Mixing
Multiplying summation with same indices and limits
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1
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What would be
$(1-sum limits_{k=0}^m x^k )(1-sum limits_{k=0}^m y^k ) ?$
I dont understand how can I multiply summation of same indices. I checked "multiplication of finite sum (inner product space)" this post but it is different than my case.
Any suggestion?
summation summation-method
asked yesterday
hakkunamattata
454
add a comment |
up vote
1
down vote
favorite
What would be
$(1-sum limits_{k=0}^m x^k )(1-sum limits_{k=0}^m y^k ) ?$
I dont understand how can I multiply summation of same indices. I checked "multiplication of finite sum (inner product space)" this post but it is different than my case.
Any suggestion?
summation summation-method
asked yesterday
hakkunamattata
454
These are both just partial sums of a geometric series, and $$sum_{k = 0}^m x^k = frac{1 - x^{m + 1}}{1 - x}.$$
– T. Bongers
yesterday
k is simply a dummy. For either sum any letter could have been used. The k's in the two sums have no relation to each other.
– herb steinberg
yesterday
As per my case, both summation have same k and same limits. I am not sure how to evaluate $sum limits_{k=0}^{m}sum limits_{k=0}^{m}x^ky^k$
– hakkunamattata
yesterday
@hakkunamattata As herb states above, it is not the "same" $k$. The summation indices are dummy variables, $sum_{k=0}^m x^k = sum_{j=0}^m x^k=sum_{heartsuit=0}^m x^{heartsuit}.$So $$left(sum_{k=0}^m x^kright)left(sum_{k=0}^m y^kright)=left(sum_{k=0}^m x^kright)left(sum_{ell=0}^m y^ellright)=sum_{k=0}^m sum_{ell=0}^m x^ky^ell$$
– Clement C.
yesterday
I suggest you work this out by hand for $m=2$ or $3$. Often summations with $Sigma$ are clearer if you do a small special case. For infinite sums write $a_1 + a_2 + cdots$.
– Ethan Bolker
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
What would be
$(1-sum limits_{k=0}^m x^k )(1-sum limits_{k=0}^m y^k ) ?$
I dont understand how can I multiply summation of same indices. I checked "multiplication of finite sum (inner product space)" this post but it is different than my case.
Any suggestion?
summation summation-method
asked yesterday
hakkunamattata
454
What would be
$(1-sum limits_{k=0}^m x^k )(1-sum limits_{k=0}^m y^k ) ?$
I dont understand how can I multiply summation of same indices. I checked "multiplication of finite sum (inner product space)" this post but it is different than my case.
Any suggestion?
summation summation-method
summation summation-method
asked yesterday
hakkunamattata
454
asked yesterday
hakkunamattata
454
asked yesterday
hakkunamattata
454
asked yesterday
hakkunamattata
454
asked yesterday
hakkunamattata
454
454
These are both just partial sums of a geometric series, and $$sum_{k = 0}^m x^k = frac{1 - x^{m + 1}}{1 - x}.$$
– T. Bongers
yesterday
k is simply a dummy. For either sum any letter could have been used. The k's in the two sums have no relation to each other.
– herb steinberg
yesterday
As per my case, both summation have same k and same limits. I am not sure how to evaluate $sum limits_{k=0}^{m}sum limits_{k=0}^{m}x^ky^k$
– hakkunamattata
yesterday
@hakkunamattata As herb states above, it is not the "same" $k$. The summation indices are dummy variables, $sum_{k=0}^m x^k = sum_{j=0}^m x^k=sum_{heartsuit=0}^m x^{heartsuit}.$So $$left(sum_{k=0}^m x^kright)left(sum_{k=0}^m y^kright)=left(sum_{k=0}^m x^kright)left(sum_{ell=0}^m y^ellright)=sum_{k=0}^m sum_{ell=0}^m x^ky^ell$$
– Clement C.
yesterday
I suggest you work this out by hand for $m=2$ or $3$. Often summations with $Sigma$ are clearer if you do a small special case. For infinite sums write $a_1 + a_2 + cdots$.
– Ethan Bolker
yesterday
add a comment |
These are both just partial sums of a geometric series, and $$sum_{k = 0}^m x^k = frac{1 - x^{m + 1}}{1 - x}.$$
– T. Bongers
yesterday
k is simply a dummy. For either sum any letter could have been used. The k's in the two sums have no relation to each other.
– herb steinberg
yesterday
As per my case, both summation have same k and same limits. I am not sure how to evaluate $sum limits_{k=0}^{m}sum limits_{k=0}^{m}x^ky^k$
– hakkunamattata
yesterday
@hakkunamattata As herb states above, it is not the "same" $k$. The summation indices are dummy variables, $sum_{k=0}^m x^k = sum_{j=0}^m x^k=sum_{heartsuit=0}^m x^{heartsuit}.$So $$left(sum_{k=0}^m x^kright)left(sum_{k=0}^m y^kright)=left(sum_{k=0}^m x^kright)left(sum_{ell=0}^m y^ellright)=sum_{k=0}^m sum_{ell=0}^m x^ky^ell$$
– Clement C.
yesterday
I suggest you work this out by hand for $m=2$ or $3$. Often summations with $Sigma$ are clearer if you do a small special case. For infinite sums write $a_1 + a_2 + cdots$.
– Ethan Bolker
yesterday
These are both just partial sums of a geometric series, and $$sum_{k = 0}^m x^k = frac{1 - x^{m + 1}}{1 - x}.$$
– T. Bongers
yesterday
These are both just partial sums of a geometric series, and $$sum_{k = 0}^m x^k = frac{1 - x^{m + 1}}{1 - x}.$$
– T. Bongers
yesterday
k is simply a dummy. For either sum any letter could have been used. The k's in the two sums have no relation to each other.
– herb steinberg
yesterday
k is simply a dummy. For either sum any letter could have been used. The k's in the two sums have no relation to each other.
– herb steinberg
yesterday
As per my case, both summation have same k and same limits. I am not sure how to evaluate $sum limits_{k=0}^{m}sum limits_{k=0}^{m}x^ky^k$
– hakkunamattata
yesterday
As per my case, both summation have same k and same limits. I am not sure how to evaluate $sum limits_{k=0}^{m}sum limits_{k=0}^{m}x^ky^k$
– hakkunamattata
yesterday
@hakkunamattata As herb states above, it is not the "same" $k$. The summation indices are dummy variables, $sum_{k=0}^m x^k = sum_{j=0}^m x^k=sum_{heartsuit=0}^m x^{heartsuit}.$So $$left(sum_{k=0}^m x^kright)left(sum_{k=0}^m y^kright)=left(sum_{k=0}^m x^kright)left(sum_{ell=0}^m y^ellright)=sum_{k=0}^m sum_{ell=0}^m x^ky^ell$$
– Clement C.
yesterday
@hakkunamattata As herb states above, it is not the "same" $k$. The summation indices are dummy variables, $sum_{k=0}^m x^k = sum_{j=0}^m x^k=sum_{heartsuit=0}^m x^{heartsuit}.$So $$left(sum_{k=0}^m x^kright)left(sum_{k=0}^m y^kright)=left(sum_{k=0}^m x^kright)left(sum_{ell=0}^m y^ellright)=sum_{k=0}^m sum_{ell=0}^m x^ky^ell$$
– Clement C.
yesterday
I suggest you work this out by hand for $m=2$ or $3$. Often summations with $Sigma$ are clearer if you do a small special case. For infinite sums write $a_1 + a_2 + cdots$.
– Ethan Bolker
yesterday
I suggest you work this out by hand for $m=2$ or $3$. Often summations with $Sigma$ are clearer if you do a small special case. For infinite sums write $a_1 + a_2 + cdots$.
– Ethan Bolker
yesterday
add a comment |
2 Answers
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0
down vote
Let's examine series multiplication first:
$$A=sum_{igeq0}a_i=a_0+a_1+a_2+dots$$
$$B=sum_{igeq0}b_i=b_0+b_1+b_2+dots$$
$$AB=big(a_0+a_1+dotsbig)B$$
$$AB=a_0S_2+big(a_1+a_2+dotsbig)B$$
$$AB=a_0S_2+a_1S_2+big(a_2+a_3+dotsbig)B$$
$$AB=a_0S_2+a_1S_2+a_2S_2+big(a_3+a_4+dotsbig)B$$
This pattern continues:
$$AB=a_0B+a_1B+a_2B+a_3B+dots$$
$$AB=sum_{igeq0}a_iB$$
Now note the following:
$$a_iB=a_isum_{kgeq0}b_k$$
$$a_iB=a_ibig(b_0+b_1+b_2+dotsbig)$$
$$a_iB=a_ib_0+a_ib_1+a_ib_2+dots$$
$$a_iB=sum_{kgeq0}a_ib_k$$
Plugging in:
$$AB=sum_{igeq0}sum_{kgeq0}a_ib_k$$
Since $i$ is independent of $k$, and they belong to the same set (namely ${xinBbb Z:xgeq0})$, we know that
$$
begin{align}
AB & = a_0b_0+a_0b_1+a_0b_2+dots \
& + a_1b_0+a_1b_1+a_1b_2+dots \
& + a_2b_0+a_2b_1+a_2b_2+dots \
& +dots
end{align}
$$
Which can be greatly abbreviated:
$$AB=sum_{i,kin S}a_ib_k$$
Where
$$S={xinBbb Z:xgeq0}={0,1,2,dots}$$
Now we can move onto something more related to your problem:
$$A=sum_{i=0}^{m}a_i$$
$$B=sum_{i=0}^{m}b_i$$
These are just like the case above:
$$AB=sum_{i,kin S}a_ib_k$$
Where $$S={xinBbb Z:0leq xleq m}$$
And of course the fact
$$(1-A)(1-B)=AB-A-B+1$$
still holds when $A$ and $B$ are series. But one should note:
$$-A=-a_0-a_1-a_2-dots$$
and not $$-A=-a_0+a_1+a_2+dots$$
answered yesterday
clathratus
1,754219
add a comment |
up vote
0
down vote
The index variables $k$ are so-called bound variables. This means that their scope (i.e. range of validity) is determined by their sigma-operator $sum$ and the operator precedence rules.
The following representations are valid
begin{align*}
left(1+sum_{k=0}^mx^kright)left(1+sum_{k=0}^my^kright)&=
left(1+color{green}{left(sum_{k=0}^mx^kright)}right)left(1+color{blue}{left(sum_{k=0}^my^kright)}right)tag{1}\
&=left(1+sum_{k=0}^mx^kright)left(1+sum_{color{blue}{j=0}}^my^{color{blue}{j}}right)tag{2}
end{align*}
Comment:
In (1) we present the scope of each of the index variables somewhat more clearly by using inner parenthesis and the colors green and blue.
In (2) we denote the index variable of the right-most sum with $j$.
Hint: It is often convenient to give different index variables different names, even if they have no overlapping scope. This usually enhances readability.
answered yesterday
Markus Scheuer
59k454140
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let's examine series multiplication first:
$$A=sum_{igeq0}a_i=a_0+a_1+a_2+dots$$
$$B=sum_{igeq0}b_i=b_0+b_1+b_2+dots$$
$$AB=big(a_0+a_1+dotsbig)B$$
$$AB=a_0S_2+big(a_1+a_2+dotsbig)B$$
$$AB=a_0S_2+a_1S_2+big(a_2+a_3+dotsbig)B$$
$$AB=a_0S_2+a_1S_2+a_2S_2+big(a_3+a_4+dotsbig)B$$
This pattern continues:
$$AB=a_0B+a_1B+a_2B+a_3B+dots$$
$$AB=sum_{igeq0}a_iB$$
Now note the following:
$$a_iB=a_isum_{kgeq0}b_k$$
$$a_iB=a_ibig(b_0+b_1+b_2+dotsbig)$$
$$a_iB=a_ib_0+a_ib_1+a_ib_2+dots$$
$$a_iB=sum_{kgeq0}a_ib_k$$
Plugging in:
$$AB=sum_{igeq0}sum_{kgeq0}a_ib_k$$
Since $i$ is independent of $k$, and they belong to the same set (namely ${xinBbb Z:xgeq0})$, we know that
$$
begin{align}
AB & = a_0b_0+a_0b_1+a_0b_2+dots \
& + a_1b_0+a_1b_1+a_1b_2+dots \
& + a_2b_0+a_2b_1+a_2b_2+dots \
& +dots
end{align}
$$
Which can be greatly abbreviated:
$$AB=sum_{i,kin S}a_ib_k$$
Where
$$S={xinBbb Z:xgeq0}={0,1,2,dots}$$
Now we can move onto something more related to your problem:
$$A=sum_{i=0}^{m}a_i$$
$$B=sum_{i=0}^{m}b_i$$
These are just like the case above:
$$AB=sum_{i,kin S}a_ib_k$$
Where $$S={xinBbb Z:0leq xleq m}$$
And of course the fact
$$(1-A)(1-B)=AB-A-B+1$$
still holds when $A$ and $B$ are series. But one should note:
$$-A=-a_0-a_1-a_2-dots$$
and not $$-A=-a_0+a_1+a_2+dots$$
answered yesterday
clathratus
1,754219
add a comment |
up vote
0
down vote
Let's examine series multiplication first:
$$A=sum_{igeq0}a_i=a_0+a_1+a_2+dots$$
$$B=sum_{igeq0}b_i=b_0+b_1+b_2+dots$$
$$AB=big(a_0+a_1+dotsbig)B$$
$$AB=a_0S_2+big(a_1+a_2+dotsbig)B$$
$$AB=a_0S_2+a_1S_2+big(a_2+a_3+dotsbig)B$$
$$AB=a_0S_2+a_1S_2+a_2S_2+big(a_3+a_4+dotsbig)B$$
This pattern continues:
$$AB=a_0B+a_1B+a_2B+a_3B+dots$$
$$AB=sum_{igeq0}a_iB$$
Now note the following:
$$a_iB=a_isum_{kgeq0}b_k$$
$$a_iB=a_ibig(b_0+b_1+b_2+dotsbig)$$
$$a_iB=a_ib_0+a_ib_1+a_ib_2+dots$$
$$a_iB=sum_{kgeq0}a_ib_k$$
Plugging in:
$$AB=sum_{igeq0}sum_{kgeq0}a_ib_k$$
Since $i$ is independent of $k$, and they belong to the same set (namely ${xinBbb Z:xgeq0})$, we know that
$$
begin{align}
AB & = a_0b_0+a_0b_1+a_0b_2+dots \
& + a_1b_0+a_1b_1+a_1b_2+dots \
& + a_2b_0+a_2b_1+a_2b_2+dots \
& +dots
end{align}
$$
Which can be greatly abbreviated:
$$AB=sum_{i,kin S}a_ib_k$$
Where
$$S={xinBbb Z:xgeq0}={0,1,2,dots}$$
Now we can move onto something more related to your problem:
$$A=sum_{i=0}^{m}a_i$$
$$B=sum_{i=0}^{m}b_i$$
These are just like the case above:
$$AB=sum_{i,kin S}a_ib_k$$
Where $$S={xinBbb Z:0leq xleq m}$$
And of course the fact
$$(1-A)(1-B)=AB-A-B+1$$
still holds when $A$ and $B$ are series. But one should note:
$$-A=-a_0-a_1-a_2-dots$$
and not $$-A=-a_0+a_1+a_2+dots$$
answered yesterday
clathratus
1,754219
add a comment |
up vote
0
down vote
up vote
0
down vote
Let's examine series multiplication first:
$$A=sum_{igeq0}a_i=a_0+a_1+a_2+dots$$
$$B=sum_{igeq0}b_i=b_0+b_1+b_2+dots$$
$$AB=big(a_0+a_1+dotsbig)B$$
$$AB=a_0S_2+big(a_1+a_2+dotsbig)B$$
$$AB=a_0S_2+a_1S_2+big(a_2+a_3+dotsbig)B$$
$$AB=a_0S_2+a_1S_2+a_2S_2+big(a_3+a_4+dotsbig)B$$
This pattern continues:
$$AB=a_0B+a_1B+a_2B+a_3B+dots$$
$$AB=sum_{igeq0}a_iB$$
Now note the following:
$$a_iB=a_isum_{kgeq0}b_k$$
$$a_iB=a_ibig(b_0+b_1+b_2+dotsbig)$$
$$a_iB=a_ib_0+a_ib_1+a_ib_2+dots$$
$$a_iB=sum_{kgeq0}a_ib_k$$
Plugging in:
$$AB=sum_{igeq0}sum_{kgeq0}a_ib_k$$
Since $i$ is independent of $k$, and they belong to the same set (namely ${xinBbb Z:xgeq0})$, we know that
$$
begin{align}
AB & = a_0b_0+a_0b_1+a_0b_2+dots \
& + a_1b_0+a_1b_1+a_1b_2+dots \
& + a_2b_0+a_2b_1+a_2b_2+dots \
& +dots
end{align}
$$
Which can be greatly abbreviated:
$$AB=sum_{i,kin S}a_ib_k$$
Where
$$S={xinBbb Z:xgeq0}={0,1,2,dots}$$
Now we can move onto something more related to your problem:
$$A=sum_{i=0}^{m}a_i$$
$$B=sum_{i=0}^{m}b_i$$
These are just like the case above:
$$AB=sum_{i,kin S}a_ib_k$$
Where $$S={xinBbb Z:0leq xleq m}$$
And of course the fact
$$(1-A)(1-B)=AB-A-B+1$$
still holds when $A$ and $B$ are series. But one should note:
$$-A=-a_0-a_1-a_2-dots$$
and not $$-A=-a_0+a_1+a_2+dots$$
answered yesterday
clathratus
1,754219
Let's examine series multiplication first:
$$A=sum_{igeq0}a_i=a_0+a_1+a_2+dots$$
$$B=sum_{igeq0}b_i=b_0+b_1+b_2+dots$$
$$AB=big(a_0+a_1+dotsbig)B$$
$$AB=a_0S_2+big(a_1+a_2+dotsbig)B$$
$$AB=a_0S_2+a_1S_2+big(a_2+a_3+dotsbig)B$$
$$AB=a_0S_2+a_1S_2+a_2S_2+big(a_3+a_4+dotsbig)B$$
This pattern continues:
$$AB=a_0B+a_1B+a_2B+a_3B+dots$$
$$AB=sum_{igeq0}a_iB$$
Now note the following:
$$a_iB=a_isum_{kgeq0}b_k$$
$$a_iB=a_ibig(b_0+b_1+b_2+dotsbig)$$
$$a_iB=a_ib_0+a_ib_1+a_ib_2+dots$$
$$a_iB=sum_{kgeq0}a_ib_k$$
Plugging in:
$$AB=sum_{igeq0}sum_{kgeq0}a_ib_k$$
Since $i$ is independent of $k$, and they belong to the same set (namely ${xinBbb Z:xgeq0})$, we know that
$$
begin{align}
AB & = a_0b_0+a_0b_1+a_0b_2+dots \
& + a_1b_0+a_1b_1+a_1b_2+dots \
& + a_2b_0+a_2b_1+a_2b_2+dots \
& +dots
end{align}
$$
Which can be greatly abbreviated:
$$AB=sum_{i,kin S}a_ib_k$$
Where
$$S={xinBbb Z:xgeq0}={0,1,2,dots}$$
Now we can move onto something more related to your problem:
$$A=sum_{i=0}^{m}a_i$$
$$B=sum_{i=0}^{m}b_i$$
These are just like the case above:
$$AB=sum_{i,kin S}a_ib_k$$
Where $$S={xinBbb Z:0leq xleq m}$$
And of course the fact
$$(1-A)(1-B)=AB-A-B+1$$
still holds when $A$ and $B$ are series. But one should note:
$$-A=-a_0-a_1-a_2-dots$$
and not $$-A=-a_0+a_1+a_2+dots$$
answered yesterday
clathratus
1,754219
answered yesterday
clathratus
1,754219
answered yesterday
clathratus
1,754219
answered yesterday
clathratus
1,754219
1,754219
add a comment |
add a comment |
up vote
0
down vote
The index variables $k$ are so-called bound variables. This means that their scope (i.e. range of validity) is determined by their sigma-operator $sum$ and the operator precedence rules.
The following representations are valid
begin{align*}
left(1+sum_{k=0}^mx^kright)left(1+sum_{k=0}^my^kright)&=
left(1+color{green}{left(sum_{k=0}^mx^kright)}right)left(1+color{blue}{left(sum_{k=0}^my^kright)}right)tag{1}\
&=left(1+sum_{k=0}^mx^kright)left(1+sum_{color{blue}{j=0}}^my^{color{blue}{j}}right)tag{2}
end{align*}
Comment:
In (1) we present the scope of each of the index variables somewhat more clearly by using inner parenthesis and the colors green and blue.
In (2) we denote the index variable of the right-most sum with $j$.
Hint: It is often convenient to give different index variables different names, even if they have no overlapping scope. This usually enhances readability.
answered yesterday
Markus Scheuer
59k454140
add a comment |
up vote
0
down vote
The index variables $k$ are so-called bound variables. This means that their scope (i.e. range of validity) is determined by their sigma-operator $sum$ and the operator precedence rules.
The following representations are valid
begin{align*}
left(1+sum_{k=0}^mx^kright)left(1+sum_{k=0}^my^kright)&=
left(1+color{green}{left(sum_{k=0}^mx^kright)}right)left(1+color{blue}{left(sum_{k=0}^my^kright)}right)tag{1}\
&=left(1+sum_{k=0}^mx^kright)left(1+sum_{color{blue}{j=0}}^my^{color{blue}{j}}right)tag{2}
end{align*}
Comment:
In (1) we present the scope of each of the index variables somewhat more clearly by using inner parenthesis and the colors green and blue.
In (2) we denote the index variable of the right-most sum with $j$.
Hint: It is often convenient to give different index variables different names, even if they have no overlapping scope. This usually enhances readability.
answered yesterday
Markus Scheuer
59k454140
add a comment |
up vote
0
down vote
up vote
0
down vote
The index variables $k$ are so-called bound variables. This means that their scope (i.e. range of validity) is determined by their sigma-operator $sum$ and the operator precedence rules.
The following representations are valid
begin{align*}
left(1+sum_{k=0}^mx^kright)left(1+sum_{k=0}^my^kright)&=
left(1+color{green}{left(sum_{k=0}^mx^kright)}right)left(1+color{blue}{left(sum_{k=0}^my^kright)}right)tag{1}\
&=left(1+sum_{k=0}^mx^kright)left(1+sum_{color{blue}{j=0}}^my^{color{blue}{j}}right)tag{2}
end{align*}
Comment:
In (1) we present the scope of each of the index variables somewhat more clearly by using inner parenthesis and the colors green and blue.
In (2) we denote the index variable of the right-most sum with $j$.
Hint: It is often convenient to give different index variables different names, even if they have no overlapping scope. This usually enhances readability.
answered yesterday
Markus Scheuer
59k454140
The index variables $k$ are so-called bound variables. This means that their scope (i.e. range of validity) is determined by their sigma-operator $sum$ and the operator precedence rules.
The following representations are valid
begin{align*}
left(1+sum_{k=0}^mx^kright)left(1+sum_{k=0}^my^kright)&=
left(1+color{green}{left(sum_{k=0}^mx^kright)}right)left(1+color{blue}{left(sum_{k=0}^my^kright)}right)tag{1}\
&=left(1+sum_{k=0}^mx^kright)left(1+sum_{color{blue}{j=0}}^my^{color{blue}{j}}right)tag{2}
end{align*}
Comment:
In (1) we present the scope of each of the index variables somewhat more clearly by using inner parenthesis and the colors green and blue.
In (2) we denote the index variable of the right-most sum with $j$.
Hint: It is often convenient to give different index variables different names, even if they have no overlapping scope. This usually enhances readability.
answered yesterday
Markus Scheuer
59k454140
edited 23 hours ago
answered yesterday
Markus Scheuer
59k454140
answered yesterday
Markus Scheuer
59k454140
answered yesterday
Markus Scheuer
59k454140
59k454140
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These are both just partial sums of a geometric series, and $$sum_{k = 0}^m x^k = frac{1 - x^{m + 1}}{1 - x}.$$
– T. Bongers
yesterday
k is simply a dummy. For either sum any letter could have been used. The k's in the two sums have no relation to each other.
– herb steinberg
yesterday
As per my case, both summation have same k and same limits. I am not sure how to evaluate $sum limits_{k=0}^{m}sum limits_{k=0}^{m}x^ky^k$
– hakkunamattata
yesterday
@hakkunamattata As herb states above, it is not the "same" $k$. The summation indices are dummy variables, $sum_{k=0}^m x^k = sum_{j=0}^m x^k=sum_{heartsuit=0}^m x^{heartsuit}.$So $$left(sum_{k=0}^m x^kright)left(sum_{k=0}^m y^kright)=left(sum_{k=0}^m x^kright)left(sum_{ell=0}^m y^ellright)=sum_{k=0}^m sum_{ell=0}^m x^ky^ell$$
– Clement C.
yesterday
I suggest you work this out by hand for $m=2$ or $3$. Often summations with $Sigma$ are clearer if you do a small special case. For infinite sums write $a_1 + a_2 + cdots$.
– Ethan Bolker
yesterday