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Finding zeros of function by integration: a novel relationship or not?
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It seems that in certain cases one can find the zero of a function by solving an integration problem instead. This surprises me, and I am wondering to what extent this (1) has been studied, and/or (2) is interesting.
For example, consider $f(x) = x^3 +ax -1$. This has a (unique) zero on the positive real axis, $x_0 in mathbb R^+$ (for any value of $a in mathbb R$). I am surprised to find that one can express this zero as follows:
$$ x_0 = frac34a left( 2 - frac1mathcal I right) qquad textrm where qquad mathcal I := frac1pi int_0^infty frac11+(y^2/3-a)^2 ; y^2/3 mathrm d y. $$
I was indeed able to prove this using some unconventional methods (which can be found in a physicist's toolbox). Is the above correspondence surprising to mathematicians?
(By the way, the fact that the $f(x)$ I chose above is a third-order polynomial is a red herring, as I found similar relationships where $f(x)$ contains, for example, logarithms.)
calculus integration roots
asked Jul 20 at 4:10
Ruben Verresen
529212
add a comment |Â
up vote
4
down vote
favorite
It seems that in certain cases one can find the zero of a function by solving an integration problem instead. This surprises me, and I am wondering to what extent this (1) has been studied, and/or (2) is interesting.
For example, consider $f(x) = x^3 +ax -1$. This has a (unique) zero on the positive real axis, $x_0 in mathbb R^+$ (for any value of $a in mathbb R$). I am surprised to find that one can express this zero as follows:
$$ x_0 = frac34a left( 2 - frac1mathcal I right) qquad textrm where qquad mathcal I := frac1pi int_0^infty frac11+(y^2/3-a)^2 ; y^2/3 mathrm d y. $$
I was indeed able to prove this using some unconventional methods (which can be found in a physicist's toolbox). Is the above correspondence surprising to mathematicians?
(By the way, the fact that the $f(x)$ I chose above is a third-order polynomial is a red herring, as I found similar relationships where $f(x)$ contains, for example, logarithms.)
calculus integration roots
asked Jul 20 at 4:10
Ruben Verresen
529212
1
You can find the roots of complex function using cauchys integral formula. chebfun.org/examples/roots/ComplexRoots.html It looks like you are doing something similar.
– RHowe
Jul 20 at 4:29
2
If I remember correctly, the roots of higher-degree polynomials ($n>5$) can be expressed via elliptic integrals (they don't have an expression in terms of radicals in general by Abel-Ruffini theorem).
– lisyarus
Jul 20 at 17:46
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
It seems that in certain cases one can find the zero of a function by solving an integration problem instead. This surprises me, and I am wondering to what extent this (1) has been studied, and/or (2) is interesting.
For example, consider $f(x) = x^3 +ax -1$. This has a (unique) zero on the positive real axis, $x_0 in mathbb R^+$ (for any value of $a in mathbb R$). I am surprised to find that one can express this zero as follows:
$$ x_0 = frac34a left( 2 - frac1mathcal I right) qquad textrm where qquad mathcal I := frac1pi int_0^infty frac11+(y^2/3-a)^2 ; y^2/3 mathrm d y. $$
I was indeed able to prove this using some unconventional methods (which can be found in a physicist's toolbox). Is the above correspondence surprising to mathematicians?
(By the way, the fact that the $f(x)$ I chose above is a third-order polynomial is a red herring, as I found similar relationships where $f(x)$ contains, for example, logarithms.)
calculus integration roots
asked Jul 20 at 4:10
Ruben Verresen
529212
It seems that in certain cases one can find the zero of a function by solving an integration problem instead. This surprises me, and I am wondering to what extent this (1) has been studied, and/or (2) is interesting.
For example, consider $f(x) = x^3 +ax -1$. This has a (unique) zero on the positive real axis, $x_0 in mathbb R^+$ (for any value of $a in mathbb R$). I am surprised to find that one can express this zero as follows:
$$ x_0 = frac34a left( 2 - frac1mathcal I right) qquad textrm where qquad mathcal I := frac1pi int_0^infty frac11+(y^2/3-a)^2 ; y^2/3 mathrm d y. $$
I was indeed able to prove this using some unconventional methods (which can be found in a physicist's toolbox). Is the above correspondence surprising to mathematicians?
(By the way, the fact that the $f(x)$ I chose above is a third-order polynomial is a red herring, as I found similar relationships where $f(x)$ contains, for example, logarithms.)
calculus integration roots
asked Jul 20 at 4:10
Ruben Verresen
529212
edited Jul 20 at 17:34
asked Jul 20 at 4:10
Ruben Verresen
529212
asked Jul 20 at 4:10
Ruben Verresen
529212
asked Jul 20 at 4:10
Ruben Verresen
529212
529212
1
You can find the roots of complex function using cauchys integral formula. chebfun.org/examples/roots/ComplexRoots.html It looks like you are doing something similar.
– RHowe
Jul 20 at 4:29
2
If I remember correctly, the roots of higher-degree polynomials ($n>5$) can be expressed via elliptic integrals (they don't have an expression in terms of radicals in general by Abel-Ruffini theorem).
– lisyarus
Jul 20 at 17:46
add a comment |Â
1
You can find the roots of complex function using cauchys integral formula. chebfun.org/examples/roots/ComplexRoots.html It looks like you are doing something similar.
– RHowe
Jul 20 at 4:29
2
If I remember correctly, the roots of higher-degree polynomials ($n>5$) can be expressed via elliptic integrals (they don't have an expression in terms of radicals in general by Abel-Ruffini theorem).
– lisyarus
Jul 20 at 17:46
1
1
You can find the roots of complex function using cauchys integral formula. chebfun.org/examples/roots/ComplexRoots.html It looks like you are doing something similar.
– RHowe
Jul 20 at 4:29
You can find the roots of complex function using cauchys integral formula. chebfun.org/examples/roots/ComplexRoots.html It looks like you are doing something similar.
– RHowe
Jul 20 at 4:29
2
2
If I remember correctly, the roots of higher-degree polynomials ($n>5$) can be expressed via elliptic integrals (they don't have an expression in terms of radicals in general by Abel-Ruffini theorem).
– lisyarus
Jul 20 at 17:46
If I remember correctly, the roots of higher-degree polynomials ($n>5$) can be expressed via elliptic integrals (they don't have an expression in terms of radicals in general by Abel-Ruffini theorem).
– lisyarus
Jul 20 at 17:46
add a comment |Â
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1
You can find the roots of complex function using cauchys integral formula. chebfun.org/examples/roots/ComplexRoots.html It looks like you are doing something similar.
– RHowe
Jul 20 at 4:29
2
If I remember correctly, the roots of higher-degree polynomials ($n>5$) can be expressed via elliptic integrals (they don't have an expression in terms of radicals in general by Abel-Ruffini theorem).
– lisyarus
Jul 20 at 17:46