Proving $int_0^n^2 leftlfloorsqrt trightrfloor ,dt=fracn(n-1)(4n+1)6$Proving that $x-leftlfloor fracx2 rightrfloor; =; leftlfloor fracx+12 rightrfloor$Show that $, 0 leq left lfloorfrac2abright rfloor - 2 left lfloorfracabright rfloor leq 1 $Calculation of value of $ int_0^ncosleft(lfloor x rfloorcdot xright)dx;,$How to solve $ left lfloor x^2 - x - 2 right rfloor = left lfloor x right rfloor $Proof for $leftlfloorfrac 1jleftlfloorfrac nkrightrfloorrightrfloor=leftlfloorfrac njkrightrfloor$$x-leftlfloor x rightrfloor +frac 1 x -leftlfloor frac 1 x rightrfloor =1implies x$ is irrational.Floor function $f(x)=lfloor 2xrfloor+lfloor 4xrfloor+lfloor 6xrfloor+lfloor 8xrfloor$, $xinmathbbR$Is there a simple way of proving that $lfloorsqrtnrfloor+lfloorsqrt4n+1rfloor = lfloorfrac32 lfloor sqrt4n+1 rfloorrfloor$?How to solve for x:$ left lfloorx right rfloor - n cdot left lfloorfracxn right rfloor = y$Proving $lfloor-xrfloor=-lfloor xrfloor-1$, where $lfloorcdotrfloor$ is the floor function
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Proving $int_0^n^2 leftlfloorsqrt trightrfloor ,dt=fracn(n-1)(4n+1)6$
Proving that $x-leftlfloor fracx2 rightrfloor; =; leftlfloor fracx+12 rightrfloor$Show that $, 0 leq left lfloorfrac2abright rfloor - 2 left lfloorfracabright rfloor leq 1 $Calculation of value of $ int_0^ncosleft(lfloor x rfloorcdot xright)dx;,$How to solve $ left lfloor x^2 - x - 2 right rfloor = left lfloor x right rfloor $Proof for $leftlfloorfrac 1jleftlfloorfrac nkrightrfloorrightrfloor=leftlfloorfrac njkrightrfloor$$x-leftlfloor x rightrfloor +frac 1 x -leftlfloor frac 1 x rightrfloor =1implies x$ is irrational.Floor function $f(x)=lfloor 2xrfloor+lfloor 4xrfloor+lfloor 6xrfloor+lfloor 8xrfloor$, $xinmathbbR$Is there a simple way of proving that $lfloorsqrtnrfloor+lfloorsqrt4n+1rfloor = lfloorfrac32 lfloor sqrt4n+1 rfloorrfloor$?How to solve for x:$ left lfloorx right rfloor - n cdot left lfloorfracxn right rfloor = y$Proving $lfloor-xrfloor=-lfloor xrfloor-1$, where $lfloorcdotrfloor$ is the floor function
$begingroup$
I want to prove
$$int_0^n^2 leftlfloor sqrtt rightrfloor ,dt=fracn(n-1)(4n+1)6$$
Is it correct to say that $leftlfloor sqrtt rightrfloor=sqrt(k-1)$ and $(k-1)^2 < t < k^2$, where
$$int_0^n^2 leftlfloor sqrtt rightrfloor ,dt =sum_k=1^n sqrt(k-1) cdot (k^2 - (k-1)^2)$$
I would know how to do the rest, but how did we know that $leftlfloor sqrtt rightrfloor=sqrt(k-1)$, is it something you assume depending on the function? Because for a previous function, there was
$$int_0^n leftlfloor t^2 rightrfloor ,dt$$
and it was set $leftlfloor t^2 rightrfloor=(k-1)^2$, I would really appreciate if someone would kindly explain this to me.
calculus floor-function
edited Mar 21 at 19:49
StubbornAtom
6,29831440
asked Mar 21 at 19:41
D. QaD. Qa
1656
$endgroup$
add a comment |
$begingroup$
I want to prove
$$int_0^n^2 leftlfloor sqrtt rightrfloor ,dt=fracn(n-1)(4n+1)6$$
Is it correct to say that $leftlfloor sqrtt rightrfloor=sqrt(k-1)$ and $(k-1)^2 < t < k^2$, where
$$int_0^n^2 leftlfloor sqrtt rightrfloor ,dt =sum_k=1^n sqrt(k-1) cdot (k^2 - (k-1)^2)$$
I would know how to do the rest, but how did we know that $leftlfloor sqrtt rightrfloor=sqrt(k-1)$, is it something you assume depending on the function? Because for a previous function, there was
$$int_0^n leftlfloor t^2 rightrfloor ,dt$$
and it was set $leftlfloor t^2 rightrfloor=(k-1)^2$, I would really appreciate if someone would kindly explain this to me.
calculus floor-function
edited Mar 21 at 19:49
StubbornAtom
6,29831440
asked Mar 21 at 19:41
D. QaD. Qa
1656
$endgroup$
$begingroup$
The floor function $lfloorsqrt trfloor$ always gives an integer value; $sqrtk-1$ is rarely an integer. What you actually have is $lfloorsqrt trfloor=k-1$ if $(k-1)^2lt tlt k^2$.
$endgroup$
– Barry Cipra
Mar 21 at 19:44
add a comment |
$begingroup$
I want to prove
$$int_0^n^2 leftlfloor sqrtt rightrfloor ,dt=fracn(n-1)(4n+1)6$$
Is it correct to say that $leftlfloor sqrtt rightrfloor=sqrt(k-1)$ and $(k-1)^2 < t < k^2$, where
$$int_0^n^2 leftlfloor sqrtt rightrfloor ,dt =sum_k=1^n sqrt(k-1) cdot (k^2 - (k-1)^2)$$
I would know how to do the rest, but how did we know that $leftlfloor sqrtt rightrfloor=sqrt(k-1)$, is it something you assume depending on the function? Because for a previous function, there was
$$int_0^n leftlfloor t^2 rightrfloor ,dt$$
and it was set $leftlfloor t^2 rightrfloor=(k-1)^2$, I would really appreciate if someone would kindly explain this to me.
calculus floor-function
edited Mar 21 at 19:49
StubbornAtom
6,29831440
asked Mar 21 at 19:41
D. QaD. Qa
1656
$endgroup$
I want to prove
$$int_0^n^2 leftlfloor sqrtt rightrfloor ,dt=fracn(n-1)(4n+1)6$$
Is it correct to say that $leftlfloor sqrtt rightrfloor=sqrt(k-1)$ and $(k-1)^2 < t < k^2$, where
$$int_0^n^2 leftlfloor sqrtt rightrfloor ,dt =sum_k=1^n sqrt(k-1) cdot (k^2 - (k-1)^2)$$
I would know how to do the rest, but how did we know that $leftlfloor sqrtt rightrfloor=sqrt(k-1)$, is it something you assume depending on the function? Because for a previous function, there was
$$int_0^n leftlfloor t^2 rightrfloor ,dt$$
and it was set $leftlfloor t^2 rightrfloor=(k-1)^2$, I would really appreciate if someone would kindly explain this to me.
calculus floor-function
calculus floor-function
edited Mar 21 at 19:49
StubbornAtom
6,29831440
asked Mar 21 at 19:41
D. QaD. Qa
1656
edited Mar 21 at 19:49
StubbornAtom
6,29831440
asked Mar 21 at 19:41
D. QaD. Qa
1656
edited Mar 21 at 19:49
StubbornAtom
6,29831440
edited Mar 21 at 19:49
StubbornAtom
6,29831440
edited Mar 21 at 19:49
StubbornAtom
6,29831440
6,29831440
asked Mar 21 at 19:41
D. QaD. Qa
1656
asked Mar 21 at 19:41
D. QaD. Qa
1656
asked Mar 21 at 19:41
D. QaD. Qa
1656
1656
$begingroup$
The floor function $lfloorsqrt trfloor$ always gives an integer value; $sqrtk-1$ is rarely an integer. What you actually have is $lfloorsqrt trfloor=k-1$ if $(k-1)^2lt tlt k^2$.
$endgroup$
– Barry Cipra
Mar 21 at 19:44
add a comment |
$begingroup$
The floor function $lfloorsqrt trfloor$ always gives an integer value; $sqrtk-1$ is rarely an integer. What you actually have is $lfloorsqrt trfloor=k-1$ if $(k-1)^2lt tlt k^2$.
$endgroup$
– Barry Cipra
Mar 21 at 19:44
$begingroup$
The floor function $lfloorsqrt trfloor$ always gives an integer value; $sqrtk-1$ is rarely an integer. What you actually have is $lfloorsqrt trfloor=k-1$ if $(k-1)^2lt tlt k^2$.
$endgroup$
– Barry Cipra
Mar 21 at 19:44
$begingroup$
The floor function $lfloorsqrt trfloor$ always gives an integer value; $sqrtk-1$ is rarely an integer. What you actually have is $lfloorsqrt trfloor=k-1$ if $(k-1)^2lt tlt k^2$.
$endgroup$
– Barry Cipra
Mar 21 at 19:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have
$$int_0^n^2 lfloor sqrtt rfloor mathrmdt = sum_k=0^n-1 int_k^2^(k+1)^2 lfloor sqrtt rfloor mathrmdt$$
But if $k^2 < t < (k+1)^2$, then $lfloor sqrtt rfloor = k$. So
$$int_0^n^2 lfloor sqrtt rfloor mathrmdt = sum_k=0^n-1 k((k+1)^2-k^2) = sum_k=0^n-1 k(2k+1) = 2 fracn(n-1)(2n-1)6+frac(n-1)n2 $$
i.e.
$$int_0^n^2 lfloor sqrtt rfloor mathrmdt = fracn(n-1)(4n+1)6$$
answered Mar 21 at 19:51
TheSilverDoeTheSilverDoe
5,433216
$endgroup$
$begingroup$
$(k + 1)^2 - k^2 = 2k + 1$ so it should be the sum of $2k^2 + k$.
$endgroup$
– Trevor Gunn
Mar 21 at 19:53
$begingroup$
Yes I answered two quickly, it is fixed now. Thanks !
$endgroup$
– TheSilverDoe
Mar 21 at 19:56
$begingroup$
Wow, thank you so much, your answer helped me understand what is going on more clearly.
$endgroup$
– D. Qa
Mar 21 at 20:22
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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$begingroup$
You have
$$int_0^n^2 lfloor sqrtt rfloor mathrmdt = sum_k=0^n-1 int_k^2^(k+1)^2 lfloor sqrtt rfloor mathrmdt$$
But if $k^2 < t < (k+1)^2$, then $lfloor sqrtt rfloor = k$. So
$$int_0^n^2 lfloor sqrtt rfloor mathrmdt = sum_k=0^n-1 k((k+1)^2-k^2) = sum_k=0^n-1 k(2k+1) = 2 fracn(n-1)(2n-1)6+frac(n-1)n2 $$
i.e.
$$int_0^n^2 lfloor sqrtt rfloor mathrmdt = fracn(n-1)(4n+1)6$$
answered Mar 21 at 19:51
TheSilverDoeTheSilverDoe
5,433216
$endgroup$
$begingroup$
$(k + 1)^2 - k^2 = 2k + 1$ so it should be the sum of $2k^2 + k$.
$endgroup$
– Trevor Gunn
Mar 21 at 19:53
$begingroup$
Yes I answered two quickly, it is fixed now. Thanks !
$endgroup$
– TheSilverDoe
Mar 21 at 19:56
$begingroup$
Wow, thank you so much, your answer helped me understand what is going on more clearly.
$endgroup$
– D. Qa
Mar 21 at 20:22
add a comment |
$begingroup$
You have
$$int_0^n^2 lfloor sqrtt rfloor mathrmdt = sum_k=0^n-1 int_k^2^(k+1)^2 lfloor sqrtt rfloor mathrmdt$$
But if $k^2 < t < (k+1)^2$, then $lfloor sqrtt rfloor = k$. So
$$int_0^n^2 lfloor sqrtt rfloor mathrmdt = sum_k=0^n-1 k((k+1)^2-k^2) = sum_k=0^n-1 k(2k+1) = 2 fracn(n-1)(2n-1)6+frac(n-1)n2 $$
i.e.
$$int_0^n^2 lfloor sqrtt rfloor mathrmdt = fracn(n-1)(4n+1)6$$
answered Mar 21 at 19:51
TheSilverDoeTheSilverDoe
5,433216
$endgroup$
$begingroup$
$(k + 1)^2 - k^2 = 2k + 1$ so it should be the sum of $2k^2 + k$.
$endgroup$
– Trevor Gunn
Mar 21 at 19:53
$begingroup$
Yes I answered two quickly, it is fixed now. Thanks !
$endgroup$
– TheSilverDoe
Mar 21 at 19:56
$begingroup$
Wow, thank you so much, your answer helped me understand what is going on more clearly.
$endgroup$
– D. Qa
Mar 21 at 20:22
add a comment |
$begingroup$
You have
$$int_0^n^2 lfloor sqrtt rfloor mathrmdt = sum_k=0^n-1 int_k^2^(k+1)^2 lfloor sqrtt rfloor mathrmdt$$
But if $k^2 < t < (k+1)^2$, then $lfloor sqrtt rfloor = k$. So
$$int_0^n^2 lfloor sqrtt rfloor mathrmdt = sum_k=0^n-1 k((k+1)^2-k^2) = sum_k=0^n-1 k(2k+1) = 2 fracn(n-1)(2n-1)6+frac(n-1)n2 $$
i.e.
$$int_0^n^2 lfloor sqrtt rfloor mathrmdt = fracn(n-1)(4n+1)6$$
answered Mar 21 at 19:51
TheSilverDoeTheSilverDoe
5,433216
$endgroup$
You have
$$int_0^n^2 lfloor sqrtt rfloor mathrmdt = sum_k=0^n-1 int_k^2^(k+1)^2 lfloor sqrtt rfloor mathrmdt$$
But if $k^2 < t < (k+1)^2$, then $lfloor sqrtt rfloor = k$. So
$$int_0^n^2 lfloor sqrtt rfloor mathrmdt = sum_k=0^n-1 k((k+1)^2-k^2) = sum_k=0^n-1 k(2k+1) = 2 fracn(n-1)(2n-1)6+frac(n-1)n2 $$
i.e.
$$int_0^n^2 lfloor sqrtt rfloor mathrmdt = fracn(n-1)(4n+1)6$$
answered Mar 21 at 19:51
TheSilverDoeTheSilverDoe
5,433216
edited Mar 21 at 19:56
answered Mar 21 at 19:51
TheSilverDoeTheSilverDoe
5,433216
answered Mar 21 at 19:51
TheSilverDoeTheSilverDoe
5,433216
answered Mar 21 at 19:51
TheSilverDoeTheSilverDoe
5,433216
5,433216
$begingroup$
$(k + 1)^2 - k^2 = 2k + 1$ so it should be the sum of $2k^2 + k$.
$endgroup$
– Trevor Gunn
Mar 21 at 19:53
$begingroup$
Yes I answered two quickly, it is fixed now. Thanks !
$endgroup$
– TheSilverDoe
Mar 21 at 19:56
$begingroup$
Wow, thank you so much, your answer helped me understand what is going on more clearly.
$endgroup$
– D. Qa
Mar 21 at 20:22
add a comment |
$begingroup$
$(k + 1)^2 - k^2 = 2k + 1$ so it should be the sum of $2k^2 + k$.
$endgroup$
– Trevor Gunn
Mar 21 at 19:53
$begingroup$
Yes I answered two quickly, it is fixed now. Thanks !
$endgroup$
– TheSilverDoe
Mar 21 at 19:56
$begingroup$
Wow, thank you so much, your answer helped me understand what is going on more clearly.
$endgroup$
– D. Qa
Mar 21 at 20:22
$begingroup$
$(k + 1)^2 - k^2 = 2k + 1$ so it should be the sum of $2k^2 + k$.
$endgroup$
– Trevor Gunn
Mar 21 at 19:53
$begingroup$
$(k + 1)^2 - k^2 = 2k + 1$ so it should be the sum of $2k^2 + k$.
$endgroup$
– Trevor Gunn
Mar 21 at 19:53
$begingroup$
Yes I answered two quickly, it is fixed now. Thanks !
$endgroup$
– TheSilverDoe
Mar 21 at 19:56
$begingroup$
Yes I answered two quickly, it is fixed now. Thanks !
$endgroup$
– TheSilverDoe
Mar 21 at 19:56
$begingroup$
Wow, thank you so much, your answer helped me understand what is going on more clearly.
$endgroup$
– D. Qa
Mar 21 at 20:22
$begingroup$
Wow, thank you so much, your answer helped me understand what is going on more clearly.
$endgroup$
– D. Qa
Mar 21 at 20:22
add a comment |
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$begingroup$
The floor function $lfloorsqrt trfloor$ always gives an integer value; $sqrtk-1$ is rarely an integer. What you actually have is $lfloorsqrt trfloor=k-1$ if $(k-1)^2lt tlt k^2$.
$endgroup$
– Barry Cipra
Mar 21 at 19:44