Π Product properties
up vote
0
down vote
favorite
I would like to know why this product tends to this result: $prod_{j=0}^{n-1}2^{n-j}=2^{frac{n(n+1)}{2}}$. I just know that I should get out of the Π: $2^{n}$ but I don't know why $prod_{j=0}^{n-1}2^{n-j}=2^{n^2} prod_{j=0}^{n-1}frac{1}{2^j}$.
If someone can tell me how the term $2^{n^2}$ is out of the Π.
Thanks in advance.
discrete-mathematics
asked Nov 13 at 20:29
user587779
84
add a comment |
up vote
0
down vote
favorite
I would like to know why this product tends to this result: $prod_{j=0}^{n-1}2^{n-j}=2^{frac{n(n+1)}{2}}$. I just know that I should get out of the Π: $2^{n}$ but I don't know why $prod_{j=0}^{n-1}2^{n-j}=2^{n^2} prod_{j=0}^{n-1}frac{1}{2^j}$.
If someone can tell me how the term $2^{n^2}$ is out of the Π.
Thanks in advance.
discrete-mathematics
asked Nov 13 at 20:29
user587779
84
1
the resulting exponent of 2 is just the sum of the exponents, $sum_{j=0}^{n-1} ; (n-j) = sum_{k=n}^{1} ; k ; ; ; $
– Will Jagy
Nov 13 at 20:35
Since all terms in the product are exponents with base $2$, you can rewrite your product as $prod_{j=0}^{n-1}2^{n-j} = 2^{sum_{j=0}^{n-1}(n-j)}$. Now... what do you know about summations and triangle sums in particular. Can you figure out $sum_{j=0}^{n-1}n-j$?
– JMoravitz
Nov 13 at 20:36
Write the product you will see... $(2^n/2^j)*(2^n/2^j)*ldots*(2^n/2^j)$. In this product, you have $n$ terms, each term contains $2^n$. So you have: $(2^n*2^n*ldots*2^n)*prod_{j=0}^{n-1}(1/2^j)$.
– zdm
Nov 13 at 20:36
@zdm what you wrote is a bit misleading since $j$ changes from term to term. It would appear at first glance that you intended the same value of $j$ for each when that should not have been the case.
– JMoravitz
Nov 13 at 20:38
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I would like to know why this product tends to this result: $prod_{j=0}^{n-1}2^{n-j}=2^{frac{n(n+1)}{2}}$. I just know that I should get out of the Π: $2^{n}$ but I don't know why $prod_{j=0}^{n-1}2^{n-j}=2^{n^2} prod_{j=0}^{n-1}frac{1}{2^j}$.
If someone can tell me how the term $2^{n^2}$ is out of the Π.
Thanks in advance.
discrete-mathematics
asked Nov 13 at 20:29
user587779
84
I would like to know why this product tends to this result: $prod_{j=0}^{n-1}2^{n-j}=2^{frac{n(n+1)}{2}}$. I just know that I should get out of the Π: $2^{n}$ but I don't know why $prod_{j=0}^{n-1}2^{n-j}=2^{n^2} prod_{j=0}^{n-1}frac{1}{2^j}$.
If someone can tell me how the term $2^{n^2}$ is out of the Π.
Thanks in advance.
discrete-mathematics
discrete-mathematics
asked Nov 13 at 20:29
user587779
84
asked Nov 13 at 20:29
user587779
84
asked Nov 13 at 20:29
user587779
84
asked Nov 13 at 20:29
user587779
84
asked Nov 13 at 20:29
user587779
84
84
1
the resulting exponent of 2 is just the sum of the exponents, $sum_{j=0}^{n-1} ; (n-j) = sum_{k=n}^{1} ; k ; ; ; $
– Will Jagy
Nov 13 at 20:35
Since all terms in the product are exponents with base $2$, you can rewrite your product as $prod_{j=0}^{n-1}2^{n-j} = 2^{sum_{j=0}^{n-1}(n-j)}$. Now... what do you know about summations and triangle sums in particular. Can you figure out $sum_{j=0}^{n-1}n-j$?
– JMoravitz
Nov 13 at 20:36
Write the product you will see... $(2^n/2^j)*(2^n/2^j)*ldots*(2^n/2^j)$. In this product, you have $n$ terms, each term contains $2^n$. So you have: $(2^n*2^n*ldots*2^n)*prod_{j=0}^{n-1}(1/2^j)$.
– zdm
Nov 13 at 20:36
@zdm what you wrote is a bit misleading since $j$ changes from term to term. It would appear at first glance that you intended the same value of $j$ for each when that should not have been the case.
– JMoravitz
Nov 13 at 20:38
add a comment |
1
the resulting exponent of 2 is just the sum of the exponents, $sum_{j=0}^{n-1} ; (n-j) = sum_{k=n}^{1} ; k ; ; ; $
– Will Jagy
Nov 13 at 20:35
Since all terms in the product are exponents with base $2$, you can rewrite your product as $prod_{j=0}^{n-1}2^{n-j} = 2^{sum_{j=0}^{n-1}(n-j)}$. Now... what do you know about summations and triangle sums in particular. Can you figure out $sum_{j=0}^{n-1}n-j$?
– JMoravitz
Nov 13 at 20:36
Write the product you will see... $(2^n/2^j)*(2^n/2^j)*ldots*(2^n/2^j)$. In this product, you have $n$ terms, each term contains $2^n$. So you have: $(2^n*2^n*ldots*2^n)*prod_{j=0}^{n-1}(1/2^j)$.
– zdm
Nov 13 at 20:36
@zdm what you wrote is a bit misleading since $j$ changes from term to term. It would appear at first glance that you intended the same value of $j$ for each when that should not have been the case.
– JMoravitz
Nov 13 at 20:38
1
1
the resulting exponent of 2 is just the sum of the exponents, $sum_{j=0}^{n-1} ; (n-j) = sum_{k=n}^{1} ; k ; ; ; $
– Will Jagy
Nov 13 at 20:35
the resulting exponent of 2 is just the sum of the exponents, $sum_{j=0}^{n-1} ; (n-j) = sum_{k=n}^{1} ; k ; ; ; $
– Will Jagy
Nov 13 at 20:35
Since all terms in the product are exponents with base $2$, you can rewrite your product as $prod_{j=0}^{n-1}2^{n-j} = 2^{sum_{j=0}^{n-1}(n-j)}$. Now... what do you know about summations and triangle sums in particular. Can you figure out $sum_{j=0}^{n-1}n-j$?
– JMoravitz
Nov 13 at 20:36
Since all terms in the product are exponents with base $2$, you can rewrite your product as $prod_{j=0}^{n-1}2^{n-j} = 2^{sum_{j=0}^{n-1}(n-j)}$. Now... what do you know about summations and triangle sums in particular. Can you figure out $sum_{j=0}^{n-1}n-j$?
– JMoravitz
Nov 13 at 20:36
Write the product you will see... $(2^n/2^j)*(2^n/2^j)*ldots*(2^n/2^j)$. In this product, you have $n$ terms, each term contains $2^n$. So you have: $(2^n*2^n*ldots*2^n)*prod_{j=0}^{n-1}(1/2^j)$.
– zdm
Nov 13 at 20:36
Write the product you will see... $(2^n/2^j)*(2^n/2^j)*ldots*(2^n/2^j)$. In this product, you have $n$ terms, each term contains $2^n$. So you have: $(2^n*2^n*ldots*2^n)*prod_{j=0}^{n-1}(1/2^j)$.
– zdm
Nov 13 at 20:36
@zdm what you wrote is a bit misleading since $j$ changes from term to term. It would appear at first glance that you intended the same value of $j$ for each when that should not have been the case.
– JMoravitz
Nov 13 at 20:38
@zdm what you wrote is a bit misleading since $j$ changes from term to term. It would appear at first glance that you intended the same value of $j$ for each when that should not have been the case.
– JMoravitz
Nov 13 at 20:38
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
We have
$$begin{align}
prod_{j=0}^{n-1}2^{n-j}&=prod_{j=0}^{n-1}dfrac{2^n}{2^j}\&=dfrac{2^n}{2^0}*dfrac{2^n}{2^1}*dfrac{2^n}{2^2}*ldots*dfrac{2^n}{2^{n-2}}*dfrac{2^n}{2^{n-1}}\&=dfrac{ overbrace{2^n*2^n*2^n*ldots*2^n*2^n}^{color{red}{n text{ terms}}}}{2^0*2^1*2^2*ldots*2^{n-2}*2^{n-1}}\&=dfrac{overbrace{2^n*2^n*2^n*ldots*2^n*2^n}^{color{red}{n text{ terms}}}}{prod_{j=0}^{n-1}2^j}\&=dfrac{(2^n)^n}{prod_{j=0}^{n-1}2^j}=dfrac{2^{n^2}}{prod_{j=0}^{n-1}2^j}.
end{align}$$
answered Nov 13 at 20:49
zdm
1535
Then, you solve the product like $2^{sum_{j=0}^{n-1} j} = 2^{frac{n(n-1)}{2}}$ or is there a property? Thanks
– user587779
Nov 13 at 21:21
Yes, you could use the summation as you did.
– zdm
Nov 13 at 21:25
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
We have
$$begin{align}
prod_{j=0}^{n-1}2^{n-j}&=prod_{j=0}^{n-1}dfrac{2^n}{2^j}\&=dfrac{2^n}{2^0}*dfrac{2^n}{2^1}*dfrac{2^n}{2^2}*ldots*dfrac{2^n}{2^{n-2}}*dfrac{2^n}{2^{n-1}}\&=dfrac{ overbrace{2^n*2^n*2^n*ldots*2^n*2^n}^{color{red}{n text{ terms}}}}{2^0*2^1*2^2*ldots*2^{n-2}*2^{n-1}}\&=dfrac{overbrace{2^n*2^n*2^n*ldots*2^n*2^n}^{color{red}{n text{ terms}}}}{prod_{j=0}^{n-1}2^j}\&=dfrac{(2^n)^n}{prod_{j=0}^{n-1}2^j}=dfrac{2^{n^2}}{prod_{j=0}^{n-1}2^j}.
end{align}$$
answered Nov 13 at 20:49
zdm
1535
Then, you solve the product like $2^{sum_{j=0}^{n-1} j} = 2^{frac{n(n-1)}{2}}$ or is there a property? Thanks
– user587779
Nov 13 at 21:21
Yes, you could use the summation as you did.
– zdm
Nov 13 at 21:25
add a comment |
up vote
0
down vote
accepted
We have
$$begin{align}
prod_{j=0}^{n-1}2^{n-j}&=prod_{j=0}^{n-1}dfrac{2^n}{2^j}\&=dfrac{2^n}{2^0}*dfrac{2^n}{2^1}*dfrac{2^n}{2^2}*ldots*dfrac{2^n}{2^{n-2}}*dfrac{2^n}{2^{n-1}}\&=dfrac{ overbrace{2^n*2^n*2^n*ldots*2^n*2^n}^{color{red}{n text{ terms}}}}{2^0*2^1*2^2*ldots*2^{n-2}*2^{n-1}}\&=dfrac{overbrace{2^n*2^n*2^n*ldots*2^n*2^n}^{color{red}{n text{ terms}}}}{prod_{j=0}^{n-1}2^j}\&=dfrac{(2^n)^n}{prod_{j=0}^{n-1}2^j}=dfrac{2^{n^2}}{prod_{j=0}^{n-1}2^j}.
end{align}$$
answered Nov 13 at 20:49
zdm
1535
Then, you solve the product like $2^{sum_{j=0}^{n-1} j} = 2^{frac{n(n-1)}{2}}$ or is there a property? Thanks
– user587779
Nov 13 at 21:21
Yes, you could use the summation as you did.
– zdm
Nov 13 at 21:25
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
We have
$$begin{align}
prod_{j=0}^{n-1}2^{n-j}&=prod_{j=0}^{n-1}dfrac{2^n}{2^j}\&=dfrac{2^n}{2^0}*dfrac{2^n}{2^1}*dfrac{2^n}{2^2}*ldots*dfrac{2^n}{2^{n-2}}*dfrac{2^n}{2^{n-1}}\&=dfrac{ overbrace{2^n*2^n*2^n*ldots*2^n*2^n}^{color{red}{n text{ terms}}}}{2^0*2^1*2^2*ldots*2^{n-2}*2^{n-1}}\&=dfrac{overbrace{2^n*2^n*2^n*ldots*2^n*2^n}^{color{red}{n text{ terms}}}}{prod_{j=0}^{n-1}2^j}\&=dfrac{(2^n)^n}{prod_{j=0}^{n-1}2^j}=dfrac{2^{n^2}}{prod_{j=0}^{n-1}2^j}.
end{align}$$
answered Nov 13 at 20:49
zdm
1535
We have
$$begin{align}
prod_{j=0}^{n-1}2^{n-j}&=prod_{j=0}^{n-1}dfrac{2^n}{2^j}\&=dfrac{2^n}{2^0}*dfrac{2^n}{2^1}*dfrac{2^n}{2^2}*ldots*dfrac{2^n}{2^{n-2}}*dfrac{2^n}{2^{n-1}}\&=dfrac{ overbrace{2^n*2^n*2^n*ldots*2^n*2^n}^{color{red}{n text{ terms}}}}{2^0*2^1*2^2*ldots*2^{n-2}*2^{n-1}}\&=dfrac{overbrace{2^n*2^n*2^n*ldots*2^n*2^n}^{color{red}{n text{ terms}}}}{prod_{j=0}^{n-1}2^j}\&=dfrac{(2^n)^n}{prod_{j=0}^{n-1}2^j}=dfrac{2^{n^2}}{prod_{j=0}^{n-1}2^j}.
end{align}$$
answered Nov 13 at 20:49
zdm
1535
answered Nov 13 at 20:49
zdm
1535
answered Nov 13 at 20:49
zdm
1535
answered Nov 13 at 20:49
zdm
1535
1535
Then, you solve the product like $2^{sum_{j=0}^{n-1} j} = 2^{frac{n(n-1)}{2}}$ or is there a property? Thanks
– user587779
Nov 13 at 21:21
Yes, you could use the summation as you did.
– zdm
Nov 13 at 21:25
add a comment |
Then, you solve the product like $2^{sum_{j=0}^{n-1} j} = 2^{frac{n(n-1)}{2}}$ or is there a property? Thanks
– user587779
Nov 13 at 21:21
Yes, you could use the summation as you did.
– zdm
Nov 13 at 21:25
Then, you solve the product like $2^{sum_{j=0}^{n-1} j} = 2^{frac{n(n-1)}{2}}$ or is there a property? Thanks
– user587779
Nov 13 at 21:21
Then, you solve the product like $2^{sum_{j=0}^{n-1} j} = 2^{frac{n(n-1)}{2}}$ or is there a property? Thanks
– user587779
Nov 13 at 21:21
Yes, you could use the summation as you did.
– zdm
Nov 13 at 21:25
Yes, you could use the summation as you did.
– zdm
Nov 13 at 21:25
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2997294%2f%25ce%25a0-product-properties%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
the resulting exponent of 2 is just the sum of the exponents, $sum_{j=0}^{n-1} ; (n-j) = sum_{k=n}^{1} ; k ; ; ; $
– Will Jagy
Nov 13 at 20:35
Since all terms in the product are exponents with base $2$, you can rewrite your product as $prod_{j=0}^{n-1}2^{n-j} = 2^{sum_{j=0}^{n-1}(n-j)}$. Now... what do you know about summations and triangle sums in particular. Can you figure out $sum_{j=0}^{n-1}n-j$?
– JMoravitz
Nov 13 at 20:36
Write the product you will see... $(2^n/2^j)*(2^n/2^j)*ldots*(2^n/2^j)$. In this product, you have $n$ terms, each term contains $2^n$. So you have: $(2^n*2^n*ldots*2^n)*prod_{j=0}^{n-1}(1/2^j)$.
– zdm
Nov 13 at 20:36
@zdm what you wrote is a bit misleading since $j$ changes from term to term. It would appear at first glance that you intended the same value of $j$ for each when that should not have been the case.
– JMoravitz
Nov 13 at 20:38