Mixing
Express a function as a power series if it respects some nice conditions
$begingroup$
Find all functions $f$ such that there is sequence $(a_n)_{n in mathbb{N}}$ such that :
$$forall x in ]-1, 1[, sum_{n = 0}^infty a_nx^n = f(x)$$
$$forall k in mathbb{N}, k geq 2, fleft(frac{1}{k}right) = e^{-sqrt{k}}$$
Here are my thoughts :
First I guess we need to use that :
$$e^{-sqrt{x}} =sum_{k= 0 }^infty frac{(-1)^k x^{k/2}}{k!}$$
So we need to find a sequnece such that :
$$forall k geq 2, sum_{n = 0}^inftyfrac{ a_n }{k^n} = sum_{n= 0 }^infty frac{(-1)^n k^{n/2}}{n!}$$
But from here I don't what to do. I know that two power series are equal if there is a segment on which there are equal. Here it's not the case, but these two power series have infinitely many points where they have the same value, but it's not sufficient to conclude that these power series are equal right ?
Moreover $0$ is an accumulation point of ${1/k mid k geq 2}$ so maybe it means that the two series are equal ? Since it means we can't find an open ball around $0$ such that the series are not equals, but the problem is that the theorem I know is only about a segment... So is what I am saying is true or they are nice counter-examples ?
Thank you !
real-analysis calculus sequences-and-series power-series
edited Jan 28 at 22:45
clathratus
5,0011438
asked Jan 28 at 22:41
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
add a comment |
$begingroup$
Find all functions $f$ such that there is sequence $(a_n)_{n in mathbb{N}}$ such that :
$$forall x in ]-1, 1[, sum_{n = 0}^infty a_nx^n = f(x)$$
$$forall k in mathbb{N}, k geq 2, fleft(frac{1}{k}right) = e^{-sqrt{k}}$$
Here are my thoughts :
First I guess we need to use that :
$$e^{-sqrt{x}} =sum_{k= 0 }^infty frac{(-1)^k x^{k/2}}{k!}$$
So we need to find a sequnece such that :
$$forall k geq 2, sum_{n = 0}^inftyfrac{ a_n }{k^n} = sum_{n= 0 }^infty frac{(-1)^n k^{n/2}}{n!}$$
But from here I don't what to do. I know that two power series are equal if there is a segment on which there are equal. Here it's not the case, but these two power series have infinitely many points where they have the same value, but it's not sufficient to conclude that these power series are equal right ?
Moreover $0$ is an accumulation point of ${1/k mid k geq 2}$ so maybe it means that the two series are equal ? Since it means we can't find an open ball around $0$ such that the series are not equals, but the problem is that the theorem I know is only about a segment... So is what I am saying is true or they are nice counter-examples ?
Thank you !
real-analysis calculus sequences-and-series power-series
edited Jan 28 at 22:45
clathratus
5,0011438
asked Jan 28 at 22:41
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
add a comment |
$begingroup$
Find all functions $f$ such that there is sequence $(a_n)_{n in mathbb{N}}$ such that :
$$forall x in ]-1, 1[, sum_{n = 0}^infty a_nx^n = f(x)$$
$$forall k in mathbb{N}, k geq 2, fleft(frac{1}{k}right) = e^{-sqrt{k}}$$
Here are my thoughts :
First I guess we need to use that :
$$e^{-sqrt{x}} =sum_{k= 0 }^infty frac{(-1)^k x^{k/2}}{k!}$$
So we need to find a sequnece such that :
$$forall k geq 2, sum_{n = 0}^inftyfrac{ a_n }{k^n} = sum_{n= 0 }^infty frac{(-1)^n k^{n/2}}{n!}$$
But from here I don't what to do. I know that two power series are equal if there is a segment on which there are equal. Here it's not the case, but these two power series have infinitely many points where they have the same value, but it's not sufficient to conclude that these power series are equal right ?
Moreover $0$ is an accumulation point of ${1/k mid k geq 2}$ so maybe it means that the two series are equal ? Since it means we can't find an open ball around $0$ such that the series are not equals, but the problem is that the theorem I know is only about a segment... So is what I am saying is true or they are nice counter-examples ?
Thank you !
real-analysis calculus sequences-and-series power-series
edited Jan 28 at 22:45
clathratus
5,0011438
asked Jan 28 at 22:41
dghkgfzyukzdghkgfzyukz
16612
$endgroup$
Find all functions $f$ such that there is sequence $(a_n)_{n in mathbb{N}}$ such that :
$$forall x in ]-1, 1[, sum_{n = 0}^infty a_nx^n = f(x)$$
$$forall k in mathbb{N}, k geq 2, fleft(frac{1}{k}right) = e^{-sqrt{k}}$$
Here are my thoughts :
First I guess we need to use that :
$$e^{-sqrt{x}} =sum_{k= 0 }^infty frac{(-1)^k x^{k/2}}{k!}$$
So we need to find a sequnece such that :
$$forall k geq 2, sum_{n = 0}^inftyfrac{ a_n }{k^n} = sum_{n= 0 }^infty frac{(-1)^n k^{n/2}}{n!}$$
But from here I don't what to do. I know that two power series are equal if there is a segment on which there are equal. Here it's not the case, but these two power series have infinitely many points where they have the same value, but it's not sufficient to conclude that these power series are equal right ?
Moreover $0$ is an accumulation point of ${1/k mid k geq 2}$ so maybe it means that the two series are equal ? Since it means we can't find an open ball around $0$ such that the series are not equals, but the problem is that the theorem I know is only about a segment... So is what I am saying is true or they are nice counter-examples ?
Thank you !
real-analysis calculus sequences-and-series power-series
real-analysis calculus sequences-and-series power-series
edited Jan 28 at 22:45
clathratus
5,0011438
asked Jan 28 at 22:41
dghkgfzyukzdghkgfzyukz
16612
edited Jan 28 at 22:45
clathratus
5,0011438
asked Jan 28 at 22:41
dghkgfzyukzdghkgfzyukz
16612
edited Jan 28 at 22:45
clathratus
5,0011438
edited Jan 28 at 22:45
clathratus
5,0011438
edited Jan 28 at 22:45
clathratus
5,0011438
5,0011438
asked Jan 28 at 22:41
dghkgfzyukzdghkgfzyukz
16612
asked Jan 28 at 22:41
dghkgfzyukzdghkgfzyukz
16612
asked Jan 28 at 22:41
dghkgfzyukzdghkgfzyukz
16612
16612
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There is no such function. Of course, the null function is no such function. On the other hand,$$a_0=f(0)=lim_{ktoinfty}fleft(frac1kright)=lim_{ktoinfty}e^{-sqrt k}=0.$$So, there is a smallest $Ninmathbb N$ such that $a_Nneq0$. But then$$lim_{ktoinfty}frac{leftlvert fleft(frac1kright)rightrvert}{frac1{k^N}}=lvert a_Nrvertneq0,$$whereas$$lim_{ktoinfty}frac{e^{-sqrt k}}{frac1{k^N}}=lim_{ktoinfty}frac{k^N}{e^{sqrt k}}=0.$$
answered Jan 28 at 23:03
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
That's clever !
$endgroup$
– Thinking
Jan 28 at 23:06
add a comment |
$begingroup$
Assume the existence of such $f$, which is not identically zero.
Since $f(0)=lim_{kto infty}f(1/k)=lim_{kto infty}e^{-sqrt k}=0$, $f$ has a zero at $x=0$. Assume the zero is of order $(m-1)inmathbb N^+$.
From the series expansion, we know $$lim_{xto 0}frac{f(x)}{x^m}=1$$
Thus, we expect
$$lim_{xto 0}frac{f(x)}{x^m}=lim_{ktoinfty}k^m f(1/k)
=lim_{ktoinfty}k^m e^{-sqrt k}color{red}{=}1$$
However, there is no $m$ that can satisfy the red equality. Therefore a contradiction is created and we conclude that such $f$ does not exist.
answered Jan 28 at 23:08
SzetoSzeto
6,7162927
$endgroup$
$begingroup$
This is the same idea as José Carlos Santos, but this is really smart, thank you !
$endgroup$
– dghkgfzyukz
Jan 28 at 23:09
1
$begingroup$
@dghkgfzyukz I typed too slowly...
$endgroup$
– Szeto
Jan 28 at 23:11
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is no such function. Of course, the null function is no such function. On the other hand,$$a_0=f(0)=lim_{ktoinfty}fleft(frac1kright)=lim_{ktoinfty}e^{-sqrt k}=0.$$So, there is a smallest $Ninmathbb N$ such that $a_Nneq0$. But then$$lim_{ktoinfty}frac{leftlvert fleft(frac1kright)rightrvert}{frac1{k^N}}=lvert a_Nrvertneq0,$$whereas$$lim_{ktoinfty}frac{e^{-sqrt k}}{frac1{k^N}}=lim_{ktoinfty}frac{k^N}{e^{sqrt k}}=0.$$
answered Jan 28 at 23:03
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
That's clever !
$endgroup$
– Thinking
Jan 28 at 23:06
add a comment |
$begingroup$
There is no such function. Of course, the null function is no such function. On the other hand,$$a_0=f(0)=lim_{ktoinfty}fleft(frac1kright)=lim_{ktoinfty}e^{-sqrt k}=0.$$So, there is a smallest $Ninmathbb N$ such that $a_Nneq0$. But then$$lim_{ktoinfty}frac{leftlvert fleft(frac1kright)rightrvert}{frac1{k^N}}=lvert a_Nrvertneq0,$$whereas$$lim_{ktoinfty}frac{e^{-sqrt k}}{frac1{k^N}}=lim_{ktoinfty}frac{k^N}{e^{sqrt k}}=0.$$
answered Jan 28 at 23:03
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
That's clever !
$endgroup$
– Thinking
Jan 28 at 23:06
add a comment |
$begingroup$
There is no such function. Of course, the null function is no such function. On the other hand,$$a_0=f(0)=lim_{ktoinfty}fleft(frac1kright)=lim_{ktoinfty}e^{-sqrt k}=0.$$So, there is a smallest $Ninmathbb N$ such that $a_Nneq0$. But then$$lim_{ktoinfty}frac{leftlvert fleft(frac1kright)rightrvert}{frac1{k^N}}=lvert a_Nrvertneq0,$$whereas$$lim_{ktoinfty}frac{e^{-sqrt k}}{frac1{k^N}}=lim_{ktoinfty}frac{k^N}{e^{sqrt k}}=0.$$
answered Jan 28 at 23:03
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
There is no such function. Of course, the null function is no such function. On the other hand,$$a_0=f(0)=lim_{ktoinfty}fleft(frac1kright)=lim_{ktoinfty}e^{-sqrt k}=0.$$So, there is a smallest $Ninmathbb N$ such that $a_Nneq0$. But then$$lim_{ktoinfty}frac{leftlvert fleft(frac1kright)rightrvert}{frac1{k^N}}=lvert a_Nrvertneq0,$$whereas$$lim_{ktoinfty}frac{e^{-sqrt k}}{frac1{k^N}}=lim_{ktoinfty}frac{k^N}{e^{sqrt k}}=0.$$
answered Jan 28 at 23:03
José Carlos SantosJosé Carlos Santos
171k23132240
edited Jan 29 at 14:54
answered Jan 28 at 23:03
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 28 at 23:03
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 28 at 23:03
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
$begingroup$
That's clever !
$endgroup$
– Thinking
Jan 28 at 23:06
add a comment |
$begingroup$
That's clever !
$endgroup$
– Thinking
Jan 28 at 23:06
$begingroup$
That's clever !
$endgroup$
– Thinking
Jan 28 at 23:06
$begingroup$
That's clever !
$endgroup$
– Thinking
Jan 28 at 23:06
add a comment |
$begingroup$
Assume the existence of such $f$, which is not identically zero.
Since $f(0)=lim_{kto infty}f(1/k)=lim_{kto infty}e^{-sqrt k}=0$, $f$ has a zero at $x=0$. Assume the zero is of order $(m-1)inmathbb N^+$.
From the series expansion, we know $$lim_{xto 0}frac{f(x)}{x^m}=1$$
Thus, we expect
$$lim_{xto 0}frac{f(x)}{x^m}=lim_{ktoinfty}k^m f(1/k)
=lim_{ktoinfty}k^m e^{-sqrt k}color{red}{=}1$$
However, there is no $m$ that can satisfy the red equality. Therefore a contradiction is created and we conclude that such $f$ does not exist.
answered Jan 28 at 23:08
SzetoSzeto
6,7162927
$endgroup$
$begingroup$
This is the same idea as José Carlos Santos, but this is really smart, thank you !
$endgroup$
– dghkgfzyukz
Jan 28 at 23:09
1
$begingroup$
@dghkgfzyukz I typed too slowly...
$endgroup$
– Szeto
Jan 28 at 23:11
add a comment |
$begingroup$
Assume the existence of such $f$, which is not identically zero.
Since $f(0)=lim_{kto infty}f(1/k)=lim_{kto infty}e^{-sqrt k}=0$, $f$ has a zero at $x=0$. Assume the zero is of order $(m-1)inmathbb N^+$.
From the series expansion, we know $$lim_{xto 0}frac{f(x)}{x^m}=1$$
Thus, we expect
$$lim_{xto 0}frac{f(x)}{x^m}=lim_{ktoinfty}k^m f(1/k)
=lim_{ktoinfty}k^m e^{-sqrt k}color{red}{=}1$$
However, there is no $m$ that can satisfy the red equality. Therefore a contradiction is created and we conclude that such $f$ does not exist.
answered Jan 28 at 23:08
SzetoSzeto
6,7162927
$endgroup$
$begingroup$
This is the same idea as José Carlos Santos, but this is really smart, thank you !
$endgroup$
– dghkgfzyukz
Jan 28 at 23:09
1
$begingroup$
@dghkgfzyukz I typed too slowly...
$endgroup$
– Szeto
Jan 28 at 23:11
add a comment |
$begingroup$
Assume the existence of such $f$, which is not identically zero.
Since $f(0)=lim_{kto infty}f(1/k)=lim_{kto infty}e^{-sqrt k}=0$, $f$ has a zero at $x=0$. Assume the zero is of order $(m-1)inmathbb N^+$.
From the series expansion, we know $$lim_{xto 0}frac{f(x)}{x^m}=1$$
Thus, we expect
$$lim_{xto 0}frac{f(x)}{x^m}=lim_{ktoinfty}k^m f(1/k)
=lim_{ktoinfty}k^m e^{-sqrt k}color{red}{=}1$$
However, there is no $m$ that can satisfy the red equality. Therefore a contradiction is created and we conclude that such $f$ does not exist.
answered Jan 28 at 23:08
SzetoSzeto
6,7162927
$endgroup$
Assume the existence of such $f$, which is not identically zero.
Since $f(0)=lim_{kto infty}f(1/k)=lim_{kto infty}e^{-sqrt k}=0$, $f$ has a zero at $x=0$. Assume the zero is of order $(m-1)inmathbb N^+$.
From the series expansion, we know $$lim_{xto 0}frac{f(x)}{x^m}=1$$
Thus, we expect
$$lim_{xto 0}frac{f(x)}{x^m}=lim_{ktoinfty}k^m f(1/k)
=lim_{ktoinfty}k^m e^{-sqrt k}color{red}{=}1$$
However, there is no $m$ that can satisfy the red equality. Therefore a contradiction is created and we conclude that such $f$ does not exist.
answered Jan 28 at 23:08
SzetoSzeto
6,7162927
edited Jan 28 at 23:12
answered Jan 28 at 23:08
SzetoSzeto
6,7162927
answered Jan 28 at 23:08
SzetoSzeto
6,7162927
answered Jan 28 at 23:08
SzetoSzeto
6,7162927
6,7162927
$begingroup$
This is the same idea as José Carlos Santos, but this is really smart, thank you !
$endgroup$
– dghkgfzyukz
Jan 28 at 23:09
1
$begingroup$
@dghkgfzyukz I typed too slowly...
$endgroup$
– Szeto
Jan 28 at 23:11
add a comment |
$begingroup$
This is the same idea as José Carlos Santos, but this is really smart, thank you !
$endgroup$
– dghkgfzyukz
Jan 28 at 23:09
1
$begingroup$
@dghkgfzyukz I typed too slowly...
$endgroup$
– Szeto
Jan 28 at 23:11
$begingroup$
This is the same idea as José Carlos Santos, but this is really smart, thank you !
$endgroup$
– dghkgfzyukz
Jan 28 at 23:09
$begingroup$
This is the same idea as José Carlos Santos, but this is really smart, thank you !
$endgroup$
– dghkgfzyukz
Jan 28 at 23:09
1
1
$begingroup$
@dghkgfzyukz I typed too slowly...
$endgroup$
– Szeto
Jan 28 at 23:11
$begingroup$
@dghkgfzyukz I typed too slowly...
$endgroup$
– Szeto
Jan 28 at 23:11
add a comment |
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