Does $f(x) = sum_{i = 1}^{infty} e^{-xi}y(1-y)^{i-1}$ have a closed form?












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Does $f(x) = sum_{i = 1}^{infty} e^{-xi}y(1-y)^{i-1}$ have a closed form? $y$ is assumed to be in $[0,1]$ and $x >0$.



What I tried: I tried differentiating/integrating the function term by term to see whether those are something that can be calculated, but it seems that this method did not work.










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    1












    $begingroup$


    Does $f(x) = sum_{i = 1}^{infty} e^{-xi}y(1-y)^{i-1}$ have a closed form? $y$ is assumed to be in $[0,1]$ and $x >0$.



    What I tried: I tried differentiating/integrating the function term by term to see whether those are something that can be calculated, but it seems that this method did not work.










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Does $f(x) = sum_{i = 1}^{infty} e^{-xi}y(1-y)^{i-1}$ have a closed form? $y$ is assumed to be in $[0,1]$ and $x >0$.



      What I tried: I tried differentiating/integrating the function term by term to see whether those are something that can be calculated, but it seems that this method did not work.










      share|cite|improve this question









      $endgroup$




      Does $f(x) = sum_{i = 1}^{infty} e^{-xi}y(1-y)^{i-1}$ have a closed form? $y$ is assumed to be in $[0,1]$ and $x >0$.



      What I tried: I tried differentiating/integrating the function term by term to see whether those are something that can be calculated, but it seems that this method did not work.







      real-analysis analysis






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      asked Nov 24 '18 at 22:24









      pennypenny

      635




      635






















          1 Answer
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          6












          $begingroup$

          Observe
          begin{align}
          f(x) = frac{y}{1-y}sum^infty_{i=1} [e^{-x}(1-y)]^i
          end{align}

          then use geometric series.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            (The case $y=1$ obviously has to be dealt with separately.)
            $endgroup$
            – Clement C.
            Nov 24 '18 at 23:35










          • $begingroup$
            @ClementC. Yes. I will leave it as an exercise. =)
            $endgroup$
            – Jacky Chong
            Nov 24 '18 at 23:43










          • $begingroup$
            ^ Seems pretty easy, at least: looking at the original form, $y=0$ gives $f(x)=sum0$—am I right? (@ClementC.)
            $endgroup$
            – Chase Ryan Taylor
            Nov 24 '18 at 23:44










          • $begingroup$
            @ChaseRyanTaylor Yes, of course: for $y=0$, the infinite sum simplifies pretty easily (be slightly careful though -- there is still one non-zero term)
            $endgroup$
            – Clement C.
            Nov 25 '18 at 0:00










          • $begingroup$
            @ClementC. Which term is that?
            $endgroup$
            – Chase Ryan Taylor
            Nov 25 '18 at 0:12











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          1 Answer
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          active

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          6












          $begingroup$

          Observe
          begin{align}
          f(x) = frac{y}{1-y}sum^infty_{i=1} [e^{-x}(1-y)]^i
          end{align}

          then use geometric series.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            (The case $y=1$ obviously has to be dealt with separately.)
            $endgroup$
            – Clement C.
            Nov 24 '18 at 23:35










          • $begingroup$
            @ClementC. Yes. I will leave it as an exercise. =)
            $endgroup$
            – Jacky Chong
            Nov 24 '18 at 23:43










          • $begingroup$
            ^ Seems pretty easy, at least: looking at the original form, $y=0$ gives $f(x)=sum0$—am I right? (@ClementC.)
            $endgroup$
            – Chase Ryan Taylor
            Nov 24 '18 at 23:44










          • $begingroup$
            @ChaseRyanTaylor Yes, of course: for $y=0$, the infinite sum simplifies pretty easily (be slightly careful though -- there is still one non-zero term)
            $endgroup$
            – Clement C.
            Nov 25 '18 at 0:00










          • $begingroup$
            @ClementC. Which term is that?
            $endgroup$
            – Chase Ryan Taylor
            Nov 25 '18 at 0:12
















          6












          $begingroup$

          Observe
          begin{align}
          f(x) = frac{y}{1-y}sum^infty_{i=1} [e^{-x}(1-y)]^i
          end{align}

          then use geometric series.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            (The case $y=1$ obviously has to be dealt with separately.)
            $endgroup$
            – Clement C.
            Nov 24 '18 at 23:35










          • $begingroup$
            @ClementC. Yes. I will leave it as an exercise. =)
            $endgroup$
            – Jacky Chong
            Nov 24 '18 at 23:43










          • $begingroup$
            ^ Seems pretty easy, at least: looking at the original form, $y=0$ gives $f(x)=sum0$—am I right? (@ClementC.)
            $endgroup$
            – Chase Ryan Taylor
            Nov 24 '18 at 23:44










          • $begingroup$
            @ChaseRyanTaylor Yes, of course: for $y=0$, the infinite sum simplifies pretty easily (be slightly careful though -- there is still one non-zero term)
            $endgroup$
            – Clement C.
            Nov 25 '18 at 0:00










          • $begingroup$
            @ClementC. Which term is that?
            $endgroup$
            – Chase Ryan Taylor
            Nov 25 '18 at 0:12














          6












          6








          6





          $begingroup$

          Observe
          begin{align}
          f(x) = frac{y}{1-y}sum^infty_{i=1} [e^{-x}(1-y)]^i
          end{align}

          then use geometric series.






          share|cite|improve this answer









          $endgroup$



          Observe
          begin{align}
          f(x) = frac{y}{1-y}sum^infty_{i=1} [e^{-x}(1-y)]^i
          end{align}

          then use geometric series.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 '18 at 22:26









          Jacky ChongJacky Chong

          17.9k21128




          17.9k21128












          • $begingroup$
            (The case $y=1$ obviously has to be dealt with separately.)
            $endgroup$
            – Clement C.
            Nov 24 '18 at 23:35










          • $begingroup$
            @ClementC. Yes. I will leave it as an exercise. =)
            $endgroup$
            – Jacky Chong
            Nov 24 '18 at 23:43










          • $begingroup$
            ^ Seems pretty easy, at least: looking at the original form, $y=0$ gives $f(x)=sum0$—am I right? (@ClementC.)
            $endgroup$
            – Chase Ryan Taylor
            Nov 24 '18 at 23:44










          • $begingroup$
            @ChaseRyanTaylor Yes, of course: for $y=0$, the infinite sum simplifies pretty easily (be slightly careful though -- there is still one non-zero term)
            $endgroup$
            – Clement C.
            Nov 25 '18 at 0:00










          • $begingroup$
            @ClementC. Which term is that?
            $endgroup$
            – Chase Ryan Taylor
            Nov 25 '18 at 0:12


















          • $begingroup$
            (The case $y=1$ obviously has to be dealt with separately.)
            $endgroup$
            – Clement C.
            Nov 24 '18 at 23:35










          • $begingroup$
            @ClementC. Yes. I will leave it as an exercise. =)
            $endgroup$
            – Jacky Chong
            Nov 24 '18 at 23:43










          • $begingroup$
            ^ Seems pretty easy, at least: looking at the original form, $y=0$ gives $f(x)=sum0$—am I right? (@ClementC.)
            $endgroup$
            – Chase Ryan Taylor
            Nov 24 '18 at 23:44










          • $begingroup$
            @ChaseRyanTaylor Yes, of course: for $y=0$, the infinite sum simplifies pretty easily (be slightly careful though -- there is still one non-zero term)
            $endgroup$
            – Clement C.
            Nov 25 '18 at 0:00










          • $begingroup$
            @ClementC. Which term is that?
            $endgroup$
            – Chase Ryan Taylor
            Nov 25 '18 at 0:12
















          $begingroup$
          (The case $y=1$ obviously has to be dealt with separately.)
          $endgroup$
          – Clement C.
          Nov 24 '18 at 23:35




          $begingroup$
          (The case $y=1$ obviously has to be dealt with separately.)
          $endgroup$
          – Clement C.
          Nov 24 '18 at 23:35












          $begingroup$
          @ClementC. Yes. I will leave it as an exercise. =)
          $endgroup$
          – Jacky Chong
          Nov 24 '18 at 23:43




          $begingroup$
          @ClementC. Yes. I will leave it as an exercise. =)
          $endgroup$
          – Jacky Chong
          Nov 24 '18 at 23:43












          $begingroup$
          ^ Seems pretty easy, at least: looking at the original form, $y=0$ gives $f(x)=sum0$—am I right? (@ClementC.)
          $endgroup$
          – Chase Ryan Taylor
          Nov 24 '18 at 23:44




          $begingroup$
          ^ Seems pretty easy, at least: looking at the original form, $y=0$ gives $f(x)=sum0$—am I right? (@ClementC.)
          $endgroup$
          – Chase Ryan Taylor
          Nov 24 '18 at 23:44












          $begingroup$
          @ChaseRyanTaylor Yes, of course: for $y=0$, the infinite sum simplifies pretty easily (be slightly careful though -- there is still one non-zero term)
          $endgroup$
          – Clement C.
          Nov 25 '18 at 0:00




          $begingroup$
          @ChaseRyanTaylor Yes, of course: for $y=0$, the infinite sum simplifies pretty easily (be slightly careful though -- there is still one non-zero term)
          $endgroup$
          – Clement C.
          Nov 25 '18 at 0:00












          $begingroup$
          @ClementC. Which term is that?
          $endgroup$
          – Chase Ryan Taylor
          Nov 25 '18 at 0:12




          $begingroup$
          @ClementC. Which term is that?
          $endgroup$
          – Chase Ryan Taylor
          Nov 25 '18 at 0:12


















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