Mixing
Find $alpha in mathbb{R}$ s.t. the second derivative ($x=0$) of a function exists.
$begingroup$
Find $alpha in mathbb{R}$ s.t. the second derivative ($x=0$) of the following function exists.
$f(x) =
begin{cases}
e^{-frac{1}{x}}, & text{if $x$}gt 0 \[2ex]
sin(x)+alpha x^2-log(1+x), & text{if $x$ } leq0
end{cases}$
How to deal with this kind of exercise? Is it enough to use Taylor at second order and then looking for the differentiability of the function?
real-analysis derivatives
asked Jan 14 at 22:04
ArcticmonkeyArcticmonkey
13911
$endgroup$
add a comment |
$begingroup$
Find $alpha in mathbb{R}$ s.t. the second derivative ($x=0$) of the following function exists.
$f(x) =
begin{cases}
e^{-frac{1}{x}}, & text{if $x$}gt 0 \[2ex]
sin(x)+alpha x^2-log(1+x), & text{if $x$ } leq0
end{cases}$
How to deal with this kind of exercise? Is it enough to use Taylor at second order and then looking for the differentiability of the function?
real-analysis derivatives
asked Jan 14 at 22:04
ArcticmonkeyArcticmonkey
13911
$endgroup$
$begingroup$
$e^{-1/x} $ doesn't have a Taylor series centered at $0$.
$endgroup$
– Acccumulation
Jan 14 at 22:31
add a comment |
$begingroup$
Find $alpha in mathbb{R}$ s.t. the second derivative ($x=0$) of the following function exists.
$f(x) =
begin{cases}
e^{-frac{1}{x}}, & text{if $x$}gt 0 \[2ex]
sin(x)+alpha x^2-log(1+x), & text{if $x$ } leq0
end{cases}$
How to deal with this kind of exercise? Is it enough to use Taylor at second order and then looking for the differentiability of the function?
real-analysis derivatives
asked Jan 14 at 22:04
ArcticmonkeyArcticmonkey
13911
$endgroup$
Find $alpha in mathbb{R}$ s.t. the second derivative ($x=0$) of the following function exists.
$f(x) =
begin{cases}
e^{-frac{1}{x}}, & text{if $x$}gt 0 \[2ex]
sin(x)+alpha x^2-log(1+x), & text{if $x$ } leq0
end{cases}$
How to deal with this kind of exercise? Is it enough to use Taylor at second order and then looking for the differentiability of the function?
real-analysis derivatives
real-analysis derivatives
asked Jan 14 at 22:04
ArcticmonkeyArcticmonkey
13911
asked Jan 14 at 22:04
ArcticmonkeyArcticmonkey
13911
asked Jan 14 at 22:04
ArcticmonkeyArcticmonkey
13911
asked Jan 14 at 22:04
ArcticmonkeyArcticmonkey
13911
asked Jan 14 at 22:04
ArcticmonkeyArcticmonkey
13911
13911
$begingroup$
$e^{-1/x} $ doesn't have a Taylor series centered at $0$.
$endgroup$
– Acccumulation
Jan 14 at 22:31
add a comment |
$begingroup$
$e^{-1/x} $ doesn't have a Taylor series centered at $0$.
$endgroup$
– Acccumulation
Jan 14 at 22:31
$begingroup$
$e^{-1/x} $ doesn't have a Taylor series centered at $0$.
$endgroup$
– Acccumulation
Jan 14 at 22:31
$begingroup$
$e^{-1/x} $ doesn't have a Taylor series centered at $0$.
$endgroup$
– Acccumulation
Jan 14 at 22:31
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Taylor at second order means that you have already calculated the second derivative. Just calculate the second derivative on the left and on the right, and make them equal. $$left(e^{-frac 1x}right)''=-frac{2x-1}{x^4}e^{-frac 1x}$$
This term goes to $0$ as $xto 0$.
For the other side the second derivative is $$-sin x+2alpha+frac{1}{(x+1)^2}$$
The limit at $0$ is $2alpha+1$. So $2alpha+1=0$ should give you the answer.
answered Jan 14 at 22:22
AndreiAndrei
12.1k21126
$endgroup$
add a comment |
$begingroup$
The function is continuous at $0$, by checking the limits.
The derivative is
$$
f'(x)=begin{cases}
dfrac{e^{-1/x}}{x^2} & x>0 \[4px]
cos x+2alpha x-dfrac{1}{1+x} & x<0
end{cases}
$$
Since the limits of $f'$ for $xto0$ from the left and from the right are both equal to $0$, we can say that the function is differentiable at $0$ and that the derivative is continuous. It follows from l'Hôpital's theorem.
Now apply the same technique, with the difference that
$$
lim_{xto0^+}f'(x)=0
$$
(check it) and
$$
lim_{xto0^-}f'(x)=lim_{xto0^-}left(-sin x+2alpha+frac{1}{(1+x)^2}right)=2alpha+1
$$
A sufficient condition for differentiability of $f'$ at $0$ is $alpha=-1/2$. Is it also necessary? Note that l'Hôpital can be used only one way.
answered Jan 14 at 22:37
egregegreg
182k1485204
$endgroup$
add a comment |
$begingroup$
Hint
A function whose right and left derivatives are given in a point, is
differentiable at that point if those right and left derivatives are
equal.
Based on this, try to differentiate the function twice, each time checking for which values of $a$ are the right and left derivatives in $x=0$ equal.
answered Jan 14 at 23:04
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Taylor at second order means that you have already calculated the second derivative. Just calculate the second derivative on the left and on the right, and make them equal. $$left(e^{-frac 1x}right)''=-frac{2x-1}{x^4}e^{-frac 1x}$$
This term goes to $0$ as $xto 0$.
For the other side the second derivative is $$-sin x+2alpha+frac{1}{(x+1)^2}$$
The limit at $0$ is $2alpha+1$. So $2alpha+1=0$ should give you the answer.
answered Jan 14 at 22:22
AndreiAndrei
12.1k21126
$endgroup$
add a comment |
$begingroup$
Taylor at second order means that you have already calculated the second derivative. Just calculate the second derivative on the left and on the right, and make them equal. $$left(e^{-frac 1x}right)''=-frac{2x-1}{x^4}e^{-frac 1x}$$
This term goes to $0$ as $xto 0$.
For the other side the second derivative is $$-sin x+2alpha+frac{1}{(x+1)^2}$$
The limit at $0$ is $2alpha+1$. So $2alpha+1=0$ should give you the answer.
answered Jan 14 at 22:22
AndreiAndrei
12.1k21126
$endgroup$
add a comment |
$begingroup$
Taylor at second order means that you have already calculated the second derivative. Just calculate the second derivative on the left and on the right, and make them equal. $$left(e^{-frac 1x}right)''=-frac{2x-1}{x^4}e^{-frac 1x}$$
This term goes to $0$ as $xto 0$.
For the other side the second derivative is $$-sin x+2alpha+frac{1}{(x+1)^2}$$
The limit at $0$ is $2alpha+1$. So $2alpha+1=0$ should give you the answer.
answered Jan 14 at 22:22
AndreiAndrei
12.1k21126
$endgroup$
Taylor at second order means that you have already calculated the second derivative. Just calculate the second derivative on the left and on the right, and make them equal. $$left(e^{-frac 1x}right)''=-frac{2x-1}{x^4}e^{-frac 1x}$$
This term goes to $0$ as $xto 0$.
For the other side the second derivative is $$-sin x+2alpha+frac{1}{(x+1)^2}$$
The limit at $0$ is $2alpha+1$. So $2alpha+1=0$ should give you the answer.
answered Jan 14 at 22:22
AndreiAndrei
12.1k21126
answered Jan 14 at 22:22
AndreiAndrei
12.1k21126
answered Jan 14 at 22:22
AndreiAndrei
12.1k21126
answered Jan 14 at 22:22
AndreiAndrei
12.1k21126
12.1k21126
add a comment |
add a comment |
$begingroup$
The function is continuous at $0$, by checking the limits.
The derivative is
$$
f'(x)=begin{cases}
dfrac{e^{-1/x}}{x^2} & x>0 \[4px]
cos x+2alpha x-dfrac{1}{1+x} & x<0
end{cases}
$$
Since the limits of $f'$ for $xto0$ from the left and from the right are both equal to $0$, we can say that the function is differentiable at $0$ and that the derivative is continuous. It follows from l'Hôpital's theorem.
Now apply the same technique, with the difference that
$$
lim_{xto0^+}f'(x)=0
$$
(check it) and
$$
lim_{xto0^-}f'(x)=lim_{xto0^-}left(-sin x+2alpha+frac{1}{(1+x)^2}right)=2alpha+1
$$
A sufficient condition for differentiability of $f'$ at $0$ is $alpha=-1/2$. Is it also necessary? Note that l'Hôpital can be used only one way.
answered Jan 14 at 22:37
egregegreg
182k1485204
$endgroup$
add a comment |
$begingroup$
The function is continuous at $0$, by checking the limits.
The derivative is
$$
f'(x)=begin{cases}
dfrac{e^{-1/x}}{x^2} & x>0 \[4px]
cos x+2alpha x-dfrac{1}{1+x} & x<0
end{cases}
$$
Since the limits of $f'$ for $xto0$ from the left and from the right are both equal to $0$, we can say that the function is differentiable at $0$ and that the derivative is continuous. It follows from l'Hôpital's theorem.
Now apply the same technique, with the difference that
$$
lim_{xto0^+}f'(x)=0
$$
(check it) and
$$
lim_{xto0^-}f'(x)=lim_{xto0^-}left(-sin x+2alpha+frac{1}{(1+x)^2}right)=2alpha+1
$$
A sufficient condition for differentiability of $f'$ at $0$ is $alpha=-1/2$. Is it also necessary? Note that l'Hôpital can be used only one way.
answered Jan 14 at 22:37
egregegreg
182k1485204
$endgroup$
add a comment |
$begingroup$
The function is continuous at $0$, by checking the limits.
The derivative is
$$
f'(x)=begin{cases}
dfrac{e^{-1/x}}{x^2} & x>0 \[4px]
cos x+2alpha x-dfrac{1}{1+x} & x<0
end{cases}
$$
Since the limits of $f'$ for $xto0$ from the left and from the right are both equal to $0$, we can say that the function is differentiable at $0$ and that the derivative is continuous. It follows from l'Hôpital's theorem.
Now apply the same technique, with the difference that
$$
lim_{xto0^+}f'(x)=0
$$
(check it) and
$$
lim_{xto0^-}f'(x)=lim_{xto0^-}left(-sin x+2alpha+frac{1}{(1+x)^2}right)=2alpha+1
$$
A sufficient condition for differentiability of $f'$ at $0$ is $alpha=-1/2$. Is it also necessary? Note that l'Hôpital can be used only one way.
answered Jan 14 at 22:37
egregegreg
182k1485204
$endgroup$
The function is continuous at $0$, by checking the limits.
The derivative is
$$
f'(x)=begin{cases}
dfrac{e^{-1/x}}{x^2} & x>0 \[4px]
cos x+2alpha x-dfrac{1}{1+x} & x<0
end{cases}
$$
Since the limits of $f'$ for $xto0$ from the left and from the right are both equal to $0$, we can say that the function is differentiable at $0$ and that the derivative is continuous. It follows from l'Hôpital's theorem.
Now apply the same technique, with the difference that
$$
lim_{xto0^+}f'(x)=0
$$
(check it) and
$$
lim_{xto0^-}f'(x)=lim_{xto0^-}left(-sin x+2alpha+frac{1}{(1+x)^2}right)=2alpha+1
$$
A sufficient condition for differentiability of $f'$ at $0$ is $alpha=-1/2$. Is it also necessary? Note that l'Hôpital can be used only one way.
answered Jan 14 at 22:37
egregegreg
182k1485204
answered Jan 14 at 22:37
egregegreg
182k1485204
answered Jan 14 at 22:37
egregegreg
182k1485204
answered Jan 14 at 22:37
egregegreg
182k1485204
182k1485204
add a comment |
add a comment |
$begingroup$
Hint
A function whose right and left derivatives are given in a point, is
differentiable at that point if those right and left derivatives are
equal.
Based on this, try to differentiate the function twice, each time checking for which values of $a$ are the right and left derivatives in $x=0$ equal.
answered Jan 14 at 23:04
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
Hint
A function whose right and left derivatives are given in a point, is
differentiable at that point if those right and left derivatives are
equal.
Based on this, try to differentiate the function twice, each time checking for which values of $a$ are the right and left derivatives in $x=0$ equal.
answered Jan 14 at 23:04
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
add a comment |
$begingroup$
Hint
A function whose right and left derivatives are given in a point, is
differentiable at that point if those right and left derivatives are
equal.
Based on this, try to differentiate the function twice, each time checking for which values of $a$ are the right and left derivatives in $x=0$ equal.
answered Jan 14 at 23:04
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
Hint
A function whose right and left derivatives are given in a point, is
differentiable at that point if those right and left derivatives are
equal.
Based on this, try to differentiate the function twice, each time checking for which values of $a$ are the right and left derivatives in $x=0$ equal.
answered Jan 14 at 23:04
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 14 at 23:04
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 14 at 23:04
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 14 at 23:04
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
add a comment |
add a comment |
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$begingroup$
$e^{-1/x} $ doesn't have a Taylor series centered at $0$.
$endgroup$
– Acccumulation
Jan 14 at 22:31