For some $A in left(-frac{1}{2},frac{1}{2} right)$ and $c in (0,1), ;...
$begingroup$
Suppose $f(x)$ is continuous on $[0,1]$ and twice differentiable on $(0,1)$.
Show that there exists $A in left(-frac{1}{2},frac{1}{2} right)$ and $c in (0,1)$ such that
begin{equation}
f(1)=f(0)+f'left(frac{1}{2}right)+Af''(c)
end{equation}
By MVT, there exists $c in (0,1)$ such that
begin{equation}
f'(c)=frac{f(1)-f(0)}{1-0} = f(1)-f(0)
end{equation}
Comparing with the required statement I see that it suffices to show, for some $A in left(-frac{1}{2},frac{1}{2} right)$,
begin{equation}
f'(c)=f'left(frac{1}{2}right)+Af''(c)
end{equation}
Rearranging,
begin{equation}
frac{f'(c)-f'left(frac{1}{2}right)}{A}=f''(c)
end{equation}
This looks like an application of MVT again, but I'm not very sure how to proceed, as $c$ appears on both LHS and RHS.
Any help would be appreciated!
calculus real-analysis
asked Nov 27 '18 at 9:37
JanJan
453
$endgroup$
add a comment |
$begingroup$
Suppose $f(x)$ is continuous on $[0,1]$ and twice differentiable on $(0,1)$.
Show that there exists $A in left(-frac{1}{2},frac{1}{2} right)$ and $c in (0,1)$ such that
begin{equation}
f(1)=f(0)+f'left(frac{1}{2}right)+Af''(c)
end{equation}
By MVT, there exists $c in (0,1)$ such that
begin{equation}
f'(c)=frac{f(1)-f(0)}{1-0} = f(1)-f(0)
end{equation}
Comparing with the required statement I see that it suffices to show, for some $A in left(-frac{1}{2},frac{1}{2} right)$,
begin{equation}
f'(c)=f'left(frac{1}{2}right)+Af''(c)
end{equation}
Rearranging,
begin{equation}
frac{f'(c)-f'left(frac{1}{2}right)}{A}=f''(c)
end{equation}
This looks like an application of MVT again, but I'm not very sure how to proceed, as $c$ appears on both LHS and RHS.
Any help would be appreciated!
calculus real-analysis
asked Nov 27 '18 at 9:37
JanJan
453
$endgroup$
add a comment |
$begingroup$
Suppose $f(x)$ is continuous on $[0,1]$ and twice differentiable on $(0,1)$.
Show that there exists $A in left(-frac{1}{2},frac{1}{2} right)$ and $c in (0,1)$ such that
begin{equation}
f(1)=f(0)+f'left(frac{1}{2}right)+Af''(c)
end{equation}
By MVT, there exists $c in (0,1)$ such that
begin{equation}
f'(c)=frac{f(1)-f(0)}{1-0} = f(1)-f(0)
end{equation}
Comparing with the required statement I see that it suffices to show, for some $A in left(-frac{1}{2},frac{1}{2} right)$,
begin{equation}
f'(c)=f'left(frac{1}{2}right)+Af''(c)
end{equation}
Rearranging,
begin{equation}
frac{f'(c)-f'left(frac{1}{2}right)}{A}=f''(c)
end{equation}
This looks like an application of MVT again, but I'm not very sure how to proceed, as $c$ appears on both LHS and RHS.
Any help would be appreciated!
calculus real-analysis
asked Nov 27 '18 at 9:37
JanJan
453
$endgroup$
Suppose $f(x)$ is continuous on $[0,1]$ and twice differentiable on $(0,1)$.
Show that there exists $A in left(-frac{1}{2},frac{1}{2} right)$ and $c in (0,1)$ such that
begin{equation}
f(1)=f(0)+f'left(frac{1}{2}right)+Af''(c)
end{equation}
By MVT, there exists $c in (0,1)$ such that
begin{equation}
f'(c)=frac{f(1)-f(0)}{1-0} = f(1)-f(0)
end{equation}
Comparing with the required statement I see that it suffices to show, for some $A in left(-frac{1}{2},frac{1}{2} right)$,
begin{equation}
f'(c)=f'left(frac{1}{2}right)+Af''(c)
end{equation}
Rearranging,
begin{equation}
frac{f'(c)-f'left(frac{1}{2}right)}{A}=f''(c)
end{equation}
This looks like an application of MVT again, but I'm not very sure how to proceed, as $c$ appears on both LHS and RHS.
Any help would be appreciated!
calculus real-analysis
calculus real-analysis
asked Nov 27 '18 at 9:37
JanJan
453
asked Nov 27 '18 at 9:37
JanJan
453
asked Nov 27 '18 at 9:37
JanJan
453
asked Nov 27 '18 at 9:37
JanJan
453
asked Nov 27 '18 at 9:37
JanJan
453
453
add a comment |
add a comment |
1 Answer
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$begingroup$
$f'(c)-f'(frac 1 2) =(c-frac 1 2)f''(d)$ for some $d$ between $c$ and $frac 1 2$ Note that $|c-frac 1 2| leq frac 1 2$. Hence the required equation holds with $A=c-frac 1 2$ and $c$ changed to $d$.
answered Nov 27 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
56k42158
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$begingroup$
$f'(c)-f'(frac 1 2) =(c-frac 1 2)f''(d)$ for some $d$ between $c$ and $frac 1 2$ Note that $|c-frac 1 2| leq frac 1 2$. Hence the required equation holds with $A=c-frac 1 2$ and $c$ changed to $d$.
answered Nov 27 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
add a comment |
$begingroup$
$f'(c)-f'(frac 1 2) =(c-frac 1 2)f''(d)$ for some $d$ between $c$ and $frac 1 2$ Note that $|c-frac 1 2| leq frac 1 2$. Hence the required equation holds with $A=c-frac 1 2$ and $c$ changed to $d$.
answered Nov 27 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
add a comment |
$begingroup$
$f'(c)-f'(frac 1 2) =(c-frac 1 2)f''(d)$ for some $d$ between $c$ and $frac 1 2$ Note that $|c-frac 1 2| leq frac 1 2$. Hence the required equation holds with $A=c-frac 1 2$ and $c$ changed to $d$.
answered Nov 27 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
56k42158
$endgroup$
$f'(c)-f'(frac 1 2) =(c-frac 1 2)f''(d)$ for some $d$ between $c$ and $frac 1 2$ Note that $|c-frac 1 2| leq frac 1 2$. Hence the required equation holds with $A=c-frac 1 2$ and $c$ changed to $d$.
answered Nov 27 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Nov 27 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Nov 27 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
56k42158
answered Nov 27 '18 at 9:45
Kavi Rama MurthyKavi Rama Murthy
56k42158
56k42158
add a comment |
add a comment |
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