Mixing
Series that vanish for infintely many values
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How to prove that if i got a series $f in mathbbR[X,Y]$ that vanishes for all $X,Y in [-1,1]$ then all his coefficients must be $0$ ?
Thank you for your answers.
power-series
asked Jul 26 at 3:04
Pierre21
828
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up vote
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favorite
How to prove that if i got a series $f in mathbbR[X,Y]$ that vanishes for all $X,Y in [-1,1]$ then all his coefficients must be $0$ ?
Thank you for your answers.
power-series
asked Jul 26 at 3:04
Pierre21
828
$XinmathbbR[X,Y]$ vanishes at all the points $(X,Y)=(0,y)$, for $yinmathbbR$, but it is not zero.
– user578878
Jul 26 at 3:11
Not sure if this helps or not but I think you should only consider the case for uncountable zeros since if it is countable, we can construct the sequence $a_n=(-1)^n$ to obtain the divergence series $S_n$ with countably many zeros.
– Evan William Chandra
Jul 26 at 3:12
Yes of course you're both right, I edited my question.
– Pierre21
Jul 26 at 3:34
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to prove that if i got a series $f in mathbbR[X,Y]$ that vanishes for all $X,Y in [-1,1]$ then all his coefficients must be $0$ ?
Thank you for your answers.
power-series
asked Jul 26 at 3:04
Pierre21
828
How to prove that if i got a series $f in mathbbR[X,Y]$ that vanishes for all $X,Y in [-1,1]$ then all his coefficients must be $0$ ?
Thank you for your answers.
power-series
asked Jul 26 at 3:04
Pierre21
828
edited Jul 26 at 3:53
asked Jul 26 at 3:04
Pierre21
828
asked Jul 26 at 3:04
Pierre21
828
asked Jul 26 at 3:04
Pierre21
828
828
$XinmathbbR[X,Y]$ vanishes at all the points $(X,Y)=(0,y)$, for $yinmathbbR$, but it is not zero.
– user578878
Jul 26 at 3:11
Not sure if this helps or not but I think you should only consider the case for uncountable zeros since if it is countable, we can construct the sequence $a_n=(-1)^n$ to obtain the divergence series $S_n$ with countably many zeros.
– Evan William Chandra
Jul 26 at 3:12
Yes of course you're both right, I edited my question.
– Pierre21
Jul 26 at 3:34
add a comment |Â
$XinmathbbR[X,Y]$ vanishes at all the points $(X,Y)=(0,y)$, for $yinmathbbR$, but it is not zero.
– user578878
Jul 26 at 3:11
Not sure if this helps or not but I think you should only consider the case for uncountable zeros since if it is countable, we can construct the sequence $a_n=(-1)^n$ to obtain the divergence series $S_n$ with countably many zeros.
– Evan William Chandra
Jul 26 at 3:12
Yes of course you're both right, I edited my question.
– Pierre21
Jul 26 at 3:34
$XinmathbbR[X,Y]$ vanishes at all the points $(X,Y)=(0,y)$, for $yinmathbbR$, but it is not zero.
– user578878
Jul 26 at 3:11
$XinmathbbR[X,Y]$ vanishes at all the points $(X,Y)=(0,y)$, for $yinmathbbR$, but it is not zero.
– user578878
Jul 26 at 3:11
Not sure if this helps or not but I think you should only consider the case for uncountable zeros since if it is countable, we can construct the sequence $a_n=(-1)^n$ to obtain the divergence series $S_n$ with countably many zeros.
– Evan William Chandra
Jul 26 at 3:12
Not sure if this helps or not but I think you should only consider the case for uncountable zeros since if it is countable, we can construct the sequence $a_n=(-1)^n$ to obtain the divergence series $S_n$ with countably many zeros.
– Evan William Chandra
Jul 26 at 3:12
Yes of course you're both right, I edited my question.
– Pierre21
Jul 26 at 3:34
Yes of course you're both right, I edited my question.
– Pierre21
Jul 26 at 3:34
add a comment |Â
2 Answers
2
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oldest
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up vote
1
down vote
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See Zeros of polynomial over an infinite field. This answers the question for polynomials. If you speak about power series, converging for all $(x,y)in [-1,1] times [-1,1]$, the same holds true by the identity theorem for analytic functions in several variables.
answered Jul 26 at 7:27
Jens Schwaiger
1,102116
add a comment |Â
up vote
1
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If $f=0$ in $(0,1)^2,$ then all partial derivatives of $f$ at $(0,0)$ are $0.$ Couple that with the following: If $f(x,y) = sum_m=0^Msum_n=0^N a_mnx^my^n,$ then $D_x^mD_y^n f(0,0) = m!n!a_mn.$
answered Jul 26 at 18:06
zhw.
65.6k42870
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
See Zeros of polynomial over an infinite field. This answers the question for polynomials. If you speak about power series, converging for all $(x,y)in [-1,1] times [-1,1]$, the same holds true by the identity theorem for analytic functions in several variables.
answered Jul 26 at 7:27
Jens Schwaiger
1,102116
add a comment |Â
up vote
1
down vote
accepted
See Zeros of polynomial over an infinite field. This answers the question for polynomials. If you speak about power series, converging for all $(x,y)in [-1,1] times [-1,1]$, the same holds true by the identity theorem for analytic functions in several variables.
answered Jul 26 at 7:27
Jens Schwaiger
1,102116
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
See Zeros of polynomial over an infinite field. This answers the question for polynomials. If you speak about power series, converging for all $(x,y)in [-1,1] times [-1,1]$, the same holds true by the identity theorem for analytic functions in several variables.
answered Jul 26 at 7:27
Jens Schwaiger
1,102116
See Zeros of polynomial over an infinite field. This answers the question for polynomials. If you speak about power series, converging for all $(x,y)in [-1,1] times [-1,1]$, the same holds true by the identity theorem for analytic functions in several variables.
answered Jul 26 at 7:27
Jens Schwaiger
1,102116
answered Jul 26 at 7:27
Jens Schwaiger
1,102116
answered Jul 26 at 7:27
Jens Schwaiger
1,102116
answered Jul 26 at 7:27
Jens Schwaiger
1,102116
1,102116
add a comment |Â
add a comment |Â
up vote
1
down vote
If $f=0$ in $(0,1)^2,$ then all partial derivatives of $f$ at $(0,0)$ are $0.$ Couple that with the following: If $f(x,y) = sum_m=0^Msum_n=0^N a_mnx^my^n,$ then $D_x^mD_y^n f(0,0) = m!n!a_mn.$
answered Jul 26 at 18:06
zhw.
65.6k42870
add a comment |Â
up vote
1
down vote
If $f=0$ in $(0,1)^2,$ then all partial derivatives of $f$ at $(0,0)$ are $0.$ Couple that with the following: If $f(x,y) = sum_m=0^Msum_n=0^N a_mnx^my^n,$ then $D_x^mD_y^n f(0,0) = m!n!a_mn.$
answered Jul 26 at 18:06
zhw.
65.6k42870
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If $f=0$ in $(0,1)^2,$ then all partial derivatives of $f$ at $(0,0)$ are $0.$ Couple that with the following: If $f(x,y) = sum_m=0^Msum_n=0^N a_mnx^my^n,$ then $D_x^mD_y^n f(0,0) = m!n!a_mn.$
answered Jul 26 at 18:06
zhw.
65.6k42870
If $f=0$ in $(0,1)^2,$ then all partial derivatives of $f$ at $(0,0)$ are $0.$ Couple that with the following: If $f(x,y) = sum_m=0^Msum_n=0^N a_mnx^my^n,$ then $D_x^mD_y^n f(0,0) = m!n!a_mn.$
answered Jul 26 at 18:06
zhw.
65.6k42870
answered Jul 26 at 18:06
zhw.
65.6k42870
answered Jul 26 at 18:06
zhw.
65.6k42870
answered Jul 26 at 18:06
zhw.
65.6k42870
65.6k42870
add a comment |Â
add a comment |Â
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$XinmathbbR[X,Y]$ vanishes at all the points $(X,Y)=(0,y)$, for $yinmathbbR$, but it is not zero.
– user578878
Jul 26 at 3:11
Not sure if this helps or not but I think you should only consider the case for uncountable zeros since if it is countable, we can construct the sequence $a_n=(-1)^n$ to obtain the divergence series $S_n$ with countably many zeros.
– Evan William Chandra
Jul 26 at 3:12
Yes of course you're both right, I edited my question.
– Pierre21
Jul 26 at 3:34