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Property of summation
Summation signs confusionSummation Indices - How to interpret the zero index?Question about index of summationsReversing the Order of Integration and SummationChanging variable in summationSummation that appears to be zeroHow to explain this summation reordering?Why exchange second summation with (n-i)?How to solve summation from k=0 to n-2?Double summation identity
$begingroup$
Very short question. Could you please explain me why
$$sum_i=0^n-1 a = na$$
with $a$ a constant?
I know that
$$sum_i=1^n a = na$$
but in my case the sum starts from zero and finishes for $(n-1)$.
Thanks.
summation
edited 10 hours ago
MarianD
1,2091614
asked 10 hours ago
KolmogorovwannabeKolmogorovwannabe
257
$endgroup$
add a comment |
$begingroup$
Very short question. Could you please explain me why
$$sum_i=0^n-1 a = na$$
with $a$ a constant?
I know that
$$sum_i=1^n a = na$$
but in my case the sum starts from zero and finishes for $(n-1)$.
Thanks.
summation
edited 10 hours ago
MarianD
1,2091614
asked 10 hours ago
KolmogorovwannabeKolmogorovwannabe
257
$endgroup$
7
$begingroup$
Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
$endgroup$
– Yves Daoust
10 hours ago
$begingroup$
Recognize that $0,1,2,3,4,dots,n-1$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements $1,2,3,dots,n-1$ and also the one additional zero element $0$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
$endgroup$
– JMoravitz
10 hours ago
$begingroup$
Thanks JMoravitz, your rationale was what I needed to convince myself.
$endgroup$
– Kolmogorovwannabe
10 hours ago
add a comment |
$begingroup$
Very short question. Could you please explain me why
$$sum_i=0^n-1 a = na$$
with $a$ a constant?
I know that
$$sum_i=1^n a = na$$
but in my case the sum starts from zero and finishes for $(n-1)$.
Thanks.
summation
edited 10 hours ago
MarianD
1,2091614
asked 10 hours ago
KolmogorovwannabeKolmogorovwannabe
257
$endgroup$
Very short question. Could you please explain me why
$$sum_i=0^n-1 a = na$$
with $a$ a constant?
I know that
$$sum_i=1^n a = na$$
but in my case the sum starts from zero and finishes for $(n-1)$.
Thanks.
summation
summation
edited 10 hours ago
MarianD
1,2091614
asked 10 hours ago
KolmogorovwannabeKolmogorovwannabe
257
edited 10 hours ago
MarianD
1,2091614
asked 10 hours ago
KolmogorovwannabeKolmogorovwannabe
257
edited 10 hours ago
MarianD
1,2091614
edited 10 hours ago
MarianD
1,2091614
edited 10 hours ago
MarianD
1,2091614
1,2091614
asked 10 hours ago
KolmogorovwannabeKolmogorovwannabe
257
asked 10 hours ago
KolmogorovwannabeKolmogorovwannabe
257
asked 10 hours ago
KolmogorovwannabeKolmogorovwannabe
257
257
7
$begingroup$
Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
$endgroup$
– Yves Daoust
10 hours ago
$begingroup$
Recognize that $0,1,2,3,4,dots,n-1$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements $1,2,3,dots,n-1$ and also the one additional zero element $0$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
$endgroup$
– JMoravitz
10 hours ago
$begingroup$
Thanks JMoravitz, your rationale was what I needed to convince myself.
$endgroup$
– Kolmogorovwannabe
10 hours ago
add a comment |
7
$begingroup$
Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
$endgroup$
– Yves Daoust
10 hours ago
$begingroup$
Recognize that $0,1,2,3,4,dots,n-1$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements $1,2,3,dots,n-1$ and also the one additional zero element $0$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
$endgroup$
– JMoravitz
10 hours ago
$begingroup$
Thanks JMoravitz, your rationale was what I needed to convince myself.
$endgroup$
– Kolmogorovwannabe
10 hours ago
7
7
$begingroup$
Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
$endgroup$
– Yves Daoust
10 hours ago
$begingroup$
Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
$endgroup$
– Yves Daoust
10 hours ago
$begingroup$
Recognize that $0,1,2,3,4,dots,n-1$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements $1,2,3,dots,n-1$ and also the one additional zero element $0$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
$endgroup$
– JMoravitz
10 hours ago
$begingroup$
Recognize that $0,1,2,3,4,dots,n-1$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements $1,2,3,dots,n-1$ and also the one additional zero element $0$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
$endgroup$
– JMoravitz
10 hours ago
$begingroup$
Thanks JMoravitz, your rationale was what I needed to convince myself.
$endgroup$
– Kolmogorovwannabe
10 hours ago
$begingroup$
Thanks JMoravitz, your rationale was what I needed to convince myself.
$endgroup$
– Kolmogorovwannabe
10 hours ago
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Since you are allready convinced that $sum_i=1^na=na$ this might help:
$sum_i=0^n-1a=a+sum_i=1^n-1a=a+sum_i=1^na-a=sum_i=1^na$
answered 10 hours ago
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
In both cases - $$sum_i=0^n-1aquad text and quadsum_i=1^na$$
- there are exactly $n$ summands.
answered 10 hours ago
MarianDMarianD
1,2091614
$endgroup$
add a comment |
$begingroup$
Since $a$ is not $i$-depending one can write: $$sum_i=0^n-1a=asum_i=0^n-11$$
And $sum_i=0^n-11=1+1+cdots+1$ $n$ times which obviously is $n$.
edited 10 hours ago
MarianD
1,2091614
answered 10 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
98929
$endgroup$
add a comment |
$begingroup$
What is the definition of $sum_i=0^n-1 x_i$? It is exactly $x_0+x_1+...x_n-1$. If $x_0=x_1=...x_n-1=a$ then it means you just sum $a$ $n$ times. And that gives $na$.
answered 10 hours ago
MarkMark
10.1k622
$endgroup$
add a comment |
$begingroup$
You are summing $n$ terms, all equal to $a$. So
$$sum_i=0^n-1 a = a+ a+dotsb + a = na.$$
answered 10 hours ago
HugoHugo
8206
$endgroup$
add a comment |
$begingroup$
Hint:
$a$ times the number of terms.
answered 10 hours ago
Yves DaoustYves Daoust
130k676229
$endgroup$
add a comment |
Your Answer
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since you are allready convinced that $sum_i=1^na=na$ this might help:
$sum_i=0^n-1a=a+sum_i=1^n-1a=a+sum_i=1^na-a=sum_i=1^na$
answered 10 hours ago
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Since you are allready convinced that $sum_i=1^na=na$ this might help:
$sum_i=0^n-1a=a+sum_i=1^n-1a=a+sum_i=1^na-a=sum_i=1^na$
answered 10 hours ago
drhabdrhab
103k545136
$endgroup$
add a comment |
$begingroup$
Since you are allready convinced that $sum_i=1^na=na$ this might help:
$sum_i=0^n-1a=a+sum_i=1^n-1a=a+sum_i=1^na-a=sum_i=1^na$
answered 10 hours ago
drhabdrhab
103k545136
$endgroup$
Since you are allready convinced that $sum_i=1^na=na$ this might help:
$sum_i=0^n-1a=a+sum_i=1^n-1a=a+sum_i=1^na-a=sum_i=1^na$
answered 10 hours ago
drhabdrhab
103k545136
answered 10 hours ago
drhabdrhab
103k545136
answered 10 hours ago
drhabdrhab
103k545136
answered 10 hours ago
drhabdrhab
103k545136
103k545136
add a comment |
add a comment |
$begingroup$
In both cases - $$sum_i=0^n-1aquad text and quadsum_i=1^na$$
- there are exactly $n$ summands.
answered 10 hours ago
MarianDMarianD
1,2091614
$endgroup$
add a comment |
$begingroup$
In both cases - $$sum_i=0^n-1aquad text and quadsum_i=1^na$$
- there are exactly $n$ summands.
answered 10 hours ago
MarianDMarianD
1,2091614
$endgroup$
add a comment |
$begingroup$
In both cases - $$sum_i=0^n-1aquad text and quadsum_i=1^na$$
- there are exactly $n$ summands.
answered 10 hours ago
MarianDMarianD
1,2091614
$endgroup$
In both cases - $$sum_i=0^n-1aquad text and quadsum_i=1^na$$
- there are exactly $n$ summands.
answered 10 hours ago
MarianDMarianD
1,2091614
answered 10 hours ago
MarianDMarianD
1,2091614
answered 10 hours ago
MarianDMarianD
1,2091614
answered 10 hours ago
MarianDMarianD
1,2091614
1,2091614
add a comment |
add a comment |
$begingroup$
Since $a$ is not $i$-depending one can write: $$sum_i=0^n-1a=asum_i=0^n-11$$
And $sum_i=0^n-11=1+1+cdots+1$ $n$ times which obviously is $n$.
edited 10 hours ago
MarianD
1,2091614
answered 10 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
98929
$endgroup$
add a comment |
$begingroup$
Since $a$ is not $i$-depending one can write: $$sum_i=0^n-1a=asum_i=0^n-11$$
And $sum_i=0^n-11=1+1+cdots+1$ $n$ times which obviously is $n$.
edited 10 hours ago
MarianD
1,2091614
answered 10 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
98929
$endgroup$
add a comment |
$begingroup$
Since $a$ is not $i$-depending one can write: $$sum_i=0^n-1a=asum_i=0^n-11$$
And $sum_i=0^n-11=1+1+cdots+1$ $n$ times which obviously is $n$.
edited 10 hours ago
MarianD
1,2091614
answered 10 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
98929
$endgroup$
Since $a$ is not $i$-depending one can write: $$sum_i=0^n-1a=asum_i=0^n-11$$
And $sum_i=0^n-11=1+1+cdots+1$ $n$ times which obviously is $n$.
edited 10 hours ago
MarianD
1,2091614
answered 10 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
98929
edited 10 hours ago
MarianD
1,2091614
edited 10 hours ago
MarianD
1,2091614
edited 10 hours ago
MarianD
1,2091614
1,2091614
answered 10 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
98929
answered 10 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
98929
answered 10 hours ago
HAMIDINE SOUMAREHAMIDINE SOUMARE
98929
98929
add a comment |
add a comment |
$begingroup$
What is the definition of $sum_i=0^n-1 x_i$? It is exactly $x_0+x_1+...x_n-1$. If $x_0=x_1=...x_n-1=a$ then it means you just sum $a$ $n$ times. And that gives $na$.
answered 10 hours ago
MarkMark
10.1k622
$endgroup$
add a comment |
$begingroup$
What is the definition of $sum_i=0^n-1 x_i$? It is exactly $x_0+x_1+...x_n-1$. If $x_0=x_1=...x_n-1=a$ then it means you just sum $a$ $n$ times. And that gives $na$.
answered 10 hours ago
MarkMark
10.1k622
$endgroup$
add a comment |
$begingroup$
What is the definition of $sum_i=0^n-1 x_i$? It is exactly $x_0+x_1+...x_n-1$. If $x_0=x_1=...x_n-1=a$ then it means you just sum $a$ $n$ times. And that gives $na$.
answered 10 hours ago
MarkMark
10.1k622
$endgroup$
What is the definition of $sum_i=0^n-1 x_i$? It is exactly $x_0+x_1+...x_n-1$. If $x_0=x_1=...x_n-1=a$ then it means you just sum $a$ $n$ times. And that gives $na$.
answered 10 hours ago
MarkMark
10.1k622
edited 9 hours ago
answered 10 hours ago
MarkMark
10.1k622
answered 10 hours ago
MarkMark
10.1k622
answered 10 hours ago
MarkMark
10.1k622
10.1k622
add a comment |
add a comment |
$begingroup$
You are summing $n$ terms, all equal to $a$. So
$$sum_i=0^n-1 a = a+ a+dotsb + a = na.$$
answered 10 hours ago
HugoHugo
8206
$endgroup$
add a comment |
$begingroup$
You are summing $n$ terms, all equal to $a$. So
$$sum_i=0^n-1 a = a+ a+dotsb + a = na.$$
answered 10 hours ago
HugoHugo
8206
$endgroup$
add a comment |
$begingroup$
You are summing $n$ terms, all equal to $a$. So
$$sum_i=0^n-1 a = a+ a+dotsb + a = na.$$
answered 10 hours ago
HugoHugo
8206
$endgroup$
You are summing $n$ terms, all equal to $a$. So
$$sum_i=0^n-1 a = a+ a+dotsb + a = na.$$
answered 10 hours ago
HugoHugo
8206
answered 10 hours ago
HugoHugo
8206
answered 10 hours ago
HugoHugo
8206
answered 10 hours ago
HugoHugo
8206
8206
add a comment |
add a comment |
$begingroup$
Hint:
$a$ times the number of terms.
answered 10 hours ago
Yves DaoustYves Daoust
130k676229
$endgroup$
add a comment |
$begingroup$
Hint:
$a$ times the number of terms.
answered 10 hours ago
Yves DaoustYves Daoust
130k676229
$endgroup$
add a comment |
$begingroup$
Hint:
$a$ times the number of terms.
answered 10 hours ago
Yves DaoustYves Daoust
130k676229
$endgroup$
Hint:
$a$ times the number of terms.
answered 10 hours ago
Yves DaoustYves Daoust
130k676229
answered 10 hours ago
Yves DaoustYves Daoust
130k676229
answered 10 hours ago
Yves DaoustYves Daoust
130k676229
answered 10 hours ago
Yves DaoustYves Daoust
130k676229
130k676229
add a comment |
add a comment |
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7
$begingroup$
Sorry, I cannot figure out what is difficult to understand here. Can you enlighten me ?
$endgroup$
– Yves Daoust
10 hours ago
$begingroup$
Recognize that $0,1,2,3,4,dots,n-1$ has $n$ elements in it. If this is not immediately obvious why, then recognize that it has the $n-1$ positive elements $1,2,3,dots,n-1$ and also the one additional zero element $0$. It follows that your summation is iterated a total of $n$ times (the one time when the index is zero, and then the following $n-1$ times while the index is positive for a total of $1+(n-1)=n$ times).
$endgroup$
– JMoravitz
10 hours ago
$begingroup$
Thanks JMoravitz, your rationale was what I needed to convince myself.
$endgroup$
– Kolmogorovwannabe
10 hours ago