Minimum value of The polynomial
$begingroup$
What is the minimum value of the expression given below?
$ x^8-8x^6+19x^4-12x^3+14x^2-8x+9$
Now to solve this I have resolved the expression, like following,
$ (x^2+2x)^2.(x^2-2x)^2+3.(x^2-2x)^2+2(x-2)^2+1$, from this expression it is easy to visualize that it has a minimum value of 1 for $ (x-2)^2=0, x=2 $
Now my question is, is it a right approach or will it fit for this kind of problem? Please suggest if there is any other option because analyzing like this is very tedious, and fault prone.What can I do for very large polynomials?
Is there any particular process to solve this.
Need your help... Thanks
polynomials
asked Nov 29 '13 at 15:14
SouravSourav
1217
$endgroup$
add a comment |
$begingroup$
What is the minimum value of the expression given below?
$ x^8-8x^6+19x^4-12x^3+14x^2-8x+9$
Now to solve this I have resolved the expression, like following,
$ (x^2+2x)^2.(x^2-2x)^2+3.(x^2-2x)^2+2(x-2)^2+1$, from this expression it is easy to visualize that it has a minimum value of 1 for $ (x-2)^2=0, x=2 $
Now my question is, is it a right approach or will it fit for this kind of problem? Please suggest if there is any other option because analyzing like this is very tedious, and fault prone.What can I do for very large polynomials?
Is there any particular process to solve this.
Need your help... Thanks
polynomials
asked Nov 29 '13 at 15:14
SouravSourav
1217
$endgroup$
add a comment |
$begingroup$
What is the minimum value of the expression given below?
$ x^8-8x^6+19x^4-12x^3+14x^2-8x+9$
Now to solve this I have resolved the expression, like following,
$ (x^2+2x)^2.(x^2-2x)^2+3.(x^2-2x)^2+2(x-2)^2+1$, from this expression it is easy to visualize that it has a minimum value of 1 for $ (x-2)^2=0, x=2 $
Now my question is, is it a right approach or will it fit for this kind of problem? Please suggest if there is any other option because analyzing like this is very tedious, and fault prone.What can I do for very large polynomials?
Is there any particular process to solve this.
Need your help... Thanks
polynomials
asked Nov 29 '13 at 15:14
SouravSourav
1217
$endgroup$
What is the minimum value of the expression given below?
$ x^8-8x^6+19x^4-12x^3+14x^2-8x+9$
Now to solve this I have resolved the expression, like following,
$ (x^2+2x)^2.(x^2-2x)^2+3.(x^2-2x)^2+2(x-2)^2+1$, from this expression it is easy to visualize that it has a minimum value of 1 for $ (x-2)^2=0, x=2 $
Now my question is, is it a right approach or will it fit for this kind of problem? Please suggest if there is any other option because analyzing like this is very tedious, and fault prone.What can I do for very large polynomials?
Is there any particular process to solve this.
Need your help... Thanks
polynomials
polynomials
asked Nov 29 '13 at 15:14
SouravSourav
1217
asked Nov 29 '13 at 15:14
SouravSourav
1217
asked Nov 29 '13 at 15:14
SouravSourav
1217
asked Nov 29 '13 at 15:14
SouravSourav
1217
asked Nov 29 '13 at 15:14
SouravSourav
1217
1217
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can also write your polynomial as this:
$$1+(x-2)^2 left[2+x^2 left(3+x^2 (x+2)^2right)right]$$
And it's immediate that 2 must be a minimum, since the part inside square brackets is always positive.
But this approach or yours is equally unfit for a general polynomial: here it works because it's a special case. If you can find such a form, yes, it works, but you will only find this by luck, on certain polynomials only. An in general, the minimum, being a root of the derivative, may be impossible to find in closed form (and even whet it's possible, for lower degree, it may be overly complicated).
Usually, you have to study the variations of your function: differentiate, study sign. And if necessary, do the same on the derivative, so differentiate again, and study sign, etc. Tedious, but unless you guess something, it's your best choice.
answered Nov 29 '13 at 15:28
Jean-Claude ArbautJean-Claude Arbaut
14.8k63464
$endgroup$
add a comment |
$begingroup$
The general approach in this kind of problems is use derivative. What you shall do is $f'(x)=0$ and $f''(x)>0$
answered Jun 24 '17 at 8:23
MatMorPau22MatMorPau22
342111
$endgroup$
1
$begingroup$
Good luck finding the roots of a polynomial of degree $7$. For the polynomial given by the OP, $x=2$ is a root that may be easy to find, but there are two other real roots that have no easy representation, and you still have to decide which one corresponds to the global minimum.
$endgroup$
– lhf
Jun 24 '17 at 11:45
add a comment |
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2 Answers
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active
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2 Answers
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$begingroup$
You can also write your polynomial as this:
$$1+(x-2)^2 left[2+x^2 left(3+x^2 (x+2)^2right)right]$$
And it's immediate that 2 must be a minimum, since the part inside square brackets is always positive.
But this approach or yours is equally unfit for a general polynomial: here it works because it's a special case. If you can find such a form, yes, it works, but you will only find this by luck, on certain polynomials only. An in general, the minimum, being a root of the derivative, may be impossible to find in closed form (and even whet it's possible, for lower degree, it may be overly complicated).
Usually, you have to study the variations of your function: differentiate, study sign. And if necessary, do the same on the derivative, so differentiate again, and study sign, etc. Tedious, but unless you guess something, it's your best choice.
answered Nov 29 '13 at 15:28
Jean-Claude ArbautJean-Claude Arbaut
14.8k63464
$endgroup$
add a comment |
$begingroup$
You can also write your polynomial as this:
$$1+(x-2)^2 left[2+x^2 left(3+x^2 (x+2)^2right)right]$$
And it's immediate that 2 must be a minimum, since the part inside square brackets is always positive.
But this approach or yours is equally unfit for a general polynomial: here it works because it's a special case. If you can find such a form, yes, it works, but you will only find this by luck, on certain polynomials only. An in general, the minimum, being a root of the derivative, may be impossible to find in closed form (and even whet it's possible, for lower degree, it may be overly complicated).
Usually, you have to study the variations of your function: differentiate, study sign. And if necessary, do the same on the derivative, so differentiate again, and study sign, etc. Tedious, but unless you guess something, it's your best choice.
answered Nov 29 '13 at 15:28
Jean-Claude ArbautJean-Claude Arbaut
14.8k63464
$endgroup$
add a comment |
$begingroup$
You can also write your polynomial as this:
$$1+(x-2)^2 left[2+x^2 left(3+x^2 (x+2)^2right)right]$$
And it's immediate that 2 must be a minimum, since the part inside square brackets is always positive.
But this approach or yours is equally unfit for a general polynomial: here it works because it's a special case. If you can find such a form, yes, it works, but you will only find this by luck, on certain polynomials only. An in general, the minimum, being a root of the derivative, may be impossible to find in closed form (and even whet it's possible, for lower degree, it may be overly complicated).
Usually, you have to study the variations of your function: differentiate, study sign. And if necessary, do the same on the derivative, so differentiate again, and study sign, etc. Tedious, but unless you guess something, it's your best choice.
answered Nov 29 '13 at 15:28
Jean-Claude ArbautJean-Claude Arbaut
14.8k63464
$endgroup$
You can also write your polynomial as this:
$$1+(x-2)^2 left[2+x^2 left(3+x^2 (x+2)^2right)right]$$
And it's immediate that 2 must be a minimum, since the part inside square brackets is always positive.
But this approach or yours is equally unfit for a general polynomial: here it works because it's a special case. If you can find such a form, yes, it works, but you will only find this by luck, on certain polynomials only. An in general, the minimum, being a root of the derivative, may be impossible to find in closed form (and even whet it's possible, for lower degree, it may be overly complicated).
Usually, you have to study the variations of your function: differentiate, study sign. And if necessary, do the same on the derivative, so differentiate again, and study sign, etc. Tedious, but unless you guess something, it's your best choice.
answered Nov 29 '13 at 15:28
Jean-Claude ArbautJean-Claude Arbaut
14.8k63464
edited Nov 29 '13 at 16:28
answered Nov 29 '13 at 15:28
Jean-Claude ArbautJean-Claude Arbaut
14.8k63464
answered Nov 29 '13 at 15:28
Jean-Claude ArbautJean-Claude Arbaut
14.8k63464
answered Nov 29 '13 at 15:28
Jean-Claude ArbautJean-Claude Arbaut
14.8k63464
14.8k63464
add a comment |
add a comment |
$begingroup$
The general approach in this kind of problems is use derivative. What you shall do is $f'(x)=0$ and $f''(x)>0$
answered Jun 24 '17 at 8:23
MatMorPau22MatMorPau22
342111
$endgroup$
1
$begingroup$
Good luck finding the roots of a polynomial of degree $7$. For the polynomial given by the OP, $x=2$ is a root that may be easy to find, but there are two other real roots that have no easy representation, and you still have to decide which one corresponds to the global minimum.
$endgroup$
– lhf
Jun 24 '17 at 11:45
add a comment |
$begingroup$
The general approach in this kind of problems is use derivative. What you shall do is $f'(x)=0$ and $f''(x)>0$
answered Jun 24 '17 at 8:23
MatMorPau22MatMorPau22
342111
$endgroup$
1
$begingroup$
Good luck finding the roots of a polynomial of degree $7$. For the polynomial given by the OP, $x=2$ is a root that may be easy to find, but there are two other real roots that have no easy representation, and you still have to decide which one corresponds to the global minimum.
$endgroup$
– lhf
Jun 24 '17 at 11:45
add a comment |
$begingroup$
The general approach in this kind of problems is use derivative. What you shall do is $f'(x)=0$ and $f''(x)>0$
answered Jun 24 '17 at 8:23
MatMorPau22MatMorPau22
342111
$endgroup$
The general approach in this kind of problems is use derivative. What you shall do is $f'(x)=0$ and $f''(x)>0$
answered Jun 24 '17 at 8:23
MatMorPau22MatMorPau22
342111
answered Jun 24 '17 at 8:23
MatMorPau22MatMorPau22
342111
answered Jun 24 '17 at 8:23
MatMorPau22MatMorPau22
342111
answered Jun 24 '17 at 8:23
MatMorPau22MatMorPau22
342111
342111
1
$begingroup$
Good luck finding the roots of a polynomial of degree $7$. For the polynomial given by the OP, $x=2$ is a root that may be easy to find, but there are two other real roots that have no easy representation, and you still have to decide which one corresponds to the global minimum.
$endgroup$
– lhf
Jun 24 '17 at 11:45
add a comment |
1
$begingroup$
Good luck finding the roots of a polynomial of degree $7$. For the polynomial given by the OP, $x=2$ is a root that may be easy to find, but there are two other real roots that have no easy representation, and you still have to decide which one corresponds to the global minimum.
$endgroup$
– lhf
Jun 24 '17 at 11:45
1
1
$begingroup$
Good luck finding the roots of a polynomial of degree $7$. For the polynomial given by the OP, $x=2$ is a root that may be easy to find, but there are two other real roots that have no easy representation, and you still have to decide which one corresponds to the global minimum.
$endgroup$
– lhf
Jun 24 '17 at 11:45
$begingroup$
Good luck finding the roots of a polynomial of degree $7$. For the polynomial given by the OP, $x=2$ is a root that may be easy to find, but there are two other real roots that have no easy representation, and you still have to decide which one corresponds to the global minimum.
$endgroup$
– lhf
Jun 24 '17 at 11:45
add a comment |
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