2}{2m}(\lambda + k - 1)(\lambda + k)b_{\lambda} -A b_{\lambda - 1} + (E - Aa)b_{\lambda - 2} \right] x^{\lambda +k - 2} = 0

    From here, I said my indicial equation is (k)(k-1) = 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

    \frac{b_{\lambda - 2}}{b_{\lambda - 1}} = \frac{A - C (\lambda^2 + \lambda)}{E - Aa} b_{\lambda}

    Where A, C = \frac{h^2}{2m} , 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 re-tried it and it's clearly wrong because I've rearranged it to get my reccurance relation wrong. What do I do from:

     \displaystyle \sum_{\lambda = 2}^{\infty} \left[ \frac{h^2}{2m}(\lambda + k - 1)(\lambda + k)b_{\lambda} -A b_{\lambda - 1} + (E - Aa)b_{\lambda - 2} \right] x^{\lambda +k - 2} = 0

    I know you do:

     \frac{h^2}{2m}(\lambda + k - 1)(\lambda + k)b_{\lambda} -A b_{\lambda - 1} + (E - Aa)b_{\lambda - 2} = 0 but I don't get how to reaarange from here.

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