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Robert Israel
Robert Israel
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YouWhile a simple closed-form solution isn't available, you could try series expansions in the parameter $d$, if $d$ is small. Thus suppose you want the root that is $a$ at $d=0$. For simplicity, rewrite the equation as $$ X (X-B)(X-C) = d(X-E)(X-F)$$ where $X=x-a$, $B = b-a$, $C=c-a$, $E=e-a$, $F=f-a$. Then this root is $$\eqalign{ X &= {\frac {EF}{BC}}d+ \left( -{\frac {{E}^{2}F}{{B}^{2}{C}^{2}}}-{\frac {E{F}^{2}}{{B}^{2}{C}^{2}}}+{\frac {{E}^{2}{F}^{2}}{{B}^{2}{C}^{3}}}+{ \frac {{E}^{2}{F}^{2}}{{B}^{3}{C}^{2}}} \right) {d}^{2}\cr+ &\left( { \frac {{E}^{3}F}{{B}^{3}{C}^{3}}}+3\,{\frac {{E}^{2}{F}^{2}}{{B}^{3}{C }^{3}}}+{\frac {E{F}^{3}}{{B}^{3}{C}^{3}}}-3\,{\frac {{E}^{3}{F}^{2}}{ {B}^{3}{C}^{4}}}-3\,{\frac {{E}^{2}{F}^{3}}{{B}^{3}{C}^{4}}}+2\,{ \frac {{E}^{3}{F}^{3}}{{B}^{3}{C}^{5}}}-3\,{\frac {{E}^{3}{F}^{2}}{{B} ^{4}{C}^{3}}}-3\,{\frac {{E}^{2}{F}^{3}}{{B}^{4}{C}^{3}}}+3\,{\frac {{ E}^{3}{F}^{3}}{{B}^{4}{C}^{4}}}+2\,{\frac {{E}^{3}{F}^{3}}{{B}^{5}{C}^ {3}}} \right) {d}^{3}\cr \cr+&\ldots} $$ This (and as many terms as you want) can be obtained from the Lagrange–Bürmann formula.

You could try series expansions in the parameter $d$, if $d$ is small. Thus suppose you want the root that is $a$ at $d=0$. For simplicity, rewrite the equation as $$ X (X-B)(X-C) = d(X-E)(X-F)$$ where $X=x-a$, $B = b-a$, $C=c-a$, $E=e-a$, $F=f-a$. Then this root is $$\eqalign{ X &= {\frac {EF}{BC}}d+ \left( -{\frac {{E}^{2}F}{{B}^{2}{C}^{2}}}-{\frac {E{F}^{2}}{{B}^{2}{C}^{2}}}+{\frac {{E}^{2}{F}^{2}}{{B}^{2}{C}^{3}}}+{ \frac {{E}^{2}{F}^{2}}{{B}^{3}{C}^{2}}} \right) {d}^{2}\cr+ &\left( { \frac {{E}^{3}F}{{B}^{3}{C}^{3}}}+3\,{\frac {{E}^{2}{F}^{2}}{{B}^{3}{C }^{3}}}+{\frac {E{F}^{3}}{{B}^{3}{C}^{3}}}-3\,{\frac {{E}^{3}{F}^{2}}{ {B}^{3}{C}^{4}}}-3\,{\frac {{E}^{2}{F}^{3}}{{B}^{3}{C}^{4}}}+2\,{ \frac {{E}^{3}{F}^{3}}{{B}^{3}{C}^{5}}}-3\,{\frac {{E}^{3}{F}^{2}}{{B} ^{4}{C}^{3}}}-3\,{\frac {{E}^{2}{F}^{3}}{{B}^{4}{C}^{3}}}+3\,{\frac {{ E}^{3}{F}^{3}}{{B}^{4}{C}^{4}}}+2\,{\frac {{E}^{3}{F}^{3}}{{B}^{5}{C}^ {3}}} \right) {d}^{3}\cr \cr+&\ldots} $$ This (and as many terms as you want) can be obtained from the Lagrange–Bürmann formula.

While a simple closed-form solution isn't available, you could try series expansions in the parameter $d$, if $d$ is small. Thus suppose you want the root that is $a$ at $d=0$. For simplicity, rewrite the equation as $$ X (X-B)(X-C) = d(X-E)(X-F)$$ where $X=x-a$, $B = b-a$, $C=c-a$, $E=e-a$, $F=f-a$. Then this root is $$\eqalign{ X &= {\frac {EF}{BC}}d+ \left( -{\frac {{E}^{2}F}{{B}^{2}{C}^{2}}}-{\frac {E{F}^{2}}{{B}^{2}{C}^{2}}}+{\frac {{E}^{2}{F}^{2}}{{B}^{2}{C}^{3}}}+{ \frac {{E}^{2}{F}^{2}}{{B}^{3}{C}^{2}}} \right) {d}^{2}\cr+ &\left( { \frac {{E}^{3}F}{{B}^{3}{C}^{3}}}+3\,{\frac {{E}^{2}{F}^{2}}{{B}^{3}{C }^{3}}}+{\frac {E{F}^{3}}{{B}^{3}{C}^{3}}}-3\,{\frac {{E}^{3}{F}^{2}}{ {B}^{3}{C}^{4}}}-3\,{\frac {{E}^{2}{F}^{3}}{{B}^{3}{C}^{4}}}+2\,{ \frac {{E}^{3}{F}^{3}}{{B}^{3}{C}^{5}}}-3\,{\frac {{E}^{3}{F}^{2}}{{B} ^{4}{C}^{3}}}-3\,{\frac {{E}^{2}{F}^{3}}{{B}^{4}{C}^{3}}}+3\,{\frac {{ E}^{3}{F}^{3}}{{B}^{4}{C}^{4}}}+2\,{\frac {{E}^{3}{F}^{3}}{{B}^{5}{C}^ {3}}} \right) {d}^{3}\cr \cr+&\ldots} $$ This (and as many terms as you want) can be obtained from the Lagrange–Bürmann formula.

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Robert Israel
Robert Israel
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  • 1
  • 76
  • 152

You could try series expansions in the parameter $d$, if $d$ is small. Thus suppose you want the root that is $a$ at $d=0$. For simplicity, rewrite the equation as $$ X (X-B)(X-C) = d(X-E)(X-F)$$ where $X=x-a$, $B = b-a$, $C=c-a$, $E=e-a$, $F=f-a$. Then this root is $$\eqalign{ X &= {\frac {EF}{BC}}d+ \left( -{\frac {{E}^{2}F}{{B}^{2}{C}^{2}}}-{\frac {E{F}^{2}}{{B}^{2}{C}^{2}}}+{\frac {{E}^{2}{F}^{2}}{{B}^{2}{C}^{3}}}+{ \frac {{E}^{2}{F}^{2}}{{B}^{3}{C}^{2}}} \right) {d}^{2}\cr+ &\left( { \frac {{E}^{3}F}{{B}^{3}{C}^{3}}}+3\,{\frac {{E}^{2}{F}^{2}}{{B}^{3}{C }^{3}}}+{\frac {E{F}^{3}}{{B}^{3}{C}^{3}}}-3\,{\frac {{E}^{3}{F}^{2}}{ {B}^{3}{C}^{4}}}-3\,{\frac {{E}^{2}{F}^{3}}{{B}^{3}{C}^{4}}}+2\,{ \frac {{E}^{3}{F}^{3}}{{B}^{3}{C}^{5}}}-3\,{\frac {{E}^{3}{F}^{2}}{{B} ^{4}{C}^{3}}}-3\,{\frac {{E}^{2}{F}^{3}}{{B}^{4}{C}^{3}}}+3\,{\frac {{ E}^{3}{F}^{3}}{{B}^{4}{C}^{4}}}+2\,{\frac {{E}^{3}{F}^{3}}{{B}^{5}{C}^ {3}}} \right) {d}^{3}\cr \cr+&\ldots} $$ This (and as many terms as you want) can be obtained from the Lagrange–Bürmann formula.