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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        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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                                Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
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                                Display a custom attribute below product name in the front-end Magento 1.9.3.8

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                                0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
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