Estimate the remainder term of an unusual interpolation: $f(-1)=f(1)=f'(0)=f''(0)=0$
$begingroup$
Suppose that $f(x)in C^{4}[-1,1]$ and
$$f(-1)=f(1)=f'(0)=f''(0)=0$$
Show that for every $xin[-1,1]$, there exist a $xi_xin[-1,1]$, such that
$$f(x)=frac{x^4-1}{4!}f^{(4)}(xi_x)$$
where $f^{(4)}(x)$ denoted the forth derivative of $f(x)$.
The question is relevant to interpolation, for the only polynomial $p(x)$ satisfying
$$p(-1)=p(1)=p'(0)=p''(0)=0$$
is $x^4-1$ and the question gives an estimate of the remainder term of this interpolation.
I tried Taylor but failed because Taylor is a Hermite interpolation on one point but this question is not a pure Hermite interpolation.
real-analysis interpolation
asked Dec 11 '18 at 2:05
LauLau
517315
$endgroup$
add a comment |
$begingroup$
Suppose that $f(x)in C^{4}[-1,1]$ and
$$f(-1)=f(1)=f'(0)=f''(0)=0$$
Show that for every $xin[-1,1]$, there exist a $xi_xin[-1,1]$, such that
$$f(x)=frac{x^4-1}{4!}f^{(4)}(xi_x)$$
where $f^{(4)}(x)$ denoted the forth derivative of $f(x)$.
The question is relevant to interpolation, for the only polynomial $p(x)$ satisfying
$$p(-1)=p(1)=p'(0)=p''(0)=0$$
is $x^4-1$ and the question gives an estimate of the remainder term of this interpolation.
I tried Taylor but failed because Taylor is a Hermite interpolation on one point but this question is not a pure Hermite interpolation.
real-analysis interpolation
asked Dec 11 '18 at 2:05
LauLau
517315
$endgroup$
$begingroup$
Try using Rolle's Theorem on $f$ then $f'$ then $f''$ finally on $f^{(3)}$. You should also use the Mean Value Thoerem and the $cin[-1,1]$ should be $xi_x$
$endgroup$
– TheD0ubleT
Dec 11 '18 at 8:55
$begingroup$
@TheD0ubleT It doesn't work. Maybe you can post an answer if you've solved it.
$endgroup$
– Lau
Dec 18 '18 at 3:02
add a comment |
$begingroup$
Suppose that $f(x)in C^{4}[-1,1]$ and
$$f(-1)=f(1)=f'(0)=f''(0)=0$$
Show that for every $xin[-1,1]$, there exist a $xi_xin[-1,1]$, such that
$$f(x)=frac{x^4-1}{4!}f^{(4)}(xi_x)$$
where $f^{(4)}(x)$ denoted the forth derivative of $f(x)$.
The question is relevant to interpolation, for the only polynomial $p(x)$ satisfying
$$p(-1)=p(1)=p'(0)=p''(0)=0$$
is $x^4-1$ and the question gives an estimate of the remainder term of this interpolation.
I tried Taylor but failed because Taylor is a Hermite interpolation on one point but this question is not a pure Hermite interpolation.
real-analysis interpolation
asked Dec 11 '18 at 2:05
LauLau
517315
$endgroup$
Suppose that $f(x)in C^{4}[-1,1]$ and
$$f(-1)=f(1)=f'(0)=f''(0)=0$$
Show that for every $xin[-1,1]$, there exist a $xi_xin[-1,1]$, such that
$$f(x)=frac{x^4-1}{4!}f^{(4)}(xi_x)$$
where $f^{(4)}(x)$ denoted the forth derivative of $f(x)$.
The question is relevant to interpolation, for the only polynomial $p(x)$ satisfying
$$p(-1)=p(1)=p'(0)=p''(0)=0$$
is $x^4-1$ and the question gives an estimate of the remainder term of this interpolation.
I tried Taylor but failed because Taylor is a Hermite interpolation on one point but this question is not a pure Hermite interpolation.
real-analysis interpolation
real-analysis interpolation
asked Dec 11 '18 at 2:05
LauLau
517315
asked Dec 11 '18 at 2:05
LauLau
517315
edited Dec 11 '18 at 2:39
Lau
asked Dec 11 '18 at 2:05
LauLau
517315
asked Dec 11 '18 at 2:05
LauLau
517315
asked Dec 11 '18 at 2:05
LauLau
517315
517315
$begingroup$
Try using Rolle's Theorem on $f$ then $f'$ then $f''$ finally on $f^{(3)}$. You should also use the Mean Value Thoerem and the $cin[-1,1]$ should be $xi_x$
$endgroup$
– TheD0ubleT
Dec 11 '18 at 8:55
$begingroup$
@TheD0ubleT It doesn't work. Maybe you can post an answer if you've solved it.
$endgroup$
– Lau
Dec 18 '18 at 3:02
add a comment |
$begingroup$
Try using Rolle's Theorem on $f$ then $f'$ then $f''$ finally on $f^{(3)}$. You should also use the Mean Value Thoerem and the $cin[-1,1]$ should be $xi_x$
$endgroup$
– TheD0ubleT
Dec 11 '18 at 8:55
$begingroup$
@TheD0ubleT It doesn't work. Maybe you can post an answer if you've solved it.
$endgroup$
– Lau
Dec 18 '18 at 3:02
$begingroup$
Try using Rolle's Theorem on $f$ then $f'$ then $f''$ finally on $f^{(3)}$. You should also use the Mean Value Thoerem and the $cin[-1,1]$ should be $xi_x$
$endgroup$
– TheD0ubleT
Dec 11 '18 at 8:55
$begingroup$
Try using Rolle's Theorem on $f$ then $f'$ then $f''$ finally on $f^{(3)}$. You should also use the Mean Value Thoerem and the $cin[-1,1]$ should be $xi_x$
$endgroup$
– TheD0ubleT
Dec 11 '18 at 8:55
$begingroup$
@TheD0ubleT It doesn't work. Maybe you can post an answer if you've solved it.
$endgroup$
– Lau
Dec 18 '18 at 3:02
$begingroup$
@TheD0ubleT It doesn't work. Maybe you can post an answer if you've solved it.
$endgroup$
– Lau
Dec 18 '18 at 3:02
add a comment |
1 Answer
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$begingroup$
Emmm... I think I've worked it out on my own.
And this kind of interpolation is called Birkhoff interpolation.
For $x=1$ and $x=-1$, we're done. For $xin(-1,1)$, define
$$phi(t)=f(t)-frac{f(x)}{x^4-1}(t^4-1),quad tin[-1,1]$$
is a $C^4[-1,1]$ function and
$$phi(1)=phi(-1)=phi(x)=phi'(0)=phi''(0)=0$$
If $x=0$, then we can solve it by repeatedly using Rolle's Theorem.
If $xnotin{-1,0,1}$, say $xin(-1,0)$, by Rolle's Theorem:
There exist a $xi_1in(-1,0)$, such that $phi'(xi_1)=0$.
There exist a $xi_2in(xi_1,0)$, such that $phi''(xi_2)=0$.
There exist a $xi_3in(xi_2,0)$, such that $phi'''(xi_3)=0$.
Now, Consider the following claims:
(a) All nulls of $phi(t)$ are $-1,x,1$.
(b) All nulls of $phi'(t)$ are $xi_1,0$.
(c) All nulls of $phi''(t)$ are $xi_2,0$.
(d) All nulls of $phi'''(t)$ are $xi_3$.
If one of the claims is false, then we can reach the conclusion by repeatedly using Rolle's Theorem.
If all the claims are true, then by (b) $phi(t)$ is monotonic respectively on $(x,0)$ and $(0,1)$.
By (a), we know $phi(t)$ has different monotonicity on $(x,0)$ and $(0,1)$, so $phi'(t)$ changes sign at $0$.
By (c), $phi''(t)$ doesn't change sign at $0$, hence we know $phi'''(t)=0$, which contradicts to (d).
So we've done, the situation where $xin(0,1)$ is similar.
answered Dec 20 '18 at 1:54
LauLau
517315
$endgroup$
$begingroup$
Why should $phi(0)$ be zero? Is that a typo or does your proof rely on it?
$endgroup$
– Martin R
Dec 20 '18 at 19:34
$begingroup$
@MartinR Thanks to point it out. I've fixed it.
$endgroup$
– Lau
Dec 21 '18 at 1:58
add a comment |
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$begingroup$
Emmm... I think I've worked it out on my own.
And this kind of interpolation is called Birkhoff interpolation.
For $x=1$ and $x=-1$, we're done. For $xin(-1,1)$, define
$$phi(t)=f(t)-frac{f(x)}{x^4-1}(t^4-1),quad tin[-1,1]$$
is a $C^4[-1,1]$ function and
$$phi(1)=phi(-1)=phi(x)=phi'(0)=phi''(0)=0$$
If $x=0$, then we can solve it by repeatedly using Rolle's Theorem.
If $xnotin{-1,0,1}$, say $xin(-1,0)$, by Rolle's Theorem:
There exist a $xi_1in(-1,0)$, such that $phi'(xi_1)=0$.
There exist a $xi_2in(xi_1,0)$, such that $phi''(xi_2)=0$.
There exist a $xi_3in(xi_2,0)$, such that $phi'''(xi_3)=0$.
Now, Consider the following claims:
(a) All nulls of $phi(t)$ are $-1,x,1$.
(b) All nulls of $phi'(t)$ are $xi_1,0$.
(c) All nulls of $phi''(t)$ are $xi_2,0$.
(d) All nulls of $phi'''(t)$ are $xi_3$.
If one of the claims is false, then we can reach the conclusion by repeatedly using Rolle's Theorem.
If all the claims are true, then by (b) $phi(t)$ is monotonic respectively on $(x,0)$ and $(0,1)$.
By (a), we know $phi(t)$ has different monotonicity on $(x,0)$ and $(0,1)$, so $phi'(t)$ changes sign at $0$.
By (c), $phi''(t)$ doesn't change sign at $0$, hence we know $phi'''(t)=0$, which contradicts to (d).
So we've done, the situation where $xin(0,1)$ is similar.
answered Dec 20 '18 at 1:54
LauLau
517315
$endgroup$
$begingroup$
Why should $phi(0)$ be zero? Is that a typo or does your proof rely on it?
$endgroup$
– Martin R
Dec 20 '18 at 19:34
$begingroup$
@MartinR Thanks to point it out. I've fixed it.
$endgroup$
– Lau
Dec 21 '18 at 1:58
add a comment |
$begingroup$
Emmm... I think I've worked it out on my own.
And this kind of interpolation is called Birkhoff interpolation.
For $x=1$ and $x=-1$, we're done. For $xin(-1,1)$, define
$$phi(t)=f(t)-frac{f(x)}{x^4-1}(t^4-1),quad tin[-1,1]$$
is a $C^4[-1,1]$ function and
$$phi(1)=phi(-1)=phi(x)=phi'(0)=phi''(0)=0$$
If $x=0$, then we can solve it by repeatedly using Rolle's Theorem.
If $xnotin{-1,0,1}$, say $xin(-1,0)$, by Rolle's Theorem:
There exist a $xi_1in(-1,0)$, such that $phi'(xi_1)=0$.
There exist a $xi_2in(xi_1,0)$, such that $phi''(xi_2)=0$.
There exist a $xi_3in(xi_2,0)$, such that $phi'''(xi_3)=0$.
Now, Consider the following claims:
(a) All nulls of $phi(t)$ are $-1,x,1$.
(b) All nulls of $phi'(t)$ are $xi_1,0$.
(c) All nulls of $phi''(t)$ are $xi_2,0$.
(d) All nulls of $phi'''(t)$ are $xi_3$.
If one of the claims is false, then we can reach the conclusion by repeatedly using Rolle's Theorem.
If all the claims are true, then by (b) $phi(t)$ is monotonic respectively on $(x,0)$ and $(0,1)$.
By (a), we know $phi(t)$ has different monotonicity on $(x,0)$ and $(0,1)$, so $phi'(t)$ changes sign at $0$.
By (c), $phi''(t)$ doesn't change sign at $0$, hence we know $phi'''(t)=0$, which contradicts to (d).
So we've done, the situation where $xin(0,1)$ is similar.
answered Dec 20 '18 at 1:54
LauLau
517315
$endgroup$
$begingroup$
Why should $phi(0)$ be zero? Is that a typo or does your proof rely on it?
$endgroup$
– Martin R
Dec 20 '18 at 19:34
$begingroup$
@MartinR Thanks to point it out. I've fixed it.
$endgroup$
– Lau
Dec 21 '18 at 1:58
add a comment |
$begingroup$
Emmm... I think I've worked it out on my own.
And this kind of interpolation is called Birkhoff interpolation.
For $x=1$ and $x=-1$, we're done. For $xin(-1,1)$, define
$$phi(t)=f(t)-frac{f(x)}{x^4-1}(t^4-1),quad tin[-1,1]$$
is a $C^4[-1,1]$ function and
$$phi(1)=phi(-1)=phi(x)=phi'(0)=phi''(0)=0$$
If $x=0$, then we can solve it by repeatedly using Rolle's Theorem.
If $xnotin{-1,0,1}$, say $xin(-1,0)$, by Rolle's Theorem:
There exist a $xi_1in(-1,0)$, such that $phi'(xi_1)=0$.
There exist a $xi_2in(xi_1,0)$, such that $phi''(xi_2)=0$.
There exist a $xi_3in(xi_2,0)$, such that $phi'''(xi_3)=0$.
Now, Consider the following claims:
(a) All nulls of $phi(t)$ are $-1,x,1$.
(b) All nulls of $phi'(t)$ are $xi_1,0$.
(c) All nulls of $phi''(t)$ are $xi_2,0$.
(d) All nulls of $phi'''(t)$ are $xi_3$.
If one of the claims is false, then we can reach the conclusion by repeatedly using Rolle's Theorem.
If all the claims are true, then by (b) $phi(t)$ is monotonic respectively on $(x,0)$ and $(0,1)$.
By (a), we know $phi(t)$ has different monotonicity on $(x,0)$ and $(0,1)$, so $phi'(t)$ changes sign at $0$.
By (c), $phi''(t)$ doesn't change sign at $0$, hence we know $phi'''(t)=0$, which contradicts to (d).
So we've done, the situation where $xin(0,1)$ is similar.
answered Dec 20 '18 at 1:54
LauLau
517315
$endgroup$
Emmm... I think I've worked it out on my own.
And this kind of interpolation is called Birkhoff interpolation.
For $x=1$ and $x=-1$, we're done. For $xin(-1,1)$, define
$$phi(t)=f(t)-frac{f(x)}{x^4-1}(t^4-1),quad tin[-1,1]$$
is a $C^4[-1,1]$ function and
$$phi(1)=phi(-1)=phi(x)=phi'(0)=phi''(0)=0$$
If $x=0$, then we can solve it by repeatedly using Rolle's Theorem.
If $xnotin{-1,0,1}$, say $xin(-1,0)$, by Rolle's Theorem:
There exist a $xi_1in(-1,0)$, such that $phi'(xi_1)=0$.
There exist a $xi_2in(xi_1,0)$, such that $phi''(xi_2)=0$.
There exist a $xi_3in(xi_2,0)$, such that $phi'''(xi_3)=0$.
Now, Consider the following claims:
(a) All nulls of $phi(t)$ are $-1,x,1$.
(b) All nulls of $phi'(t)$ are $xi_1,0$.
(c) All nulls of $phi''(t)$ are $xi_2,0$.
(d) All nulls of $phi'''(t)$ are $xi_3$.
If one of the claims is false, then we can reach the conclusion by repeatedly using Rolle's Theorem.
If all the claims are true, then by (b) $phi(t)$ is monotonic respectively on $(x,0)$ and $(0,1)$.
By (a), we know $phi(t)$ has different monotonicity on $(x,0)$ and $(0,1)$, so $phi'(t)$ changes sign at $0$.
By (c), $phi''(t)$ doesn't change sign at $0$, hence we know $phi'''(t)=0$, which contradicts to (d).
So we've done, the situation where $xin(0,1)$ is similar.
answered Dec 20 '18 at 1:54
LauLau
517315
edited Dec 21 '18 at 1:57
answered Dec 20 '18 at 1:54
LauLau
517315
answered Dec 20 '18 at 1:54
LauLau
517315
answered Dec 20 '18 at 1:54
LauLau
517315
517315
$begingroup$
Why should $phi(0)$ be zero? Is that a typo or does your proof rely on it?
$endgroup$
– Martin R
Dec 20 '18 at 19:34
$begingroup$
@MartinR Thanks to point it out. I've fixed it.
$endgroup$
– Lau
Dec 21 '18 at 1:58
add a comment |
$begingroup$
Why should $phi(0)$ be zero? Is that a typo or does your proof rely on it?
$endgroup$
– Martin R
Dec 20 '18 at 19:34
$begingroup$
@MartinR Thanks to point it out. I've fixed it.
$endgroup$
– Lau
Dec 21 '18 at 1:58
$begingroup$
Why should $phi(0)$ be zero? Is that a typo or does your proof rely on it?
$endgroup$
– Martin R
Dec 20 '18 at 19:34
$begingroup$
Why should $phi(0)$ be zero? Is that a typo or does your proof rely on it?
$endgroup$
– Martin R
Dec 20 '18 at 19:34
$begingroup$
@MartinR Thanks to point it out. I've fixed it.
$endgroup$
– Lau
Dec 21 '18 at 1:58
$begingroup$
@MartinR Thanks to point it out. I've fixed it.
$endgroup$
– Lau
Dec 21 '18 at 1:58
add a comment |
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$begingroup$
Try using Rolle's Theorem on $f$ then $f'$ then $f''$ finally on $f^{(3)}$. You should also use the Mean Value Thoerem and the $cin[-1,1]$ should be $xi_x$
$endgroup$
– TheD0ubleT
Dec 11 '18 at 8:55
$begingroup$
@TheD0ubleT It doesn't work. Maybe you can post an answer if you've solved it.
$endgroup$
– Lau
Dec 18 '18 at 3:02