Mixing
Prove that this expression is real and positive
$begingroup$
I need to show that a function (related to enzyme kinetics) has two real, positive roots. Without giving the entire kinetics model, I have the following differential equation. By the constraints of the kinetics model this is derived from, all of the constants involved in the above equation are strictly positive.
$$dot{C} = k_{+}(E_{0}-C)(S_{0} - C) - (k+k{-})C. $$
I have decided to define
$$frac{dot{C}}{k_{+}} = f(c), $$
$$frac{k+k_{-}}{k_{+}} = r, mathrm{and thus}$$
$$f(c) = (E_{0}-C)(S_{0} - C) - rC$$
To hopefully make the problem easier to deal with. My next step was to expand the polynomial, and set it equal to 0 in order to solve for the system equilibria.
$$0 = C^2+(E_0-S_0-r)C + E_0S_0$$
Using the quadratic formula to get roots, I get that the solutions are
$$frac{-(E_0-S_0-r) pm sqrt{(E_0-S_0-r)^2-4E_0S_0}}{2}. $$
I need to show that both roots are real and positive. I know that to show they are real, all I need to do is show that the discriminant is positive, which I do not know how to do.
After showing that the discriminant is positive, I need to show that the whole numerator is positive to verify positivity, which I also am not sure how to do (perhaps it will become more obvious after verifying realness).
algebra-precalculus real-numbers
asked Feb 1 at 4:23
jeanquiltjeanquilt
301213
$endgroup$
add a comment |
$begingroup$
I need to show that a function (related to enzyme kinetics) has two real, positive roots. Without giving the entire kinetics model, I have the following differential equation. By the constraints of the kinetics model this is derived from, all of the constants involved in the above equation are strictly positive.
$$dot{C} = k_{+}(E_{0}-C)(S_{0} - C) - (k+k{-})C. $$
I have decided to define
$$frac{dot{C}}{k_{+}} = f(c), $$
$$frac{k+k_{-}}{k_{+}} = r, mathrm{and thus}$$
$$f(c) = (E_{0}-C)(S_{0} - C) - rC$$
To hopefully make the problem easier to deal with. My next step was to expand the polynomial, and set it equal to 0 in order to solve for the system equilibria.
$$0 = C^2+(E_0-S_0-r)C + E_0S_0$$
Using the quadratic formula to get roots, I get that the solutions are
$$frac{-(E_0-S_0-r) pm sqrt{(E_0-S_0-r)^2-4E_0S_0}}{2}. $$
I need to show that both roots are real and positive. I know that to show they are real, all I need to do is show that the discriminant is positive, which I do not know how to do.
After showing that the discriminant is positive, I need to show that the whole numerator is positive to verify positivity, which I also am not sure how to do (perhaps it will become more obvious after verifying realness).
algebra-precalculus real-numbers
asked Feb 1 at 4:23
jeanquiltjeanquilt
301213
$endgroup$
1
$begingroup$
This is not true for all real numbers $r,S_0,E_0$ (try $r=S_0=E_0=1$) so if the statement you're trying to prove is true, there must be additional conditions you haven't told us.
$endgroup$
– saulspatz
Feb 1 at 4:27
$begingroup$
@saulspatz The only constraint given to me is the original quadratic equation, and that all constants are strictly positive. I will post the original quadratic so my work can be checked.
$endgroup$
– jeanquilt
Feb 1 at 4:36
$begingroup$
@jeanquilt the quadratic equation was mistaken, see my answer bellow.
$endgroup$
– user376343
Feb 1 at 22:41
add a comment |
$begingroup$
I need to show that a function (related to enzyme kinetics) has two real, positive roots. Without giving the entire kinetics model, I have the following differential equation. By the constraints of the kinetics model this is derived from, all of the constants involved in the above equation are strictly positive.
$$dot{C} = k_{+}(E_{0}-C)(S_{0} - C) - (k+k{-})C. $$
I have decided to define
$$frac{dot{C}}{k_{+}} = f(c), $$
$$frac{k+k_{-}}{k_{+}} = r, mathrm{and thus}$$
$$f(c) = (E_{0}-C)(S_{0} - C) - rC$$
To hopefully make the problem easier to deal with. My next step was to expand the polynomial, and set it equal to 0 in order to solve for the system equilibria.
$$0 = C^2+(E_0-S_0-r)C + E_0S_0$$
Using the quadratic formula to get roots, I get that the solutions are
$$frac{-(E_0-S_0-r) pm sqrt{(E_0-S_0-r)^2-4E_0S_0}}{2}. $$
I need to show that both roots are real and positive. I know that to show they are real, all I need to do is show that the discriminant is positive, which I do not know how to do.
After showing that the discriminant is positive, I need to show that the whole numerator is positive to verify positivity, which I also am not sure how to do (perhaps it will become more obvious after verifying realness).
algebra-precalculus real-numbers
asked Feb 1 at 4:23
jeanquiltjeanquilt
301213
$endgroup$
I need to show that a function (related to enzyme kinetics) has two real, positive roots. Without giving the entire kinetics model, I have the following differential equation. By the constraints of the kinetics model this is derived from, all of the constants involved in the above equation are strictly positive.
$$dot{C} = k_{+}(E_{0}-C)(S_{0} - C) - (k+k{-})C. $$
I have decided to define
$$frac{dot{C}}{k_{+}} = f(c), $$
$$frac{k+k_{-}}{k_{+}} = r, mathrm{and thus}$$
$$f(c) = (E_{0}-C)(S_{0} - C) - rC$$
To hopefully make the problem easier to deal with. My next step was to expand the polynomial, and set it equal to 0 in order to solve for the system equilibria.
$$0 = C^2+(E_0-S_0-r)C + E_0S_0$$
Using the quadratic formula to get roots, I get that the solutions are
$$frac{-(E_0-S_0-r) pm sqrt{(E_0-S_0-r)^2-4E_0S_0}}{2}. $$
I need to show that both roots are real and positive. I know that to show they are real, all I need to do is show that the discriminant is positive, which I do not know how to do.
After showing that the discriminant is positive, I need to show that the whole numerator is positive to verify positivity, which I also am not sure how to do (perhaps it will become more obvious after verifying realness).
algebra-precalculus real-numbers
algebra-precalculus real-numbers
asked Feb 1 at 4:23
jeanquiltjeanquilt
301213
asked Feb 1 at 4:23
jeanquiltjeanquilt
301213
edited Feb 1 at 4:46
jeanquilt
asked Feb 1 at 4:23
jeanquiltjeanquilt
301213
asked Feb 1 at 4:23
jeanquiltjeanquilt
301213
asked Feb 1 at 4:23
jeanquiltjeanquilt
301213
301213
1
$begingroup$
This is not true for all real numbers $r,S_0,E_0$ (try $r=S_0=E_0=1$) so if the statement you're trying to prove is true, there must be additional conditions you haven't told us.
$endgroup$
– saulspatz
Feb 1 at 4:27
$begingroup$
@saulspatz The only constraint given to me is the original quadratic equation, and that all constants are strictly positive. I will post the original quadratic so my work can be checked.
$endgroup$
– jeanquilt
Feb 1 at 4:36
$begingroup$
@jeanquilt the quadratic equation was mistaken, see my answer bellow.
$endgroup$
– user376343
Feb 1 at 22:41
add a comment |
1
$begingroup$
This is not true for all real numbers $r,S_0,E_0$ (try $r=S_0=E_0=1$) so if the statement you're trying to prove is true, there must be additional conditions you haven't told us.
$endgroup$
– saulspatz
Feb 1 at 4:27
$begingroup$
@saulspatz The only constraint given to me is the original quadratic equation, and that all constants are strictly positive. I will post the original quadratic so my work can be checked.
$endgroup$
– jeanquilt
Feb 1 at 4:36
$begingroup$
@jeanquilt the quadratic equation was mistaken, see my answer bellow.
$endgroup$
– user376343
Feb 1 at 22:41
1
1
$begingroup$
This is not true for all real numbers $r,S_0,E_0$ (try $r=S_0=E_0=1$) so if the statement you're trying to prove is true, there must be additional conditions you haven't told us.
$endgroup$
– saulspatz
Feb 1 at 4:27
$begingroup$
This is not true for all real numbers $r,S_0,E_0$ (try $r=S_0=E_0=1$) so if the statement you're trying to prove is true, there must be additional conditions you haven't told us.
$endgroup$
– saulspatz
Feb 1 at 4:27
$begingroup$
@saulspatz The only constraint given to me is the original quadratic equation, and that all constants are strictly positive. I will post the original quadratic so my work can be checked.
$endgroup$
– jeanquilt
Feb 1 at 4:36
$begingroup$
@saulspatz The only constraint given to me is the original quadratic equation, and that all constants are strictly positive. I will post the original quadratic so my work can be checked.
$endgroup$
– jeanquilt
Feb 1 at 4:36
$begingroup$
@jeanquilt the quadratic equation was mistaken, see my answer bellow.
$endgroup$
– user376343
Feb 1 at 22:41
$begingroup$
@jeanquilt the quadratic equation was mistaken, see my answer bellow.
$endgroup$
– user376343
Feb 1 at 22:41
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Presumably, those are the roots of the quadratic equation $$x^2 + (E_0 - S_0 - r)x
+ E_0S_0 = 0$$
As you noted, the roots are real if the discriminant is non-negative. The condition for them to be positive is that $E_0 - S_0 - r < 0$ and $E_0S_0 >0$,
answered Feb 1 at 4:33
Michael BiroMichael Biro
11.6k21831
$endgroup$
$begingroup$
Yes, this is the equation I am solving. $E_0S_0$ is positive due to the constraint that all coeffs are strictly positive, but I do not know if the other constraint is always true.
$endgroup$
– jeanquilt
Feb 1 at 4:48
$begingroup$
If $E_0 - S_0 - r geq 0$, then there's no chance for both roots to be positive, since you then have a negative number (or zero) minus a positive number (or zero) for one of the roots, which is necessarily negative (or zero).
$endgroup$
– Michael Biro
Feb 1 at 4:54
add a comment |
$begingroup$
The quadratic equation obtained from $;f(c) = (E_{0}-C)(S_{0} - C) - rC;$
is $$0 = C^2+(-E_0-S_0-r)C + E_0S_0.$$
The discriminant is $$(E_0+S_0+r)^2-4E_0S_0=(E_0-S_0+r)^2+4S_0r>0$$ because
$S_0>0,;r>0.$
Thus the solutions are real, equal to
$$frac{E_0+S_0+r pm sqrt{(E_0+S_0+r)^2-4E_0S_0}}{2},$$
and are obviously positive.
answered Feb 1 at 21:57
user376343user376343
3,9584829
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Presumably, those are the roots of the quadratic equation $$x^2 + (E_0 - S_0 - r)x
+ E_0S_0 = 0$$
As you noted, the roots are real if the discriminant is non-negative. The condition for them to be positive is that $E_0 - S_0 - r < 0$ and $E_0S_0 >0$,
answered Feb 1 at 4:33
Michael BiroMichael Biro
11.6k21831
$endgroup$
$begingroup$
Yes, this is the equation I am solving. $E_0S_0$ is positive due to the constraint that all coeffs are strictly positive, but I do not know if the other constraint is always true.
$endgroup$
– jeanquilt
Feb 1 at 4:48
$begingroup$
If $E_0 - S_0 - r geq 0$, then there's no chance for both roots to be positive, since you then have a negative number (or zero) minus a positive number (or zero) for one of the roots, which is necessarily negative (or zero).
$endgroup$
– Michael Biro
Feb 1 at 4:54
add a comment |
$begingroup$
Presumably, those are the roots of the quadratic equation $$x^2 + (E_0 - S_0 - r)x
+ E_0S_0 = 0$$
As you noted, the roots are real if the discriminant is non-negative. The condition for them to be positive is that $E_0 - S_0 - r < 0$ and $E_0S_0 >0$,
answered Feb 1 at 4:33
Michael BiroMichael Biro
11.6k21831
$endgroup$
$begingroup$
Yes, this is the equation I am solving. $E_0S_0$ is positive due to the constraint that all coeffs are strictly positive, but I do not know if the other constraint is always true.
$endgroup$
– jeanquilt
Feb 1 at 4:48
$begingroup$
If $E_0 - S_0 - r geq 0$, then there's no chance for both roots to be positive, since you then have a negative number (or zero) minus a positive number (or zero) for one of the roots, which is necessarily negative (or zero).
$endgroup$
– Michael Biro
Feb 1 at 4:54
add a comment |
$begingroup$
Presumably, those are the roots of the quadratic equation $$x^2 + (E_0 - S_0 - r)x
+ E_0S_0 = 0$$
As you noted, the roots are real if the discriminant is non-negative. The condition for them to be positive is that $E_0 - S_0 - r < 0$ and $E_0S_0 >0$,
answered Feb 1 at 4:33
Michael BiroMichael Biro
11.6k21831
$endgroup$
Presumably, those are the roots of the quadratic equation $$x^2 + (E_0 - S_0 - r)x
+ E_0S_0 = 0$$
As you noted, the roots are real if the discriminant is non-negative. The condition for them to be positive is that $E_0 - S_0 - r < 0$ and $E_0S_0 >0$,
answered Feb 1 at 4:33
Michael BiroMichael Biro
11.6k21831
edited Feb 1 at 4:39
answered Feb 1 at 4:33
Michael BiroMichael Biro
11.6k21831
answered Feb 1 at 4:33
Michael BiroMichael Biro
11.6k21831
answered Feb 1 at 4:33
Michael BiroMichael Biro
11.6k21831
11.6k21831
$begingroup$
Yes, this is the equation I am solving. $E_0S_0$ is positive due to the constraint that all coeffs are strictly positive, but I do not know if the other constraint is always true.
$endgroup$
– jeanquilt
Feb 1 at 4:48
$begingroup$
If $E_0 - S_0 - r geq 0$, then there's no chance for both roots to be positive, since you then have a negative number (or zero) minus a positive number (or zero) for one of the roots, which is necessarily negative (or zero).
$endgroup$
– Michael Biro
Feb 1 at 4:54
add a comment |
$begingroup$
Yes, this is the equation I am solving. $E_0S_0$ is positive due to the constraint that all coeffs are strictly positive, but I do not know if the other constraint is always true.
$endgroup$
– jeanquilt
Feb 1 at 4:48
$begingroup$
If $E_0 - S_0 - r geq 0$, then there's no chance for both roots to be positive, since you then have a negative number (or zero) minus a positive number (or zero) for one of the roots, which is necessarily negative (or zero).
$endgroup$
– Michael Biro
Feb 1 at 4:54
$begingroup$
Yes, this is the equation I am solving. $E_0S_0$ is positive due to the constraint that all coeffs are strictly positive, but I do not know if the other constraint is always true.
$endgroup$
– jeanquilt
Feb 1 at 4:48
$begingroup$
Yes, this is the equation I am solving. $E_0S_0$ is positive due to the constraint that all coeffs are strictly positive, but I do not know if the other constraint is always true.
$endgroup$
– jeanquilt
Feb 1 at 4:48
$begingroup$
If $E_0 - S_0 - r geq 0$, then there's no chance for both roots to be positive, since you then have a negative number (or zero) minus a positive number (or zero) for one of the roots, which is necessarily negative (or zero).
$endgroup$
– Michael Biro
Feb 1 at 4:54
$begingroup$
If $E_0 - S_0 - r geq 0$, then there's no chance for both roots to be positive, since you then have a negative number (or zero) minus a positive number (or zero) for one of the roots, which is necessarily negative (or zero).
$endgroup$
– Michael Biro
Feb 1 at 4:54
add a comment |
$begingroup$
The quadratic equation obtained from $;f(c) = (E_{0}-C)(S_{0} - C) - rC;$
is $$0 = C^2+(-E_0-S_0-r)C + E_0S_0.$$
The discriminant is $$(E_0+S_0+r)^2-4E_0S_0=(E_0-S_0+r)^2+4S_0r>0$$ because
$S_0>0,;r>0.$
Thus the solutions are real, equal to
$$frac{E_0+S_0+r pm sqrt{(E_0+S_0+r)^2-4E_0S_0}}{2},$$
and are obviously positive.
answered Feb 1 at 21:57
user376343user376343
3,9584829
$endgroup$
add a comment |
$begingroup$
The quadratic equation obtained from $;f(c) = (E_{0}-C)(S_{0} - C) - rC;$
is $$0 = C^2+(-E_0-S_0-r)C + E_0S_0.$$
The discriminant is $$(E_0+S_0+r)^2-4E_0S_0=(E_0-S_0+r)^2+4S_0r>0$$ because
$S_0>0,;r>0.$
Thus the solutions are real, equal to
$$frac{E_0+S_0+r pm sqrt{(E_0+S_0+r)^2-4E_0S_0}}{2},$$
and are obviously positive.
answered Feb 1 at 21:57
user376343user376343
3,9584829
$endgroup$
add a comment |
$begingroup$
The quadratic equation obtained from $;f(c) = (E_{0}-C)(S_{0} - C) - rC;$
is $$0 = C^2+(-E_0-S_0-r)C + E_0S_0.$$
The discriminant is $$(E_0+S_0+r)^2-4E_0S_0=(E_0-S_0+r)^2+4S_0r>0$$ because
$S_0>0,;r>0.$
Thus the solutions are real, equal to
$$frac{E_0+S_0+r pm sqrt{(E_0+S_0+r)^2-4E_0S_0}}{2},$$
and are obviously positive.
answered Feb 1 at 21:57
user376343user376343
3,9584829
$endgroup$
The quadratic equation obtained from $;f(c) = (E_{0}-C)(S_{0} - C) - rC;$
is $$0 = C^2+(-E_0-S_0-r)C + E_0S_0.$$
The discriminant is $$(E_0+S_0+r)^2-4E_0S_0=(E_0-S_0+r)^2+4S_0r>0$$ because
$S_0>0,;r>0.$
Thus the solutions are real, equal to
$$frac{E_0+S_0+r pm sqrt{(E_0+S_0+r)^2-4E_0S_0}}{2},$$
and are obviously positive.
answered Feb 1 at 21:57
user376343user376343
3,9584829
edited Feb 1 at 22:07
answered Feb 1 at 21:57
user376343user376343
3,9584829
answered Feb 1 at 21:57
user376343user376343
3,9584829
answered Feb 1 at 21:57
user376343user376343
3,9584829
3,9584829
add a comment |
add a comment |
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$begingroup$
This is not true for all real numbers $r,S_0,E_0$ (try $r=S_0=E_0=1$) so if the statement you're trying to prove is true, there must be additional conditions you haven't told us.
$endgroup$
– saulspatz
Feb 1 at 4:27
$begingroup$
@saulspatz The only constraint given to me is the original quadratic equation, and that all constants are strictly positive. I will post the original quadratic so my work can be checked.
$endgroup$
– jeanquilt
Feb 1 at 4:36
$begingroup$
@jeanquilt the quadratic equation was mistaken, see my answer bellow.
$endgroup$
– user376343
Feb 1 at 22:41