inverse of polynomial as sum of two power series












0












$begingroup$


given polynomial $$g(x) =ax^2+bx+c$$

I try to find $g(x)^{-1}$ as a sum of two power series.



I wrote $g(x) = alphabeta(1-frac{x}{alpha})(1-frac{x}{alpha})$ when $ alpha,beta = frac{ - bpmsqrt{b^2 - 4ac}}{2a} in C$

then $$g(x)^{-1} = frac{1}{alphabeta} frac{1}{(1-frac{x}{alpha})(1-frac{x}{beta})} = frac{1}{alphabeta} left[frac{A}{(1-frac{x}{alpha})} + frac{B}{(1-frac{x}{beta})} right]$$

then I found $$A = frac{-beta}{alpha - beta} ,B = frac{alpha }{alpha - beta}$$

so I can get $$g(x)^{-1} = frac{1}{alphabeta}left[frac{-beta}{alpha - beta} sum_{n=0}^{n=infty} {left(frac{x}{alpha }right)}^n + frac{alpha }{alpha - beta} sum_{n=0}^{n=infty} {left(frac{x}{beta}right)}^nright]$$



my question is, is there any way to assure that the each power series is Real or change it into sum of two Real power series?






polynomials power-series inverse inverse-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are $alpha,beta$ real?
    $endgroup$
    – lab bhattacharjee
    Dec 29 '18 at 11:04












  • $begingroup$
    not necessarily
    $endgroup$
    – MSm
    Dec 29 '18 at 12:55
















0












$begingroup$


given polynomial $$g(x) =ax^2+bx+c$$

I try to find $g(x)^{-1}$ as a sum of two power series.



I wrote $g(x) = alphabeta(1-frac{x}{alpha})(1-frac{x}{alpha})$ when $ alpha,beta = frac{ - bpmsqrt{b^2 - 4ac}}{2a} in C$

then $$g(x)^{-1} = frac{1}{alphabeta} frac{1}{(1-frac{x}{alpha})(1-frac{x}{beta})} = frac{1}{alphabeta} left[frac{A}{(1-frac{x}{alpha})} + frac{B}{(1-frac{x}{beta})} right]$$

then I found $$A = frac{-beta}{alpha - beta} ,B = frac{alpha }{alpha - beta}$$

so I can get $$g(x)^{-1} = frac{1}{alphabeta}left[frac{-beta}{alpha - beta} sum_{n=0}^{n=infty} {left(frac{x}{alpha }right)}^n + frac{alpha }{alpha - beta} sum_{n=0}^{n=infty} {left(frac{x}{beta}right)}^nright]$$



my question is, is there any way to assure that the each power series is Real or change it into sum of two Real power series?






polynomials power-series inverse inverse-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are $alpha,beta$ real?
    $endgroup$
    – lab bhattacharjee
    Dec 29 '18 at 11:04












  • $begingroup$
    not necessarily
    $endgroup$
    – MSm
    Dec 29 '18 at 12:55














0












0








0





$begingroup$


given polynomial $$g(x) =ax^2+bx+c$$

I try to find $g(x)^{-1}$ as a sum of two power series.



I wrote $g(x) = alphabeta(1-frac{x}{alpha})(1-frac{x}{alpha})$ when $ alpha,beta = frac{ - bpmsqrt{b^2 - 4ac}}{2a} in C$

then $$g(x)^{-1} = frac{1}{alphabeta} frac{1}{(1-frac{x}{alpha})(1-frac{x}{beta})} = frac{1}{alphabeta} left[frac{A}{(1-frac{x}{alpha})} + frac{B}{(1-frac{x}{beta})} right]$$

then I found $$A = frac{-beta}{alpha - beta} ,B = frac{alpha }{alpha - beta}$$

so I can get $$g(x)^{-1} = frac{1}{alphabeta}left[frac{-beta}{alpha - beta} sum_{n=0}^{n=infty} {left(frac{x}{alpha }right)}^n + frac{alpha }{alpha - beta} sum_{n=0}^{n=infty} {left(frac{x}{beta}right)}^nright]$$



my question is, is there any way to assure that the each power series is Real or change it into sum of two Real power series?






polynomials power-series inverse inverse-function







share|cite|improve this question











$endgroup$




given polynomial $$g(x) =ax^2+bx+c$$

I try to find $g(x)^{-1}$ as a sum of two power series.



I wrote $g(x) = alphabeta(1-frac{x}{alpha})(1-frac{x}{alpha})$ when $ alpha,beta = frac{ - bpmsqrt{b^2 - 4ac}}{2a} in C$

then $$g(x)^{-1} = frac{1}{alphabeta} frac{1}{(1-frac{x}{alpha})(1-frac{x}{beta})} = frac{1}{alphabeta} left[frac{A}{(1-frac{x}{alpha})} + frac{B}{(1-frac{x}{beta})} right]$$

then I found $$A = frac{-beta}{alpha - beta} ,B = frac{alpha }{alpha - beta}$$

so I can get $$g(x)^{-1} = frac{1}{alphabeta}left[frac{-beta}{alpha - beta} sum_{n=0}^{n=infty} {left(frac{x}{alpha }right)}^n + frac{alpha }{alpha - beta} sum_{n=0}^{n=infty} {left(frac{x}{beta}right)}^nright]$$



my question is, is there any way to assure that the each power series is Real or change it into sum of two Real power series?






polynomials power-series inverse inverse-function




polynomials power-series inverse inverse-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 13:18







MSm

















asked Dec 29 '18 at 10:56









MSmMSm

35719




35719












  • $begingroup$
    Are $alpha,beta$ real?
    $endgroup$
    – lab bhattacharjee
    Dec 29 '18 at 11:04












  • $begingroup$
    not necessarily
    $endgroup$
    – MSm
    Dec 29 '18 at 12:55


















  • $begingroup$
    Are $alpha,beta$ real?
    $endgroup$
    – lab bhattacharjee
    Dec 29 '18 at 11:04












  • $begingroup$
    not necessarily
    $endgroup$
    – MSm
    Dec 29 '18 at 12:55
















$begingroup$
Are $alpha,beta$ real?
$endgroup$
– lab bhattacharjee
Dec 29 '18 at 11:04






$begingroup$
Are $alpha,beta$ real?
$endgroup$
– lab bhattacharjee
Dec 29 '18 at 11:04














$begingroup$
not necessarily
$endgroup$
– MSm
Dec 29 '18 at 12:55




$begingroup$
not necessarily
$endgroup$
– MSm
Dec 29 '18 at 12:55










1 Answer
1






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oldest

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2












$begingroup$

The coefficient of $x^n$ will be $$dfrac1{alphabeta(alpha-beta)}left(dfracalpha{beta^n}-dfracbeta{alpha^n}right)=dfrac{alpha^{n+1}-beta^{n+1}}{(alphabeta)^{n+1}(alpha-beta)}$$



As



$$(alpha+beta)(alpha^n-beta^n)=alpha^{n+1}-beta^{n+1}+alphabeta(alpha^{n-1}-beta^{n-1})$$



If $f(m)=dfrac{alpha^m-beta^m}{alpha-beta},$ $$f(n+1)=(alpha+beta)f(n)-alphabeta f(n-1)$$



Use strong induction to establish that if $f(m)$ is real for $mle n,f(n+1)$ will be real



Assumption: $alpha+beta,alphabeta$ are real






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks for the answer. is it possible to show that each power series is Real?
    $endgroup$
    – MSm
    Dec 29 '18 at 12:55













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

The coefficient of $x^n$ will be $$dfrac1{alphabeta(alpha-beta)}left(dfracalpha{beta^n}-dfracbeta{alpha^n}right)=dfrac{alpha^{n+1}-beta^{n+1}}{(alphabeta)^{n+1}(alpha-beta)}$$



As



$$(alpha+beta)(alpha^n-beta^n)=alpha^{n+1}-beta^{n+1}+alphabeta(alpha^{n-1}-beta^{n-1})$$



If $f(m)=dfrac{alpha^m-beta^m}{alpha-beta},$ $$f(n+1)=(alpha+beta)f(n)-alphabeta f(n-1)$$



Use strong induction to establish that if $f(m)$ is real for $mle n,f(n+1)$ will be real



Assumption: $alpha+beta,alphabeta$ are real






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks for the answer. is it possible to show that each power series is Real?
    $endgroup$
    – MSm
    Dec 29 '18 at 12:55


















2












$begingroup$

The coefficient of $x^n$ will be $$dfrac1{alphabeta(alpha-beta)}left(dfracalpha{beta^n}-dfracbeta{alpha^n}right)=dfrac{alpha^{n+1}-beta^{n+1}}{(alphabeta)^{n+1}(alpha-beta)}$$



As



$$(alpha+beta)(alpha^n-beta^n)=alpha^{n+1}-beta^{n+1}+alphabeta(alpha^{n-1}-beta^{n-1})$$



If $f(m)=dfrac{alpha^m-beta^m}{alpha-beta},$ $$f(n+1)=(alpha+beta)f(n)-alphabeta f(n-1)$$



Use strong induction to establish that if $f(m)$ is real for $mle n,f(n+1)$ will be real



Assumption: $alpha+beta,alphabeta$ are real






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks for the answer. is it possible to show that each power series is Real?
    $endgroup$
    – MSm
    Dec 29 '18 at 12:55
















2












2








2





$begingroup$

The coefficient of $x^n$ will be $$dfrac1{alphabeta(alpha-beta)}left(dfracalpha{beta^n}-dfracbeta{alpha^n}right)=dfrac{alpha^{n+1}-beta^{n+1}}{(alphabeta)^{n+1}(alpha-beta)}$$



As



$$(alpha+beta)(alpha^n-beta^n)=alpha^{n+1}-beta^{n+1}+alphabeta(alpha^{n-1}-beta^{n-1})$$



If $f(m)=dfrac{alpha^m-beta^m}{alpha-beta},$ $$f(n+1)=(alpha+beta)f(n)-alphabeta f(n-1)$$



Use strong induction to establish that if $f(m)$ is real for $mle n,f(n+1)$ will be real



Assumption: $alpha+beta,alphabeta$ are real






share|cite|improve this answer









$endgroup$



The coefficient of $x^n$ will be $$dfrac1{alphabeta(alpha-beta)}left(dfracalpha{beta^n}-dfracbeta{alpha^n}right)=dfrac{alpha^{n+1}-beta^{n+1}}{(alphabeta)^{n+1}(alpha-beta)}$$



As



$$(alpha+beta)(alpha^n-beta^n)=alpha^{n+1}-beta^{n+1}+alphabeta(alpha^{n-1}-beta^{n-1})$$



If $f(m)=dfrac{alpha^m-beta^m}{alpha-beta},$ $$f(n+1)=(alpha+beta)f(n)-alphabeta f(n-1)$$



Use strong induction to establish that if $f(m)$ is real for $mle n,f(n+1)$ will be real



Assumption: $alpha+beta,alphabeta$ are real







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 11:11









lab bhattacharjeelab bhattacharjee

226k15157275




226k15157275












  • $begingroup$
    thanks for the answer. is it possible to show that each power series is Real?
    $endgroup$
    – MSm
    Dec 29 '18 at 12:55




















  • $begingroup$
    thanks for the answer. is it possible to show that each power series is Real?
    $endgroup$
    – MSm
    Dec 29 '18 at 12:55


















$begingroup$
thanks for the answer. is it possible to show that each power series is Real?
$endgroup$
– MSm
Dec 29 '18 at 12:55






$begingroup$
thanks for the answer. is it possible to show that each power series is Real?
$endgroup$
– MSm
Dec 29 '18 at 12:55




















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