sequence that adds its previous results












0














Let $x = 0.3$.
The first number of the sequence is $x$.
The second number is the first number + $(0.3cdot 0.3)$.
The third number is the second number + $(0.3cdot 0.3cdot 0.3)$.



This is a recursive formula:



$a_1 = 0.3$
$a_{n+1} = a_n + 0.3^{n-1}$



Is it possible to write this equation without recursion?

I want to write a programming function in JavaScript that encodes this without using recursion if possible.






sequences-and-series recursion computability recursive-algorithms







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penfold1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $0.428571, -0.428571 times 0.3^n$
    – David G. Stork
    Jan 4 at 18:49








  • 1




    Is the second of your recursion equations supposed to be $a_{n+1}=a_n + 0.3^{n-1}$ or $a_{n+1}=a_n + 0.3^{n+1}$? The latter would match better with your $a_1$.
    – Henning Makholm
    Jan 4 at 18:50










  • Slight discrepancy in your text and your recursive equation. I have posted an answer based on your recursion, which can be easily modified to match your text if needed.
    – Aditya Dua
    Jan 4 at 19:18










  • Google geometric series.
    – fleablood
    Jan 4 at 19:57
















0














Let $x = 0.3$.
The first number of the sequence is $x$.
The second number is the first number + $(0.3cdot 0.3)$.
The third number is the second number + $(0.3cdot 0.3cdot 0.3)$.



This is a recursive formula:



$a_1 = 0.3$
$a_{n+1} = a_n + 0.3^{n-1}$



Is it possible to write this equation without recursion?

I want to write a programming function in JavaScript that encodes this without using recursion if possible.






sequences-and-series recursion computability recursive-algorithms







share|cite|improve this question









New contributor




penfold1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • $0.428571, -0.428571 times 0.3^n$
    – David G. Stork
    Jan 4 at 18:49








  • 1




    Is the second of your recursion equations supposed to be $a_{n+1}=a_n + 0.3^{n-1}$ or $a_{n+1}=a_n + 0.3^{n+1}$? The latter would match better with your $a_1$.
    – Henning Makholm
    Jan 4 at 18:50










  • Slight discrepancy in your text and your recursive equation. I have posted an answer based on your recursion, which can be easily modified to match your text if needed.
    – Aditya Dua
    Jan 4 at 19:18










  • Google geometric series.
    – fleablood
    Jan 4 at 19:57














0












0








0







Let $x = 0.3$.
The first number of the sequence is $x$.
The second number is the first number + $(0.3cdot 0.3)$.
The third number is the second number + $(0.3cdot 0.3cdot 0.3)$.



This is a recursive formula:



$a_1 = 0.3$
$a_{n+1} = a_n + 0.3^{n-1}$



Is it possible to write this equation without recursion?

I want to write a programming function in JavaScript that encodes this without using recursion if possible.






sequences-and-series recursion computability recursive-algorithms







share|cite|improve this question









New contributor




penfold1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Let $x = 0.3$.
The first number of the sequence is $x$.
The second number is the first number + $(0.3cdot 0.3)$.
The third number is the second number + $(0.3cdot 0.3cdot 0.3)$.



This is a recursive formula:



$a_1 = 0.3$
$a_{n+1} = a_n + 0.3^{n-1}$



Is it possible to write this equation without recursion?

I want to write a programming function in JavaScript that encodes this without using recursion if possible.






sequences-and-series recursion computability recursive-algorithms




sequences-and-series recursion computability recursive-algorithms






share|cite|improve this question









New contributor




penfold1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




penfold1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 18:51









amWhy

192k28225439




192k28225439






New contributor




penfold1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked Jan 4 at 18:41









penfold1992penfold1992

1




1




New contributor




penfold1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





penfold1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






penfold1992 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $0.428571, -0.428571 times 0.3^n$
    – David G. Stork
    Jan 4 at 18:49








  • 1




    Is the second of your recursion equations supposed to be $a_{n+1}=a_n + 0.3^{n-1}$ or $a_{n+1}=a_n + 0.3^{n+1}$? The latter would match better with your $a_1$.
    – Henning Makholm
    Jan 4 at 18:50










  • Slight discrepancy in your text and your recursive equation. I have posted an answer based on your recursion, which can be easily modified to match your text if needed.
    – Aditya Dua
    Jan 4 at 19:18










  • Google geometric series.
    – fleablood
    Jan 4 at 19:57


















  • $0.428571, -0.428571 times 0.3^n$
    – David G. Stork
    Jan 4 at 18:49








  • 1




    Is the second of your recursion equations supposed to be $a_{n+1}=a_n + 0.3^{n-1}$ or $a_{n+1}=a_n + 0.3^{n+1}$? The latter would match better with your $a_1$.
    – Henning Makholm
    Jan 4 at 18:50










  • Slight discrepancy in your text and your recursive equation. I have posted an answer based on your recursion, which can be easily modified to match your text if needed.
    – Aditya Dua
    Jan 4 at 19:18










  • Google geometric series.
    – fleablood
    Jan 4 at 19:57
















$0.428571, -0.428571 times 0.3^n$
– David G. Stork
Jan 4 at 18:49






$0.428571, -0.428571 times 0.3^n$
– David G. Stork
Jan 4 at 18:49






1




1




Is the second of your recursion equations supposed to be $a_{n+1}=a_n + 0.3^{n-1}$ or $a_{n+1}=a_n + 0.3^{n+1}$? The latter would match better with your $a_1$.
– Henning Makholm
Jan 4 at 18:50




Is the second of your recursion equations supposed to be $a_{n+1}=a_n + 0.3^{n-1}$ or $a_{n+1}=a_n + 0.3^{n+1}$? The latter would match better with your $a_1$.
– Henning Makholm
Jan 4 at 18:50












Slight discrepancy in your text and your recursive equation. I have posted an answer based on your recursion, which can be easily modified to match your text if needed.
– Aditya Dua
Jan 4 at 19:18




Slight discrepancy in your text and your recursive equation. I have posted an answer based on your recursion, which can be easily modified to match your text if needed.
– Aditya Dua
Jan 4 at 19:18












Google geometric series.
– fleablood
Jan 4 at 19:57




Google geometric series.
– fleablood
Jan 4 at 19:57










3 Answers
3






active

oldest

votes


















2














Your sequence seems to be $a_n = x + x^2 + x^3 + ..... + x^n$



$= (1 + x + x^2 + x^3 + ...... + x^n) - 1$



$= frac {1-x^{n+1}}{1-x}-1$



$=frac {1-.3^{n+1}}{.7} - 1$



Google geometric series



Or maybe more straightforward:



$a_n = x + x^2 + ...... + x^n =$



$x(1 + ..... + x^{n-1}) =$



$x frac {1-x^n}{1-x} = frac {x - x^{n+1}}{1-x}=$



$frac {.3-.3^{n+1}}{.7}$



Which can be what $frac {1 - .3^{n+1}}{.7} -1 = frac {1 - .3^{n+1}}{.7} -frac {.7}{.7} = frac {.3-.3^{n+1}}{.7}$ is also equal to.



You can also express it as $frac {3 - 10*(.3)^{n+1}}{7}$






share|cite|improve this answer































    0














    You can write:



    $a_{n+1} = a_n + 0.3^{n-1}$



    $= a_{n-1} + 0.3^{n-2} + 0.3^{n-1}$



    $ = a_{n-2} + 0.3^{n-3} + 0.3^{n-2} + 0.3^{n-1}$



    $ ... $



    $ = a_1 + sum_{k=1}^n 0.3^{n-k+1}$



    $ = a_1 + sum_{k=1}^n 0.3^k$



    $ = a_1 + frac{3}{7}(1-0.3^n)$



    $ = 0.3 + frac{3}{7}(1-0.3^n)$



    $ = 0.7286 - 0.4286 times 0.3^n$






    share|cite|improve this answer





























      0














      If we write $$a_n=1+0.3+(0.3)^2+cdots + (0.3)^{n-2}$$then we have $$a_{n+1}=a_n+(0.3)^{n-1}=1+0.3+(0.3)^2+cdots + (0.3)^{n-2}+(0.3)^{n-1}$$also $$1+0.3+(0.3)^2+cdots + (0.3)^{n-2}={1-(0.3)^{n-1}over 1-0.3}$$therefore $$a_n={1-(0.3)^{n-1}over 0.7}$$






      share|cite|improve this answer





















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        3 Answers
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        active

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        3 Answers
        3






        active

        oldest

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        active

        oldest

        votes






        active

        oldest

        votes









        2














        Your sequence seems to be $a_n = x + x^2 + x^3 + ..... + x^n$



        $= (1 + x + x^2 + x^3 + ...... + x^n) - 1$



        $= frac {1-x^{n+1}}{1-x}-1$



        $=frac {1-.3^{n+1}}{.7} - 1$



        Google geometric series



        Or maybe more straightforward:



        $a_n = x + x^2 + ...... + x^n =$



        $x(1 + ..... + x^{n-1}) =$



        $x frac {1-x^n}{1-x} = frac {x - x^{n+1}}{1-x}=$



        $frac {.3-.3^{n+1}}{.7}$



        Which can be what $frac {1 - .3^{n+1}}{.7} -1 = frac {1 - .3^{n+1}}{.7} -frac {.7}{.7} = frac {.3-.3^{n+1}}{.7}$ is also equal to.



        You can also express it as $frac {3 - 10*(.3)^{n+1}}{7}$






        share|cite|improve this answer




























          2














          Your sequence seems to be $a_n = x + x^2 + x^3 + ..... + x^n$



          $= (1 + x + x^2 + x^3 + ...... + x^n) - 1$



          $= frac {1-x^{n+1}}{1-x}-1$



          $=frac {1-.3^{n+1}}{.7} - 1$



          Google geometric series



          Or maybe more straightforward:



          $a_n = x + x^2 + ...... + x^n =$



          $x(1 + ..... + x^{n-1}) =$



          $x frac {1-x^n}{1-x} = frac {x - x^{n+1}}{1-x}=$



          $frac {.3-.3^{n+1}}{.7}$



          Which can be what $frac {1 - .3^{n+1}}{.7} -1 = frac {1 - .3^{n+1}}{.7} -frac {.7}{.7} = frac {.3-.3^{n+1}}{.7}$ is also equal to.



          You can also express it as $frac {3 - 10*(.3)^{n+1}}{7}$






          share|cite|improve this answer


























            2












            2








            2






            Your sequence seems to be $a_n = x + x^2 + x^3 + ..... + x^n$



            $= (1 + x + x^2 + x^3 + ...... + x^n) - 1$



            $= frac {1-x^{n+1}}{1-x}-1$



            $=frac {1-.3^{n+1}}{.7} - 1$



            Google geometric series



            Or maybe more straightforward:



            $a_n = x + x^2 + ...... + x^n =$



            $x(1 + ..... + x^{n-1}) =$



            $x frac {1-x^n}{1-x} = frac {x - x^{n+1}}{1-x}=$



            $frac {.3-.3^{n+1}}{.7}$



            Which can be what $frac {1 - .3^{n+1}}{.7} -1 = frac {1 - .3^{n+1}}{.7} -frac {.7}{.7} = frac {.3-.3^{n+1}}{.7}$ is also equal to.



            You can also express it as $frac {3 - 10*(.3)^{n+1}}{7}$






            share|cite|improve this answer














            Your sequence seems to be $a_n = x + x^2 + x^3 + ..... + x^n$



            $= (1 + x + x^2 + x^3 + ...... + x^n) - 1$



            $= frac {1-x^{n+1}}{1-x}-1$



            $=frac {1-.3^{n+1}}{.7} - 1$



            Google geometric series



            Or maybe more straightforward:



            $a_n = x + x^2 + ...... + x^n =$



            $x(1 + ..... + x^{n-1}) =$



            $x frac {1-x^n}{1-x} = frac {x - x^{n+1}}{1-x}=$



            $frac {.3-.3^{n+1}}{.7}$



            Which can be what $frac {1 - .3^{n+1}}{.7} -1 = frac {1 - .3^{n+1}}{.7} -frac {.7}{.7} = frac {.3-.3^{n+1}}{.7}$ is also equal to.



            You can also express it as $frac {3 - 10*(.3)^{n+1}}{7}$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 4 at 20:06

























            answered Jan 4 at 19:58









            fleabloodfleablood

            68.7k22685




            68.7k22685























                0














                You can write:



                $a_{n+1} = a_n + 0.3^{n-1}$



                $= a_{n-1} + 0.3^{n-2} + 0.3^{n-1}$



                $ = a_{n-2} + 0.3^{n-3} + 0.3^{n-2} + 0.3^{n-1}$



                $ ... $



                $ = a_1 + sum_{k=1}^n 0.3^{n-k+1}$



                $ = a_1 + sum_{k=1}^n 0.3^k$



                $ = a_1 + frac{3}{7}(1-0.3^n)$



                $ = 0.3 + frac{3}{7}(1-0.3^n)$



                $ = 0.7286 - 0.4286 times 0.3^n$






                share|cite|improve this answer


























                  0














                  You can write:



                  $a_{n+1} = a_n + 0.3^{n-1}$



                  $= a_{n-1} + 0.3^{n-2} + 0.3^{n-1}$



                  $ = a_{n-2} + 0.3^{n-3} + 0.3^{n-2} + 0.3^{n-1}$



                  $ ... $



                  $ = a_1 + sum_{k=1}^n 0.3^{n-k+1}$



                  $ = a_1 + sum_{k=1}^n 0.3^k$



                  $ = a_1 + frac{3}{7}(1-0.3^n)$



                  $ = 0.3 + frac{3}{7}(1-0.3^n)$



                  $ = 0.7286 - 0.4286 times 0.3^n$






                  share|cite|improve this answer
























                    0












                    0








                    0






                    You can write:



                    $a_{n+1} = a_n + 0.3^{n-1}$



                    $= a_{n-1} + 0.3^{n-2} + 0.3^{n-1}$



                    $ = a_{n-2} + 0.3^{n-3} + 0.3^{n-2} + 0.3^{n-1}$



                    $ ... $



                    $ = a_1 + sum_{k=1}^n 0.3^{n-k+1}$



                    $ = a_1 + sum_{k=1}^n 0.3^k$



                    $ = a_1 + frac{3}{7}(1-0.3^n)$



                    $ = 0.3 + frac{3}{7}(1-0.3^n)$



                    $ = 0.7286 - 0.4286 times 0.3^n$






                    share|cite|improve this answer












                    You can write:



                    $a_{n+1} = a_n + 0.3^{n-1}$



                    $= a_{n-1} + 0.3^{n-2} + 0.3^{n-1}$



                    $ = a_{n-2} + 0.3^{n-3} + 0.3^{n-2} + 0.3^{n-1}$



                    $ ... $



                    $ = a_1 + sum_{k=1}^n 0.3^{n-k+1}$



                    $ = a_1 + sum_{k=1}^n 0.3^k$



                    $ = a_1 + frac{3}{7}(1-0.3^n)$



                    $ = 0.3 + frac{3}{7}(1-0.3^n)$



                    $ = 0.7286 - 0.4286 times 0.3^n$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 4 at 19:17









                    Aditya DuaAditya Dua

                    90418




                    90418























                        0














                        If we write $$a_n=1+0.3+(0.3)^2+cdots + (0.3)^{n-2}$$then we have $$a_{n+1}=a_n+(0.3)^{n-1}=1+0.3+(0.3)^2+cdots + (0.3)^{n-2}+(0.3)^{n-1}$$also $$1+0.3+(0.3)^2+cdots + (0.3)^{n-2}={1-(0.3)^{n-1}over 1-0.3}$$therefore $$a_n={1-(0.3)^{n-1}over 0.7}$$






                        share|cite|improve this answer


























                          0














                          If we write $$a_n=1+0.3+(0.3)^2+cdots + (0.3)^{n-2}$$then we have $$a_{n+1}=a_n+(0.3)^{n-1}=1+0.3+(0.3)^2+cdots + (0.3)^{n-2}+(0.3)^{n-1}$$also $$1+0.3+(0.3)^2+cdots + (0.3)^{n-2}={1-(0.3)^{n-1}over 1-0.3}$$therefore $$a_n={1-(0.3)^{n-1}over 0.7}$$






                          share|cite|improve this answer
























                            0












                            0








                            0






                            If we write $$a_n=1+0.3+(0.3)^2+cdots + (0.3)^{n-2}$$then we have $$a_{n+1}=a_n+(0.3)^{n-1}=1+0.3+(0.3)^2+cdots + (0.3)^{n-2}+(0.3)^{n-1}$$also $$1+0.3+(0.3)^2+cdots + (0.3)^{n-2}={1-(0.3)^{n-1}over 1-0.3}$$therefore $$a_n={1-(0.3)^{n-1}over 0.7}$$






                            share|cite|improve this answer












                            If we write $$a_n=1+0.3+(0.3)^2+cdots + (0.3)^{n-2}$$then we have $$a_{n+1}=a_n+(0.3)^{n-1}=1+0.3+(0.3)^2+cdots + (0.3)^{n-2}+(0.3)^{n-1}$$also $$1+0.3+(0.3)^2+cdots + (0.3)^{n-2}={1-(0.3)^{n-1}over 1-0.3}$$therefore $$a_n={1-(0.3)^{n-1}over 0.7}$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 5 at 6:40









                            Mostafa AyazMostafa Ayaz

                            14.5k3937




                            14.5k3937






















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                                Has there ever been an instance of an active nuclear power plant within or near a war zone?

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                                36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
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