Is this an empty product?
$begingroup$
Assume that we have:
$$Psi_{j,k}=(t_{j+1}-tau)times...times(t_{j+k-1}-tau)$$
that can be written as:
$$Pi_{i=j+1}^{j+k-1} (t_i-tau)$$
If we have $k=1$ , my text-book says that this is an empty product and equals $1$ by definition.
It seems to make sense because in this case the "upper limit" of the product is lower than the lower one, but following the first expression, couldn't we write
$$Psi_{j,1}=(t_{j+1}-tau)times(t_{j}-tau)$$
Is this wrong? I am not familiar with this notation and it may be very trivial.
algebra-precalculus products
edited Dec 7 '18 at 11:13
dallonsi
1187
asked Dec 7 '18 at 10:58
RScrlliRScrlli
649114
$endgroup$
add a comment |
$begingroup$
Assume that we have:
$$Psi_{j,k}=(t_{j+1}-tau)times...times(t_{j+k-1}-tau)$$
that can be written as:
$$Pi_{i=j+1}^{j+k-1} (t_i-tau)$$
If we have $k=1$ , my text-book says that this is an empty product and equals $1$ by definition.
It seems to make sense because in this case the "upper limit" of the product is lower than the lower one, but following the first expression, couldn't we write
$$Psi_{j,1}=(t_{j+1}-tau)times(t_{j}-tau)$$
Is this wrong? I am not familiar with this notation and it may be very trivial.
algebra-precalculus products
edited Dec 7 '18 at 11:13
dallonsi
1187
asked Dec 7 '18 at 10:58
RScrlliRScrlli
649114
$endgroup$
add a comment |
$begingroup$
Assume that we have:
$$Psi_{j,k}=(t_{j+1}-tau)times...times(t_{j+k-1}-tau)$$
that can be written as:
$$Pi_{i=j+1}^{j+k-1} (t_i-tau)$$
If we have $k=1$ , my text-book says that this is an empty product and equals $1$ by definition.
It seems to make sense because in this case the "upper limit" of the product is lower than the lower one, but following the first expression, couldn't we write
$$Psi_{j,1}=(t_{j+1}-tau)times(t_{j}-tau)$$
Is this wrong? I am not familiar with this notation and it may be very trivial.
algebra-precalculus products
edited Dec 7 '18 at 11:13
dallonsi
1187
asked Dec 7 '18 at 10:58
RScrlliRScrlli
649114
$endgroup$
Assume that we have:
$$Psi_{j,k}=(t_{j+1}-tau)times...times(t_{j+k-1}-tau)$$
that can be written as:
$$Pi_{i=j+1}^{j+k-1} (t_i-tau)$$
If we have $k=1$ , my text-book says that this is an empty product and equals $1$ by definition.
It seems to make sense because in this case the "upper limit" of the product is lower than the lower one, but following the first expression, couldn't we write
$$Psi_{j,1}=(t_{j+1}-tau)times(t_{j}-tau)$$
Is this wrong? I am not familiar with this notation and it may be very trivial.
algebra-precalculus products
algebra-precalculus products
edited Dec 7 '18 at 11:13
dallonsi
1187
asked Dec 7 '18 at 10:58
RScrlliRScrlli
649114
edited Dec 7 '18 at 11:13
dallonsi
1187
asked Dec 7 '18 at 10:58
RScrlliRScrlli
649114
edited Dec 7 '18 at 11:13
dallonsi
1187
edited Dec 7 '18 at 11:13
dallonsi
1187
edited Dec 7 '18 at 11:13
dallonsi
1187
1187
asked Dec 7 '18 at 10:58
RScrlliRScrlli
649114
asked Dec 7 '18 at 10:58
RScrlliRScrlli
649114
asked Dec 7 '18 at 10:58
RScrlliRScrlli
649114
649114
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Read : $Pi_{i=j+1}^{j+k-1} (t_i-tau)$ as
"The product of $t_i - tau$ FROM $j+1$ TO $j+k-1$"
Now if $j+1$ is less than $j+k-1$, than the previous assertion makes no sense and by convention we replace the product by $1$.
answered Dec 7 '18 at 11:15
dallonsidallonsi
1187
$endgroup$
$begingroup$
Using the same logic if $k=2$ we also have an empty product?
$endgroup$
– RScrlli
Dec 7 '18 at 11:18
$begingroup$
as I explained, no, you would there have then a single factor (by popping in the definition via the product), which is $psi_{j+1}-tau$
$endgroup$
– Enkidu
Dec 7 '18 at 11:28
add a comment |
$begingroup$
no, you can't, since the above definition is suggestive to make clear that youmultiply all between $j+1$ and $j+k-1$. I understand that the first definition would look like $psi_{j,1}=(t_{j+1}-tau)times (t_j - tau)$. I myself prefer to exclude those boundary cases in my definitions and define them seperately, but sometimes you just do not have the time, space or simply forgot.
in my opinion your second equation would be the better definition, and your top line a remark about what that means for $k> 2$
answered Dec 7 '18 at 11:09
EnkiduEnkidu
1,39419
$endgroup$
1
$begingroup$
makes completely sense, but I am not sure where $k>j+2$ comes from, if we set $k=3$ for instance the line above works perfectly: $(t_{j+1}-tau)times( t_{j+3-1}-tau)$ irrespective of the value of $j$
$endgroup$
– RScrlli
Dec 7 '18 at 11:21
$begingroup$
you are right, I am sorry, I messed it up with notations I usually use (if I got sums like that I usually either index from 0 by force or run between two one variable ends). also, thanks a lot, I edited my answer now!
$endgroup$
– Enkidu
Dec 7 '18 at 11:27
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
$begingroup$
Read : $Pi_{i=j+1}^{j+k-1} (t_i-tau)$ as
"The product of $t_i - tau$ FROM $j+1$ TO $j+k-1$"
Now if $j+1$ is less than $j+k-1$, than the previous assertion makes no sense and by convention we replace the product by $1$.
answered Dec 7 '18 at 11:15
dallonsidallonsi
1187
$endgroup$
$begingroup$
Using the same logic if $k=2$ we also have an empty product?
$endgroup$
– RScrlli
Dec 7 '18 at 11:18
$begingroup$
as I explained, no, you would there have then a single factor (by popping in the definition via the product), which is $psi_{j+1}-tau$
$endgroup$
– Enkidu
Dec 7 '18 at 11:28
add a comment |
$begingroup$
Read : $Pi_{i=j+1}^{j+k-1} (t_i-tau)$ as
"The product of $t_i - tau$ FROM $j+1$ TO $j+k-1$"
Now if $j+1$ is less than $j+k-1$, than the previous assertion makes no sense and by convention we replace the product by $1$.
answered Dec 7 '18 at 11:15
dallonsidallonsi
1187
$endgroup$
$begingroup$
Using the same logic if $k=2$ we also have an empty product?
$endgroup$
– RScrlli
Dec 7 '18 at 11:18
$begingroup$
as I explained, no, you would there have then a single factor (by popping in the definition via the product), which is $psi_{j+1}-tau$
$endgroup$
– Enkidu
Dec 7 '18 at 11:28
add a comment |
$begingroup$
Read : $Pi_{i=j+1}^{j+k-1} (t_i-tau)$ as
"The product of $t_i - tau$ FROM $j+1$ TO $j+k-1$"
Now if $j+1$ is less than $j+k-1$, than the previous assertion makes no sense and by convention we replace the product by $1$.
answered Dec 7 '18 at 11:15
dallonsidallonsi
1187
$endgroup$
Read : $Pi_{i=j+1}^{j+k-1} (t_i-tau)$ as
"The product of $t_i - tau$ FROM $j+1$ TO $j+k-1$"
Now if $j+1$ is less than $j+k-1$, than the previous assertion makes no sense and by convention we replace the product by $1$.
answered Dec 7 '18 at 11:15
dallonsidallonsi
1187
answered Dec 7 '18 at 11:15
dallonsidallonsi
1187
answered Dec 7 '18 at 11:15
dallonsidallonsi
1187
answered Dec 7 '18 at 11:15
dallonsidallonsi
1187
1187
$begingroup$
Using the same logic if $k=2$ we also have an empty product?
$endgroup$
– RScrlli
Dec 7 '18 at 11:18
$begingroup$
as I explained, no, you would there have then a single factor (by popping in the definition via the product), which is $psi_{j+1}-tau$
$endgroup$
– Enkidu
Dec 7 '18 at 11:28
add a comment |
$begingroup$
Using the same logic if $k=2$ we also have an empty product?
$endgroup$
– RScrlli
Dec 7 '18 at 11:18
$begingroup$
as I explained, no, you would there have then a single factor (by popping in the definition via the product), which is $psi_{j+1}-tau$
$endgroup$
– Enkidu
Dec 7 '18 at 11:28
$begingroup$
Using the same logic if $k=2$ we also have an empty product?
$endgroup$
– RScrlli
Dec 7 '18 at 11:18
$begingroup$
Using the same logic if $k=2$ we also have an empty product?
$endgroup$
– RScrlli
Dec 7 '18 at 11:18
$begingroup$
as I explained, no, you would there have then a single factor (by popping in the definition via the product), which is $psi_{j+1}-tau$
$endgroup$
– Enkidu
Dec 7 '18 at 11:28
$begingroup$
as I explained, no, you would there have then a single factor (by popping in the definition via the product), which is $psi_{j+1}-tau$
$endgroup$
– Enkidu
Dec 7 '18 at 11:28
add a comment |
$begingroup$
no, you can't, since the above definition is suggestive to make clear that youmultiply all between $j+1$ and $j+k-1$. I understand that the first definition would look like $psi_{j,1}=(t_{j+1}-tau)times (t_j - tau)$. I myself prefer to exclude those boundary cases in my definitions and define them seperately, but sometimes you just do not have the time, space or simply forgot.
in my opinion your second equation would be the better definition, and your top line a remark about what that means for $k> 2$
answered Dec 7 '18 at 11:09
EnkiduEnkidu
1,39419
$endgroup$
1
$begingroup$
makes completely sense, but I am not sure where $k>j+2$ comes from, if we set $k=3$ for instance the line above works perfectly: $(t_{j+1}-tau)times( t_{j+3-1}-tau)$ irrespective of the value of $j$
$endgroup$
– RScrlli
Dec 7 '18 at 11:21
$begingroup$
you are right, I am sorry, I messed it up with notations I usually use (if I got sums like that I usually either index from 0 by force or run between two one variable ends). also, thanks a lot, I edited my answer now!
$endgroup$
– Enkidu
Dec 7 '18 at 11:27
add a comment |
$begingroup$
no, you can't, since the above definition is suggestive to make clear that youmultiply all between $j+1$ and $j+k-1$. I understand that the first definition would look like $psi_{j,1}=(t_{j+1}-tau)times (t_j - tau)$. I myself prefer to exclude those boundary cases in my definitions and define them seperately, but sometimes you just do not have the time, space or simply forgot.
in my opinion your second equation would be the better definition, and your top line a remark about what that means for $k> 2$
answered Dec 7 '18 at 11:09
EnkiduEnkidu
1,39419
$endgroup$
1
$begingroup$
makes completely sense, but I am not sure where $k>j+2$ comes from, if we set $k=3$ for instance the line above works perfectly: $(t_{j+1}-tau)times( t_{j+3-1}-tau)$ irrespective of the value of $j$
$endgroup$
– RScrlli
Dec 7 '18 at 11:21
$begingroup$
you are right, I am sorry, I messed it up with notations I usually use (if I got sums like that I usually either index from 0 by force or run between two one variable ends). also, thanks a lot, I edited my answer now!
$endgroup$
– Enkidu
Dec 7 '18 at 11:27
add a comment |
$begingroup$
no, you can't, since the above definition is suggestive to make clear that youmultiply all between $j+1$ and $j+k-1$. I understand that the first definition would look like $psi_{j,1}=(t_{j+1}-tau)times (t_j - tau)$. I myself prefer to exclude those boundary cases in my definitions and define them seperately, but sometimes you just do not have the time, space or simply forgot.
in my opinion your second equation would be the better definition, and your top line a remark about what that means for $k> 2$
answered Dec 7 '18 at 11:09
EnkiduEnkidu
1,39419
$endgroup$
no, you can't, since the above definition is suggestive to make clear that youmultiply all between $j+1$ and $j+k-1$. I understand that the first definition would look like $psi_{j,1}=(t_{j+1}-tau)times (t_j - tau)$. I myself prefer to exclude those boundary cases in my definitions and define them seperately, but sometimes you just do not have the time, space or simply forgot.
in my opinion your second equation would be the better definition, and your top line a remark about what that means for $k> 2$
answered Dec 7 '18 at 11:09
EnkiduEnkidu
1,39419
edited Dec 7 '18 at 11:25
answered Dec 7 '18 at 11:09
EnkiduEnkidu
1,39419
answered Dec 7 '18 at 11:09
EnkiduEnkidu
1,39419
answered Dec 7 '18 at 11:09
EnkiduEnkidu
1,39419
1,39419
1
$begingroup$
makes completely sense, but I am not sure where $k>j+2$ comes from, if we set $k=3$ for instance the line above works perfectly: $(t_{j+1}-tau)times( t_{j+3-1}-tau)$ irrespective of the value of $j$
$endgroup$
– RScrlli
Dec 7 '18 at 11:21
$begingroup$
you are right, I am sorry, I messed it up with notations I usually use (if I got sums like that I usually either index from 0 by force or run between two one variable ends). also, thanks a lot, I edited my answer now!
$endgroup$
– Enkidu
Dec 7 '18 at 11:27
add a comment |
1
$begingroup$
makes completely sense, but I am not sure where $k>j+2$ comes from, if we set $k=3$ for instance the line above works perfectly: $(t_{j+1}-tau)times( t_{j+3-1}-tau)$ irrespective of the value of $j$
$endgroup$
– RScrlli
Dec 7 '18 at 11:21
$begingroup$
you are right, I am sorry, I messed it up with notations I usually use (if I got sums like that I usually either index from 0 by force or run between two one variable ends). also, thanks a lot, I edited my answer now!
$endgroup$
– Enkidu
Dec 7 '18 at 11:27
1
1
$begingroup$
makes completely sense, but I am not sure where $k>j+2$ comes from, if we set $k=3$ for instance the line above works perfectly: $(t_{j+1}-tau)times( t_{j+3-1}-tau)$ irrespective of the value of $j$
$endgroup$
– RScrlli
Dec 7 '18 at 11:21
$begingroup$
makes completely sense, but I am not sure where $k>j+2$ comes from, if we set $k=3$ for instance the line above works perfectly: $(t_{j+1}-tau)times( t_{j+3-1}-tau)$ irrespective of the value of $j$
$endgroup$
– RScrlli
Dec 7 '18 at 11:21
$begingroup$
you are right, I am sorry, I messed it up with notations I usually use (if I got sums like that I usually either index from 0 by force or run between two one variable ends). also, thanks a lot, I edited my answer now!
$endgroup$
– Enkidu
Dec 7 '18 at 11:27
$begingroup$
you are right, I am sorry, I messed it up with notations I usually use (if I got sums like that I usually either index from 0 by force or run between two one variable ends). also, thanks a lot, I edited my answer now!
$endgroup$
– Enkidu
Dec 7 '18 at 11:27
add a comment |
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