is this a valid proof ? and is there a simpler proof?
$begingroup$
Let $f$ be twice differential function and that $2f(0) = f(1)+f(-1)$, prove that there exists $c in (-1,1)$ such that $f''(c)=0$
My attempt : Using Taylor series we get that :
$f(1) = f(0) +frac{f'(0)}{1!} (1-0)^1 +frac{f''(a)}{2!} (1-0)^2$
$f(-1) = f(0)+frac{f'(0)}{1!} (-1-0)^1 + frac{f''(b)}{2!}(-1-0)^2$
Adding together gives $f(1)+f(-1) = 2f(0) +frac{f''(a)+f''(b)}{2}$
Which is $0 = frac{f''(a)+f''(b)}{2}$ leading to $f''(a)=-f''(b)$ , now if $f''(a)=0$ then let $c=a$, otherwise WLG $f''(a) < 0 < f''(b)$ so by Darboux theorem the derivative must have the intermediate value property so there is $c$ such that $f''(c)=0$ , first is this proof correct ?
secondly , is there a simpler proof ?
Thanks
real-analysis calculus derivatives
asked Dec 23 '18 at 17:05
AhmadAhmad
2,5911725
$endgroup$
|
show 1 more comment
$begingroup$
Let $f$ be twice differential function and that $2f(0) = f(1)+f(-1)$, prove that there exists $c in (-1,1)$ such that $f''(c)=0$
My attempt : Using Taylor series we get that :
$f(1) = f(0) +frac{f'(0)}{1!} (1-0)^1 +frac{f''(a)}{2!} (1-0)^2$
$f(-1) = f(0)+frac{f'(0)}{1!} (-1-0)^1 + frac{f''(b)}{2!}(-1-0)^2$
Adding together gives $f(1)+f(-1) = 2f(0) +frac{f''(a)+f''(b)}{2}$
Which is $0 = frac{f''(a)+f''(b)}{2}$ leading to $f''(a)=-f''(b)$ , now if $f''(a)=0$ then let $c=a$, otherwise WLG $f''(a) < 0 < f''(b)$ so by Darboux theorem the derivative must have the intermediate value property so there is $c$ such that $f''(c)=0$ , first is this proof correct ?
secondly , is there a simpler proof ?
Thanks
real-analysis calculus derivatives
asked Dec 23 '18 at 17:05
AhmadAhmad
2,5911725
$endgroup$
1
$begingroup$
If it is twice differentiable, you only know that $f''$ exists, but you do not have information about continuity of $f''$
$endgroup$
– Tito Eliatron
Dec 23 '18 at 17:09
2
$begingroup$
@TitoEliatron true but $f''$ is the derivative of $f'$ (a function) so by Darboux it must satisfy the intermediate value property, it does not need to be continues.
$endgroup$
– Ahmad
Dec 23 '18 at 17:10
$begingroup$
Hint: consider $g(x)=f(x)+f(-x)$. Argue that there must be a solution to $g''(c)=0$. That $c$ doesn't give you a solution directly, but you can work from there.
$endgroup$
– lulu
Dec 23 '18 at 17:11
$begingroup$
@lulu so my proof is correct ?
$endgroup$
– Ahmad
Dec 23 '18 at 17:23
$begingroup$
Yes, it looks good.
$endgroup$
– lulu
Dec 23 '18 at 18:46
|
show 1 more comment
$begingroup$
Let $f$ be twice differential function and that $2f(0) = f(1)+f(-1)$, prove that there exists $c in (-1,1)$ such that $f''(c)=0$
My attempt : Using Taylor series we get that :
$f(1) = f(0) +frac{f'(0)}{1!} (1-0)^1 +frac{f''(a)}{2!} (1-0)^2$
$f(-1) = f(0)+frac{f'(0)}{1!} (-1-0)^1 + frac{f''(b)}{2!}(-1-0)^2$
Adding together gives $f(1)+f(-1) = 2f(0) +frac{f''(a)+f''(b)}{2}$
Which is $0 = frac{f''(a)+f''(b)}{2}$ leading to $f''(a)=-f''(b)$ , now if $f''(a)=0$ then let $c=a$, otherwise WLG $f''(a) < 0 < f''(b)$ so by Darboux theorem the derivative must have the intermediate value property so there is $c$ such that $f''(c)=0$ , first is this proof correct ?
secondly , is there a simpler proof ?
Thanks
real-analysis calculus derivatives
asked Dec 23 '18 at 17:05
AhmadAhmad
2,5911725
$endgroup$
Let $f$ be twice differential function and that $2f(0) = f(1)+f(-1)$, prove that there exists $c in (-1,1)$ such that $f''(c)=0$
My attempt : Using Taylor series we get that :
$f(1) = f(0) +frac{f'(0)}{1!} (1-0)^1 +frac{f''(a)}{2!} (1-0)^2$
$f(-1) = f(0)+frac{f'(0)}{1!} (-1-0)^1 + frac{f''(b)}{2!}(-1-0)^2$
Adding together gives $f(1)+f(-1) = 2f(0) +frac{f''(a)+f''(b)}{2}$
Which is $0 = frac{f''(a)+f''(b)}{2}$ leading to $f''(a)=-f''(b)$ , now if $f''(a)=0$ then let $c=a$, otherwise WLG $f''(a) < 0 < f''(b)$ so by Darboux theorem the derivative must have the intermediate value property so there is $c$ such that $f''(c)=0$ , first is this proof correct ?
secondly , is there a simpler proof ?
Thanks
real-analysis calculus derivatives
real-analysis calculus derivatives
asked Dec 23 '18 at 17:05
AhmadAhmad
2,5911725
asked Dec 23 '18 at 17:05
AhmadAhmad
2,5911725
asked Dec 23 '18 at 17:05
AhmadAhmad
2,5911725
asked Dec 23 '18 at 17:05
AhmadAhmad
2,5911725
asked Dec 23 '18 at 17:05
AhmadAhmad
2,5911725
2,5911725
1
$begingroup$
If it is twice differentiable, you only know that $f''$ exists, but you do not have information about continuity of $f''$
$endgroup$
– Tito Eliatron
Dec 23 '18 at 17:09
2
$begingroup$
@TitoEliatron true but $f''$ is the derivative of $f'$ (a function) so by Darboux it must satisfy the intermediate value property, it does not need to be continues.
$endgroup$
– Ahmad
Dec 23 '18 at 17:10
$begingroup$
Hint: consider $g(x)=f(x)+f(-x)$. Argue that there must be a solution to $g''(c)=0$. That $c$ doesn't give you a solution directly, but you can work from there.
$endgroup$
– lulu
Dec 23 '18 at 17:11
$begingroup$
@lulu so my proof is correct ?
$endgroup$
– Ahmad
Dec 23 '18 at 17:23
$begingroup$
Yes, it looks good.
$endgroup$
– lulu
Dec 23 '18 at 18:46
|
show 1 more comment
1
$begingroup$
If it is twice differentiable, you only know that $f''$ exists, but you do not have information about continuity of $f''$
$endgroup$
– Tito Eliatron
Dec 23 '18 at 17:09
2
$begingroup$
@TitoEliatron true but $f''$ is the derivative of $f'$ (a function) so by Darboux it must satisfy the intermediate value property, it does not need to be continues.
$endgroup$
– Ahmad
Dec 23 '18 at 17:10
$begingroup$
Hint: consider $g(x)=f(x)+f(-x)$. Argue that there must be a solution to $g''(c)=0$. That $c$ doesn't give you a solution directly, but you can work from there.
$endgroup$
– lulu
Dec 23 '18 at 17:11
$begingroup$
@lulu so my proof is correct ?
$endgroup$
– Ahmad
Dec 23 '18 at 17:23
$begingroup$
Yes, it looks good.
$endgroup$
– lulu
Dec 23 '18 at 18:46
1
1
$begingroup$
If it is twice differentiable, you only know that $f''$ exists, but you do not have information about continuity of $f''$
$endgroup$
– Tito Eliatron
Dec 23 '18 at 17:09
$begingroup$
If it is twice differentiable, you only know that $f''$ exists, but you do not have information about continuity of $f''$
$endgroup$
– Tito Eliatron
Dec 23 '18 at 17:09
2
2
$begingroup$
@TitoEliatron true but $f''$ is the derivative of $f'$ (a function) so by Darboux it must satisfy the intermediate value property, it does not need to be continues.
$endgroup$
– Ahmad
Dec 23 '18 at 17:10
$begingroup$
@TitoEliatron true but $f''$ is the derivative of $f'$ (a function) so by Darboux it must satisfy the intermediate value property, it does not need to be continues.
$endgroup$
– Ahmad
Dec 23 '18 at 17:10
$begingroup$
Hint: consider $g(x)=f(x)+f(-x)$. Argue that there must be a solution to $g''(c)=0$. That $c$ doesn't give you a solution directly, but you can work from there.
$endgroup$
– lulu
Dec 23 '18 at 17:11
$begingroup$
Hint: consider $g(x)=f(x)+f(-x)$. Argue that there must be a solution to $g''(c)=0$. That $c$ doesn't give you a solution directly, but you can work from there.
$endgroup$
– lulu
Dec 23 '18 at 17:11
$begingroup$
@lulu so my proof is correct ?
$endgroup$
– Ahmad
Dec 23 '18 at 17:23
$begingroup$
@lulu so my proof is correct ?
$endgroup$
– Ahmad
Dec 23 '18 at 17:23
$begingroup$
Yes, it looks good.
$endgroup$
– lulu
Dec 23 '18 at 18:46
$begingroup$
Yes, it looks good.
$endgroup$
– lulu
Dec 23 '18 at 18:46
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Your proof looks OK to me.
Attempt.
MVT:
$f(1)-f(0)+f(-1)-f(0)=0;$
$dfrac{f(1)-f(0)}{1-0} - dfrac{f(-1) -f(0)}{-1-0}=0;$
$f'(b)-f'(a)=0$, where
$b in (0,1), a in (-1,0).$
With $a <b:$
$dfrac{f'(b)-f'(a)}{b-a}= f''(c)=0$, $c in (a,b)$
answered Dec 23 '18 at 17:52
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
add a comment |
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$begingroup$
Your proof looks OK to me.
Attempt.
MVT:
$f(1)-f(0)+f(-1)-f(0)=0;$
$dfrac{f(1)-f(0)}{1-0} - dfrac{f(-1) -f(0)}{-1-0}=0;$
$f'(b)-f'(a)=0$, where
$b in (0,1), a in (-1,0).$
With $a <b:$
$dfrac{f'(b)-f'(a)}{b-a}= f''(c)=0$, $c in (a,b)$
answered Dec 23 '18 at 17:52
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
add a comment |
$begingroup$
Your proof looks OK to me.
Attempt.
MVT:
$f(1)-f(0)+f(-1)-f(0)=0;$
$dfrac{f(1)-f(0)}{1-0} - dfrac{f(-1) -f(0)}{-1-0}=0;$
$f'(b)-f'(a)=0$, where
$b in (0,1), a in (-1,0).$
With $a <b:$
$dfrac{f'(b)-f'(a)}{b-a}= f''(c)=0$, $c in (a,b)$
answered Dec 23 '18 at 17:52
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
add a comment |
$begingroup$
Your proof looks OK to me.
Attempt.
MVT:
$f(1)-f(0)+f(-1)-f(0)=0;$
$dfrac{f(1)-f(0)}{1-0} - dfrac{f(-1) -f(0)}{-1-0}=0;$
$f'(b)-f'(a)=0$, where
$b in (0,1), a in (-1,0).$
With $a <b:$
$dfrac{f'(b)-f'(a)}{b-a}= f''(c)=0$, $c in (a,b)$
answered Dec 23 '18 at 17:52
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
Your proof looks OK to me.
Attempt.
MVT:
$f(1)-f(0)+f(-1)-f(0)=0;$
$dfrac{f(1)-f(0)}{1-0} - dfrac{f(-1) -f(0)}{-1-0}=0;$
$f'(b)-f'(a)=0$, where
$b in (0,1), a in (-1,0).$
With $a <b:$
$dfrac{f'(b)-f'(a)}{b-a}= f''(c)=0$, $c in (a,b)$
answered Dec 23 '18 at 17:52
Peter SzilasPeter Szilas
11.5k2822
answered Dec 23 '18 at 17:52
Peter SzilasPeter Szilas
11.5k2822
answered Dec 23 '18 at 17:52
Peter SzilasPeter Szilas
11.5k2822
answered Dec 23 '18 at 17:52
Peter SzilasPeter Szilas
11.5k2822
11.5k2822
add a comment |
add a comment |
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1
$begingroup$
If it is twice differentiable, you only know that $f''$ exists, but you do not have information about continuity of $f''$
$endgroup$
– Tito Eliatron
Dec 23 '18 at 17:09
2
$begingroup$
@TitoEliatron true but $f''$ is the derivative of $f'$ (a function) so by Darboux it must satisfy the intermediate value property, it does not need to be continues.
$endgroup$
– Ahmad
Dec 23 '18 at 17:10
$begingroup$
Hint: consider $g(x)=f(x)+f(-x)$. Argue that there must be a solution to $g''(c)=0$. That $c$ doesn't give you a solution directly, but you can work from there.
$endgroup$
– lulu
Dec 23 '18 at 17:11
$begingroup$
@lulu so my proof is correct ?
$endgroup$
– Ahmad
Dec 23 '18 at 17:23
$begingroup$
Yes, it looks good.
$endgroup$
– lulu
Dec 23 '18 at 18:46