Mixing
Question on l'Hospital's rule
$begingroup$
Let $f:textbf{R} rightarrow textbf{R}$ be twice differentiable. Suppose that $f''(0) not= 0$. By MVT it follows that:
$forall x >0 $ there exists some $ lambda(x) in (0,1)$ such that $f(x)-f(0)=xf^{prime}(xlambda(x))$
Show that
$$displaystylelim_{xrightarrow0}frac{f(x)-f(0)-xf^{prime}(0)}{x^2} = frac{f^{primeprime}(0)}{2}$$
and hence that:
$$displaystylelim_{xrightarrow0}lambda(x) = frac{1}{2}$$
I have tried substituting the given equation and using l'Hospital's rule, however all attempts have lead to failures.
real-analysis
edited Jan 19 at 16:40
user1337
46110
asked Jan 19 at 16:14
LibabolLibabol
183
$endgroup$
|
show 5 more comments
$begingroup$
Let $f:textbf{R} rightarrow textbf{R}$ be twice differentiable. Suppose that $f''(0) not= 0$. By MVT it follows that:
$forall x >0 $ there exists some $ lambda(x) in (0,1)$ such that $f(x)-f(0)=xf^{prime}(xlambda(x))$
Show that
$$displaystylelim_{xrightarrow0}frac{f(x)-f(0)-xf^{prime}(0)}{x^2} = frac{f^{primeprime}(0)}{2}$$
and hence that:
$$displaystylelim_{xrightarrow0}lambda(x) = frac{1}{2}$$
I have tried substituting the given equation and using l'Hospital's rule, however all attempts have lead to failures.
real-analysis
edited Jan 19 at 16:40
user1337
46110
asked Jan 19 at 16:14
LibabolLibabol
183
$endgroup$
1
$begingroup$
It should be $f(x)-f(0)=xf^prime(xlambda(x))$ and not $f(x)-f(0)=xf(xlambda(x))$. Can you confirm?
$endgroup$
– mathcounterexamples.net
Jan 19 at 16:20
1
$begingroup$
Welcome to Maths SX! Where are you stuck the first or the second limit?
$endgroup$
– Bernard
Jan 19 at 16:21
$begingroup$
@mathcounterexamples.net Actually it's $f(x)−f(0)=xf(xλ(x))$. Have you ever seen this question before? This was in a problem set for an analysis class. Perhaps it was a typo.
$endgroup$
– Libabol
Jan 19 at 16:29
$begingroup$
@Libabol It must be $;f'(xlambda(x));$ and it is even written that "by the MVT"...otherwise it doesn't make much sense.
$endgroup$
– DonAntonio
Jan 19 at 16:31
$begingroup$
It was a typo I think since it is quite straight forward with $f(x)−f(0)=xf′(xλ(x))$.
$endgroup$
– Libabol
Jan 19 at 16:41
|
show 5 more comments
$begingroup$
Let $f:textbf{R} rightarrow textbf{R}$ be twice differentiable. Suppose that $f''(0) not= 0$. By MVT it follows that:
$forall x >0 $ there exists some $ lambda(x) in (0,1)$ such that $f(x)-f(0)=xf^{prime}(xlambda(x))$
Show that
$$displaystylelim_{xrightarrow0}frac{f(x)-f(0)-xf^{prime}(0)}{x^2} = frac{f^{primeprime}(0)}{2}$$
and hence that:
$$displaystylelim_{xrightarrow0}lambda(x) = frac{1}{2}$$
I have tried substituting the given equation and using l'Hospital's rule, however all attempts have lead to failures.
real-analysis
edited Jan 19 at 16:40
user1337
46110
asked Jan 19 at 16:14
LibabolLibabol
183
$endgroup$
Let $f:textbf{R} rightarrow textbf{R}$ be twice differentiable. Suppose that $f''(0) not= 0$. By MVT it follows that:
$forall x >0 $ there exists some $ lambda(x) in (0,1)$ such that $f(x)-f(0)=xf^{prime}(xlambda(x))$
Show that
$$displaystylelim_{xrightarrow0}frac{f(x)-f(0)-xf^{prime}(0)}{x^2} = frac{f^{primeprime}(0)}{2}$$
and hence that:
$$displaystylelim_{xrightarrow0}lambda(x) = frac{1}{2}$$
I have tried substituting the given equation and using l'Hospital's rule, however all attempts have lead to failures.
real-analysis
real-analysis
edited Jan 19 at 16:40
user1337
46110
asked Jan 19 at 16:14
LibabolLibabol
183
edited Jan 19 at 16:40
user1337
46110
asked Jan 19 at 16:14
LibabolLibabol
183
edited Jan 19 at 16:40
user1337
46110
edited Jan 19 at 16:40
user1337
46110
edited Jan 19 at 16:40
user1337
46110
46110
asked Jan 19 at 16:14
LibabolLibabol
183
asked Jan 19 at 16:14
LibabolLibabol
183
asked Jan 19 at 16:14
LibabolLibabol
183
183
1
$begingroup$
It should be $f(x)-f(0)=xf^prime(xlambda(x))$ and not $f(x)-f(0)=xf(xlambda(x))$. Can you confirm?
$endgroup$
– mathcounterexamples.net
Jan 19 at 16:20
1
$begingroup$
Welcome to Maths SX! Where are you stuck the first or the second limit?
$endgroup$
– Bernard
Jan 19 at 16:21
$begingroup$
@mathcounterexamples.net Actually it's $f(x)−f(0)=xf(xλ(x))$. Have you ever seen this question before? This was in a problem set for an analysis class. Perhaps it was a typo.
$endgroup$
– Libabol
Jan 19 at 16:29
$begingroup$
@Libabol It must be $;f'(xlambda(x));$ and it is even written that "by the MVT"...otherwise it doesn't make much sense.
$endgroup$
– DonAntonio
Jan 19 at 16:31
$begingroup$
It was a typo I think since it is quite straight forward with $f(x)−f(0)=xf′(xλ(x))$.
$endgroup$
– Libabol
Jan 19 at 16:41
|
show 5 more comments
1
$begingroup$
It should be $f(x)-f(0)=xf^prime(xlambda(x))$ and not $f(x)-f(0)=xf(xlambda(x))$. Can you confirm?
$endgroup$
– mathcounterexamples.net
Jan 19 at 16:20
1
$begingroup$
Welcome to Maths SX! Where are you stuck the first or the second limit?
$endgroup$
– Bernard
Jan 19 at 16:21
$begingroup$
@mathcounterexamples.net Actually it's $f(x)−f(0)=xf(xλ(x))$. Have you ever seen this question before? This was in a problem set for an analysis class. Perhaps it was a typo.
$endgroup$
– Libabol
Jan 19 at 16:29
$begingroup$
@Libabol It must be $;f'(xlambda(x));$ and it is even written that "by the MVT"...otherwise it doesn't make much sense.
$endgroup$
– DonAntonio
Jan 19 at 16:31
$begingroup$
It was a typo I think since it is quite straight forward with $f(x)−f(0)=xf′(xλ(x))$.
$endgroup$
– Libabol
Jan 19 at 16:41
1
1
$begingroup$
It should be $f(x)-f(0)=xf^prime(xlambda(x))$ and not $f(x)-f(0)=xf(xlambda(x))$. Can you confirm?
$endgroup$
– mathcounterexamples.net
Jan 19 at 16:20
$begingroup$
It should be $f(x)-f(0)=xf^prime(xlambda(x))$ and not $f(x)-f(0)=xf(xlambda(x))$. Can you confirm?
$endgroup$
– mathcounterexamples.net
Jan 19 at 16:20
1
1
$begingroup$
Welcome to Maths SX! Where are you stuck the first or the second limit?
$endgroup$
– Bernard
Jan 19 at 16:21
$begingroup$
Welcome to Maths SX! Where are you stuck the first or the second limit?
$endgroup$
– Bernard
Jan 19 at 16:21
$begingroup$
@mathcounterexamples.net Actually it's $f(x)−f(0)=xf(xλ(x))$. Have you ever seen this question before? This was in a problem set for an analysis class. Perhaps it was a typo.
$endgroup$
– Libabol
Jan 19 at 16:29
$begingroup$
@mathcounterexamples.net Actually it's $f(x)−f(0)=xf(xλ(x))$. Have you ever seen this question before? This was in a problem set for an analysis class. Perhaps it was a typo.
$endgroup$
– Libabol
Jan 19 at 16:29
$begingroup$
@Libabol It must be $;f'(xlambda(x));$ and it is even written that "by the MVT"...otherwise it doesn't make much sense.
$endgroup$
– DonAntonio
Jan 19 at 16:31
$begingroup$
@Libabol It must be $;f'(xlambda(x));$ and it is even written that "by the MVT"...otherwise it doesn't make much sense.
$endgroup$
– DonAntonio
Jan 19 at 16:31
$begingroup$
It was a typo I think since it is quite straight forward with $f(x)−f(0)=xf′(xλ(x))$.
$endgroup$
– Libabol
Jan 19 at 16:41
$begingroup$
It was a typo I think since it is quite straight forward with $f(x)−f(0)=xf′(xλ(x))$.
$endgroup$
– Libabol
Jan 19 at 16:41
|
show 5 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Let us compute the limit
$$
lim_{xto 0} frac{f(x) - f(0) - x f'(0)}{x^2},.
$$
By de l'Hopital's rule, this limit is equal to
$$
lim_{xto 0} frac{f'(x) - f'(0)}{2x},,
$$
provided that this second limit exists. On the other hand, by the definition of derivative, this second limit equals $f''(0)/2$.
Now, let $lambda(x)in (0,1)$ be such that $f(x) - f(0) = x f'(x lambda(x))$.
We have that
$$
g(x) := frac{f(x) - f(0) - x f'(0)}{x^2}=
frac{f'(xlambda(x)) - f'(0)}{x}=
frac{f'(xlambda(x)) - f'(0)}{xlambda(x)}, lambda(x)=: varphi(x), lambda(x).
$$
Clearly, by definition of derivative,
$$
lim_{xto 0}varphi(x) =
lim_{xto 0} frac{f'(xlambda(x)) - f'(0)}{xlambda(x)} = f''(0) neq 0,
$$
whereas, by the previous step, $lim_{xto 0} g(x) = f''(0)/2$.
Finally,
$$
lim_{xto 0}lambda (x) = lim_{xto 0} frac{g(x)}{varphi(x)} = frac{1}{2}.
$$
answered Jan 19 at 16:44
RigelRigel
11.3k11320
$endgroup$
add a comment |
$begingroup$
With the due correction as pointed out by Mathcounterexamples in the first comment, we get
using the Maclaurin series of order $1$ for $;f;$ (with Lagrange remainder) :
$$f(x)=f(0)+f'(0)x+frac{f''(xlambda(x))x^2}{2}implies f(x)-f(0)=f'(0)x+frac{f''(xlambda(x))x^2}{2}implies$$
$$frac{f(x)-f(0)-xf'(0)}{x^2}=frac{f''(xlambda(x))}2xrightarrow[xto0]{}frac{f''(0)}2 tag{1}$$
and on the other hand
$$frac{f(x)-f(0)-xf'(0)}{x^2}=frac{xf'(xlambda(x))}{x^2}-frac{f'(0)}x=frac{f'(xlambda(x))-f'(0)}x=$$
$$=frac{f'(xlambda(x))-f'
(0)}{xlambda(x)}cdotlambda(x)xrightarrow[xto0]{}f''(0)lambda(0) tag{2}$$
The above is assuming the limit of $;lambda(x);$ when $;xto0;$ exists.
edited Jan 19 at 16:47
mathcounterexamples.net
27k22157
answered Jan 19 at 16:37
DonAntonioDonAntonio
179k1494230
$endgroup$
$begingroup$
This argument assumes the existence of the limit $lim_{xto 0}lambda(x)$.
$endgroup$
– Rigel
Jan 19 at 16:39
$begingroup$
In the second part, we must deduce $limx→0λ(x)$ from the first part.
$endgroup$
– Libabol
Jan 19 at 16:44
$begingroup$
@Rogel Formally you're right. However this is immediate from DonAntonio equations $(1)$ and $(2)$ with a simple division... which is what you did in your answer.
$endgroup$
– mathcounterexamples.net
Jan 19 at 16:48
$begingroup$
@Rigel Yes, that's exactly what I wrote there...and as Libabol wrote, from the first part we can deduce that. There, he completed his own question's answer...:)
$endgroup$
– DonAntonio
Jan 19 at 16:48
add a comment |
$begingroup$
The key fact is that $f''(0)ne0$. The Inverse Function Theorem then gives you that $f'$ is invertible on some neighbourhood of $0$, so there exists $g$ with $g(f'(x))=x$ and
$$
g'(f'(0))=frac1{f''(g(f'(0)))}=frac1{f''(0)}.
$$ So there exists $g$, differentiable on some interval $Ini0$, with $g(f'(x))=x$. From $f(x)-f(0)=xf'(xlambda(x))$ we get, for $xin I$,
$$
lambda(x)=frac{gleft(frac{f(x)-f(0)}xright)}x=frac{gleft(frac{f(x)-f(0)}xright)-g(f'(0))}x=frac{gleft(frac{f(x)-f(0)}xright)-g(f'(0))}{frac{f(x)-f(0)}x-f'(0)},frac{ {f(x)-f(0)}-xf'(0)}{x^2}.
$$
Using the Mean Value Theorem,
$$
lambda(x)=g'(c(x)),frac{ {f(x)-f(0)}-xf'(0)}{x^2}
$$
with $c(x)$ between $f'(0)$ and $frac{f(x)-f(0)}x$. Then
$$
lim_{xto0}lambda(x)=g'(f'(0)),frac{f''(0)}2=frac1{f''(0)},frac{f''(0)}2=frac12.
$$
answered Jan 19 at 17:23
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let us compute the limit
$$
lim_{xto 0} frac{f(x) - f(0) - x f'(0)}{x^2},.
$$
By de l'Hopital's rule, this limit is equal to
$$
lim_{xto 0} frac{f'(x) - f'(0)}{2x},,
$$
provided that this second limit exists. On the other hand, by the definition of derivative, this second limit equals $f''(0)/2$.
Now, let $lambda(x)in (0,1)$ be such that $f(x) - f(0) = x f'(x lambda(x))$.
We have that
$$
g(x) := frac{f(x) - f(0) - x f'(0)}{x^2}=
frac{f'(xlambda(x)) - f'(0)}{x}=
frac{f'(xlambda(x)) - f'(0)}{xlambda(x)}, lambda(x)=: varphi(x), lambda(x).
$$
Clearly, by definition of derivative,
$$
lim_{xto 0}varphi(x) =
lim_{xto 0} frac{f'(xlambda(x)) - f'(0)}{xlambda(x)} = f''(0) neq 0,
$$
whereas, by the previous step, $lim_{xto 0} g(x) = f''(0)/2$.
Finally,
$$
lim_{xto 0}lambda (x) = lim_{xto 0} frac{g(x)}{varphi(x)} = frac{1}{2}.
$$
answered Jan 19 at 16:44
RigelRigel
11.3k11320
$endgroup$
add a comment |
$begingroup$
Let us compute the limit
$$
lim_{xto 0} frac{f(x) - f(0) - x f'(0)}{x^2},.
$$
By de l'Hopital's rule, this limit is equal to
$$
lim_{xto 0} frac{f'(x) - f'(0)}{2x},,
$$
provided that this second limit exists. On the other hand, by the definition of derivative, this second limit equals $f''(0)/2$.
Now, let $lambda(x)in (0,1)$ be such that $f(x) - f(0) = x f'(x lambda(x))$.
We have that
$$
g(x) := frac{f(x) - f(0) - x f'(0)}{x^2}=
frac{f'(xlambda(x)) - f'(0)}{x}=
frac{f'(xlambda(x)) - f'(0)}{xlambda(x)}, lambda(x)=: varphi(x), lambda(x).
$$
Clearly, by definition of derivative,
$$
lim_{xto 0}varphi(x) =
lim_{xto 0} frac{f'(xlambda(x)) - f'(0)}{xlambda(x)} = f''(0) neq 0,
$$
whereas, by the previous step, $lim_{xto 0} g(x) = f''(0)/2$.
Finally,
$$
lim_{xto 0}lambda (x) = lim_{xto 0} frac{g(x)}{varphi(x)} = frac{1}{2}.
$$
answered Jan 19 at 16:44
RigelRigel
11.3k11320
$endgroup$
add a comment |
$begingroup$
Let us compute the limit
$$
lim_{xto 0} frac{f(x) - f(0) - x f'(0)}{x^2},.
$$
By de l'Hopital's rule, this limit is equal to
$$
lim_{xto 0} frac{f'(x) - f'(0)}{2x},,
$$
provided that this second limit exists. On the other hand, by the definition of derivative, this second limit equals $f''(0)/2$.
Now, let $lambda(x)in (0,1)$ be such that $f(x) - f(0) = x f'(x lambda(x))$.
We have that
$$
g(x) := frac{f(x) - f(0) - x f'(0)}{x^2}=
frac{f'(xlambda(x)) - f'(0)}{x}=
frac{f'(xlambda(x)) - f'(0)}{xlambda(x)}, lambda(x)=: varphi(x), lambda(x).
$$
Clearly, by definition of derivative,
$$
lim_{xto 0}varphi(x) =
lim_{xto 0} frac{f'(xlambda(x)) - f'(0)}{xlambda(x)} = f''(0) neq 0,
$$
whereas, by the previous step, $lim_{xto 0} g(x) = f''(0)/2$.
Finally,
$$
lim_{xto 0}lambda (x) = lim_{xto 0} frac{g(x)}{varphi(x)} = frac{1}{2}.
$$
answered Jan 19 at 16:44
RigelRigel
11.3k11320
$endgroup$
Let us compute the limit
$$
lim_{xto 0} frac{f(x) - f(0) - x f'(0)}{x^2},.
$$
By de l'Hopital's rule, this limit is equal to
$$
lim_{xto 0} frac{f'(x) - f'(0)}{2x},,
$$
provided that this second limit exists. On the other hand, by the definition of derivative, this second limit equals $f''(0)/2$.
Now, let $lambda(x)in (0,1)$ be such that $f(x) - f(0) = x f'(x lambda(x))$.
We have that
$$
g(x) := frac{f(x) - f(0) - x f'(0)}{x^2}=
frac{f'(xlambda(x)) - f'(0)}{x}=
frac{f'(xlambda(x)) - f'(0)}{xlambda(x)}, lambda(x)=: varphi(x), lambda(x).
$$
Clearly, by definition of derivative,
$$
lim_{xto 0}varphi(x) =
lim_{xto 0} frac{f'(xlambda(x)) - f'(0)}{xlambda(x)} = f''(0) neq 0,
$$
whereas, by the previous step, $lim_{xto 0} g(x) = f''(0)/2$.
Finally,
$$
lim_{xto 0}lambda (x) = lim_{xto 0} frac{g(x)}{varphi(x)} = frac{1}{2}.
$$
answered Jan 19 at 16:44
RigelRigel
11.3k11320
answered Jan 19 at 16:44
RigelRigel
11.3k11320
answered Jan 19 at 16:44
RigelRigel
11.3k11320
answered Jan 19 at 16:44
RigelRigel
11.3k11320
11.3k11320
add a comment |
add a comment |
$begingroup$
With the due correction as pointed out by Mathcounterexamples in the first comment, we get
using the Maclaurin series of order $1$ for $;f;$ (with Lagrange remainder) :
$$f(x)=f(0)+f'(0)x+frac{f''(xlambda(x))x^2}{2}implies f(x)-f(0)=f'(0)x+frac{f''(xlambda(x))x^2}{2}implies$$
$$frac{f(x)-f(0)-xf'(0)}{x^2}=frac{f''(xlambda(x))}2xrightarrow[xto0]{}frac{f''(0)}2 tag{1}$$
and on the other hand
$$frac{f(x)-f(0)-xf'(0)}{x^2}=frac{xf'(xlambda(x))}{x^2}-frac{f'(0)}x=frac{f'(xlambda(x))-f'(0)}x=$$
$$=frac{f'(xlambda(x))-f'
(0)}{xlambda(x)}cdotlambda(x)xrightarrow[xto0]{}f''(0)lambda(0) tag{2}$$
The above is assuming the limit of $;lambda(x);$ when $;xto0;$ exists.
edited Jan 19 at 16:47
mathcounterexamples.net
27k22157
answered Jan 19 at 16:37
DonAntonioDonAntonio
179k1494230
$endgroup$
$begingroup$
This argument assumes the existence of the limit $lim_{xto 0}lambda(x)$.
$endgroup$
– Rigel
Jan 19 at 16:39
$begingroup$
In the second part, we must deduce $limx→0λ(x)$ from the first part.
$endgroup$
– Libabol
Jan 19 at 16:44
$begingroup$
@Rogel Formally you're right. However this is immediate from DonAntonio equations $(1)$ and $(2)$ with a simple division... which is what you did in your answer.
$endgroup$
– mathcounterexamples.net
Jan 19 at 16:48
$begingroup$
@Rigel Yes, that's exactly what I wrote there...and as Libabol wrote, from the first part we can deduce that. There, he completed his own question's answer...:)
$endgroup$
– DonAntonio
Jan 19 at 16:48
add a comment |
$begingroup$
With the due correction as pointed out by Mathcounterexamples in the first comment, we get
using the Maclaurin series of order $1$ for $;f;$ (with Lagrange remainder) :
$$f(x)=f(0)+f'(0)x+frac{f''(xlambda(x))x^2}{2}implies f(x)-f(0)=f'(0)x+frac{f''(xlambda(x))x^2}{2}implies$$
$$frac{f(x)-f(0)-xf'(0)}{x^2}=frac{f''(xlambda(x))}2xrightarrow[xto0]{}frac{f''(0)}2 tag{1}$$
and on the other hand
$$frac{f(x)-f(0)-xf'(0)}{x^2}=frac{xf'(xlambda(x))}{x^2}-frac{f'(0)}x=frac{f'(xlambda(x))-f'(0)}x=$$
$$=frac{f'(xlambda(x))-f'
(0)}{xlambda(x)}cdotlambda(x)xrightarrow[xto0]{}f''(0)lambda(0) tag{2}$$
The above is assuming the limit of $;lambda(x);$ when $;xto0;$ exists.
edited Jan 19 at 16:47
mathcounterexamples.net
27k22157
answered Jan 19 at 16:37
DonAntonioDonAntonio
179k1494230
$endgroup$
$begingroup$
This argument assumes the existence of the limit $lim_{xto 0}lambda(x)$.
$endgroup$
– Rigel
Jan 19 at 16:39
$begingroup$
In the second part, we must deduce $limx→0λ(x)$ from the first part.
$endgroup$
– Libabol
Jan 19 at 16:44
$begingroup$
@Rogel Formally you're right. However this is immediate from DonAntonio equations $(1)$ and $(2)$ with a simple division... which is what you did in your answer.
$endgroup$
– mathcounterexamples.net
Jan 19 at 16:48
$begingroup$
@Rigel Yes, that's exactly what I wrote there...and as Libabol wrote, from the first part we can deduce that. There, he completed his own question's answer...:)
$endgroup$
– DonAntonio
Jan 19 at 16:48
add a comment |
$begingroup$
With the due correction as pointed out by Mathcounterexamples in the first comment, we get
using the Maclaurin series of order $1$ for $;f;$ (with Lagrange remainder) :
$$f(x)=f(0)+f'(0)x+frac{f''(xlambda(x))x^2}{2}implies f(x)-f(0)=f'(0)x+frac{f''(xlambda(x))x^2}{2}implies$$
$$frac{f(x)-f(0)-xf'(0)}{x^2}=frac{f''(xlambda(x))}2xrightarrow[xto0]{}frac{f''(0)}2 tag{1}$$
and on the other hand
$$frac{f(x)-f(0)-xf'(0)}{x^2}=frac{xf'(xlambda(x))}{x^2}-frac{f'(0)}x=frac{f'(xlambda(x))-f'(0)}x=$$
$$=frac{f'(xlambda(x))-f'
(0)}{xlambda(x)}cdotlambda(x)xrightarrow[xto0]{}f''(0)lambda(0) tag{2}$$
The above is assuming the limit of $;lambda(x);$ when $;xto0;$ exists.
edited Jan 19 at 16:47
mathcounterexamples.net
27k22157
answered Jan 19 at 16:37
DonAntonioDonAntonio
179k1494230
$endgroup$
With the due correction as pointed out by Mathcounterexamples in the first comment, we get
using the Maclaurin series of order $1$ for $;f;$ (with Lagrange remainder) :
$$f(x)=f(0)+f'(0)x+frac{f''(xlambda(x))x^2}{2}implies f(x)-f(0)=f'(0)x+frac{f''(xlambda(x))x^2}{2}implies$$
$$frac{f(x)-f(0)-xf'(0)}{x^2}=frac{f''(xlambda(x))}2xrightarrow[xto0]{}frac{f''(0)}2 tag{1}$$
and on the other hand
$$frac{f(x)-f(0)-xf'(0)}{x^2}=frac{xf'(xlambda(x))}{x^2}-frac{f'(0)}x=frac{f'(xlambda(x))-f'(0)}x=$$
$$=frac{f'(xlambda(x))-f'
(0)}{xlambda(x)}cdotlambda(x)xrightarrow[xto0]{}f''(0)lambda(0) tag{2}$$
The above is assuming the limit of $;lambda(x);$ when $;xto0;$ exists.
edited Jan 19 at 16:47
mathcounterexamples.net
27k22157
answered Jan 19 at 16:37
DonAntonioDonAntonio
179k1494230
edited Jan 19 at 16:47
mathcounterexamples.net
27k22157
edited Jan 19 at 16:47
mathcounterexamples.net
27k22157
edited Jan 19 at 16:47
mathcounterexamples.net
27k22157
27k22157
answered Jan 19 at 16:37
DonAntonioDonAntonio
179k1494230
answered Jan 19 at 16:37
DonAntonioDonAntonio
179k1494230
answered Jan 19 at 16:37
DonAntonioDonAntonio
179k1494230
179k1494230
$begingroup$
This argument assumes the existence of the limit $lim_{xto 0}lambda(x)$.
$endgroup$
– Rigel
Jan 19 at 16:39
$begingroup$
In the second part, we must deduce $limx→0λ(x)$ from the first part.
$endgroup$
– Libabol
Jan 19 at 16:44
$begingroup$
@Rogel Formally you're right. However this is immediate from DonAntonio equations $(1)$ and $(2)$ with a simple division... which is what you did in your answer.
$endgroup$
– mathcounterexamples.net
Jan 19 at 16:48
$begingroup$
@Rigel Yes, that's exactly what I wrote there...and as Libabol wrote, from the first part we can deduce that. There, he completed his own question's answer...:)
$endgroup$
– DonAntonio
Jan 19 at 16:48
add a comment |
$begingroup$
This argument assumes the existence of the limit $lim_{xto 0}lambda(x)$.
$endgroup$
– Rigel
Jan 19 at 16:39
$begingroup$
In the second part, we must deduce $limx→0λ(x)$ from the first part.
$endgroup$
– Libabol
Jan 19 at 16:44
$begingroup$
@Rogel Formally you're right. However this is immediate from DonAntonio equations $(1)$ and $(2)$ with a simple division... which is what you did in your answer.
$endgroup$
– mathcounterexamples.net
Jan 19 at 16:48
$begingroup$
@Rigel Yes, that's exactly what I wrote there...and as Libabol wrote, from the first part we can deduce that. There, he completed his own question's answer...:)
$endgroup$
– DonAntonio
Jan 19 at 16:48
$begingroup$
This argument assumes the existence of the limit $lim_{xto 0}lambda(x)$.
$endgroup$
– Rigel
Jan 19 at 16:39
$begingroup$
This argument assumes the existence of the limit $lim_{xto 0}lambda(x)$.
$endgroup$
– Rigel
Jan 19 at 16:39
$begingroup$
In the second part, we must deduce $limx→0λ(x)$ from the first part.
$endgroup$
– Libabol
Jan 19 at 16:44
$begingroup$
In the second part, we must deduce $limx→0λ(x)$ from the first part.
$endgroup$
– Libabol
Jan 19 at 16:44
$begingroup$
@Rogel Formally you're right. However this is immediate from DonAntonio equations $(1)$ and $(2)$ with a simple division... which is what you did in your answer.
$endgroup$
– mathcounterexamples.net
Jan 19 at 16:48
$begingroup$
@Rogel Formally you're right. However this is immediate from DonAntonio equations $(1)$ and $(2)$ with a simple division... which is what you did in your answer.
$endgroup$
– mathcounterexamples.net
Jan 19 at 16:48
$begingroup$
@Rigel Yes, that's exactly what I wrote there...and as Libabol wrote, from the first part we can deduce that. There, he completed his own question's answer...:)
$endgroup$
– DonAntonio
Jan 19 at 16:48
$begingroup$
@Rigel Yes, that's exactly what I wrote there...and as Libabol wrote, from the first part we can deduce that. There, he completed his own question's answer...:)
$endgroup$
– DonAntonio
Jan 19 at 16:48
add a comment |
$begingroup$
The key fact is that $f''(0)ne0$. The Inverse Function Theorem then gives you that $f'$ is invertible on some neighbourhood of $0$, so there exists $g$ with $g(f'(x))=x$ and
$$
g'(f'(0))=frac1{f''(g(f'(0)))}=frac1{f''(0)}.
$$ So there exists $g$, differentiable on some interval $Ini0$, with $g(f'(x))=x$. From $f(x)-f(0)=xf'(xlambda(x))$ we get, for $xin I$,
$$
lambda(x)=frac{gleft(frac{f(x)-f(0)}xright)}x=frac{gleft(frac{f(x)-f(0)}xright)-g(f'(0))}x=frac{gleft(frac{f(x)-f(0)}xright)-g(f'(0))}{frac{f(x)-f(0)}x-f'(0)},frac{ {f(x)-f(0)}-xf'(0)}{x^2}.
$$
Using the Mean Value Theorem,
$$
lambda(x)=g'(c(x)),frac{ {f(x)-f(0)}-xf'(0)}{x^2}
$$
with $c(x)$ between $f'(0)$ and $frac{f(x)-f(0)}x$. Then
$$
lim_{xto0}lambda(x)=g'(f'(0)),frac{f''(0)}2=frac1{f''(0)},frac{f''(0)}2=frac12.
$$
answered Jan 19 at 17:23
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
The key fact is that $f''(0)ne0$. The Inverse Function Theorem then gives you that $f'$ is invertible on some neighbourhood of $0$, so there exists $g$ with $g(f'(x))=x$ and
$$
g'(f'(0))=frac1{f''(g(f'(0)))}=frac1{f''(0)}.
$$ So there exists $g$, differentiable on some interval $Ini0$, with $g(f'(x))=x$. From $f(x)-f(0)=xf'(xlambda(x))$ we get, for $xin I$,
$$
lambda(x)=frac{gleft(frac{f(x)-f(0)}xright)}x=frac{gleft(frac{f(x)-f(0)}xright)-g(f'(0))}x=frac{gleft(frac{f(x)-f(0)}xright)-g(f'(0))}{frac{f(x)-f(0)}x-f'(0)},frac{ {f(x)-f(0)}-xf'(0)}{x^2}.
$$
Using the Mean Value Theorem,
$$
lambda(x)=g'(c(x)),frac{ {f(x)-f(0)}-xf'(0)}{x^2}
$$
with $c(x)$ between $f'(0)$ and $frac{f(x)-f(0)}x$. Then
$$
lim_{xto0}lambda(x)=g'(f'(0)),frac{f''(0)}2=frac1{f''(0)},frac{f''(0)}2=frac12.
$$
answered Jan 19 at 17:23
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
The key fact is that $f''(0)ne0$. The Inverse Function Theorem then gives you that $f'$ is invertible on some neighbourhood of $0$, so there exists $g$ with $g(f'(x))=x$ and
$$
g'(f'(0))=frac1{f''(g(f'(0)))}=frac1{f''(0)}.
$$ So there exists $g$, differentiable on some interval $Ini0$, with $g(f'(x))=x$. From $f(x)-f(0)=xf'(xlambda(x))$ we get, for $xin I$,
$$
lambda(x)=frac{gleft(frac{f(x)-f(0)}xright)}x=frac{gleft(frac{f(x)-f(0)}xright)-g(f'(0))}x=frac{gleft(frac{f(x)-f(0)}xright)-g(f'(0))}{frac{f(x)-f(0)}x-f'(0)},frac{ {f(x)-f(0)}-xf'(0)}{x^2}.
$$
Using the Mean Value Theorem,
$$
lambda(x)=g'(c(x)),frac{ {f(x)-f(0)}-xf'(0)}{x^2}
$$
with $c(x)$ between $f'(0)$ and $frac{f(x)-f(0)}x$. Then
$$
lim_{xto0}lambda(x)=g'(f'(0)),frac{f''(0)}2=frac1{f''(0)},frac{f''(0)}2=frac12.
$$
answered Jan 19 at 17:23
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
The key fact is that $f''(0)ne0$. The Inverse Function Theorem then gives you that $f'$ is invertible on some neighbourhood of $0$, so there exists $g$ with $g(f'(x))=x$ and
$$
g'(f'(0))=frac1{f''(g(f'(0)))}=frac1{f''(0)}.
$$ So there exists $g$, differentiable on some interval $Ini0$, with $g(f'(x))=x$. From $f(x)-f(0)=xf'(xlambda(x))$ we get, for $xin I$,
$$
lambda(x)=frac{gleft(frac{f(x)-f(0)}xright)}x=frac{gleft(frac{f(x)-f(0)}xright)-g(f'(0))}x=frac{gleft(frac{f(x)-f(0)}xright)-g(f'(0))}{frac{f(x)-f(0)}x-f'(0)},frac{ {f(x)-f(0)}-xf'(0)}{x^2}.
$$
Using the Mean Value Theorem,
$$
lambda(x)=g'(c(x)),frac{ {f(x)-f(0)}-xf'(0)}{x^2}
$$
with $c(x)$ between $f'(0)$ and $frac{f(x)-f(0)}x$. Then
$$
lim_{xto0}lambda(x)=g'(f'(0)),frac{f''(0)}2=frac1{f''(0)},frac{f''(0)}2=frac12.
$$
answered Jan 19 at 17:23
Martin ArgeramiMartin Argerami
127k1182183
edited Jan 20 at 19:41
answered Jan 19 at 17:23
Martin ArgeramiMartin Argerami
127k1182183
answered Jan 19 at 17:23
Martin ArgeramiMartin Argerami
127k1182183
answered Jan 19 at 17:23
Martin ArgeramiMartin Argerami
127k1182183
127k1182183
add a comment |
add a comment |
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1
$begingroup$
It should be $f(x)-f(0)=xf^prime(xlambda(x))$ and not $f(x)-f(0)=xf(xlambda(x))$. Can you confirm?
$endgroup$
– mathcounterexamples.net
Jan 19 at 16:20
1
$begingroup$
Welcome to Maths SX! Where are you stuck the first or the second limit?
$endgroup$
– Bernard
Jan 19 at 16:21
$begingroup$
@mathcounterexamples.net Actually it's $f(x)−f(0)=xf(xλ(x))$. Have you ever seen this question before? This was in a problem set for an analysis class. Perhaps it was a typo.
$endgroup$
– Libabol
Jan 19 at 16:29
$begingroup$
@Libabol It must be $;f'(xlambda(x));$ and it is even written that "by the MVT"...otherwise it doesn't make much sense.
$endgroup$
– DonAntonio
Jan 19 at 16:31
$begingroup$
It was a typo I think since it is quite straight forward with $f(x)−f(0)=xf′(xλ(x))$.
$endgroup$
– Libabol
Jan 19 at 16:41