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Math Help - Congruence

  1. #1
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    Congruence

    Can somebody help with this question please?

    Prove that

    (e^x)(e^x2/2)(e^x^3/3)...=1+x+x^2+... when |x|<1.

    Show that the coefficient of x^19 in the power series expansion on the LHS has the form

    1/19! + 1/19 + r/s,

    where 19 does not divide s.

    Deduce that 18!= -1(mod 19).
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  2. #2
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    Hi

    \prod_{k=1}^{+\infty}e^{\frac{x^k}{k}} = e^{\sum_{k=1}^{+\infty}\frac{x^k}{k}}

    To find \sum_{k=1}^{+\infty}\frac{x^k}{k} let's take the derivative

    \sum_{k=0}^{+\infty}x^k = \frac{1}{1-x}

    Therefore \sum_{k=1}^{+\infty}\frac{x^k}{k} is the antiderivative of \frac{1}{1-x} that is 0 for x=0

    \sum_{k=1}^{+\infty}\frac{x^k}{k} = -\ln|1-x|

    Therefore \prod_{k=1}^{+\infty}e^{\frac{x^k}{k}} = e^{\sum_{k=1}^{+\infty}\frac{x^k}{k}} = e^{-\ln|1-x|} = \frac{1}{|1-x|}
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  3. #3
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    Thanks.

    Is anybody able to help with the second part of the question please?
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  4. #4
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    e^{x} = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots  + \frac{x^n}{n!} + \cdots

    e^{\frac{x^2}{2}} = 1 + \frac{x^2}{2} + \frac{x^4}{8} + \frac{x^6}{8\cdot 3!} + \cdots  + \frac{x^{2n}}{2^n\cdot n!} + \cdots

    e^{\frac{x^3}{3}} = 1 + \frac{x^3}{3} + \frac{x^6}{18} + \frac{x^9}{27\cdot 3!} + \cdots  + \frac{x^{3n}}{3^n\cdot n!} + \cdots

    etc

    To find the coefficient of x^19 in the power series expansion you need to multiply all these identities and find the coefficient of x^19
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  5. #5
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    surely there must be an easier way to do this?
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  6. #6
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    I hope so

    If you consider each term of the expansion of e^x from x^19 down to 1
    - x^19/19! must be multiplied by a constant which is 1 therefore 1/19!
    - x^18/18! must be multiplied by something times x, but this is not possible since in the other series are involved terms starting from x, therefore 0
    - x^17/17! must be multiplied by something times x, which is x/2 coming from the expansion of e^(x/2) therefore 1/(2x17!)
    - x^16/16! must be multiplied by something times x^3, which is x^3/3 coming from the expansion of e^(x^3/3) therefore 1/(3x16!)
    - x^15/15! must be multiplied by something times x^4, which is x^4/4 coming from the expansion of e^(x^4/4) but also x^4/8 coming from the expansion of e^(x^2/2) therefore (1/4+1/8)/15! = 3/(8x15!)
    - x^14/14! must be multiplied by something times x^5, which is x^5/5 coming from the expansion of e^(x^5/5) but also x^5/6 coming from the product of the expansion of e^(x^2/2) and the expansion of e^(x^3/3) therefore (1/5+1/6)/14! = 11/(30x14!)

    and so on ... but it becomes more and more difficult
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  7. #7
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    You guys are so close...

    Yes, this is the ticket. Running-gag has the winning strategy, but don't think for a minute you have to find the actual coefficient. All the problem calls for is showing that it is of the form \frac{1}{19!}+\frac{1}{19}+\frac{r}{s} , for 19 not dividing s.

    For reference, call the series expansion e^{\frac{x^k}{k}} by a_k, and the coefficient preceding x^{kn} by a_k(n) . For example, a_1(3)=\frac{1}{3!} and a_3(2)=\frac{1}{18} .

    Now the coefficient on x^{19} is the sum of all possible products of terms whose exponents sum to 19, for example, one of which is a_1(4)*a_2(3)*a_3(3) = \frac{1}{4!}*\frac{1}{2^3 3!}*\frac{1}{3^3 3!}

    (i) a_1(19)=\frac{1}{19!}
    (ii) a_{19}(1)=\frac{1}{19}
    (iii) Since 19 is prime, nowhere else in any coefficient a_k(n) does 19 appear, for k<19 and n<19.

    Therefore, the sum of all such terms must be of the form \frac{1}{19!}+\frac{1}{19}+\frac{r}{s} , for 19 not dividing s.

    QED

    EDIT: The coefficient is actually \frac{1}{19!}+\frac{1}{19}+\frac{31546534213457683  52}{15687942664565435561256}
    Last edited by Media_Man; May 11th 2009 at 08:58 PM.
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  8. #8
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    Thanks

    How do we deduce that 18! = -1 (mod 19) from all this?
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