When does a Taylor expansion have only nonnegative coefficients

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Maybe the question is obvious, but I don't know the answer



Let $p(x_1,...,x_r)$ be a polynomial with nonnegative coefficients and consider
$$
F_n(x) = sum_k=0^n(p(x))^k
$$
where $x = (x_1,...,x_r) in mathbbR^r$. Suppose $F_n$ converges uniformly to some function $f$. Now the coefficient of the taylor expansion of $F_n$ around $0$ are nonnegative. Can I conclude that the coefficients of the taylor epansion of $f$ around 0 are also nonnegative?










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  • 1




    What do you mean by "all coefficients of $f$ are nonnegative"?
    – Sobi
    Sep 10 at 12:28










  • I have edited my question. Sorry!
    – love_math
    Sep 10 at 12:32











  • Well the exponential function is one example. So is fx = 1.
    – Paul Childs
    Sep 10 at 12:35










  • but is there a general result for rational functions?
    – love_math
    Sep 10 at 12:36














up vote
0
down vote

favorite












Maybe the question is obvious, but I don't know the answer



Let $p(x_1,...,x_r)$ be a polynomial with nonnegative coefficients and consider
$$
F_n(x) = sum_k=0^n(p(x))^k
$$
where $x = (x_1,...,x_r) in mathbbR^r$. Suppose $F_n$ converges uniformly to some function $f$. Now the coefficient of the taylor expansion of $F_n$ around $0$ are nonnegative. Can I conclude that the coefficients of the taylor epansion of $f$ around 0 are also nonnegative?










share|cite|improve this question



















  • 1




    What do you mean by "all coefficients of $f$ are nonnegative"?
    – Sobi
    Sep 10 at 12:28










  • I have edited my question. Sorry!
    – love_math
    Sep 10 at 12:32











  • Well the exponential function is one example. So is fx = 1.
    – Paul Childs
    Sep 10 at 12:35










  • but is there a general result for rational functions?
    – love_math
    Sep 10 at 12:36












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Maybe the question is obvious, but I don't know the answer



Let $p(x_1,...,x_r)$ be a polynomial with nonnegative coefficients and consider
$$
F_n(x) = sum_k=0^n(p(x))^k
$$
where $x = (x_1,...,x_r) in mathbbR^r$. Suppose $F_n$ converges uniformly to some function $f$. Now the coefficient of the taylor expansion of $F_n$ around $0$ are nonnegative. Can I conclude that the coefficients of the taylor epansion of $f$ around 0 are also nonnegative?










share|cite|improve this question















Maybe the question is obvious, but I don't know the answer



Let $p(x_1,...,x_r)$ be a polynomial with nonnegative coefficients and consider
$$
F_n(x) = sum_k=0^n(p(x))^k
$$
where $x = (x_1,...,x_r) in mathbbR^r$. Suppose $F_n$ converges uniformly to some function $f$. Now the coefficient of the taylor expansion of $F_n$ around $0$ are nonnegative. Can I conclude that the coefficients of the taylor epansion of $f$ around 0 are also nonnegative?







real-analysis sequences-and-series taylor-expansion






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edited Sep 10 at 13:30

























asked Sep 10 at 12:27









love_math

1455




1455







  • 1




    What do you mean by "all coefficients of $f$ are nonnegative"?
    – Sobi
    Sep 10 at 12:28










  • I have edited my question. Sorry!
    – love_math
    Sep 10 at 12:32











  • Well the exponential function is one example. So is fx = 1.
    – Paul Childs
    Sep 10 at 12:35










  • but is there a general result for rational functions?
    – love_math
    Sep 10 at 12:36












  • 1




    What do you mean by "all coefficients of $f$ are nonnegative"?
    – Sobi
    Sep 10 at 12:28










  • I have edited my question. Sorry!
    – love_math
    Sep 10 at 12:32











  • Well the exponential function is one example. So is fx = 1.
    – Paul Childs
    Sep 10 at 12:35










  • but is there a general result for rational functions?
    – love_math
    Sep 10 at 12:36







1




1




What do you mean by "all coefficients of $f$ are nonnegative"?
– Sobi
Sep 10 at 12:28




What do you mean by "all coefficients of $f$ are nonnegative"?
– Sobi
Sep 10 at 12:28












I have edited my question. Sorry!
– love_math
Sep 10 at 12:32





I have edited my question. Sorry!
– love_math
Sep 10 at 12:32













Well the exponential function is one example. So is fx = 1.
– Paul Childs
Sep 10 at 12:35




Well the exponential function is one example. So is fx = 1.
– Paul Childs
Sep 10 at 12:35












but is there a general result for rational functions?
– love_math
Sep 10 at 12:36




but is there a general result for rational functions?
– love_math
Sep 10 at 12:36










1 Answer
1






active

oldest

votes

















up vote
1
down vote













EDIT: This was an answer to the question before it was edited.



No. Take for example $f : mathbbR to mathbbR$ defined by
$$ f(x) = frac11+x^2. $$
Then by the geometric series formula
$$ f(x) = frac11-(-x^2) = sum_k=0^infty (-1)^kx^2k, $$
whenever $|x| < 1$.
By the uniqueness of Taylor expansions, we know that this is the Taylor expansion of $f$ at $0$, but its coefficients are certainly not nonnegative.






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  • The reason why I'm asking this question is that I read a paper and there is given a rational function $f : (-1,1)^r xrightarrow mathbbR$ which is the uniform limit of some functions $f_n$ on the same domain. Then it was stated that all the coefficients of the taylor expansion of $f$ around 0 are nonnegtive since the coefficients of $f$ are nonnegative
    – love_math
    Sep 10 at 12:43











  • @love_math I would guess that some further restrictions are needed on $f$ to obtain such a result
    – Sobi
    Sep 10 at 12:46











  • can you mention some restrictions?
    – love_math
    Sep 10 at 12:49










  • @love_math I cannot come up with anything out of the blue. However, you could of course just require all derivatives of $f$ at $0$ to be nonnegative.
    – Sobi
    Sep 10 at 12:52










  • Say we have a polynomial $f(x_1,...,x_r) = sum_i_1,...,i_ra_i_1,...,i_rx_1^i_1...x_r^r$. Suppose all coefficients are not zero. What can I say now about the coefficient of the taylor expansion around $0$?
    – love_math
    Sep 10 at 12:57










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote













EDIT: This was an answer to the question before it was edited.



No. Take for example $f : mathbbR to mathbbR$ defined by
$$ f(x) = frac11+x^2. $$
Then by the geometric series formula
$$ f(x) = frac11-(-x^2) = sum_k=0^infty (-1)^kx^2k, $$
whenever $|x| < 1$.
By the uniqueness of Taylor expansions, we know that this is the Taylor expansion of $f$ at $0$, but its coefficients are certainly not nonnegative.






share|cite|improve this answer






















  • The reason why I'm asking this question is that I read a paper and there is given a rational function $f : (-1,1)^r xrightarrow mathbbR$ which is the uniform limit of some functions $f_n$ on the same domain. Then it was stated that all the coefficients of the taylor expansion of $f$ around 0 are nonnegtive since the coefficients of $f$ are nonnegative
    – love_math
    Sep 10 at 12:43











  • @love_math I would guess that some further restrictions are needed on $f$ to obtain such a result
    – Sobi
    Sep 10 at 12:46











  • can you mention some restrictions?
    – love_math
    Sep 10 at 12:49










  • @love_math I cannot come up with anything out of the blue. However, you could of course just require all derivatives of $f$ at $0$ to be nonnegative.
    – Sobi
    Sep 10 at 12:52










  • Say we have a polynomial $f(x_1,...,x_r) = sum_i_1,...,i_ra_i_1,...,i_rx_1^i_1...x_r^r$. Suppose all coefficients are not zero. What can I say now about the coefficient of the taylor expansion around $0$?
    – love_math
    Sep 10 at 12:57














up vote
1
down vote













EDIT: This was an answer to the question before it was edited.



No. Take for example $f : mathbbR to mathbbR$ defined by
$$ f(x) = frac11+x^2. $$
Then by the geometric series formula
$$ f(x) = frac11-(-x^2) = sum_k=0^infty (-1)^kx^2k, $$
whenever $|x| < 1$.
By the uniqueness of Taylor expansions, we know that this is the Taylor expansion of $f$ at $0$, but its coefficients are certainly not nonnegative.






share|cite|improve this answer






















  • The reason why I'm asking this question is that I read a paper and there is given a rational function $f : (-1,1)^r xrightarrow mathbbR$ which is the uniform limit of some functions $f_n$ on the same domain. Then it was stated that all the coefficients of the taylor expansion of $f$ around 0 are nonnegtive since the coefficients of $f$ are nonnegative
    – love_math
    Sep 10 at 12:43











  • @love_math I would guess that some further restrictions are needed on $f$ to obtain such a result
    – Sobi
    Sep 10 at 12:46











  • can you mention some restrictions?
    – love_math
    Sep 10 at 12:49










  • @love_math I cannot come up with anything out of the blue. However, you could of course just require all derivatives of $f$ at $0$ to be nonnegative.
    – Sobi
    Sep 10 at 12:52










  • Say we have a polynomial $f(x_1,...,x_r) = sum_i_1,...,i_ra_i_1,...,i_rx_1^i_1...x_r^r$. Suppose all coefficients are not zero. What can I say now about the coefficient of the taylor expansion around $0$?
    – love_math
    Sep 10 at 12:57












up vote
1
down vote










up vote
1
down vote









EDIT: This was an answer to the question before it was edited.



No. Take for example $f : mathbbR to mathbbR$ defined by
$$ f(x) = frac11+x^2. $$
Then by the geometric series formula
$$ f(x) = frac11-(-x^2) = sum_k=0^infty (-1)^kx^2k, $$
whenever $|x| < 1$.
By the uniqueness of Taylor expansions, we know that this is the Taylor expansion of $f$ at $0$, but its coefficients are certainly not nonnegative.






share|cite|improve this answer














EDIT: This was an answer to the question before it was edited.



No. Take for example $f : mathbbR to mathbbR$ defined by
$$ f(x) = frac11+x^2. $$
Then by the geometric series formula
$$ f(x) = frac11-(-x^2) = sum_k=0^infty (-1)^kx^2k, $$
whenever $|x| < 1$.
By the uniqueness of Taylor expansions, we know that this is the Taylor expansion of $f$ at $0$, but its coefficients are certainly not nonnegative.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 10 at 13:31

























answered Sep 10 at 12:36









Sobi

2,925517




2,925517











  • The reason why I'm asking this question is that I read a paper and there is given a rational function $f : (-1,1)^r xrightarrow mathbbR$ which is the uniform limit of some functions $f_n$ on the same domain. Then it was stated that all the coefficients of the taylor expansion of $f$ around 0 are nonnegtive since the coefficients of $f$ are nonnegative
    – love_math
    Sep 10 at 12:43











  • @love_math I would guess that some further restrictions are needed on $f$ to obtain such a result
    – Sobi
    Sep 10 at 12:46











  • can you mention some restrictions?
    – love_math
    Sep 10 at 12:49










  • @love_math I cannot come up with anything out of the blue. However, you could of course just require all derivatives of $f$ at $0$ to be nonnegative.
    – Sobi
    Sep 10 at 12:52










  • Say we have a polynomial $f(x_1,...,x_r) = sum_i_1,...,i_ra_i_1,...,i_rx_1^i_1...x_r^r$. Suppose all coefficients are not zero. What can I say now about the coefficient of the taylor expansion around $0$?
    – love_math
    Sep 10 at 12:57
















  • The reason why I'm asking this question is that I read a paper and there is given a rational function $f : (-1,1)^r xrightarrow mathbbR$ which is the uniform limit of some functions $f_n$ on the same domain. Then it was stated that all the coefficients of the taylor expansion of $f$ around 0 are nonnegtive since the coefficients of $f$ are nonnegative
    – love_math
    Sep 10 at 12:43











  • @love_math I would guess that some further restrictions are needed on $f$ to obtain such a result
    – Sobi
    Sep 10 at 12:46











  • can you mention some restrictions?
    – love_math
    Sep 10 at 12:49










  • @love_math I cannot come up with anything out of the blue. However, you could of course just require all derivatives of $f$ at $0$ to be nonnegative.
    – Sobi
    Sep 10 at 12:52










  • Say we have a polynomial $f(x_1,...,x_r) = sum_i_1,...,i_ra_i_1,...,i_rx_1^i_1...x_r^r$. Suppose all coefficients are not zero. What can I say now about the coefficient of the taylor expansion around $0$?
    – love_math
    Sep 10 at 12:57















The reason why I'm asking this question is that I read a paper and there is given a rational function $f : (-1,1)^r xrightarrow mathbbR$ which is the uniform limit of some functions $f_n$ on the same domain. Then it was stated that all the coefficients of the taylor expansion of $f$ around 0 are nonnegtive since the coefficients of $f$ are nonnegative
– love_math
Sep 10 at 12:43





The reason why I'm asking this question is that I read a paper and there is given a rational function $f : (-1,1)^r xrightarrow mathbbR$ which is the uniform limit of some functions $f_n$ on the same domain. Then it was stated that all the coefficients of the taylor expansion of $f$ around 0 are nonnegtive since the coefficients of $f$ are nonnegative
– love_math
Sep 10 at 12:43













@love_math I would guess that some further restrictions are needed on $f$ to obtain such a result
– Sobi
Sep 10 at 12:46





@love_math I would guess that some further restrictions are needed on $f$ to obtain such a result
– Sobi
Sep 10 at 12:46













can you mention some restrictions?
– love_math
Sep 10 at 12:49




can you mention some restrictions?
– love_math
Sep 10 at 12:49












@love_math I cannot come up with anything out of the blue. However, you could of course just require all derivatives of $f$ at $0$ to be nonnegative.
– Sobi
Sep 10 at 12:52




@love_math I cannot come up with anything out of the blue. However, you could of course just require all derivatives of $f$ at $0$ to be nonnegative.
– Sobi
Sep 10 at 12:52












Say we have a polynomial $f(x_1,...,x_r) = sum_i_1,...,i_ra_i_1,...,i_rx_1^i_1...x_r^r$. Suppose all coefficients are not zero. What can I say now about the coefficient of the taylor expansion around $0$?
– love_math
Sep 10 at 12:57




Say we have a polynomial $f(x_1,...,x_r) = sum_i_1,...,i_ra_i_1,...,i_rx_1^i_1...x_r^r$. Suppose all coefficients are not zero. What can I say now about the coefficient of the taylor expansion around $0$?
– love_math
Sep 10 at 12:57

















 

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