Is my solution correct? Something feels wrong with it.
$begingroup$
I have the following question. Suppose $f$ is a function which satisfies for some $x$: 1. $f'$ exists in $(x-epsilon,x+epsilon)$ and 2. $f''(x)$ exists.
Show that $f''(x) = lim_{h to 0} (frac{f(x+h)-2f(x)+f(x-h)}{h^2})$.
I used L'hopital rule, and that means the limit is: $lim_{h to 0} (frac{f'(x+h)-f'(x-h)}{2h})=lim_{h to 0}(frac{f'(x+h)-f'(x)}{2h})+lim_{h to 0} (frac{f'(x)-f'(x-h)}{2h})=f''(x)/2 + f''(x)/2=f''(x)$
I am a little bit confused, because while taking the derivative of the numerator, I treated $x$ as a constant. My first intuition was to write $-2f'(x)$, and then I realized x is like a constant, and thus so is $f(x)$, so we need to write $(f(x))'=0$. That is why I'm a bit confused. Is this correct? Thanks!
calculus limits derivatives
asked Jan 9 at 20:07
OmerOmer
3138
$endgroup$
add a comment |
$begingroup$
I have the following question. Suppose $f$ is a function which satisfies for some $x$: 1. $f'$ exists in $(x-epsilon,x+epsilon)$ and 2. $f''(x)$ exists.
Show that $f''(x) = lim_{h to 0} (frac{f(x+h)-2f(x)+f(x-h)}{h^2})$.
I used L'hopital rule, and that means the limit is: $lim_{h to 0} (frac{f'(x+h)-f'(x-h)}{2h})=lim_{h to 0}(frac{f'(x+h)-f'(x)}{2h})+lim_{h to 0} (frac{f'(x)-f'(x-h)}{2h})=f''(x)/2 + f''(x)/2=f''(x)$
I am a little bit confused, because while taking the derivative of the numerator, I treated $x$ as a constant. My first intuition was to write $-2f'(x)$, and then I realized x is like a constant, and thus so is $f(x)$, so we need to write $(f(x))'=0$. That is why I'm a bit confused. Is this correct? Thanks!
calculus limits derivatives
asked Jan 9 at 20:07
OmerOmer
3138
$endgroup$
2
$begingroup$
Treating $x$ as a constant here is fine, because it is, in fact, constant.
$endgroup$
– user3482749
Jan 9 at 20:16
$begingroup$
Since your limit variable is $h$ any other variable (not dependent on $h$) is to be treated like a constant during the evaluation of limit.
$endgroup$
– Paramanand Singh
Jan 10 at 3:08
add a comment |
$begingroup$
I have the following question. Suppose $f$ is a function which satisfies for some $x$: 1. $f'$ exists in $(x-epsilon,x+epsilon)$ and 2. $f''(x)$ exists.
Show that $f''(x) = lim_{h to 0} (frac{f(x+h)-2f(x)+f(x-h)}{h^2})$.
I used L'hopital rule, and that means the limit is: $lim_{h to 0} (frac{f'(x+h)-f'(x-h)}{2h})=lim_{h to 0}(frac{f'(x+h)-f'(x)}{2h})+lim_{h to 0} (frac{f'(x)-f'(x-h)}{2h})=f''(x)/2 + f''(x)/2=f''(x)$
I am a little bit confused, because while taking the derivative of the numerator, I treated $x$ as a constant. My first intuition was to write $-2f'(x)$, and then I realized x is like a constant, and thus so is $f(x)$, so we need to write $(f(x))'=0$. That is why I'm a bit confused. Is this correct? Thanks!
calculus limits derivatives
asked Jan 9 at 20:07
OmerOmer
3138
$endgroup$
I have the following question. Suppose $f$ is a function which satisfies for some $x$: 1. $f'$ exists in $(x-epsilon,x+epsilon)$ and 2. $f''(x)$ exists.
Show that $f''(x) = lim_{h to 0} (frac{f(x+h)-2f(x)+f(x-h)}{h^2})$.
I used L'hopital rule, and that means the limit is: $lim_{h to 0} (frac{f'(x+h)-f'(x-h)}{2h})=lim_{h to 0}(frac{f'(x+h)-f'(x)}{2h})+lim_{h to 0} (frac{f'(x)-f'(x-h)}{2h})=f''(x)/2 + f''(x)/2=f''(x)$
I am a little bit confused, because while taking the derivative of the numerator, I treated $x$ as a constant. My first intuition was to write $-2f'(x)$, and then I realized x is like a constant, and thus so is $f(x)$, so we need to write $(f(x))'=0$. That is why I'm a bit confused. Is this correct? Thanks!
calculus limits derivatives
calculus limits derivatives
asked Jan 9 at 20:07
OmerOmer
3138
asked Jan 9 at 20:07
OmerOmer
3138
edited Jan 9 at 20:28
Omer
asked Jan 9 at 20:07
OmerOmer
3138
asked Jan 9 at 20:07
OmerOmer
3138
asked Jan 9 at 20:07
OmerOmer
3138
3138
2
$begingroup$
Treating $x$ as a constant here is fine, because it is, in fact, constant.
$endgroup$
– user3482749
Jan 9 at 20:16
$begingroup$
Since your limit variable is $h$ any other variable (not dependent on $h$) is to be treated like a constant during the evaluation of limit.
$endgroup$
– Paramanand Singh
Jan 10 at 3:08
add a comment |
2
$begingroup$
Treating $x$ as a constant here is fine, because it is, in fact, constant.
$endgroup$
– user3482749
Jan 9 at 20:16
$begingroup$
Since your limit variable is $h$ any other variable (not dependent on $h$) is to be treated like a constant during the evaluation of limit.
$endgroup$
– Paramanand Singh
Jan 10 at 3:08
2
2
$begingroup$
Treating $x$ as a constant here is fine, because it is, in fact, constant.
$endgroup$
– user3482749
Jan 9 at 20:16
$begingroup$
Treating $x$ as a constant here is fine, because it is, in fact, constant.
$endgroup$
– user3482749
Jan 9 at 20:16
$begingroup$
Since your limit variable is $h$ any other variable (not dependent on $h$) is to be treated like a constant during the evaluation of limit.
$endgroup$
– Paramanand Singh
Jan 10 at 3:08
$begingroup$
Since your limit variable is $h$ any other variable (not dependent on $h$) is to be treated like a constant during the evaluation of limit.
$endgroup$
– Paramanand Singh
Jan 10 at 3:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To elaborate on a comment:
The premise of the theorem is that you are given a value of $x$ such that
$f'$ exists in $(x - varepsilon, x + varepsilon)$ and $f''(x)$ exists
for that value of $x$.
It is not given that $f$ has these properties for any other value of $x.$ Only for that one given value.
So if you started the proof with something like, "Let $x$ be a real number such that (various given facts about $f$),"
from that point until the end of the proof $x$ can be assumed to have one fixed value that satisfies the conditions.
Furthermore, when you write out the formal definition of $f''(x)$ in terms of a limit of an expression involving $f',$ as you did,
then while you are evaluating $f''(x)$ at any value of $x$
it is necessarily true that $x$ is the same value throughout the evaluation of $f''(x)$--namely, it's the point in the domain at which you're trying to find the derivative of $f'$.
That's generally true when you attempt take the derivative of any function
(in this case $f'$) at any point in its domain using this definition of the derivative.
Even if you had not already fixed a value of $x,$ this would fix its value while you took a limit as $hto 0$.
Hence it is correct to treat $x$ as a constant while taking your limit.
It's a good thing that you're thinking about things like this, however,
because sometimes people assume something is constant during some step of a proof when it actually isn't constant in that context, and then they get wrong results.
answered Jan 9 at 20:59
David KDavid K
53.2k341115
$endgroup$
add a comment |
$begingroup$
hint
$f''(x)$ exists, thus by Taylor-Young formula,
$$f(y)=f(x)+(y-x)f'(x)+frac{(y-x)^2}{2}f''(x)+(y-x)^2epsilon(y)$$
with $$lim_{yto x}epsilon(y)=0$$
replace $y$ by $x+h$ and by $x-h$ and sum.
answered Jan 9 at 20:17
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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votes
active
oldest
votes
$begingroup$
To elaborate on a comment:
The premise of the theorem is that you are given a value of $x$ such that
$f'$ exists in $(x - varepsilon, x + varepsilon)$ and $f''(x)$ exists
for that value of $x$.
It is not given that $f$ has these properties for any other value of $x.$ Only for that one given value.
So if you started the proof with something like, "Let $x$ be a real number such that (various given facts about $f$),"
from that point until the end of the proof $x$ can be assumed to have one fixed value that satisfies the conditions.
Furthermore, when you write out the formal definition of $f''(x)$ in terms of a limit of an expression involving $f',$ as you did,
then while you are evaluating $f''(x)$ at any value of $x$
it is necessarily true that $x$ is the same value throughout the evaluation of $f''(x)$--namely, it's the point in the domain at which you're trying to find the derivative of $f'$.
That's generally true when you attempt take the derivative of any function
(in this case $f'$) at any point in its domain using this definition of the derivative.
Even if you had not already fixed a value of $x,$ this would fix its value while you took a limit as $hto 0$.
Hence it is correct to treat $x$ as a constant while taking your limit.
It's a good thing that you're thinking about things like this, however,
because sometimes people assume something is constant during some step of a proof when it actually isn't constant in that context, and then they get wrong results.
answered Jan 9 at 20:59
David KDavid K
53.2k341115
$endgroup$
add a comment |
$begingroup$
To elaborate on a comment:
The premise of the theorem is that you are given a value of $x$ such that
$f'$ exists in $(x - varepsilon, x + varepsilon)$ and $f''(x)$ exists
for that value of $x$.
It is not given that $f$ has these properties for any other value of $x.$ Only for that one given value.
So if you started the proof with something like, "Let $x$ be a real number such that (various given facts about $f$),"
from that point until the end of the proof $x$ can be assumed to have one fixed value that satisfies the conditions.
Furthermore, when you write out the formal definition of $f''(x)$ in terms of a limit of an expression involving $f',$ as you did,
then while you are evaluating $f''(x)$ at any value of $x$
it is necessarily true that $x$ is the same value throughout the evaluation of $f''(x)$--namely, it's the point in the domain at which you're trying to find the derivative of $f'$.
That's generally true when you attempt take the derivative of any function
(in this case $f'$) at any point in its domain using this definition of the derivative.
Even if you had not already fixed a value of $x,$ this would fix its value while you took a limit as $hto 0$.
Hence it is correct to treat $x$ as a constant while taking your limit.
It's a good thing that you're thinking about things like this, however,
because sometimes people assume something is constant during some step of a proof when it actually isn't constant in that context, and then they get wrong results.
answered Jan 9 at 20:59
David KDavid K
53.2k341115
$endgroup$
add a comment |
$begingroup$
To elaborate on a comment:
The premise of the theorem is that you are given a value of $x$ such that
$f'$ exists in $(x - varepsilon, x + varepsilon)$ and $f''(x)$ exists
for that value of $x$.
It is not given that $f$ has these properties for any other value of $x.$ Only for that one given value.
So if you started the proof with something like, "Let $x$ be a real number such that (various given facts about $f$),"
from that point until the end of the proof $x$ can be assumed to have one fixed value that satisfies the conditions.
Furthermore, when you write out the formal definition of $f''(x)$ in terms of a limit of an expression involving $f',$ as you did,
then while you are evaluating $f''(x)$ at any value of $x$
it is necessarily true that $x$ is the same value throughout the evaluation of $f''(x)$--namely, it's the point in the domain at which you're trying to find the derivative of $f'$.
That's generally true when you attempt take the derivative of any function
(in this case $f'$) at any point in its domain using this definition of the derivative.
Even if you had not already fixed a value of $x,$ this would fix its value while you took a limit as $hto 0$.
Hence it is correct to treat $x$ as a constant while taking your limit.
It's a good thing that you're thinking about things like this, however,
because sometimes people assume something is constant during some step of a proof when it actually isn't constant in that context, and then they get wrong results.
answered Jan 9 at 20:59
David KDavid K
53.2k341115
$endgroup$
To elaborate on a comment:
The premise of the theorem is that you are given a value of $x$ such that
$f'$ exists in $(x - varepsilon, x + varepsilon)$ and $f''(x)$ exists
for that value of $x$.
It is not given that $f$ has these properties for any other value of $x.$ Only for that one given value.
So if you started the proof with something like, "Let $x$ be a real number such that (various given facts about $f$),"
from that point until the end of the proof $x$ can be assumed to have one fixed value that satisfies the conditions.
Furthermore, when you write out the formal definition of $f''(x)$ in terms of a limit of an expression involving $f',$ as you did,
then while you are evaluating $f''(x)$ at any value of $x$
it is necessarily true that $x$ is the same value throughout the evaluation of $f''(x)$--namely, it's the point in the domain at which you're trying to find the derivative of $f'$.
That's generally true when you attempt take the derivative of any function
(in this case $f'$) at any point in its domain using this definition of the derivative.
Even if you had not already fixed a value of $x,$ this would fix its value while you took a limit as $hto 0$.
Hence it is correct to treat $x$ as a constant while taking your limit.
It's a good thing that you're thinking about things like this, however,
because sometimes people assume something is constant during some step of a proof when it actually isn't constant in that context, and then they get wrong results.
answered Jan 9 at 20:59
David KDavid K
53.2k341115
answered Jan 9 at 20:59
David KDavid K
53.2k341115
answered Jan 9 at 20:59
David KDavid K
53.2k341115
answered Jan 9 at 20:59
David KDavid K
53.2k341115
53.2k341115
add a comment |
add a comment |
$begingroup$
hint
$f''(x)$ exists, thus by Taylor-Young formula,
$$f(y)=f(x)+(y-x)f'(x)+frac{(y-x)^2}{2}f''(x)+(y-x)^2epsilon(y)$$
with $$lim_{yto x}epsilon(y)=0$$
replace $y$ by $x+h$ and by $x-h$ and sum.
answered Jan 9 at 20:17
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
$begingroup$
hint
$f''(x)$ exists, thus by Taylor-Young formula,
$$f(y)=f(x)+(y-x)f'(x)+frac{(y-x)^2}{2}f''(x)+(y-x)^2epsilon(y)$$
with $$lim_{yto x}epsilon(y)=0$$
replace $y$ by $x+h$ and by $x-h$ and sum.
answered Jan 9 at 20:17
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
add a comment |
$begingroup$
hint
$f''(x)$ exists, thus by Taylor-Young formula,
$$f(y)=f(x)+(y-x)f'(x)+frac{(y-x)^2}{2}f''(x)+(y-x)^2epsilon(y)$$
with $$lim_{yto x}epsilon(y)=0$$
replace $y$ by $x+h$ and by $x-h$ and sum.
answered Jan 9 at 20:17
hamam_Abdallahhamam_Abdallah
38k21634
$endgroup$
hint
$f''(x)$ exists, thus by Taylor-Young formula,
$$f(y)=f(x)+(y-x)f'(x)+frac{(y-x)^2}{2}f''(x)+(y-x)^2epsilon(y)$$
with $$lim_{yto x}epsilon(y)=0$$
replace $y$ by $x+h$ and by $x-h$ and sum.
answered Jan 9 at 20:17
hamam_Abdallahhamam_Abdallah
38k21634
answered Jan 9 at 20:17
hamam_Abdallahhamam_Abdallah
38k21634
answered Jan 9 at 20:17
hamam_Abdallahhamam_Abdallah
38k21634
answered Jan 9 at 20:17
hamam_Abdallahhamam_Abdallah
38k21634
38k21634
add a comment |
add a comment |
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2
$begingroup$
Treating $x$ as a constant here is fine, because it is, in fact, constant.
$endgroup$
– user3482749
Jan 9 at 20:16
$begingroup$
Since your limit variable is $h$ any other variable (not dependent on $h$) is to be treated like a constant during the evaluation of limit.
$endgroup$
– Paramanand Singh
Jan 10 at 3:08