Creating exponential decay function $f$ such that $int_0^1 f = 1$












0












$begingroup$


I am working on a project where I would like to take a weighted average of values in a set, with the weights decreasing exponentially. To do this, I am trying to find a function $f$ such that $int_0^1 f = 1$. $f$ will be used in the following formula:



$$sum_{n=0}^{|S|}S_n * frac{fleft(frac{n}{|S|}right) + fleft(frac{n + 1}{|S|}right)}{2} * frac{1}{|S|} $$



I would appreciate help with two aspects of this.




  1. It has been a while since I took calculus, so I am at a bit of a loss as to how to move forward with finding a function $f$ suitable for these purposes, generally. I have been able to find a few functions which have integrals close to 1, but have only been able to use guess-and-check to find these approximations.

  2. I wonder if I am on the right track here with doing this weighted average, and am curious if I am over-complicating it.


Thanks!






definite-integrals







share|cite|improve this question









$endgroup$












  • $begingroup$
    I am probably confused, so please forgive me for asking, but doesn't $int_0^1{e^{-x}} = -frac{1}{e} + 1$?
    $endgroup$
    – Erik
    Dec 1 '18 at 0:08






  • 1




    $begingroup$
    Indeed it does! Hence I removed my comment and (hopefully) answered your question below.
    $endgroup$
    – Aditya Dua
    Dec 1 '18 at 0:11










  • $begingroup$
    If you do it this way, the weights will not sum to 1.
    $endgroup$
    – eyeballfrog
    Dec 1 '18 at 0:25










  • $begingroup$
    Can you elaborate @eyeballfrog? I understand that it is only an approximation of the integral, but with larger values of $|S|$, the error should be reasonably small, no? Is there another approach that you can recommend?
    $endgroup$
    – Erik
    Dec 1 '18 at 0:31








  • 1




    $begingroup$
    The error is proportional to $1/|S|$, which may not be good enough unless $S$ is very large. Still, you can just use the fact that $sum_0^n a^i = (1 - a^n)/(1-a)$ to quickly normalize the weights without needing the integral at all.
    $endgroup$
    – eyeballfrog
    Dec 1 '18 at 0:52


















0












$begingroup$


I am working on a project where I would like to take a weighted average of values in a set, with the weights decreasing exponentially. To do this, I am trying to find a function $f$ such that $int_0^1 f = 1$. $f$ will be used in the following formula:



$$sum_{n=0}^{|S|}S_n * frac{fleft(frac{n}{|S|}right) + fleft(frac{n + 1}{|S|}right)}{2} * frac{1}{|S|} $$



I would appreciate help with two aspects of this.




  1. It has been a while since I took calculus, so I am at a bit of a loss as to how to move forward with finding a function $f$ suitable for these purposes, generally. I have been able to find a few functions which have integrals close to 1, but have only been able to use guess-and-check to find these approximations.

  2. I wonder if I am on the right track here with doing this weighted average, and am curious if I am over-complicating it.


Thanks!






definite-integrals







share|cite|improve this question









$endgroup$












  • $begingroup$
    I am probably confused, so please forgive me for asking, but doesn't $int_0^1{e^{-x}} = -frac{1}{e} + 1$?
    $endgroup$
    – Erik
    Dec 1 '18 at 0:08






  • 1




    $begingroup$
    Indeed it does! Hence I removed my comment and (hopefully) answered your question below.
    $endgroup$
    – Aditya Dua
    Dec 1 '18 at 0:11










  • $begingroup$
    If you do it this way, the weights will not sum to 1.
    $endgroup$
    – eyeballfrog
    Dec 1 '18 at 0:25










  • $begingroup$
    Can you elaborate @eyeballfrog? I understand that it is only an approximation of the integral, but with larger values of $|S|$, the error should be reasonably small, no? Is there another approach that you can recommend?
    $endgroup$
    – Erik
    Dec 1 '18 at 0:31








  • 1




    $begingroup$
    The error is proportional to $1/|S|$, which may not be good enough unless $S$ is very large. Still, you can just use the fact that $sum_0^n a^i = (1 - a^n)/(1-a)$ to quickly normalize the weights without needing the integral at all.
    $endgroup$
    – eyeballfrog
    Dec 1 '18 at 0:52
















0












0








0





$begingroup$


I am working on a project where I would like to take a weighted average of values in a set, with the weights decreasing exponentially. To do this, I am trying to find a function $f$ such that $int_0^1 f = 1$. $f$ will be used in the following formula:



$$sum_{n=0}^{|S|}S_n * frac{fleft(frac{n}{|S|}right) + fleft(frac{n + 1}{|S|}right)}{2} * frac{1}{|S|} $$



I would appreciate help with two aspects of this.




  1. It has been a while since I took calculus, so I am at a bit of a loss as to how to move forward with finding a function $f$ suitable for these purposes, generally. I have been able to find a few functions which have integrals close to 1, but have only been able to use guess-and-check to find these approximations.

  2. I wonder if I am on the right track here with doing this weighted average, and am curious if I am over-complicating it.


Thanks!






definite-integrals







share|cite|improve this question









$endgroup$




I am working on a project where I would like to take a weighted average of values in a set, with the weights decreasing exponentially. To do this, I am trying to find a function $f$ such that $int_0^1 f = 1$. $f$ will be used in the following formula:



$$sum_{n=0}^{|S|}S_n * frac{fleft(frac{n}{|S|}right) + fleft(frac{n + 1}{|S|}right)}{2} * frac{1}{|S|} $$



I would appreciate help with two aspects of this.




  1. It has been a while since I took calculus, so I am at a bit of a loss as to how to move forward with finding a function $f$ suitable for these purposes, generally. I have been able to find a few functions which have integrals close to 1, but have only been able to use guess-and-check to find these approximations.

  2. I wonder if I am on the right track here with doing this weighted average, and am curious if I am over-complicating it.


Thanks!






definite-integrals




definite-integrals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 30 '18 at 23:58









ErikErik

1033




1033












  • $begingroup$
    I am probably confused, so please forgive me for asking, but doesn't $int_0^1{e^{-x}} = -frac{1}{e} + 1$?
    $endgroup$
    – Erik
    Dec 1 '18 at 0:08






  • 1




    $begingroup$
    Indeed it does! Hence I removed my comment and (hopefully) answered your question below.
    $endgroup$
    – Aditya Dua
    Dec 1 '18 at 0:11










  • $begingroup$
    If you do it this way, the weights will not sum to 1.
    $endgroup$
    – eyeballfrog
    Dec 1 '18 at 0:25










  • $begingroup$
    Can you elaborate @eyeballfrog? I understand that it is only an approximation of the integral, but with larger values of $|S|$, the error should be reasonably small, no? Is there another approach that you can recommend?
    $endgroup$
    – Erik
    Dec 1 '18 at 0:31








  • 1




    $begingroup$
    The error is proportional to $1/|S|$, which may not be good enough unless $S$ is very large. Still, you can just use the fact that $sum_0^n a^i = (1 - a^n)/(1-a)$ to quickly normalize the weights without needing the integral at all.
    $endgroup$
    – eyeballfrog
    Dec 1 '18 at 0:52




















  • $begingroup$
    I am probably confused, so please forgive me for asking, but doesn't $int_0^1{e^{-x}} = -frac{1}{e} + 1$?
    $endgroup$
    – Erik
    Dec 1 '18 at 0:08






  • 1




    $begingroup$
    Indeed it does! Hence I removed my comment and (hopefully) answered your question below.
    $endgroup$
    – Aditya Dua
    Dec 1 '18 at 0:11










  • $begingroup$
    If you do it this way, the weights will not sum to 1.
    $endgroup$
    – eyeballfrog
    Dec 1 '18 at 0:25










  • $begingroup$
    Can you elaborate @eyeballfrog? I understand that it is only an approximation of the integral, but with larger values of $|S|$, the error should be reasonably small, no? Is there another approach that you can recommend?
    $endgroup$
    – Erik
    Dec 1 '18 at 0:31








  • 1




    $begingroup$
    The error is proportional to $1/|S|$, which may not be good enough unless $S$ is very large. Still, you can just use the fact that $sum_0^n a^i = (1 - a^n)/(1-a)$ to quickly normalize the weights without needing the integral at all.
    $endgroup$
    – eyeballfrog
    Dec 1 '18 at 0:52


















$begingroup$
I am probably confused, so please forgive me for asking, but doesn't $int_0^1{e^{-x}} = -frac{1}{e} + 1$?
$endgroup$
– Erik
Dec 1 '18 at 0:08




$begingroup$
I am probably confused, so please forgive me for asking, but doesn't $int_0^1{e^{-x}} = -frac{1}{e} + 1$?
$endgroup$
– Erik
Dec 1 '18 at 0:08




1




1




$begingroup$
Indeed it does! Hence I removed my comment and (hopefully) answered your question below.
$endgroup$
– Aditya Dua
Dec 1 '18 at 0:11




$begingroup$
Indeed it does! Hence I removed my comment and (hopefully) answered your question below.
$endgroup$
– Aditya Dua
Dec 1 '18 at 0:11












$begingroup$
If you do it this way, the weights will not sum to 1.
$endgroup$
– eyeballfrog
Dec 1 '18 at 0:25




$begingroup$
If you do it this way, the weights will not sum to 1.
$endgroup$
– eyeballfrog
Dec 1 '18 at 0:25












$begingroup$
Can you elaborate @eyeballfrog? I understand that it is only an approximation of the integral, but with larger values of $|S|$, the error should be reasonably small, no? Is there another approach that you can recommend?
$endgroup$
– Erik
Dec 1 '18 at 0:31






$begingroup$
Can you elaborate @eyeballfrog? I understand that it is only an approximation of the integral, but with larger values of $|S|$, the error should be reasonably small, no? Is there another approach that you can recommend?
$endgroup$
– Erik
Dec 1 '18 at 0:31






1




1




$begingroup$
The error is proportional to $1/|S|$, which may not be good enough unless $S$ is very large. Still, you can just use the fact that $sum_0^n a^i = (1 - a^n)/(1-a)$ to quickly normalize the weights without needing the integral at all.
$endgroup$
– eyeballfrog
Dec 1 '18 at 0:52






$begingroup$
The error is proportional to $1/|S|$, which may not be good enough unless $S$ is very large. Still, you can just use the fact that $sum_0^n a^i = (1 - a^n)/(1-a)$ to quickly normalize the weights without needing the integral at all.
$endgroup$
– eyeballfrog
Dec 1 '18 at 0:52












1 Answer
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$begingroup$

You can start with the function $f(x) = ae^{-x}$ for some $a > 0$. Now, you have:



$int_0^1 f(x)dx = aint_0^1 e^{-x} dx = aleft(1 - frac{1}{e} right) = 1$. This implies a = $frac{e}{e-1}$ and



$f(x) = frac{e^{1-x}}{e-1}$.






share|cite|improve this answer









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    $begingroup$

    You can start with the function $f(x) = ae^{-x}$ for some $a > 0$. Now, you have:



    $int_0^1 f(x)dx = aint_0^1 e^{-x} dx = aleft(1 - frac{1}{e} right) = 1$. This implies a = $frac{e}{e-1}$ and



    $f(x) = frac{e^{1-x}}{e-1}$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      You can start with the function $f(x) = ae^{-x}$ for some $a > 0$. Now, you have:



      $int_0^1 f(x)dx = aint_0^1 e^{-x} dx = aleft(1 - frac{1}{e} right) = 1$. This implies a = $frac{e}{e-1}$ and



      $f(x) = frac{e^{1-x}}{e-1}$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        You can start with the function $f(x) = ae^{-x}$ for some $a > 0$. Now, you have:



        $int_0^1 f(x)dx = aint_0^1 e^{-x} dx = aleft(1 - frac{1}{e} right) = 1$. This implies a = $frac{e}{e-1}$ and



        $f(x) = frac{e^{1-x}}{e-1}$.






        share|cite|improve this answer









        $endgroup$



        You can start with the function $f(x) = ae^{-x}$ for some $a > 0$. Now, you have:



        $int_0^1 f(x)dx = aint_0^1 e^{-x} dx = aleft(1 - frac{1}{e} right) = 1$. This implies a = $frac{e}{e-1}$ and



        $f(x) = frac{e^{1-x}}{e-1}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 '18 at 0:08









        Aditya DuaAditya Dua

        1,00418




        1,00418






























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