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Assuming that $f''(a)$ exists, show that: $f''(a)=lim_{hto 0}frac{f(a+h)-2f(a)+f(a-h)}{h^2}$ [duplicate]
$begingroup$
This question already has an answer here:
Second derivative “formula derivation”
3 answers
Assuming that $f''(a)$ exists, show that:
$$f''(a)=lim_{hto 0}frac{f(a+h)-2f(a)+f(a-h)}{h^2}$$
I got the first derivative by using the limit rule $(f(a+h)-f(a))/h$. But when taking the second derivative, I thought of doing the same thing on $f(a)$ of the equation, but I couldn't reach this answer.
calculus limits
edited Jan 23 at 0:14
Blue
49k870156
asked Jan 23 at 0:08
Hami the PenguinHami the Penguin
474
$endgroup$
marked as duplicate by Lord Shark the Unknown, Martin Argerami, Kemono Chen, Jyrki Lahtonen, mrtaurho Jan 23 at 6:43
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
|
show 12 more comments
$begingroup$
This question already has an answer here:
Second derivative “formula derivation”
3 answers
Assuming that $f''(a)$ exists, show that:
$$f''(a)=lim_{hto 0}frac{f(a+h)-2f(a)+f(a-h)}{h^2}$$
I got the first derivative by using the limit rule $(f(a+h)-f(a))/h$. But when taking the second derivative, I thought of doing the same thing on $f(a)$ of the equation, but I couldn't reach this answer.
calculus limits
edited Jan 23 at 0:14
Blue
49k870156
asked Jan 23 at 0:08
Hami the PenguinHami the Penguin
474
$endgroup$
marked as duplicate by Lord Shark the Unknown, Martin Argerami, Kemono Chen, Jyrki Lahtonen, mrtaurho Jan 23 at 6:43
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Do you have Taylor's formula in your toolkit?
$endgroup$
– Bernard
Jan 23 at 0:13
$begingroup$
No, I don't have.
$endgroup$
– Hami the Penguin
Jan 23 at 0:15
1
$begingroup$
Then, apply it twice.
$endgroup$
– Bernard
Jan 23 at 0:20
1
$begingroup$
@Bernard L'Hopital rule assumes that $f$ is differentiable in some neighbourhood of $a$, and we also need continuity of the second derivative
$endgroup$
– Jakobian
Jan 23 at 0:22
1
$begingroup$
I know you didn't mean it. I am myself from time to time a victim of one of these brave anonymous downvoters.I quite agree with you for a correct way of downvoting. Personally, when I see a mistake in an answer, I just make a comment for the author to point it out, so he/she may modify the answer.
$endgroup$
– Bernard
Jan 23 at 23:05
|
show 12 more comments
$begingroup$
This question already has an answer here:
Second derivative “formula derivation”
3 answers
Assuming that $f''(a)$ exists, show that:
$$f''(a)=lim_{hto 0}frac{f(a+h)-2f(a)+f(a-h)}{h^2}$$
I got the first derivative by using the limit rule $(f(a+h)-f(a))/h$. But when taking the second derivative, I thought of doing the same thing on $f(a)$ of the equation, but I couldn't reach this answer.
calculus limits
edited Jan 23 at 0:14
Blue
49k870156
asked Jan 23 at 0:08
Hami the PenguinHami the Penguin
474
$endgroup$
This question already has an answer here:
Second derivative “formula derivation”
3 answers
Assuming that $f''(a)$ exists, show that:
$$f''(a)=lim_{hto 0}frac{f(a+h)-2f(a)+f(a-h)}{h^2}$$
I got the first derivative by using the limit rule $(f(a+h)-f(a))/h$. But when taking the second derivative, I thought of doing the same thing on $f(a)$ of the equation, but I couldn't reach this answer.
This question already has an answer here:
Second derivative “formula derivation”
3 answers
calculus limits
calculus limits
edited Jan 23 at 0:14
Blue
49k870156
asked Jan 23 at 0:08
Hami the PenguinHami the Penguin
474
edited Jan 23 at 0:14
Blue
49k870156
asked Jan 23 at 0:08
Hami the PenguinHami the Penguin
474
edited Jan 23 at 0:14
Blue
49k870156
edited Jan 23 at 0:14
Blue
49k870156
edited Jan 23 at 0:14
Blue
49k870156
49k870156
asked Jan 23 at 0:08
Hami the PenguinHami the Penguin
474
asked Jan 23 at 0:08
Hami the PenguinHami the Penguin
474
asked Jan 23 at 0:08
Hami the PenguinHami the Penguin
474
474
marked as duplicate by Lord Shark the Unknown, Martin Argerami, Kemono Chen, Jyrki Lahtonen, mrtaurho Jan 23 at 6:43
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Lord Shark the Unknown, Martin Argerami, Kemono Chen, Jyrki Lahtonen, mrtaurho Jan 23 at 6:43
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Do you have Taylor's formula in your toolkit?
$endgroup$
– Bernard
Jan 23 at 0:13
$begingroup$
No, I don't have.
$endgroup$
– Hami the Penguin
Jan 23 at 0:15
1
$begingroup$
Then, apply it twice.
$endgroup$
– Bernard
Jan 23 at 0:20
1
$begingroup$
@Bernard L'Hopital rule assumes that $f$ is differentiable in some neighbourhood of $a$, and we also need continuity of the second derivative
$endgroup$
– Jakobian
Jan 23 at 0:22
1
$begingroup$
I know you didn't mean it. I am myself from time to time a victim of one of these brave anonymous downvoters.I quite agree with you for a correct way of downvoting. Personally, when I see a mistake in an answer, I just make a comment for the author to point it out, so he/she may modify the answer.
$endgroup$
– Bernard
Jan 23 at 23:05
|
show 12 more comments
1
$begingroup$
Do you have Taylor's formula in your toolkit?
$endgroup$
– Bernard
Jan 23 at 0:13
$begingroup$
No, I don't have.
$endgroup$
– Hami the Penguin
Jan 23 at 0:15
1
$begingroup$
Then, apply it twice.
$endgroup$
– Bernard
Jan 23 at 0:20
1
$begingroup$
@Bernard L'Hopital rule assumes that $f$ is differentiable in some neighbourhood of $a$, and we also need continuity of the second derivative
$endgroup$
– Jakobian
Jan 23 at 0:22
1
$begingroup$
I know you didn't mean it. I am myself from time to time a victim of one of these brave anonymous downvoters.I quite agree with you for a correct way of downvoting. Personally, when I see a mistake in an answer, I just make a comment for the author to point it out, so he/she may modify the answer.
$endgroup$
– Bernard
Jan 23 at 23:05
1
1
$begingroup$
Do you have Taylor's formula in your toolkit?
$endgroup$
– Bernard
Jan 23 at 0:13
$begingroup$
Do you have Taylor's formula in your toolkit?
$endgroup$
– Bernard
Jan 23 at 0:13
$begingroup$
No, I don't have.
$endgroup$
– Hami the Penguin
Jan 23 at 0:15
$begingroup$
No, I don't have.
$endgroup$
– Hami the Penguin
Jan 23 at 0:15
1
1
$begingroup$
Then, apply it twice.
$endgroup$
– Bernard
Jan 23 at 0:20
$begingroup$
Then, apply it twice.
$endgroup$
– Bernard
Jan 23 at 0:20
1
1
$begingroup$
@Bernard L'Hopital rule assumes that $f$ is differentiable in some neighbourhood of $a$, and we also need continuity of the second derivative
$endgroup$
– Jakobian
Jan 23 at 0:22
$begingroup$
@Bernard L'Hopital rule assumes that $f$ is differentiable in some neighbourhood of $a$, and we also need continuity of the second derivative
$endgroup$
– Jakobian
Jan 23 at 0:22
1
1
$begingroup$
I know you didn't mean it. I am myself from time to time a victim of one of these brave anonymous downvoters.I quite agree with you for a correct way of downvoting. Personally, when I see a mistake in an answer, I just make a comment for the author to point it out, so he/she may modify the answer.
$endgroup$
– Bernard
Jan 23 at 23:05
$begingroup$
I know you didn't mean it. I am myself from time to time a victim of one of these brave anonymous downvoters.I quite agree with you for a correct way of downvoting. Personally, when I see a mistake in an answer, I just make a comment for the author to point it out, so he/she may modify the answer.
$endgroup$
– Bernard
Jan 23 at 23:05
|
show 12 more comments
1 Answer
1
active
oldest
votes
$begingroup$
To be precise, let me re-phrase the question: Let $I$ be an open
interval, $ain I$, and $f:Irightarrowmathbb{R}$. Suppose that
$f''(a)$ exists, then
$$
f''(a)=lim_{hrightarrow0}frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}.
$$
Proof: That $f''(a)$ exists implies that there exists $delta>0$ such
that $f'$ exists on $(a-delta,a+delta)$ and $f'$ is continuous
at $a$. Define functions $phi:(0,delta)rightarrowmathbb{R}$
and $psi:(0,delta)rightarrowmathbb{R}$ by $phi(t)=f(a+t)+f(a-t)-2f(a)$
and $psi(t)=t^{2}$. Clearly $phi$ and $psi$ are differentiable.
Note that
$$
lim_{trightarrow0+}frac{phi(t)}{psi(t)}
$$
is a $0/0$-indeterminate form. By L'Hospital rule, we have
$$
lim_{trightarrow0+}frac{phi(t)}{psi(t)}=lim_{trightarrow0+}frac{phi'(t)}{psi'(t)},
$$
provided that the limit on the RHS exists. But
begin{eqnarray*}
lim_{trightarrow0+}frac{phi'(t)}{psi'(t)} & = & lim_{trightarrow0+}frac{f'(a+t)-f'(a-t)}{2t}\
& = & lim_{trightarrow0+}left[frac{1}{2}frac{f'(a+t)-f'(a)}{t}+frac{1}{2}frac{f'(a)-f'(a-t)}{t}right]\
& = & frac{1}{2}f''(a)+frac{1}{2}f''(a)\
& = & f''(a).
end{eqnarray*}
Therefore, we conclude that
$$
f''(a)=lim_{hrightarrow0+}frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}.
$$
The case that $hrightarrow0-$ can be proved similarly.
answered Jan 23 at 0:45
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,46638
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To be precise, let me re-phrase the question: Let $I$ be an open
interval, $ain I$, and $f:Irightarrowmathbb{R}$. Suppose that
$f''(a)$ exists, then
$$
f''(a)=lim_{hrightarrow0}frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}.
$$
Proof: That $f''(a)$ exists implies that there exists $delta>0$ such
that $f'$ exists on $(a-delta,a+delta)$ and $f'$ is continuous
at $a$. Define functions $phi:(0,delta)rightarrowmathbb{R}$
and $psi:(0,delta)rightarrowmathbb{R}$ by $phi(t)=f(a+t)+f(a-t)-2f(a)$
and $psi(t)=t^{2}$. Clearly $phi$ and $psi$ are differentiable.
Note that
$$
lim_{trightarrow0+}frac{phi(t)}{psi(t)}
$$
is a $0/0$-indeterminate form. By L'Hospital rule, we have
$$
lim_{trightarrow0+}frac{phi(t)}{psi(t)}=lim_{trightarrow0+}frac{phi'(t)}{psi'(t)},
$$
provided that the limit on the RHS exists. But
begin{eqnarray*}
lim_{trightarrow0+}frac{phi'(t)}{psi'(t)} & = & lim_{trightarrow0+}frac{f'(a+t)-f'(a-t)}{2t}\
& = & lim_{trightarrow0+}left[frac{1}{2}frac{f'(a+t)-f'(a)}{t}+frac{1}{2}frac{f'(a)-f'(a-t)}{t}right]\
& = & frac{1}{2}f''(a)+frac{1}{2}f''(a)\
& = & f''(a).
end{eqnarray*}
Therefore, we conclude that
$$
f''(a)=lim_{hrightarrow0+}frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}.
$$
The case that $hrightarrow0-$ can be proved similarly.
answered Jan 23 at 0:45
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,46638
$endgroup$
add a comment |
$begingroup$
To be precise, let me re-phrase the question: Let $I$ be an open
interval, $ain I$, and $f:Irightarrowmathbb{R}$. Suppose that
$f''(a)$ exists, then
$$
f''(a)=lim_{hrightarrow0}frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}.
$$
Proof: That $f''(a)$ exists implies that there exists $delta>0$ such
that $f'$ exists on $(a-delta,a+delta)$ and $f'$ is continuous
at $a$. Define functions $phi:(0,delta)rightarrowmathbb{R}$
and $psi:(0,delta)rightarrowmathbb{R}$ by $phi(t)=f(a+t)+f(a-t)-2f(a)$
and $psi(t)=t^{2}$. Clearly $phi$ and $psi$ are differentiable.
Note that
$$
lim_{trightarrow0+}frac{phi(t)}{psi(t)}
$$
is a $0/0$-indeterminate form. By L'Hospital rule, we have
$$
lim_{trightarrow0+}frac{phi(t)}{psi(t)}=lim_{trightarrow0+}frac{phi'(t)}{psi'(t)},
$$
provided that the limit on the RHS exists. But
begin{eqnarray*}
lim_{trightarrow0+}frac{phi'(t)}{psi'(t)} & = & lim_{trightarrow0+}frac{f'(a+t)-f'(a-t)}{2t}\
& = & lim_{trightarrow0+}left[frac{1}{2}frac{f'(a+t)-f'(a)}{t}+frac{1}{2}frac{f'(a)-f'(a-t)}{t}right]\
& = & frac{1}{2}f''(a)+frac{1}{2}f''(a)\
& = & f''(a).
end{eqnarray*}
Therefore, we conclude that
$$
f''(a)=lim_{hrightarrow0+}frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}.
$$
The case that $hrightarrow0-$ can be proved similarly.
answered Jan 23 at 0:45
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,46638
$endgroup$
add a comment |
$begingroup$
To be precise, let me re-phrase the question: Let $I$ be an open
interval, $ain I$, and $f:Irightarrowmathbb{R}$. Suppose that
$f''(a)$ exists, then
$$
f''(a)=lim_{hrightarrow0}frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}.
$$
Proof: That $f''(a)$ exists implies that there exists $delta>0$ such
that $f'$ exists on $(a-delta,a+delta)$ and $f'$ is continuous
at $a$. Define functions $phi:(0,delta)rightarrowmathbb{R}$
and $psi:(0,delta)rightarrowmathbb{R}$ by $phi(t)=f(a+t)+f(a-t)-2f(a)$
and $psi(t)=t^{2}$. Clearly $phi$ and $psi$ are differentiable.
Note that
$$
lim_{trightarrow0+}frac{phi(t)}{psi(t)}
$$
is a $0/0$-indeterminate form. By L'Hospital rule, we have
$$
lim_{trightarrow0+}frac{phi(t)}{psi(t)}=lim_{trightarrow0+}frac{phi'(t)}{psi'(t)},
$$
provided that the limit on the RHS exists. But
begin{eqnarray*}
lim_{trightarrow0+}frac{phi'(t)}{psi'(t)} & = & lim_{trightarrow0+}frac{f'(a+t)-f'(a-t)}{2t}\
& = & lim_{trightarrow0+}left[frac{1}{2}frac{f'(a+t)-f'(a)}{t}+frac{1}{2}frac{f'(a)-f'(a-t)}{t}right]\
& = & frac{1}{2}f''(a)+frac{1}{2}f''(a)\
& = & f''(a).
end{eqnarray*}
Therefore, we conclude that
$$
f''(a)=lim_{hrightarrow0+}frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}.
$$
The case that $hrightarrow0-$ can be proved similarly.
answered Jan 23 at 0:45
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,46638
$endgroup$
To be precise, let me re-phrase the question: Let $I$ be an open
interval, $ain I$, and $f:Irightarrowmathbb{R}$. Suppose that
$f''(a)$ exists, then
$$
f''(a)=lim_{hrightarrow0}frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}.
$$
Proof: That $f''(a)$ exists implies that there exists $delta>0$ such
that $f'$ exists on $(a-delta,a+delta)$ and $f'$ is continuous
at $a$. Define functions $phi:(0,delta)rightarrowmathbb{R}$
and $psi:(0,delta)rightarrowmathbb{R}$ by $phi(t)=f(a+t)+f(a-t)-2f(a)$
and $psi(t)=t^{2}$. Clearly $phi$ and $psi$ are differentiable.
Note that
$$
lim_{trightarrow0+}frac{phi(t)}{psi(t)}
$$
is a $0/0$-indeterminate form. By L'Hospital rule, we have
$$
lim_{trightarrow0+}frac{phi(t)}{psi(t)}=lim_{trightarrow0+}frac{phi'(t)}{psi'(t)},
$$
provided that the limit on the RHS exists. But
begin{eqnarray*}
lim_{trightarrow0+}frac{phi'(t)}{psi'(t)} & = & lim_{trightarrow0+}frac{f'(a+t)-f'(a-t)}{2t}\
& = & lim_{trightarrow0+}left[frac{1}{2}frac{f'(a+t)-f'(a)}{t}+frac{1}{2}frac{f'(a)-f'(a-t)}{t}right]\
& = & frac{1}{2}f''(a)+frac{1}{2}f''(a)\
& = & f''(a).
end{eqnarray*}
Therefore, we conclude that
$$
f''(a)=lim_{hrightarrow0+}frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}.
$$
The case that $hrightarrow0-$ can be proved similarly.
answered Jan 23 at 0:45
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,46638
answered Jan 23 at 0:45
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,46638
answered Jan 23 at 0:45
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,46638
answered Jan 23 at 0:45
Danny Pak-Keung ChanDanny Pak-Keung Chan
2,46638
2,46638
add a comment |
add a comment |
1
$begingroup$
Do you have Taylor's formula in your toolkit?
$endgroup$
– Bernard
Jan 23 at 0:13
$begingroup$
No, I don't have.
$endgroup$
– Hami the Penguin
Jan 23 at 0:15
1
$begingroup$
Then, apply it twice.
$endgroup$
– Bernard
Jan 23 at 0:20
1
$begingroup$
@Bernard L'Hopital rule assumes that $f$ is differentiable in some neighbourhood of $a$, and we also need continuity of the second derivative
$endgroup$
– Jakobian
Jan 23 at 0:22
1
$begingroup$
I know you didn't mean it. I am myself from time to time a victim of one of these brave anonymous downvoters.I quite agree with you for a correct way of downvoting. Personally, when I see a mistake in an answer, I just make a comment for the author to point it out, so he/she may modify the answer.
$endgroup$
– Bernard
Jan 23 at 23:05