shifting indices of summation
how come you can rewrite $$sumlimits_{j=i}^{n} j = sumlimits_{j=0}^{n-i} (i+j) = sumlimits_{j=0}^{n-i} i + sumlimits_{j=0}^{n-i} j$$ I don't understand how the indices were shifted and why you're able to do this. Is there a formula? How do you reason about the newly written summation here?
sequences-and-series summation
asked Dec 6 at 17:02
randomvalue
164
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how come you can rewrite $$sumlimits_{j=i}^{n} j = sumlimits_{j=0}^{n-i} (i+j) = sumlimits_{j=0}^{n-i} i + sumlimits_{j=0}^{n-i} j$$ I don't understand how the indices were shifted and why you're able to do this. Is there a formula? How do you reason about the newly written summation here?
sequences-and-series summation
asked Dec 6 at 17:02
randomvalue
164
add a comment |
how come you can rewrite $$sumlimits_{j=i}^{n} j = sumlimits_{j=0}^{n-i} (i+j) = sumlimits_{j=0}^{n-i} i + sumlimits_{j=0}^{n-i} j$$ I don't understand how the indices were shifted and why you're able to do this. Is there a formula? How do you reason about the newly written summation here?
sequences-and-series summation
asked Dec 6 at 17:02
randomvalue
164
how come you can rewrite $$sumlimits_{j=i}^{n} j = sumlimits_{j=0}^{n-i} (i+j) = sumlimits_{j=0}^{n-i} i + sumlimits_{j=0}^{n-i} j$$ I don't understand how the indices were shifted and why you're able to do this. Is there a formula? How do you reason about the newly written summation here?
sequences-and-series summation
sequences-and-series summation
asked Dec 6 at 17:02
randomvalue
164
asked Dec 6 at 17:02
randomvalue
164
asked Dec 6 at 17:02
randomvalue
164
asked Dec 6 at 17:02
randomvalue
164
asked Dec 6 at 17:02
randomvalue
164
164
add a comment |
add a comment |
3 Answers
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Let $k=j-i$. When $j=i,k=0$ and when $j=n,k=n-i$. So we can write $$sumlimits_{j=i}^{n} j =sumlimits_{k=0}^{n-i} i+k $$.
Since the summation is independent of change of variable, we can replace the dummy variable $k$ by $j$ itself. Last equation follows from the distributive rule of summation.
answered Dec 6 at 17:44
Thomas Shelby
1,185116
add a comment |
Just write out what the summations mean.
$$sumlimits_{j=i}^{n} j = i+(i+1)+(i+2)+…+(i+n)$$
$$sumlimits_{j=0}^{n-i} (i+j) = (i+0)+(i+1)+(i+2)+…+(i+n)$$
$$ = i+(i+1)+(i+2)+…+(i+n)$$
If you think about it for a second, it all becomes clear. Usually, writing examples and noticing a pattern is a good way to go. (This isn’t anything “formal” but it often helps you to get the basic idea.) For example, take the following sum:
$$sum_limits{j = 5}^{12} j = 5+6+7+8+9+10+11$$
Imagine you wanted to rewrite the sum by shifting the starting point to, say, $3$.
$$3+4+5+6+7+8+9$$
But all the numbers are shifted down by $2$, so you have to write
$$(3+2)+(4+2)+(5+2)+(6+2)+(7+2)+(8+2)+(9+2)$$
$$= sumlimits_{j = 3}^{9} j+2$$
You can rewrite the summation in yet another way:
$$ = underbrace{3+4+5+6+7+8+9}_{j = 3 text{ to } j = 9}+underbrace{2+2+2+2+2+2+2}_{{j = 3 text{ to } j = 9}} = sumlimits_{j = 3}^{9} j+sumlimits_{j = 3}^{9} 2$$
which is exactly what the summation means.
$$sumlimits_{j=i}^{n} j = sumlimits_{j=i-m}^{n-m} (m+j) = sumlimits_{j=i-m}^{n-m} m + sumlimits_{j=i-m}^{n-m} j$$
For shifting down to $0$, you can generalize this as
$$sumlimits_{j=i}^{n} j = sumlimits_{j=0}^{n-i} (i+j) = sumlimits_{j=0}^{n-i} i + sumlimits_{j=0}^{n-i} j$$
answered Dec 6 at 17:46
KM101
3,958417
add a comment |
$$n = n + 0 = n + (i-i) = i + (n-i), text{so}$$
$$ sum_{j=i}^n j = i + (i+1) + ldots + n = (i+0) + (i+1) + ldots + (i + (n - i)) = sum_{j=0}^{n-i} (i+j) $$.
answered Dec 6 at 17:40
Andrew Whelan
1,358415
add a comment |
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3 Answers
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3 Answers
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Let $k=j-i$. When $j=i,k=0$ and when $j=n,k=n-i$. So we can write $$sumlimits_{j=i}^{n} j =sumlimits_{k=0}^{n-i} i+k $$.
Since the summation is independent of change of variable, we can replace the dummy variable $k$ by $j$ itself. Last equation follows from the distributive rule of summation.
answered Dec 6 at 17:44
Thomas Shelby
1,185116
add a comment |
Let $k=j-i$. When $j=i,k=0$ and when $j=n,k=n-i$. So we can write $$sumlimits_{j=i}^{n} j =sumlimits_{k=0}^{n-i} i+k $$.
Since the summation is independent of change of variable, we can replace the dummy variable $k$ by $j$ itself. Last equation follows from the distributive rule of summation.
answered Dec 6 at 17:44
Thomas Shelby
1,185116
add a comment |
Let $k=j-i$. When $j=i,k=0$ and when $j=n,k=n-i$. So we can write $$sumlimits_{j=i}^{n} j =sumlimits_{k=0}^{n-i} i+k $$.
Since the summation is independent of change of variable, we can replace the dummy variable $k$ by $j$ itself. Last equation follows from the distributive rule of summation.
answered Dec 6 at 17:44
Thomas Shelby
1,185116
Let $k=j-i$. When $j=i,k=0$ and when $j=n,k=n-i$. So we can write $$sumlimits_{j=i}^{n} j =sumlimits_{k=0}^{n-i} i+k $$.
Since the summation is independent of change of variable, we can replace the dummy variable $k$ by $j$ itself. Last equation follows from the distributive rule of summation.
answered Dec 6 at 17:44
Thomas Shelby
1,185116
answered Dec 6 at 17:44
Thomas Shelby
1,185116
answered Dec 6 at 17:44
Thomas Shelby
1,185116
answered Dec 6 at 17:44
Thomas Shelby
1,185116
1,185116
add a comment |
add a comment |
Just write out what the summations mean.
$$sumlimits_{j=i}^{n} j = i+(i+1)+(i+2)+…+(i+n)$$
$$sumlimits_{j=0}^{n-i} (i+j) = (i+0)+(i+1)+(i+2)+…+(i+n)$$
$$ = i+(i+1)+(i+2)+…+(i+n)$$
If you think about it for a second, it all becomes clear. Usually, writing examples and noticing a pattern is a good way to go. (This isn’t anything “formal” but it often helps you to get the basic idea.) For example, take the following sum:
$$sum_limits{j = 5}^{12} j = 5+6+7+8+9+10+11$$
Imagine you wanted to rewrite the sum by shifting the starting point to, say, $3$.
$$3+4+5+6+7+8+9$$
But all the numbers are shifted down by $2$, so you have to write
$$(3+2)+(4+2)+(5+2)+(6+2)+(7+2)+(8+2)+(9+2)$$
$$= sumlimits_{j = 3}^{9} j+2$$
You can rewrite the summation in yet another way:
$$ = underbrace{3+4+5+6+7+8+9}_{j = 3 text{ to } j = 9}+underbrace{2+2+2+2+2+2+2}_{{j = 3 text{ to } j = 9}} = sumlimits_{j = 3}^{9} j+sumlimits_{j = 3}^{9} 2$$
which is exactly what the summation means.
$$sumlimits_{j=i}^{n} j = sumlimits_{j=i-m}^{n-m} (m+j) = sumlimits_{j=i-m}^{n-m} m + sumlimits_{j=i-m}^{n-m} j$$
For shifting down to $0$, you can generalize this as
$$sumlimits_{j=i}^{n} j = sumlimits_{j=0}^{n-i} (i+j) = sumlimits_{j=0}^{n-i} i + sumlimits_{j=0}^{n-i} j$$
answered Dec 6 at 17:46
KM101
3,958417
add a comment |
Just write out what the summations mean.
$$sumlimits_{j=i}^{n} j = i+(i+1)+(i+2)+…+(i+n)$$
$$sumlimits_{j=0}^{n-i} (i+j) = (i+0)+(i+1)+(i+2)+…+(i+n)$$
$$ = i+(i+1)+(i+2)+…+(i+n)$$
If you think about it for a second, it all becomes clear. Usually, writing examples and noticing a pattern is a good way to go. (This isn’t anything “formal” but it often helps you to get the basic idea.) For example, take the following sum:
$$sum_limits{j = 5}^{12} j = 5+6+7+8+9+10+11$$
Imagine you wanted to rewrite the sum by shifting the starting point to, say, $3$.
$$3+4+5+6+7+8+9$$
But all the numbers are shifted down by $2$, so you have to write
$$(3+2)+(4+2)+(5+2)+(6+2)+(7+2)+(8+2)+(9+2)$$
$$= sumlimits_{j = 3}^{9} j+2$$
You can rewrite the summation in yet another way:
$$ = underbrace{3+4+5+6+7+8+9}_{j = 3 text{ to } j = 9}+underbrace{2+2+2+2+2+2+2}_{{j = 3 text{ to } j = 9}} = sumlimits_{j = 3}^{9} j+sumlimits_{j = 3}^{9} 2$$
which is exactly what the summation means.
$$sumlimits_{j=i}^{n} j = sumlimits_{j=i-m}^{n-m} (m+j) = sumlimits_{j=i-m}^{n-m} m + sumlimits_{j=i-m}^{n-m} j$$
For shifting down to $0$, you can generalize this as
$$sumlimits_{j=i}^{n} j = sumlimits_{j=0}^{n-i} (i+j) = sumlimits_{j=0}^{n-i} i + sumlimits_{j=0}^{n-i} j$$
answered Dec 6 at 17:46
KM101
3,958417
add a comment |
Just write out what the summations mean.
$$sumlimits_{j=i}^{n} j = i+(i+1)+(i+2)+…+(i+n)$$
$$sumlimits_{j=0}^{n-i} (i+j) = (i+0)+(i+1)+(i+2)+…+(i+n)$$
$$ = i+(i+1)+(i+2)+…+(i+n)$$
If you think about it for a second, it all becomes clear. Usually, writing examples and noticing a pattern is a good way to go. (This isn’t anything “formal” but it often helps you to get the basic idea.) For example, take the following sum:
$$sum_limits{j = 5}^{12} j = 5+6+7+8+9+10+11$$
Imagine you wanted to rewrite the sum by shifting the starting point to, say, $3$.
$$3+4+5+6+7+8+9$$
But all the numbers are shifted down by $2$, so you have to write
$$(3+2)+(4+2)+(5+2)+(6+2)+(7+2)+(8+2)+(9+2)$$
$$= sumlimits_{j = 3}^{9} j+2$$
You can rewrite the summation in yet another way:
$$ = underbrace{3+4+5+6+7+8+9}_{j = 3 text{ to } j = 9}+underbrace{2+2+2+2+2+2+2}_{{j = 3 text{ to } j = 9}} = sumlimits_{j = 3}^{9} j+sumlimits_{j = 3}^{9} 2$$
which is exactly what the summation means.
$$sumlimits_{j=i}^{n} j = sumlimits_{j=i-m}^{n-m} (m+j) = sumlimits_{j=i-m}^{n-m} m + sumlimits_{j=i-m}^{n-m} j$$
For shifting down to $0$, you can generalize this as
$$sumlimits_{j=i}^{n} j = sumlimits_{j=0}^{n-i} (i+j) = sumlimits_{j=0}^{n-i} i + sumlimits_{j=0}^{n-i} j$$
answered Dec 6 at 17:46
KM101
3,958417
Just write out what the summations mean.
$$sumlimits_{j=i}^{n} j = i+(i+1)+(i+2)+…+(i+n)$$
$$sumlimits_{j=0}^{n-i} (i+j) = (i+0)+(i+1)+(i+2)+…+(i+n)$$
$$ = i+(i+1)+(i+2)+…+(i+n)$$
If you think about it for a second, it all becomes clear. Usually, writing examples and noticing a pattern is a good way to go. (This isn’t anything “formal” but it often helps you to get the basic idea.) For example, take the following sum:
$$sum_limits{j = 5}^{12} j = 5+6+7+8+9+10+11$$
Imagine you wanted to rewrite the sum by shifting the starting point to, say, $3$.
$$3+4+5+6+7+8+9$$
But all the numbers are shifted down by $2$, so you have to write
$$(3+2)+(4+2)+(5+2)+(6+2)+(7+2)+(8+2)+(9+2)$$
$$= sumlimits_{j = 3}^{9} j+2$$
You can rewrite the summation in yet another way:
$$ = underbrace{3+4+5+6+7+8+9}_{j = 3 text{ to } j = 9}+underbrace{2+2+2+2+2+2+2}_{{j = 3 text{ to } j = 9}} = sumlimits_{j = 3}^{9} j+sumlimits_{j = 3}^{9} 2$$
which is exactly what the summation means.
$$sumlimits_{j=i}^{n} j = sumlimits_{j=i-m}^{n-m} (m+j) = sumlimits_{j=i-m}^{n-m} m + sumlimits_{j=i-m}^{n-m} j$$
For shifting down to $0$, you can generalize this as
$$sumlimits_{j=i}^{n} j = sumlimits_{j=0}^{n-i} (i+j) = sumlimits_{j=0}^{n-i} i + sumlimits_{j=0}^{n-i} j$$
answered Dec 6 at 17:46
KM101
3,958417
edited Dec 6 at 19:27
answered Dec 6 at 17:46
KM101
3,958417
answered Dec 6 at 17:46
KM101
3,958417
answered Dec 6 at 17:46
KM101
3,958417
3,958417
add a comment |
add a comment |
$$n = n + 0 = n + (i-i) = i + (n-i), text{so}$$
$$ sum_{j=i}^n j = i + (i+1) + ldots + n = (i+0) + (i+1) + ldots + (i + (n - i)) = sum_{j=0}^{n-i} (i+j) $$.
answered Dec 6 at 17:40
Andrew Whelan
1,358415
add a comment |
$$n = n + 0 = n + (i-i) = i + (n-i), text{so}$$
$$ sum_{j=i}^n j = i + (i+1) + ldots + n = (i+0) + (i+1) + ldots + (i + (n - i)) = sum_{j=0}^{n-i} (i+j) $$.
answered Dec 6 at 17:40
Andrew Whelan
1,358415
add a comment |
$$n = n + 0 = n + (i-i) = i + (n-i), text{so}$$
$$ sum_{j=i}^n j = i + (i+1) + ldots + n = (i+0) + (i+1) + ldots + (i + (n - i)) = sum_{j=0}^{n-i} (i+j) $$.
answered Dec 6 at 17:40
Andrew Whelan
1,358415
$$n = n + 0 = n + (i-i) = i + (n-i), text{so}$$
$$ sum_{j=i}^n j = i + (i+1) + ldots + n = (i+0) + (i+1) + ldots + (i + (n - i)) = sum_{j=0}^{n-i} (i+j) $$.
answered Dec 6 at 17:40
Andrew Whelan
1,358415
answered Dec 6 at 17:40
Andrew Whelan
1,358415
answered Dec 6 at 17:40
Andrew Whelan
1,358415
answered Dec 6 at 17:40
Andrew Whelan
1,358415
1,358415
add a comment |
add a comment |
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