Show that $f(x)-P(x)=frac{x^4-1}{4!}f^{4}(c) $












0












$begingroup$



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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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












  • $begingroup$
    @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
















0












$begingroup$



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.










share|cite|improve this question











$endgroup$












  • $begingroup$
    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












  • $begingroup$
    @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














0












0








0


1



$begingroup$



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.










share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








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


















  • $begingroup$
    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










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