Uniqueness of an Integral eq.











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I’m trying to solve



$$int_0^1|(t^2 - at -b)|,dt =1/12$$



for $a$ and $b$.



I got one solution when $a=1$ and $b=-1/4$, how I can prove that solution is unique?!










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  • 1




    Is that the absolute value of $t^2-at-b$? If so, you'll need to consider the values of $t$ such that $t^2-at-b < 0$
    – Mattos
    Nov 26 at 3:15










  • Yes, it’s the absolute value.
    – Jim
    Nov 26 at 3:19















up vote
0
down vote

favorite












I’m trying to solve



$$int_0^1|(t^2 - at -b)|,dt =1/12$$



for $a$ and $b$.



I got one solution when $a=1$ and $b=-1/4$, how I can prove that solution is unique?!










share|cite|improve this question


















  • 1




    Is that the absolute value of $t^2-at-b$? If so, you'll need to consider the values of $t$ such that $t^2-at-b < 0$
    – Mattos
    Nov 26 at 3:15










  • Yes, it’s the absolute value.
    – Jim
    Nov 26 at 3:19













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I’m trying to solve



$$int_0^1|(t^2 - at -b)|,dt =1/12$$



for $a$ and $b$.



I got one solution when $a=1$ and $b=-1/4$, how I can prove that solution is unique?!










share|cite|improve this question













I’m trying to solve



$$int_0^1|(t^2 - at -b)|,dt =1/12$$



for $a$ and $b$.



I got one solution when $a=1$ and $b=-1/4$, how I can prove that solution is unique?!







real-analysis integration analysis multivariable-calculus






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 26 at 3:05









Jim

94




94








  • 1




    Is that the absolute value of $t^2-at-b$? If so, you'll need to consider the values of $t$ such that $t^2-at-b < 0$
    – Mattos
    Nov 26 at 3:15










  • Yes, it’s the absolute value.
    – Jim
    Nov 26 at 3:19














  • 1




    Is that the absolute value of $t^2-at-b$? If so, you'll need to consider the values of $t$ such that $t^2-at-b < 0$
    – Mattos
    Nov 26 at 3:15










  • Yes, it’s the absolute value.
    – Jim
    Nov 26 at 3:19








1




1




Is that the absolute value of $t^2-at-b$? If so, you'll need to consider the values of $t$ such that $t^2-at-b < 0$
– Mattos
Nov 26 at 3:15




Is that the absolute value of $t^2-at-b$? If so, you'll need to consider the values of $t$ such that $t^2-at-b < 0$
– Mattos
Nov 26 at 3:15












Yes, it’s the absolute value.
– Jim
Nov 26 at 3:19




Yes, it’s the absolute value.
– Jim
Nov 26 at 3:19










1 Answer
1






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up vote
1
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This solution is not unique. Try $a=1$ and $b=-7/64$ for example. There are even other solutions with different values of $a$ and $b$!






share|cite|improve this answer





















  • Any idea to finding the other solutions form?!
    – Jim
    Nov 26 at 3:31










  • I don't think the equation will look that nice. In order to evaluate $a$'s and $b$'s exactly, you'll have to split up the integration into parts where $t^2-at-b$ is positive and parts where it is negative. You can probably do that integration out on your own if you really need to. The general idea is just integrate it then solve for $a$ and $b$ after setting the resulting expression equal to $1/12$
    – Isaac Browne
    Nov 26 at 3:33












  • Maybe I am wrong, but shouldn't the range of $t$ must be between $0$ and $1$?
    – PradyumanDixit
    Nov 26 at 3:41










  • @PradyumanDixit Yes, it should. I don't think I said otherwise anywhere. Did I?
    – Isaac Browne
    Nov 26 at 3:50










  • No, I was just proposing a way to see the problem a bit differently and maybe this point could help.
    – PradyumanDixit
    Nov 26 at 3:53











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1 Answer
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1 Answer
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active

oldest

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active

oldest

votes








up vote
1
down vote













This solution is not unique. Try $a=1$ and $b=-7/64$ for example. There are even other solutions with different values of $a$ and $b$!






share|cite|improve this answer





















  • Any idea to finding the other solutions form?!
    – Jim
    Nov 26 at 3:31










  • I don't think the equation will look that nice. In order to evaluate $a$'s and $b$'s exactly, you'll have to split up the integration into parts where $t^2-at-b$ is positive and parts where it is negative. You can probably do that integration out on your own if you really need to. The general idea is just integrate it then solve for $a$ and $b$ after setting the resulting expression equal to $1/12$
    – Isaac Browne
    Nov 26 at 3:33












  • Maybe I am wrong, but shouldn't the range of $t$ must be between $0$ and $1$?
    – PradyumanDixit
    Nov 26 at 3:41










  • @PradyumanDixit Yes, it should. I don't think I said otherwise anywhere. Did I?
    – Isaac Browne
    Nov 26 at 3:50










  • No, I was just proposing a way to see the problem a bit differently and maybe this point could help.
    – PradyumanDixit
    Nov 26 at 3:53















up vote
1
down vote













This solution is not unique. Try $a=1$ and $b=-7/64$ for example. There are even other solutions with different values of $a$ and $b$!






share|cite|improve this answer





















  • Any idea to finding the other solutions form?!
    – Jim
    Nov 26 at 3:31










  • I don't think the equation will look that nice. In order to evaluate $a$'s and $b$'s exactly, you'll have to split up the integration into parts where $t^2-at-b$ is positive and parts where it is negative. You can probably do that integration out on your own if you really need to. The general idea is just integrate it then solve for $a$ and $b$ after setting the resulting expression equal to $1/12$
    – Isaac Browne
    Nov 26 at 3:33












  • Maybe I am wrong, but shouldn't the range of $t$ must be between $0$ and $1$?
    – PradyumanDixit
    Nov 26 at 3:41










  • @PradyumanDixit Yes, it should. I don't think I said otherwise anywhere. Did I?
    – Isaac Browne
    Nov 26 at 3:50










  • No, I was just proposing a way to see the problem a bit differently and maybe this point could help.
    – PradyumanDixit
    Nov 26 at 3:53













up vote
1
down vote










up vote
1
down vote









This solution is not unique. Try $a=1$ and $b=-7/64$ for example. There are even other solutions with different values of $a$ and $b$!






share|cite|improve this answer












This solution is not unique. Try $a=1$ and $b=-7/64$ for example. There are even other solutions with different values of $a$ and $b$!







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 at 3:21









Isaac Browne

4,59731032




4,59731032












  • Any idea to finding the other solutions form?!
    – Jim
    Nov 26 at 3:31










  • I don't think the equation will look that nice. In order to evaluate $a$'s and $b$'s exactly, you'll have to split up the integration into parts where $t^2-at-b$ is positive and parts where it is negative. You can probably do that integration out on your own if you really need to. The general idea is just integrate it then solve for $a$ and $b$ after setting the resulting expression equal to $1/12$
    – Isaac Browne
    Nov 26 at 3:33












  • Maybe I am wrong, but shouldn't the range of $t$ must be between $0$ and $1$?
    – PradyumanDixit
    Nov 26 at 3:41










  • @PradyumanDixit Yes, it should. I don't think I said otherwise anywhere. Did I?
    – Isaac Browne
    Nov 26 at 3:50










  • No, I was just proposing a way to see the problem a bit differently and maybe this point could help.
    – PradyumanDixit
    Nov 26 at 3:53


















  • Any idea to finding the other solutions form?!
    – Jim
    Nov 26 at 3:31










  • I don't think the equation will look that nice. In order to evaluate $a$'s and $b$'s exactly, you'll have to split up the integration into parts where $t^2-at-b$ is positive and parts where it is negative. You can probably do that integration out on your own if you really need to. The general idea is just integrate it then solve for $a$ and $b$ after setting the resulting expression equal to $1/12$
    – Isaac Browne
    Nov 26 at 3:33












  • Maybe I am wrong, but shouldn't the range of $t$ must be between $0$ and $1$?
    – PradyumanDixit
    Nov 26 at 3:41










  • @PradyumanDixit Yes, it should. I don't think I said otherwise anywhere. Did I?
    – Isaac Browne
    Nov 26 at 3:50










  • No, I was just proposing a way to see the problem a bit differently and maybe this point could help.
    – PradyumanDixit
    Nov 26 at 3:53
















Any idea to finding the other solutions form?!
– Jim
Nov 26 at 3:31




Any idea to finding the other solutions form?!
– Jim
Nov 26 at 3:31












I don't think the equation will look that nice. In order to evaluate $a$'s and $b$'s exactly, you'll have to split up the integration into parts where $t^2-at-b$ is positive and parts where it is negative. You can probably do that integration out on your own if you really need to. The general idea is just integrate it then solve for $a$ and $b$ after setting the resulting expression equal to $1/12$
– Isaac Browne
Nov 26 at 3:33






I don't think the equation will look that nice. In order to evaluate $a$'s and $b$'s exactly, you'll have to split up the integration into parts where $t^2-at-b$ is positive and parts where it is negative. You can probably do that integration out on your own if you really need to. The general idea is just integrate it then solve for $a$ and $b$ after setting the resulting expression equal to $1/12$
– Isaac Browne
Nov 26 at 3:33














Maybe I am wrong, but shouldn't the range of $t$ must be between $0$ and $1$?
– PradyumanDixit
Nov 26 at 3:41




Maybe I am wrong, but shouldn't the range of $t$ must be between $0$ and $1$?
– PradyumanDixit
Nov 26 at 3:41












@PradyumanDixit Yes, it should. I don't think I said otherwise anywhere. Did I?
– Isaac Browne
Nov 26 at 3:50




@PradyumanDixit Yes, it should. I don't think I said otherwise anywhere. Did I?
– Isaac Browne
Nov 26 at 3:50












No, I was just proposing a way to see the problem a bit differently and maybe this point could help.
– PradyumanDixit
Nov 26 at 3:53




No, I was just proposing a way to see the problem a bit differently and maybe this point could help.
– PradyumanDixit
Nov 26 at 3:53


















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