Convert expression to standard geometric progression












0














How can I transform a formula describing a curve (a discrete series) to the standard formula for a geometric progression?



(Note: Maths is still very much a foreign language to me. I could be using any of these terms incorrectly. I could be using mathematical notation incorrectly. I could be using MathJax incorrectly. I am trying to describe computer code in mathematical language, and may have any of the translation wrong.)



The existing function computes a sum that is expressed as (I think) the following:



$$
sum_{k=1}^{n} ({x}+({y}({k-1})^{z}))
$$




  • $n$ is an input to the function. It is always an integer > 0.



  • $x$, $y$, $z$ are known (static) parameters in the function.




    • In this specific program, $z$ is typically a negative number, such as $-1.1$.



  • The function produces the sum of the series, for a given $n$.



Currently I have a function that does this via iterative addition. I want instead to use the sum of geometric series formula:



$$
sum_{k=1}^{n} ar^{k-1} = frac{a(1-r^{n})}{1-r}
$$



What I can't figure out is how to transform this so that:




  • The formula meets the geometric series formula, so that $x$, $y$, $z$ are encapsulated instead as appropriate values for $a$ and $r$.


  • The $n$ retains the same value i the geometric series formula, so that existing users of this function can feed the same $n$ and get the same result.



If there's some standard transformations that will make this easier, I simply don't know them. I can noodle around in a spreadsheet and watch the plotted results change for n=[1..1000000], but don't understand the mathematical basis of those effects.



How (preferably explaining what the transformation means) do I transform the $x$, $y$, $z$ from the initial expression, to the $a$ and $r$ of the geometric series expression?






geometric-series







share|cite|improve this question
























  • Isn´t it obvious that $sumlimits_{k=1}^{n} {x}=ncdot x$?
    – callculus
    Dec 31 '18 at 5:48












  • Also note that $sum_{k=1}^{n} ar^{color{red}k} = frac{acdot color{red}r cdot (1-r^{n})}{1-r}$
    – callculus
    Dec 31 '18 at 5:55












  • @callculus, that's not the same expression though? I think maybe my initial expression is ambiguous; I've added more parentheses to better show what's grouped with what.
    – bignose
    Dec 31 '18 at 5:56










  • But you still just add $x$ n-times. You can write $sum_{k=1}^{n} {x}+ ycdot sum_{k=1}^{n} ({k-1})^{z}$ as well.
    – callculus
    Dec 31 '18 at 5:57












  • @callculus, care to set out an answer? I'm going to need more explanation, I fear :-)
    – bignose
    Dec 31 '18 at 6:00
















0














How can I transform a formula describing a curve (a discrete series) to the standard formula for a geometric progression?



(Note: Maths is still very much a foreign language to me. I could be using any of these terms incorrectly. I could be using mathematical notation incorrectly. I could be using MathJax incorrectly. I am trying to describe computer code in mathematical language, and may have any of the translation wrong.)



The existing function computes a sum that is expressed as (I think) the following:



$$
sum_{k=1}^{n} ({x}+({y}({k-1})^{z}))
$$




  • $n$ is an input to the function. It is always an integer > 0.



  • $x$, $y$, $z$ are known (static) parameters in the function.




    • In this specific program, $z$ is typically a negative number, such as $-1.1$.



  • The function produces the sum of the series, for a given $n$.



Currently I have a function that does this via iterative addition. I want instead to use the sum of geometric series formula:



$$
sum_{k=1}^{n} ar^{k-1} = frac{a(1-r^{n})}{1-r}
$$



What I can't figure out is how to transform this so that:




  • The formula meets the geometric series formula, so that $x$, $y$, $z$ are encapsulated instead as appropriate values for $a$ and $r$.


  • The $n$ retains the same value i the geometric series formula, so that existing users of this function can feed the same $n$ and get the same result.



If there's some standard transformations that will make this easier, I simply don't know them. I can noodle around in a spreadsheet and watch the plotted results change for n=[1..1000000], but don't understand the mathematical basis of those effects.



How (preferably explaining what the transformation means) do I transform the $x$, $y$, $z$ from the initial expression, to the $a$ and $r$ of the geometric series expression?






geometric-series







share|cite|improve this question
























  • Isn´t it obvious that $sumlimits_{k=1}^{n} {x}=ncdot x$?
    – callculus
    Dec 31 '18 at 5:48












  • Also note that $sum_{k=1}^{n} ar^{color{red}k} = frac{acdot color{red}r cdot (1-r^{n})}{1-r}$
    – callculus
    Dec 31 '18 at 5:55












  • @callculus, that's not the same expression though? I think maybe my initial expression is ambiguous; I've added more parentheses to better show what's grouped with what.
    – bignose
    Dec 31 '18 at 5:56










  • But you still just add $x$ n-times. You can write $sum_{k=1}^{n} {x}+ ycdot sum_{k=1}^{n} ({k-1})^{z}$ as well.
    – callculus
    Dec 31 '18 at 5:57












  • @callculus, care to set out an answer? I'm going to need more explanation, I fear :-)
    – bignose
    Dec 31 '18 at 6:00














0












0








0







How can I transform a formula describing a curve (a discrete series) to the standard formula for a geometric progression?



(Note: Maths is still very much a foreign language to me. I could be using any of these terms incorrectly. I could be using mathematical notation incorrectly. I could be using MathJax incorrectly. I am trying to describe computer code in mathematical language, and may have any of the translation wrong.)



The existing function computes a sum that is expressed as (I think) the following:



$$
sum_{k=1}^{n} ({x}+({y}({k-1})^{z}))
$$




  • $n$ is an input to the function. It is always an integer > 0.



  • $x$, $y$, $z$ are known (static) parameters in the function.




    • In this specific program, $z$ is typically a negative number, such as $-1.1$.



  • The function produces the sum of the series, for a given $n$.



Currently I have a function that does this via iterative addition. I want instead to use the sum of geometric series formula:



$$
sum_{k=1}^{n} ar^{k-1} = frac{a(1-r^{n})}{1-r}
$$



What I can't figure out is how to transform this so that:




  • The formula meets the geometric series formula, so that $x$, $y$, $z$ are encapsulated instead as appropriate values for $a$ and $r$.


  • The $n$ retains the same value i the geometric series formula, so that existing users of this function can feed the same $n$ and get the same result.



If there's some standard transformations that will make this easier, I simply don't know them. I can noodle around in a spreadsheet and watch the plotted results change for n=[1..1000000], but don't understand the mathematical basis of those effects.



How (preferably explaining what the transformation means) do I transform the $x$, $y$, $z$ from the initial expression, to the $a$ and $r$ of the geometric series expression?






geometric-series







share|cite|improve this question















How can I transform a formula describing a curve (a discrete series) to the standard formula for a geometric progression?



(Note: Maths is still very much a foreign language to me. I could be using any of these terms incorrectly. I could be using mathematical notation incorrectly. I could be using MathJax incorrectly. I am trying to describe computer code in mathematical language, and may have any of the translation wrong.)



The existing function computes a sum that is expressed as (I think) the following:



$$
sum_{k=1}^{n} ({x}+({y}({k-1})^{z}))
$$




  • $n$ is an input to the function. It is always an integer > 0.



  • $x$, $y$, $z$ are known (static) parameters in the function.




    • In this specific program, $z$ is typically a negative number, such as $-1.1$.



  • The function produces the sum of the series, for a given $n$.



Currently I have a function that does this via iterative addition. I want instead to use the sum of geometric series formula:



$$
sum_{k=1}^{n} ar^{k-1} = frac{a(1-r^{n})}{1-r}
$$



What I can't figure out is how to transform this so that:




  • The formula meets the geometric series formula, so that $x$, $y$, $z$ are encapsulated instead as appropriate values for $a$ and $r$.


  • The $n$ retains the same value i the geometric series formula, so that existing users of this function can feed the same $n$ and get the same result.



If there's some standard transformations that will make this easier, I simply don't know them. I can noodle around in a spreadsheet and watch the plotted results change for n=[1..1000000], but don't understand the mathematical basis of those effects.



How (preferably explaining what the transformation means) do I transform the $x$, $y$, $z$ from the initial expression, to the $a$ and $r$ of the geometric series expression?






geometric-series




geometric-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 21:48

























asked Dec 31 '18 at 5:42









bignose

1175




1175












  • Isn´t it obvious that $sumlimits_{k=1}^{n} {x}=ncdot x$?
    – callculus
    Dec 31 '18 at 5:48












  • Also note that $sum_{k=1}^{n} ar^{color{red}k} = frac{acdot color{red}r cdot (1-r^{n})}{1-r}$
    – callculus
    Dec 31 '18 at 5:55












  • @callculus, that's not the same expression though? I think maybe my initial expression is ambiguous; I've added more parentheses to better show what's grouped with what.
    – bignose
    Dec 31 '18 at 5:56










  • But you still just add $x$ n-times. You can write $sum_{k=1}^{n} {x}+ ycdot sum_{k=1}^{n} ({k-1})^{z}$ as well.
    – callculus
    Dec 31 '18 at 5:57












  • @callculus, care to set out an answer? I'm going to need more explanation, I fear :-)
    – bignose
    Dec 31 '18 at 6:00


















  • Isn´t it obvious that $sumlimits_{k=1}^{n} {x}=ncdot x$?
    – callculus
    Dec 31 '18 at 5:48












  • Also note that $sum_{k=1}^{n} ar^{color{red}k} = frac{acdot color{red}r cdot (1-r^{n})}{1-r}$
    – callculus
    Dec 31 '18 at 5:55












  • @callculus, that's not the same expression though? I think maybe my initial expression is ambiguous; I've added more parentheses to better show what's grouped with what.
    – bignose
    Dec 31 '18 at 5:56










  • But you still just add $x$ n-times. You can write $sum_{k=1}^{n} {x}+ ycdot sum_{k=1}^{n} ({k-1})^{z}$ as well.
    – callculus
    Dec 31 '18 at 5:57












  • @callculus, care to set out an answer? I'm going to need more explanation, I fear :-)
    – bignose
    Dec 31 '18 at 6:00
















Isn´t it obvious that $sumlimits_{k=1}^{n} {x}=ncdot x$?
– callculus
Dec 31 '18 at 5:48






Isn´t it obvious that $sumlimits_{k=1}^{n} {x}=ncdot x$?
– callculus
Dec 31 '18 at 5:48














Also note that $sum_{k=1}^{n} ar^{color{red}k} = frac{acdot color{red}r cdot (1-r^{n})}{1-r}$
– callculus
Dec 31 '18 at 5:55






Also note that $sum_{k=1}^{n} ar^{color{red}k} = frac{acdot color{red}r cdot (1-r^{n})}{1-r}$
– callculus
Dec 31 '18 at 5:55














@callculus, that's not the same expression though? I think maybe my initial expression is ambiguous; I've added more parentheses to better show what's grouped with what.
– bignose
Dec 31 '18 at 5:56




@callculus, that's not the same expression though? I think maybe my initial expression is ambiguous; I've added more parentheses to better show what's grouped with what.
– bignose
Dec 31 '18 at 5:56












But you still just add $x$ n-times. You can write $sum_{k=1}^{n} {x}+ ycdot sum_{k=1}^{n} ({k-1})^{z}$ as well.
– callculus
Dec 31 '18 at 5:57






But you still just add $x$ n-times. You can write $sum_{k=1}^{n} {x}+ ycdot sum_{k=1}^{n} ({k-1})^{z}$ as well.
– callculus
Dec 31 '18 at 5:57














@callculus, care to set out an answer? I'm going to need more explanation, I fear :-)
– bignose
Dec 31 '18 at 6:00




@callculus, care to set out an answer? I'm going to need more explanation, I fear :-)
– bignose
Dec 31 '18 at 6:00










2 Answers
2






active

oldest

votes


















2














If your formula
$$S=sum_{k=1}^{n} ({x}+({y}({k-1})^{z}))$$ is correct, I do not see what you can do except writing (assuming $z >0$)
$$S=sum_{k=1}^{n} x+ y sum_{k=1}^{n} (k-1)^z=nx+ysum_{k=2}^{n} (k-1)^z=nx+y, H_{n-1}^{(-z)}$$ where appear generalized harmonic numbers.



For sure, for a given integer value of $z$, you can use Faulhaber's formulae. If this is the case, you will find the expressions for $z=1,2,cdots,10$. If you need for larger values of $z$, it is possible to generate them for you in explicit forms easy to code.






share|cite|improve this answer























  • Thanks. In case it's important, $z$ is negative; I have updated the question to clarify this.
    – bignose
    Jan 2 at 21:50



















1














Synthesising from @Claude-Leibovici and @Mark-S:



This is not a Geometric Series since the base is changing in each term of the sum, not the exponent. So, the expression cannot be transformed to the standard geometric progression.






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    2














    If your formula
    $$S=sum_{k=1}^{n} ({x}+({y}({k-1})^{z}))$$ is correct, I do not see what you can do except writing (assuming $z >0$)
    $$S=sum_{k=1}^{n} x+ y sum_{k=1}^{n} (k-1)^z=nx+ysum_{k=2}^{n} (k-1)^z=nx+y, H_{n-1}^{(-z)}$$ where appear generalized harmonic numbers.



    For sure, for a given integer value of $z$, you can use Faulhaber's formulae. If this is the case, you will find the expressions for $z=1,2,cdots,10$. If you need for larger values of $z$, it is possible to generate them for you in explicit forms easy to code.






    share|cite|improve this answer























    • Thanks. In case it's important, $z$ is negative; I have updated the question to clarify this.
      – bignose
      Jan 2 at 21:50
















    2














    If your formula
    $$S=sum_{k=1}^{n} ({x}+({y}({k-1})^{z}))$$ is correct, I do not see what you can do except writing (assuming $z >0$)
    $$S=sum_{k=1}^{n} x+ y sum_{k=1}^{n} (k-1)^z=nx+ysum_{k=2}^{n} (k-1)^z=nx+y, H_{n-1}^{(-z)}$$ where appear generalized harmonic numbers.



    For sure, for a given integer value of $z$, you can use Faulhaber's formulae. If this is the case, you will find the expressions for $z=1,2,cdots,10$. If you need for larger values of $z$, it is possible to generate them for you in explicit forms easy to code.






    share|cite|improve this answer























    • Thanks. In case it's important, $z$ is negative; I have updated the question to clarify this.
      – bignose
      Jan 2 at 21:50














    2












    2








    2






    If your formula
    $$S=sum_{k=1}^{n} ({x}+({y}({k-1})^{z}))$$ is correct, I do not see what you can do except writing (assuming $z >0$)
    $$S=sum_{k=1}^{n} x+ y sum_{k=1}^{n} (k-1)^z=nx+ysum_{k=2}^{n} (k-1)^z=nx+y, H_{n-1}^{(-z)}$$ where appear generalized harmonic numbers.



    For sure, for a given integer value of $z$, you can use Faulhaber's formulae. If this is the case, you will find the expressions for $z=1,2,cdots,10$. If you need for larger values of $z$, it is possible to generate them for you in explicit forms easy to code.






    share|cite|improve this answer














    If your formula
    $$S=sum_{k=1}^{n} ({x}+({y}({k-1})^{z}))$$ is correct, I do not see what you can do except writing (assuming $z >0$)
    $$S=sum_{k=1}^{n} x+ y sum_{k=1}^{n} (k-1)^z=nx+ysum_{k=2}^{n} (k-1)^z=nx+y, H_{n-1}^{(-z)}$$ where appear generalized harmonic numbers.



    For sure, for a given integer value of $z$, you can use Faulhaber's formulae. If this is the case, you will find the expressions for $z=1,2,cdots,10$. If you need for larger values of $z$, it is possible to generate them for you in explicit forms easy to code.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 31 '18 at 6:47

























    answered Dec 31 '18 at 6:28









    Claude Leibovici

    119k1157132




    119k1157132












    • Thanks. In case it's important, $z$ is negative; I have updated the question to clarify this.
      – bignose
      Jan 2 at 21:50


















    • Thanks. In case it's important, $z$ is negative; I have updated the question to clarify this.
      – bignose
      Jan 2 at 21:50
















    Thanks. In case it's important, $z$ is negative; I have updated the question to clarify this.
    – bignose
    Jan 2 at 21:50




    Thanks. In case it's important, $z$ is negative; I have updated the question to clarify this.
    – bignose
    Jan 2 at 21:50











    1














    Synthesising from @Claude-Leibovici and @Mark-S:



    This is not a Geometric Series since the base is changing in each term of the sum, not the exponent. So, the expression cannot be transformed to the standard geometric progression.






    share|cite|improve this answer


























      1














      Synthesising from @Claude-Leibovici and @Mark-S:



      This is not a Geometric Series since the base is changing in each term of the sum, not the exponent. So, the expression cannot be transformed to the standard geometric progression.






      share|cite|improve this answer
























        1












        1








        1






        Synthesising from @Claude-Leibovici and @Mark-S:



        This is not a Geometric Series since the base is changing in each term of the sum, not the exponent. So, the expression cannot be transformed to the standard geometric progression.






        share|cite|improve this answer












        Synthesising from @Claude-Leibovici and @Mark-S:



        This is not a Geometric Series since the base is changing in each term of the sum, not the exponent. So, the expression cannot be transformed to the standard geometric progression.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 20:05









        bignose

        1175




        1175






























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