Given four real numbers $a_1<a_2<a_3<a_4$, rearrange them in such an order $a_i_1$, $a_i_2$, $a_i_3$, $a_i_4$ that the sum The Next CEO of Stack OverflowProve that $f(a,c,b,d) > f(a,b,c,d) > f(a,b,d,c).$Given $a_1,a_100, a_i=a_i-1a_i+1$, what's $a_1+a_2$?Largest value of $alpha$ to satisfy a given inequalityShowing the convergence of $a_1+a_2+a_3…$ given $(a_1+a_2)+(a_3+a_4)…=S, a_n to 0$Prove by mathematical induction from “Axiom II” that if $a_1,a_2,a_3,…,a_n$ are positive, then the sum $a_1+a_2+…a_n$ is positive.Show that the sequence $a_nleq a_2n+a_2n+1$ divergesAre there any known criteria in order for $sqrta_1+sqrta_2-sqrta_3+sqrta_4-sqrta_5+ldots$ to converge?$a_n$ is the smallest positive integer number such that $sqrta_n+sqrta_n-1+…+sqrta_1$ is positive integerProve that $a_1 > (a_1 + a_2 + a_3 + a_4)/4$ implies $a_i < (a_1 + a_2 + a_3 + a_4)/4$ for $i in 2, 3, 4$.Finding $sum_i=1^100 a_i$ given that $sqrta_1+sqrta_2-1+sqrta_3-2+dots+sqrta_n-(n-1)=frac12(a_1+a_2+dots+a_n)=fracn(n-3)4$Finding the difference $(a_2 + a_4 + … + a_100) - (a_1 + a_3 + … + a_99)$
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$begingroup$
Given four real numbers $a_1 lt a_2 lt a_3 lt a_4$, rearrange them in such an order
$a_i_1
, a_i_2
, a_i_3
, a_i_4 $
that the sum
$$S = (a_i_1-a_i_2)^2 + (a_i_2-a_i_3)^2 + (a_i_3-a_i_4)^2 + (a_i_4-a_i_1)^2$$ has the least possible value?
Can you generalise for $n$ real numbers?
I opened up the equation but it was not of much use.
We can observe that the $3$ inequalities hold:
1) $a_4-a_1 gt a_3-a_1 gt a_2-a_1 $
2) $a_4-a_1 gt a_4-a_2 gt a_4-a_3 $ .
3) $a_4-a_2 gt a_3-a_2$
And I believe the order giving the smallest $S$ will be given by $ a_4, a_2, a_1, a_3 $ which is equivalent to the sum if the order is $ a_4, a_3, a_1, a_2 $,
so
$$S = (a_i_4-a_i_2)^2 + (a_i_2-a_i_1)^2 + (a_i_1-a_i_3)^2 + (a_i_3-a_i_4)^2$$
and
$$S = (a_i_4-a_i_3)^2 + (a_i_3-a_i_1)^2 + (a_i_1-a_i_2)^2 + (a_i_2-a_i_4)^2$$
Although I am not sure on how to prove this, any help would be appreciated, thank you.
sequences-and-series inequality
edited Mar 19 at 12:25
Rócherz
3,0013821
asked Mar 19 at 12:02
user101user101
304
$endgroup$
add a comment |
$begingroup$
Given four real numbers $a_1 lt a_2 lt a_3 lt a_4$, rearrange them in such an order
$a_i_1
, a_i_2
, a_i_3
, a_i_4 $
that the sum
$$S = (a_i_1-a_i_2)^2 + (a_i_2-a_i_3)^2 + (a_i_3-a_i_4)^2 + (a_i_4-a_i_1)^2$$ has the least possible value?
Can you generalise for $n$ real numbers?
I opened up the equation but it was not of much use.
We can observe that the $3$ inequalities hold:
1) $a_4-a_1 gt a_3-a_1 gt a_2-a_1 $
2) $a_4-a_1 gt a_4-a_2 gt a_4-a_3 $ .
3) $a_4-a_2 gt a_3-a_2$
And I believe the order giving the smallest $S$ will be given by $ a_4, a_2, a_1, a_3 $ which is equivalent to the sum if the order is $ a_4, a_3, a_1, a_2 $,
so
$$S = (a_i_4-a_i_2)^2 + (a_i_2-a_i_1)^2 + (a_i_1-a_i_3)^2 + (a_i_3-a_i_4)^2$$
and
$$S = (a_i_4-a_i_3)^2 + (a_i_3-a_i_1)^2 + (a_i_1-a_i_2)^2 + (a_i_2-a_i_4)^2$$
Although I am not sure on how to prove this, any help would be appreciated, thank you.
sequences-and-series inequality
edited Mar 19 at 12:25
Rócherz
3,0013821
asked Mar 19 at 12:02
user101user101
304
$endgroup$
1
$begingroup$
Possible duplicate of Prove that $f(a,c,b,d) > f(a,b,c,d) > f(a,b,d,c).$
$endgroup$
– Martin R
Mar 19 at 12:06
1
$begingroup$
Just adding that the above link indeed contains all rearrangements one needs to consider. In general, there are $4! = 24$ rearrangements of these numbers, but firstly since $abcd$ yields the same sum as the cyclic permutations $bcda,cdab,dabc$, for every rearrangement there are four redundant ones, so 24/4 = 6 rearrangements up to these cyclic ones. But additionally, for every arrangement $abcd$ the reverse arrangement $dcba$ also yields the same sum, so perform another division 6/2 = 3. So the above three options $acbd, abcd, abdc$ are the only ones one needs to care about.
$endgroup$
– Lukas Miristwhisky
Mar 19 at 14:36
add a comment |
$begingroup$
Given four real numbers $a_1 lt a_2 lt a_3 lt a_4$, rearrange them in such an order
$a_i_1
, a_i_2
, a_i_3
, a_i_4 $
that the sum
$$S = (a_i_1-a_i_2)^2 + (a_i_2-a_i_3)^2 + (a_i_3-a_i_4)^2 + (a_i_4-a_i_1)^2$$ has the least possible value?
Can you generalise for $n$ real numbers?
I opened up the equation but it was not of much use.
We can observe that the $3$ inequalities hold:
1) $a_4-a_1 gt a_3-a_1 gt a_2-a_1 $
2) $a_4-a_1 gt a_4-a_2 gt a_4-a_3 $ .
3) $a_4-a_2 gt a_3-a_2$
And I believe the order giving the smallest $S$ will be given by $ a_4, a_2, a_1, a_3 $ which is equivalent to the sum if the order is $ a_4, a_3, a_1, a_2 $,
so
$$S = (a_i_4-a_i_2)^2 + (a_i_2-a_i_1)^2 + (a_i_1-a_i_3)^2 + (a_i_3-a_i_4)^2$$
and
$$S = (a_i_4-a_i_3)^2 + (a_i_3-a_i_1)^2 + (a_i_1-a_i_2)^2 + (a_i_2-a_i_4)^2$$
Although I am not sure on how to prove this, any help would be appreciated, thank you.
sequences-and-series inequality
edited Mar 19 at 12:25
Rócherz
3,0013821
asked Mar 19 at 12:02
user101user101
304
$endgroup$
Given four real numbers $a_1 lt a_2 lt a_3 lt a_4$, rearrange them in such an order
$a_i_1
, a_i_2
, a_i_3
, a_i_4 $
that the sum
$$S = (a_i_1-a_i_2)^2 + (a_i_2-a_i_3)^2 + (a_i_3-a_i_4)^2 + (a_i_4-a_i_1)^2$$ has the least possible value?
Can you generalise for $n$ real numbers?
I opened up the equation but it was not of much use.
We can observe that the $3$ inequalities hold:
1) $a_4-a_1 gt a_3-a_1 gt a_2-a_1 $
2) $a_4-a_1 gt a_4-a_2 gt a_4-a_3 $ .
3) $a_4-a_2 gt a_3-a_2$
And I believe the order giving the smallest $S$ will be given by $ a_4, a_2, a_1, a_3 $ which is equivalent to the sum if the order is $ a_4, a_3, a_1, a_2 $,
so
$$S = (a_i_4-a_i_2)^2 + (a_i_2-a_i_1)^2 + (a_i_1-a_i_3)^2 + (a_i_3-a_i_4)^2$$
and
$$S = (a_i_4-a_i_3)^2 + (a_i_3-a_i_1)^2 + (a_i_1-a_i_2)^2 + (a_i_2-a_i_4)^2$$
Although I am not sure on how to prove this, any help would be appreciated, thank you.
sequences-and-series inequality
sequences-and-series inequality
edited Mar 19 at 12:25
Rócherz
3,0013821
asked Mar 19 at 12:02
user101user101
304
edited Mar 19 at 12:25
Rócherz
3,0013821
asked Mar 19 at 12:02
user101user101
304
edited Mar 19 at 12:25
Rócherz
3,0013821
edited Mar 19 at 12:25
Rócherz
3,0013821
edited Mar 19 at 12:25
Rócherz
3,0013821
3,0013821
asked Mar 19 at 12:02
user101user101
304
asked Mar 19 at 12:02
user101user101
304
asked Mar 19 at 12:02
user101user101
304
304
1
$begingroup$
Possible duplicate of Prove that $f(a,c,b,d) > f(a,b,c,d) > f(a,b,d,c).$
$endgroup$
– Martin R
Mar 19 at 12:06
1
$begingroup$
Just adding that the above link indeed contains all rearrangements one needs to consider. In general, there are $4! = 24$ rearrangements of these numbers, but firstly since $abcd$ yields the same sum as the cyclic permutations $bcda,cdab,dabc$, for every rearrangement there are four redundant ones, so 24/4 = 6 rearrangements up to these cyclic ones. But additionally, for every arrangement $abcd$ the reverse arrangement $dcba$ also yields the same sum, so perform another division 6/2 = 3. So the above three options $acbd, abcd, abdc$ are the only ones one needs to care about.
$endgroup$
– Lukas Miristwhisky
Mar 19 at 14:36
add a comment |
1
$begingroup$
Possible duplicate of Prove that $f(a,c,b,d) > f(a,b,c,d) > f(a,b,d,c).$
$endgroup$
– Martin R
Mar 19 at 12:06
1
$begingroup$
Just adding that the above link indeed contains all rearrangements one needs to consider. In general, there are $4! = 24$ rearrangements of these numbers, but firstly since $abcd$ yields the same sum as the cyclic permutations $bcda,cdab,dabc$, for every rearrangement there are four redundant ones, so 24/4 = 6 rearrangements up to these cyclic ones. But additionally, for every arrangement $abcd$ the reverse arrangement $dcba$ also yields the same sum, so perform another division 6/2 = 3. So the above three options $acbd, abcd, abdc$ are the only ones one needs to care about.
$endgroup$
– Lukas Miristwhisky
Mar 19 at 14:36
1
1
$begingroup$
Possible duplicate of Prove that $f(a,c,b,d) > f(a,b,c,d) > f(a,b,d,c).$
$endgroup$
– Martin R
Mar 19 at 12:06
$begingroup$
Possible duplicate of Prove that $f(a,c,b,d) > f(a,b,c,d) > f(a,b,d,c).$
$endgroup$
– Martin R
Mar 19 at 12:06
1
1
$begingroup$
Just adding that the above link indeed contains all rearrangements one needs to consider. In general, there are $4! = 24$ rearrangements of these numbers, but firstly since $abcd$ yields the same sum as the cyclic permutations $bcda,cdab,dabc$, for every rearrangement there are four redundant ones, so 24/4 = 6 rearrangements up to these cyclic ones. But additionally, for every arrangement $abcd$ the reverse arrangement $dcba$ also yields the same sum, so perform another division 6/2 = 3. So the above three options $acbd, abcd, abdc$ are the only ones one needs to care about.
$endgroup$
– Lukas Miristwhisky
Mar 19 at 14:36
$begingroup$
Just adding that the above link indeed contains all rearrangements one needs to consider. In general, there are $4! = 24$ rearrangements of these numbers, but firstly since $abcd$ yields the same sum as the cyclic permutations $bcda,cdab,dabc$, for every rearrangement there are four redundant ones, so 24/4 = 6 rearrangements up to these cyclic ones. But additionally, for every arrangement $abcd$ the reverse arrangement $dcba$ also yields the same sum, so perform another division 6/2 = 3. So the above three options $acbd, abcd, abdc$ are the only ones one needs to care about.
$endgroup$
– Lukas Miristwhisky
Mar 19 at 14:36
add a comment |
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$begingroup$
Possible duplicate of Prove that $f(a,c,b,d) > f(a,b,c,d) > f(a,b,d,c).$
$endgroup$
– Martin R
Mar 19 at 12:06
1
$begingroup$
Just adding that the above link indeed contains all rearrangements one needs to consider. In general, there are $4! = 24$ rearrangements of these numbers, but firstly since $abcd$ yields the same sum as the cyclic permutations $bcda,cdab,dabc$, for every rearrangement there are four redundant ones, so 24/4 = 6 rearrangements up to these cyclic ones. But additionally, for every arrangement $abcd$ the reverse arrangement $dcba$ also yields the same sum, so perform another division 6/2 = 3. So the above three options $acbd, abcd, abdc$ are the only ones one needs to care about.
$endgroup$
– Lukas Miristwhisky
Mar 19 at 14:36