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Taylor expansion for formal power series
Clash Royale CLAN TAG#URR8PPP
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Let $P = p_1x + p_2x^2 + dots$, $Y = y_1x + y_2x^2 + dots$, and $V = v_1x + v_2x^2 + dots$ all be formal power series with indeterminate $x$ and coefficients in some field $mathbbF$, satisfying $p_1,y_1,v_1 neq 0$. How can we prove that
$$P(Y) = P(V) + P'(V)(Y-V) + frac12!P''(V)(Y-V)^2 + frac13!P'''(V)(Y-V)^3 + dots$$
holds formally, i.e. the coefficients of $x^k$ on both sides are equal for all $k$? This is the Taylor expansion for formal power series, but I can't seem to find any reference proof of this result.
power-series taylor-expansion formal-power-series
asked Jul 27 at 17:53
Wakaka
621510
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Let $P = p_1x + p_2x^2 + dots$, $Y = y_1x + y_2x^2 + dots$, and $V = v_1x + v_2x^2 + dots$ all be formal power series with indeterminate $x$ and coefficients in some field $mathbbF$, satisfying $p_1,y_1,v_1 neq 0$. How can we prove that
$$P(Y) = P(V) + P'(V)(Y-V) + frac12!P''(V)(Y-V)^2 + frac13!P'''(V)(Y-V)^3 + dots$$
holds formally, i.e. the coefficients of $x^k$ on both sides are equal for all $k$? This is the Taylor expansion for formal power series, but I can't seem to find any reference proof of this result.
power-series taylor-expansion formal-power-series
asked Jul 27 at 17:53
Wakaka
621510
1
How are Taylor series expansions proved in Calculus?
– Somos
Jul 27 at 19:31
add a comment |Â
up vote
0
down vote
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up vote
0
down vote
favorite
Let $P = p_1x + p_2x^2 + dots$, $Y = y_1x + y_2x^2 + dots$, and $V = v_1x + v_2x^2 + dots$ all be formal power series with indeterminate $x$ and coefficients in some field $mathbbF$, satisfying $p_1,y_1,v_1 neq 0$. How can we prove that
$$P(Y) = P(V) + P'(V)(Y-V) + frac12!P''(V)(Y-V)^2 + frac13!P'''(V)(Y-V)^3 + dots$$
holds formally, i.e. the coefficients of $x^k$ on both sides are equal for all $k$? This is the Taylor expansion for formal power series, but I can't seem to find any reference proof of this result.
power-series taylor-expansion formal-power-series
asked Jul 27 at 17:53
Wakaka
621510
Let $P = p_1x + p_2x^2 + dots$, $Y = y_1x + y_2x^2 + dots$, and $V = v_1x + v_2x^2 + dots$ all be formal power series with indeterminate $x$ and coefficients in some field $mathbbF$, satisfying $p_1,y_1,v_1 neq 0$. How can we prove that
$$P(Y) = P(V) + P'(V)(Y-V) + frac12!P''(V)(Y-V)^2 + frac13!P'''(V)(Y-V)^3 + dots$$
holds formally, i.e. the coefficients of $x^k$ on both sides are equal for all $k$? This is the Taylor expansion for formal power series, but I can't seem to find any reference proof of this result.
power-series taylor-expansion formal-power-series
asked Jul 27 at 17:53
Wakaka
621510
asked Jul 27 at 17:53
Wakaka
621510
asked Jul 27 at 17:53
Wakaka
621510
asked Jul 27 at 17:53
Wakaka
621510
621510
1
How are Taylor series expansions proved in Calculus?
– Somos
Jul 27 at 19:31
add a comment |Â
1
How are Taylor series expansions proved in Calculus?
– Somos
Jul 27 at 19:31
1
1
How are Taylor series expansions proved in Calculus?
– Somos
Jul 27 at 19:31
How are Taylor series expansions proved in Calculus?
– Somos
Jul 27 at 19:31
add a comment |Â
1 Answer
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A general reference for formal power series (f.p.s.) in one and several variables would be Wikipedia. But substitution in the context of f.p.s in several variables is not treated there.
So, following Somos' ideas, we can do the following. We start with two independent indeterminates $u,v$ to get
$$ P(v)=P(u+(v-u))=sum_n a_n (v-u)^l=sum_na_nsum_lbinomnlu^n-l(v-u)^l=\sum_lleft(frac1l!sum_n (n)_l a_n u^n-lright)(v-u)^l=sum_lfracP^(l)(u)l!(v-u)^l,$$ where $(n)_l:=prod_j=0^l-1 (n-j)$.
Then the result follows by substituting $V$ for $u$ and $Y$ for $v$ in this formula.
answered Jul 28 at 8:55
Jens Schwaiger
1,092116
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
A general reference for formal power series (f.p.s.) in one and several variables would be Wikipedia. But substitution in the context of f.p.s in several variables is not treated there.
So, following Somos' ideas, we can do the following. We start with two independent indeterminates $u,v$ to get
$$ P(v)=P(u+(v-u))=sum_n a_n (v-u)^l=sum_na_nsum_lbinomnlu^n-l(v-u)^l=\sum_lleft(frac1l!sum_n (n)_l a_n u^n-lright)(v-u)^l=sum_lfracP^(l)(u)l!(v-u)^l,$$ where $(n)_l:=prod_j=0^l-1 (n-j)$.
Then the result follows by substituting $V$ for $u$ and $Y$ for $v$ in this formula.
answered Jul 28 at 8:55
Jens Schwaiger
1,092116
add a comment |Â
up vote
1
down vote
accepted
A general reference for formal power series (f.p.s.) in one and several variables would be Wikipedia. But substitution in the context of f.p.s in several variables is not treated there.
So, following Somos' ideas, we can do the following. We start with two independent indeterminates $u,v$ to get
$$ P(v)=P(u+(v-u))=sum_n a_n (v-u)^l=sum_na_nsum_lbinomnlu^n-l(v-u)^l=\sum_lleft(frac1l!sum_n (n)_l a_n u^n-lright)(v-u)^l=sum_lfracP^(l)(u)l!(v-u)^l,$$ where $(n)_l:=prod_j=0^l-1 (n-j)$.
Then the result follows by substituting $V$ for $u$ and $Y$ for $v$ in this formula.
answered Jul 28 at 8:55
Jens Schwaiger
1,092116
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
A general reference for formal power series (f.p.s.) in one and several variables would be Wikipedia. But substitution in the context of f.p.s in several variables is not treated there.
So, following Somos' ideas, we can do the following. We start with two independent indeterminates $u,v$ to get
$$ P(v)=P(u+(v-u))=sum_n a_n (v-u)^l=sum_na_nsum_lbinomnlu^n-l(v-u)^l=\sum_lleft(frac1l!sum_n (n)_l a_n u^n-lright)(v-u)^l=sum_lfracP^(l)(u)l!(v-u)^l,$$ where $(n)_l:=prod_j=0^l-1 (n-j)$.
Then the result follows by substituting $V$ for $u$ and $Y$ for $v$ in this formula.
answered Jul 28 at 8:55
Jens Schwaiger
1,092116
A general reference for formal power series (f.p.s.) in one and several variables would be Wikipedia. But substitution in the context of f.p.s in several variables is not treated there.
So, following Somos' ideas, we can do the following. We start with two independent indeterminates $u,v$ to get
$$ P(v)=P(u+(v-u))=sum_n a_n (v-u)^l=sum_na_nsum_lbinomnlu^n-l(v-u)^l=\sum_lleft(frac1l!sum_n (n)_l a_n u^n-lright)(v-u)^l=sum_lfracP^(l)(u)l!(v-u)^l,$$ where $(n)_l:=prod_j=0^l-1 (n-j)$.
Then the result follows by substituting $V$ for $u$ and $Y$ for $v$ in this formula.
answered Jul 28 at 8:55
Jens Schwaiger
1,092116
answered Jul 28 at 8:55
Jens Schwaiger
1,092116
answered Jul 28 at 8:55
Jens Schwaiger
1,092116
answered Jul 28 at 8:55
Jens Schwaiger
1,092116
1,092116
add a comment |Â
add a comment |Â
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1
How are Taylor series expansions proved in Calculus?
– Somos
Jul 27 at 19:31