Uniqueness of an Integral eq.
0
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
asked Nov 26 at 3:05
Jim
94
add a comment |
0
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
asked Nov 26 at 3:05
Jim
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
add a comment |
0
0
0
0
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
asked Nov 26 at 3:05
Jim
94
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
real-analysis integration analysis multivariable-calculus
asked Nov 26 at 3:05
Jim
94
asked Nov 26 at 3:05
Jim
94
asked Nov 26 at 3:05
Jim
94
asked Nov 26 at 3:05
Jim
94
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
add a comment |
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
add a comment |
1 Answer
1
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1
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 $b$!
answered Nov 26 at 3:21
Isaac Browne
4,60731132
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
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
1
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 $b$!
answered Nov 26 at 3:21
Isaac Browne
4,60731132
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
add a comment |
1
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 $b$!
answered Nov 26 at 3:21
Isaac Browne
4,60731132
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
add a comment |
1
1
1
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 $b$!
answered Nov 26 at 3:21
Isaac Browne
4,60731132
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 $b$!
answered Nov 26 at 3:21
Isaac Browne
4,60731132
answered Nov 26 at 3:21
Isaac Browne
4,60731132
answered Nov 26 at 3:21
Isaac Browne
4,60731132
answered Nov 26 at 3:21
Isaac Browne
4,60731132
4,60731132
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
add a comment |
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
add a comment |
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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