Mixing
Prove $[sqrt{x}+sqrt{x+1}]=[sqrt{4x+2}], x in Bbb N$
$begingroup$
prove that $[sqrt{x}+sqrt{x+1}]=[sqrt{4x+2}], x in Bbb N$. Could someone help me to solve. I try many ways but I can't solve it ( I tried : Let $x=n^2 +k$ and don't know what to do afterwards).
calculus floor-function
edited Feb 4 at 14:11
Martin Sleziak
44.8k10119272
asked Nov 23 '15 at 13:28
TuanleeTuanlee
724
$endgroup$
|
show 10 more comments
$begingroup$
prove that $[sqrt{x}+sqrt{x+1}]=[sqrt{4x+2}], x in Bbb N$. Could someone help me to solve. I try many ways but I can't solve it ( I tried : Let $x=n^2 +k$ and don't know what to do afterwards).
calculus floor-function
edited Feb 4 at 14:11
Martin Sleziak
44.8k10119272
asked Nov 23 '15 at 13:28
TuanleeTuanlee
724
$endgroup$
1
$begingroup$
is there any significance of the square brackets?
$endgroup$
– Chinny84
Nov 23 '15 at 13:29
2
$begingroup$
@Chinny84: they usually signify "greatest integer less than contained value" or some similar wording...
$endgroup$
– abiessu
Nov 23 '15 at 13:31
3
$begingroup$
I assume that what you actually tried was $x=n^2+k$...
$endgroup$
– abiessu
Nov 23 '15 at 13:32
3
$begingroup$
The equation is not correct for general positive $x in mathbb{R}$. Try $x=frac{1}{2}$
$endgroup$
– gammatester
Nov 23 '15 at 13:43
1
$begingroup$
if it's the floor function, then it's not true. Check $1/2leq x< 9/16$, for example. Similarly, the ceiling function also fails.
$endgroup$
– A Simmons
Nov 23 '15 at 13:45
|
show 10 more comments
$begingroup$
prove that $[sqrt{x}+sqrt{x+1}]=[sqrt{4x+2}], x in Bbb N$. Could someone help me to solve. I try many ways but I can't solve it ( I tried : Let $x=n^2 +k$ and don't know what to do afterwards).
calculus floor-function
edited Feb 4 at 14:11
Martin Sleziak
44.8k10119272
asked Nov 23 '15 at 13:28
TuanleeTuanlee
724
$endgroup$
prove that $[sqrt{x}+sqrt{x+1}]=[sqrt{4x+2}], x in Bbb N$. Could someone help me to solve. I try many ways but I can't solve it ( I tried : Let $x=n^2 +k$ and don't know what to do afterwards).
calculus floor-function
calculus floor-function
edited Feb 4 at 14:11
Martin Sleziak
44.8k10119272
asked Nov 23 '15 at 13:28
TuanleeTuanlee
724
edited Feb 4 at 14:11
Martin Sleziak
44.8k10119272
asked Nov 23 '15 at 13:28
TuanleeTuanlee
724
edited Feb 4 at 14:11
Martin Sleziak
44.8k10119272
edited Feb 4 at 14:11
Martin Sleziak
44.8k10119272
edited Feb 4 at 14:11
Martin Sleziak
44.8k10119272
44.8k10119272
asked Nov 23 '15 at 13:28
TuanleeTuanlee
724
asked Nov 23 '15 at 13:28
TuanleeTuanlee
724
asked Nov 23 '15 at 13:28
TuanleeTuanlee
724
724
1
$begingroup$
is there any significance of the square brackets?
$endgroup$
– Chinny84
Nov 23 '15 at 13:29
2
$begingroup$
@Chinny84: they usually signify "greatest integer less than contained value" or some similar wording...
$endgroup$
– abiessu
Nov 23 '15 at 13:31
3
$begingroup$
I assume that what you actually tried was $x=n^2+k$...
$endgroup$
– abiessu
Nov 23 '15 at 13:32
3
$begingroup$
The equation is not correct for general positive $x in mathbb{R}$. Try $x=frac{1}{2}$
$endgroup$
– gammatester
Nov 23 '15 at 13:43
1
$begingroup$
if it's the floor function, then it's not true. Check $1/2leq x< 9/16$, for example. Similarly, the ceiling function also fails.
$endgroup$
– A Simmons
Nov 23 '15 at 13:45
|
show 10 more comments
1
$begingroup$
is there any significance of the square brackets?
$endgroup$
– Chinny84
Nov 23 '15 at 13:29
2
$begingroup$
@Chinny84: they usually signify "greatest integer less than contained value" or some similar wording...
$endgroup$
– abiessu
Nov 23 '15 at 13:31
3
$begingroup$
I assume that what you actually tried was $x=n^2+k$...
$endgroup$
– abiessu
Nov 23 '15 at 13:32
3
$begingroup$
The equation is not correct for general positive $x in mathbb{R}$. Try $x=frac{1}{2}$
$endgroup$
– gammatester
Nov 23 '15 at 13:43
1
$begingroup$
if it's the floor function, then it's not true. Check $1/2leq x< 9/16$, for example. Similarly, the ceiling function also fails.
$endgroup$
– A Simmons
Nov 23 '15 at 13:45
1
1
$begingroup$
is there any significance of the square brackets?
$endgroup$
– Chinny84
Nov 23 '15 at 13:29
$begingroup$
is there any significance of the square brackets?
$endgroup$
– Chinny84
Nov 23 '15 at 13:29
2
2
$begingroup$
@Chinny84: they usually signify "greatest integer less than contained value" or some similar wording...
$endgroup$
– abiessu
Nov 23 '15 at 13:31
$begingroup$
@Chinny84: they usually signify "greatest integer less than contained value" or some similar wording...
$endgroup$
– abiessu
Nov 23 '15 at 13:31
3
3
$begingroup$
I assume that what you actually tried was $x=n^2+k$...
$endgroup$
– abiessu
Nov 23 '15 at 13:32
$begingroup$
I assume that what you actually tried was $x=n^2+k$...
$endgroup$
– abiessu
Nov 23 '15 at 13:32
3
3
$begingroup$
The equation is not correct for general positive $x in mathbb{R}$. Try $x=frac{1}{2}$
$endgroup$
– gammatester
Nov 23 '15 at 13:43
$begingroup$
The equation is not correct for general positive $x in mathbb{R}$. Try $x=frac{1}{2}$
$endgroup$
– gammatester
Nov 23 '15 at 13:43
1
1
$begingroup$
if it's the floor function, then it's not true. Check $1/2leq x< 9/16$, for example. Similarly, the ceiling function also fails.
$endgroup$
– A Simmons
Nov 23 '15 at 13:45
$begingroup$
if it's the floor function, then it's not true. Check $1/2leq x< 9/16$, for example. Similarly, the ceiling function also fails.
$endgroup$
– A Simmons
Nov 23 '15 at 13:45
|
show 10 more comments
3 Answers
3
active
oldest
votes
$begingroup$
First a kooky Lemma:
For $x > 1$ then $sqrt{1 - frac 1 x} + sqrt{1 + frac 3 x} > 2$
[$sqrt{1 - frac 1 x} + sqrt{1 + frac 3 x} > 2$ iff
$1 - frac 1 x + 1 + frac 3 x + 2sqrt{(1 - frac 1 x)(1 + frac 3 x)} > 4$ iff
$2 + frac 2 x + 2sqrt{(1 + frac 2 x -frac 2 {x^2}} > 4$ iff
$1 + frac 2 x -frac 2 {x^2} > 1$ iff
$frac 2 x -frac 2 {x^2} > 0$ ]
================
So now to answer your question:
Let $n^2 le x < (n+1)^2$ or in other words $n^2 le x le n^2 + 2n$.
It's obvious:
$2n < sqrt x + sqrt {x + 1} < (n + 1) + (n + 1) = 2n + 2$.
So $lfloor sqrt x + sqrt {x + 1} rfloor = {2n, 2n + 1}$.
Likewise $2n = sqrt{4n} < sqrt{4x + 2} le sqrt{4n^2 + 8n + 2} < sqrt{4^2 + 8 + 4} = 2(n + 1) = 2n + 2$.
So $lfloor sqrt {4x + 2} rfloor = {2n, 2n + 1}$.
====================
Case 1:
$n^2 le x < x + 1 le n^2 + n < n^2 + n + frac 1 4 = (n + frac 1 2)^2$
Then $ sqrt x + sqrt {x + 1} < n + frac 1 2 + n + frac 1 2 < 2n + 1$
So $lfloor sqrt x + sqrt {x + 1} rfloor = 2n$.
Likewise $ sqrt{4x + 2} le sqrt{4n^2 + 4(n -1) + 2} < sqrt{4n^2 + 4n + 1} = 2(n + frac 1 2) = 2n + 1$.
So $lfloor sqrt {4x + 2} rfloor = 2n = lfloor sqrt x + sqrt {x + 1} rfloor$.
================
Case 2:
$n^2 + 2n + 1 ge x + 1 > x ge n^2 + n$.
Then $sqrt{4x + 2} ge sqrt{4n^2 + 4n + 2} = 2(n + frac 12) = 2n + 1$.
So $lfloor sqrt {4x + 2} rfloor = 2n + 1$.
Now $sqrt x + sqrt {x + 1} ge sqrt{n^2 + n} + sqrt{n^2 + n + 1} = sqrt{n^2 + n + frac 1 4 - frac 1 4} + sqrt{n^2 + n + frac 1 4 + frac 3 4}$
$= (n + frac 1 2)[sqrt{1 - frac 1 {4(n + frac 1 2)^2}} + sqrt{1 + frac 3 {4(n + frac 1 2)^2}}]$
(remember the kooky lemma?)
$ > (n + frac 1 2)2 = 2n + 1$.
So $lfloor sqrt x + sqrt {x + 1} rfloor = 2n + 1 = lfloor sqrt{4x +2} rfloor$.
answered Nov 23 '15 at 22:17
fleabloodfleablood
71.7k22686
$endgroup$
$begingroup$
That's better :-)
$endgroup$
– abiessu
Nov 24 '15 at 3:19
add a comment |
$begingroup$
Case 1: $n^2le xle(n+1)^2-2$ (for $nge 3$), then
$$ nle sqrt{x}lesqrt{(n+1)^2-2}<n+1, n<sqrt{n^2+1}le sqrt{x+1}<sqrt{(n+1)^2-1}=n+1$$
and
$$ 2n<sqrt{4n^2+2}le sqrt{4x+2}<sqrt{4(n+1)^2-2}<2n+1$$
So
$$ 2n<sqrt{x}+sqrt{x+1}<2n+2, sqrt{4x+2}=2n $$
and
$$ 2n+1le[sqrt{x}+sqrt{x+1}]le2n+2, [sqrt{4x+2}]=2n+1. $$
Now we show
$$ sqrt{x}+sqrt{x+1}<2n+1. $$
Note $sqrt{x}+sqrt{x+1}$ is increasing and hence
$$sqrt{x}+sqrt{x+1}lesqrt{(n+1)^2-2}+sqrt{(n+1)^2-1}.$$
Note
begin{eqnarray}
&&sqrt{(n+1)^2-2}+sqrt{(n+1)^2-1}-2n-1\
&=&frac{(n+1)^2-2-n^2}{sqrt{(n+1)^2-2}+n^2}+frac{(n+1)^2-1-n^2}{sqrt{(n+1)^2-1}+n^2}-1\
&<&frac{4n-1}{sqrt{(n+1)^2-2}+n^2}-1\
&=&frac{4n-1}{sqrt{n^2+2n-1}+n^2}-1\
&<&frac{4n-1}{n+n^2}-1\
&=&-frac{n^2-3n+1}{n+n^2}\
&<0
end{eqnarray}
for $nge 3$.
Thus
$$2n<sqrt{(n+1)^2-2}+sqrt{(n+1)^2-1}<2n+1$$
or
$$[sqrt{x}+sqrt{x+1}=2n+1$$
Case 2: $x=(n+1)^2-1$.
begin{eqnarray}
&&sqrt{x}+sqrt{x+1}\
&=&sqrt{(n+1)^2-1}+n+1\
&<&2(n+1)
end{eqnarray}
and
begin{eqnarray}
&&sqrt{x}+sqrt{x+1}\
&=&sqrt{(n+1)^2-1}+n+1\
&>&2n+1
end{eqnarray}
and hence
$$ [sqrt{x}+sqrt{x+1}]=2(n+1). $$
On the other hand
$$ sqrt{4x+2}=sqrt{4(n+1)^2-2}<2(n+1) $$
and
$$ sqrt{4x+2}=sqrt{4(n+1)^2-2}>2n+1 $$
and hence
$$ [sqrt{4x+2}]=2n+2 $$
In both cases,
$$ [sqrt{x}+sqrt{x+1}]=[sqrt{4x+2}]. $$
Case 3: for $n=1,2$, it is easy to check the inequality is true.
answered Nov 23 '15 at 16:45
xpaulxpaul
23k24455
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add a comment |
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Taking the listed approach, we get
$$[sqrt{n^2+k}+sqrt{n^2+k+1}]=[sqrt{4n^2+4k +2}]$$
As $k$ relates to $n$, we have a floor-function boundary for the RHS when $kin{-1,0}$ and also at $kin{n-1,n}$, so these values need to be checked on the LHS to verify that these are consecutive boundaries there as well. For $kin{-1,0}$, the LHS shows the same behavior as the RHS, so we only need to show that the LHS also has a boundary at $kin{n-1,n}$ by comparing with $2n+1$ from the RHS:
$$left[sqrt{n^2+n-1}+sqrt{n^2+n}right], R, 2n+1 \ text{ vs. }\ left[sqrt{n^2+n}+sqrt{n^2+n+1}right], R, 2n+1$$
Squaring yields
$$left[2n^2+2n-1+2sqrt{(n^2+n)(n^2+n-1)}right], R, 4n^2+4n+1\
left[2sqrt{(n^2+n)(n^2+n-1)}right], R, 2n^2+2n+2$$
and it should be clear that $[sqrt{n^2+n-1}+sqrt{n^2+n}]lt 2n+1$. Next we have
$$left[2n^2+2n+1+2sqrt{(n^2+n)(n^2+n+1)}right], R, 4n^2+4n+1\
left[2sqrt{(n^2+n)(n^2+n+1)}right], R, 2n^2+2n$$
and it should also be clear that $[sqrt{n^2+n}+sqrt{n^2+n+1}]ge 2n+1$, completing the proof.
answered Nov 23 '15 at 16:59
abiessuabiessu
6,68721541
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add a comment |
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3 Answers
3
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3 Answers
3
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
First a kooky Lemma:
For $x > 1$ then $sqrt{1 - frac 1 x} + sqrt{1 + frac 3 x} > 2$
[$sqrt{1 - frac 1 x} + sqrt{1 + frac 3 x} > 2$ iff
$1 - frac 1 x + 1 + frac 3 x + 2sqrt{(1 - frac 1 x)(1 + frac 3 x)} > 4$ iff
$2 + frac 2 x + 2sqrt{(1 + frac 2 x -frac 2 {x^2}} > 4$ iff
$1 + frac 2 x -frac 2 {x^2} > 1$ iff
$frac 2 x -frac 2 {x^2} > 0$ ]
================
So now to answer your question:
Let $n^2 le x < (n+1)^2$ or in other words $n^2 le x le n^2 + 2n$.
It's obvious:
$2n < sqrt x + sqrt {x + 1} < (n + 1) + (n + 1) = 2n + 2$.
So $lfloor sqrt x + sqrt {x + 1} rfloor = {2n, 2n + 1}$.
Likewise $2n = sqrt{4n} < sqrt{4x + 2} le sqrt{4n^2 + 8n + 2} < sqrt{4^2 + 8 + 4} = 2(n + 1) = 2n + 2$.
So $lfloor sqrt {4x + 2} rfloor = {2n, 2n + 1}$.
====================
Case 1:
$n^2 le x < x + 1 le n^2 + n < n^2 + n + frac 1 4 = (n + frac 1 2)^2$
Then $ sqrt x + sqrt {x + 1} < n + frac 1 2 + n + frac 1 2 < 2n + 1$
So $lfloor sqrt x + sqrt {x + 1} rfloor = 2n$.
Likewise $ sqrt{4x + 2} le sqrt{4n^2 + 4(n -1) + 2} < sqrt{4n^2 + 4n + 1} = 2(n + frac 1 2) = 2n + 1$.
So $lfloor sqrt {4x + 2} rfloor = 2n = lfloor sqrt x + sqrt {x + 1} rfloor$.
================
Case 2:
$n^2 + 2n + 1 ge x + 1 > x ge n^2 + n$.
Then $sqrt{4x + 2} ge sqrt{4n^2 + 4n + 2} = 2(n + frac 12) = 2n + 1$.
So $lfloor sqrt {4x + 2} rfloor = 2n + 1$.
Now $sqrt x + sqrt {x + 1} ge sqrt{n^2 + n} + sqrt{n^2 + n + 1} = sqrt{n^2 + n + frac 1 4 - frac 1 4} + sqrt{n^2 + n + frac 1 4 + frac 3 4}$
$= (n + frac 1 2)[sqrt{1 - frac 1 {4(n + frac 1 2)^2}} + sqrt{1 + frac 3 {4(n + frac 1 2)^2}}]$
(remember the kooky lemma?)
$ > (n + frac 1 2)2 = 2n + 1$.
So $lfloor sqrt x + sqrt {x + 1} rfloor = 2n + 1 = lfloor sqrt{4x +2} rfloor$.
answered Nov 23 '15 at 22:17
fleabloodfleablood
71.7k22686
$endgroup$
$begingroup$
That's better :-)
$endgroup$
– abiessu
Nov 24 '15 at 3:19
add a comment |
$begingroup$
First a kooky Lemma:
For $x > 1$ then $sqrt{1 - frac 1 x} + sqrt{1 + frac 3 x} > 2$
[$sqrt{1 - frac 1 x} + sqrt{1 + frac 3 x} > 2$ iff
$1 - frac 1 x + 1 + frac 3 x + 2sqrt{(1 - frac 1 x)(1 + frac 3 x)} > 4$ iff
$2 + frac 2 x + 2sqrt{(1 + frac 2 x -frac 2 {x^2}} > 4$ iff
$1 + frac 2 x -frac 2 {x^2} > 1$ iff
$frac 2 x -frac 2 {x^2} > 0$ ]
================
So now to answer your question:
Let $n^2 le x < (n+1)^2$ or in other words $n^2 le x le n^2 + 2n$.
It's obvious:
$2n < sqrt x + sqrt {x + 1} < (n + 1) + (n + 1) = 2n + 2$.
So $lfloor sqrt x + sqrt {x + 1} rfloor = {2n, 2n + 1}$.
Likewise $2n = sqrt{4n} < sqrt{4x + 2} le sqrt{4n^2 + 8n + 2} < sqrt{4^2 + 8 + 4} = 2(n + 1) = 2n + 2$.
So $lfloor sqrt {4x + 2} rfloor = {2n, 2n + 1}$.
====================
Case 1:
$n^2 le x < x + 1 le n^2 + n < n^2 + n + frac 1 4 = (n + frac 1 2)^2$
Then $ sqrt x + sqrt {x + 1} < n + frac 1 2 + n + frac 1 2 < 2n + 1$
So $lfloor sqrt x + sqrt {x + 1} rfloor = 2n$.
Likewise $ sqrt{4x + 2} le sqrt{4n^2 + 4(n -1) + 2} < sqrt{4n^2 + 4n + 1} = 2(n + frac 1 2) = 2n + 1$.
So $lfloor sqrt {4x + 2} rfloor = 2n = lfloor sqrt x + sqrt {x + 1} rfloor$.
================
Case 2:
$n^2 + 2n + 1 ge x + 1 > x ge n^2 + n$.
Then $sqrt{4x + 2} ge sqrt{4n^2 + 4n + 2} = 2(n + frac 12) = 2n + 1$.
So $lfloor sqrt {4x + 2} rfloor = 2n + 1$.
Now $sqrt x + sqrt {x + 1} ge sqrt{n^2 + n} + sqrt{n^2 + n + 1} = sqrt{n^2 + n + frac 1 4 - frac 1 4} + sqrt{n^2 + n + frac 1 4 + frac 3 4}$
$= (n + frac 1 2)[sqrt{1 - frac 1 {4(n + frac 1 2)^2}} + sqrt{1 + frac 3 {4(n + frac 1 2)^2}}]$
(remember the kooky lemma?)
$ > (n + frac 1 2)2 = 2n + 1$.
So $lfloor sqrt x + sqrt {x + 1} rfloor = 2n + 1 = lfloor sqrt{4x +2} rfloor$.
answered Nov 23 '15 at 22:17
fleabloodfleablood
71.7k22686
$endgroup$
$begingroup$
That's better :-)
$endgroup$
– abiessu
Nov 24 '15 at 3:19
add a comment |
$begingroup$
First a kooky Lemma:
For $x > 1$ then $sqrt{1 - frac 1 x} + sqrt{1 + frac 3 x} > 2$
[$sqrt{1 - frac 1 x} + sqrt{1 + frac 3 x} > 2$ iff
$1 - frac 1 x + 1 + frac 3 x + 2sqrt{(1 - frac 1 x)(1 + frac 3 x)} > 4$ iff
$2 + frac 2 x + 2sqrt{(1 + frac 2 x -frac 2 {x^2}} > 4$ iff
$1 + frac 2 x -frac 2 {x^2} > 1$ iff
$frac 2 x -frac 2 {x^2} > 0$ ]
================
So now to answer your question:
Let $n^2 le x < (n+1)^2$ or in other words $n^2 le x le n^2 + 2n$.
It's obvious:
$2n < sqrt x + sqrt {x + 1} < (n + 1) + (n + 1) = 2n + 2$.
So $lfloor sqrt x + sqrt {x + 1} rfloor = {2n, 2n + 1}$.
Likewise $2n = sqrt{4n} < sqrt{4x + 2} le sqrt{4n^2 + 8n + 2} < sqrt{4^2 + 8 + 4} = 2(n + 1) = 2n + 2$.
So $lfloor sqrt {4x + 2} rfloor = {2n, 2n + 1}$.
====================
Case 1:
$n^2 le x < x + 1 le n^2 + n < n^2 + n + frac 1 4 = (n + frac 1 2)^2$
Then $ sqrt x + sqrt {x + 1} < n + frac 1 2 + n + frac 1 2 < 2n + 1$
So $lfloor sqrt x + sqrt {x + 1} rfloor = 2n$.
Likewise $ sqrt{4x + 2} le sqrt{4n^2 + 4(n -1) + 2} < sqrt{4n^2 + 4n + 1} = 2(n + frac 1 2) = 2n + 1$.
So $lfloor sqrt {4x + 2} rfloor = 2n = lfloor sqrt x + sqrt {x + 1} rfloor$.
================
Case 2:
$n^2 + 2n + 1 ge x + 1 > x ge n^2 + n$.
Then $sqrt{4x + 2} ge sqrt{4n^2 + 4n + 2} = 2(n + frac 12) = 2n + 1$.
So $lfloor sqrt {4x + 2} rfloor = 2n + 1$.
Now $sqrt x + sqrt {x + 1} ge sqrt{n^2 + n} + sqrt{n^2 + n + 1} = sqrt{n^2 + n + frac 1 4 - frac 1 4} + sqrt{n^2 + n + frac 1 4 + frac 3 4}$
$= (n + frac 1 2)[sqrt{1 - frac 1 {4(n + frac 1 2)^2}} + sqrt{1 + frac 3 {4(n + frac 1 2)^2}}]$
(remember the kooky lemma?)
$ > (n + frac 1 2)2 = 2n + 1$.
So $lfloor sqrt x + sqrt {x + 1} rfloor = 2n + 1 = lfloor sqrt{4x +2} rfloor$.
answered Nov 23 '15 at 22:17
fleabloodfleablood
71.7k22686
$endgroup$
First a kooky Lemma:
For $x > 1$ then $sqrt{1 - frac 1 x} + sqrt{1 + frac 3 x} > 2$
[$sqrt{1 - frac 1 x} + sqrt{1 + frac 3 x} > 2$ iff
$1 - frac 1 x + 1 + frac 3 x + 2sqrt{(1 - frac 1 x)(1 + frac 3 x)} > 4$ iff
$2 + frac 2 x + 2sqrt{(1 + frac 2 x -frac 2 {x^2}} > 4$ iff
$1 + frac 2 x -frac 2 {x^2} > 1$ iff
$frac 2 x -frac 2 {x^2} > 0$ ]
================
So now to answer your question:
Let $n^2 le x < (n+1)^2$ or in other words $n^2 le x le n^2 + 2n$.
It's obvious:
$2n < sqrt x + sqrt {x + 1} < (n + 1) + (n + 1) = 2n + 2$.
So $lfloor sqrt x + sqrt {x + 1} rfloor = {2n, 2n + 1}$.
Likewise $2n = sqrt{4n} < sqrt{4x + 2} le sqrt{4n^2 + 8n + 2} < sqrt{4^2 + 8 + 4} = 2(n + 1) = 2n + 2$.
So $lfloor sqrt {4x + 2} rfloor = {2n, 2n + 1}$.
====================
Case 1:
$n^2 le x < x + 1 le n^2 + n < n^2 + n + frac 1 4 = (n + frac 1 2)^2$
Then $ sqrt x + sqrt {x + 1} < n + frac 1 2 + n + frac 1 2 < 2n + 1$
So $lfloor sqrt x + sqrt {x + 1} rfloor = 2n$.
Likewise $ sqrt{4x + 2} le sqrt{4n^2 + 4(n -1) + 2} < sqrt{4n^2 + 4n + 1} = 2(n + frac 1 2) = 2n + 1$.
So $lfloor sqrt {4x + 2} rfloor = 2n = lfloor sqrt x + sqrt {x + 1} rfloor$.
================
Case 2:
$n^2 + 2n + 1 ge x + 1 > x ge n^2 + n$.
Then $sqrt{4x + 2} ge sqrt{4n^2 + 4n + 2} = 2(n + frac 12) = 2n + 1$.
So $lfloor sqrt {4x + 2} rfloor = 2n + 1$.
Now $sqrt x + sqrt {x + 1} ge sqrt{n^2 + n} + sqrt{n^2 + n + 1} = sqrt{n^2 + n + frac 1 4 - frac 1 4} + sqrt{n^2 + n + frac 1 4 + frac 3 4}$
$= (n + frac 1 2)[sqrt{1 - frac 1 {4(n + frac 1 2)^2}} + sqrt{1 + frac 3 {4(n + frac 1 2)^2}}]$
(remember the kooky lemma?)
$ > (n + frac 1 2)2 = 2n + 1$.
So $lfloor sqrt x + sqrt {x + 1} rfloor = 2n + 1 = lfloor sqrt{4x +2} rfloor$.
answered Nov 23 '15 at 22:17
fleabloodfleablood
71.7k22686
answered Nov 23 '15 at 22:17
fleabloodfleablood
71.7k22686
answered Nov 23 '15 at 22:17
fleabloodfleablood
71.7k22686
answered Nov 23 '15 at 22:17
fleabloodfleablood
71.7k22686
71.7k22686
$begingroup$
That's better :-)
$endgroup$
– abiessu
Nov 24 '15 at 3:19
add a comment |
$begingroup$
That's better :-)
$endgroup$
– abiessu
Nov 24 '15 at 3:19
$begingroup$
That's better :-)
$endgroup$
– abiessu
Nov 24 '15 at 3:19
$begingroup$
That's better :-)
$endgroup$
– abiessu
Nov 24 '15 at 3:19
add a comment |
$begingroup$
Case 1: $n^2le xle(n+1)^2-2$ (for $nge 3$), then
$$ nle sqrt{x}lesqrt{(n+1)^2-2}<n+1, n<sqrt{n^2+1}le sqrt{x+1}<sqrt{(n+1)^2-1}=n+1$$
and
$$ 2n<sqrt{4n^2+2}le sqrt{4x+2}<sqrt{4(n+1)^2-2}<2n+1$$
So
$$ 2n<sqrt{x}+sqrt{x+1}<2n+2, sqrt{4x+2}=2n $$
and
$$ 2n+1le[sqrt{x}+sqrt{x+1}]le2n+2, [sqrt{4x+2}]=2n+1. $$
Now we show
$$ sqrt{x}+sqrt{x+1}<2n+1. $$
Note $sqrt{x}+sqrt{x+1}$ is increasing and hence
$$sqrt{x}+sqrt{x+1}lesqrt{(n+1)^2-2}+sqrt{(n+1)^2-1}.$$
Note
begin{eqnarray}
&&sqrt{(n+1)^2-2}+sqrt{(n+1)^2-1}-2n-1\
&=&frac{(n+1)^2-2-n^2}{sqrt{(n+1)^2-2}+n^2}+frac{(n+1)^2-1-n^2}{sqrt{(n+1)^2-1}+n^2}-1\
&<&frac{4n-1}{sqrt{(n+1)^2-2}+n^2}-1\
&=&frac{4n-1}{sqrt{n^2+2n-1}+n^2}-1\
&<&frac{4n-1}{n+n^2}-1\
&=&-frac{n^2-3n+1}{n+n^2}\
&<0
end{eqnarray}
for $nge 3$.
Thus
$$2n<sqrt{(n+1)^2-2}+sqrt{(n+1)^2-1}<2n+1$$
or
$$[sqrt{x}+sqrt{x+1}=2n+1$$
Case 2: $x=(n+1)^2-1$.
begin{eqnarray}
&&sqrt{x}+sqrt{x+1}\
&=&sqrt{(n+1)^2-1}+n+1\
&<&2(n+1)
end{eqnarray}
and
begin{eqnarray}
&&sqrt{x}+sqrt{x+1}\
&=&sqrt{(n+1)^2-1}+n+1\
&>&2n+1
end{eqnarray}
and hence
$$ [sqrt{x}+sqrt{x+1}]=2(n+1). $$
On the other hand
$$ sqrt{4x+2}=sqrt{4(n+1)^2-2}<2(n+1) $$
and
$$ sqrt{4x+2}=sqrt{4(n+1)^2-2}>2n+1 $$
and hence
$$ [sqrt{4x+2}]=2n+2 $$
In both cases,
$$ [sqrt{x}+sqrt{x+1}]=[sqrt{4x+2}]. $$
Case 3: for $n=1,2$, it is easy to check the inequality is true.
answered Nov 23 '15 at 16:45
xpaulxpaul
23k24455
$endgroup$
add a comment |
$begingroup$
Case 1: $n^2le xle(n+1)^2-2$ (for $nge 3$), then
$$ nle sqrt{x}lesqrt{(n+1)^2-2}<n+1, n<sqrt{n^2+1}le sqrt{x+1}<sqrt{(n+1)^2-1}=n+1$$
and
$$ 2n<sqrt{4n^2+2}le sqrt{4x+2}<sqrt{4(n+1)^2-2}<2n+1$$
So
$$ 2n<sqrt{x}+sqrt{x+1}<2n+2, sqrt{4x+2}=2n $$
and
$$ 2n+1le[sqrt{x}+sqrt{x+1}]le2n+2, [sqrt{4x+2}]=2n+1. $$
Now we show
$$ sqrt{x}+sqrt{x+1}<2n+1. $$
Note $sqrt{x}+sqrt{x+1}$ is increasing and hence
$$sqrt{x}+sqrt{x+1}lesqrt{(n+1)^2-2}+sqrt{(n+1)^2-1}.$$
Note
begin{eqnarray}
&&sqrt{(n+1)^2-2}+sqrt{(n+1)^2-1}-2n-1\
&=&frac{(n+1)^2-2-n^2}{sqrt{(n+1)^2-2}+n^2}+frac{(n+1)^2-1-n^2}{sqrt{(n+1)^2-1}+n^2}-1\
&<&frac{4n-1}{sqrt{(n+1)^2-2}+n^2}-1\
&=&frac{4n-1}{sqrt{n^2+2n-1}+n^2}-1\
&<&frac{4n-1}{n+n^2}-1\
&=&-frac{n^2-3n+1}{n+n^2}\
&<0
end{eqnarray}
for $nge 3$.
Thus
$$2n<sqrt{(n+1)^2-2}+sqrt{(n+1)^2-1}<2n+1$$
or
$$[sqrt{x}+sqrt{x+1}=2n+1$$
Case 2: $x=(n+1)^2-1$.
begin{eqnarray}
&&sqrt{x}+sqrt{x+1}\
&=&sqrt{(n+1)^2-1}+n+1\
&<&2(n+1)
end{eqnarray}
and
begin{eqnarray}
&&sqrt{x}+sqrt{x+1}\
&=&sqrt{(n+1)^2-1}+n+1\
&>&2n+1
end{eqnarray}
and hence
$$ [sqrt{x}+sqrt{x+1}]=2(n+1). $$
On the other hand
$$ sqrt{4x+2}=sqrt{4(n+1)^2-2}<2(n+1) $$
and
$$ sqrt{4x+2}=sqrt{4(n+1)^2-2}>2n+1 $$
and hence
$$ [sqrt{4x+2}]=2n+2 $$
In both cases,
$$ [sqrt{x}+sqrt{x+1}]=[sqrt{4x+2}]. $$
Case 3: for $n=1,2$, it is easy to check the inequality is true.
answered Nov 23 '15 at 16:45
xpaulxpaul
23k24455
$endgroup$
add a comment |
$begingroup$
Case 1: $n^2le xle(n+1)^2-2$ (for $nge 3$), then
$$ nle sqrt{x}lesqrt{(n+1)^2-2}<n+1, n<sqrt{n^2+1}le sqrt{x+1}<sqrt{(n+1)^2-1}=n+1$$
and
$$ 2n<sqrt{4n^2+2}le sqrt{4x+2}<sqrt{4(n+1)^2-2}<2n+1$$
So
$$ 2n<sqrt{x}+sqrt{x+1}<2n+2, sqrt{4x+2}=2n $$
and
$$ 2n+1le[sqrt{x}+sqrt{x+1}]le2n+2, [sqrt{4x+2}]=2n+1. $$
Now we show
$$ sqrt{x}+sqrt{x+1}<2n+1. $$
Note $sqrt{x}+sqrt{x+1}$ is increasing and hence
$$sqrt{x}+sqrt{x+1}lesqrt{(n+1)^2-2}+sqrt{(n+1)^2-1}.$$
Note
begin{eqnarray}
&&sqrt{(n+1)^2-2}+sqrt{(n+1)^2-1}-2n-1\
&=&frac{(n+1)^2-2-n^2}{sqrt{(n+1)^2-2}+n^2}+frac{(n+1)^2-1-n^2}{sqrt{(n+1)^2-1}+n^2}-1\
&<&frac{4n-1}{sqrt{(n+1)^2-2}+n^2}-1\
&=&frac{4n-1}{sqrt{n^2+2n-1}+n^2}-1\
&<&frac{4n-1}{n+n^2}-1\
&=&-frac{n^2-3n+1}{n+n^2}\
&<0
end{eqnarray}
for $nge 3$.
Thus
$$2n<sqrt{(n+1)^2-2}+sqrt{(n+1)^2-1}<2n+1$$
or
$$[sqrt{x}+sqrt{x+1}=2n+1$$
Case 2: $x=(n+1)^2-1$.
begin{eqnarray}
&&sqrt{x}+sqrt{x+1}\
&=&sqrt{(n+1)^2-1}+n+1\
&<&2(n+1)
end{eqnarray}
and
begin{eqnarray}
&&sqrt{x}+sqrt{x+1}\
&=&sqrt{(n+1)^2-1}+n+1\
&>&2n+1
end{eqnarray}
and hence
$$ [sqrt{x}+sqrt{x+1}]=2(n+1). $$
On the other hand
$$ sqrt{4x+2}=sqrt{4(n+1)^2-2}<2(n+1) $$
and
$$ sqrt{4x+2}=sqrt{4(n+1)^2-2}>2n+1 $$
and hence
$$ [sqrt{4x+2}]=2n+2 $$
In both cases,
$$ [sqrt{x}+sqrt{x+1}]=[sqrt{4x+2}]. $$
Case 3: for $n=1,2$, it is easy to check the inequality is true.
answered Nov 23 '15 at 16:45
xpaulxpaul
23k24455
$endgroup$
Case 1: $n^2le xle(n+1)^2-2$ (for $nge 3$), then
$$ nle sqrt{x}lesqrt{(n+1)^2-2}<n+1, n<sqrt{n^2+1}le sqrt{x+1}<sqrt{(n+1)^2-1}=n+1$$
and
$$ 2n<sqrt{4n^2+2}le sqrt{4x+2}<sqrt{4(n+1)^2-2}<2n+1$$
So
$$ 2n<sqrt{x}+sqrt{x+1}<2n+2, sqrt{4x+2}=2n $$
and
$$ 2n+1le[sqrt{x}+sqrt{x+1}]le2n+2, [sqrt{4x+2}]=2n+1. $$
Now we show
$$ sqrt{x}+sqrt{x+1}<2n+1. $$
Note $sqrt{x}+sqrt{x+1}$ is increasing and hence
$$sqrt{x}+sqrt{x+1}lesqrt{(n+1)^2-2}+sqrt{(n+1)^2-1}.$$
Note
begin{eqnarray}
&&sqrt{(n+1)^2-2}+sqrt{(n+1)^2-1}-2n-1\
&=&frac{(n+1)^2-2-n^2}{sqrt{(n+1)^2-2}+n^2}+frac{(n+1)^2-1-n^2}{sqrt{(n+1)^2-1}+n^2}-1\
&<&frac{4n-1}{sqrt{(n+1)^2-2}+n^2}-1\
&=&frac{4n-1}{sqrt{n^2+2n-1}+n^2}-1\
&<&frac{4n-1}{n+n^2}-1\
&=&-frac{n^2-3n+1}{n+n^2}\
&<0
end{eqnarray}
for $nge 3$.
Thus
$$2n<sqrt{(n+1)^2-2}+sqrt{(n+1)^2-1}<2n+1$$
or
$$[sqrt{x}+sqrt{x+1}=2n+1$$
Case 2: $x=(n+1)^2-1$.
begin{eqnarray}
&&sqrt{x}+sqrt{x+1}\
&=&sqrt{(n+1)^2-1}+n+1\
&<&2(n+1)
end{eqnarray}
and
begin{eqnarray}
&&sqrt{x}+sqrt{x+1}\
&=&sqrt{(n+1)^2-1}+n+1\
&>&2n+1
end{eqnarray}
and hence
$$ [sqrt{x}+sqrt{x+1}]=2(n+1). $$
On the other hand
$$ sqrt{4x+2}=sqrt{4(n+1)^2-2}<2(n+1) $$
and
$$ sqrt{4x+2}=sqrt{4(n+1)^2-2}>2n+1 $$
and hence
$$ [sqrt{4x+2}]=2n+2 $$
In both cases,
$$ [sqrt{x}+sqrt{x+1}]=[sqrt{4x+2}]. $$
Case 3: for $n=1,2$, it is easy to check the inequality is true.
answered Nov 23 '15 at 16:45
xpaulxpaul
23k24455
answered Nov 23 '15 at 16:45
xpaulxpaul
23k24455
answered Nov 23 '15 at 16:45
xpaulxpaul
23k24455
answered Nov 23 '15 at 16:45
xpaulxpaul
23k24455
23k24455
add a comment |
add a comment |
$begingroup$
Taking the listed approach, we get
$$[sqrt{n^2+k}+sqrt{n^2+k+1}]=[sqrt{4n^2+4k +2}]$$
As $k$ relates to $n$, we have a floor-function boundary for the RHS when $kin{-1,0}$ and also at $kin{n-1,n}$, so these values need to be checked on the LHS to verify that these are consecutive boundaries there as well. For $kin{-1,0}$, the LHS shows the same behavior as the RHS, so we only need to show that the LHS also has a boundary at $kin{n-1,n}$ by comparing with $2n+1$ from the RHS:
$$left[sqrt{n^2+n-1}+sqrt{n^2+n}right], R, 2n+1 \ text{ vs. }\ left[sqrt{n^2+n}+sqrt{n^2+n+1}right], R, 2n+1$$
Squaring yields
$$left[2n^2+2n-1+2sqrt{(n^2+n)(n^2+n-1)}right], R, 4n^2+4n+1\
left[2sqrt{(n^2+n)(n^2+n-1)}right], R, 2n^2+2n+2$$
and it should be clear that $[sqrt{n^2+n-1}+sqrt{n^2+n}]lt 2n+1$. Next we have
$$left[2n^2+2n+1+2sqrt{(n^2+n)(n^2+n+1)}right], R, 4n^2+4n+1\
left[2sqrt{(n^2+n)(n^2+n+1)}right], R, 2n^2+2n$$
and it should also be clear that $[sqrt{n^2+n}+sqrt{n^2+n+1}]ge 2n+1$, completing the proof.
answered Nov 23 '15 at 16:59
abiessuabiessu
6,68721541
$endgroup$
add a comment |
$begingroup$
Taking the listed approach, we get
$$[sqrt{n^2+k}+sqrt{n^2+k+1}]=[sqrt{4n^2+4k +2}]$$
As $k$ relates to $n$, we have a floor-function boundary for the RHS when $kin{-1,0}$ and also at $kin{n-1,n}$, so these values need to be checked on the LHS to verify that these are consecutive boundaries there as well. For $kin{-1,0}$, the LHS shows the same behavior as the RHS, so we only need to show that the LHS also has a boundary at $kin{n-1,n}$ by comparing with $2n+1$ from the RHS:
$$left[sqrt{n^2+n-1}+sqrt{n^2+n}right], R, 2n+1 \ text{ vs. }\ left[sqrt{n^2+n}+sqrt{n^2+n+1}right], R, 2n+1$$
Squaring yields
$$left[2n^2+2n-1+2sqrt{(n^2+n)(n^2+n-1)}right], R, 4n^2+4n+1\
left[2sqrt{(n^2+n)(n^2+n-1)}right], R, 2n^2+2n+2$$
and it should be clear that $[sqrt{n^2+n-1}+sqrt{n^2+n}]lt 2n+1$. Next we have
$$left[2n^2+2n+1+2sqrt{(n^2+n)(n^2+n+1)}right], R, 4n^2+4n+1\
left[2sqrt{(n^2+n)(n^2+n+1)}right], R, 2n^2+2n$$
and it should also be clear that $[sqrt{n^2+n}+sqrt{n^2+n+1}]ge 2n+1$, completing the proof.
answered Nov 23 '15 at 16:59
abiessuabiessu
6,68721541
$endgroup$
add a comment |
$begingroup$
Taking the listed approach, we get
$$[sqrt{n^2+k}+sqrt{n^2+k+1}]=[sqrt{4n^2+4k +2}]$$
As $k$ relates to $n$, we have a floor-function boundary for the RHS when $kin{-1,0}$ and also at $kin{n-1,n}$, so these values need to be checked on the LHS to verify that these are consecutive boundaries there as well. For $kin{-1,0}$, the LHS shows the same behavior as the RHS, so we only need to show that the LHS also has a boundary at $kin{n-1,n}$ by comparing with $2n+1$ from the RHS:
$$left[sqrt{n^2+n-1}+sqrt{n^2+n}right], R, 2n+1 \ text{ vs. }\ left[sqrt{n^2+n}+sqrt{n^2+n+1}right], R, 2n+1$$
Squaring yields
$$left[2n^2+2n-1+2sqrt{(n^2+n)(n^2+n-1)}right], R, 4n^2+4n+1\
left[2sqrt{(n^2+n)(n^2+n-1)}right], R, 2n^2+2n+2$$
and it should be clear that $[sqrt{n^2+n-1}+sqrt{n^2+n}]lt 2n+1$. Next we have
$$left[2n^2+2n+1+2sqrt{(n^2+n)(n^2+n+1)}right], R, 4n^2+4n+1\
left[2sqrt{(n^2+n)(n^2+n+1)}right], R, 2n^2+2n$$
and it should also be clear that $[sqrt{n^2+n}+sqrt{n^2+n+1}]ge 2n+1$, completing the proof.
answered Nov 23 '15 at 16:59
abiessuabiessu
6,68721541
$endgroup$
Taking the listed approach, we get
$$[sqrt{n^2+k}+sqrt{n^2+k+1}]=[sqrt{4n^2+4k +2}]$$
As $k$ relates to $n$, we have a floor-function boundary for the RHS when $kin{-1,0}$ and also at $kin{n-1,n}$, so these values need to be checked on the LHS to verify that these are consecutive boundaries there as well. For $kin{-1,0}$, the LHS shows the same behavior as the RHS, so we only need to show that the LHS also has a boundary at $kin{n-1,n}$ by comparing with $2n+1$ from the RHS:
$$left[sqrt{n^2+n-1}+sqrt{n^2+n}right], R, 2n+1 \ text{ vs. }\ left[sqrt{n^2+n}+sqrt{n^2+n+1}right], R, 2n+1$$
Squaring yields
$$left[2n^2+2n-1+2sqrt{(n^2+n)(n^2+n-1)}right], R, 4n^2+4n+1\
left[2sqrt{(n^2+n)(n^2+n-1)}right], R, 2n^2+2n+2$$
and it should be clear that $[sqrt{n^2+n-1}+sqrt{n^2+n}]lt 2n+1$. Next we have
$$left[2n^2+2n+1+2sqrt{(n^2+n)(n^2+n+1)}right], R, 4n^2+4n+1\
left[2sqrt{(n^2+n)(n^2+n+1)}right], R, 2n^2+2n$$
and it should also be clear that $[sqrt{n^2+n}+sqrt{n^2+n+1}]ge 2n+1$, completing the proof.
answered Nov 23 '15 at 16:59
abiessuabiessu
6,68721541
edited Nov 23 '15 at 19:43
answered Nov 23 '15 at 16:59
abiessuabiessu
6,68721541
answered Nov 23 '15 at 16:59
abiessuabiessu
6,68721541
answered Nov 23 '15 at 16:59
abiessuabiessu
6,68721541
6,68721541
add a comment |
add a comment |
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1
$begingroup$
is there any significance of the square brackets?
$endgroup$
– Chinny84
Nov 23 '15 at 13:29
2
$begingroup$
@Chinny84: they usually signify "greatest integer less than contained value" or some similar wording...
$endgroup$
– abiessu
Nov 23 '15 at 13:31
3
$begingroup$
I assume that what you actually tried was $x=n^2+k$...
$endgroup$
– abiessu
Nov 23 '15 at 13:32
3
$begingroup$
The equation is not correct for general positive $x in mathbb{R}$. Try $x=frac{1}{2}$
$endgroup$
– gammatester
Nov 23 '15 at 13:43
1
$begingroup$
if it's the floor function, then it's not true. Check $1/2leq x< 9/16$, for example. Similarly, the ceiling function also fails.
$endgroup$
– A Simmons
Nov 23 '15 at 13:45