Mixing
Cauchy-Hadamard formula proof
$begingroup$
So I'm looking at the proof of the Cauchy-Hadamard formula in my textbook and I understand the first part, but the second one I'm confused about. Here goes:
$$DeclareMathOperator*uplim{overline{lim}} text{Let } sum_{n=0}^{infty}a_n(x-x_0) text{ be a power series, and let } rho=uplim_{n to infty}{sqrt[n]{|a_n|}} text{. Then}$$
(1) if $rho in(0, infty)$ $R = frac{1}{rho}$
(2) if $rho = +infty$ $R=0$
(3) if $rho=0$ $R = +infty $
The proofs for (2) and (3) are beyond the scope of our course, so we only did proof for (1). This is the first part:
Let $tin mathbb{R}$, and suppose $|t-x_0|>frac{1}{rho}$. Then $frac{1}{|t-x_o|}<rho$, therefore we can say $(existsepsilon>0)$ such that $frac{1}{|t-x_0|} < rho-epsilon Rightarrow (rho-epsilon)|t-x_0|>1$.
Since $rho=uplim_{n to infty}{sqrt[n]{|a_n|}}$ then $sqrt[n]{|a_n|}>rho-epsilon$ for all $ngeq n_o$, $n_0 in mathbb{N}$. Therefore:
$$|a_n(t-x_0)^n| = bigg|sqrt[n]{|a_n|}(t-x_0)bigg|^n>((rho-epsilon)|t-x_0|)^n > 1$$
So we have that $|a_n(t-x_0)^n| > 1$ meaning that it cannot converge to zero, meaning that the series diverges. So we conclude that:
$$|t-x_0|>frac{1}{rho} Rightarrow text{the series diverges}$$
Taking the counterposition of this gives us
$$text{the series converges} Rightarrow |t-x_0|leqfrac{1}{rho}Rightarrow Rleqfrac{1}{rho}$$
And all this is clear so far. Now we have to show that also $Rgeqfrac{1}{rho}$ which will get us to the final conclusion that $R = frac{1}{rho}$
Let's suppose otherwise $R < frac{1}{rho}$
Now here's the problematic part (I'm quoting now):
then we know $(exists tin mathbb{R})$ such that $R<|t-x_0|<frac{1}{rho}$. I have no idea how this was concluded.
After this we have something similar to the first part:
$$frac{1}{|t-x_0|}>rho Rightarrow (exists epsilon>0) quad frac{1}{|t-x_0|} >rho + epsilon Rightarrow (rho+epsilon)|t-x_0|<1$$
Because of how we defined $rho$ we got:
$$(exists n_0 in mathbb{N})(forall n geq n_0)(sqrt[n]{|a_n|}<rho +epsilon)$$
Then:
$$sqrt[n]{|a_n|(t-x_0)^n} = sqrt[n]{|a_n|}|t-x_0| < (rho + epsilon)|t-x_0| = q$$ where $|q|<1$.
Now we have that $a_n(t-x_0)^n < q^n$ where $|q|<1$ meaning $q$ converges, meaning $a_n(t-x_0)^n$ converges (by the comparison criteria). Since $R$ is defined as
$$R = sup{bigg{|t-x_0| text{ the sequence converges for }tbigg}}$$
And since our sequence indeed converges for $t$, we have (the above mentioned set we'll call $E$).
$$|t-x_0| in E Rightarrow |t-x_0| leq sup{E} = R $$
which gives us a contradiction with one of the beginning statements (exactly the one that I don't understand how it got there). So could anyone clarify? Thanks.
real-analysis power-series
asked Jan 29 at 10:38
KoyKoy
35116
$endgroup$
add a comment |
$begingroup$
So I'm looking at the proof of the Cauchy-Hadamard formula in my textbook and I understand the first part, but the second one I'm confused about. Here goes:
$$DeclareMathOperator*uplim{overline{lim}} text{Let } sum_{n=0}^{infty}a_n(x-x_0) text{ be a power series, and let } rho=uplim_{n to infty}{sqrt[n]{|a_n|}} text{. Then}$$
(1) if $rho in(0, infty)$ $R = frac{1}{rho}$
(2) if $rho = +infty$ $R=0$
(3) if $rho=0$ $R = +infty $
The proofs for (2) and (3) are beyond the scope of our course, so we only did proof for (1). This is the first part:
Let $tin mathbb{R}$, and suppose $|t-x_0|>frac{1}{rho}$. Then $frac{1}{|t-x_o|}<rho$, therefore we can say $(existsepsilon>0)$ such that $frac{1}{|t-x_0|} < rho-epsilon Rightarrow (rho-epsilon)|t-x_0|>1$.
Since $rho=uplim_{n to infty}{sqrt[n]{|a_n|}}$ then $sqrt[n]{|a_n|}>rho-epsilon$ for all $ngeq n_o$, $n_0 in mathbb{N}$. Therefore:
$$|a_n(t-x_0)^n| = bigg|sqrt[n]{|a_n|}(t-x_0)bigg|^n>((rho-epsilon)|t-x_0|)^n > 1$$
So we have that $|a_n(t-x_0)^n| > 1$ meaning that it cannot converge to zero, meaning that the series diverges. So we conclude that:
$$|t-x_0|>frac{1}{rho} Rightarrow text{the series diverges}$$
Taking the counterposition of this gives us
$$text{the series converges} Rightarrow |t-x_0|leqfrac{1}{rho}Rightarrow Rleqfrac{1}{rho}$$
And all this is clear so far. Now we have to show that also $Rgeqfrac{1}{rho}$ which will get us to the final conclusion that $R = frac{1}{rho}$
Let's suppose otherwise $R < frac{1}{rho}$
Now here's the problematic part (I'm quoting now):
then we know $(exists tin mathbb{R})$ such that $R<|t-x_0|<frac{1}{rho}$. I have no idea how this was concluded.
After this we have something similar to the first part:
$$frac{1}{|t-x_0|}>rho Rightarrow (exists epsilon>0) quad frac{1}{|t-x_0|} >rho + epsilon Rightarrow (rho+epsilon)|t-x_0|<1$$
Because of how we defined $rho$ we got:
$$(exists n_0 in mathbb{N})(forall n geq n_0)(sqrt[n]{|a_n|}<rho +epsilon)$$
Then:
$$sqrt[n]{|a_n|(t-x_0)^n} = sqrt[n]{|a_n|}|t-x_0| < (rho + epsilon)|t-x_0| = q$$ where $|q|<1$.
Now we have that $a_n(t-x_0)^n < q^n$ where $|q|<1$ meaning $q$ converges, meaning $a_n(t-x_0)^n$ converges (by the comparison criteria). Since $R$ is defined as
$$R = sup{bigg{|t-x_0| text{ the sequence converges for }tbigg}}$$
And since our sequence indeed converges for $t$, we have (the above mentioned set we'll call $E$).
$$|t-x_0| in E Rightarrow |t-x_0| leq sup{E} = R $$
which gives us a contradiction with one of the beginning statements (exactly the one that I don't understand how it got there). So could anyone clarify? Thanks.
real-analysis power-series
asked Jan 29 at 10:38
KoyKoy
35116
$endgroup$
add a comment |
$begingroup$
So I'm looking at the proof of the Cauchy-Hadamard formula in my textbook and I understand the first part, but the second one I'm confused about. Here goes:
$$DeclareMathOperator*uplim{overline{lim}} text{Let } sum_{n=0}^{infty}a_n(x-x_0) text{ be a power series, and let } rho=uplim_{n to infty}{sqrt[n]{|a_n|}} text{. Then}$$
(1) if $rho in(0, infty)$ $R = frac{1}{rho}$
(2) if $rho = +infty$ $R=0$
(3) if $rho=0$ $R = +infty $
The proofs for (2) and (3) are beyond the scope of our course, so we only did proof for (1). This is the first part:
Let $tin mathbb{R}$, and suppose $|t-x_0|>frac{1}{rho}$. Then $frac{1}{|t-x_o|}<rho$, therefore we can say $(existsepsilon>0)$ such that $frac{1}{|t-x_0|} < rho-epsilon Rightarrow (rho-epsilon)|t-x_0|>1$.
Since $rho=uplim_{n to infty}{sqrt[n]{|a_n|}}$ then $sqrt[n]{|a_n|}>rho-epsilon$ for all $ngeq n_o$, $n_0 in mathbb{N}$. Therefore:
$$|a_n(t-x_0)^n| = bigg|sqrt[n]{|a_n|}(t-x_0)bigg|^n>((rho-epsilon)|t-x_0|)^n > 1$$
So we have that $|a_n(t-x_0)^n| > 1$ meaning that it cannot converge to zero, meaning that the series diverges. So we conclude that:
$$|t-x_0|>frac{1}{rho} Rightarrow text{the series diverges}$$
Taking the counterposition of this gives us
$$text{the series converges} Rightarrow |t-x_0|leqfrac{1}{rho}Rightarrow Rleqfrac{1}{rho}$$
And all this is clear so far. Now we have to show that also $Rgeqfrac{1}{rho}$ which will get us to the final conclusion that $R = frac{1}{rho}$
Let's suppose otherwise $R < frac{1}{rho}$
Now here's the problematic part (I'm quoting now):
then we know $(exists tin mathbb{R})$ such that $R<|t-x_0|<frac{1}{rho}$. I have no idea how this was concluded.
After this we have something similar to the first part:
$$frac{1}{|t-x_0|}>rho Rightarrow (exists epsilon>0) quad frac{1}{|t-x_0|} >rho + epsilon Rightarrow (rho+epsilon)|t-x_0|<1$$
Because of how we defined $rho$ we got:
$$(exists n_0 in mathbb{N})(forall n geq n_0)(sqrt[n]{|a_n|}<rho +epsilon)$$
Then:
$$sqrt[n]{|a_n|(t-x_0)^n} = sqrt[n]{|a_n|}|t-x_0| < (rho + epsilon)|t-x_0| = q$$ where $|q|<1$.
Now we have that $a_n(t-x_0)^n < q^n$ where $|q|<1$ meaning $q$ converges, meaning $a_n(t-x_0)^n$ converges (by the comparison criteria). Since $R$ is defined as
$$R = sup{bigg{|t-x_0| text{ the sequence converges for }tbigg}}$$
And since our sequence indeed converges for $t$, we have (the above mentioned set we'll call $E$).
$$|t-x_0| in E Rightarrow |t-x_0| leq sup{E} = R $$
which gives us a contradiction with one of the beginning statements (exactly the one that I don't understand how it got there). So could anyone clarify? Thanks.
real-analysis power-series
asked Jan 29 at 10:38
KoyKoy
35116
$endgroup$
So I'm looking at the proof of the Cauchy-Hadamard formula in my textbook and I understand the first part, but the second one I'm confused about. Here goes:
$$DeclareMathOperator*uplim{overline{lim}} text{Let } sum_{n=0}^{infty}a_n(x-x_0) text{ be a power series, and let } rho=uplim_{n to infty}{sqrt[n]{|a_n|}} text{. Then}$$
(1) if $rho in(0, infty)$ $R = frac{1}{rho}$
(2) if $rho = +infty$ $R=0$
(3) if $rho=0$ $R = +infty $
The proofs for (2) and (3) are beyond the scope of our course, so we only did proof for (1). This is the first part:
Let $tin mathbb{R}$, and suppose $|t-x_0|>frac{1}{rho}$. Then $frac{1}{|t-x_o|}<rho$, therefore we can say $(existsepsilon>0)$ such that $frac{1}{|t-x_0|} < rho-epsilon Rightarrow (rho-epsilon)|t-x_0|>1$.
Since $rho=uplim_{n to infty}{sqrt[n]{|a_n|}}$ then $sqrt[n]{|a_n|}>rho-epsilon$ for all $ngeq n_o$, $n_0 in mathbb{N}$. Therefore:
$$|a_n(t-x_0)^n| = bigg|sqrt[n]{|a_n|}(t-x_0)bigg|^n>((rho-epsilon)|t-x_0|)^n > 1$$
So we have that $|a_n(t-x_0)^n| > 1$ meaning that it cannot converge to zero, meaning that the series diverges. So we conclude that:
$$|t-x_0|>frac{1}{rho} Rightarrow text{the series diverges}$$
Taking the counterposition of this gives us
$$text{the series converges} Rightarrow |t-x_0|leqfrac{1}{rho}Rightarrow Rleqfrac{1}{rho}$$
And all this is clear so far. Now we have to show that also $Rgeqfrac{1}{rho}$ which will get us to the final conclusion that $R = frac{1}{rho}$
Let's suppose otherwise $R < frac{1}{rho}$
Now here's the problematic part (I'm quoting now):
then we know $(exists tin mathbb{R})$ such that $R<|t-x_0|<frac{1}{rho}$. I have no idea how this was concluded.
After this we have something similar to the first part:
$$frac{1}{|t-x_0|}>rho Rightarrow (exists epsilon>0) quad frac{1}{|t-x_0|} >rho + epsilon Rightarrow (rho+epsilon)|t-x_0|<1$$
Because of how we defined $rho$ we got:
$$(exists n_0 in mathbb{N})(forall n geq n_0)(sqrt[n]{|a_n|}<rho +epsilon)$$
Then:
$$sqrt[n]{|a_n|(t-x_0)^n} = sqrt[n]{|a_n|}|t-x_0| < (rho + epsilon)|t-x_0| = q$$ where $|q|<1$.
Now we have that $a_n(t-x_0)^n < q^n$ where $|q|<1$ meaning $q$ converges, meaning $a_n(t-x_0)^n$ converges (by the comparison criteria). Since $R$ is defined as
$$R = sup{bigg{|t-x_0| text{ the sequence converges for }tbigg}}$$
And since our sequence indeed converges for $t$, we have (the above mentioned set we'll call $E$).
$$|t-x_0| in E Rightarrow |t-x_0| leq sup{E} = R $$
which gives us a contradiction with one of the beginning statements (exactly the one that I don't understand how it got there). So could anyone clarify? Thanks.
real-analysis power-series
real-analysis power-series
asked Jan 29 at 10:38
KoyKoy
35116
asked Jan 29 at 10:38
KoyKoy
35116
edited Jan 29 at 11:10
Koy
asked Jan 29 at 10:38
KoyKoy
35116
asked Jan 29 at 10:38
KoyKoy
35116
asked Jan 29 at 10:38
KoyKoy
35116
35116
add a comment |
add a comment |
1 Answer
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$begingroup$
Let $R, rho, x_0$ be any reals such that $R< frac{1}{rho}$. Pick any $a in mathbb{R}$ such that $R<a<frac{1}{rho}$ and set $|t-x_0| = a$ (for example $t=a+x_0$). Then there exists a $t>0$ such that $R < |t-x_0| < frac{1}{rho}$.
This is just the density of the real numbers: between any two reals there is an interval of real numbers.
answered Jan 29 at 11:39
Sorin TircSorin Tirc
1,855213
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$begingroup$
Let $R, rho, x_0$ be any reals such that $R< frac{1}{rho}$. Pick any $a in mathbb{R}$ such that $R<a<frac{1}{rho}$ and set $|t-x_0| = a$ (for example $t=a+x_0$). Then there exists a $t>0$ such that $R < |t-x_0| < frac{1}{rho}$.
This is just the density of the real numbers: between any two reals there is an interval of real numbers.
answered Jan 29 at 11:39
Sorin TircSorin Tirc
1,855213
$endgroup$
add a comment |
$begingroup$
Let $R, rho, x_0$ be any reals such that $R< frac{1}{rho}$. Pick any $a in mathbb{R}$ such that $R<a<frac{1}{rho}$ and set $|t-x_0| = a$ (for example $t=a+x_0$). Then there exists a $t>0$ such that $R < |t-x_0| < frac{1}{rho}$.
This is just the density of the real numbers: between any two reals there is an interval of real numbers.
answered Jan 29 at 11:39
Sorin TircSorin Tirc
1,855213
$endgroup$
add a comment |
$begingroup$
Let $R, rho, x_0$ be any reals such that $R< frac{1}{rho}$. Pick any $a in mathbb{R}$ such that $R<a<frac{1}{rho}$ and set $|t-x_0| = a$ (for example $t=a+x_0$). Then there exists a $t>0$ such that $R < |t-x_0| < frac{1}{rho}$.
This is just the density of the real numbers: between any two reals there is an interval of real numbers.
answered Jan 29 at 11:39
Sorin TircSorin Tirc
1,855213
$endgroup$
Let $R, rho, x_0$ be any reals such that $R< frac{1}{rho}$. Pick any $a in mathbb{R}$ such that $R<a<frac{1}{rho}$ and set $|t-x_0| = a$ (for example $t=a+x_0$). Then there exists a $t>0$ such that $R < |t-x_0| < frac{1}{rho}$.
This is just the density of the real numbers: between any two reals there is an interval of real numbers.
answered Jan 29 at 11:39
Sorin TircSorin Tirc
1,855213
answered Jan 29 at 11:39
Sorin TircSorin Tirc
1,855213
answered Jan 29 at 11:39
Sorin TircSorin Tirc
1,855213
answered Jan 29 at 11:39
Sorin TircSorin Tirc
1,855213
1,855213
add a comment |
add a comment |
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