Mixing
Baby Rudin Limit of Logarithmic Function
$begingroup$
On page 181 of Rudin's Principles of Mathematical Analysis, the following statement about the infinite limit of the logarithmic function is given:
And here is Theorem 8.6(e):
How do I arrive at the conclusion that $displaystylelim_{xto+infty}L(x)=+infty$ and $displaystylelim_{xto0}L(x)=-infty$ from Theorem 8.6(r)?
(In Rudin's notation, $E(x)=exp x$ and $L(x)=log x$)
Here’s my attempt in the middle of which I am stuck.
First, by definition, $displaystylelim_{xto-infty}E(x)=0$ is to say
$$
forall M.exists N.forall x.x<Nimplies E(x)<M.
$$
Now take the contrapositive of the inner proposition, we have
$$
forall M.exists N.forall x.E(x)geq Mimplies xgeq N.
$$
Since $L(E(x))=x$, it is possible to write $y=E(x)$ ($y>0$) and rewrite the previous proposition as
$$
forall M.exists N.forall y.(y>0land ygeq M)implies L(y)geq N.
$$
To show $displaystylelim_{yto+infty}L(y)=+infty$, one has to show that
$$
forall N.exists M.forall y.(y>0land ygeq M)implies L(y)geq N;
$$
the order of the quantifier is incorrect.
Here I noticed that the two symbols bound by a universal quantifier and an existential quantifier can be interchanged with a condition:
$$
forall ain A.exists b(a)in B.P(a,b)iffforall b’intext{Range}(b).exists a=b^{-1}(b’).P(a,b’),
$$
where $b(a)$ is a multivalued function (it is defined as a multivalued function in order to capture the case where there exists more than one $b(a)$ satisfying $P$) that denotes the dependency of the existential quantified to the universal quantified and $b^{-1}(b’)$ denotes one arbitrary element of the preimage of $b$ (it is defined so that the injectivity of $b$ is not required, and such choice of element is always possible since $b'in text{Range}(b)$).
And here is where I am stuck. To use the fact I discovered for interchanging the order of the quantified symbols to acquire the desired conclusion, one must show that $text{Range}(N)=mathbb{R}$ (or at least a neighborhood of positive infinity), and I cannot do that without triggering a circular argument where the conclusion itself is used.
Frankly, at this point, I'm considering that my approach may not be the correct approach, or at least not the simplest one, so if you have a better idea, please tell me. Thanks.
real-analysis logarithms
asked Jan 20 at 17:26
Yutong ZhangYutong Zhang
397
$endgroup$
add a comment |
$begingroup$
On page 181 of Rudin's Principles of Mathematical Analysis, the following statement about the infinite limit of the logarithmic function is given:
And here is Theorem 8.6(e):
How do I arrive at the conclusion that $displaystylelim_{xto+infty}L(x)=+infty$ and $displaystylelim_{xto0}L(x)=-infty$ from Theorem 8.6(r)?
(In Rudin's notation, $E(x)=exp x$ and $L(x)=log x$)
Here’s my attempt in the middle of which I am stuck.
First, by definition, $displaystylelim_{xto-infty}E(x)=0$ is to say
$$
forall M.exists N.forall x.x<Nimplies E(x)<M.
$$
Now take the contrapositive of the inner proposition, we have
$$
forall M.exists N.forall x.E(x)geq Mimplies xgeq N.
$$
Since $L(E(x))=x$, it is possible to write $y=E(x)$ ($y>0$) and rewrite the previous proposition as
$$
forall M.exists N.forall y.(y>0land ygeq M)implies L(y)geq N.
$$
To show $displaystylelim_{yto+infty}L(y)=+infty$, one has to show that
$$
forall N.exists M.forall y.(y>0land ygeq M)implies L(y)geq N;
$$
the order of the quantifier is incorrect.
Here I noticed that the two symbols bound by a universal quantifier and an existential quantifier can be interchanged with a condition:
$$
forall ain A.exists b(a)in B.P(a,b)iffforall b’intext{Range}(b).exists a=b^{-1}(b’).P(a,b’),
$$
where $b(a)$ is a multivalued function (it is defined as a multivalued function in order to capture the case where there exists more than one $b(a)$ satisfying $P$) that denotes the dependency of the existential quantified to the universal quantified and $b^{-1}(b’)$ denotes one arbitrary element of the preimage of $b$ (it is defined so that the injectivity of $b$ is not required, and such choice of element is always possible since $b'in text{Range}(b)$).
And here is where I am stuck. To use the fact I discovered for interchanging the order of the quantified symbols to acquire the desired conclusion, one must show that $text{Range}(N)=mathbb{R}$ (or at least a neighborhood of positive infinity), and I cannot do that without triggering a circular argument where the conclusion itself is used.
Frankly, at this point, I'm considering that my approach may not be the correct approach, or at least not the simplest one, so if you have a better idea, please tell me. Thanks.
real-analysis logarithms
asked Jan 20 at 17:26
Yutong ZhangYutong Zhang
397
$endgroup$
add a comment |
$begingroup$
On page 181 of Rudin's Principles of Mathematical Analysis, the following statement about the infinite limit of the logarithmic function is given:
And here is Theorem 8.6(e):
How do I arrive at the conclusion that $displaystylelim_{xto+infty}L(x)=+infty$ and $displaystylelim_{xto0}L(x)=-infty$ from Theorem 8.6(r)?
(In Rudin's notation, $E(x)=exp x$ and $L(x)=log x$)
Here’s my attempt in the middle of which I am stuck.
First, by definition, $displaystylelim_{xto-infty}E(x)=0$ is to say
$$
forall M.exists N.forall x.x<Nimplies E(x)<M.
$$
Now take the contrapositive of the inner proposition, we have
$$
forall M.exists N.forall x.E(x)geq Mimplies xgeq N.
$$
Since $L(E(x))=x$, it is possible to write $y=E(x)$ ($y>0$) and rewrite the previous proposition as
$$
forall M.exists N.forall y.(y>0land ygeq M)implies L(y)geq N.
$$
To show $displaystylelim_{yto+infty}L(y)=+infty$, one has to show that
$$
forall N.exists M.forall y.(y>0land ygeq M)implies L(y)geq N;
$$
the order of the quantifier is incorrect.
Here I noticed that the two symbols bound by a universal quantifier and an existential quantifier can be interchanged with a condition:
$$
forall ain A.exists b(a)in B.P(a,b)iffforall b’intext{Range}(b).exists a=b^{-1}(b’).P(a,b’),
$$
where $b(a)$ is a multivalued function (it is defined as a multivalued function in order to capture the case where there exists more than one $b(a)$ satisfying $P$) that denotes the dependency of the existential quantified to the universal quantified and $b^{-1}(b’)$ denotes one arbitrary element of the preimage of $b$ (it is defined so that the injectivity of $b$ is not required, and such choice of element is always possible since $b'in text{Range}(b)$).
And here is where I am stuck. To use the fact I discovered for interchanging the order of the quantified symbols to acquire the desired conclusion, one must show that $text{Range}(N)=mathbb{R}$ (or at least a neighborhood of positive infinity), and I cannot do that without triggering a circular argument where the conclusion itself is used.
Frankly, at this point, I'm considering that my approach may not be the correct approach, or at least not the simplest one, so if you have a better idea, please tell me. Thanks.
real-analysis logarithms
asked Jan 20 at 17:26
Yutong ZhangYutong Zhang
397
$endgroup$
On page 181 of Rudin's Principles of Mathematical Analysis, the following statement about the infinite limit of the logarithmic function is given:
And here is Theorem 8.6(e):
How do I arrive at the conclusion that $displaystylelim_{xto+infty}L(x)=+infty$ and $displaystylelim_{xto0}L(x)=-infty$ from Theorem 8.6(r)?
(In Rudin's notation, $E(x)=exp x$ and $L(x)=log x$)
Here’s my attempt in the middle of which I am stuck.
First, by definition, $displaystylelim_{xto-infty}E(x)=0$ is to say
$$
forall M.exists N.forall x.x<Nimplies E(x)<M.
$$
Now take the contrapositive of the inner proposition, we have
$$
forall M.exists N.forall x.E(x)geq Mimplies xgeq N.
$$
Since $L(E(x))=x$, it is possible to write $y=E(x)$ ($y>0$) and rewrite the previous proposition as
$$
forall M.exists N.forall y.(y>0land ygeq M)implies L(y)geq N.
$$
To show $displaystylelim_{yto+infty}L(y)=+infty$, one has to show that
$$
forall N.exists M.forall y.(y>0land ygeq M)implies L(y)geq N;
$$
the order of the quantifier is incorrect.
Here I noticed that the two symbols bound by a universal quantifier and an existential quantifier can be interchanged with a condition:
$$
forall ain A.exists b(a)in B.P(a,b)iffforall b’intext{Range}(b).exists a=b^{-1}(b’).P(a,b’),
$$
where $b(a)$ is a multivalued function (it is defined as a multivalued function in order to capture the case where there exists more than one $b(a)$ satisfying $P$) that denotes the dependency of the existential quantified to the universal quantified and $b^{-1}(b’)$ denotes one arbitrary element of the preimage of $b$ (it is defined so that the injectivity of $b$ is not required, and such choice of element is always possible since $b'in text{Range}(b)$).
And here is where I am stuck. To use the fact I discovered for interchanging the order of the quantified symbols to acquire the desired conclusion, one must show that $text{Range}(N)=mathbb{R}$ (or at least a neighborhood of positive infinity), and I cannot do that without triggering a circular argument where the conclusion itself is used.
Frankly, at this point, I'm considering that my approach may not be the correct approach, or at least not the simplest one, so if you have a better idea, please tell me. Thanks.
real-analysis logarithms
real-analysis logarithms
asked Jan 20 at 17:26
Yutong ZhangYutong Zhang
397
asked Jan 20 at 17:26
Yutong ZhangYutong Zhang
397
asked Jan 20 at 17:26
Yutong ZhangYutong Zhang
397
asked Jan 20 at 17:26
Yutong ZhangYutong Zhang
397
asked Jan 20 at 17:26
Yutong ZhangYutong Zhang
397
397
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This has been solved through another method.
The goal is to prove
$$
forall M.exists N.y>Nimplies L(y)>M.
$$
By the monotonicity of $E(x)$,
$$
left(forall x_1,x_2.x_1geq x_2implies E(x_1)geq E(x_2)right)impliesleft(forall x_1,x_2.E(x_1)<E(x_2)implies x_1<x_2right).
$$
For $y_1<y_2$, represent it with
$$
y_1=E(x_1)<E(x_2)y_2
$$
where
$$
x_1<x_2
$$
by previous argument. Thus
$$
L(y_1)=L(E(x_1))=x_1<x_2=L(E(x_2))=L(y_2)
$$
Take $N=E(M)$, then by the monotonicity of $L$ we have
$$
y>E(M)implies L(y)>L(E(M))geq M,
$$
as desired.
answered Jan 26 at 10:34
Yutong ZhangYutong Zhang
397
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
votes
$begingroup$
This has been solved through another method.
The goal is to prove
$$
forall M.exists N.y>Nimplies L(y)>M.
$$
By the monotonicity of $E(x)$,
$$
left(forall x_1,x_2.x_1geq x_2implies E(x_1)geq E(x_2)right)impliesleft(forall x_1,x_2.E(x_1)<E(x_2)implies x_1<x_2right).
$$
For $y_1<y_2$, represent it with
$$
y_1=E(x_1)<E(x_2)y_2
$$
where
$$
x_1<x_2
$$
by previous argument. Thus
$$
L(y_1)=L(E(x_1))=x_1<x_2=L(E(x_2))=L(y_2)
$$
Take $N=E(M)$, then by the monotonicity of $L$ we have
$$
y>E(M)implies L(y)>L(E(M))geq M,
$$
as desired.
answered Jan 26 at 10:34
Yutong ZhangYutong Zhang
397
$endgroup$
add a comment |
$begingroup$
This has been solved through another method.
The goal is to prove
$$
forall M.exists N.y>Nimplies L(y)>M.
$$
By the monotonicity of $E(x)$,
$$
left(forall x_1,x_2.x_1geq x_2implies E(x_1)geq E(x_2)right)impliesleft(forall x_1,x_2.E(x_1)<E(x_2)implies x_1<x_2right).
$$
For $y_1<y_2$, represent it with
$$
y_1=E(x_1)<E(x_2)y_2
$$
where
$$
x_1<x_2
$$
by previous argument. Thus
$$
L(y_1)=L(E(x_1))=x_1<x_2=L(E(x_2))=L(y_2)
$$
Take $N=E(M)$, then by the monotonicity of $L$ we have
$$
y>E(M)implies L(y)>L(E(M))geq M,
$$
as desired.
answered Jan 26 at 10:34
Yutong ZhangYutong Zhang
397
$endgroup$
add a comment |
$begingroup$
This has been solved through another method.
The goal is to prove
$$
forall M.exists N.y>Nimplies L(y)>M.
$$
By the monotonicity of $E(x)$,
$$
left(forall x_1,x_2.x_1geq x_2implies E(x_1)geq E(x_2)right)impliesleft(forall x_1,x_2.E(x_1)<E(x_2)implies x_1<x_2right).
$$
For $y_1<y_2$, represent it with
$$
y_1=E(x_1)<E(x_2)y_2
$$
where
$$
x_1<x_2
$$
by previous argument. Thus
$$
L(y_1)=L(E(x_1))=x_1<x_2=L(E(x_2))=L(y_2)
$$
Take $N=E(M)$, then by the monotonicity of $L$ we have
$$
y>E(M)implies L(y)>L(E(M))geq M,
$$
as desired.
answered Jan 26 at 10:34
Yutong ZhangYutong Zhang
397
$endgroup$
This has been solved through another method.
The goal is to prove
$$
forall M.exists N.y>Nimplies L(y)>M.
$$
By the monotonicity of $E(x)$,
$$
left(forall x_1,x_2.x_1geq x_2implies E(x_1)geq E(x_2)right)impliesleft(forall x_1,x_2.E(x_1)<E(x_2)implies x_1<x_2right).
$$
For $y_1<y_2$, represent it with
$$
y_1=E(x_1)<E(x_2)y_2
$$
where
$$
x_1<x_2
$$
by previous argument. Thus
$$
L(y_1)=L(E(x_1))=x_1<x_2=L(E(x_2))=L(y_2)
$$
Take $N=E(M)$, then by the monotonicity of $L$ we have
$$
y>E(M)implies L(y)>L(E(M))geq M,
$$
as desired.
answered Jan 26 at 10:34
Yutong ZhangYutong Zhang
397
answered Jan 26 at 10:34
Yutong ZhangYutong Zhang
397
answered Jan 26 at 10:34
Yutong ZhangYutong Zhang
397
answered Jan 26 at 10:34
Yutong ZhangYutong Zhang
397
397
add a comment |
add a comment |
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