show that we can write $ [B_{frac{1}{N}}(x)]^N=(1+x)$, similar to Binomial expansion
$begingroup$
We know binomial expansion $$B_{a}(x)=(1+x)^a=sum_{n=0}^{infty} frac{a(a-1)(a-2) cdots (a-n+1)}{n!}x^n$$ For any $a in mathbb{R} $ or $mathbb{C}$, this series converges in $mathbb{R}$ or $mathbb{C}$ if $|x|<1$.
Now consider an arbitrary complete field $K$ and let $a in K$ and define
$$ B_{a}(X)=sum_{n=0}^{infty} frac{a(a-1)(a-2) cdots (a-n+1)}{n!}X^n in K[[X]],$$ where $K[[X]]$ is the ring of formal power series.
$ underline{text{My question}:}$
If $ a=frac{1}{N}, N in mathbb{Z}$ be a rational number and $ x in {x: |x|<1 }$, then show that we can write $$ [B_{frac{1}{N}}(x)]^N=(1+x)$$.
That is to show $left[ B_{frac{1}{N}}(x) right]$ is an $N^{th}$ root of $1+x$.
Answer:
There is a principle as below:
$text{Any identity between real or complex power series involving addition , multiplication is }$
$ text{also an identity in the ring of formal power series}$
I think using this prove the result.
Let $F(X)$ be difference between $ [B_{1/N}(x)]^N$ and $1+x$, i.e., $F(X)=[B_{1/N}(X)]^N-(1+X)$, where $$B_{1/N}(X)=sum_{n=0}^{infty} frac{1/N(1/N-1)(1/N-2) cdots (1/N-n+1)}{n!} X^n in mathbb{Q}[[X]]$$
Then $F(X)$ must vanish for all real values of $X$ in the region $|x|<1$.
If we gather together all $X^n$-terms in this expression, its coefficient must always be $0$.
Since the binomial series converges, $ left[ B_{frac{1}{N}}(x) right]=(1+x)$ can be written as Taylor's series like $$ sum_{n=0}^{infty} c_n x^n$$
Now if we show that $c_n=0 forall n$, then we are done.
Since the formal identity $F(X)$
in $ mathbb{Q}[[X]]$ tells that all coefficients $c_n=0$.
Hence $$left[B_{1/N}(x) right]^N=1+x$$
Is this conclusion true?
power-series formal-power-series
edited Jan 1 at 16:23
Larry
2,53031131
asked Jan 1 at 13:55
M. A. SARKARM. A. SARKAR
2,4551820
$endgroup$
add a comment |
$begingroup$
We know binomial expansion $$B_{a}(x)=(1+x)^a=sum_{n=0}^{infty} frac{a(a-1)(a-2) cdots (a-n+1)}{n!}x^n$$ For any $a in mathbb{R} $ or $mathbb{C}$, this series converges in $mathbb{R}$ or $mathbb{C}$ if $|x|<1$.
Now consider an arbitrary complete field $K$ and let $a in K$ and define
$$ B_{a}(X)=sum_{n=0}^{infty} frac{a(a-1)(a-2) cdots (a-n+1)}{n!}X^n in K[[X]],$$ where $K[[X]]$ is the ring of formal power series.
$ underline{text{My question}:}$
If $ a=frac{1}{N}, N in mathbb{Z}$ be a rational number and $ x in {x: |x|<1 }$, then show that we can write $$ [B_{frac{1}{N}}(x)]^N=(1+x)$$.
That is to show $left[ B_{frac{1}{N}}(x) right]$ is an $N^{th}$ root of $1+x$.
Answer:
There is a principle as below:
$text{Any identity between real or complex power series involving addition , multiplication is }$
$ text{also an identity in the ring of formal power series}$
I think using this prove the result.
Let $F(X)$ be difference between $ [B_{1/N}(x)]^N$ and $1+x$, i.e., $F(X)=[B_{1/N}(X)]^N-(1+X)$, where $$B_{1/N}(X)=sum_{n=0}^{infty} frac{1/N(1/N-1)(1/N-2) cdots (1/N-n+1)}{n!} X^n in mathbb{Q}[[X]]$$
Then $F(X)$ must vanish for all real values of $X$ in the region $|x|<1$.
If we gather together all $X^n$-terms in this expression, its coefficient must always be $0$.
Since the binomial series converges, $ left[ B_{frac{1}{N}}(x) right]=(1+x)$ can be written as Taylor's series like $$ sum_{n=0}^{infty} c_n x^n$$
Now if we show that $c_n=0 forall n$, then we are done.
Since the formal identity $F(X)$
in $ mathbb{Q}[[X]]$ tells that all coefficients $c_n=0$.
Hence $$left[B_{1/N}(x) right]^N=1+x$$
Is this conclusion true?
power-series formal-power-series
edited Jan 1 at 16:23
Larry
2,53031131
asked Jan 1 at 13:55
M. A. SARKARM. A. SARKAR
2,4551820
$endgroup$
add a comment |
$begingroup$
We know binomial expansion $$B_{a}(x)=(1+x)^a=sum_{n=0}^{infty} frac{a(a-1)(a-2) cdots (a-n+1)}{n!}x^n$$ For any $a in mathbb{R} $ or $mathbb{C}$, this series converges in $mathbb{R}$ or $mathbb{C}$ if $|x|<1$.
Now consider an arbitrary complete field $K$ and let $a in K$ and define
$$ B_{a}(X)=sum_{n=0}^{infty} frac{a(a-1)(a-2) cdots (a-n+1)}{n!}X^n in K[[X]],$$ where $K[[X]]$ is the ring of formal power series.
$ underline{text{My question}:}$
If $ a=frac{1}{N}, N in mathbb{Z}$ be a rational number and $ x in {x: |x|<1 }$, then show that we can write $$ [B_{frac{1}{N}}(x)]^N=(1+x)$$.
That is to show $left[ B_{frac{1}{N}}(x) right]$ is an $N^{th}$ root of $1+x$.
Answer:
There is a principle as below:
$text{Any identity between real or complex power series involving addition , multiplication is }$
$ text{also an identity in the ring of formal power series}$
I think using this prove the result.
Let $F(X)$ be difference between $ [B_{1/N}(x)]^N$ and $1+x$, i.e., $F(X)=[B_{1/N}(X)]^N-(1+X)$, where $$B_{1/N}(X)=sum_{n=0}^{infty} frac{1/N(1/N-1)(1/N-2) cdots (1/N-n+1)}{n!} X^n in mathbb{Q}[[X]]$$
Then $F(X)$ must vanish for all real values of $X$ in the region $|x|<1$.
If we gather together all $X^n$-terms in this expression, its coefficient must always be $0$.
Since the binomial series converges, $ left[ B_{frac{1}{N}}(x) right]=(1+x)$ can be written as Taylor's series like $$ sum_{n=0}^{infty} c_n x^n$$
Now if we show that $c_n=0 forall n$, then we are done.
Since the formal identity $F(X)$
in $ mathbb{Q}[[X]]$ tells that all coefficients $c_n=0$.
Hence $$left[B_{1/N}(x) right]^N=1+x$$
Is this conclusion true?
power-series formal-power-series
edited Jan 1 at 16:23
Larry
2,53031131
asked Jan 1 at 13:55
M. A. SARKARM. A. SARKAR
2,4551820
$endgroup$
We know binomial expansion $$B_{a}(x)=(1+x)^a=sum_{n=0}^{infty} frac{a(a-1)(a-2) cdots (a-n+1)}{n!}x^n$$ For any $a in mathbb{R} $ or $mathbb{C}$, this series converges in $mathbb{R}$ or $mathbb{C}$ if $|x|<1$.
Now consider an arbitrary complete field $K$ and let $a in K$ and define
$$ B_{a}(X)=sum_{n=0}^{infty} frac{a(a-1)(a-2) cdots (a-n+1)}{n!}X^n in K[[X]],$$ where $K[[X]]$ is the ring of formal power series.
$ underline{text{My question}:}$
If $ a=frac{1}{N}, N in mathbb{Z}$ be a rational number and $ x in {x: |x|<1 }$, then show that we can write $$ [B_{frac{1}{N}}(x)]^N=(1+x)$$.
That is to show $left[ B_{frac{1}{N}}(x) right]$ is an $N^{th}$ root of $1+x$.
Answer:
There is a principle as below:
$text{Any identity between real or complex power series involving addition , multiplication is }$
$ text{also an identity in the ring of formal power series}$
I think using this prove the result.
Let $F(X)$ be difference between $ [B_{1/N}(x)]^N$ and $1+x$, i.e., $F(X)=[B_{1/N}(X)]^N-(1+X)$, where $$B_{1/N}(X)=sum_{n=0}^{infty} frac{1/N(1/N-1)(1/N-2) cdots (1/N-n+1)}{n!} X^n in mathbb{Q}[[X]]$$
Then $F(X)$ must vanish for all real values of $X$ in the region $|x|<1$.
If we gather together all $X^n$-terms in this expression, its coefficient must always be $0$.
Since the binomial series converges, $ left[ B_{frac{1}{N}}(x) right]=(1+x)$ can be written as Taylor's series like $$ sum_{n=0}^{infty} c_n x^n$$
Now if we show that $c_n=0 forall n$, then we are done.
Since the formal identity $F(X)$
in $ mathbb{Q}[[X]]$ tells that all coefficients $c_n=0$.
Hence $$left[B_{1/N}(x) right]^N=1+x$$
Is this conclusion true?
power-series formal-power-series
power-series formal-power-series
edited Jan 1 at 16:23
Larry
2,53031131
asked Jan 1 at 13:55
M. A. SARKARM. A. SARKAR
2,4551820
edited Jan 1 at 16:23
Larry
2,53031131
asked Jan 1 at 13:55
M. A. SARKARM. A. SARKAR
2,4551820
edited Jan 1 at 16:23
Larry
2,53031131
edited Jan 1 at 16:23
Larry
2,53031131
edited Jan 1 at 16:23
Larry
2,53031131
2,53031131
asked Jan 1 at 13:55
M. A. SARKARM. A. SARKAR
2,4551820
asked Jan 1 at 13:55
M. A. SARKARM. A. SARKAR
2,4551820
asked Jan 1 at 13:55
M. A. SARKARM. A. SARKAR
2,4551820
2,4551820
add a comment |
add a comment |
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