Finding $f$ such that $f(xy) = xf(y)+yf(x)-2xy$ given $f'(1)=3$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that no function exists such that…Proving that the relation from the null set to the null set is a functionProve $lim_xtoinfty left( sqrtx+1 - sqrtx right) = 0$Then the value of $ [f(2)] $ where [.] represents the greatest integer function is?Finding a line tangent to two points of a graph WITHOUT calculusFind the minimum roots of $f'(x)cdot f'''(x)+(f''(x))^2 =0$ given certain conditions on $f(x)$.$f(x+yf(x))+f(xf(y)-y) = f(x)-f(y)+2xy^2$For the given function find k such that f(x)≠f(x+k) for any value of xWhen will the function be identically zeroProving the required condition for $f(x)$ from given information
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Finding $f$ such that $f(xy) = xf(y)+yf(x)-2xy$ given $f'(1)=3$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that no function exists such that…Proving that the relation from the null set to the null set is a functionProve $lim_xtoinfty left( sqrtx+1 - sqrtx right) = 0$Then the value of $ [f(2)] $ where [.] represents the greatest integer function is?Finding a line tangent to two points of a graph WITHOUT calculusFind the minimum roots of $f'(x)cdot f'''(x)+(f''(x))^2 =0$ given certain conditions on $f(x)$.$f(x+yf(x))+f(xf(y)-y) = f(x)-f(y)+2xy^2$For the given function find k such that f(x)≠f(x+k) for any value of xWhen will the function be identically zeroProving the required condition for $f(x)$ from given information
$begingroup$
Let $f$ be a differentiable function satisfying the relation $$f(xy) = xf(y)+yf(x)-2xy$$ where $x, y>0$ and $f'(1)=3$ then prove that the equation f(x) = k has two solutions in
$kin(-e^-3, 0)$
I tried differentiating this function but couldn't get anything from it. How to proceed here?
functions
edited Apr 1 at 8:41
N. F. Taussig
45.4k103358
asked Apr 1 at 5:59
GENESECT GENESECT
768
$endgroup$
add a comment |
$begingroup$
Let $f$ be a differentiable function satisfying the relation $$f(xy) = xf(y)+yf(x)-2xy$$ where $x, y>0$ and $f'(1)=3$ then prove that the equation f(x) = k has two solutions in
$kin(-e^-3, 0)$
I tried differentiating this function but couldn't get anything from it. How to proceed here?
functions
edited Apr 1 at 8:41
N. F. Taussig
45.4k103358
asked Apr 1 at 5:59
GENESECT GENESECT
768
$endgroup$
1
$begingroup$
please edit the equation correctly, i don't understand what it says
$endgroup$
– gt6989b
Apr 1 at 6:01
$begingroup$
Is it legible now?
$endgroup$
– GENESECT
Apr 1 at 6:12
add a comment |
$begingroup$
Let $f$ be a differentiable function satisfying the relation $$f(xy) = xf(y)+yf(x)-2xy$$ where $x, y>0$ and $f'(1)=3$ then prove that the equation f(x) = k has two solutions in
$kin(-e^-3, 0)$
I tried differentiating this function but couldn't get anything from it. How to proceed here?
functions
edited Apr 1 at 8:41
N. F. Taussig
45.4k103358
asked Apr 1 at 5:59
GENESECT GENESECT
768
$endgroup$
Let $f$ be a differentiable function satisfying the relation $$f(xy) = xf(y)+yf(x)-2xy$$ where $x, y>0$ and $f'(1)=3$ then prove that the equation f(x) = k has two solutions in
$kin(-e^-3, 0)$
I tried differentiating this function but couldn't get anything from it. How to proceed here?
functions
functions
edited Apr 1 at 8:41
N. F. Taussig
45.4k103358
asked Apr 1 at 5:59
GENESECT GENESECT
768
edited Apr 1 at 8:41
N. F. Taussig
45.4k103358
asked Apr 1 at 5:59
GENESECT GENESECT
768
edited Apr 1 at 8:41
N. F. Taussig
45.4k103358
edited Apr 1 at 8:41
N. F. Taussig
45.4k103358
edited Apr 1 at 8:41
N. F. Taussig
45.4k103358
45.4k103358
asked Apr 1 at 5:59
GENESECT GENESECT
768
asked Apr 1 at 5:59
GENESECT GENESECT
768
asked Apr 1 at 5:59
GENESECT GENESECT
768
768
1
$begingroup$
please edit the equation correctly, i don't understand what it says
$endgroup$
– gt6989b
Apr 1 at 6:01
$begingroup$
Is it legible now?
$endgroup$
– GENESECT
Apr 1 at 6:12
add a comment |
1
$begingroup$
please edit the equation correctly, i don't understand what it says
$endgroup$
– gt6989b
Apr 1 at 6:01
$begingroup$
Is it legible now?
$endgroup$
– GENESECT
Apr 1 at 6:12
1
1
$begingroup$
please edit the equation correctly, i don't understand what it says
$endgroup$
– gt6989b
Apr 1 at 6:01
$begingroup$
please edit the equation correctly, i don't understand what it says
$endgroup$
– gt6989b
Apr 1 at 6:01
$begingroup$
Is it legible now?
$endgroup$
– GENESECT
Apr 1 at 6:12
$begingroup$
Is it legible now?
$endgroup$
– GENESECT
Apr 1 at 6:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: you can compute $f$ explicitly. Let $g(x)=frac f(x) x -2$ and verify that $g(xy)=g(x)+g(y)$. Do you know how to find all continuous functions satisfying this equation?. [$f(x)=x(clog, x+2)$].
answered Apr 1 at 6:23
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
$endgroup$
$begingroup$
I am ashamed to say I don't understand what you did. Can you tell me more about how you wrote the equation
$endgroup$
– GENESECT
Apr 1 at 6:30
$begingroup$
@GENESECT You can easily guess that dividing the given equation by $xy$ brings it to a more manageable form. After this I just adjusted for the the constant term.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 6:34
$begingroup$
@kavi ram murty Sir can i find this kind of promblems in the book of functional equation written by 'b j venkatachala'
$endgroup$
– NewBornMATH
Apr 1 at 8:11
$begingroup$
I am sure you will find similar problem in Vektachala's book.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 8:12
$begingroup$
I deleted my answer. It was a Calculation mistake,no D.E. can be obtained by just differentiating i guess. By the way in your answer how do you find all the continuos solution? (I dont know how to do that , i am sorry )
$endgroup$
– NewBornMATH
Apr 1 at 8:39
add a comment |
$begingroup$
Hint: you can compute f explicitly. Let g(x)=f(x)x−2 and verify that g(xy)=g(x)+g(y). Do you know how to find all continuous functions satisfying this equation?. [f(x)=x(clogx+2)].
answered Apr 2 at 7:39
user660100user660100
1
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: you can compute $f$ explicitly. Let $g(x)=frac f(x) x -2$ and verify that $g(xy)=g(x)+g(y)$. Do you know how to find all continuous functions satisfying this equation?. [$f(x)=x(clog, x+2)$].
answered Apr 1 at 6:23
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
$endgroup$
$begingroup$
I am ashamed to say I don't understand what you did. Can you tell me more about how you wrote the equation
$endgroup$
– GENESECT
Apr 1 at 6:30
$begingroup$
@GENESECT You can easily guess that dividing the given equation by $xy$ brings it to a more manageable form. After this I just adjusted for the the constant term.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 6:34
$begingroup$
@kavi ram murty Sir can i find this kind of promblems in the book of functional equation written by 'b j venkatachala'
$endgroup$
– NewBornMATH
Apr 1 at 8:11
$begingroup$
I am sure you will find similar problem in Vektachala's book.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 8:12
$begingroup$
I deleted my answer. It was a Calculation mistake,no D.E. can be obtained by just differentiating i guess. By the way in your answer how do you find all the continuos solution? (I dont know how to do that , i am sorry )
$endgroup$
– NewBornMATH
Apr 1 at 8:39
add a comment |
$begingroup$
Hint: you can compute $f$ explicitly. Let $g(x)=frac f(x) x -2$ and verify that $g(xy)=g(x)+g(y)$. Do you know how to find all continuous functions satisfying this equation?. [$f(x)=x(clog, x+2)$].
answered Apr 1 at 6:23
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
$endgroup$
$begingroup$
I am ashamed to say I don't understand what you did. Can you tell me more about how you wrote the equation
$endgroup$
– GENESECT
Apr 1 at 6:30
$begingroup$
@GENESECT You can easily guess that dividing the given equation by $xy$ brings it to a more manageable form. After this I just adjusted for the the constant term.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 6:34
$begingroup$
@kavi ram murty Sir can i find this kind of promblems in the book of functional equation written by 'b j venkatachala'
$endgroup$
– NewBornMATH
Apr 1 at 8:11
$begingroup$
I am sure you will find similar problem in Vektachala's book.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 8:12
$begingroup$
I deleted my answer. It was a Calculation mistake,no D.E. can be obtained by just differentiating i guess. By the way in your answer how do you find all the continuos solution? (I dont know how to do that , i am sorry )
$endgroup$
– NewBornMATH
Apr 1 at 8:39
add a comment |
$begingroup$
Hint: you can compute $f$ explicitly. Let $g(x)=frac f(x) x -2$ and verify that $g(xy)=g(x)+g(y)$. Do you know how to find all continuous functions satisfying this equation?. [$f(x)=x(clog, x+2)$].
answered Apr 1 at 6:23
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
$endgroup$
Hint: you can compute $f$ explicitly. Let $g(x)=frac f(x) x -2$ and verify that $g(xy)=g(x)+g(y)$. Do you know how to find all continuous functions satisfying this equation?. [$f(x)=x(clog, x+2)$].
answered Apr 1 at 6:23
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
edited Apr 1 at 8:13
answered Apr 1 at 6:23
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
answered Apr 1 at 6:23
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
answered Apr 1 at 6:23
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
75.3k53270
$begingroup$
I am ashamed to say I don't understand what you did. Can you tell me more about how you wrote the equation
$endgroup$
– GENESECT
Apr 1 at 6:30
$begingroup$
@GENESECT You can easily guess that dividing the given equation by $xy$ brings it to a more manageable form. After this I just adjusted for the the constant term.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 6:34
$begingroup$
@kavi ram murty Sir can i find this kind of promblems in the book of functional equation written by 'b j venkatachala'
$endgroup$
– NewBornMATH
Apr 1 at 8:11
$begingroup$
I am sure you will find similar problem in Vektachala's book.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 8:12
$begingroup$
I deleted my answer. It was a Calculation mistake,no D.E. can be obtained by just differentiating i guess. By the way in your answer how do you find all the continuos solution? (I dont know how to do that , i am sorry )
$endgroup$
– NewBornMATH
Apr 1 at 8:39
add a comment |
$begingroup$
I am ashamed to say I don't understand what you did. Can you tell me more about how you wrote the equation
$endgroup$
– GENESECT
Apr 1 at 6:30
$begingroup$
@GENESECT You can easily guess that dividing the given equation by $xy$ brings it to a more manageable form. After this I just adjusted for the the constant term.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 6:34
$begingroup$
@kavi ram murty Sir can i find this kind of promblems in the book of functional equation written by 'b j venkatachala'
$endgroup$
– NewBornMATH
Apr 1 at 8:11
$begingroup$
I am sure you will find similar problem in Vektachala's book.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 8:12
$begingroup$
I deleted my answer. It was a Calculation mistake,no D.E. can be obtained by just differentiating i guess. By the way in your answer how do you find all the continuos solution? (I dont know how to do that , i am sorry )
$endgroup$
– NewBornMATH
Apr 1 at 8:39
$begingroup$
I am ashamed to say I don't understand what you did. Can you tell me more about how you wrote the equation
$endgroup$
– GENESECT
Apr 1 at 6:30
$begingroup$
I am ashamed to say I don't understand what you did. Can you tell me more about how you wrote the equation
$endgroup$
– GENESECT
Apr 1 at 6:30
$begingroup$
@GENESECT You can easily guess that dividing the given equation by $xy$ brings it to a more manageable form. After this I just adjusted for the the constant term.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 6:34
$begingroup$
@GENESECT You can easily guess that dividing the given equation by $xy$ brings it to a more manageable form. After this I just adjusted for the the constant term.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 6:34
$begingroup$
@kavi ram murty Sir can i find this kind of promblems in the book of functional equation written by 'b j venkatachala'
$endgroup$
– NewBornMATH
Apr 1 at 8:11
$begingroup$
@kavi ram murty Sir can i find this kind of promblems in the book of functional equation written by 'b j venkatachala'
$endgroup$
– NewBornMATH
Apr 1 at 8:11
$begingroup$
I am sure you will find similar problem in Vektachala's book.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 8:12
$begingroup$
I am sure you will find similar problem in Vektachala's book.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 8:12
$begingroup$
I deleted my answer. It was a Calculation mistake,no D.E. can be obtained by just differentiating i guess. By the way in your answer how do you find all the continuos solution? (I dont know how to do that , i am sorry )
$endgroup$
– NewBornMATH
Apr 1 at 8:39
$begingroup$
I deleted my answer. It was a Calculation mistake,no D.E. can be obtained by just differentiating i guess. By the way in your answer how do you find all the continuos solution? (I dont know how to do that , i am sorry )
$endgroup$
– NewBornMATH
Apr 1 at 8:39
add a comment |
$begingroup$
Hint: you can compute f explicitly. Let g(x)=f(x)x−2 and verify that g(xy)=g(x)+g(y). Do you know how to find all continuous functions satisfying this equation?. [f(x)=x(clogx+2)].
answered Apr 2 at 7:39
user660100user660100
1
$endgroup$
add a comment |
$begingroup$
Hint: you can compute f explicitly. Let g(x)=f(x)x−2 and verify that g(xy)=g(x)+g(y). Do you know how to find all continuous functions satisfying this equation?. [f(x)=x(clogx+2)].
answered Apr 2 at 7:39
user660100user660100
1
$endgroup$
add a comment |
$begingroup$
Hint: you can compute f explicitly. Let g(x)=f(x)x−2 and verify that g(xy)=g(x)+g(y). Do you know how to find all continuous functions satisfying this equation?. [f(x)=x(clogx+2)].
answered Apr 2 at 7:39
user660100user660100
1
$endgroup$
Hint: you can compute f explicitly. Let g(x)=f(x)x−2 and verify that g(xy)=g(x)+g(y). Do you know how to find all continuous functions satisfying this equation?. [f(x)=x(clogx+2)].
answered Apr 2 at 7:39
user660100user660100
1
answered Apr 2 at 7:39
user660100user660100
1
answered Apr 2 at 7:39
user660100user660100
1
answered Apr 2 at 7:39
user660100user660100
1
1
add a comment |
add a comment |
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$begingroup$
please edit the equation correctly, i don't understand what it says
$endgroup$
– gt6989b
Apr 1 at 6:01
$begingroup$
Is it legible now?
$endgroup$
– GENESECT
Apr 1 at 6:12