Multiplying summation with same indices and limits
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 Nov 19 '18 at 0:15
hakkunamattata
475
add a comment |
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 Nov 19 '18 at 0:15
hakkunamattata
475
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
Nov 19 '18 at 0:23
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
Nov 19 '18 at 0:42
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
Nov 19 '18 at 0:54
@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.
Nov 19 '18 at 1:29
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
Nov 19 '18 at 1:57
add a comment |
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 Nov 19 '18 at 0:15
hakkunamattata
475
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 Nov 19 '18 at 0:15
hakkunamattata
475
asked Nov 19 '18 at 0:15
hakkunamattata
475
asked Nov 19 '18 at 0:15
hakkunamattata
475
asked Nov 19 '18 at 0:15
hakkunamattata
475
asked Nov 19 '18 at 0:15
hakkunamattata
475
475
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
Nov 19 '18 at 0:23
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
Nov 19 '18 at 0:42
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
Nov 19 '18 at 0:54
@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.
Nov 19 '18 at 1:29
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
Nov 19 '18 at 1:57
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
Nov 19 '18 at 0:23
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
Nov 19 '18 at 0:42
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
Nov 19 '18 at 0:54
@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.
Nov 19 '18 at 1:29
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
Nov 19 '18 at 1:57
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
Nov 19 '18 at 0:23
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
Nov 19 '18 at 0:23
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
Nov 19 '18 at 0:42
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
Nov 19 '18 at 0:42
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
Nov 19 '18 at 0:54
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
Nov 19 '18 at 0:54
@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.
Nov 19 '18 at 1:29
@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.
Nov 19 '18 at 1:29
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
Nov 19 '18 at 1:57
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
Nov 19 '18 at 1:57
add a comment |
2 Answers
2
active
oldest
votes
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 Nov 19 '18 at 1:52
clathratus
3,126331
add a comment |
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 Nov 19 '18 at 12:52
Markus Scheuer
60k455143
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 Nov 19 '18 at 1:52
clathratus
3,126331
add a comment |
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 Nov 19 '18 at 1:52
clathratus
3,126331
add a comment |
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 Nov 19 '18 at 1:52
clathratus
3,126331
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 Nov 19 '18 at 1:52
clathratus
3,126331
answered Nov 19 '18 at 1:52
clathratus
3,126331
answered Nov 19 '18 at 1:52
clathratus
3,126331
answered Nov 19 '18 at 1:52
clathratus
3,126331
3,126331
add a comment |
add a comment |
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 Nov 19 '18 at 12:52
Markus Scheuer
60k455143
add a comment |
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 Nov 19 '18 at 12:52
Markus Scheuer
60k455143
add a comment |
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 Nov 19 '18 at 12:52
Markus Scheuer
60k455143
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 Nov 19 '18 at 12:52
Markus Scheuer
60k455143
edited Nov 19 '18 at 16:38
answered Nov 19 '18 at 12:52
Markus Scheuer
60k455143
answered Nov 19 '18 at 12:52
Markus Scheuer
60k455143
answered Nov 19 '18 at 12:52
Markus Scheuer
60k455143
60k455143
add a comment |
add a comment |
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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
Nov 19 '18 at 0:23
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
Nov 19 '18 at 0:42
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
Nov 19 '18 at 0:54
@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.
Nov 19 '18 at 1:29
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
Nov 19 '18 at 1:57