Why is $ prod_{n=0}^{N-1} u[x_n + theta] - u[x_n-theta] = u[theta - max(|x_n| )]$?
$begingroup$
I'm self-studying math, and came across a problem:
$$
prod_{n=0}^{N-1} left(u[x_n + theta] - u[x_n-theta]right) = u[theta - max(|x_n| )]
$$
where $u$ is a unit step function, $x_n$ are sample points, $theta$ is a constant. So basically, the two unit steps functions form a rectangular function, and the centers $x_n$ of the rectangular functions are different and if they don't overlap then the whole product is zero. And we are to find how to have this product not equal to zero.
So by the explanation from the book, if $-theta < x[n] < theta$, this product equals to $u(theta - max(|x[n]| )$. Why is that?
calculus probability statistics products step-function
edited Dec 19 '18 at 20:52
Lorenzo B.
1,8602520
asked Dec 19 '18 at 6:42
drerDdrerD
1609
$endgroup$
add a comment |
$begingroup$
I'm self-studying math, and came across a problem:
$$
prod_{n=0}^{N-1} left(u[x_n + theta] - u[x_n-theta]right) = u[theta - max(|x_n| )]
$$
where $u$ is a unit step function, $x_n$ are sample points, $theta$ is a constant. So basically, the two unit steps functions form a rectangular function, and the centers $x_n$ of the rectangular functions are different and if they don't overlap then the whole product is zero. And we are to find how to have this product not equal to zero.
So by the explanation from the book, if $-theta < x[n] < theta$, this product equals to $u(theta - max(|x[n]| )$. Why is that?
calculus probability statistics products step-function
edited Dec 19 '18 at 20:52
Lorenzo B.
1,8602520
asked Dec 19 '18 at 6:42
drerDdrerD
1609
$endgroup$
1
$begingroup$
So as you said, the LHS equals to $1$ if and only if $-theta < x_n < theta$ for $n = 0, 1, 2 ldots, N-1$, which is equivalent to $|x_n| < theta$. Since it holds for all $n$, it is equivalent to say that the $max |x_n| < theta$. So the unit step function in the RHS satisfy this and thus the equality holds.
$endgroup$
– BGM
Dec 19 '18 at 10:54
$begingroup$
@BGM thanks for trying to help. I understand up the the max part, but I don't know how that translates to $u(theta-max(|x_n|))$
$endgroup$
– drerD
Dec 19 '18 at 20:09
add a comment |
$begingroup$
I'm self-studying math, and came across a problem:
$$
prod_{n=0}^{N-1} left(u[x_n + theta] - u[x_n-theta]right) = u[theta - max(|x_n| )]
$$
where $u$ is a unit step function, $x_n$ are sample points, $theta$ is a constant. So basically, the two unit steps functions form a rectangular function, and the centers $x_n$ of the rectangular functions are different and if they don't overlap then the whole product is zero. And we are to find how to have this product not equal to zero.
So by the explanation from the book, if $-theta < x[n] < theta$, this product equals to $u(theta - max(|x[n]| )$. Why is that?
calculus probability statistics products step-function
edited Dec 19 '18 at 20:52
Lorenzo B.
1,8602520
asked Dec 19 '18 at 6:42
drerDdrerD
1609
$endgroup$
I'm self-studying math, and came across a problem:
$$
prod_{n=0}^{N-1} left(u[x_n + theta] - u[x_n-theta]right) = u[theta - max(|x_n| )]
$$
where $u$ is a unit step function, $x_n$ are sample points, $theta$ is a constant. So basically, the two unit steps functions form a rectangular function, and the centers $x_n$ of the rectangular functions are different and if they don't overlap then the whole product is zero. And we are to find how to have this product not equal to zero.
So by the explanation from the book, if $-theta < x[n] < theta$, this product equals to $u(theta - max(|x[n]| )$. Why is that?
calculus probability statistics products step-function
calculus probability statistics products step-function
edited Dec 19 '18 at 20:52
Lorenzo B.
1,8602520
asked Dec 19 '18 at 6:42
drerDdrerD
1609
edited Dec 19 '18 at 20:52
Lorenzo B.
1,8602520
asked Dec 19 '18 at 6:42
drerDdrerD
1609
edited Dec 19 '18 at 20:52
Lorenzo B.
1,8602520
edited Dec 19 '18 at 20:52
Lorenzo B.
1,8602520
edited Dec 19 '18 at 20:52
Lorenzo B.
1,8602520
1,8602520
asked Dec 19 '18 at 6:42
drerDdrerD
1609
asked Dec 19 '18 at 6:42
drerDdrerD
1609
asked Dec 19 '18 at 6:42
drerDdrerD
1609
1609
1
$begingroup$
So as you said, the LHS equals to $1$ if and only if $-theta < x_n < theta$ for $n = 0, 1, 2 ldots, N-1$, which is equivalent to $|x_n| < theta$. Since it holds for all $n$, it is equivalent to say that the $max |x_n| < theta$. So the unit step function in the RHS satisfy this and thus the equality holds.
$endgroup$
– BGM
Dec 19 '18 at 10:54
$begingroup$
@BGM thanks for trying to help. I understand up the the max part, but I don't know how that translates to $u(theta-max(|x_n|))$
$endgroup$
– drerD
Dec 19 '18 at 20:09
add a comment |
1
$begingroup$
So as you said, the LHS equals to $1$ if and only if $-theta < x_n < theta$ for $n = 0, 1, 2 ldots, N-1$, which is equivalent to $|x_n| < theta$. Since it holds for all $n$, it is equivalent to say that the $max |x_n| < theta$. So the unit step function in the RHS satisfy this and thus the equality holds.
$endgroup$
– BGM
Dec 19 '18 at 10:54
$begingroup$
@BGM thanks for trying to help. I understand up the the max part, but I don't know how that translates to $u(theta-max(|x_n|))$
$endgroup$
– drerD
Dec 19 '18 at 20:09
1
1
$begingroup$
So as you said, the LHS equals to $1$ if and only if $-theta < x_n < theta$ for $n = 0, 1, 2 ldots, N-1$, which is equivalent to $|x_n| < theta$. Since it holds for all $n$, it is equivalent to say that the $max |x_n| < theta$. So the unit step function in the RHS satisfy this and thus the equality holds.
$endgroup$
– BGM
Dec 19 '18 at 10:54
$begingroup$
So as you said, the LHS equals to $1$ if and only if $-theta < x_n < theta$ for $n = 0, 1, 2 ldots, N-1$, which is equivalent to $|x_n| < theta$. Since it holds for all $n$, it is equivalent to say that the $max |x_n| < theta$. So the unit step function in the RHS satisfy this and thus the equality holds.
$endgroup$
– BGM
Dec 19 '18 at 10:54
$begingroup$
@BGM thanks for trying to help. I understand up the the max part, but I don't know how that translates to $u(theta-max(|x_n|))$
$endgroup$
– drerD
Dec 19 '18 at 20:09
$begingroup$
@BGM thanks for trying to help. I understand up the the max part, but I don't know how that translates to $u(theta-max(|x_n|))$
$endgroup$
– drerD
Dec 19 '18 at 20:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The step function $u[t]$ equals $1$ if its argument $t$ is greater than zero, and equals $0$ otherwise. Another way to write this is $$u[t] = I(t>0),tag1$$ where $I(cdot)$ denotes the indicator function, which is one when its argument is true and zero otherwise. Your book argues that the expression
$$prod_{n=0}^{N-1} left(u[x_n + theta] - u[x_n-theta]right)tag2$$
equals one if $max|x_n|<theta$, and equals zero otherwise. This means expression (2) is equal to
$$I(max|x_n|<theta),$$
which can be rewritten $I(theta-max|x_n|>0)$, which by (1) can be rewritten $u[theta-max|x_n|]$.
answered Dec 19 '18 at 21:21
grand_chatgrand_chat
20.3k11326
$endgroup$
$begingroup$
wow, thank you. It has puzzled me for more than week.
$endgroup$
– drerD
Dec 20 '18 at 2:43
add a comment |
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$begingroup$
The step function $u[t]$ equals $1$ if its argument $t$ is greater than zero, and equals $0$ otherwise. Another way to write this is $$u[t] = I(t>0),tag1$$ where $I(cdot)$ denotes the indicator function, which is one when its argument is true and zero otherwise. Your book argues that the expression
$$prod_{n=0}^{N-1} left(u[x_n + theta] - u[x_n-theta]right)tag2$$
equals one if $max|x_n|<theta$, and equals zero otherwise. This means expression (2) is equal to
$$I(max|x_n|<theta),$$
which can be rewritten $I(theta-max|x_n|>0)$, which by (1) can be rewritten $u[theta-max|x_n|]$.
answered Dec 19 '18 at 21:21
grand_chatgrand_chat
20.3k11326
$endgroup$
$begingroup$
wow, thank you. It has puzzled me for more than week.
$endgroup$
– drerD
Dec 20 '18 at 2:43
add a comment |
$begingroup$
The step function $u[t]$ equals $1$ if its argument $t$ is greater than zero, and equals $0$ otherwise. Another way to write this is $$u[t] = I(t>0),tag1$$ where $I(cdot)$ denotes the indicator function, which is one when its argument is true and zero otherwise. Your book argues that the expression
$$prod_{n=0}^{N-1} left(u[x_n + theta] - u[x_n-theta]right)tag2$$
equals one if $max|x_n|<theta$, and equals zero otherwise. This means expression (2) is equal to
$$I(max|x_n|<theta),$$
which can be rewritten $I(theta-max|x_n|>0)$, which by (1) can be rewritten $u[theta-max|x_n|]$.
answered Dec 19 '18 at 21:21
grand_chatgrand_chat
20.3k11326
$endgroup$
$begingroup$
wow, thank you. It has puzzled me for more than week.
$endgroup$
– drerD
Dec 20 '18 at 2:43
add a comment |
$begingroup$
The step function $u[t]$ equals $1$ if its argument $t$ is greater than zero, and equals $0$ otherwise. Another way to write this is $$u[t] = I(t>0),tag1$$ where $I(cdot)$ denotes the indicator function, which is one when its argument is true and zero otherwise. Your book argues that the expression
$$prod_{n=0}^{N-1} left(u[x_n + theta] - u[x_n-theta]right)tag2$$
equals one if $max|x_n|<theta$, and equals zero otherwise. This means expression (2) is equal to
$$I(max|x_n|<theta),$$
which can be rewritten $I(theta-max|x_n|>0)$, which by (1) can be rewritten $u[theta-max|x_n|]$.
answered Dec 19 '18 at 21:21
grand_chatgrand_chat
20.3k11326
$endgroup$
The step function $u[t]$ equals $1$ if its argument $t$ is greater than zero, and equals $0$ otherwise. Another way to write this is $$u[t] = I(t>0),tag1$$ where $I(cdot)$ denotes the indicator function, which is one when its argument is true and zero otherwise. Your book argues that the expression
$$prod_{n=0}^{N-1} left(u[x_n + theta] - u[x_n-theta]right)tag2$$
equals one if $max|x_n|<theta$, and equals zero otherwise. This means expression (2) is equal to
$$I(max|x_n|<theta),$$
which can be rewritten $I(theta-max|x_n|>0)$, which by (1) can be rewritten $u[theta-max|x_n|]$.
answered Dec 19 '18 at 21:21
grand_chatgrand_chat
20.3k11326
answered Dec 19 '18 at 21:21
grand_chatgrand_chat
20.3k11326
answered Dec 19 '18 at 21:21
grand_chatgrand_chat
20.3k11326
answered Dec 19 '18 at 21:21
grand_chatgrand_chat
20.3k11326
20.3k11326
$begingroup$
wow, thank you. It has puzzled me for more than week.
$endgroup$
– drerD
Dec 20 '18 at 2:43
add a comment |
$begingroup$
wow, thank you. It has puzzled me for more than week.
$endgroup$
– drerD
Dec 20 '18 at 2:43
$begingroup$
wow, thank you. It has puzzled me for more than week.
$endgroup$
– drerD
Dec 20 '18 at 2:43
$begingroup$
wow, thank you. It has puzzled me for more than week.
$endgroup$
– drerD
Dec 20 '18 at 2:43
add a comment |
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$begingroup$
So as you said, the LHS equals to $1$ if and only if $-theta < x_n < theta$ for $n = 0, 1, 2 ldots, N-1$, which is equivalent to $|x_n| < theta$. Since it holds for all $n$, it is equivalent to say that the $max |x_n| < theta$. So the unit step function in the RHS satisfy this and thus the equality holds.
$endgroup$
– BGM
Dec 19 '18 at 10:54
$begingroup$
@BGM thanks for trying to help. I understand up the the max part, but I don't know how that translates to $u(theta-max(|x_n|))$
$endgroup$
– drerD
Dec 19 '18 at 20:09