Show that two binary expansion are equal [closed]

Clash Royale CLAN TAG#URR8PPP

up vote
-1
down vote

favorite

I am new to analysis and dyadic expansion. One of the problem I came across is

If $x_n$, $y_n$ are two sequences of zeros and ones, show that

$sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$

if and only if there is an integer $n$ such that $x_k=0$ and $y_k=1$ for all $kgeq n$.

I understand that every number has a unique binary expansion, but this problem just not intuitively make sense to me.

asked Sep 10 at 18:23

Neyo Yang

closed as off-topic by Did, JosÃ© Carlos Santos, Theoretical Economist, Adrian Keister, amWhy Sep 11 at 0:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â€“ Did, JosÃƒÂ© Carlos Santos, Theoretical Economist, Adrian Keister, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

add a commentÂ |Â

up vote
-1
down vote

favorite

I am new to analysis and dyadic expansion. One of the problem I came across is

If $x_n$, $y_n$ are two sequences of zeros and ones, show that

$sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$

if and only if there is an integer $n$ such that $x_k=0$ and $y_k=1$ for all $kgeq n$.

I understand that every number has a unique binary expansion, but this problem just not intuitively make sense to me.

asked Sep 10 at 18:23

Neyo Yang

closed as off-topic by Did, JosÃ© Carlos Santos, Theoretical Economist, Adrian Keister, amWhy Sep 11 at 0:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â€“ Did, JosÃƒÂ© Carlos Santos, Theoretical Economist, Adrian Keister, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

add a commentÂ |Â

up vote
-1
down vote

favorite

I am new to analysis and dyadic expansion. One of the problem I came across is

If $x_n$, $y_n$ are two sequences of zeros and ones, show that

$sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$

if and only if there is an integer $n$ such that $x_k=0$ and $y_k=1$ for all $kgeq n$.

I understand that every number has a unique binary expansion, but this problem just not intuitively make sense to me.

asked Sep 10 at 18:23

Neyo Yang

I am new to analysis and dyadic expansion. One of the problem I came across is

If $x_n$, $y_n$ are two sequences of zeros and ones, show that

$sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$

if and only if there is an integer $n$ such that $x_k=0$ and $y_k=1$ for all $kgeq n$.

I understand that every number has a unique binary expansion, but this problem just not intuitively make sense to me.

real-analysis binary

asked Sep 10 at 18:23

Neyo Yang

asked Sep 10 at 18:23

Neyo Yang

asked Sep 10 at 18:23

Neyo Yang

asked Sep 10 at 18:23

Neyo Yang

asked Sep 10 at 18:23

Neyo Yang

closed as off-topic by Did, JosÃ© Carlos Santos, Theoretical Economist, Adrian Keister, amWhy Sep 11 at 0:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â€“ Did, JosÃƒÂ© Carlos Santos, Theoretical Economist, Adrian Keister, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Did, JosÃ© Carlos Santos, Theoretical Economist, Adrian Keister, amWhy Sep 11 at 0:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â€“ Did, JosÃƒÂ© Carlos Santos, Theoretical Economist, Adrian Keister, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

add a commentÂ |Â

1 Answer
1

active

oldest

votes

up vote
1
down vote

$$sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$$ will not hold if you have all your $x_n =1$ and all you $y_n=0$ for $nge 1$

On the other hand if you have $x_1=0, x_n=1$ for $nge 2$ we will get

$$sum_n=1^inftyx_n/2^n=1/2$$

And if we let $y_1=0$ and $y_n=1$ for all $nge 2$ we get $$ sum_n=1^inftyy_n/2^n=1/2$$

Thus we have to modify the problem so it makes sense.

answered Sep 10 at 18:43

Mohammad Riazi-Kermani

32.2k41853

add a commentÂ |Â

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
1
down vote

$$sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$$ will not hold if you have all your $x_n =1$ and all you $y_n=0$ for $nge 1$

On the other hand if you have $x_1=0, x_n=1$ for $nge 2$ we will get

$$sum_n=1^inftyx_n/2^n=1/2$$

And if we let $y_1=0$ and $y_n=1$ for all $nge 2$ we get $$ sum_n=1^inftyy_n/2^n=1/2$$

Thus we have to modify the problem so it makes sense.

answered Sep 10 at 18:43

Mohammad Riazi-Kermani

32.2k41853

add a commentÂ |Â

up vote
1
down vote

$$sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$$ will not hold if you have all your $x_n =1$ and all you $y_n=0$ for $nge 1$

On the other hand if you have $x_1=0, x_n=1$ for $nge 2$ we will get

$$sum_n=1^inftyx_n/2^n=1/2$$

And if we let $y_1=0$ and $y_n=1$ for all $nge 2$ we get $$ sum_n=1^inftyy_n/2^n=1/2$$

Thus we have to modify the problem so it makes sense.

answered Sep 10 at 18:43

Mohammad Riazi-Kermani

32.2k41853

add a commentÂ |Â

up vote
1
down vote

$$sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$$ will not hold if you have all your $x_n =1$ and all you $y_n=0$ for $nge 1$

On the other hand if you have $x_1=0, x_n=1$ for $nge 2$ we will get

$$sum_n=1^inftyx_n/2^n=1/2$$

And if we let $y_1=0$ and $y_n=1$ for all $nge 2$ we get $$ sum_n=1^inftyy_n/2^n=1/2$$

Thus we have to modify the problem so it makes sense.

answered Sep 10 at 18:43

Mohammad Riazi-Kermani

32.2k41853

$$sum_n=1^inftyx_n/2^n = sum_n=1^inftyy_n/2^n$$ will not hold if you have all your $x_n =1$ and all you $y_n=0$ for $nge 1$

On the other hand if you have $x_1=0, x_n=1$ for $nge 2$ we will get

$$sum_n=1^inftyx_n/2^n=1/2$$

And if we let $y_1=0$ and $y_n=1$ for all $nge 2$ we get $$ sum_n=1^inftyy_n/2^n=1/2$$

Thus we have to modify the problem so it makes sense.

answered Sep 10 at 18:43

Mohammad Riazi-Kermani

32.2k41853

answered Sep 10 at 18:43

Mohammad Riazi-Kermani

32.2k41853

answered Sep 10 at 18:43

Mohammad Riazi-Kermani

32.2k41853

answered Sep 10 at 18:43

Mohammad Riazi-Kermani

32.2k41853

add a commentÂ |Â

搜尋此網誌

Vtyjkyuk