Mathematically expressing the sum of odd and even numbers [closed]
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I am aware that if I want to write the product of a function $2N^2$ starting with numbers from $1$ to $n$, I write something like this $prod_{N=1}^{n} 2N^2$, what if I want $N$ to start from 1 and step through odd numbers or even numbers? How do I express this? It is a simple question but I am getting somethings a bit mixed up.
soft-question
asked Dec 30 '18 at 4:35
AbdulhameedAbdulhameed
105112
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closed as unclear what you're asking by Eevee Trainer, mrtaurho, Shailesh, Paul Frost, amWhy Dec 30 '18 at 15:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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show 3 more comments
$begingroup$
I am aware that if I want to write the product of a function $2N^2$ starting with numbers from $1$ to $n$, I write something like this $prod_{N=1}^{n} 2N^2$, what if I want $N$ to start from 1 and step through odd numbers or even numbers? How do I express this? It is a simple question but I am getting somethings a bit mixed up.
soft-question
asked Dec 30 '18 at 4:35
AbdulhameedAbdulhameed
105112
$endgroup$
closed as unclear what you're asking by Eevee Trainer, mrtaurho, Shailesh, Paul Frost, amWhy Dec 30 '18 at 15:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
That product notation doesn't even make sense. Your question in general is also rather unclear. An explicit example would help a lot...
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– Eevee Trainer
Dec 30 '18 at 4:37
3
$begingroup$
$prod_{k=0}^N 2(2k+1)^2$
$endgroup$
– Zachary Hunter
Dec 30 '18 at 4:38
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Its been edited. Thanks
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:38
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@Zachary Hunter, I guess that is meant for the even function. Cant one use the $prod$ sign?
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:40
1
$begingroup$
Your title refers to a sum while the body refers to a product. The question is the same, but they should be consistent. When you write $prod_{N=1}^{N} 2N^2$ you are using $N$ in two different ways. The top one is the upper limit of the dummy variable while the other two are the dummy variable. It would be much better to write $prod_{N=1}^{n} 2N^2$ The result will depend on $n$ but not $N$.
$endgroup$
– Ross Millikan
Dec 30 '18 at 4:44
|
show 3 more comments
$begingroup$
I am aware that if I want to write the product of a function $2N^2$ starting with numbers from $1$ to $n$, I write something like this $prod_{N=1}^{n} 2N^2$, what if I want $N$ to start from 1 and step through odd numbers or even numbers? How do I express this? It is a simple question but I am getting somethings a bit mixed up.
soft-question
asked Dec 30 '18 at 4:35
AbdulhameedAbdulhameed
105112
$endgroup$
I am aware that if I want to write the product of a function $2N^2$ starting with numbers from $1$ to $n$, I write something like this $prod_{N=1}^{n} 2N^2$, what if I want $N$ to start from 1 and step through odd numbers or even numbers? How do I express this? It is a simple question but I am getting somethings a bit mixed up.
soft-question
soft-question
asked Dec 30 '18 at 4:35
AbdulhameedAbdulhameed
105112
asked Dec 30 '18 at 4:35
AbdulhameedAbdulhameed
105112
edited Dec 30 '18 at 5:06
Abdulhameed
asked Dec 30 '18 at 4:35
AbdulhameedAbdulhameed
105112
asked Dec 30 '18 at 4:35
AbdulhameedAbdulhameed
105112
asked Dec 30 '18 at 4:35
AbdulhameedAbdulhameed
105112
105112
closed as unclear what you're asking by Eevee Trainer, mrtaurho, Shailesh, Paul Frost, amWhy Dec 30 '18 at 15:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Eevee Trainer, mrtaurho, Shailesh, Paul Frost, amWhy Dec 30 '18 at 15:02
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
That product notation doesn't even make sense. Your question in general is also rather unclear. An explicit example would help a lot...
$endgroup$
– Eevee Trainer
Dec 30 '18 at 4:37
3
$begingroup$
$prod_{k=0}^N 2(2k+1)^2$
$endgroup$
– Zachary Hunter
Dec 30 '18 at 4:38
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Its been edited. Thanks
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:38
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@Zachary Hunter, I guess that is meant for the even function. Cant one use the $prod$ sign?
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:40
1
$begingroup$
Your title refers to a sum while the body refers to a product. The question is the same, but they should be consistent. When you write $prod_{N=1}^{N} 2N^2$ you are using $N$ in two different ways. The top one is the upper limit of the dummy variable while the other two are the dummy variable. It would be much better to write $prod_{N=1}^{n} 2N^2$ The result will depend on $n$ but not $N$.
$endgroup$
– Ross Millikan
Dec 30 '18 at 4:44
|
show 3 more comments
$begingroup$
That product notation doesn't even make sense. Your question in general is also rather unclear. An explicit example would help a lot...
$endgroup$
– Eevee Trainer
Dec 30 '18 at 4:37
3
$begingroup$
$prod_{k=0}^N 2(2k+1)^2$
$endgroup$
– Zachary Hunter
Dec 30 '18 at 4:38
$begingroup$
Its been edited. Thanks
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:38
$begingroup$
@Zachary Hunter, I guess that is meant for the even function. Cant one use the $prod$ sign?
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:40
1
$begingroup$
Your title refers to a sum while the body refers to a product. The question is the same, but they should be consistent. When you write $prod_{N=1}^{N} 2N^2$ you are using $N$ in two different ways. The top one is the upper limit of the dummy variable while the other two are the dummy variable. It would be much better to write $prod_{N=1}^{n} 2N^2$ The result will depend on $n$ but not $N$.
$endgroup$
– Ross Millikan
Dec 30 '18 at 4:44
$begingroup$
That product notation doesn't even make sense. Your question in general is also rather unclear. An explicit example would help a lot...
$endgroup$
– Eevee Trainer
Dec 30 '18 at 4:37
$begingroup$
That product notation doesn't even make sense. Your question in general is also rather unclear. An explicit example would help a lot...
$endgroup$
– Eevee Trainer
Dec 30 '18 at 4:37
3
3
$begingroup$
$prod_{k=0}^N 2(2k+1)^2$
$endgroup$
– Zachary Hunter
Dec 30 '18 at 4:38
$begingroup$
$prod_{k=0}^N 2(2k+1)^2$
$endgroup$
– Zachary Hunter
Dec 30 '18 at 4:38
$begingroup$
Its been edited. Thanks
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:38
$begingroup$
Its been edited. Thanks
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:38
$begingroup$
@Zachary Hunter, I guess that is meant for the even function. Cant one use the $prod$ sign?
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:40
$begingroup$
@Zachary Hunter, I guess that is meant for the even function. Cant one use the $prod$ sign?
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:40
1
1
$begingroup$
Your title refers to a sum while the body refers to a product. The question is the same, but they should be consistent. When you write $prod_{N=1}^{N} 2N^2$ you are using $N$ in two different ways. The top one is the upper limit of the dummy variable while the other two are the dummy variable. It would be much better to write $prod_{N=1}^{n} 2N^2$ The result will depend on $n$ but not $N$.
$endgroup$
– Ross Millikan
Dec 30 '18 at 4:44
$begingroup$
Your title refers to a sum while the body refers to a product. The question is the same, but they should be consistent. When you write $prod_{N=1}^{N} 2N^2$ you are using $N$ in two different ways. The top one is the upper limit of the dummy variable while the other two are the dummy variable. It would be much better to write $prod_{N=1}^{n} 2N^2$ The result will depend on $n$ but not $N$.
$endgroup$
– Ross Millikan
Dec 30 '18 at 4:44
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
If you want to step through the even numbers, you can write $prod _{n=1}^k(2(2n)^2)$. If you want to step through the odd numbers, you can write $prod _{n=0}^k(2(2n+1)^2)$. I don't know of a mathematical equivalent to the computer step as in for i=1 to 11 step 2 other than this.
I have seen the notation $$sum_{stackrel {p le n}{p text { prime}}}$$ for the sum of some expression over all the primes less than or equal to $n$.
Using the formula for even numbers makes it easier to deal with. In your example we can recognize the factorial and write
$$prod _{n=1}^k(2(2n)^2)=2^k(k!)^2\
prod _{n=0}^k(2(2n+1)^2)=2^{k+1}left(frac {(2n+1)!}{2^nn!}right)^2$$
answered Dec 30 '18 at 4:46
Ross MillikanRoss Millikan
293k23197372
$endgroup$
$begingroup$
@Thanks Ross!, but with the square you have just included, I guess it would not start from 1 and 2 this time.
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:48
$begingroup$
We can just start with $n=0$ or use $2n-1$ if you want to start from $1$. I'll fix. You need to check the limits each time to make sure you get what you want and no more. The even one does start with $2$.
$endgroup$
– Ross Millikan
Dec 30 '18 at 4:50
$begingroup$
I want to start with 1, I guess it looks like $prod _{n=1}^k(2(2n-1)^2)$ and for the even $prod _{n=1}^k((2n))$
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:54
$begingroup$
I would think, as in my answer, that for the evens you want $2(2n)^2$ because you said you wanted the product of $2N^2$ for $N$ even. $2n$ will just give you the product of all the even numbers. That is fine if that is what you want. $2n$ starts $2,4,6,8$ while $2(2n)^2 starts $8,32,72$
$endgroup$
– Ross Millikan
Dec 30 '18 at 5:04
$begingroup$
Thanks for the clarification @Ross Milikan
$endgroup$
– Abdulhameed
Dec 30 '18 at 5:07
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you want to step through the even numbers, you can write $prod _{n=1}^k(2(2n)^2)$. If you want to step through the odd numbers, you can write $prod _{n=0}^k(2(2n+1)^2)$. I don't know of a mathematical equivalent to the computer step as in for i=1 to 11 step 2 other than this.
I have seen the notation $$sum_{stackrel {p le n}{p text { prime}}}$$ for the sum of some expression over all the primes less than or equal to $n$.
Using the formula for even numbers makes it easier to deal with. In your example we can recognize the factorial and write
$$prod _{n=1}^k(2(2n)^2)=2^k(k!)^2\
prod _{n=0}^k(2(2n+1)^2)=2^{k+1}left(frac {(2n+1)!}{2^nn!}right)^2$$
answered Dec 30 '18 at 4:46
Ross MillikanRoss Millikan
293k23197372
$endgroup$
$begingroup$
@Thanks Ross!, but with the square you have just included, I guess it would not start from 1 and 2 this time.
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:48
$begingroup$
We can just start with $n=0$ or use $2n-1$ if you want to start from $1$. I'll fix. You need to check the limits each time to make sure you get what you want and no more. The even one does start with $2$.
$endgroup$
– Ross Millikan
Dec 30 '18 at 4:50
$begingroup$
I want to start with 1, I guess it looks like $prod _{n=1}^k(2(2n-1)^2)$ and for the even $prod _{n=1}^k((2n))$
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:54
$begingroup$
I would think, as in my answer, that for the evens you want $2(2n)^2$ because you said you wanted the product of $2N^2$ for $N$ even. $2n$ will just give you the product of all the even numbers. That is fine if that is what you want. $2n$ starts $2,4,6,8$ while $2(2n)^2 starts $8,32,72$
$endgroup$
– Ross Millikan
Dec 30 '18 at 5:04
$begingroup$
Thanks for the clarification @Ross Milikan
$endgroup$
– Abdulhameed
Dec 30 '18 at 5:07
add a comment |
$begingroup$
If you want to step through the even numbers, you can write $prod _{n=1}^k(2(2n)^2)$. If you want to step through the odd numbers, you can write $prod _{n=0}^k(2(2n+1)^2)$. I don't know of a mathematical equivalent to the computer step as in for i=1 to 11 step 2 other than this.
I have seen the notation $$sum_{stackrel {p le n}{p text { prime}}}$$ for the sum of some expression over all the primes less than or equal to $n$.
Using the formula for even numbers makes it easier to deal with. In your example we can recognize the factorial and write
$$prod _{n=1}^k(2(2n)^2)=2^k(k!)^2\
prod _{n=0}^k(2(2n+1)^2)=2^{k+1}left(frac {(2n+1)!}{2^nn!}right)^2$$
answered Dec 30 '18 at 4:46
Ross MillikanRoss Millikan
293k23197372
$endgroup$
$begingroup$
@Thanks Ross!, but with the square you have just included, I guess it would not start from 1 and 2 this time.
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:48
$begingroup$
We can just start with $n=0$ or use $2n-1$ if you want to start from $1$. I'll fix. You need to check the limits each time to make sure you get what you want and no more. The even one does start with $2$.
$endgroup$
– Ross Millikan
Dec 30 '18 at 4:50
$begingroup$
I want to start with 1, I guess it looks like $prod _{n=1}^k(2(2n-1)^2)$ and for the even $prod _{n=1}^k((2n))$
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:54
$begingroup$
I would think, as in my answer, that for the evens you want $2(2n)^2$ because you said you wanted the product of $2N^2$ for $N$ even. $2n$ will just give you the product of all the even numbers. That is fine if that is what you want. $2n$ starts $2,4,6,8$ while $2(2n)^2 starts $8,32,72$
$endgroup$
– Ross Millikan
Dec 30 '18 at 5:04
$begingroup$
Thanks for the clarification @Ross Milikan
$endgroup$
– Abdulhameed
Dec 30 '18 at 5:07
add a comment |
$begingroup$
If you want to step through the even numbers, you can write $prod _{n=1}^k(2(2n)^2)$. If you want to step through the odd numbers, you can write $prod _{n=0}^k(2(2n+1)^2)$. I don't know of a mathematical equivalent to the computer step as in for i=1 to 11 step 2 other than this.
I have seen the notation $$sum_{stackrel {p le n}{p text { prime}}}$$ for the sum of some expression over all the primes less than or equal to $n$.
Using the formula for even numbers makes it easier to deal with. In your example we can recognize the factorial and write
$$prod _{n=1}^k(2(2n)^2)=2^k(k!)^2\
prod _{n=0}^k(2(2n+1)^2)=2^{k+1}left(frac {(2n+1)!}{2^nn!}right)^2$$
answered Dec 30 '18 at 4:46
Ross MillikanRoss Millikan
293k23197372
$endgroup$
If you want to step through the even numbers, you can write $prod _{n=1}^k(2(2n)^2)$. If you want to step through the odd numbers, you can write $prod _{n=0}^k(2(2n+1)^2)$. I don't know of a mathematical equivalent to the computer step as in for i=1 to 11 step 2 other than this.
I have seen the notation $$sum_{stackrel {p le n}{p text { prime}}}$$ for the sum of some expression over all the primes less than or equal to $n$.
Using the formula for even numbers makes it easier to deal with. In your example we can recognize the factorial and write
$$prod _{n=1}^k(2(2n)^2)=2^k(k!)^2\
prod _{n=0}^k(2(2n+1)^2)=2^{k+1}left(frac {(2n+1)!}{2^nn!}right)^2$$
answered Dec 30 '18 at 4:46
Ross MillikanRoss Millikan
293k23197372
edited Dec 30 '18 at 14:17
answered Dec 30 '18 at 4:46
Ross MillikanRoss Millikan
293k23197372
answered Dec 30 '18 at 4:46
Ross MillikanRoss Millikan
293k23197372
answered Dec 30 '18 at 4:46
Ross MillikanRoss Millikan
293k23197372
293k23197372
$begingroup$
@Thanks Ross!, but with the square you have just included, I guess it would not start from 1 and 2 this time.
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:48
$begingroup$
We can just start with $n=0$ or use $2n-1$ if you want to start from $1$. I'll fix. You need to check the limits each time to make sure you get what you want and no more. The even one does start with $2$.
$endgroup$
– Ross Millikan
Dec 30 '18 at 4:50
$begingroup$
I want to start with 1, I guess it looks like $prod _{n=1}^k(2(2n-1)^2)$ and for the even $prod _{n=1}^k((2n))$
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:54
$begingroup$
I would think, as in my answer, that for the evens you want $2(2n)^2$ because you said you wanted the product of $2N^2$ for $N$ even. $2n$ will just give you the product of all the even numbers. That is fine if that is what you want. $2n$ starts $2,4,6,8$ while $2(2n)^2 starts $8,32,72$
$endgroup$
– Ross Millikan
Dec 30 '18 at 5:04
$begingroup$
Thanks for the clarification @Ross Milikan
$endgroup$
– Abdulhameed
Dec 30 '18 at 5:07
add a comment |
$begingroup$
@Thanks Ross!, but with the square you have just included, I guess it would not start from 1 and 2 this time.
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:48
$begingroup$
We can just start with $n=0$ or use $2n-1$ if you want to start from $1$. I'll fix. You need to check the limits each time to make sure you get what you want and no more. The even one does start with $2$.
$endgroup$
– Ross Millikan
Dec 30 '18 at 4:50
$begingroup$
I want to start with 1, I guess it looks like $prod _{n=1}^k(2(2n-1)^2)$ and for the even $prod _{n=1}^k((2n))$
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:54
$begingroup$
I would think, as in my answer, that for the evens you want $2(2n)^2$ because you said you wanted the product of $2N^2$ for $N$ even. $2n$ will just give you the product of all the even numbers. That is fine if that is what you want. $2n$ starts $2,4,6,8$ while $2(2n)^2 starts $8,32,72$
$endgroup$
– Ross Millikan
Dec 30 '18 at 5:04
$begingroup$
Thanks for the clarification @Ross Milikan
$endgroup$
– Abdulhameed
Dec 30 '18 at 5:07
$begingroup$
@Thanks Ross!, but with the square you have just included, I guess it would not start from 1 and 2 this time.
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:48
$begingroup$
@Thanks Ross!, but with the square you have just included, I guess it would not start from 1 and 2 this time.
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:48
$begingroup$
We can just start with $n=0$ or use $2n-1$ if you want to start from $1$. I'll fix. You need to check the limits each time to make sure you get what you want and no more. The even one does start with $2$.
$endgroup$
– Ross Millikan
Dec 30 '18 at 4:50
$begingroup$
We can just start with $n=0$ or use $2n-1$ if you want to start from $1$. I'll fix. You need to check the limits each time to make sure you get what you want and no more. The even one does start with $2$.
$endgroup$
– Ross Millikan
Dec 30 '18 at 4:50
$begingroup$
I want to start with 1, I guess it looks like $prod _{n=1}^k(2(2n-1)^2)$ and for the even $prod _{n=1}^k((2n))$
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:54
$begingroup$
I want to start with 1, I guess it looks like $prod _{n=1}^k(2(2n-1)^2)$ and for the even $prod _{n=1}^k((2n))$
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:54
$begingroup$
I would think, as in my answer, that for the evens you want $2(2n)^2$ because you said you wanted the product of $2N^2$ for $N$ even. $2n$ will just give you the product of all the even numbers. That is fine if that is what you want. $2n$ starts $2,4,6,8$ while $2(2n)^2 starts $8,32,72$
$endgroup$
– Ross Millikan
Dec 30 '18 at 5:04
$begingroup$
I would think, as in my answer, that for the evens you want $2(2n)^2$ because you said you wanted the product of $2N^2$ for $N$ even. $2n$ will just give you the product of all the even numbers. That is fine if that is what you want. $2n$ starts $2,4,6,8$ while $2(2n)^2 starts $8,32,72$
$endgroup$
– Ross Millikan
Dec 30 '18 at 5:04
$begingroup$
Thanks for the clarification @Ross Milikan
$endgroup$
– Abdulhameed
Dec 30 '18 at 5:07
$begingroup$
Thanks for the clarification @Ross Milikan
$endgroup$
– Abdulhameed
Dec 30 '18 at 5:07
add a comment |
$begingroup$
That product notation doesn't even make sense. Your question in general is also rather unclear. An explicit example would help a lot...
$endgroup$
– Eevee Trainer
Dec 30 '18 at 4:37
3
$begingroup$
$prod_{k=0}^N 2(2k+1)^2$
$endgroup$
– Zachary Hunter
Dec 30 '18 at 4:38
$begingroup$
Its been edited. Thanks
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:38
$begingroup$
@Zachary Hunter, I guess that is meant for the even function. Cant one use the $prod$ sign?
$endgroup$
– Abdulhameed
Dec 30 '18 at 4:40
1
$begingroup$
Your title refers to a sum while the body refers to a product. The question is the same, but they should be consistent. When you write $prod_{N=1}^{N} 2N^2$ you are using $N$ in two different ways. The top one is the upper limit of the dummy variable while the other two are the dummy variable. It would be much better to write $prod_{N=1}^{n} 2N^2$ The result will depend on $n$ but not $N$.
$endgroup$
– Ross Millikan
Dec 30 '18 at 4:44