Computing the value of a function based on the geometric series of 1/2^n
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So in John B. Conway's First Course in Analysis Book we're told to consider the interval $[a,b],$ and let ${r_n}$ denote a sequence of all rationals in said interval. Next, we define a map $alpha:[a,b]tomathbb{R}$ by $$alpha(t)=sum_{substack{ninmathbb{N}\r_n<t}}dfrac{1}{2^n}.$$
Totally fine with that definition. Where I'm confused is when he says that $alpha(b)=1.$ If $b$ is irrational surely that would be the case as $alpha(b)$ would just be a geometric sum which converges to $1$. If $b$ is rational, however, it seems like $alpha(b)$ would never be $1$. Consider $r_n=b$ for some $ninmathbb{N}$, hence $alpha(b)=1-1/2^n<1.$ So I'm confused as to why Conway says it's $1$. Could anyone explain? Thank you.
real-analysis analysis geometric-series
asked Nov 23 at 18:53
Melody
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So in John B. Conway's First Course in Analysis Book we're told to consider the interval $[a,b],$ and let ${r_n}$ denote a sequence of all rationals in said interval. Next, we define a map $alpha:[a,b]tomathbb{R}$ by $$alpha(t)=sum_{substack{ninmathbb{N}\r_n<t}}dfrac{1}{2^n}.$$
Totally fine with that definition. Where I'm confused is when he says that $alpha(b)=1.$ If $b$ is irrational surely that would be the case as $alpha(b)$ would just be a geometric sum which converges to $1$. If $b$ is rational, however, it seems like $alpha(b)$ would never be $1$. Consider $r_n=b$ for some $ninmathbb{N}$, hence $alpha(b)=1-1/2^n<1.$ So I'm confused as to why Conway says it's $1$. Could anyone explain? Thank you.
real-analysis analysis geometric-series
asked Nov 23 at 18:53
Melody
63212
3
I don't know the context of the question, but could it be possible that ${r_n}$ is to be a sequence of all rationals in $(a,b)$ rather than $[a, b]$?
– angryavian
Nov 23 at 18:59
1
I agree with what you say. Is the interval really closed? Also just to be sure,is the sequence rn supposed to be a bijection?
– Keen
Nov 23 at 19:08
You know, Conway doesn't really specify. However, I've deduced that the interval must be closed, as we call it $J$ with implicit endpoints $a,b$. We define $alpha:Jtomathbb{R}$. We then ask for $alpha(a),alpha(b)$ so already seems implied the interval is closed. Next, we're given that it is discontinuous at each rational. This implies we mean to choose the $r_n$ from the closed interval, as otherwise we have no right Jump discontinuity at $a$.
– Melody
Nov 23 at 19:21
1
Just sum over $r_nleq t$ if you prefer. This gives a cadlag function
– Federico
Nov 23 at 19:21
@Federico I thought about that, only Conway also claims we have $alpha(a)=0$, which fails if we change it to $r_nleq t$ when $a$ is rational.
– Melody
Nov 23 at 19:23
|
show 7 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
So in John B. Conway's First Course in Analysis Book we're told to consider the interval $[a,b],$ and let ${r_n}$ denote a sequence of all rationals in said interval. Next, we define a map $alpha:[a,b]tomathbb{R}$ by $$alpha(t)=sum_{substack{ninmathbb{N}\r_n<t}}dfrac{1}{2^n}.$$
Totally fine with that definition. Where I'm confused is when he says that $alpha(b)=1.$ If $b$ is irrational surely that would be the case as $alpha(b)$ would just be a geometric sum which converges to $1$. If $b$ is rational, however, it seems like $alpha(b)$ would never be $1$. Consider $r_n=b$ for some $ninmathbb{N}$, hence $alpha(b)=1-1/2^n<1.$ So I'm confused as to why Conway says it's $1$. Could anyone explain? Thank you.
real-analysis analysis geometric-series
asked Nov 23 at 18:53
Melody
63212
So in John B. Conway's First Course in Analysis Book we're told to consider the interval $[a,b],$ and let ${r_n}$ denote a sequence of all rationals in said interval. Next, we define a map $alpha:[a,b]tomathbb{R}$ by $$alpha(t)=sum_{substack{ninmathbb{N}\r_n<t}}dfrac{1}{2^n}.$$
Totally fine with that definition. Where I'm confused is when he says that $alpha(b)=1.$ If $b$ is irrational surely that would be the case as $alpha(b)$ would just be a geometric sum which converges to $1$. If $b$ is rational, however, it seems like $alpha(b)$ would never be $1$. Consider $r_n=b$ for some $ninmathbb{N}$, hence $alpha(b)=1-1/2^n<1.$ So I'm confused as to why Conway says it's $1$. Could anyone explain? Thank you.
real-analysis analysis geometric-series
real-analysis analysis geometric-series
asked Nov 23 at 18:53
Melody
63212
asked Nov 23 at 18:53
Melody
63212
asked Nov 23 at 18:53
Melody
63212
asked Nov 23 at 18:53
Melody
63212
asked Nov 23 at 18:53
Melody
63212
63212
3
I don't know the context of the question, but could it be possible that ${r_n}$ is to be a sequence of all rationals in $(a,b)$ rather than $[a, b]$?
– angryavian
Nov 23 at 18:59
1
I agree with what you say. Is the interval really closed? Also just to be sure,is the sequence rn supposed to be a bijection?
– Keen
Nov 23 at 19:08
You know, Conway doesn't really specify. However, I've deduced that the interval must be closed, as we call it $J$ with implicit endpoints $a,b$. We define $alpha:Jtomathbb{R}$. We then ask for $alpha(a),alpha(b)$ so already seems implied the interval is closed. Next, we're given that it is discontinuous at each rational. This implies we mean to choose the $r_n$ from the closed interval, as otherwise we have no right Jump discontinuity at $a$.
– Melody
Nov 23 at 19:21
1
Just sum over $r_nleq t$ if you prefer. This gives a cadlag function
– Federico
Nov 23 at 19:21
@Federico I thought about that, only Conway also claims we have $alpha(a)=0$, which fails if we change it to $r_nleq t$ when $a$ is rational.
– Melody
Nov 23 at 19:23
|
show 7 more comments
3
I don't know the context of the question, but could it be possible that ${r_n}$ is to be a sequence of all rationals in $(a,b)$ rather than $[a, b]$?
– angryavian
Nov 23 at 18:59
1
I agree with what you say. Is the interval really closed? Also just to be sure,is the sequence rn supposed to be a bijection?
– Keen
Nov 23 at 19:08
You know, Conway doesn't really specify. However, I've deduced that the interval must be closed, as we call it $J$ with implicit endpoints $a,b$. We define $alpha:Jtomathbb{R}$. We then ask for $alpha(a),alpha(b)$ so already seems implied the interval is closed. Next, we're given that it is discontinuous at each rational. This implies we mean to choose the $r_n$ from the closed interval, as otherwise we have no right Jump discontinuity at $a$.
– Melody
Nov 23 at 19:21
1
Just sum over $r_nleq t$ if you prefer. This gives a cadlag function
– Federico
Nov 23 at 19:21
@Federico I thought about that, only Conway also claims we have $alpha(a)=0$, which fails if we change it to $r_nleq t$ when $a$ is rational.
– Melody
Nov 23 at 19:23
3
3
I don't know the context of the question, but could it be possible that ${r_n}$ is to be a sequence of all rationals in $(a,b)$ rather than $[a, b]$?
– angryavian
Nov 23 at 18:59
I don't know the context of the question, but could it be possible that ${r_n}$ is to be a sequence of all rationals in $(a,b)$ rather than $[a, b]$?
– angryavian
Nov 23 at 18:59
1
1
I agree with what you say. Is the interval really closed? Also just to be sure,is the sequence rn supposed to be a bijection?
– Keen
Nov 23 at 19:08
I agree with what you say. Is the interval really closed? Also just to be sure,is the sequence rn supposed to be a bijection?
– Keen
Nov 23 at 19:08
You know, Conway doesn't really specify. However, I've deduced that the interval must be closed, as we call it $J$ with implicit endpoints $a,b$. We define $alpha:Jtomathbb{R}$. We then ask for $alpha(a),alpha(b)$ so already seems implied the interval is closed. Next, we're given that it is discontinuous at each rational. This implies we mean to choose the $r_n$ from the closed interval, as otherwise we have no right Jump discontinuity at $a$.
– Melody
Nov 23 at 19:21
You know, Conway doesn't really specify. However, I've deduced that the interval must be closed, as we call it $J$ with implicit endpoints $a,b$. We define $alpha:Jtomathbb{R}$. We then ask for $alpha(a),alpha(b)$ so already seems implied the interval is closed. Next, we're given that it is discontinuous at each rational. This implies we mean to choose the $r_n$ from the closed interval, as otherwise we have no right Jump discontinuity at $a$.
– Melody
Nov 23 at 19:21
1
1
Just sum over $r_nleq t$ if you prefer. This gives a cadlag function
– Federico
Nov 23 at 19:21
Just sum over $r_nleq t$ if you prefer. This gives a cadlag function
– Federico
Nov 23 at 19:21
@Federico I thought about that, only Conway also claims we have $alpha(a)=0$, which fails if we change it to $r_nleq t$ when $a$ is rational.
– Melody
Nov 23 at 19:23
@Federico I thought about that, only Conway also claims we have $alpha(a)=0$, which fails if we change it to $r_nleq t$ when $a$ is rational.
– Melody
Nov 23 at 19:23
|
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3
I don't know the context of the question, but could it be possible that ${r_n}$ is to be a sequence of all rationals in $(a,b)$ rather than $[a, b]$?
– angryavian
Nov 23 at 18:59
1
I agree with what you say. Is the interval really closed? Also just to be sure,is the sequence rn supposed to be a bijection?
– Keen
Nov 23 at 19:08
You know, Conway doesn't really specify. However, I've deduced that the interval must be closed, as we call it $J$ with implicit endpoints $a,b$. We define $alpha:Jtomathbb{R}$. We then ask for $alpha(a),alpha(b)$ so already seems implied the interval is closed. Next, we're given that it is discontinuous at each rational. This implies we mean to choose the $r_n$ from the closed interval, as otherwise we have no right Jump discontinuity at $a$.
– Melody
Nov 23 at 19:21
1
Just sum over $r_nleq t$ if you prefer. This gives a cadlag function
– Federico
Nov 23 at 19:21
@Federico I thought about that, only Conway also claims we have $alpha(a)=0$, which fails if we change it to $r_nleq t$ when $a$ is rational.
– Melody
Nov 23 at 19:23