How to calculate this kind of summation?












1












$begingroup$


I'm now on the first chapter of the book 'Mathematical Methods For Physics and Engineering' and I came across this summation.



$displaystylesumlimits_{n=1}^{n}$$displaystylesumlimits_{k>j}^{n} a_ja_k$



How to do this computation ?I tried searching for an explanation in the internet but I got none.






polynomials summation







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is the equal sign supposed to be there?
    $endgroup$
    – fleablood
    Jan 16 at 17:43










  • $begingroup$
    @fleablood nope. It was a mistake i removed it
    $endgroup$
    – Luke_hog
    Jan 16 at 17:59
















1












$begingroup$


I'm now on the first chapter of the book 'Mathematical Methods For Physics and Engineering' and I came across this summation.



$displaystylesumlimits_{n=1}^{n}$$displaystylesumlimits_{k>j}^{n} a_ja_k$



How to do this computation ?I tried searching for an explanation in the internet but I got none.






polynomials summation







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is the equal sign supposed to be there?
    $endgroup$
    – fleablood
    Jan 16 at 17:43










  • $begingroup$
    @fleablood nope. It was a mistake i removed it
    $endgroup$
    – Luke_hog
    Jan 16 at 17:59














1












1








1





$begingroup$


I'm now on the first chapter of the book 'Mathematical Methods For Physics and Engineering' and I came across this summation.



$displaystylesumlimits_{n=1}^{n}$$displaystylesumlimits_{k>j}^{n} a_ja_k$



How to do this computation ?I tried searching for an explanation in the internet but I got none.






polynomials summation







share|cite|improve this question











$endgroup$




I'm now on the first chapter of the book 'Mathematical Methods For Physics and Engineering' and I came across this summation.



$displaystylesumlimits_{n=1}^{n}$$displaystylesumlimits_{k>j}^{n} a_ja_k$



How to do this computation ?I tried searching for an explanation in the internet but I got none.






polynomials summation




polynomials summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 17:59







Luke_hog

















asked Jan 16 at 16:39









Luke_hogLuke_hog

83




83












  • $begingroup$
    Is the equal sign supposed to be there?
    $endgroup$
    – fleablood
    Jan 16 at 17:43










  • $begingroup$
    @fleablood nope. It was a mistake i removed it
    $endgroup$
    – Luke_hog
    Jan 16 at 17:59


















  • $begingroup$
    Is the equal sign supposed to be there?
    $endgroup$
    – fleablood
    Jan 16 at 17:43










  • $begingroup$
    @fleablood nope. It was a mistake i removed it
    $endgroup$
    – Luke_hog
    Jan 16 at 17:59
















$begingroup$
Is the equal sign supposed to be there?
$endgroup$
– fleablood
Jan 16 at 17:43




$begingroup$
Is the equal sign supposed to be there?
$endgroup$
– fleablood
Jan 16 at 17:43












$begingroup$
@fleablood nope. It was a mistake i removed it
$endgroup$
– Luke_hog
Jan 16 at 17:59




$begingroup$
@fleablood nope. It was a mistake i removed it
$endgroup$
– Luke_hog
Jan 16 at 17:59










2 Answers
2






active

oldest

votes


















1












$begingroup$

Downloaded the book. https://www.zuj.edu.jo/download/mathematical-methods-for-physics-and-engineering-riley-hobson-pdf/



Are you doing equation 1.14?



$sumlimits_{j=1}^nsumlimits_{k>j}^n alpha_jalpha_k = frac {a_{n-2}}{a_n}$?



In the case the summation means:



$sumlimits_{j=1}^nsumlimits_{k>j}^n alpha_jalpha_k=$



$sumlimits_{j=1}^n(sumlimits_{k>j}^n alpha_jalpha_k)=$



$sumlimits_{j=1}^n(color{blue}{sumlimits_{ktext{ is any natural number} > jtext{ up to the final summand of } n}^n} alpha_jalpha_k)=$



$sumlimits_{j=1}^n(sumlimits_{k=j+1}^n alpha_jalpha_k)=$



$sumlimits_{j=1}^n(alpha_jalpha_{j+1} + alpha_jalpha_{j+2} + ..... + alpha_jalpha_n) = $



$color{blue}{sumlimits_{jtext { is every natural number from } 1text{ to } n}^n}(alpha_jalpha_{j+1} + alpha_jalpha_{j+2} + ..... + alpha_jalpha_n) = $



$(alpha_1alpha_2 + ................... + alpha_1alpha_n) + $



$(alpha_2alpha_3 + ........... + alpha_2alpha_n) + $



.........



$(alpha_{n-2}alpha_{n-1} + alpha_{n-2}alpha_n) + $



$(alpha_{n-1}alpha_{n})$



....



In short this is the way we write "the sum of all possible $alpha_jalpha_k$ where $k > j$ and and $j,k le n$".






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes , that is the equation i was seeking the explanation for. Thanks for your help
    $endgroup$
    – Luke_hog
    Jan 16 at 18:13





















1












$begingroup$

$$sum_{j=1}^nsum_{k=j+1}^n1=sum_{j=1}^n(n-(j+1)+1)=sum_{j=1}^n(j-1)$$






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    You have a good sense of divination to understand what the sum of the OP is!
    $endgroup$
    – mathcounterexamples.net
    Jan 16 at 16:46










  • $begingroup$
    I am still not clarified but I would like to know how you wrote the summation sign. Do you know a way to write it in android
    $endgroup$
    – Luke_hog
    Jan 16 at 16:48










  • $begingroup$
    @Luke_hog, What is your question? Do you want to calculate the sum ?
    $endgroup$
    – lab bhattacharjee
    Jan 16 at 16:52










  • $begingroup$
    @mathcounterexamples.net if only i could send you the picture , it would be more clear to you. I am just a beginner so i cant post picture.
    $endgroup$
    – Luke_hog
    Jan 16 at 16:57






  • 1




    $begingroup$
    @manooooh thanks for you advice i corrected it.
    $endgroup$
    – Luke_hog
    Jan 16 at 17:38














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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Downloaded the book. https://www.zuj.edu.jo/download/mathematical-methods-for-physics-and-engineering-riley-hobson-pdf/



Are you doing equation 1.14?



$sumlimits_{j=1}^nsumlimits_{k>j}^n alpha_jalpha_k = frac {a_{n-2}}{a_n}$?



In the case the summation means:



$sumlimits_{j=1}^nsumlimits_{k>j}^n alpha_jalpha_k=$



$sumlimits_{j=1}^n(sumlimits_{k>j}^n alpha_jalpha_k)=$



$sumlimits_{j=1}^n(color{blue}{sumlimits_{ktext{ is any natural number} > jtext{ up to the final summand of } n}^n} alpha_jalpha_k)=$



$sumlimits_{j=1}^n(sumlimits_{k=j+1}^n alpha_jalpha_k)=$



$sumlimits_{j=1}^n(alpha_jalpha_{j+1} + alpha_jalpha_{j+2} + ..... + alpha_jalpha_n) = $



$color{blue}{sumlimits_{jtext { is every natural number from } 1text{ to } n}^n}(alpha_jalpha_{j+1} + alpha_jalpha_{j+2} + ..... + alpha_jalpha_n) = $



$(alpha_1alpha_2 + ................... + alpha_1alpha_n) + $



$(alpha_2alpha_3 + ........... + alpha_2alpha_n) + $



.........



$(alpha_{n-2}alpha_{n-1} + alpha_{n-2}alpha_n) + $



$(alpha_{n-1}alpha_{n})$



....



In short this is the way we write "the sum of all possible $alpha_jalpha_k$ where $k > j$ and and $j,k le n$".






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes , that is the equation i was seeking the explanation for. Thanks for your help
    $endgroup$
    – Luke_hog
    Jan 16 at 18:13


















1












$begingroup$

Downloaded the book. https://www.zuj.edu.jo/download/mathematical-methods-for-physics-and-engineering-riley-hobson-pdf/



Are you doing equation 1.14?



$sumlimits_{j=1}^nsumlimits_{k>j}^n alpha_jalpha_k = frac {a_{n-2}}{a_n}$?



In the case the summation means:



$sumlimits_{j=1}^nsumlimits_{k>j}^n alpha_jalpha_k=$



$sumlimits_{j=1}^n(sumlimits_{k>j}^n alpha_jalpha_k)=$



$sumlimits_{j=1}^n(color{blue}{sumlimits_{ktext{ is any natural number} > jtext{ up to the final summand of } n}^n} alpha_jalpha_k)=$



$sumlimits_{j=1}^n(sumlimits_{k=j+1}^n alpha_jalpha_k)=$



$sumlimits_{j=1}^n(alpha_jalpha_{j+1} + alpha_jalpha_{j+2} + ..... + alpha_jalpha_n) = $



$color{blue}{sumlimits_{jtext { is every natural number from } 1text{ to } n}^n}(alpha_jalpha_{j+1} + alpha_jalpha_{j+2} + ..... + alpha_jalpha_n) = $



$(alpha_1alpha_2 + ................... + alpha_1alpha_n) + $



$(alpha_2alpha_3 + ........... + alpha_2alpha_n) + $



.........



$(alpha_{n-2}alpha_{n-1} + alpha_{n-2}alpha_n) + $



$(alpha_{n-1}alpha_{n})$



....



In short this is the way we write "the sum of all possible $alpha_jalpha_k$ where $k > j$ and and $j,k le n$".






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Yes , that is the equation i was seeking the explanation for. Thanks for your help
    $endgroup$
    – Luke_hog
    Jan 16 at 18:13
















1












1








1





$begingroup$

Downloaded the book. https://www.zuj.edu.jo/download/mathematical-methods-for-physics-and-engineering-riley-hobson-pdf/



Are you doing equation 1.14?



$sumlimits_{j=1}^nsumlimits_{k>j}^n alpha_jalpha_k = frac {a_{n-2}}{a_n}$?



In the case the summation means:



$sumlimits_{j=1}^nsumlimits_{k>j}^n alpha_jalpha_k=$



$sumlimits_{j=1}^n(sumlimits_{k>j}^n alpha_jalpha_k)=$



$sumlimits_{j=1}^n(color{blue}{sumlimits_{ktext{ is any natural number} > jtext{ up to the final summand of } n}^n} alpha_jalpha_k)=$



$sumlimits_{j=1}^n(sumlimits_{k=j+1}^n alpha_jalpha_k)=$



$sumlimits_{j=1}^n(alpha_jalpha_{j+1} + alpha_jalpha_{j+2} + ..... + alpha_jalpha_n) = $



$color{blue}{sumlimits_{jtext { is every natural number from } 1text{ to } n}^n}(alpha_jalpha_{j+1} + alpha_jalpha_{j+2} + ..... + alpha_jalpha_n) = $



$(alpha_1alpha_2 + ................... + alpha_1alpha_n) + $



$(alpha_2alpha_3 + ........... + alpha_2alpha_n) + $



.........



$(alpha_{n-2}alpha_{n-1} + alpha_{n-2}alpha_n) + $



$(alpha_{n-1}alpha_{n})$



....



In short this is the way we write "the sum of all possible $alpha_jalpha_k$ where $k > j$ and and $j,k le n$".






share|cite|improve this answer









$endgroup$



Downloaded the book. https://www.zuj.edu.jo/download/mathematical-methods-for-physics-and-engineering-riley-hobson-pdf/



Are you doing equation 1.14?



$sumlimits_{j=1}^nsumlimits_{k>j}^n alpha_jalpha_k = frac {a_{n-2}}{a_n}$?



In the case the summation means:



$sumlimits_{j=1}^nsumlimits_{k>j}^n alpha_jalpha_k=$



$sumlimits_{j=1}^n(sumlimits_{k>j}^n alpha_jalpha_k)=$



$sumlimits_{j=1}^n(color{blue}{sumlimits_{ktext{ is any natural number} > jtext{ up to the final summand of } n}^n} alpha_jalpha_k)=$



$sumlimits_{j=1}^n(sumlimits_{k=j+1}^n alpha_jalpha_k)=$



$sumlimits_{j=1}^n(alpha_jalpha_{j+1} + alpha_jalpha_{j+2} + ..... + alpha_jalpha_n) = $



$color{blue}{sumlimits_{jtext { is every natural number from } 1text{ to } n}^n}(alpha_jalpha_{j+1} + alpha_jalpha_{j+2} + ..... + alpha_jalpha_n) = $



$(alpha_1alpha_2 + ................... + alpha_1alpha_n) + $



$(alpha_2alpha_3 + ........... + alpha_2alpha_n) + $



.........



$(alpha_{n-2}alpha_{n-1} + alpha_{n-2}alpha_n) + $



$(alpha_{n-1}alpha_{n})$



....



In short this is the way we write "the sum of all possible $alpha_jalpha_k$ where $k > j$ and and $j,k le n$".







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 16 at 18:01









fleabloodfleablood

73.8k22891




73.8k22891












  • $begingroup$
    Yes , that is the equation i was seeking the explanation for. Thanks for your help
    $endgroup$
    – Luke_hog
    Jan 16 at 18:13




















  • $begingroup$
    Yes , that is the equation i was seeking the explanation for. Thanks for your help
    $endgroup$
    – Luke_hog
    Jan 16 at 18:13


















$begingroup$
Yes , that is the equation i was seeking the explanation for. Thanks for your help
$endgroup$
– Luke_hog
Jan 16 at 18:13






$begingroup$
Yes , that is the equation i was seeking the explanation for. Thanks for your help
$endgroup$
– Luke_hog
Jan 16 at 18:13













1












$begingroup$

$$sum_{j=1}^nsum_{k=j+1}^n1=sum_{j=1}^n(n-(j+1)+1)=sum_{j=1}^n(j-1)$$






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    You have a good sense of divination to understand what the sum of the OP is!
    $endgroup$
    – mathcounterexamples.net
    Jan 16 at 16:46










  • $begingroup$
    I am still not clarified but I would like to know how you wrote the summation sign. Do you know a way to write it in android
    $endgroup$
    – Luke_hog
    Jan 16 at 16:48










  • $begingroup$
    @Luke_hog, What is your question? Do you want to calculate the sum ?
    $endgroup$
    – lab bhattacharjee
    Jan 16 at 16:52










  • $begingroup$
    @mathcounterexamples.net if only i could send you the picture , it would be more clear to you. I am just a beginner so i cant post picture.
    $endgroup$
    – Luke_hog
    Jan 16 at 16:57






  • 1




    $begingroup$
    @manooooh thanks for you advice i corrected it.
    $endgroup$
    – Luke_hog
    Jan 16 at 17:38


















1












$begingroup$

$$sum_{j=1}^nsum_{k=j+1}^n1=sum_{j=1}^n(n-(j+1)+1)=sum_{j=1}^n(j-1)$$






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    You have a good sense of divination to understand what the sum of the OP is!
    $endgroup$
    – mathcounterexamples.net
    Jan 16 at 16:46










  • $begingroup$
    I am still not clarified but I would like to know how you wrote the summation sign. Do you know a way to write it in android
    $endgroup$
    – Luke_hog
    Jan 16 at 16:48










  • $begingroup$
    @Luke_hog, What is your question? Do you want to calculate the sum ?
    $endgroup$
    – lab bhattacharjee
    Jan 16 at 16:52










  • $begingroup$
    @mathcounterexamples.net if only i could send you the picture , it would be more clear to you. I am just a beginner so i cant post picture.
    $endgroup$
    – Luke_hog
    Jan 16 at 16:57






  • 1




    $begingroup$
    @manooooh thanks for you advice i corrected it.
    $endgroup$
    – Luke_hog
    Jan 16 at 17:38
















1












1








1





$begingroup$

$$sum_{j=1}^nsum_{k=j+1}^n1=sum_{j=1}^n(n-(j+1)+1)=sum_{j=1}^n(j-1)$$






share|cite|improve this answer











$endgroup$



$$sum_{j=1}^nsum_{k=j+1}^n1=sum_{j=1}^n(n-(j+1)+1)=sum_{j=1}^n(j-1)$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 16 at 16:47

























answered Jan 16 at 16:45









lab bhattacharjeelab bhattacharjee

228k15158279




228k15158279








  • 3




    $begingroup$
    You have a good sense of divination to understand what the sum of the OP is!
    $endgroup$
    – mathcounterexamples.net
    Jan 16 at 16:46










  • $begingroup$
    I am still not clarified but I would like to know how you wrote the summation sign. Do you know a way to write it in android
    $endgroup$
    – Luke_hog
    Jan 16 at 16:48










  • $begingroup$
    @Luke_hog, What is your question? Do you want to calculate the sum ?
    $endgroup$
    – lab bhattacharjee
    Jan 16 at 16:52










  • $begingroup$
    @mathcounterexamples.net if only i could send you the picture , it would be more clear to you. I am just a beginner so i cant post picture.
    $endgroup$
    – Luke_hog
    Jan 16 at 16:57






  • 1




    $begingroup$
    @manooooh thanks for you advice i corrected it.
    $endgroup$
    – Luke_hog
    Jan 16 at 17:38
















  • 3




    $begingroup$
    You have a good sense of divination to understand what the sum of the OP is!
    $endgroup$
    – mathcounterexamples.net
    Jan 16 at 16:46










  • $begingroup$
    I am still not clarified but I would like to know how you wrote the summation sign. Do you know a way to write it in android
    $endgroup$
    – Luke_hog
    Jan 16 at 16:48










  • $begingroup$
    @Luke_hog, What is your question? Do you want to calculate the sum ?
    $endgroup$
    – lab bhattacharjee
    Jan 16 at 16:52










  • $begingroup$
    @mathcounterexamples.net if only i could send you the picture , it would be more clear to you. I am just a beginner so i cant post picture.
    $endgroup$
    – Luke_hog
    Jan 16 at 16:57






  • 1




    $begingroup$
    @manooooh thanks for you advice i corrected it.
    $endgroup$
    – Luke_hog
    Jan 16 at 17:38










3




3




$begingroup$
You have a good sense of divination to understand what the sum of the OP is!
$endgroup$
– mathcounterexamples.net
Jan 16 at 16:46




$begingroup$
You have a good sense of divination to understand what the sum of the OP is!
$endgroup$
– mathcounterexamples.net
Jan 16 at 16:46












$begingroup$
I am still not clarified but I would like to know how you wrote the summation sign. Do you know a way to write it in android
$endgroup$
– Luke_hog
Jan 16 at 16:48




$begingroup$
I am still not clarified but I would like to know how you wrote the summation sign. Do you know a way to write it in android
$endgroup$
– Luke_hog
Jan 16 at 16:48












$begingroup$
@Luke_hog, What is your question? Do you want to calculate the sum ?
$endgroup$
– lab bhattacharjee
Jan 16 at 16:52




$begingroup$
@Luke_hog, What is your question? Do you want to calculate the sum ?
$endgroup$
– lab bhattacharjee
Jan 16 at 16:52












$begingroup$
@mathcounterexamples.net if only i could send you the picture , it would be more clear to you. I am just a beginner so i cant post picture.
$endgroup$
– Luke_hog
Jan 16 at 16:57




$begingroup$
@mathcounterexamples.net if only i could send you the picture , it would be more clear to you. I am just a beginner so i cant post picture.
$endgroup$
– Luke_hog
Jan 16 at 16:57




1




1




$begingroup$
@manooooh thanks for you advice i corrected it.
$endgroup$
– Luke_hog
Jan 16 at 17:38






$begingroup$
@manooooh thanks for you advice i corrected it.
$endgroup$
– Luke_hog
Jan 16 at 17:38




















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張江高科駅

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張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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