Mixing
Expressing a step function
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Take the interval $I=[0,1]$ and the sequence $x_n_ninBbbN$ defined as follows:
$$x_n=sum_k=1^n frac12^k, forall n in BbbN$$
and $x_0=0$.
Now, that sequence defines a partition for the interval $I$. Let's define the function $f$ such that $f(0)=0, f(1)=1$ and for all other $xin I$, if $x_i-1<xleq x_i$ then $f(x)=x_i$, where $i in BbbN$. This funcion should look like this:
Did I define it correctly? Is there another way to define this function in a more elegant/short way? Bonus points if someone could post a plot of this function where the area "under the curve" is shown!
Edit: Is the area under the function equal to $frac12$?
Thanks in advance!
functions graphing-functions step-function
asked Jul 16 at 4:59
NotAMathematician
33011
add a comment |Â
up vote
0
down vote
favorite
Take the interval $I=[0,1]$ and the sequence $x_n_ninBbbN$ defined as follows:
$$x_n=sum_k=1^n frac12^k, forall n in BbbN$$
and $x_0=0$.
Now, that sequence defines a partition for the interval $I$. Let's define the function $f$ such that $f(0)=0, f(1)=1$ and for all other $xin I$, if $x_i-1<xleq x_i$ then $f(x)=x_i$, where $i in BbbN$. This funcion should look like this:
Did I define it correctly? Is there another way to define this function in a more elegant/short way? Bonus points if someone could post a plot of this function where the area "under the curve" is shown!
Edit: Is the area under the function equal to $frac12$?
Thanks in advance!
functions graphing-functions step-function
asked Jul 16 at 4:59
NotAMathematician
33011
1
In my experience, step functions are defined so as to only take on finitely many values, so I would hesitate to call that a step function
– Brevan Ellefsen
Jul 16 at 5:02
As for what you seem to intend... your function appears to be the one which takes an input of $0.a_1a_2a_3a_4a_5dots$ in binary and searches for the left-most occurrence of zero after the decimal point, changes it to a $1$, and truncates all the remaining digits after it. So, for example it would map $0.11colorred01000dots$ to $0.111$ in binary. (interesting to note is that choice of representation issues like $0.011111dots = 0.10000dots$ don't seem to matter here since they both map to the same output anyways).
– JMoravitz
Jul 16 at 5:05
The rest looks OK to me, at least in that the function you describe matches the picture you have. As for "Did I define it correctly?" - I'm not sure what you mean. Did you define what correctly? If you are asking if your function matches the picture, then yes (except that your picture doesn't have $f(0)=0$ and $f(1)=1$)
– Brevan Ellefsen
Jul 16 at 5:05
3
The summation for the area under the curve: $sumlimits_k=1^infty frac12^k(1-frac12^k)=sumfrac12^k-sumfrac14^k=1-frac13=frac23$
– JMoravitz
Jul 16 at 5:11
... and the area under the curve is clearly greater than $frac12$ because all sections of the curve lie above the diagonal line $y=x$.
– gandalf61
Jul 16 at 10:24
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Take the interval $I=[0,1]$ and the sequence $x_n_ninBbbN$ defined as follows:
$$x_n=sum_k=1^n frac12^k, forall n in BbbN$$
and $x_0=0$.
Now, that sequence defines a partition for the interval $I$. Let's define the function $f$ such that $f(0)=0, f(1)=1$ and for all other $xin I$, if $x_i-1<xleq x_i$ then $f(x)=x_i$, where $i in BbbN$. This funcion should look like this:
Did I define it correctly? Is there another way to define this function in a more elegant/short way? Bonus points if someone could post a plot of this function where the area "under the curve" is shown!
Edit: Is the area under the function equal to $frac12$?
Thanks in advance!
functions graphing-functions step-function
asked Jul 16 at 4:59
NotAMathematician
33011
Take the interval $I=[0,1]$ and the sequence $x_n_ninBbbN$ defined as follows:
$$x_n=sum_k=1^n frac12^k, forall n in BbbN$$
and $x_0=0$.
Now, that sequence defines a partition for the interval $I$. Let's define the function $f$ such that $f(0)=0, f(1)=1$ and for all other $xin I$, if $x_i-1<xleq x_i$ then $f(x)=x_i$, where $i in BbbN$. This funcion should look like this:
Did I define it correctly? Is there another way to define this function in a more elegant/short way? Bonus points if someone could post a plot of this function where the area "under the curve" is shown!
Edit: Is the area under the function equal to $frac12$?
Thanks in advance!
functions graphing-functions step-function
asked Jul 16 at 4:59
NotAMathematician
33011
edited Jul 16 at 5:05
asked Jul 16 at 4:59
NotAMathematician
33011
asked Jul 16 at 4:59
NotAMathematician
33011
asked Jul 16 at 4:59
NotAMathematician
33011
33011
1
In my experience, step functions are defined so as to only take on finitely many values, so I would hesitate to call that a step function
– Brevan Ellefsen
Jul 16 at 5:02
As for what you seem to intend... your function appears to be the one which takes an input of $0.a_1a_2a_3a_4a_5dots$ in binary and searches for the left-most occurrence of zero after the decimal point, changes it to a $1$, and truncates all the remaining digits after it. So, for example it would map $0.11colorred01000dots$ to $0.111$ in binary. (interesting to note is that choice of representation issues like $0.011111dots = 0.10000dots$ don't seem to matter here since they both map to the same output anyways).
– JMoravitz
Jul 16 at 5:05
The rest looks OK to me, at least in that the function you describe matches the picture you have. As for "Did I define it correctly?" - I'm not sure what you mean. Did you define what correctly? If you are asking if your function matches the picture, then yes (except that your picture doesn't have $f(0)=0$ and $f(1)=1$)
– Brevan Ellefsen
Jul 16 at 5:05
3
The summation for the area under the curve: $sumlimits_k=1^infty frac12^k(1-frac12^k)=sumfrac12^k-sumfrac14^k=1-frac13=frac23$
– JMoravitz
Jul 16 at 5:11
... and the area under the curve is clearly greater than $frac12$ because all sections of the curve lie above the diagonal line $y=x$.
– gandalf61
Jul 16 at 10:24
add a comment |Â
1
In my experience, step functions are defined so as to only take on finitely many values, so I would hesitate to call that a step function
– Brevan Ellefsen
Jul 16 at 5:02
As for what you seem to intend... your function appears to be the one which takes an input of $0.a_1a_2a_3a_4a_5dots$ in binary and searches for the left-most occurrence of zero after the decimal point, changes it to a $1$, and truncates all the remaining digits after it. So, for example it would map $0.11colorred01000dots$ to $0.111$ in binary. (interesting to note is that choice of representation issues like $0.011111dots = 0.10000dots$ don't seem to matter here since they both map to the same output anyways).
– JMoravitz
Jul 16 at 5:05
The rest looks OK to me, at least in that the function you describe matches the picture you have. As for "Did I define it correctly?" - I'm not sure what you mean. Did you define what correctly? If you are asking if your function matches the picture, then yes (except that your picture doesn't have $f(0)=0$ and $f(1)=1$)
– Brevan Ellefsen
Jul 16 at 5:05
3
The summation for the area under the curve: $sumlimits_k=1^infty frac12^k(1-frac12^k)=sumfrac12^k-sumfrac14^k=1-frac13=frac23$
– JMoravitz
Jul 16 at 5:11
... and the area under the curve is clearly greater than $frac12$ because all sections of the curve lie above the diagonal line $y=x$.
– gandalf61
Jul 16 at 10:24
1
1
In my experience, step functions are defined so as to only take on finitely many values, so I would hesitate to call that a step function
– Brevan Ellefsen
Jul 16 at 5:02
In my experience, step functions are defined so as to only take on finitely many values, so I would hesitate to call that a step function
– Brevan Ellefsen
Jul 16 at 5:02
As for what you seem to intend... your function appears to be the one which takes an input of $0.a_1a_2a_3a_4a_5dots$ in binary and searches for the left-most occurrence of zero after the decimal point, changes it to a $1$, and truncates all the remaining digits after it. So, for example it would map $0.11colorred01000dots$ to $0.111$ in binary. (interesting to note is that choice of representation issues like $0.011111dots = 0.10000dots$ don't seem to matter here since they both map to the same output anyways).
– JMoravitz
Jul 16 at 5:05
As for what you seem to intend... your function appears to be the one which takes an input of $0.a_1a_2a_3a_4a_5dots$ in binary and searches for the left-most occurrence of zero after the decimal point, changes it to a $1$, and truncates all the remaining digits after it. So, for example it would map $0.11colorred01000dots$ to $0.111$ in binary. (interesting to note is that choice of representation issues like $0.011111dots = 0.10000dots$ don't seem to matter here since they both map to the same output anyways).
– JMoravitz
Jul 16 at 5:05
The rest looks OK to me, at least in that the function you describe matches the picture you have. As for "Did I define it correctly?" - I'm not sure what you mean. Did you define what correctly? If you are asking if your function matches the picture, then yes (except that your picture doesn't have $f(0)=0$ and $f(1)=1$)
– Brevan Ellefsen
Jul 16 at 5:05
The rest looks OK to me, at least in that the function you describe matches the picture you have. As for "Did I define it correctly?" - I'm not sure what you mean. Did you define what correctly? If you are asking if your function matches the picture, then yes (except that your picture doesn't have $f(0)=0$ and $f(1)=1$)
– Brevan Ellefsen
Jul 16 at 5:05
3
3
The summation for the area under the curve: $sumlimits_k=1^infty frac12^k(1-frac12^k)=sumfrac12^k-sumfrac14^k=1-frac13=frac23$
– JMoravitz
Jul 16 at 5:11
The summation for the area under the curve: $sumlimits_k=1^infty frac12^k(1-frac12^k)=sumfrac12^k-sumfrac14^k=1-frac13=frac23$
– JMoravitz
Jul 16 at 5:11
... and the area under the curve is clearly greater than $frac12$ because all sections of the curve lie above the diagonal line $y=x$.
– gandalf61
Jul 16 at 10:24
... and the area under the curve is clearly greater than $frac12$ because all sections of the curve lie above the diagonal line $y=x$.
– gandalf61
Jul 16 at 10:24
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
$$x_n=sum_k=1^n frac12^k=1-frac12^n tag 1$$
$$2^n=frac11-x_n$$
$$n=frac-ln(1-x_n)ln(2) tag 2$$
Eq.$(2)$ is only valid for the discret values of $x$ defined by Eq.$(1)$.
If we want to extend to continuous $x$ we have to use the celling or the floor functions, respectively noted $lceil:rceil$ and $lfloor:rfloor$ :
http://mathworld.wolfram.com/CeilingFunction.html
http://mathworld.wolfram.com/FloorFunction.html
$$n=leftlceilfrac-ln(1-x)ln(2)rightrceil=1+leftlfloorfrac-ln(1-x)ln(2)rightrfloor$$
Thus, the equation represented on your graph is :
$$f(x)=1-frac12^1+leftlfloorfrac-ln(1-x)ln(2)rightrfloor$$
$$A=int_0^1 f(x)dx=sum_n=1^infty f(x_n)(x_n-x_n-1)$$
$$A = sum_k=1^n left(1-frac12^nright)left((1-frac12^n)-(1-frac12^n-1) right) =sum_k=1^n left(1-frac12^nright)frac12^n$$
$$A = sum_k=1^nfrac12^n-sum_k=1^nfrac12^2n=1-frac13$$
$$int_0^1 f(x)dx=frac23$$
answered Jul 16 at 9:40
JJacquelin
40.1k21649
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$$x_n=sum_k=1^n frac12^k=1-frac12^n tag 1$$
$$2^n=frac11-x_n$$
$$n=frac-ln(1-x_n)ln(2) tag 2$$
Eq.$(2)$ is only valid for the discret values of $x$ defined by Eq.$(1)$.
If we want to extend to continuous $x$ we have to use the celling or the floor functions, respectively noted $lceil:rceil$ and $lfloor:rfloor$ :
http://mathworld.wolfram.com/CeilingFunction.html
http://mathworld.wolfram.com/FloorFunction.html
$$n=leftlceilfrac-ln(1-x)ln(2)rightrceil=1+leftlfloorfrac-ln(1-x)ln(2)rightrfloor$$
Thus, the equation represented on your graph is :
$$f(x)=1-frac12^1+leftlfloorfrac-ln(1-x)ln(2)rightrfloor$$
$$A=int_0^1 f(x)dx=sum_n=1^infty f(x_n)(x_n-x_n-1)$$
$$A = sum_k=1^n left(1-frac12^nright)left((1-frac12^n)-(1-frac12^n-1) right) =sum_k=1^n left(1-frac12^nright)frac12^n$$
$$A = sum_k=1^nfrac12^n-sum_k=1^nfrac12^2n=1-frac13$$
$$int_0^1 f(x)dx=frac23$$
answered Jul 16 at 9:40
JJacquelin
40.1k21649
add a comment |Â
up vote
3
down vote
accepted
$$x_n=sum_k=1^n frac12^k=1-frac12^n tag 1$$
$$2^n=frac11-x_n$$
$$n=frac-ln(1-x_n)ln(2) tag 2$$
Eq.$(2)$ is only valid for the discret values of $x$ defined by Eq.$(1)$.
If we want to extend to continuous $x$ we have to use the celling or the floor functions, respectively noted $lceil:rceil$ and $lfloor:rfloor$ :
http://mathworld.wolfram.com/CeilingFunction.html
http://mathworld.wolfram.com/FloorFunction.html
$$n=leftlceilfrac-ln(1-x)ln(2)rightrceil=1+leftlfloorfrac-ln(1-x)ln(2)rightrfloor$$
Thus, the equation represented on your graph is :
$$f(x)=1-frac12^1+leftlfloorfrac-ln(1-x)ln(2)rightrfloor$$
$$A=int_0^1 f(x)dx=sum_n=1^infty f(x_n)(x_n-x_n-1)$$
$$A = sum_k=1^n left(1-frac12^nright)left((1-frac12^n)-(1-frac12^n-1) right) =sum_k=1^n left(1-frac12^nright)frac12^n$$
$$A = sum_k=1^nfrac12^n-sum_k=1^nfrac12^2n=1-frac13$$
$$int_0^1 f(x)dx=frac23$$
answered Jul 16 at 9:40
JJacquelin
40.1k21649
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$$x_n=sum_k=1^n frac12^k=1-frac12^n tag 1$$
$$2^n=frac11-x_n$$
$$n=frac-ln(1-x_n)ln(2) tag 2$$
Eq.$(2)$ is only valid for the discret values of $x$ defined by Eq.$(1)$.
If we want to extend to continuous $x$ we have to use the celling or the floor functions, respectively noted $lceil:rceil$ and $lfloor:rfloor$ :
http://mathworld.wolfram.com/CeilingFunction.html
http://mathworld.wolfram.com/FloorFunction.html
$$n=leftlceilfrac-ln(1-x)ln(2)rightrceil=1+leftlfloorfrac-ln(1-x)ln(2)rightrfloor$$
Thus, the equation represented on your graph is :
$$f(x)=1-frac12^1+leftlfloorfrac-ln(1-x)ln(2)rightrfloor$$
$$A=int_0^1 f(x)dx=sum_n=1^infty f(x_n)(x_n-x_n-1)$$
$$A = sum_k=1^n left(1-frac12^nright)left((1-frac12^n)-(1-frac12^n-1) right) =sum_k=1^n left(1-frac12^nright)frac12^n$$
$$A = sum_k=1^nfrac12^n-sum_k=1^nfrac12^2n=1-frac13$$
$$int_0^1 f(x)dx=frac23$$
answered Jul 16 at 9:40
JJacquelin
40.1k21649
$$x_n=sum_k=1^n frac12^k=1-frac12^n tag 1$$
$$2^n=frac11-x_n$$
$$n=frac-ln(1-x_n)ln(2) tag 2$$
Eq.$(2)$ is only valid for the discret values of $x$ defined by Eq.$(1)$.
If we want to extend to continuous $x$ we have to use the celling or the floor functions, respectively noted $lceil:rceil$ and $lfloor:rfloor$ :
http://mathworld.wolfram.com/CeilingFunction.html
http://mathworld.wolfram.com/FloorFunction.html
$$n=leftlceilfrac-ln(1-x)ln(2)rightrceil=1+leftlfloorfrac-ln(1-x)ln(2)rightrfloor$$
Thus, the equation represented on your graph is :
$$f(x)=1-frac12^1+leftlfloorfrac-ln(1-x)ln(2)rightrfloor$$
$$A=int_0^1 f(x)dx=sum_n=1^infty f(x_n)(x_n-x_n-1)$$
$$A = sum_k=1^n left(1-frac12^nright)left((1-frac12^n)-(1-frac12^n-1) right) =sum_k=1^n left(1-frac12^nright)frac12^n$$
$$A = sum_k=1^nfrac12^n-sum_k=1^nfrac12^2n=1-frac13$$
$$int_0^1 f(x)dx=frac23$$
answered Jul 16 at 9:40
JJacquelin
40.1k21649
edited Jul 16 at 10:25
answered Jul 16 at 9:40
JJacquelin
40.1k21649
answered Jul 16 at 9:40
JJacquelin
40.1k21649
answered Jul 16 at 9:40
JJacquelin
40.1k21649
40.1k21649
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2853109%2fexpressing-a-step-function%23new-answer', 'question_page');
);
Post as a guest
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
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
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
1
In my experience, step functions are defined so as to only take on finitely many values, so I would hesitate to call that a step function
– Brevan Ellefsen
Jul 16 at 5:02
As for what you seem to intend... your function appears to be the one which takes an input of $0.a_1a_2a_3a_4a_5dots$ in binary and searches for the left-most occurrence of zero after the decimal point, changes it to a $1$, and truncates all the remaining digits after it. So, for example it would map $0.11colorred01000dots$ to $0.111$ in binary. (interesting to note is that choice of representation issues like $0.011111dots = 0.10000dots$ don't seem to matter here since they both map to the same output anyways).
– JMoravitz
Jul 16 at 5:05
The rest looks OK to me, at least in that the function you describe matches the picture you have. As for "Did I define it correctly?" - I'm not sure what you mean. Did you define what correctly? If you are asking if your function matches the picture, then yes (except that your picture doesn't have $f(0)=0$ and $f(1)=1$)
– Brevan Ellefsen
Jul 16 at 5:05
3
The summation for the area under the curve: $sumlimits_k=1^infty frac12^k(1-frac12^k)=sumfrac12^k-sumfrac14^k=1-frac13=frac23$
– JMoravitz
Jul 16 at 5:11
... and the area under the curve is clearly greater than $frac12$ because all sections of the curve lie above the diagonal line $y=x$.
– gandalf61
Jul 16 at 10:24