Mixing
A power series which is one to one
$begingroup$
Let $(a_n)_{n in mathbb{N}}$ be a sequence of complex numbers with $a_1 ne 0$. Moreover we assume that $sum na_n$ converges absolutely, and that $sum_{n = 2}^infty n mid a_n mid leq mid a_1mid$. Prove that $f$ is one-to-one on the open unit disk (ie.${z in mathbb{C}, mid z mid < 1 }$), where $f(x) = sum_{n = 0}^infty a_nx^n$.
Here is what I've done :
We have $f'(x) = a_1 + sum_{n = 2}^{infty} a_nnx^{n-1}$. Moreover, since we have
$$mid sum_{ n= 2}^infty a_nnx^{n-1} mid leq sum_{ n = 2}^infty mid a_n mid mid nx^{n-1} mid leq sum_{ n = 2}^infty n mid a_n mid leq mid a_1 mid$$
Hence we have (we can assume $a_1geq 0$ since when $a_1 < 0$ it is the same argument in reverse) :
$$-a_1 leq sum_{n = 2}^ infty a_nnx^{n-1} leq a_1$$
So we have
$0 leq f'(x) leq 2a_1$ (if $a_1 < 0$ then we will get $-2a_1 leq f'(x) leq 0$).
Thus to prove that $f$ is one-to-one it is sufficient to prove that $f'(x) ne 0$ for all $x$. But this clearly is true, because to have $f'(x) = 0$ for some $x$ we need
$$sum_{n = 2}^infty n mid a_n mid =mid a_ 1 mid$$
Yet since we clearly have
$$mid sum_{n = 2}^infty a_nnx^{n-1} mid leq sum_{n = 2}^infty mid a_n mid mid nx^{n-1} mid < sum_{n = 2}^infty n mid a_n mid $$
So it's impossible to find such an $x$. So to conclude we get $f'(x) > 0$ ($f'(x) < 0$ if $a_1 < 0$) for all $x$ in the open unit disk, so $f$ is one-to-one.
I don't know if what I've done is correct, because my book is doing something far more complicated using a proof by contradiction which is not intuitive.
real-analysis calculus integration proof-verification power-series
edited Feb 1 at 2:21
J. W. Tanner
4,6291320
asked Jan 31 at 23:13
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
add a comment |
$begingroup$
Let $(a_n)_{n in mathbb{N}}$ be a sequence of complex numbers with $a_1 ne 0$. Moreover we assume that $sum na_n$ converges absolutely, and that $sum_{n = 2}^infty n mid a_n mid leq mid a_1mid$. Prove that $f$ is one-to-one on the open unit disk (ie.${z in mathbb{C}, mid z mid < 1 }$), where $f(x) = sum_{n = 0}^infty a_nx^n$.
Here is what I've done :
We have $f'(x) = a_1 + sum_{n = 2}^{infty} a_nnx^{n-1}$. Moreover, since we have
$$mid sum_{ n= 2}^infty a_nnx^{n-1} mid leq sum_{ n = 2}^infty mid a_n mid mid nx^{n-1} mid leq sum_{ n = 2}^infty n mid a_n mid leq mid a_1 mid$$
Hence we have (we can assume $a_1geq 0$ since when $a_1 < 0$ it is the same argument in reverse) :
$$-a_1 leq sum_{n = 2}^ infty a_nnx^{n-1} leq a_1$$
So we have
$0 leq f'(x) leq 2a_1$ (if $a_1 < 0$ then we will get $-2a_1 leq f'(x) leq 0$).
Thus to prove that $f$ is one-to-one it is sufficient to prove that $f'(x) ne 0$ for all $x$. But this clearly is true, because to have $f'(x) = 0$ for some $x$ we need
$$sum_{n = 2}^infty n mid a_n mid =mid a_ 1 mid$$
Yet since we clearly have
$$mid sum_{n = 2}^infty a_nnx^{n-1} mid leq sum_{n = 2}^infty mid a_n mid mid nx^{n-1} mid < sum_{n = 2}^infty n mid a_n mid $$
So it's impossible to find such an $x$. So to conclude we get $f'(x) > 0$ ($f'(x) < 0$ if $a_1 < 0$) for all $x$ in the open unit disk, so $f$ is one-to-one.
I don't know if what I've done is correct, because my book is doing something far more complicated using a proof by contradiction which is not intuitive.
real-analysis calculus integration proof-verification power-series
edited Feb 1 at 2:21
J. W. Tanner
4,6291320
asked Jan 31 at 23:13
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
$begingroup$
You cannot write inequalities for complex numbers. The entire proof makes no sense.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:20
$begingroup$
@KaviRamaMurthy That's why I am taking the $mid . mid$
$endgroup$
– dghkgfzyukz
Jan 31 at 23:21
1
$begingroup$
Just read your own answer. You have written lots of inequalities for complex numbers. What do you mean by $a_1 geq 0$, fro example.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:23
$begingroup$
@KaviRamaMurthy Yes you are totally right sorry...
$endgroup$
– dghkgfzyukz
Jan 31 at 23:24
add a comment |
$begingroup$
Let $(a_n)_{n in mathbb{N}}$ be a sequence of complex numbers with $a_1 ne 0$. Moreover we assume that $sum na_n$ converges absolutely, and that $sum_{n = 2}^infty n mid a_n mid leq mid a_1mid$. Prove that $f$ is one-to-one on the open unit disk (ie.${z in mathbb{C}, mid z mid < 1 }$), where $f(x) = sum_{n = 0}^infty a_nx^n$.
Here is what I've done :
We have $f'(x) = a_1 + sum_{n = 2}^{infty} a_nnx^{n-1}$. Moreover, since we have
$$mid sum_{ n= 2}^infty a_nnx^{n-1} mid leq sum_{ n = 2}^infty mid a_n mid mid nx^{n-1} mid leq sum_{ n = 2}^infty n mid a_n mid leq mid a_1 mid$$
Hence we have (we can assume $a_1geq 0$ since when $a_1 < 0$ it is the same argument in reverse) :
$$-a_1 leq sum_{n = 2}^ infty a_nnx^{n-1} leq a_1$$
So we have
$0 leq f'(x) leq 2a_1$ (if $a_1 < 0$ then we will get $-2a_1 leq f'(x) leq 0$).
Thus to prove that $f$ is one-to-one it is sufficient to prove that $f'(x) ne 0$ for all $x$. But this clearly is true, because to have $f'(x) = 0$ for some $x$ we need
$$sum_{n = 2}^infty n mid a_n mid =mid a_ 1 mid$$
Yet since we clearly have
$$mid sum_{n = 2}^infty a_nnx^{n-1} mid leq sum_{n = 2}^infty mid a_n mid mid nx^{n-1} mid < sum_{n = 2}^infty n mid a_n mid $$
So it's impossible to find such an $x$. So to conclude we get $f'(x) > 0$ ($f'(x) < 0$ if $a_1 < 0$) for all $x$ in the open unit disk, so $f$ is one-to-one.
I don't know if what I've done is correct, because my book is doing something far more complicated using a proof by contradiction which is not intuitive.
real-analysis calculus integration proof-verification power-series
edited Feb 1 at 2:21
J. W. Tanner
4,6291320
asked Jan 31 at 23:13
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
Let $(a_n)_{n in mathbb{N}}$ be a sequence of complex numbers with $a_1 ne 0$. Moreover we assume that $sum na_n$ converges absolutely, and that $sum_{n = 2}^infty n mid a_n mid leq mid a_1mid$. Prove that $f$ is one-to-one on the open unit disk (ie.${z in mathbb{C}, mid z mid < 1 }$), where $f(x) = sum_{n = 0}^infty a_nx^n$.
Here is what I've done :
We have $f'(x) = a_1 + sum_{n = 2}^{infty} a_nnx^{n-1}$. Moreover, since we have
$$mid sum_{ n= 2}^infty a_nnx^{n-1} mid leq sum_{ n = 2}^infty mid a_n mid mid nx^{n-1} mid leq sum_{ n = 2}^infty n mid a_n mid leq mid a_1 mid$$
Hence we have (we can assume $a_1geq 0$ since when $a_1 < 0$ it is the same argument in reverse) :
$$-a_1 leq sum_{n = 2}^ infty a_nnx^{n-1} leq a_1$$
So we have
$0 leq f'(x) leq 2a_1$ (if $a_1 < 0$ then we will get $-2a_1 leq f'(x) leq 0$).
Thus to prove that $f$ is one-to-one it is sufficient to prove that $f'(x) ne 0$ for all $x$. But this clearly is true, because to have $f'(x) = 0$ for some $x$ we need
$$sum_{n = 2}^infty n mid a_n mid =mid a_ 1 mid$$
Yet since we clearly have
$$mid sum_{n = 2}^infty a_nnx^{n-1} mid leq sum_{n = 2}^infty mid a_n mid mid nx^{n-1} mid < sum_{n = 2}^infty n mid a_n mid $$
So it's impossible to find such an $x$. So to conclude we get $f'(x) > 0$ ($f'(x) < 0$ if $a_1 < 0$) for all $x$ in the open unit disk, so $f$ is one-to-one.
I don't know if what I've done is correct, because my book is doing something far more complicated using a proof by contradiction which is not intuitive.
real-analysis calculus integration proof-verification power-series
real-analysis calculus integration proof-verification power-series
edited Feb 1 at 2:21
J. W. Tanner
4,6291320
asked Jan 31 at 23:13
dghkgfzyukzdghkgfzyukz
16612
edited Feb 1 at 2:21
J. W. Tanner
4,6291320
asked Jan 31 at 23:13
dghkgfzyukzdghkgfzyukz
16612
edited Feb 1 at 2:21
J. W. Tanner
4,6291320
edited Feb 1 at 2:21
J. W. Tanner
4,6291320
edited Feb 1 at 2:21
J. W. Tanner
4,6291320
4,6291320
asked Jan 31 at 23:13
dghkgfzyukzdghkgfzyukz
16612
asked Jan 31 at 23:13
dghkgfzyukzdghkgfzyukz
16612
asked Jan 31 at 23:13
dghkgfzyukzdghkgfzyukz
16612
16612
$begingroup$
You cannot write inequalities for complex numbers. The entire proof makes no sense.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:20
$begingroup$
@KaviRamaMurthy That's why I am taking the $mid . mid$
$endgroup$
– dghkgfzyukz
Jan 31 at 23:21
1
$begingroup$
Just read your own answer. You have written lots of inequalities for complex numbers. What do you mean by $a_1 geq 0$, fro example.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:23
$begingroup$
@KaviRamaMurthy Yes you are totally right sorry...
$endgroup$
– dghkgfzyukz
Jan 31 at 23:24
add a comment |
$begingroup$
You cannot write inequalities for complex numbers. The entire proof makes no sense.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:20
$begingroup$
@KaviRamaMurthy That's why I am taking the $mid . mid$
$endgroup$
– dghkgfzyukz
Jan 31 at 23:21
1
$begingroup$
Just read your own answer. You have written lots of inequalities for complex numbers. What do you mean by $a_1 geq 0$, fro example.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:23
$begingroup$
@KaviRamaMurthy Yes you are totally right sorry...
$endgroup$
– dghkgfzyukz
Jan 31 at 23:24
$begingroup$
You cannot write inequalities for complex numbers. The entire proof makes no sense.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:20
$begingroup$
You cannot write inequalities for complex numbers. The entire proof makes no sense.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:20
$begingroup$
@KaviRamaMurthy That's why I am taking the $mid . mid$
$endgroup$
– dghkgfzyukz
Jan 31 at 23:21
$begingroup$
@KaviRamaMurthy That's why I am taking the $mid . mid$
$endgroup$
– dghkgfzyukz
Jan 31 at 23:21
1
1
$begingroup$
Just read your own answer. You have written lots of inequalities for complex numbers. What do you mean by $a_1 geq 0$, fro example.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:23
$begingroup$
Just read your own answer. You have written lots of inequalities for complex numbers. What do you mean by $a_1 geq 0$, fro example.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:23
$begingroup$
@KaviRamaMurthy Yes you are totally right sorry...
$endgroup$
– dghkgfzyukz
Jan 31 at 23:24
$begingroup$
@KaviRamaMurthy Yes you are totally right sorry...
$endgroup$
– dghkgfzyukz
Jan 31 at 23:24
add a comment |
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$begingroup$
You cannot write inequalities for complex numbers. The entire proof makes no sense.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:20
$begingroup$
@KaviRamaMurthy That's why I am taking the $mid . mid$
$endgroup$
– dghkgfzyukz
Jan 31 at 23:21
1
$begingroup$
Just read your own answer. You have written lots of inequalities for complex numbers. What do you mean by $a_1 geq 0$, fro example.
$endgroup$
– Kavi Rama Murthy
Jan 31 at 23:23
$begingroup$
@KaviRamaMurthy Yes you are totally right sorry...
$endgroup$
– dghkgfzyukz
Jan 31 at 23:24