Determining if $x=c$ is an extreme point of a function $f:mathbb{R}tomathbb{R}$, given its Taylor polynomial...
$begingroup$
Let $f:mathbb{R}tomathbb{R}$ be a function such that $fin C^3(mathbb{R})$. Its quadratic Taylor polynomial at $x=1$ is $5-4(x-1)-(x-1)^2$. Prove/disprove: $x=0$ is not an extreme point of $f$.
Hey everyone. I'm trying to brush up on my calculus. I've come across this simple question and I'm not sure how to approach it. I've been looking into posts about Taylor polynomials and its applications, in regards to extreme points, but I haven't found anything helpful with this question.
Does the function $f$ necessarily share its extreme points with its Taylor polynomial of order $k$? I have some counter examples so no. I thought this was a classic disproof question, but I'm not sure how to construct such function.
I would be very happy to hear your thoughts since I'm lost. Thank you
real-analysis calculus taylor-expansion
asked Jan 20 at 23:52
Noy PerelNoy Perel
478213
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb{R}tomathbb{R}$ be a function such that $fin C^3(mathbb{R})$. Its quadratic Taylor polynomial at $x=1$ is $5-4(x-1)-(x-1)^2$. Prove/disprove: $x=0$ is not an extreme point of $f$.
Hey everyone. I'm trying to brush up on my calculus. I've come across this simple question and I'm not sure how to approach it. I've been looking into posts about Taylor polynomials and its applications, in regards to extreme points, but I haven't found anything helpful with this question.
Does the function $f$ necessarily share its extreme points with its Taylor polynomial of order $k$? I have some counter examples so no. I thought this was a classic disproof question, but I'm not sure how to construct such function.
I would be very happy to hear your thoughts since I'm lost. Thank you
real-analysis calculus taylor-expansion
asked Jan 20 at 23:52
Noy PerelNoy Perel
478213
$endgroup$
$begingroup$
Definitely not.
$endgroup$
– William Elliot
Jan 21 at 3:12
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@WilliamElliot Can you please give me a clue on how to disprove this? thank you
$endgroup$
– Noy Perel
Jan 21 at 6:49
$begingroup$
A counterexample is sufficient.
$endgroup$
– Raskolnikov
Jan 21 at 6:51
$begingroup$
$x = 0$ is not even an extreme point of the Taylor polynomial! And even if it would be: your last sentence is not true -- just add something like $(x-1)^4$ to the Taylor polynomial to obtain a polynomial which does not posses an extreme point in $x = 0$.
$endgroup$
– gerw
Jan 21 at 6:56
1
$begingroup$
@gerw I know, that would be a counter-example for the statement that says $f$ has got an extreme point at $x=0$, but I’m required to give a counterexample for the claim that $f$ hasn’t got an extreme point at $x=0$..
$endgroup$
– Noy Perel
Jan 21 at 9:26
add a comment |
$begingroup$
Let $f:mathbb{R}tomathbb{R}$ be a function such that $fin C^3(mathbb{R})$. Its quadratic Taylor polynomial at $x=1$ is $5-4(x-1)-(x-1)^2$. Prove/disprove: $x=0$ is not an extreme point of $f$.
Hey everyone. I'm trying to brush up on my calculus. I've come across this simple question and I'm not sure how to approach it. I've been looking into posts about Taylor polynomials and its applications, in regards to extreme points, but I haven't found anything helpful with this question.
Does the function $f$ necessarily share its extreme points with its Taylor polynomial of order $k$? I have some counter examples so no. I thought this was a classic disproof question, but I'm not sure how to construct such function.
I would be very happy to hear your thoughts since I'm lost. Thank you
real-analysis calculus taylor-expansion
asked Jan 20 at 23:52
Noy PerelNoy Perel
478213
$endgroup$
Let $f:mathbb{R}tomathbb{R}$ be a function such that $fin C^3(mathbb{R})$. Its quadratic Taylor polynomial at $x=1$ is $5-4(x-1)-(x-1)^2$. Prove/disprove: $x=0$ is not an extreme point of $f$.
Hey everyone. I'm trying to brush up on my calculus. I've come across this simple question and I'm not sure how to approach it. I've been looking into posts about Taylor polynomials and its applications, in regards to extreme points, but I haven't found anything helpful with this question.
Does the function $f$ necessarily share its extreme points with its Taylor polynomial of order $k$? I have some counter examples so no. I thought this was a classic disproof question, but I'm not sure how to construct such function.
I would be very happy to hear your thoughts since I'm lost. Thank you
real-analysis calculus taylor-expansion
real-analysis calculus taylor-expansion
asked Jan 20 at 23:52
Noy PerelNoy Perel
478213
asked Jan 20 at 23:52
Noy PerelNoy Perel
478213
edited Jan 21 at 9:30
Noy Perel
asked Jan 20 at 23:52
Noy PerelNoy Perel
478213
asked Jan 20 at 23:52
Noy PerelNoy Perel
478213
asked Jan 20 at 23:52
Noy PerelNoy Perel
478213
478213
$begingroup$
Definitely not.
$endgroup$
– William Elliot
Jan 21 at 3:12
$begingroup$
@WilliamElliot Can you please give me a clue on how to disprove this? thank you
$endgroup$
– Noy Perel
Jan 21 at 6:49
$begingroup$
A counterexample is sufficient.
$endgroup$
– Raskolnikov
Jan 21 at 6:51
$begingroup$
$x = 0$ is not even an extreme point of the Taylor polynomial! And even if it would be: your last sentence is not true -- just add something like $(x-1)^4$ to the Taylor polynomial to obtain a polynomial which does not posses an extreme point in $x = 0$.
$endgroup$
– gerw
Jan 21 at 6:56
1
$begingroup$
@gerw I know, that would be a counter-example for the statement that says $f$ has got an extreme point at $x=0$, but I’m required to give a counterexample for the claim that $f$ hasn’t got an extreme point at $x=0$..
$endgroup$
– Noy Perel
Jan 21 at 9:26
add a comment |
$begingroup$
Definitely not.
$endgroup$
– William Elliot
Jan 21 at 3:12
$begingroup$
@WilliamElliot Can you please give me a clue on how to disprove this? thank you
$endgroup$
– Noy Perel
Jan 21 at 6:49
$begingroup$
A counterexample is sufficient.
$endgroup$
– Raskolnikov
Jan 21 at 6:51
$begingroup$
$x = 0$ is not even an extreme point of the Taylor polynomial! And even if it would be: your last sentence is not true -- just add something like $(x-1)^4$ to the Taylor polynomial to obtain a polynomial which does not posses an extreme point in $x = 0$.
$endgroup$
– gerw
Jan 21 at 6:56
1
$begingroup$
@gerw I know, that would be a counter-example for the statement that says $f$ has got an extreme point at $x=0$, but I’m required to give a counterexample for the claim that $f$ hasn’t got an extreme point at $x=0$..
$endgroup$
– Noy Perel
Jan 21 at 9:26
$begingroup$
Definitely not.
$endgroup$
– William Elliot
Jan 21 at 3:12
$begingroup$
Definitely not.
$endgroup$
– William Elliot
Jan 21 at 3:12
$begingroup$
@WilliamElliot Can you please give me a clue on how to disprove this? thank you
$endgroup$
– Noy Perel
Jan 21 at 6:49
$begingroup$
@WilliamElliot Can you please give me a clue on how to disprove this? thank you
$endgroup$
– Noy Perel
Jan 21 at 6:49
$begingroup$
A counterexample is sufficient.
$endgroup$
– Raskolnikov
Jan 21 at 6:51
$begingroup$
A counterexample is sufficient.
$endgroup$
– Raskolnikov
Jan 21 at 6:51
$begingroup$
$x = 0$ is not even an extreme point of the Taylor polynomial! And even if it would be: your last sentence is not true -- just add something like $(x-1)^4$ to the Taylor polynomial to obtain a polynomial which does not posses an extreme point in $x = 0$.
$endgroup$
– gerw
Jan 21 at 6:56
$begingroup$
$x = 0$ is not even an extreme point of the Taylor polynomial! And even if it would be: your last sentence is not true -- just add something like $(x-1)^4$ to the Taylor polynomial to obtain a polynomial which does not posses an extreme point in $x = 0$.
$endgroup$
– gerw
Jan 21 at 6:56
1
1
$begingroup$
@gerw I know, that would be a counter-example for the statement that says $f$ has got an extreme point at $x=0$, but I’m required to give a counterexample for the claim that $f$ hasn’t got an extreme point at $x=0$..
$endgroup$
– Noy Perel
Jan 21 at 9:26
$begingroup$
@gerw I know, that would be a counter-example for the statement that says $f$ has got an extreme point at $x=0$, but I’m required to give a counterexample for the claim that $f$ hasn’t got an extreme point at $x=0$..
$endgroup$
– Noy Perel
Jan 21 at 9:26
add a comment |
1 Answer
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Consider the following polynomial function:
$$f:mathbb{R}to mathbb{R}:x mapsto 5-4left(x-1right)-left(x-1right)^2+frac{2}{3}left(x-1right)^3 ; .$$
This clearly has the required second order Taylor polynomial and it also possesses an extremum in $x=0$.
answered Jan 21 at 7:04
RaskolnikovRaskolnikov
12.6k23571
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1 Answer
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1 Answer
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active
oldest
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oldest
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$begingroup$
Consider the following polynomial function:
$$f:mathbb{R}to mathbb{R}:x mapsto 5-4left(x-1right)-left(x-1right)^2+frac{2}{3}left(x-1right)^3 ; .$$
This clearly has the required second order Taylor polynomial and it also possesses an extremum in $x=0$.
answered Jan 21 at 7:04
RaskolnikovRaskolnikov
12.6k23571
$endgroup$
add a comment |
$begingroup$
Consider the following polynomial function:
$$f:mathbb{R}to mathbb{R}:x mapsto 5-4left(x-1right)-left(x-1right)^2+frac{2}{3}left(x-1right)^3 ; .$$
This clearly has the required second order Taylor polynomial and it also possesses an extremum in $x=0$.
answered Jan 21 at 7:04
RaskolnikovRaskolnikov
12.6k23571
$endgroup$
add a comment |
$begingroup$
Consider the following polynomial function:
$$f:mathbb{R}to mathbb{R}:x mapsto 5-4left(x-1right)-left(x-1right)^2+frac{2}{3}left(x-1right)^3 ; .$$
This clearly has the required second order Taylor polynomial and it also possesses an extremum in $x=0$.
answered Jan 21 at 7:04
RaskolnikovRaskolnikov
12.6k23571
$endgroup$
Consider the following polynomial function:
$$f:mathbb{R}to mathbb{R}:x mapsto 5-4left(x-1right)-left(x-1right)^2+frac{2}{3}left(x-1right)^3 ; .$$
This clearly has the required second order Taylor polynomial and it also possesses an extremum in $x=0$.
answered Jan 21 at 7:04
RaskolnikovRaskolnikov
12.6k23571
answered Jan 21 at 7:04
RaskolnikovRaskolnikov
12.6k23571
answered Jan 21 at 7:04
RaskolnikovRaskolnikov
12.6k23571
answered Jan 21 at 7:04
RaskolnikovRaskolnikov
12.6k23571
12.6k23571
add a comment |
add a comment |
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$begingroup$
Definitely not.
$endgroup$
– William Elliot
Jan 21 at 3:12
$begingroup$
@WilliamElliot Can you please give me a clue on how to disprove this? thank you
$endgroup$
– Noy Perel
Jan 21 at 6:49
$begingroup$
A counterexample is sufficient.
$endgroup$
– Raskolnikov
Jan 21 at 6:51
$begingroup$
$x = 0$ is not even an extreme point of the Taylor polynomial! And even if it would be: your last sentence is not true -- just add something like $(x-1)^4$ to the Taylor polynomial to obtain a polynomial which does not posses an extreme point in $x = 0$.
$endgroup$
– gerw
Jan 21 at 6:56
1
$begingroup$
@gerw I know, that would be a counter-example for the statement that says $f$ has got an extreme point at $x=0$, but I’m required to give a counterexample for the claim that $f$ hasn’t got an extreme point at $x=0$..
$endgroup$
– Noy Perel
Jan 21 at 9:26