Show that $f(x)-P(x)=frac{x^4-1}{4!}f^{4}(c) $
$begingroup$
If $P(x)$ is a unique cubic polynomial for which $P(x_0)=f(x_0),P(x_2)=f(x_2),P^{'}(x_1)=f^{'}(x_1),P^{''}(x_1)=f^{''}(x_1)$,$f(x)$ is a given function differentiable $4$ times.
Show that $f(x)-P(x)=dfrac{x^4-1}{4!}f^{4}(c) $ where $x_0=-1,x_1=0,x_2=1,cin (-1,1) $
MY TRY:
Let us assume that $P(x)=a_0+a_1x+a_2x^2+a_3x^3.$
Then $$begin{aligned}P(-1)=a_0-a_1+a_2-a_3&=f(-1)quad &(1)\
P(1)=a_0+a_1+a_2+a_3&=f(1) quad &(2)\
P^{'}(0)=a_1&=f^{'}(0)quad &(3)\
P^{''}(0)=2a_2&=f^{''}(0)quad &(4)
end{aligned}$$
How can I prove the given fact using the $4$ relations I have found here?
Please help.
real-analysis functions derivatives polynomials
$endgroup$
add a comment |
$begingroup$
If $P(x)$ is a unique cubic polynomial for which $P(x_0)=f(x_0),P(x_2)=f(x_2),P^{'}(x_1)=f^{'}(x_1),P^{''}(x_1)=f^{''}(x_1)$,$f(x)$ is a given function differentiable $4$ times.
Show that $f(x)-P(x)=dfrac{x^4-1}{4!}f^{4}(c) $ where $x_0=-1,x_1=0,x_2=1,cin (-1,1) $
MY TRY:
Let us assume that $P(x)=a_0+a_1x+a_2x^2+a_3x^3.$
Then $$begin{aligned}P(-1)=a_0-a_1+a_2-a_3&=f(-1)quad &(1)\
P(1)=a_0+a_1+a_2+a_3&=f(1) quad &(2)\
P^{'}(0)=a_1&=f^{'}(0)quad &(3)\
P^{''}(0)=2a_2&=f^{''}(0)quad &(4)
end{aligned}$$
How can I prove the given fact using the $4$ relations I have found here?
Please help.
real-analysis functions derivatives polynomials
$endgroup$
$begingroup$
You presumably have some typo in the last line of the box, I guess $x_0=-1,x_1=0,x_2=1$? Also presumably $P''(x)=f''(x)$ should also only be at one point, but you seem to have sorted that out in the main body of the post.
$endgroup$
– Ian
Dec 16 '18 at 15:13
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@Ian,Thanks for the info.I edited it
$endgroup$
– user596656
Dec 17 '18 at 7:21
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you've got typos.. $x_0,x_1,x_2$ do not correspond to the ones in the problem and equations 1 thru 4.
$endgroup$
– Ahmad Bazzi
Dec 17 '18 at 7:24
$begingroup$
@AhmadBazzi;I think it is fixed now
$endgroup$
– user596656
Dec 17 '18 at 8:26
add a comment |
$begingroup$
If $P(x)$ is a unique cubic polynomial for which $P(x_0)=f(x_0),P(x_2)=f(x_2),P^{'}(x_1)=f^{'}(x_1),P^{''}(x_1)=f^{''}(x_1)$,$f(x)$ is a given function differentiable $4$ times.
Show that $f(x)-P(x)=dfrac{x^4-1}{4!}f^{4}(c) $ where $x_0=-1,x_1=0,x_2=1,cin (-1,1) $
MY TRY:
Let us assume that $P(x)=a_0+a_1x+a_2x^2+a_3x^3.$
Then $$begin{aligned}P(-1)=a_0-a_1+a_2-a_3&=f(-1)quad &(1)\
P(1)=a_0+a_1+a_2+a_3&=f(1) quad &(2)\
P^{'}(0)=a_1&=f^{'}(0)quad &(3)\
P^{''}(0)=2a_2&=f^{''}(0)quad &(4)
end{aligned}$$
How can I prove the given fact using the $4$ relations I have found here?
Please help.
real-analysis functions derivatives polynomials
$endgroup$
If $P(x)$ is a unique cubic polynomial for which $P(x_0)=f(x_0),P(x_2)=f(x_2),P^{'}(x_1)=f^{'}(x_1),P^{''}(x_1)=f^{''}(x_1)$,$f(x)$ is a given function differentiable $4$ times.
Show that $f(x)-P(x)=dfrac{x^4-1}{4!}f^{4}(c) $ where $x_0=-1,x_1=0,x_2=1,cin (-1,1) $
MY TRY:
Let us assume that $P(x)=a_0+a_1x+a_2x^2+a_3x^3.$
Then $$begin{aligned}P(-1)=a_0-a_1+a_2-a_3&=f(-1)quad &(1)\
P(1)=a_0+a_1+a_2+a_3&=f(1) quad &(2)\
P^{'}(0)=a_1&=f^{'}(0)quad &(3)\
P^{''}(0)=2a_2&=f^{''}(0)quad &(4)
end{aligned}$$
How can I prove the given fact using the $4$ relations I have found here?
Please help.
real-analysis functions derivatives polynomials
real-analysis functions derivatives polynomials
edited Dec 17 '18 at 8:26
asked Dec 16 '18 at 6:11
user596656
$begingroup$
You presumably have some typo in the last line of the box, I guess $x_0=-1,x_1=0,x_2=1$? Also presumably $P''(x)=f''(x)$ should also only be at one point, but you seem to have sorted that out in the main body of the post.
$endgroup$
– Ian
Dec 16 '18 at 15:13
$begingroup$
@Ian,Thanks for the info.I edited it
$endgroup$
– user596656
Dec 17 '18 at 7:21
$begingroup$
you've got typos.. $x_0,x_1,x_2$ do not correspond to the ones in the problem and equations 1 thru 4.
$endgroup$
– Ahmad Bazzi
Dec 17 '18 at 7:24
$begingroup$
@AhmadBazzi;I think it is fixed now
$endgroup$
– user596656
Dec 17 '18 at 8:26
add a comment |
$begingroup$
You presumably have some typo in the last line of the box, I guess $x_0=-1,x_1=0,x_2=1$? Also presumably $P''(x)=f''(x)$ should also only be at one point, but you seem to have sorted that out in the main body of the post.
$endgroup$
– Ian
Dec 16 '18 at 15:13
$begingroup$
@Ian,Thanks for the info.I edited it
$endgroup$
– user596656
Dec 17 '18 at 7:21
$begingroup$
you've got typos.. $x_0,x_1,x_2$ do not correspond to the ones in the problem and equations 1 thru 4.
$endgroup$
– Ahmad Bazzi
Dec 17 '18 at 7:24
$begingroup$
@AhmadBazzi;I think it is fixed now
$endgroup$
– user596656
Dec 17 '18 at 8:26
$begingroup$
You presumably have some typo in the last line of the box, I guess $x_0=-1,x_1=0,x_2=1$? Also presumably $P''(x)=f''(x)$ should also only be at one point, but you seem to have sorted that out in the main body of the post.
$endgroup$
– Ian
Dec 16 '18 at 15:13
$begingroup$
You presumably have some typo in the last line of the box, I guess $x_0=-1,x_1=0,x_2=1$? Also presumably $P''(x)=f''(x)$ should also only be at one point, but you seem to have sorted that out in the main body of the post.
$endgroup$
– Ian
Dec 16 '18 at 15:13
$begingroup$
@Ian,Thanks for the info.I edited it
$endgroup$
– user596656
Dec 17 '18 at 7:21
$begingroup$
@Ian,Thanks for the info.I edited it
$endgroup$
– user596656
Dec 17 '18 at 7:21
$begingroup$
you've got typos.. $x_0,x_1,x_2$ do not correspond to the ones in the problem and equations 1 thru 4.
$endgroup$
– Ahmad Bazzi
Dec 17 '18 at 7:24
$begingroup$
you've got typos.. $x_0,x_1,x_2$ do not correspond to the ones in the problem and equations 1 thru 4.
$endgroup$
– Ahmad Bazzi
Dec 17 '18 at 7:24
$begingroup$
@AhmadBazzi;I think it is fixed now
$endgroup$
– user596656
Dec 17 '18 at 8:26
$begingroup$
@AhmadBazzi;I think it is fixed now
$endgroup$
– user596656
Dec 17 '18 at 8:26
add a comment |
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$begingroup$
You presumably have some typo in the last line of the box, I guess $x_0=-1,x_1=0,x_2=1$? Also presumably $P''(x)=f''(x)$ should also only be at one point, but you seem to have sorted that out in the main body of the post.
$endgroup$
– Ian
Dec 16 '18 at 15:13
$begingroup$
@Ian,Thanks for the info.I edited it
$endgroup$
– user596656
Dec 17 '18 at 7:21
$begingroup$
you've got typos.. $x_0,x_1,x_2$ do not correspond to the ones in the problem and equations 1 thru 4.
$endgroup$
– Ahmad Bazzi
Dec 17 '18 at 7:24
$begingroup$
@AhmadBazzi;I think it is fixed now
$endgroup$
– user596656
Dec 17 '18 at 8:26