Inequality that I believe can be conquered with Cauchy Schwarz I.E
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There are ten real numbers $x_0, . . . , x_9$ with $x_0 = 0, x_9 = 9$. What is the smallest possible value of the expression
$$(x_1 − x_0)^2/1 +(x_2 − x_1)^2/2+(x_3 − x_2)^2/3+ · · · +(x_9 − x_8)^2/9$$?
I thought this problem had cauchy schwartz I.E written all over it , I was able to get the $x_i$'s to telescope down to $9^2$ or $ 81$ but then Im getting confused what will be the bi in the sum. will it be the sum of $1+1/2+1/3...+1/9$ or just $1+2+3...+9$ ? please help :(
inequality
asked Nov 13 at 7:51
Randin
259116
|
show 2 more comments
up vote
-2
down vote
favorite
There are ten real numbers $x_0, . . . , x_9$ with $x_0 = 0, x_9 = 9$. What is the smallest possible value of the expression
$$(x_1 − x_0)^2/1 +(x_2 − x_1)^2/2+(x_3 − x_2)^2/3+ · · · +(x_9 − x_8)^2/9$$?
I thought this problem had cauchy schwartz I.E written all over it , I was able to get the $x_i$'s to telescope down to $9^2$ or $ 81$ but then Im getting confused what will be the bi in the sum. will it be the sum of $1+1/2+1/3...+1/9$ or just $1+2+3...+9$ ? please help :(
inequality
asked Nov 13 at 7:51
Randin
259116
2
well the expression is just $1+(1-1/2)x_1+(1/2-1/3)x_2+cdots+(1/8-1/9)x_8$ so you can make it as small as you like, as there are no constraints on $x_1,cdots,x_8$
– TheSimpliFire
Nov 13 at 7:56
maybe $x_i$'s are ordered?
– pointguard0
Nov 13 at 8:00
Please clarify how $x_i$ are selected among 0,1,...,9
– gimusi
Nov 13 at 8:18
thats exactly how contest author worded it .. xi are real
– Randin
Nov 13 at 8:31
1
@Randin: With $(x_0, ldots, x_9) = (0, a, ldots, a, 9)$ the expression becomes $frac 89 a$, that can be arbitrary large or small.
– Martin R
Nov 13 at 8:41
|
show 2 more comments
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
There are ten real numbers $x_0, . . . , x_9$ with $x_0 = 0, x_9 = 9$. What is the smallest possible value of the expression
$$(x_1 − x_0)^2/1 +(x_2 − x_1)^2/2+(x_3 − x_2)^2/3+ · · · +(x_9 − x_8)^2/9$$?
I thought this problem had cauchy schwartz I.E written all over it , I was able to get the $x_i$'s to telescope down to $9^2$ or $ 81$ but then Im getting confused what will be the bi in the sum. will it be the sum of $1+1/2+1/3...+1/9$ or just $1+2+3...+9$ ? please help :(
inequality
asked Nov 13 at 7:51
Randin
259116
There are ten real numbers $x_0, . . . , x_9$ with $x_0 = 0, x_9 = 9$. What is the smallest possible value of the expression
$$(x_1 − x_0)^2/1 +(x_2 − x_1)^2/2+(x_3 − x_2)^2/3+ · · · +(x_9 − x_8)^2/9$$?
I thought this problem had cauchy schwartz I.E written all over it , I was able to get the $x_i$'s to telescope down to $9^2$ or $ 81$ but then Im getting confused what will be the bi in the sum. will it be the sum of $1+1/2+1/3...+1/9$ or just $1+2+3...+9$ ? please help :(
inequality
inequality
asked Nov 13 at 7:51
Randin
259116
asked Nov 13 at 7:51
Randin
259116
edited 2 days ago
asked Nov 13 at 7:51
Randin
259116
asked Nov 13 at 7:51
Randin
259116
asked Nov 13 at 7:51
Randin
259116
259116
2
well the expression is just $1+(1-1/2)x_1+(1/2-1/3)x_2+cdots+(1/8-1/9)x_8$ so you can make it as small as you like, as there are no constraints on $x_1,cdots,x_8$
– TheSimpliFire
Nov 13 at 7:56
maybe $x_i$'s are ordered?
– pointguard0
Nov 13 at 8:00
Please clarify how $x_i$ are selected among 0,1,...,9
– gimusi
Nov 13 at 8:18
thats exactly how contest author worded it .. xi are real
– Randin
Nov 13 at 8:31
1
@Randin: With $(x_0, ldots, x_9) = (0, a, ldots, a, 9)$ the expression becomes $frac 89 a$, that can be arbitrary large or small.
– Martin R
Nov 13 at 8:41
|
show 2 more comments
2
well the expression is just $1+(1-1/2)x_1+(1/2-1/3)x_2+cdots+(1/8-1/9)x_8$ so you can make it as small as you like, as there are no constraints on $x_1,cdots,x_8$
– TheSimpliFire
Nov 13 at 7:56
maybe $x_i$'s are ordered?
– pointguard0
Nov 13 at 8:00
Please clarify how $x_i$ are selected among 0,1,...,9
– gimusi
Nov 13 at 8:18
thats exactly how contest author worded it .. xi are real
– Randin
Nov 13 at 8:31
1
@Randin: With $(x_0, ldots, x_9) = (0, a, ldots, a, 9)$ the expression becomes $frac 89 a$, that can be arbitrary large or small.
– Martin R
Nov 13 at 8:41
2
2
well the expression is just $1+(1-1/2)x_1+(1/2-1/3)x_2+cdots+(1/8-1/9)x_8$ so you can make it as small as you like, as there are no constraints on $x_1,cdots,x_8$
– TheSimpliFire
Nov 13 at 7:56
well the expression is just $1+(1-1/2)x_1+(1/2-1/3)x_2+cdots+(1/8-1/9)x_8$ so you can make it as small as you like, as there are no constraints on $x_1,cdots,x_8$
– TheSimpliFire
Nov 13 at 7:56
maybe $x_i$'s are ordered?
– pointguard0
Nov 13 at 8:00
maybe $x_i$'s are ordered?
– pointguard0
Nov 13 at 8:00
Please clarify how $x_i$ are selected among 0,1,...,9
– gimusi
Nov 13 at 8:18
Please clarify how $x_i$ are selected among 0,1,...,9
– gimusi
Nov 13 at 8:18
thats exactly how contest author worded it .. xi are real
– Randin
Nov 13 at 8:31
thats exactly how contest author worded it .. xi are real
– Randin
Nov 13 at 8:31
1
1
@Randin: With $(x_0, ldots, x_9) = (0, a, ldots, a, 9)$ the expression becomes $frac 89 a$, that can be arbitrary large or small.
– Martin R
Nov 13 at 8:41
@Randin: With $(x_0, ldots, x_9) = (0, a, ldots, a, 9)$ the expression becomes $frac 89 a$, that can be arbitrary large or small.
– Martin R
Nov 13 at 8:41
|
show 2 more comments
3 Answers
3
active
oldest
votes
up vote
0
down vote
cauchy schwartz inequality with writing the denominators as a sequence of radicals ]1
answered 16 hours ago
Randin
259116
add a comment |
up vote
-1
down vote
As noticed by TheSimpliFire we have
$$(x_1 − x_0)/1 +(x_2 − x_1)/2+(x_3 − x_2)/3+ · · · +(x_9 − x_8)/9=$$$$=(-1)cdot x_0+left(1-frac12right)x_1+left(frac12-frac13right)x_2+ldots+left(frac18-frac19right)x_8+frac19cdot x_9=$$
and by $x_0=0$ and $x_9=9$
$$frac12x_1+frac16x_2+frac1{12}x_3+frac1{20}x_4+frac1{30}x_5+frac1{42}x_6+frac1{56}x_7+frac1{72}x_8+1$$
which can assume any value if $x_1,...,x_8$ are free and if we assume $x_0le x_1le ldots le x_8le x_9$ it is bounded between $1$ and $frac{33}4$.
answered Nov 13 at 8:00
gimusi
84.7k74292
For $(x_0, ldots, x_8, x_9) = (0, ldots, 0, 9)$ the given expression evaluates to $1$, that is smaller than your alleged minimum.
– Martin R
Nov 13 at 8:12
I thought we needed to select 9 different values among $0,1,...,9$. Isn’t it?
– gimusi
Nov 13 at 8:14
Where do you see that in the question? The $x_i$ are only described as “ten real numbers.”
– Martin R
Nov 13 at 8:16
In the first line they seem defined in that way. I ask for a clarification on that.
– gimusi
Nov 13 at 8:17
Even if the $x_i$ are a permutation of $0, ldots, 9$: The question is about $sum frac{x_i - x_{i-1}}{i}$, not about $sum frac{x_i}{i}$.
– Martin R
Nov 13 at 8:33
|
show 3 more comments
up vote
-1
down vote
[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/fXUVO.jpg by cauchy schwarz the (xi+1 -x i ) telescope leaving 9^2=81 but then we have the denominators which are the sum of integers from 1 to 9 , this gives a total of 9/5
answered Nov 13 at 13:55
Randin
259116
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
cauchy schwartz inequality with writing the denominators as a sequence of radicals ]1
answered 16 hours ago
Randin
259116
add a comment |
up vote
0
down vote
cauchy schwartz inequality with writing the denominators as a sequence of radicals ]1
answered 16 hours ago
Randin
259116
add a comment |
up vote
0
down vote
up vote
0
down vote
cauchy schwartz inequality with writing the denominators as a sequence of radicals ]1
answered 16 hours ago
Randin
259116
cauchy schwartz inequality with writing the denominators as a sequence of radicals ]1
answered 16 hours ago
Randin
259116
answered 16 hours ago
Randin
259116
answered 16 hours ago
Randin
259116
answered 16 hours ago
Randin
259116
259116
add a comment |
add a comment |
up vote
-1
down vote
As noticed by TheSimpliFire we have
$$(x_1 − x_0)/1 +(x_2 − x_1)/2+(x_3 − x_2)/3+ · · · +(x_9 − x_8)/9=$$$$=(-1)cdot x_0+left(1-frac12right)x_1+left(frac12-frac13right)x_2+ldots+left(frac18-frac19right)x_8+frac19cdot x_9=$$
and by $x_0=0$ and $x_9=9$
$$frac12x_1+frac16x_2+frac1{12}x_3+frac1{20}x_4+frac1{30}x_5+frac1{42}x_6+frac1{56}x_7+frac1{72}x_8+1$$
which can assume any value if $x_1,...,x_8$ are free and if we assume $x_0le x_1le ldots le x_8le x_9$ it is bounded between $1$ and $frac{33}4$.
answered Nov 13 at 8:00
gimusi
84.7k74292
For $(x_0, ldots, x_8, x_9) = (0, ldots, 0, 9)$ the given expression evaluates to $1$, that is smaller than your alleged minimum.
– Martin R
Nov 13 at 8:12
I thought we needed to select 9 different values among $0,1,...,9$. Isn’t it?
– gimusi
Nov 13 at 8:14
Where do you see that in the question? The $x_i$ are only described as “ten real numbers.”
– Martin R
Nov 13 at 8:16
In the first line they seem defined in that way. I ask for a clarification on that.
– gimusi
Nov 13 at 8:17
Even if the $x_i$ are a permutation of $0, ldots, 9$: The question is about $sum frac{x_i - x_{i-1}}{i}$, not about $sum frac{x_i}{i}$.
– Martin R
Nov 13 at 8:33
|
show 3 more comments
up vote
-1
down vote
As noticed by TheSimpliFire we have
$$(x_1 − x_0)/1 +(x_2 − x_1)/2+(x_3 − x_2)/3+ · · · +(x_9 − x_8)/9=$$$$=(-1)cdot x_0+left(1-frac12right)x_1+left(frac12-frac13right)x_2+ldots+left(frac18-frac19right)x_8+frac19cdot x_9=$$
and by $x_0=0$ and $x_9=9$
$$frac12x_1+frac16x_2+frac1{12}x_3+frac1{20}x_4+frac1{30}x_5+frac1{42}x_6+frac1{56}x_7+frac1{72}x_8+1$$
which can assume any value if $x_1,...,x_8$ are free and if we assume $x_0le x_1le ldots le x_8le x_9$ it is bounded between $1$ and $frac{33}4$.
answered Nov 13 at 8:00
gimusi
84.7k74292
For $(x_0, ldots, x_8, x_9) = (0, ldots, 0, 9)$ the given expression evaluates to $1$, that is smaller than your alleged minimum.
– Martin R
Nov 13 at 8:12
I thought we needed to select 9 different values among $0,1,...,9$. Isn’t it?
– gimusi
Nov 13 at 8:14
Where do you see that in the question? The $x_i$ are only described as “ten real numbers.”
– Martin R
Nov 13 at 8:16
In the first line they seem defined in that way. I ask for a clarification on that.
– gimusi
Nov 13 at 8:17
Even if the $x_i$ are a permutation of $0, ldots, 9$: The question is about $sum frac{x_i - x_{i-1}}{i}$, not about $sum frac{x_i}{i}$.
– Martin R
Nov 13 at 8:33
|
show 3 more comments
up vote
-1
down vote
up vote
-1
down vote
As noticed by TheSimpliFire we have
$$(x_1 − x_0)/1 +(x_2 − x_1)/2+(x_3 − x_2)/3+ · · · +(x_9 − x_8)/9=$$$$=(-1)cdot x_0+left(1-frac12right)x_1+left(frac12-frac13right)x_2+ldots+left(frac18-frac19right)x_8+frac19cdot x_9=$$
and by $x_0=0$ and $x_9=9$
$$frac12x_1+frac16x_2+frac1{12}x_3+frac1{20}x_4+frac1{30}x_5+frac1{42}x_6+frac1{56}x_7+frac1{72}x_8+1$$
which can assume any value if $x_1,...,x_8$ are free and if we assume $x_0le x_1le ldots le x_8le x_9$ it is bounded between $1$ and $frac{33}4$.
answered Nov 13 at 8:00
gimusi
84.7k74292
As noticed by TheSimpliFire we have
$$(x_1 − x_0)/1 +(x_2 − x_1)/2+(x_3 − x_2)/3+ · · · +(x_9 − x_8)/9=$$$$=(-1)cdot x_0+left(1-frac12right)x_1+left(frac12-frac13right)x_2+ldots+left(frac18-frac19right)x_8+frac19cdot x_9=$$
and by $x_0=0$ and $x_9=9$
$$frac12x_1+frac16x_2+frac1{12}x_3+frac1{20}x_4+frac1{30}x_5+frac1{42}x_6+frac1{56}x_7+frac1{72}x_8+1$$
which can assume any value if $x_1,...,x_8$ are free and if we assume $x_0le x_1le ldots le x_8le x_9$ it is bounded between $1$ and $frac{33}4$.
answered Nov 13 at 8:00
gimusi
84.7k74292
edited Nov 13 at 13:13
answered Nov 13 at 8:00
gimusi
84.7k74292
answered Nov 13 at 8:00
gimusi
84.7k74292
answered Nov 13 at 8:00
gimusi
84.7k74292
84.7k74292
For $(x_0, ldots, x_8, x_9) = (0, ldots, 0, 9)$ the given expression evaluates to $1$, that is smaller than your alleged minimum.
– Martin R
Nov 13 at 8:12
I thought we needed to select 9 different values among $0,1,...,9$. Isn’t it?
– gimusi
Nov 13 at 8:14
Where do you see that in the question? The $x_i$ are only described as “ten real numbers.”
– Martin R
Nov 13 at 8:16
In the first line they seem defined in that way. I ask for a clarification on that.
– gimusi
Nov 13 at 8:17
Even if the $x_i$ are a permutation of $0, ldots, 9$: The question is about $sum frac{x_i - x_{i-1}}{i}$, not about $sum frac{x_i}{i}$.
– Martin R
Nov 13 at 8:33
|
show 3 more comments
For $(x_0, ldots, x_8, x_9) = (0, ldots, 0, 9)$ the given expression evaluates to $1$, that is smaller than your alleged minimum.
– Martin R
Nov 13 at 8:12
I thought we needed to select 9 different values among $0,1,...,9$. Isn’t it?
– gimusi
Nov 13 at 8:14
Where do you see that in the question? The $x_i$ are only described as “ten real numbers.”
– Martin R
Nov 13 at 8:16
In the first line they seem defined in that way. I ask for a clarification on that.
– gimusi
Nov 13 at 8:17
Even if the $x_i$ are a permutation of $0, ldots, 9$: The question is about $sum frac{x_i - x_{i-1}}{i}$, not about $sum frac{x_i}{i}$.
– Martin R
Nov 13 at 8:33
For $(x_0, ldots, x_8, x_9) = (0, ldots, 0, 9)$ the given expression evaluates to $1$, that is smaller than your alleged minimum.
– Martin R
Nov 13 at 8:12
For $(x_0, ldots, x_8, x_9) = (0, ldots, 0, 9)$ the given expression evaluates to $1$, that is smaller than your alleged minimum.
– Martin R
Nov 13 at 8:12
I thought we needed to select 9 different values among $0,1,...,9$. Isn’t it?
– gimusi
Nov 13 at 8:14
I thought we needed to select 9 different values among $0,1,...,9$. Isn’t it?
– gimusi
Nov 13 at 8:14
Where do you see that in the question? The $x_i$ are only described as “ten real numbers.”
– Martin R
Nov 13 at 8:16
Where do you see that in the question? The $x_i$ are only described as “ten real numbers.”
– Martin R
Nov 13 at 8:16
In the first line they seem defined in that way. I ask for a clarification on that.
– gimusi
Nov 13 at 8:17
In the first line they seem defined in that way. I ask for a clarification on that.
– gimusi
Nov 13 at 8:17
Even if the $x_i$ are a permutation of $0, ldots, 9$: The question is about $sum frac{x_i - x_{i-1}}{i}$, not about $sum frac{x_i}{i}$.
– Martin R
Nov 13 at 8:33
Even if the $x_i$ are a permutation of $0, ldots, 9$: The question is about $sum frac{x_i - x_{i-1}}{i}$, not about $sum frac{x_i}{i}$.
– Martin R
Nov 13 at 8:33
|
show 3 more comments
up vote
-1
down vote
[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/fXUVO.jpg by cauchy schwarz the (xi+1 -x i ) telescope leaving 9^2=81 but then we have the denominators which are the sum of integers from 1 to 9 , this gives a total of 9/5
answered Nov 13 at 13:55
Randin
259116
add a comment |
up vote
-1
down vote
[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/fXUVO.jpg by cauchy schwarz the (xi+1 -x i ) telescope leaving 9^2=81 but then we have the denominators which are the sum of integers from 1 to 9 , this gives a total of 9/5
answered Nov 13 at 13:55
Randin
259116
add a comment |
up vote
-1
down vote
up vote
-1
down vote
[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/fXUVO.jpg by cauchy schwarz the (xi+1 -x i ) telescope leaving 9^2=81 but then we have the denominators which are the sum of integers from 1 to 9 , this gives a total of 9/5
answered Nov 13 at 13:55
Randin
259116
[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/fXUVO.jpg by cauchy schwarz the (xi+1 -x i ) telescope leaving 9^2=81 but then we have the denominators which are the sum of integers from 1 to 9 , this gives a total of 9/5
answered Nov 13 at 13:55
Randin
259116
answered Nov 13 at 13:55
Randin
259116
answered Nov 13 at 13:55
Randin
259116
answered Nov 13 at 13:55
Randin
259116
259116
add a comment |
add a comment |
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2
well the expression is just $1+(1-1/2)x_1+(1/2-1/3)x_2+cdots+(1/8-1/9)x_8$ so you can make it as small as you like, as there are no constraints on $x_1,cdots,x_8$
– TheSimpliFire
Nov 13 at 7:56
maybe $x_i$'s are ordered?
– pointguard0
Nov 13 at 8:00
Please clarify how $x_i$ are selected among 0,1,...,9
– gimusi
Nov 13 at 8:18
thats exactly how contest author worded it .. xi are real
– Randin
Nov 13 at 8:31
1
@Randin: With $(x_0, ldots, x_9) = (0, a, ldots, a, 9)$ the expression becomes $frac 89 a$, that can be arbitrary large or small.
– Martin R
Nov 13 at 8:41