Multiplication of sums
$begingroup$
In this set of online math notes the following is stipulated:
$$left(sum_{n=0}^infty a_nright)left(sum_{n=0}^infty b_nright) =sum_{n=0}^infty (c_n) $$
Where apparently $c_n = sum_{i=0}^n a_ib_{n-i}$. Replacing $c_n$ into above equation we get
$$left(sum_{n=0}^infty a_nright)left(sum_{n=0}^infty b_nright) =sum_{n=0}^infty left(sum_{i=0}^n a_ib_{n-i}right) $$
I tried to expand a few terms of $c_n$ using sympy:
$$ {c}_{0} = sum_{i=0}^{0} {a}_{i} {b}_{- i} = {a}_{0} {b}_{0} $$
$$ {c}_{1} = sum_{i=0}^{1} {a}_{i} {b}_{- i + 1} = {a}_{0} {b}_{1} + {a}_{1} {b}_{0} $$
$$ {c}_{2} = sum_{i=0}^{2} {a}_{i} {b}_{- i + 2} = {a}_{0} {b}_{2} + {a}_{1} {b}_{1} + {a}_{2} {b}_{0} $$
Then I ran this example on a verifiable multiplication of sums as follows:
$$(2+x+2x^2)(3-5x+x^2) = 6-7x+3x^2-9x^3+2x^4$$
By setting
begin{equation*}
a_0 = 2, a_1 = x, a_2 = 2x^2
end{equation*}
and
begin{equation*}
b_0 = 3, b_1 = -5x, b_2 = x^2
end{equation*}
Following the online reference this should equal
$$sum_{n=0}^2 c_n =c_0+c_1+c_2 =a_0b_0+a_0b_1+a_1b_0+a_0b_2+a_1b_1+a_2b_0 $$
However when you substitute in the values you get
$$ 3 x^{2} - 7 x + 6 $$
Which is not equal to $6-7x+3x^2-9x^3+2x^4$.
I have a feeling the notes are wrong here. Could anyone confirm this? THanks.
summation
asked Dec 20 '18 at 15:46
user32882user32882
391114
$endgroup$
add a comment |
$begingroup$
In this set of online math notes the following is stipulated:
$$left(sum_{n=0}^infty a_nright)left(sum_{n=0}^infty b_nright) =sum_{n=0}^infty (c_n) $$
Where apparently $c_n = sum_{i=0}^n a_ib_{n-i}$. Replacing $c_n$ into above equation we get
$$left(sum_{n=0}^infty a_nright)left(sum_{n=0}^infty b_nright) =sum_{n=0}^infty left(sum_{i=0}^n a_ib_{n-i}right) $$
I tried to expand a few terms of $c_n$ using sympy:
$$ {c}_{0} = sum_{i=0}^{0} {a}_{i} {b}_{- i} = {a}_{0} {b}_{0} $$
$$ {c}_{1} = sum_{i=0}^{1} {a}_{i} {b}_{- i + 1} = {a}_{0} {b}_{1} + {a}_{1} {b}_{0} $$
$$ {c}_{2} = sum_{i=0}^{2} {a}_{i} {b}_{- i + 2} = {a}_{0} {b}_{2} + {a}_{1} {b}_{1} + {a}_{2} {b}_{0} $$
Then I ran this example on a verifiable multiplication of sums as follows:
$$(2+x+2x^2)(3-5x+x^2) = 6-7x+3x^2-9x^3+2x^4$$
By setting
begin{equation*}
a_0 = 2, a_1 = x, a_2 = 2x^2
end{equation*}
and
begin{equation*}
b_0 = 3, b_1 = -5x, b_2 = x^2
end{equation*}
Following the online reference this should equal
$$sum_{n=0}^2 c_n =c_0+c_1+c_2 =a_0b_0+a_0b_1+a_1b_0+a_0b_2+a_1b_1+a_2b_0 $$
However when you substitute in the values you get
$$ 3 x^{2} - 7 x + 6 $$
Which is not equal to $6-7x+3x^2-9x^3+2x^4$.
I have a feeling the notes are wrong here. Could anyone confirm this? THanks.
summation
asked Dec 20 '18 at 15:46
user32882user32882
391114
$endgroup$
$begingroup$
You forgot some term... at least $a_2b_2$. $3 text { terms } times 3 text { terms } = 9$.
$endgroup$
– Mauro ALLEGRANZA
Dec 20 '18 at 15:55
$begingroup$
The missing terms are : $a_1b_2+a_2b_1+a_2b_2$.
$endgroup$
– Mauro ALLEGRANZA
Dec 20 '18 at 15:59
add a comment |
$begingroup$
In this set of online math notes the following is stipulated:
$$left(sum_{n=0}^infty a_nright)left(sum_{n=0}^infty b_nright) =sum_{n=0}^infty (c_n) $$
Where apparently $c_n = sum_{i=0}^n a_ib_{n-i}$. Replacing $c_n$ into above equation we get
$$left(sum_{n=0}^infty a_nright)left(sum_{n=0}^infty b_nright) =sum_{n=0}^infty left(sum_{i=0}^n a_ib_{n-i}right) $$
I tried to expand a few terms of $c_n$ using sympy:
$$ {c}_{0} = sum_{i=0}^{0} {a}_{i} {b}_{- i} = {a}_{0} {b}_{0} $$
$$ {c}_{1} = sum_{i=0}^{1} {a}_{i} {b}_{- i + 1} = {a}_{0} {b}_{1} + {a}_{1} {b}_{0} $$
$$ {c}_{2} = sum_{i=0}^{2} {a}_{i} {b}_{- i + 2} = {a}_{0} {b}_{2} + {a}_{1} {b}_{1} + {a}_{2} {b}_{0} $$
Then I ran this example on a verifiable multiplication of sums as follows:
$$(2+x+2x^2)(3-5x+x^2) = 6-7x+3x^2-9x^3+2x^4$$
By setting
begin{equation*}
a_0 = 2, a_1 = x, a_2 = 2x^2
end{equation*}
and
begin{equation*}
b_0 = 3, b_1 = -5x, b_2 = x^2
end{equation*}
Following the online reference this should equal
$$sum_{n=0}^2 c_n =c_0+c_1+c_2 =a_0b_0+a_0b_1+a_1b_0+a_0b_2+a_1b_1+a_2b_0 $$
However when you substitute in the values you get
$$ 3 x^{2} - 7 x + 6 $$
Which is not equal to $6-7x+3x^2-9x^3+2x^4$.
I have a feeling the notes are wrong here. Could anyone confirm this? THanks.
summation
asked Dec 20 '18 at 15:46
user32882user32882
391114
$endgroup$
In this set of online math notes the following is stipulated:
$$left(sum_{n=0}^infty a_nright)left(sum_{n=0}^infty b_nright) =sum_{n=0}^infty (c_n) $$
Where apparently $c_n = sum_{i=0}^n a_ib_{n-i}$. Replacing $c_n$ into above equation we get
$$left(sum_{n=0}^infty a_nright)left(sum_{n=0}^infty b_nright) =sum_{n=0}^infty left(sum_{i=0}^n a_ib_{n-i}right) $$
I tried to expand a few terms of $c_n$ using sympy:
$$ {c}_{0} = sum_{i=0}^{0} {a}_{i} {b}_{- i} = {a}_{0} {b}_{0} $$
$$ {c}_{1} = sum_{i=0}^{1} {a}_{i} {b}_{- i + 1} = {a}_{0} {b}_{1} + {a}_{1} {b}_{0} $$
$$ {c}_{2} = sum_{i=0}^{2} {a}_{i} {b}_{- i + 2} = {a}_{0} {b}_{2} + {a}_{1} {b}_{1} + {a}_{2} {b}_{0} $$
Then I ran this example on a verifiable multiplication of sums as follows:
$$(2+x+2x^2)(3-5x+x^2) = 6-7x+3x^2-9x^3+2x^4$$
By setting
begin{equation*}
a_0 = 2, a_1 = x, a_2 = 2x^2
end{equation*}
and
begin{equation*}
b_0 = 3, b_1 = -5x, b_2 = x^2
end{equation*}
Following the online reference this should equal
$$sum_{n=0}^2 c_n =c_0+c_1+c_2 =a_0b_0+a_0b_1+a_1b_0+a_0b_2+a_1b_1+a_2b_0 $$
However when you substitute in the values you get
$$ 3 x^{2} - 7 x + 6 $$
Which is not equal to $6-7x+3x^2-9x^3+2x^4$.
I have a feeling the notes are wrong here. Could anyone confirm this? THanks.
summation
summation
asked Dec 20 '18 at 15:46
user32882user32882
391114
asked Dec 20 '18 at 15:46
user32882user32882
391114
asked Dec 20 '18 at 15:46
user32882user32882
391114
asked Dec 20 '18 at 15:46
user32882user32882
391114
asked Dec 20 '18 at 15:46
user32882user32882
391114
391114
$begingroup$
You forgot some term... at least $a_2b_2$. $3 text { terms } times 3 text { terms } = 9$.
$endgroup$
– Mauro ALLEGRANZA
Dec 20 '18 at 15:55
$begingroup$
The missing terms are : $a_1b_2+a_2b_1+a_2b_2$.
$endgroup$
– Mauro ALLEGRANZA
Dec 20 '18 at 15:59
add a comment |
$begingroup$
You forgot some term... at least $a_2b_2$. $3 text { terms } times 3 text { terms } = 9$.
$endgroup$
– Mauro ALLEGRANZA
Dec 20 '18 at 15:55
$begingroup$
The missing terms are : $a_1b_2+a_2b_1+a_2b_2$.
$endgroup$
– Mauro ALLEGRANZA
Dec 20 '18 at 15:59
$begingroup$
You forgot some term... at least $a_2b_2$. $3 text { terms } times 3 text { terms } = 9$.
$endgroup$
– Mauro ALLEGRANZA
Dec 20 '18 at 15:55
$begingroup$
You forgot some term... at least $a_2b_2$. $3 text { terms } times 3 text { terms } = 9$.
$endgroup$
– Mauro ALLEGRANZA
Dec 20 '18 at 15:55
$begingroup$
The missing terms are : $a_1b_2+a_2b_1+a_2b_2$.
$endgroup$
– Mauro ALLEGRANZA
Dec 20 '18 at 15:59
$begingroup$
The missing terms are : $a_1b_2+a_2b_1+a_2b_2$.
$endgroup$
– Mauro ALLEGRANZA
Dec 20 '18 at 15:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The added terms, $-9x^3+2x^4$, will come from $c_3$ and $c_4$, in particular $a_1b_2+a_2b_1$ for $c_3$ and $a_2b_2$ for $c_4$. You have truncated the $c$ sum too early to see them.
answered Dec 20 '18 at 15:53
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
but $n$ in my case goes from $0$ to $2$.... What is the correct expression for finite sums?
$endgroup$
– user32882
Dec 20 '18 at 15:55
$begingroup$
For finite sums you need the $c$ sum to go up to the sum of the upper limits of the $a$ and $b$ sums. Your example shows why. Without that you miss some terms. When the sums are infinite this is not so obvious.
$endgroup$
– Ross Millikan
Dec 20 '18 at 15:58
$begingroup$
Could you provide an expression which gives the correct answer for finite sums?
$endgroup$
– user32882
Dec 21 '18 at 14:47
$begingroup$
Yours does as long as the $c$ sum goes up to $2n$, not to $n$ and you regard $a_i,b_i=0$ for $i gt n$
$endgroup$
– Ross Millikan
Dec 21 '18 at 14:51
add a comment |
$begingroup$
The first identity should read $sum_{nge 0}a_n x^nsum_{nge 0}b_n x^n=sum_{nge 0}(sum_{i= 0}^n a_i b_{n-i})x^n$, or even better $sum_{jge 0}a_j x^jsum_{kge 0}b_k x^k=sum_{nge 0}(sum_{i= 0}^n a_i b_{n-i})x^n$. The proof is simple: just find all contributions to the $x^n$ coefficient on the left-hand side. When multiplying polynomials, which have finite degrees, all sufficiently late terms in each sequence are $0$. In this case $a_n=0$ for $n>2$, $b_n=0$ for $n>2$, $c_n=0$ for $n>4$.
answered Dec 20 '18 at 15:58
J.G.J.G.
28.1k22844
$endgroup$
$begingroup$
The convolution works with x=1 just as easily, so nothing wrong with the first identity just as stated.
$endgroup$
– Macavity
Dec 21 '18 at 3:18
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The added terms, $-9x^3+2x^4$, will come from $c_3$ and $c_4$, in particular $a_1b_2+a_2b_1$ for $c_3$ and $a_2b_2$ for $c_4$. You have truncated the $c$ sum too early to see them.
answered Dec 20 '18 at 15:53
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
but $n$ in my case goes from $0$ to $2$.... What is the correct expression for finite sums?
$endgroup$
– user32882
Dec 20 '18 at 15:55
$begingroup$
For finite sums you need the $c$ sum to go up to the sum of the upper limits of the $a$ and $b$ sums. Your example shows why. Without that you miss some terms. When the sums are infinite this is not so obvious.
$endgroup$
– Ross Millikan
Dec 20 '18 at 15:58
$begingroup$
Could you provide an expression which gives the correct answer for finite sums?
$endgroup$
– user32882
Dec 21 '18 at 14:47
$begingroup$
Yours does as long as the $c$ sum goes up to $2n$, not to $n$ and you regard $a_i,b_i=0$ for $i gt n$
$endgroup$
– Ross Millikan
Dec 21 '18 at 14:51
add a comment |
$begingroup$
The added terms, $-9x^3+2x^4$, will come from $c_3$ and $c_4$, in particular $a_1b_2+a_2b_1$ for $c_3$ and $a_2b_2$ for $c_4$. You have truncated the $c$ sum too early to see them.
answered Dec 20 '18 at 15:53
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
but $n$ in my case goes from $0$ to $2$.... What is the correct expression for finite sums?
$endgroup$
– user32882
Dec 20 '18 at 15:55
$begingroup$
For finite sums you need the $c$ sum to go up to the sum of the upper limits of the $a$ and $b$ sums. Your example shows why. Without that you miss some terms. When the sums are infinite this is not so obvious.
$endgroup$
– Ross Millikan
Dec 20 '18 at 15:58
$begingroup$
Could you provide an expression which gives the correct answer for finite sums?
$endgroup$
– user32882
Dec 21 '18 at 14:47
$begingroup$
Yours does as long as the $c$ sum goes up to $2n$, not to $n$ and you regard $a_i,b_i=0$ for $i gt n$
$endgroup$
– Ross Millikan
Dec 21 '18 at 14:51
add a comment |
$begingroup$
The added terms, $-9x^3+2x^4$, will come from $c_3$ and $c_4$, in particular $a_1b_2+a_2b_1$ for $c_3$ and $a_2b_2$ for $c_4$. You have truncated the $c$ sum too early to see them.
answered Dec 20 '18 at 15:53
Ross MillikanRoss Millikan
297k23198371
$endgroup$
The added terms, $-9x^3+2x^4$, will come from $c_3$ and $c_4$, in particular $a_1b_2+a_2b_1$ for $c_3$ and $a_2b_2$ for $c_4$. You have truncated the $c$ sum too early to see them.
answered Dec 20 '18 at 15:53
Ross MillikanRoss Millikan
297k23198371
answered Dec 20 '18 at 15:53
Ross MillikanRoss Millikan
297k23198371
answered Dec 20 '18 at 15:53
Ross MillikanRoss Millikan
297k23198371
answered Dec 20 '18 at 15:53
Ross MillikanRoss Millikan
297k23198371
297k23198371
$begingroup$
but $n$ in my case goes from $0$ to $2$.... What is the correct expression for finite sums?
$endgroup$
– user32882
Dec 20 '18 at 15:55
$begingroup$
For finite sums you need the $c$ sum to go up to the sum of the upper limits of the $a$ and $b$ sums. Your example shows why. Without that you miss some terms. When the sums are infinite this is not so obvious.
$endgroup$
– Ross Millikan
Dec 20 '18 at 15:58
$begingroup$
Could you provide an expression which gives the correct answer for finite sums?
$endgroup$
– user32882
Dec 21 '18 at 14:47
$begingroup$
Yours does as long as the $c$ sum goes up to $2n$, not to $n$ and you regard $a_i,b_i=0$ for $i gt n$
$endgroup$
– Ross Millikan
Dec 21 '18 at 14:51
add a comment |
$begingroup$
but $n$ in my case goes from $0$ to $2$.... What is the correct expression for finite sums?
$endgroup$
– user32882
Dec 20 '18 at 15:55
$begingroup$
For finite sums you need the $c$ sum to go up to the sum of the upper limits of the $a$ and $b$ sums. Your example shows why. Without that you miss some terms. When the sums are infinite this is not so obvious.
$endgroup$
– Ross Millikan
Dec 20 '18 at 15:58
$begingroup$
Could you provide an expression which gives the correct answer for finite sums?
$endgroup$
– user32882
Dec 21 '18 at 14:47
$begingroup$
Yours does as long as the $c$ sum goes up to $2n$, not to $n$ and you regard $a_i,b_i=0$ for $i gt n$
$endgroup$
– Ross Millikan
Dec 21 '18 at 14:51
$begingroup$
but $n$ in my case goes from $0$ to $2$.... What is the correct expression for finite sums?
$endgroup$
– user32882
Dec 20 '18 at 15:55
$begingroup$
but $n$ in my case goes from $0$ to $2$.... What is the correct expression for finite sums?
$endgroup$
– user32882
Dec 20 '18 at 15:55
$begingroup$
For finite sums you need the $c$ sum to go up to the sum of the upper limits of the $a$ and $b$ sums. Your example shows why. Without that you miss some terms. When the sums are infinite this is not so obvious.
$endgroup$
– Ross Millikan
Dec 20 '18 at 15:58
$begingroup$
For finite sums you need the $c$ sum to go up to the sum of the upper limits of the $a$ and $b$ sums. Your example shows why. Without that you miss some terms. When the sums are infinite this is not so obvious.
$endgroup$
– Ross Millikan
Dec 20 '18 at 15:58
$begingroup$
Could you provide an expression which gives the correct answer for finite sums?
$endgroup$
– user32882
Dec 21 '18 at 14:47
$begingroup$
Could you provide an expression which gives the correct answer for finite sums?
$endgroup$
– user32882
Dec 21 '18 at 14:47
$begingroup$
Yours does as long as the $c$ sum goes up to $2n$, not to $n$ and you regard $a_i,b_i=0$ for $i gt n$
$endgroup$
– Ross Millikan
Dec 21 '18 at 14:51
$begingroup$
Yours does as long as the $c$ sum goes up to $2n$, not to $n$ and you regard $a_i,b_i=0$ for $i gt n$
$endgroup$
– Ross Millikan
Dec 21 '18 at 14:51
add a comment |
$begingroup$
The first identity should read $sum_{nge 0}a_n x^nsum_{nge 0}b_n x^n=sum_{nge 0}(sum_{i= 0}^n a_i b_{n-i})x^n$, or even better $sum_{jge 0}a_j x^jsum_{kge 0}b_k x^k=sum_{nge 0}(sum_{i= 0}^n a_i b_{n-i})x^n$. The proof is simple: just find all contributions to the $x^n$ coefficient on the left-hand side. When multiplying polynomials, which have finite degrees, all sufficiently late terms in each sequence are $0$. In this case $a_n=0$ for $n>2$, $b_n=0$ for $n>2$, $c_n=0$ for $n>4$.
answered Dec 20 '18 at 15:58
J.G.J.G.
28.1k22844
$endgroup$
$begingroup$
The convolution works with x=1 just as easily, so nothing wrong with the first identity just as stated.
$endgroup$
– Macavity
Dec 21 '18 at 3:18
add a comment |
$begingroup$
The first identity should read $sum_{nge 0}a_n x^nsum_{nge 0}b_n x^n=sum_{nge 0}(sum_{i= 0}^n a_i b_{n-i})x^n$, or even better $sum_{jge 0}a_j x^jsum_{kge 0}b_k x^k=sum_{nge 0}(sum_{i= 0}^n a_i b_{n-i})x^n$. The proof is simple: just find all contributions to the $x^n$ coefficient on the left-hand side. When multiplying polynomials, which have finite degrees, all sufficiently late terms in each sequence are $0$. In this case $a_n=0$ for $n>2$, $b_n=0$ for $n>2$, $c_n=0$ for $n>4$.
answered Dec 20 '18 at 15:58
J.G.J.G.
28.1k22844
$endgroup$
$begingroup$
The convolution works with x=1 just as easily, so nothing wrong with the first identity just as stated.
$endgroup$
– Macavity
Dec 21 '18 at 3:18
add a comment |
$begingroup$
The first identity should read $sum_{nge 0}a_n x^nsum_{nge 0}b_n x^n=sum_{nge 0}(sum_{i= 0}^n a_i b_{n-i})x^n$, or even better $sum_{jge 0}a_j x^jsum_{kge 0}b_k x^k=sum_{nge 0}(sum_{i= 0}^n a_i b_{n-i})x^n$. The proof is simple: just find all contributions to the $x^n$ coefficient on the left-hand side. When multiplying polynomials, which have finite degrees, all sufficiently late terms in each sequence are $0$. In this case $a_n=0$ for $n>2$, $b_n=0$ for $n>2$, $c_n=0$ for $n>4$.
answered Dec 20 '18 at 15:58
J.G.J.G.
28.1k22844
$endgroup$
The first identity should read $sum_{nge 0}a_n x^nsum_{nge 0}b_n x^n=sum_{nge 0}(sum_{i= 0}^n a_i b_{n-i})x^n$, or even better $sum_{jge 0}a_j x^jsum_{kge 0}b_k x^k=sum_{nge 0}(sum_{i= 0}^n a_i b_{n-i})x^n$. The proof is simple: just find all contributions to the $x^n$ coefficient on the left-hand side. When multiplying polynomials, which have finite degrees, all sufficiently late terms in each sequence are $0$. In this case $a_n=0$ for $n>2$, $b_n=0$ for $n>2$, $c_n=0$ for $n>4$.
answered Dec 20 '18 at 15:58
J.G.J.G.
28.1k22844
answered Dec 20 '18 at 15:58
J.G.J.G.
28.1k22844
answered Dec 20 '18 at 15:58
J.G.J.G.
28.1k22844
answered Dec 20 '18 at 15:58
J.G.J.G.
28.1k22844
28.1k22844
$begingroup$
The convolution works with x=1 just as easily, so nothing wrong with the first identity just as stated.
$endgroup$
– Macavity
Dec 21 '18 at 3:18
add a comment |
$begingroup$
The convolution works with x=1 just as easily, so nothing wrong with the first identity just as stated.
$endgroup$
– Macavity
Dec 21 '18 at 3:18
$begingroup$
The convolution works with x=1 just as easily, so nothing wrong with the first identity just as stated.
$endgroup$
– Macavity
Dec 21 '18 at 3:18
$begingroup$
The convolution works with x=1 just as easily, so nothing wrong with the first identity just as stated.
$endgroup$
– Macavity
Dec 21 '18 at 3:18
add a comment |
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$begingroup$
You forgot some term... at least $a_2b_2$. $3 text { terms } times 3 text { terms } = 9$.
$endgroup$
– Mauro ALLEGRANZA
Dec 20 '18 at 15:55
$begingroup$
The missing terms are : $a_1b_2+a_2b_1+a_2b_2$.
$endgroup$
– Mauro ALLEGRANZA
Dec 20 '18 at 15:59