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How one can show that this polynomial is negative
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How one can show that this polynomial.
$$-2+2x^2k+1-x^3k+2+x^k-1$$
is negative for all integer $k>1$ and real $x>2$.
I have no idea to start.
real-analysis algebra-precalculus polynomials
edited 49 mins ago
greedoid
31.2k94287
asked 1 hour ago
DER
1,656918
 |Â
show 3 more comments
up vote
2
down vote
favorite
How one can show that this polynomial.
$$-2+2x^2k+1-x^3k+2+x^k-1$$
is negative for all integer $k>1$ and real $x>2$.
I have no idea to start.
real-analysis algebra-precalculus polynomials
edited 49 mins ago
greedoid
31.2k94287
asked 1 hour ago
DER
1,656918
@greedoid: No it is $k-1$.
– DER
1 hour ago
Any reason why it should be $x>5$ specifically? desmos.com/calculator/esk3my2jfc
– Mohammad Zuhair Khan
1 hour ago
That will likely become clear in the answer
– HackerBoss
1 hour ago
@HackerBoss I only meant that because it seems that for $k>1,$ $-2+2x^2k+1-x^3k+2+x^k-1$ is negative for all values $x>1.18.$
– Mohammad Zuhair Khan
57 mins ago
@MohammadZuhairKhan: How you get this result
– DER
56 mins ago
 |Â
show 3 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
How one can show that this polynomial.
$$-2+2x^2k+1-x^3k+2+x^k-1$$
is negative for all integer $k>1$ and real $x>2$.
I have no idea to start.
real-analysis algebra-precalculus polynomials
edited 49 mins ago
greedoid
31.2k94287
asked 1 hour ago
DER
1,656918
How one can show that this polynomial.
$$-2+2x^2k+1-x^3k+2+x^k-1$$
is negative for all integer $k>1$ and real $x>2$.
I have no idea to start.
real-analysis algebra-precalculus polynomials
real-analysis algebra-precalculus polynomials
edited 49 mins ago
greedoid
31.2k94287
asked 1 hour ago
DER
1,656918
edited 49 mins ago
greedoid
31.2k94287
asked 1 hour ago
DER
1,656918
edited 49 mins ago
greedoid
31.2k94287
edited 49 mins ago
greedoid
31.2k94287
edited 49 mins ago
greedoid
31.2k94287
31.2k94287
asked 1 hour ago
DER
1,656918
asked 1 hour ago
DER
1,656918
asked 1 hour ago
DER
1,656918
1,656918
@greedoid: No it is $k-1$.
– DER
1 hour ago
Any reason why it should be $x>5$ specifically? desmos.com/calculator/esk3my2jfc
– Mohammad Zuhair Khan
1 hour ago
That will likely become clear in the answer
– HackerBoss
1 hour ago
@HackerBoss I only meant that because it seems that for $k>1,$ $-2+2x^2k+1-x^3k+2+x^k-1$ is negative for all values $x>1.18.$
– Mohammad Zuhair Khan
57 mins ago
@MohammadZuhairKhan: How you get this result
– DER
56 mins ago
 |Â
show 3 more comments
@greedoid: No it is $k-1$.
– DER
1 hour ago
Any reason why it should be $x>5$ specifically? desmos.com/calculator/esk3my2jfc
– Mohammad Zuhair Khan
1 hour ago
That will likely become clear in the answer
– HackerBoss
1 hour ago
@HackerBoss I only meant that because it seems that for $k>1,$ $-2+2x^2k+1-x^3k+2+x^k-1$ is negative for all values $x>1.18.$
– Mohammad Zuhair Khan
57 mins ago
@MohammadZuhairKhan: How you get this result
– DER
56 mins ago
@greedoid: No it is $k-1$.
– DER
1 hour ago
@greedoid: No it is $k-1$.
– DER
1 hour ago
Any reason why it should be $x>5$ specifically? desmos.com/calculator/esk3my2jfc
– Mohammad Zuhair Khan
1 hour ago
Any reason why it should be $x>5$ specifically? desmos.com/calculator/esk3my2jfc
– Mohammad Zuhair Khan
1 hour ago
That will likely become clear in the answer
– HackerBoss
1 hour ago
That will likely become clear in the answer
– HackerBoss
1 hour ago
@HackerBoss I only meant that because it seems that for $k>1,$ $-2+2x^2k+1-x^3k+2+x^k-1$ is negative for all values $x>1.18.$
– Mohammad Zuhair Khan
57 mins ago
@HackerBoss I only meant that because it seems that for $k>1,$ $-2+2x^2k+1-x^3k+2+x^k-1$ is negative for all values $x>1.18.$
– Mohammad Zuhair Khan
57 mins ago
@MohammadZuhairKhan: How you get this result
– DER
56 mins ago
@MohammadZuhairKhan: How you get this result
– DER
56 mins ago
 |Â
show 3 more comments
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
For $x>2$ we have $x^k+1 >4x^k-1>x^k-1$
$$x^2k+1(2-x^k+1)-(2-x^k-1)<x^2k+1(2-x^k-1)-(2-x^k-1)= underbrace(x^2k+1-1)_>0underbrace(2-x^k-1)_<0$$
answered 53 mins ago
greedoid
31.2k94287
add a comment |Â
up vote
1
down vote
Well I can change all the signs and assert that the result is positive. If I multiply by $x$ I don't change the sign, because $x$ is positive and I then put $y=x^k$ to simplify the expression. Then I am looking at $$x^3y^3-2x^2y^2-y+2gt 0$$
I am then guessing that $-2$ is irrelevant once $x$ and $k$ are large, so I drop it and divide through by $y$ (positive) and I expect to find $$x^3y^2-2x^2y-1gt 0$$ perhaps with some extra cases to consider. Now I drop the $-1$ as small in relation to the other terms and divide through by $x^2y$ (positive) and consider $$xy-2$$
Now with $kgt 1$ and $xgt 5$ we have also $ygt 5$ so that $$xy-2gt23$$
Now we see how to reverse the process with crude estimates.
Multiply through by $x^2y^gt 1$ (very crude estimate) and subtract $1$ to obtain $$x^3y^2-2x^2y-1gt 23x^2y-1gt 22$$
Multiply by $ygt 1$and add $2$ to obtain $$x^3y^3-2x^2y^2-y+2gt22y+2gt 24gt 0$$
which is equivalent to the original inequality. As you see, only very crude estimates are required. I used $y$ to see if there were any obvious squares hiding in he background.
answered 47 mins ago
Mark Bennet
78.6k777177
add a comment |Â
up vote
0
down vote
Note that
$$-2+2x^2k+1-x^3k+2+x^k-1=-2+left(2x^k+2-x^2k+3+1right)x^k-1$$
and
$$2x^k+2-x^2k+3+1=1+left(2-x^k+1right)x^k+2$$
If $x > 5$, and $k>1$, then $2-x^k+1<-1$, $x^k+2>1$, and $x^k-1>1$, so that $$-2+left(1+left(2-x^k+1right)x^k+2right)x^k-1<0$$
Informally,
$$-2+(1+(<-1)(>1))(>1)rightarrow-2+(1+(<-1))(>1)rightarrow-2+(<0)(>1)rightarrow-2+(<0)<0$$
answered 47 mins ago
HackerBoss
1515
New contributor
HackerBoss is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
For $x>2$ we have $x^k+1 >4x^k-1>x^k-1$
$$x^2k+1(2-x^k+1)-(2-x^k-1)<x^2k+1(2-x^k-1)-(2-x^k-1)= underbrace(x^2k+1-1)_>0underbrace(2-x^k-1)_<0$$
answered 53 mins ago
greedoid
31.2k94287
add a comment |Â
up vote
4
down vote
accepted
For $x>2$ we have $x^k+1 >4x^k-1>x^k-1$
$$x^2k+1(2-x^k+1)-(2-x^k-1)<x^2k+1(2-x^k-1)-(2-x^k-1)= underbrace(x^2k+1-1)_>0underbrace(2-x^k-1)_<0$$
answered 53 mins ago
greedoid
31.2k94287
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
For $x>2$ we have $x^k+1 >4x^k-1>x^k-1$
$$x^2k+1(2-x^k+1)-(2-x^k-1)<x^2k+1(2-x^k-1)-(2-x^k-1)= underbrace(x^2k+1-1)_>0underbrace(2-x^k-1)_<0$$
answered 53 mins ago
greedoid
31.2k94287
For $x>2$ we have $x^k+1 >4x^k-1>x^k-1$
$$x^2k+1(2-x^k+1)-(2-x^k-1)<x^2k+1(2-x^k-1)-(2-x^k-1)= underbrace(x^2k+1-1)_>0underbrace(2-x^k-1)_<0$$
answered 53 mins ago
greedoid
31.2k94287
answered 53 mins ago
greedoid
31.2k94287
answered 53 mins ago
greedoid
31.2k94287
answered 53 mins ago
greedoid
31.2k94287
31.2k94287
add a comment |Â
add a comment |Â
up vote
1
down vote
Well I can change all the signs and assert that the result is positive. If I multiply by $x$ I don't change the sign, because $x$ is positive and I then put $y=x^k$ to simplify the expression. Then I am looking at $$x^3y^3-2x^2y^2-y+2gt 0$$
I am then guessing that $-2$ is irrelevant once $x$ and $k$ are large, so I drop it and divide through by $y$ (positive) and I expect to find $$x^3y^2-2x^2y-1gt 0$$ perhaps with some extra cases to consider. Now I drop the $-1$ as small in relation to the other terms and divide through by $x^2y$ (positive) and consider $$xy-2$$
Now with $kgt 1$ and $xgt 5$ we have also $ygt 5$ so that $$xy-2gt23$$
Now we see how to reverse the process with crude estimates.
Multiply through by $x^2y^gt 1$ (very crude estimate) and subtract $1$ to obtain $$x^3y^2-2x^2y-1gt 23x^2y-1gt 22$$
Multiply by $ygt 1$and add $2$ to obtain $$x^3y^3-2x^2y^2-y+2gt22y+2gt 24gt 0$$
which is equivalent to the original inequality. As you see, only very crude estimates are required. I used $y$ to see if there were any obvious squares hiding in he background.
answered 47 mins ago
Mark Bennet
78.6k777177
add a comment |Â
up vote
1
down vote
Well I can change all the signs and assert that the result is positive. If I multiply by $x$ I don't change the sign, because $x$ is positive and I then put $y=x^k$ to simplify the expression. Then I am looking at $$x^3y^3-2x^2y^2-y+2gt 0$$
I am then guessing that $-2$ is irrelevant once $x$ and $k$ are large, so I drop it and divide through by $y$ (positive) and I expect to find $$x^3y^2-2x^2y-1gt 0$$ perhaps with some extra cases to consider. Now I drop the $-1$ as small in relation to the other terms and divide through by $x^2y$ (positive) and consider $$xy-2$$
Now with $kgt 1$ and $xgt 5$ we have also $ygt 5$ so that $$xy-2gt23$$
Now we see how to reverse the process with crude estimates.
Multiply through by $x^2y^gt 1$ (very crude estimate) and subtract $1$ to obtain $$x^3y^2-2x^2y-1gt 23x^2y-1gt 22$$
Multiply by $ygt 1$and add $2$ to obtain $$x^3y^3-2x^2y^2-y+2gt22y+2gt 24gt 0$$
which is equivalent to the original inequality. As you see, only very crude estimates are required. I used $y$ to see if there were any obvious squares hiding in he background.
answered 47 mins ago
Mark Bennet
78.6k777177
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Well I can change all the signs and assert that the result is positive. If I multiply by $x$ I don't change the sign, because $x$ is positive and I then put $y=x^k$ to simplify the expression. Then I am looking at $$x^3y^3-2x^2y^2-y+2gt 0$$
I am then guessing that $-2$ is irrelevant once $x$ and $k$ are large, so I drop it and divide through by $y$ (positive) and I expect to find $$x^3y^2-2x^2y-1gt 0$$ perhaps with some extra cases to consider. Now I drop the $-1$ as small in relation to the other terms and divide through by $x^2y$ (positive) and consider $$xy-2$$
Now with $kgt 1$ and $xgt 5$ we have also $ygt 5$ so that $$xy-2gt23$$
Now we see how to reverse the process with crude estimates.
Multiply through by $x^2y^gt 1$ (very crude estimate) and subtract $1$ to obtain $$x^3y^2-2x^2y-1gt 23x^2y-1gt 22$$
Multiply by $ygt 1$and add $2$ to obtain $$x^3y^3-2x^2y^2-y+2gt22y+2gt 24gt 0$$
which is equivalent to the original inequality. As you see, only very crude estimates are required. I used $y$ to see if there were any obvious squares hiding in he background.
answered 47 mins ago
Mark Bennet
78.6k777177
Well I can change all the signs and assert that the result is positive. If I multiply by $x$ I don't change the sign, because $x$ is positive and I then put $y=x^k$ to simplify the expression. Then I am looking at $$x^3y^3-2x^2y^2-y+2gt 0$$
I am then guessing that $-2$ is irrelevant once $x$ and $k$ are large, so I drop it and divide through by $y$ (positive) and I expect to find $$x^3y^2-2x^2y-1gt 0$$ perhaps with some extra cases to consider. Now I drop the $-1$ as small in relation to the other terms and divide through by $x^2y$ (positive) and consider $$xy-2$$
Now with $kgt 1$ and $xgt 5$ we have also $ygt 5$ so that $$xy-2gt23$$
Now we see how to reverse the process with crude estimates.
Multiply through by $x^2y^gt 1$ (very crude estimate) and subtract $1$ to obtain $$x^3y^2-2x^2y-1gt 23x^2y-1gt 22$$
Multiply by $ygt 1$and add $2$ to obtain $$x^3y^3-2x^2y^2-y+2gt22y+2gt 24gt 0$$
which is equivalent to the original inequality. As you see, only very crude estimates are required. I used $y$ to see if there were any obvious squares hiding in he background.
answered 47 mins ago
Mark Bennet
78.6k777177
answered 47 mins ago
Mark Bennet
78.6k777177
answered 47 mins ago
Mark Bennet
78.6k777177
answered 47 mins ago
Mark Bennet
78.6k777177
78.6k777177
add a comment |Â
add a comment |Â
up vote
0
down vote
Note that
$$-2+2x^2k+1-x^3k+2+x^k-1=-2+left(2x^k+2-x^2k+3+1right)x^k-1$$
and
$$2x^k+2-x^2k+3+1=1+left(2-x^k+1right)x^k+2$$
If $x > 5$, and $k>1$, then $2-x^k+1<-1$, $x^k+2>1$, and $x^k-1>1$, so that $$-2+left(1+left(2-x^k+1right)x^k+2right)x^k-1<0$$
Informally,
$$-2+(1+(<-1)(>1))(>1)rightarrow-2+(1+(<-1))(>1)rightarrow-2+(<0)(>1)rightarrow-2+(<0)<0$$
answered 47 mins ago
HackerBoss
1515
New contributor
HackerBoss is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
Note that
$$-2+2x^2k+1-x^3k+2+x^k-1=-2+left(2x^k+2-x^2k+3+1right)x^k-1$$
and
$$2x^k+2-x^2k+3+1=1+left(2-x^k+1right)x^k+2$$
If $x > 5$, and $k>1$, then $2-x^k+1<-1$, $x^k+2>1$, and $x^k-1>1$, so that $$-2+left(1+left(2-x^k+1right)x^k+2right)x^k-1<0$$
Informally,
$$-2+(1+(<-1)(>1))(>1)rightarrow-2+(1+(<-1))(>1)rightarrow-2+(<0)(>1)rightarrow-2+(<0)<0$$
answered 47 mins ago
HackerBoss
1515
New contributor
HackerBoss is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Note that
$$-2+2x^2k+1-x^3k+2+x^k-1=-2+left(2x^k+2-x^2k+3+1right)x^k-1$$
and
$$2x^k+2-x^2k+3+1=1+left(2-x^k+1right)x^k+2$$
If $x > 5$, and $k>1$, then $2-x^k+1<-1$, $x^k+2>1$, and $x^k-1>1$, so that $$-2+left(1+left(2-x^k+1right)x^k+2right)x^k-1<0$$
Informally,
$$-2+(1+(<-1)(>1))(>1)rightarrow-2+(1+(<-1))(>1)rightarrow-2+(<0)(>1)rightarrow-2+(<0)<0$$
answered 47 mins ago
HackerBoss
1515
New contributor
HackerBoss is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Note that
$$-2+2x^2k+1-x^3k+2+x^k-1=-2+left(2x^k+2-x^2k+3+1right)x^k-1$$
and
$$2x^k+2-x^2k+3+1=1+left(2-x^k+1right)x^k+2$$
If $x > 5$, and $k>1$, then $2-x^k+1<-1$, $x^k+2>1$, and $x^k-1>1$, so that $$-2+left(1+left(2-x^k+1right)x^k+2right)x^k-1<0$$
Informally,
$$-2+(1+(<-1)(>1))(>1)rightarrow-2+(1+(<-1))(>1)rightarrow-2+(<0)(>1)rightarrow-2+(<0)<0$$
answered 47 mins ago
HackerBoss
1515
New contributor
HackerBoss is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 47 mins ago
HackerBoss
1515
New contributor
HackerBoss is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 47 mins ago
HackerBoss
1515
answered 47 mins ago
HackerBoss
1515
1515
New contributor
HackerBoss is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
HackerBoss is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
HackerBoss is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
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@greedoid: No it is $k-1$.
– DER
1 hour ago
Any reason why it should be $x>5$ specifically? desmos.com/calculator/esk3my2jfc
– Mohammad Zuhair Khan
1 hour ago
That will likely become clear in the answer
– HackerBoss
1 hour ago
@HackerBoss I only meant that because it seems that for $k>1,$ $-2+2x^2k+1-x^3k+2+x^k-1$ is negative for all values $x>1.18.$
– Mohammad Zuhair Khan
57 mins ago
@MohammadZuhairKhan: How you get this result
– DER
56 mins ago