Why isn't this true for $x<0$?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Prove that
$$lfloorxrfloor=bigglfloorfracx+12biggrfloor+bigglfloorfracx+22^2biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots$$
$xgeq0$
I asked this question here yesterday. Although I was able to come up with a proof, the second part was left unanswered.
I proved it using,
Subtract $lfloorxrfloor$ from both sides.
Then we need to prove that,
$$-lfloorxrfloor+bigglfloorfracx+12biggrfloor+bigglfloorfracx+22^2biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots=0$$
We know that,
$lfloorxrfloor=bigglfloorfracx2biggrfloor+bigglfloorfracx+12biggrfloor$
Therefore,
$-lfloorxrfloor+bigglfloorfracx+12biggrfloor=-bigglfloorfracx2biggrfloor$
Subsituting this,
$$-bigglfloorfracx2biggrfloor+bigglfloorfracx+22^2biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots$$
$$-bigglfloorfracx2biggrfloor+bigglfloorfracfracx2+12biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots$$
Similarly, I reach,
$$lim_ntoinfty-bigglfloorfracx2^nbiggrfloor+bigglfloorfracfracx2^n+12biggrfloor$$
i.e. $$=0$$
My question is,as $lfloorxrfloor=bigglfloorfracx2biggrfloor+bigglfloorfracx+12biggrfloor$ is true $forall xinmathbbR$.
Why isn't
$$lfloorxrfloor=bigglfloorfracx+12biggrfloor+bigglfloorfracx+22^2biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots$$ true $forall xinmathbbR$.
Also how would the expression change for $xlt0$.
Thank you.
functions floor-function
asked Aug 15 at 8:48
prog_SAHIL
1,015318
add a comment |Â
up vote
2
down vote
favorite
Prove that
$$lfloorxrfloor=bigglfloorfracx+12biggrfloor+bigglfloorfracx+22^2biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots$$
$xgeq0$
I asked this question here yesterday. Although I was able to come up with a proof, the second part was left unanswered.
I proved it using,
Subtract $lfloorxrfloor$ from both sides.
Then we need to prove that,
$$-lfloorxrfloor+bigglfloorfracx+12biggrfloor+bigglfloorfracx+22^2biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots=0$$
We know that,
$lfloorxrfloor=bigglfloorfracx2biggrfloor+bigglfloorfracx+12biggrfloor$
Therefore,
$-lfloorxrfloor+bigglfloorfracx+12biggrfloor=-bigglfloorfracx2biggrfloor$
Subsituting this,
$$-bigglfloorfracx2biggrfloor+bigglfloorfracx+22^2biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots$$
$$-bigglfloorfracx2biggrfloor+bigglfloorfracfracx2+12biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots$$
Similarly, I reach,
$$lim_ntoinfty-bigglfloorfracx2^nbiggrfloor+bigglfloorfracfracx2^n+12biggrfloor$$
i.e. $$=0$$
My question is,as $lfloorxrfloor=bigglfloorfracx2biggrfloor+bigglfloorfracx+12biggrfloor$ is true $forall xinmathbbR$.
Why isn't
$$lfloorxrfloor=bigglfloorfracx+12biggrfloor+bigglfloorfracx+22^2biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots$$ true $forall xinmathbbR$.
Also how would the expression change for $xlt0$.
Thank you.
functions floor-function
asked Aug 15 at 8:48
prog_SAHIL
1,015318
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Prove that
$$lfloorxrfloor=bigglfloorfracx+12biggrfloor+bigglfloorfracx+22^2biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots$$
$xgeq0$
I asked this question here yesterday. Although I was able to come up with a proof, the second part was left unanswered.
I proved it using,
Subtract $lfloorxrfloor$ from both sides.
Then we need to prove that,
$$-lfloorxrfloor+bigglfloorfracx+12biggrfloor+bigglfloorfracx+22^2biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots=0$$
We know that,
$lfloorxrfloor=bigglfloorfracx2biggrfloor+bigglfloorfracx+12biggrfloor$
Therefore,
$-lfloorxrfloor+bigglfloorfracx+12biggrfloor=-bigglfloorfracx2biggrfloor$
Subsituting this,
$$-bigglfloorfracx2biggrfloor+bigglfloorfracx+22^2biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots$$
$$-bigglfloorfracx2biggrfloor+bigglfloorfracfracx2+12biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots$$
Similarly, I reach,
$$lim_ntoinfty-bigglfloorfracx2^nbiggrfloor+bigglfloorfracfracx2^n+12biggrfloor$$
i.e. $$=0$$
My question is,as $lfloorxrfloor=bigglfloorfracx2biggrfloor+bigglfloorfracx+12biggrfloor$ is true $forall xinmathbbR$.
Why isn't
$$lfloorxrfloor=bigglfloorfracx+12biggrfloor+bigglfloorfracx+22^2biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots$$ true $forall xinmathbbR$.
Also how would the expression change for $xlt0$.
Thank you.
functions floor-function
asked Aug 15 at 8:48
prog_SAHIL
1,015318
Prove that
$$lfloorxrfloor=bigglfloorfracx+12biggrfloor+bigglfloorfracx+22^2biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots$$
$xgeq0$
I asked this question here yesterday. Although I was able to come up with a proof, the second part was left unanswered.
I proved it using,
Subtract $lfloorxrfloor$ from both sides.
Then we need to prove that,
$$-lfloorxrfloor+bigglfloorfracx+12biggrfloor+bigglfloorfracx+22^2biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots=0$$
We know that,
$lfloorxrfloor=bigglfloorfracx2biggrfloor+bigglfloorfracx+12biggrfloor$
Therefore,
$-lfloorxrfloor+bigglfloorfracx+12biggrfloor=-bigglfloorfracx2biggrfloor$
Subsituting this,
$$-bigglfloorfracx2biggrfloor+bigglfloorfracx+22^2biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots$$
$$-bigglfloorfracx2biggrfloor+bigglfloorfracfracx2+12biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots$$
Similarly, I reach,
$$lim_ntoinfty-bigglfloorfracx2^nbiggrfloor+bigglfloorfracfracx2^n+12biggrfloor$$
i.e. $$=0$$
My question is,as $lfloorxrfloor=bigglfloorfracx2biggrfloor+bigglfloorfracx+12biggrfloor$ is true $forall xinmathbbR$.
Why isn't
$$lfloorxrfloor=bigglfloorfracx+12biggrfloor+bigglfloorfracx+22^2biggrfloor+bigglfloorfracx+2^22^3biggrfloor+
ldots$$ true $forall xinmathbbR$.
Also how would the expression change for $xlt0$.
Thank you.
functions floor-function
asked Aug 15 at 8:48
prog_SAHIL
1,015318
edited Aug 15 at 10:36
asked Aug 15 at 8:48
prog_SAHIL
1,015318
asked Aug 15 at 8:48
prog_SAHIL
1,015318
asked Aug 15 at 8:48
prog_SAHIL
1,015318
1,015318
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
For $x=-1$ the lest side is $-1$ and the right side is $0$.
answered Aug 15 at 8:53
Kavi Rama Murthy
22.5k2933
Yes, I figured that out. But why is that?
– prog_SAHIL
Aug 15 at 10:19
$frac -1+2^n 2^n+1 =frac 1 2 -frac 1 2^n+1 $. This number lies between $0$ and $1$ (strictly less than $1$ so the integer part is $0$. Thus every term on the right side of the equation is $0$.
– Kavi Rama Murthy
Aug 15 at 10:24
What is my wrong with my prove then?
– prog_SAHIL
Aug 15 at 10:39
@prog_SAHIL: At the very least, your error is that you haven't actually checked what your proof says when $x<0$; the number at the end is $-1$, not $0$.
– Hurkyl
Aug 15 at 10:52
@Hurkyl, understood. Thank you.
– prog_SAHIL
Aug 15 at 10:58
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
For $x=-1$ the lest side is $-1$ and the right side is $0$.
answered Aug 15 at 8:53
Kavi Rama Murthy
22.5k2933
Yes, I figured that out. But why is that?
– prog_SAHIL
Aug 15 at 10:19
$frac -1+2^n 2^n+1 =frac 1 2 -frac 1 2^n+1 $. This number lies between $0$ and $1$ (strictly less than $1$ so the integer part is $0$. Thus every term on the right side of the equation is $0$.
– Kavi Rama Murthy
Aug 15 at 10:24
What is my wrong with my prove then?
– prog_SAHIL
Aug 15 at 10:39
@prog_SAHIL: At the very least, your error is that you haven't actually checked what your proof says when $x<0$; the number at the end is $-1$, not $0$.
– Hurkyl
Aug 15 at 10:52
@Hurkyl, understood. Thank you.
– prog_SAHIL
Aug 15 at 10:58
add a comment |Â
up vote
2
down vote
accepted
For $x=-1$ the lest side is $-1$ and the right side is $0$.
answered Aug 15 at 8:53
Kavi Rama Murthy
22.5k2933
Yes, I figured that out. But why is that?
– prog_SAHIL
Aug 15 at 10:19
$frac -1+2^n 2^n+1 =frac 1 2 -frac 1 2^n+1 $. This number lies between $0$ and $1$ (strictly less than $1$ so the integer part is $0$. Thus every term on the right side of the equation is $0$.
– Kavi Rama Murthy
Aug 15 at 10:24
What is my wrong with my prove then?
– prog_SAHIL
Aug 15 at 10:39
@prog_SAHIL: At the very least, your error is that you haven't actually checked what your proof says when $x<0$; the number at the end is $-1$, not $0$.
– Hurkyl
Aug 15 at 10:52
@Hurkyl, understood. Thank you.
– prog_SAHIL
Aug 15 at 10:58
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
For $x=-1$ the lest side is $-1$ and the right side is $0$.
answered Aug 15 at 8:53
Kavi Rama Murthy
22.5k2933
For $x=-1$ the lest side is $-1$ and the right side is $0$.
answered Aug 15 at 8:53
Kavi Rama Murthy
22.5k2933
answered Aug 15 at 8:53
Kavi Rama Murthy
22.5k2933
answered Aug 15 at 8:53
Kavi Rama Murthy
22.5k2933
answered Aug 15 at 8:53
Kavi Rama Murthy
22.5k2933
22.5k2933
Yes, I figured that out. But why is that?
– prog_SAHIL
Aug 15 at 10:19
$frac -1+2^n 2^n+1 =frac 1 2 -frac 1 2^n+1 $. This number lies between $0$ and $1$ (strictly less than $1$ so the integer part is $0$. Thus every term on the right side of the equation is $0$.
– Kavi Rama Murthy
Aug 15 at 10:24
What is my wrong with my prove then?
– prog_SAHIL
Aug 15 at 10:39
@prog_SAHIL: At the very least, your error is that you haven't actually checked what your proof says when $x<0$; the number at the end is $-1$, not $0$.
– Hurkyl
Aug 15 at 10:52
@Hurkyl, understood. Thank you.
– prog_SAHIL
Aug 15 at 10:58
add a comment |Â
Yes, I figured that out. But why is that?
– prog_SAHIL
Aug 15 at 10:19
$frac -1+2^n 2^n+1 =frac 1 2 -frac 1 2^n+1 $. This number lies between $0$ and $1$ (strictly less than $1$ so the integer part is $0$. Thus every term on the right side of the equation is $0$.
– Kavi Rama Murthy
Aug 15 at 10:24
What is my wrong with my prove then?
– prog_SAHIL
Aug 15 at 10:39
@prog_SAHIL: At the very least, your error is that you haven't actually checked what your proof says when $x<0$; the number at the end is $-1$, not $0$.
– Hurkyl
Aug 15 at 10:52
@Hurkyl, understood. Thank you.
– prog_SAHIL
Aug 15 at 10:58
Yes, I figured that out. But why is that?
– prog_SAHIL
Aug 15 at 10:19
Yes, I figured that out. But why is that?
– prog_SAHIL
Aug 15 at 10:19
$frac -1+2^n 2^n+1 =frac 1 2 -frac 1 2^n+1 $. This number lies between $0$ and $1$ (strictly less than $1$ so the integer part is $0$. Thus every term on the right side of the equation is $0$.
– Kavi Rama Murthy
Aug 15 at 10:24
$frac -1+2^n 2^n+1 =frac 1 2 -frac 1 2^n+1 $. This number lies between $0$ and $1$ (strictly less than $1$ so the integer part is $0$. Thus every term on the right side of the equation is $0$.
– Kavi Rama Murthy
Aug 15 at 10:24
What is my wrong with my prove then?
– prog_SAHIL
Aug 15 at 10:39
What is my wrong with my prove then?
– prog_SAHIL
Aug 15 at 10:39
@prog_SAHIL: At the very least, your error is that you haven't actually checked what your proof says when $x<0$; the number at the end is $-1$, not $0$.
– Hurkyl
Aug 15 at 10:52
@prog_SAHIL: At the very least, your error is that you haven't actually checked what your proof says when $x<0$; the number at the end is $-1$, not $0$.
– Hurkyl
Aug 15 at 10:52
@Hurkyl, understood. Thank you.
– prog_SAHIL
Aug 15 at 10:58
@Hurkyl, understood. Thank you.
– prog_SAHIL
Aug 15 at 10:58
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%2f2883364%2fwhy-isnt-this-true-for-x0%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