Find the coefficient of $x^{24}$ in the power series of $e^{2x}(e^x-1)$.












0












$begingroup$


I tried to work through the problem algebraically, and got to $frac{2^{24}}{24!}(2^{24}-1)$, but comparing with the Taylor series generated for the function by Wolfram Alpha, it seems to be incorrect. I plugged in the power series for $e^x$ and used the formula for multiplication of series.



As a follow up question, is a power series converging to a function singular? Or can there be multiple correct answers?






calculus combinatorics power-series generating-functions







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  • 2




    $begingroup$
    You seem to have accidentally computed the $x^{24}$ coefficient for $e^{2x}(e^{2x}-1)$ instead.
    $endgroup$
    – J.G.
    Jan 24 at 8:37


















0












$begingroup$


I tried to work through the problem algebraically, and got to $frac{2^{24}}{24!}(2^{24}-1)$, but comparing with the Taylor series generated for the function by Wolfram Alpha, it seems to be incorrect. I plugged in the power series for $e^x$ and used the formula for multiplication of series.



As a follow up question, is a power series converging to a function singular? Or can there be multiple correct answers?






calculus combinatorics power-series generating-functions







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    You seem to have accidentally computed the $x^{24}$ coefficient for $e^{2x}(e^{2x}-1)$ instead.
    $endgroup$
    – J.G.
    Jan 24 at 8:37
















0












0








0





$begingroup$


I tried to work through the problem algebraically, and got to $frac{2^{24}}{24!}(2^{24}-1)$, but comparing with the Taylor series generated for the function by Wolfram Alpha, it seems to be incorrect. I plugged in the power series for $e^x$ and used the formula for multiplication of series.



As a follow up question, is a power series converging to a function singular? Or can there be multiple correct answers?






calculus combinatorics power-series generating-functions







share|cite|improve this question









$endgroup$




I tried to work through the problem algebraically, and got to $frac{2^{24}}{24!}(2^{24}-1)$, but comparing with the Taylor series generated for the function by Wolfram Alpha, it seems to be incorrect. I plugged in the power series for $e^x$ and used the formula for multiplication of series.



As a follow up question, is a power series converging to a function singular? Or can there be multiple correct answers?






calculus combinatorics power-series generating-functions




calculus combinatorics power-series generating-functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 24 at 8:25









Amit LevyAmit Levy

747




747








  • 2




    $begingroup$
    You seem to have accidentally computed the $x^{24}$ coefficient for $e^{2x}(e^{2x}-1)$ instead.
    $endgroup$
    – J.G.
    Jan 24 at 8:37
















  • 2




    $begingroup$
    You seem to have accidentally computed the $x^{24}$ coefficient for $e^{2x}(e^{2x}-1)$ instead.
    $endgroup$
    – J.G.
    Jan 24 at 8:37










2




2




$begingroup$
You seem to have accidentally computed the $x^{24}$ coefficient for $e^{2x}(e^{2x}-1)$ instead.
$endgroup$
– J.G.
Jan 24 at 8:37






$begingroup$
You seem to have accidentally computed the $x^{24}$ coefficient for $e^{2x}(e^{2x}-1)$ instead.
$endgroup$
– J.G.
Jan 24 at 8:37












1 Answer
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$begingroup$

Just write $e^{2x}(e^{x}-1)$ as $e^{3x}-e^{2x}$. The coefficident is $frac {3^{24}} {(24)!} -frac {2^{24}} {(24)!}$.






share|cite|improve this answer









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    1 Answer
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    1






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    active

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    10












    $begingroup$

    Just write $e^{2x}(e^{x}-1)$ as $e^{3x}-e^{2x}$. The coefficident is $frac {3^{24}} {(24)!} -frac {2^{24}} {(24)!}$.






    share|cite|improve this answer









    $endgroup$


















      10












      $begingroup$

      Just write $e^{2x}(e^{x}-1)$ as $e^{3x}-e^{2x}$. The coefficident is $frac {3^{24}} {(24)!} -frac {2^{24}} {(24)!}$.






      share|cite|improve this answer









      $endgroup$
















        10












        10








        10





        $begingroup$

        Just write $e^{2x}(e^{x}-1)$ as $e^{3x}-e^{2x}$. The coefficident is $frac {3^{24}} {(24)!} -frac {2^{24}} {(24)!}$.






        share|cite|improve this answer









        $endgroup$



        Just write $e^{2x}(e^{x}-1)$ as $e^{3x}-e^{2x}$. The coefficident is $frac {3^{24}} {(24)!} -frac {2^{24}} {(24)!}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 24 at 8:30









        Kavi Rama MurthyKavi Rama Murthy

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